m
14.2
CH 3 CH=CHCH=CH 2
1,3-Pentadiene
Name
Product
Results from :
CI I
CH 3 CH= CHCHCH 3
4-Chloro-2-pentene
1,2 addition 1
CI
CHgCh^CHCH _ — CH2
addition
,4
I
3-Chloro- 1 -pentene
1,2 addition
-Chloro-2-pentene
1,4 addition
CH 3 CH 2 CH=CHCH 2CI
1
14.3 6+
5+
CH3CH2CH D
CH
CH2
protonation
on carbon 4 protonation
H+
/ \ CH=CHCH=CH CH 3
J
on carbon 3
CHgCHCH^CH — CH2 C
6+
6+
E-J
CH 3 CH— CH— CHCH3 protonation A on carbon 2
.
1
protonation
on carbon 2
N
H+
»
+
CH 3 CH= CHCH 2 CH 2 B
A and D, which are resonance-stabilized, are formed in preference to B and C, which are not. The positive charge of allylic carbocation A is delocalized over two secondary carbons, while the positive charge of carbocation D is delocalized over one secondary and one primary carbon. We therefore predict that carbocation A is the major intermediate formed, and that 4-chloro-2-pentene predominates. Note that 1,2
and
1,4 addition.
this
product results from both
322
Chapter 14
14.5
CH 3 CHCH=CH 2 ^t|CH 3 CHCH= CH
]^ CH CH= CHCH
CH 3 CH=CHCH2
;
3-Bromo- 1 -butene
3
2 Br
l-Bromo-2-butene
Br~
(l,2adduct)
(1,4 adduct)
Ally lie halides can undergo slow dissociation to form stabilized carbocations (SnI reaction). Both 3 -bromo-1 -butene and l-bromo-2-butene form the same allylic carbocation, pictured above, on dissociation. Addition of bromide ion to the allylic carbocation then occurs to form a mixture of bromobutenes. Since the reaction is run under equilibrium conditions, the thermodynamically more stable l-bromo-2-butene predominates.
14.6
X 1
1
2
/ H 2 C^~ CHCH^- CH 2
+
1
H 2 CCHCH=CH 2
monosubstituted double bond
X
X2 *
14 \V-^-V HoCCH= CCH = CHCH 2
2
disubstituted double
bond
X 1,4 adducts are
more
stable than 1,2 adducts because disubstituted double
stable than monosubstituted double
14.7
bonds are more
bonds (see Chapter 7).
Draw the reactants in an orientation that shows where the new bonds will form. Form the new bonds by connecting the two reactants, removing two double bonds, and relocating bond so that it lies between carbon 2 and carbon 3 of the diene. The on the dienophile retain their trans relationship in the product. The product is a
the remaining double substituents
racemic mixture.
Conjugated Compounds and Ultraviolet Spectroscopy
pCH 3
323
OCH, OCH,
CH 3
H
14.8
Good dienophiles have an electron-withdrawing group Good dienophiles:
conjugated with a double bond.
(d)
(a)
O II
H 2 C=CHCCI
Poor dienophiles:
(c)
(b)
(e)
O
O
H 2 C= CHCH 2 CH 2 COCH 3
Compound
(a) and (d) are good dienophiles because they have electron- withdrawing groups conjugated with a carbon-carbon double bond. Alkene (c) is a poor dienophile because it has no electron- withdrawing functional group. Compounds (b) and (e) are poor dienophiles because their electron- withdrawing groups are not conjugated with the double bond.
14.9
(a)
This diene has an s-cis conformation and should undergo Diels-Alder cycloaddition.
an s-trans conformation. Because the double bonds are in a fused ring not possible for them to rotate to an s-cis conformation.
(b) This diene has
system, (c)
it is
Rotation can occur about the single bond of this s-trans diene. The resulting s-cis conformation, however, has an unfavorable steric interaction of a methyl group with a hydrogen at carbon 1 Rotation to the s-cis conformation is possible but not favored. .
H
f C H 3 C'
C
C
*C{3 *CH;
H3C'
V I
H
H s-trans
(more
stable)
5-C1S (less stable)
324
Chapter 14
14.10
Rotation of the diene to the s-cis conformation must occur in order for reaction to take place.
O """"•-tiv.
-C
s
OCH 3 OCH,
o 5-trans
14.11 The
5-C1S
may be either a radical or a cation. Diene polymerization is a 1 ,4 addition that forms a polymer whose monomer units have a 4 carbon chain that contains
initiator
process
a
double bond every 4 bonds.
14.12
etc.
14.13 200 x 10-9 m =
nm 400 nm 200
forx
400
x 10-9
2x
10- 7
1.20 x
E
m
2x = 4x
10-7 10-7
m:
KHkJ/mol
1.20 x
10^kJ/mol
=
6.0 x 10 2 kJ/mol
=
3.0 x 10 2 kJ/mol
2.0 x 10- 7
k (in m) 4 x 10-7 m:
forX
1.20 x
E
10^kJ/mol
X (in m) The energy of electromagnetic nm is 300-600 kJ/mol.
(in
The energy required
NMR transitions.
for
10^ kJ/mol -7
radiation in the region of the spectrum
300-600
kJ/mol)
1.20 x
4.0 x 10
UV Energy
m
IR 4.8-48
*H
from 200
nm to 400
NMR (at 200 MHz) 8.0 x 10-5
UV transitions is greater than the energy required for IR or *H
Conjugated Compounds and Ultraviolet Spectroscopy
14.14 8
A
=
CX
Where
8
= molar
absorptivity (in L/mol-cm)
/
A =
absorbance
In this problem: 8 /
= 50,100 =
=
A =
1.00
5.01 x 10 4 L/mol
/
cm c
cm
-
concentration (in mol/L)
0.735
0.735 8
14.15
= sample pathlength (in cm)
X
/
5.01 x 104
=
L/mol-cm x 1.00
1.47 x 10- 5
M
cm
All compounds having alternating single and multiple bonds should show ultraviolet absorption in the range 200^100 nm. Only compound (a) is not UV-active. All of the
compounds pictured below show
UV absorptions. O C0H
H 2 C=CHC
=N
cx
OCCHo3 II
o
325
Width: 612 Height: 792
326
Chapter 14
14.17
5-trans
\
/
5-C1S
H ? C= CH
c=o HgC
14.18
would have to rotate to an s-cis arrangement. In an s-cis conformation, however, the two circled methyl groups experience steric strain by being too close to each other, preventing the molecule from adopting this conformation. Thus, Diels-Alder reaction doesn't occur. In order to undergo Diels-Alder reaction, this s-trans diene
H3C CH, 5-C1S
5-trans
14.19 CH,
CHo3
CH,
I
/CH 3
HqC
or
or
HoC-
? H
C^ "CH 2 CH 3
I
H
4-Methyl- 1 ,3-hexadiene
3-Methyl-2,4-hexadiene
H
/
+
\ CH,
I
I
CH2CH3
C
"I H
2-Ethyl- 1 ,3-pentadiene
CH 3
H
Br
\ /
f
Br"
HgC
C
H3
H HoC 6 I
I
C
CH2CHg
CH2CHg
+
Br
>
HgC
CH2CH3
C
H
H
HgC
HqC 3
I
HoC
C.s.
Br
\ /
*
C«^
CH2CHg
H 2-Bromo-4-methyl-3-hexene
4-Bromo-4-methyl-2-hexene
Conjugated Compounds and Ultraviolet Spectroscopy
Additional Problems Conjugated Dienes
14.20
All of these (a>
compounds can
exhibit
E/Z isomerism. (b)
CH 3 I
CH 3 CH=CCH=CHCH 3
H 2 C=CHCH =CHCH =CHCH 3
3-Methyl-2,4-hexadiene
1,3,5-Heptatriene
(d)
(c)
CH 3 CH= C= CHCH= CHCH 3
CH 3 CH=CCH=CH 2 3-Propyl- 1 ,3-pentadiene
2,3,5-Heptatriene
14.21 Excluding double-bond
CH2CH2CH 3
isomers:
CH 3
Conjugated dienes:
CH 3 CH= CHCH= CH 2
H2C
— CHC — CH2
2-Methyl- 1 ,3-butadiene
1.3- Pcntadiene
Cumulated dienes:
CH 3 CH= C= CHCH 3
Ch^Ch^CH^- C^— CH2
2,3-Pentadiene
1,2-Pentadiene
H2C
— C — C(CH 3
3-Methyl- 1 ,2-butadiene
Nonconjugated diene:
— CHCH2CH — CH2 1.4- Pentadiene
14.22
a
(a)
1
mol Br2
(b) 1.
O.
2. Zn,
H3
O.
+
H^O H
(c)
1
mol HCI Ether
)2
Br
Br
327
328
Chapter 14
HO
H
enantiomers
14.23 CHo 5+1
BrCH 2
— C— CH— CH A
tertiary/primary allylic
CH 3
+ Br
6+
*
2
addition to
carbon
H3
— C — CH _ — CH 2
Br+
I
*
6+
CH 2
?
carbon 4
carbocation
Br
B
addition to
1
6+
— C— *CH— CH 2
secondary/primary allylic
carbocation
A
Tertiary/primary allylic carbocation is more stable than secondary/primary allylic carbocation B. Since the products formed from the more stable intermediate predominate, 3,4-dibromo-3-methyl-l-butene is the major product of 1,2 addition of bromine to isoprene. In both cases, the product with the more substituted double bond (1,4 addition
product) predominates.
14.24 Any
unsubstituted cyclic 1,3-diene cyclic diene gives the
addition.
For example:
same product from
1,2-
and 1,4
Conjugated Compounds and Ultraviolet Spectroscopy
329
14.25 2
1
3
4
CH=CHCH=CH2 1
Protonation of carbon
-Phenyl- 1 ,3-butadiene
ci
1:
CH2CHCH _ — CH2 I
6+
CH2CH
5
CH
+
cr 3- Chloro-4-phenyl-
CH2
1- butene
CH 2 CH= CHCH 2 CI cr
A
allylic
isomers 1
Protonation of carbon
E/Z
-Chloro-4-phenyl2- butene CI
2:
CHCH2CH _ — CH2
+
I
CHCh^CH^— CH2 cr
4- Chloro-4-phenyl-
1-butene
Protonation of carbon 3: +
aCH—
CH= CHCH 2 CH 2 CI
CHCH2CH2
E/Z isomers
cr 4-Chloro- 1 -phenyl1-butene
C Protonation of carbon 4:
6+
? CH=CHCHCH 3
cr
5+
E/Z isomers
3-Chloro- 1 -phenyl1- butene
CH— CH— CHCH 3
CI I
CHCH= CHCH 3 D
allylic
1
D
E/Z isomers
cr -Chloro- 1 -pheny 12- butene
Carbocation is most stable because it can use the x systems of both the benzene ring and the side chain to delocalize positive charge. 3-Chloro-l -phenyl- 1-butene is the major product because it results from cation and because its double bond can be conjugated with the benzene ring to provide extra stability.
D
330
Chapter 14
Diels-Alder Reaction
14.26
If
two equivalents of cyclohexadiene are present for each equivalent of dienophile, you can
also obtain a second product:
14.27 This conformation of 2,3-di-te/t-butyl-l,3-butadiene, cis relationship, suffers
in which the tert-butyl groups have a due to the bulky substituents. Instead, the conformation, which relieves the strain but does not allow
from
steric strain
molecule adopts the s-trans Diels-Alder reaction to take place.
14.28 The diene
rotates to the s-cis conformation.
The
in the dienophile is preserved in the product.
trans relationship of the
two
ester groups
Conjugated Compounds and Ultraviolet Spectroscopy
331
14.29 H
H \
HoC /
C=CH 2
C=CH 2
H
c=c /
/
\
c=c\
H3 C
H
H
ds-l,3-Pentadiene
H
trans- 1 ,3-Pentadiene
Both pentadienes are more stable in s-trans conformations. To undergo Diels-Alder must rotate about the single bond between the double bonds assume s-cis conformations.
reactions, however, they
H \
H
H
\
/
C=C
\
/
/
—C
to
H /
C=C\
H
H
\\
C— H
c/
H ds-l,3-Pentadiene
When
H 3C trans- 1 ,3-Pentadiene
ds-l,3-pentadiene rotates to the s-cis conformation,
steric interaction
occurs
between the methyl-group protons and a hydrogen on CI. Since it's more difficult for 1,3-pentadiene to assume the s-cis conformation, it is less reactive in the Diels-Alder
cis-
reaction.
14.30 HC=CC=CH
can't be used as a Diels-Alder diene because it is linear. The end carbons are too far apart to be able to react with a dienophile in a cyclic transition state. Furthermore, the product of Diels-Alder addition would be impossibly strained, with two sp-hybridized
carbons in a six-membered ring.
14.31
och 2 ch3
och 2 ch 3
Two different orientations of the dienophile ester group are possible, and two different products can form.
332
Chapter 14
14.32 The most reactive dienophiles contain
electron-withdrawing groups.
Most reactive
Least reactive
(NC) 2 C=C(CN) 2
>
Four electronwithdrawing groups
H 2 C=CHCHO
One
>
electron-
withdrawing group
The methyl groups of 2,3-dimethyl-2-butene
H 2 C=CHCH 3
>
(CH 3 2 )
One electron-
C= C(CH 3 2 )
Four electrondonating groups
donating group
also decrease reactivity for steric reasons.
14.33 The difference
in reactivity of the three cyclic dienes is due to steric factors. As the nondiene part of the molecule becomes larger, the carbon atoms at the end of the diene portion of the ring are forced farther apart. Overlap with the n system of the dienophile in the cyclic transition state is poorer, and reaction is slower.
14.34 Although an electron-withdrawing group increases
the reactivity of a dienophile,
it
decreases the reactivity of a diene.
14.35
First, find the
cyclohexene ring formed by the Diels-Alder reaction. After you locate the able to identify the diene and the dienophile.
new bonds, you should then be
A
.0
O
V bonds formed
diene
dienophile
CN
r diene
dienophile
A V 1
I
o
diene
dienophile
(d)
C0 2 CH 3
C0 2 CH 3
? H dienophile
diene
Conjugated Compounds and Ultraviolet Spectroscopy
333
Diene Polymers
14.36
A vinyl branch in a diene polymer is the result of an occasional polymer chain, rather than the usual polymerization for the same reason. the
1
,4 addition.
1,2 double bond addition to Branching can also occur in cationic
14.37
Ozone causes
oxidative cleavage of the double bonds in rubber and breaks the polymer
chain.
14.38
Polycyclopentadiene to a
is
the product of successive Diels-Alder additions of cyclopentadiene
growing polymer chain. Strong heat causes depolymerization of the chain and
reformation of cyclopentadiene
UV
monomer units.
Spectroscopy
14.39 Only compounds having alternating multiple bonds show st -* nt* ultraviolet absorptions in the 200-400 nm range. Of the compounds shown, only pyridine (b) absorbs in this range. 14.40 To absorb
in the
bonds of allene
200—400
nm range, an alkene must be conjugated.
aren't conjugated, allene doesn't
absorb light in the
Since the double
UV region.
334
Chapter 14
14.41 The
value of
in the ultraviolet
spectrum of dienes becomes larger with increasing
inversely related to ^max> the energy needed to produce ultraviolet absorption decreases with increasing substitution.
energy
substitution. Since
is
#
of-CH3 groups
Diene
V
ax (
nm )
Vax ~ Kim (butadiene)
H
H 2C*
C
^
H H ? C^
217
CH 2
CH 3
.C^
1
220
3
1
223
6
2
226
9
2
227
10
232
15
240
23
H
v
r
/C^ ^C^ CH2
C
HgC
H
CHo3 I
/C.
H 2 C.
CH 2 CH 3
H
H
I
I
.C^
.C^.
.
H 3 C^ I
H CHq
CHo3
I
H CHq
,3,3
HoC
C f
CH'
H
CH 3
H I
I
^C.% ^C. ^CH3
HoC 3
C I
H Each
4
^C I
CH 3
alkyl substituent causes an increase in
V
of 3-6 nm.
Conjugated Compounds and Ultraviolet Spectroscopy
335
14.42 H
H I
H
CH,
I
<>^v.
HoC^ 2
H2 C I
^.
C
C
I
I
H
H
CH, 2,3-Dimethyl- 1 ,3,5-hexatriene
1,3,5-Hexatriene
\nax = 258
CH 2
nm
^max
ra
268
nm
In Problem 14.41, we concluded that one alkyl group increases Xmax of a conjugated diene by approximately 5 nm. Since 2,3-dimethyl-l,3,5-hexatriene has two methyl substituents, its
14.43
(a)
UV ^max should be about
nm longer than the Xmax of 1 ,3,5-hexatriene.
10
B-Ocimene, Q0H16, has three degrees of unsaturation. Catalytic hydrogenation yields a hydrocarbon of formula C10H22. 6-Ocimene thus contains three double bonds and no rings.
(b) (c)
The ultraviolet absorption at 232 nm indicates that 6-ocimene The carbon skeleton, as determined from hydrogenation, is:
is
conjugated.
CH 3
CH3
CHgCHpCHCHpCHpCHpCHCHg 2,6-Dimethyloctane
Ozonolysis data are used to determine the location of the double bonds. The acetone fragment, which comes from carbon atoms 1 and 2 of 2,6-dimethyloctane, fixes the position of one double bond. Formaldehyde results from ozonolysis of a double bond at the other end of fi-ocimene. Placement of the other fragments to conform to the carbon skeleton yields the following structural formula for 8-ocimene.
CHg
CH3
H 2 C= CHC= CHCH 2 CH= CCH 3 p-Ocimene (d)
I
H 2 C= CHC= CHCH 2 CH= CCH 3 1.
H2
1
3
H3
+
I
CH3CH 2 CHCH 2 CH 2 CH 2 CHCH3
Pd
p-Ocimene
2. Zn,
CHo3
CHo
9H3
CH,v
2,6-Dimethyloctane
CH 3
(b)
Phenol
> benzene >
(c)
Aniline
> benzene > bromobenzene >
toluene
> benzene >
nitrobenzene
chlorobenzene
>
benzoic acid
benzaldehyde
1 in the text for the directing effects of substituents. You should the effects of the most important groups. As in Worked Example 16.2, identify the directing effect of the substituent, and draw the product.
Refer to Figure 16. 1
memorize
The -NO2 group
is
a meta-director.
CI
Chemistry of Benzene:
Electrophilic Aromatic Substitution
367
Br
No catalyst is necessary because aniline is highly 16.10 An
acyl substituent
is
deactivating.
reactive to further substitution. substituted ring
is
more
An
activating.
Once an aromatic alkyl substituent
ring has been acylated,
is
activating,
it is
much
less
however, so an alkyl-
reactive than an unsubstituted ring, and polysubstitution occurs
readily.
16.11
(Trifluoromethyl)benzene is less reactive toward electrophilic substitution than toluene. electronegativity of the three fluorine atoms causes the trifluoromethyl group to be electron-withdrawing and deactivating toward electrophilic substitution. The electrostatic potential map shows that the aromatic ring of (trifluoromethyl)benzene is more electronpoor, and thus less reactive, than the ring of toluene (red).
The
16.12
more favored For acetanilide, resonance derealization of the nitrogen lone pair electrons to the aromatic ring is less favored because the positive charge on nitrogen is next to the positively polarized carbonyl group. Resonance derealization to the carbonyl oxygen is favored because of the electronegativity of oxygen. Since the nitrogen lone pair electrons are less available to the ring than in aniline, the reactivity of the ring toward electrophilic substitution is decreased, and acetanilide is less reactive than aniline toward electrophilic substitution.
368
Chapter 16
16.13 Ortho
attack:
fi
H
cf
H —
-,.
i
N II
Least
:o:
:0:
N
N
:o:
stable
Meta attack: :
't
+
:
.?
e
N
Para attack:
E
,
\\
Least
s
/
II
II
:0:
:o:
stable
The
x<
,
two positive charges thus favored.
circled resonance forms are unfavorable, because they place
adjacent to each other.
The
intermediate from meta attack
is
16.14 (a)
OCH<
OCH
ix
Br
Both groups are ortho, para directors and direct substitution two groups for steric reasons.
to the
same
positions. Attack
doesn't occur between the
NH, Br
E
+
Both groups are ortho, para directors, but direct to different positions. Because group is a more powerful activator, substitution occurs ortho and para to it.
-NH2
Chemistry of Benzene:
Electrophilic Aromatic Substitution
369
Both groups are deactivating, but they orient substitution toward the same positions.
16.15
N0 2 Although both groups are ortho, para directors, the methyl group directs the orientation of the substituents because it is a stronger activating group than bromine.
N0 2 The methoxyl group
directs substitution to the positions ortho
and para to
it.
370
Chapter 16
16.16 Hydroxide
is
used to form the nucleophilic phenoxide anion.
OCH 2 CH 3
'•?.'•
Oxyfluorfen
Meisenheimer complex Step 1: Addition of the nucleophile. Step 2: Elimination of fluoride ion.
The
nitro
group makes the ring electron-poor and vulnerable to attack by the nucleophilic It also stabilizes the negatively charged Meisenheimer complex.
RCT group. 16.17
H-
HoO
"OH
OH +
Br
Br"
p-Methylphenol
/7-Bromotoluene
CH<
CH<
HoO
"OH
6r-
Br
m-Bromotoluene
+
m-Methylphenol
o-Methylphenol
OH /n-Methylphenol
CH.
+ Br"
/?-Methylphenol
Treatment of m-bromotoluene with NaOH leads to two possible benzyne intermediates, which react with water to yield three methylphenol products.
Chemistry of Benzene:
16.18
Oxidation takes place
at the
Electrophilic Aromatic Substitution
371
benzylic position.
/?-ter?-Butylbenzoic acid
Treatment with untouched.
KMn04 oxidizes the methyl group but leaves the terf-butyl group
16.19
Bond
Bond dissociation energy
CH 3 CH 2
—H
.CH 2
421 kJ/mol
—H
375 kJ/mol
H 2 C=CHCH 2
—H
369kJ/mol
Bond dissociation energies measure the amount of energy that must be supplied to cleave a bond into two radical fragments. A radical is thus higher in energy and less stable than the compound from which it came. Since the C-H bond dissociation energy is 421 kJ/mol for ethane and 375 kJ/mol for a methyl group C-H bond of toluene, less energy is required to form a benzyl radical than to form an ethyl radical. A benzyl radical is thus more stable than a primary alkyl radical by 46 kJ/mol. The bond dissociation energy of an allyl C-H bond is 369 kJ/mol, indicating that a benzyl radical is nearly as stable as an allyl radical.
16.20 Br
16.21
372
Chapter 16
16.22
(a) In
order to synthesize the product with the correct orientation of substituents, benzene nitrated before it is chlorinated.
must be
HNO,
Clc
H 2 S0 4
FeCI< CI
m-Chloronitrobenzene (b) Chlorine
can be introduced into the correct position
if benzene is first
acylated.
The
chlorination product can then be reduced. .
CHo
CH2CH3
O II
CH3CCI AICI3
m-Chloroethylbenzene followed by chlorination, reduction, and nitration, is the only route that gives a product in which the alkyl group and chlorine have a meta relationship. (c) Friedel-Crafts acylation,
O II
CH3 CH 2 CCI AICI<
H2 Pd/C
4-Chloro- 1 -nitro-2-propy lbenzene
pathway, remember that the ring must be sulfonated after Friedel-Crafts alkylation because a sulfonated ring is too deactivated for alkylation to occur. Performing the reactions in this order allows the first two groups to direct bromine (d) In planning this
to the
same
position.
SO3H
—
CHoCI AlClr
3-Bromo-2-methylbenzenesulfonic acid
Chemistry of Benzene:
16.23
Electrophilic Aromatic Substitution
(a) Friedel-Crafts acylation, like Friedel-Crafts alkylation,
does not occur
at
an aromatic
ring carrying a strongly electron-withdrawing group. (b)
There are two problems with this synthesis as it is written: Rearrangement often occurs during Friedel-Crafts alkylations using primary 1 .
halides. 2.
Even if /?-chloropropylbenzene could be synthesized, introduction of the second -CI group would occur ortho, not meta, to the alkyl group.
A possible route to this compound: ,
o
Ch^CHg
CH 2 CH 3
O^.
Ch^Ch^CHg
i 1
ii
CH3 CH 2 CCI
r
AIGIg
\
ci 2
Jl
FeCI 3
T CI
Visualizing Chemistry
16.24
(a)
(1)
The methoxyl group
is
an ortho-para director.
^^OCH
^S^OCHg
OCH<
3
Br c
FeBr, Br
p-Bromomethoxybenzene (2)
a
OCH< CH3CCI CH<
/7-Methoxyacetophenone (b)
o-B romomethoxybenzene
AICk
Both functional groups
direct substituents to the
o-Methoxyacetophenone
same position.
(1)
Br 2
FeBrc
3-Bromo-4-methylbenzaldehyde
373
374
Chapter 16
(2)
H.
H.
O
O
3-Acetyl-4-methylbenzaldehyde
16.25
In the lowest-energy conformation of this biphenyl, the aromatic rings are tilted. If the rings had a planar relationship, steric strain between the methyl groups and the ring
hydrogens on the second ring would occur. Complete rotation around the single bond doesn't take place because the repulsive interaction between the methyl groups causes a barrier to rotation.
16.26
C0 2 H
CH3 KMn0 4
CHoCI *-9 AlCIo
OoN
16.27 Imagine two
routes for synthesis of m-nitrotoluene: Alkylation of benzene, followed by nitration, doesn't succeed because an alkyl group is (1) an o,p-director. (2) Nitration of benzene, followed by alkylation, doesn't succeed because nitrobenzene is unreactive to Friedel-Crafts alkylation. Thus, it isn't possible to synthesize m-nitrotoluene by any route that we have studied in this chapter.
Additional Problems Reactivity and Orientation of Electrophilic Substitutions
16.28 Group: (a)
-j-N(CH 3
)
Identification:
2
e>,/7-activator
Reason: Reaction intermediates are stabilized by electron donation
by the amine (b)
nitrogen.
0,p-activator
Reaction intermediates are stabilized by the electron donating inductive effect of the alkyl group.
0,/?-activator
Reaction intermediates are stabilized by electron donation by the ether oxygen.
m-deactivator
Reaction intermediates are destabilized
by electron with-
drawal by the carbonyl oxygen.
Chemistry of Benzene:
Electrophilic Aromatic Substitution
16.29
o-B romonitrobenzene
/?-Bromonitrobenzene
m-Dinitrobenzene
m-Nitrobenzenesulfonic acid
o-Methoxynitrobenzene
Only methoxybenzene
16.30
Most reactive
reacts faster than
/?-Methoxynitrobenzene
benzene (See Figure 16.1
1).
> Least reactive
Benzene > Chlorobenzene > o-Dichlorobenzene Phenol > Nitrobenzene > p-Bromonitrobenzene (c) o-Xylene > Fluorobenzene > Benzaldehyde (d) p-Methoxybenzonitrile > p-Methylbenzonitrile > Benzonitrile (a)
(b)
375
Width: 612 Height: 792
376
Chapter 16
16.31 (a)
a
1
ch 3 ci
a:-
AICk
o-Bromotoluene
^s
HON
s
/7-Bromotoluene
H0\
.Br
^\ /
Y<
Br
Y
CH<
5-Bromo-2-methylphenol
Both groups
direct substitution to the
same
3-Bromo-4-methylphenol
position.
(c)
No
reaction.
A1C1 3 combines with
form a complex
-NH2 t0
that deactivates the ring
toward Friedel-Crafts alkylation. (d)
CH3CI
No
reaction.
The
ring
is
deactivated.
AICI. CI
(e)
OH
XX
OH
v XX
CH3CI
AICk
cr
CI
ci
2,4-Dichloro-6-methylphenol
C0 2 H
(f)
CH3CI AICk (g)
No
reaction.The ring
is
deactivated.
No reaction.The ring
is
deactivated.
^\^S0 Hr
fir
3
7T
CH3CI
AICk
(h)
CH<
Br 1 ,4-Dibromo-2,5-dimethylbenzene
Alkylation occurs in the indicated position because the methyl group is more activating than bromine, and because substitution rarely takes place between two groups.
Electrophilic Aromatic Substitution
Chemistry of Benzene:
16.32
OsN^^Ts^OH CI,
FeCI-
v XJ
2N *
^
cr
4-Chloro-3-nitrophenol
The -OH group
ci
2-Chloro-5-nitrophenol
directs the orientation of substitution,
(b)
^s.CH
CH,
3 ci.
FeCI,
CH,
CH 3
CI
4-Chloro-l,2dhnethylbenzene (c)
CI Clr
W\/
l-Chloro-2,3dimethylbenzene
C ° 2H
C0 2 H
FeCI,
2N
2N
3-Chloro-4-nitro-
benzoic acid
Both groups are deactivating
to a similar extent,
2-Chloro-4-nitrobenzoic acid
and both possible products form.
S0 3 H CI,
FeCI-
4-B romo-3 -chlorobenzenesulfonic acid
16.33 (a)
SO,
H 2 S0 4
HO3S
v
p-Fluorobenzenesulfonic acid
v
SO3H
o-Fluorobenzenesulfonic acid
(b)
H 2 S0 4
HO3S
SO3H
2-Bromo-4-hydroxy-
4-Bromo-2-hydroxy-
benzenesulfonic acid
benzenesulfonic acid
377
378
Chapter 16
(c)
CI
SO,
H 2 S0 4
H0 3
\^^
CI
S^^
2,4-Dichlorobenzenesulfonic acid (d)
A S0 3 H
OH SO,
H 2 S0 4 Br
Br
"
Br
3,5-Dibromo-2-hydroxybenzenesulfonic acid
>
16.34 Most reactive
Least reactive
Phenol > Toluene > p-Bromotoluene > Bromobenzene Aniline and nitrobenzene don't undergo Friedel-Crafts alkylations.
16.35 (a) f?
Pd
NH 2
NOc
o-Ethylaniline Catalytic hydrogenation reduces both the aromatic ketone and the nitro group.
(b)
Br
^\
1 |1
HNO3, H 2 SQ 4
2. Fe,
H3 +
Br
3,4-Dibromoaniline Nitration, followed
by reduction with Fe, produces substituted
KMnO, H2
or
C0 2 H
oBenzenedicarboxylic acid (Phthalic acid)
Aqueous
2,3-Dibromoaniline
KMn04 oxidizes alkyl side chains to benzoic acids.
anilines.
Chemistry of Benzene:
Electrophilic Aromatic Substitution
379
(d)
5-Chloro-2-methoxypropylbenzene
4-Chloro-2-isopropyl-
methoxybenzene
directs substitution because it is a more powerful activating group. Rearranged and unrearranged side chains are present in the products.
The methoxyl group
16.36
(d)
'^Tss
.N(CH 2 CH 3
)
2
N(CH 2 CH 3 2 )
SO
N(CH 2 CH 3 2 )
+
H 2 S0 4
H0 3 S
SO3H
380
Chapter 16
Mechanisms of Electrophilic Substitutions 16.37 1 ^ :
|_|^
Base
H 6+
8"
—
CI because chlorine is a more electronegative element than IC1 can be represented as I iodine. Iodine can act as an electrophile in electrophilic aromatic substitution reactions.
16.38
S0 3 H k
H
fa
H— OSO3H
2>
+
is
the reverse of the sulfonation
S0 3 H
H
+ "OSO3H This mechanism
+
mechanism
illustrated in the text.
H+ is
the electrophile in this reaction.
16.39
H— OPO3H2 CH 3 C= CH 2
(CH 3 3 C+ )
C(CH 3 3
+
~OP0 3 H 2
)
C(CH 3
):
V :Base Phosphoric acid protonates 2-methylpropene, forming a tert-butyl carbocation. This carbocation acts as an electrophile in a Friedel-Crafts reaction to yield rm-butylbenzene.
Chemistry of Benzene:
16.40 When an
Electrophilic Aromatic Substitution
electrophile reacts with an aromatic ring bearing a
381
+
(CH3)3N - group:
This is a destabilizing resonance form because two positive charges are next to each other.
Meta
attack:
This form
is
destabilizing.
The AWAf-trimethylarnmonium group has no electron-withdrawing resonance effect because it has no vacant p orbitals to overlap with the n orbital system of the aromatic The (CH3)3N +- group is inductively deactivating, however, because it is positively charged.
It is
ring.
meta-directing because the cationic intermediate resulting from meta attack
somewhat more
16.41 The
stable than those resulting
from ortho or para
is
attack.
aromatic ring is deactivated toward electrophilic aromatic substitution by the combined electron-withdrawing inductive effect of electronegative nitrogen and oxygen. The lone pair of electrons of nitrogen can, however, stabilize by resonance the ortho and para substituted intermediates but not the meta intermediate.
Ortho attack:
382
Chapter 16
Meta
attack: :o:
:o:
HE
HE
:o:
HE
HE
16.42 CHCI 3 + AICI3
«
*
+
CHCI 2 AICI4"
(Dichloromethyl)benzene
(Dichloromethyl)benzene can react with two additional equivalents of benzene by the same to produce triphenylmethane.
mechanism
Chemistry of Benzene:
Electrophilic Aromatic Substitution
383
16.43
structures show that bromination occurs in the ortho and para positions of the The positively charged intermediate formed from ortho or para attack can be stabilized by resonance contributions from the second ring of biphenyl, but this
Resonance rings.
stabilization is not possible for
meta
attack.
384
Chapter 16
16.44
HO— OH
—
acid ,
,
»
catalyst
HO— OHo*
reactive electrophile
PH
HO-OHo
H
^
:OH 2
OH
Attack of jt
electrons
Loss of proton
on reactive electrophile
The
reactive electrophile (protonated
H2O2)
is
equivalent to
+
OH.
Organic Synthesis
16.45
C0 2 H
CH 3
IH 3
CH3C1 FeBr<
AlCIo
oBromobenzoic (b)
acid
OH CH3CI
1
AlCIo
2.
NaOH CH 3 Br
CH 3 p-Methoxytoluene
The
reactions in (b) can be performed in either order,
(c)
CH<
C0 2 H
CH<
N0 2
OoN
2N
CH3CI
HNO<
KMnO/i
AlCIo
H 2 S0 4
HoO NOc
N0 2 2,4,6-Trinitrobenzoic acid
Chemistry of Benzene:
Electrophilic Aromatic Substitution
385
m-Bromoaniline
16.46 When
synthesizing substituted aromatic rings, it is necessary to introduce substituents in group that is introduced out of order will not have the proper directing
the proper order. effect.
A
Remember that in many of these reactions
a mixture of ortho and para isomers
be formed.
/7-Chloroacetophenone
ra-Bromonitrobenzene
THF H 2 2 "OH ,
X
CHoCHo'
is a substituted phenol, whose -OH group directs the orientation of the -C(CH3)3 groups. The precursor to MON-0585 is synthesized by a Friedel-Crafts
16.56 The product
alkylation of phenol
by
by the appropriate hydrocarbon
halide. This
compound
is
synthesized
NBS bromination of the product of alkylation of benzene with 2-chloropropane.
MON-0585
C(CH 3) 3
Electrophilic Aromatic Substitution
Chemistry of Benzene:
391
16.57 +
(1) '
OH
:OH
^H— OS03 H
11
H
H'
Formaldehyde
is
H
H
+
H
protonated to form a carbocation.
(2)
H
? clv
CI-
CH 2 OH
Ylb)
CH 2 OH
Hv:Base
CI
jf^ CI
cation acts as the electrophile in a substitution reaction at the "6" position of 2,4,5-trichlorophenol.
The formaldehyde
H
CH 2 — OH 2
CI
CI
H-rOS0 3 H CI
CI
The product from
CI
step 2
is
+ ~:OS0 3 Hj
protonated by strong acid to produce a cation.
(4)
Hexachlorophene This cation is attacked by a second molecule of 2,4,5-trichlorophenol to produce hexachlorophene.
392
Chapter 16
16.58 :o: II
heat
a?
9
N=N Benzyne +
The
boron atom
C0 2
+
N2
phenylboronic acid has only six outer-shell electrons and is a resonance forms for phenylboronic acid in which an electron pair from the phenyl ring is delocalized onto boron. In these resonance forms, the ortho and para positions of phenylboronic acid are the most electron-deficient, and substitutions occur primarily at the meta position. trivalent
Lewis
in
acid. It is possible to write
Electrophilic Aromatic Substitution
Chemistry of Benzene:
16.60 Resonance forms Br
for the intermediate
H
Br
Br
from attack
at
H
C
1
Br
H
Br
393
H
Br
H
Br
H
H
+
Resonance forms
for the intermediate
Br
from attack
Br
at
C2:
Br
+
Br
There are seven resonance forms for attack at CI and six for attack at C2. Look carefully at the forms, however. In the first four resonance structures for CI attack, the second ring is still fully aromatic. In the other three forms, however, the positive charge has been delocalized into the second ring, destroying the ring's aromaticity. For C2 attack, only the first two resonance structures have a fully aromatic second ring. Since stabilization is lost when aromaticity is disrupted, the intermediate from C2 attack is less stable than the intermediate from CI attack, and CI attack is favored.
394
Chapter 16
16.61
I
:OCH 3
:o:
ci
OCH 3
L.
Meisenheimer complex Step 1: Addition of the nucleophile -OCH3. Step 2: Elimination of -Cl~.
The carbonyl oxygens make the chlorine-containing ring electron-poor and open to attack by the nucleophile ""OCH3. They also stabilize the negatively charged Meisenheimer complex.
16.62 •:Base
+ :
cr
N(CH 3) 2
(CH 3 2 NH )
3.
2.
N
N'
Step 1: Attack of the nucleophile diethylamine. Step 2: Loss of proton. Step 3: Loss of CI". This reaction
is
an example of nucleophilic aromatic substitution. Dimethylamine
a
is
nucleophile, and the pyridine nitrogen acts as an electron- withdrawing group that can stabilize the negatively-charged intermediate.
16.63
Step Step
The
1: 1:
Abstraction of proton and elimination of Br Addition of NH3 to the benzyne intermediate to form two aniline products. .
reaction of an aryl halide with potassium amide proceeds through a benzyne Ammonia can then add to either end of the triple bond to produce the
intermediate.
methylanilines observed.
two
Chemistry of Benzene:
Electrophilic Aromatic Substitution
16.64
Protonation of the cyclic ether creates a carbocation intermediate that can react in a Friedel-Crafts alkylation.
OH The intermediate
alkylates benzene, forming an alcohol product.
Protonation of the alcohol, followed by loss of water, generates a second carbocation.
(d)
This carbocation undergoes internal alkylation to yield the observed product.
395
Width: 612 Height: 792
396
Chapter 16
16.65
Step 1: Formation of primary carbocation. Step 2: Rearrangement to a secondary carbocation. Step 3: Attack of ring n electrons on the carbocation. + Step 4: Loss of H .
This reaction takes place despite the fact that an electron- withdrawing group the ring. Apparently, the cyclization reaction is strongly favored.
is
16.66
W_
..
:C=0 +
*
Carbon monoxide
The
HCI,AICI 3
is
» :
H— C=0+
" AICI 4
protonated to form an acyl cation.
acyl cation reacts with benzene
by a Friedel-Crafts acylation mechanism.
attached to
Electrophilic Aromatic Substitution
Chemistry of Benzene:
397
16.67
OH
H
H
f~
:
Base
+ (CH 3 ) 2 C JC3H 2
(CH 3 2 C^~ CH2 )
Loss of
H
C(CH 3) 3
)
3
proton
Loss of
Protonation of aromatic ring
16.68 Both
C(CH 3
tert-butyl carbocation
of these syntheses
test
^
your
ability to carry out steps in the correct order.
CHo
CHo
CH< Br
HNO 3^ H 2 S0 4
CH 3 CI AlCIo
Br 2
FeBrc
N0 2
NOc
+ ortho isomer (b)
0>,
C
^CH(CH 3
)
2
c
o II
(CH 3 2 CHCQ * AiCk )
FeCI
CH 2 CH(CH 3
)
2
S0 3 H
CI
^CH(CH 3
)
2
398
Chapter 16
16.69 Problem 16.51 shows the mechanism of the addition of HBr to 1-phenylpropene and shows how the aromatic ring stabilizes the carbocation intermediate. For the methoxylform can be drawn in which the cation is by the electron-donating resonance effect of the oxygen atom. For the nitrosubstituted styrene, the cation is destabilized by the electron-withdrawing effect of the substituted styrene, an additional resonance
stabilized
nitro
group.
Thus, the intermediate resulting from addition of HBr to the methoxyl- substituted styrene is more stable, and reaction of /7-methoxystyrene is faster.
16.70 HoC
— S — CH
(CH 3 2 S )
arc -
H 2.
Base
+ Br
1: Sn2 displacement takes place when the negatively charged oxygen of dimethyl sulfoxide attacks the benzylic carbon of benzyl bromide, displacing Br~.
Step
Step
2:
Base removes a benzylic proton, and dimethyl
sulfide
is
eliminated in an
reaction.
16.71 NH,
H = 1.52
D
strong electron-withdrawing
-NH 2 has
a strong
inductive effect.
resonance
n
=
1.53
D
-Br has a
electron-donating effect.
NH,
Br
[i
=
D
2.91
The polarities of the two groups add to produce a net dipole moment almost equal to the sum of the individual
moments.
E2
Chemistry of Benzene:
Electrophilic Aromatic Substitution
399
16.72
O
(a)
CH3CH2COCI,
AICI3; (b)
H2
,
Pd/C;
(c)
Br2 FeBr 3 ,
;
(d)
NBS, (PhC0 2 ) 2
;
(e)
KOH,
ethanol
16.73
An electron-withdrawing
substituent destabilizes a positively charged intermediate (as in
electrophilic aromatic substitution) but stabilizes a negatively charged intermediate.
dissociation of a phenol, an
For the
-N02 group stabilizes the phenoxide anion by resonance, thus
lowering AG and pKa In the starred resonance form for /7-nitrophenol, the negative charge has been delocalized onto the oxygens of the nitro group. .
16.74 For the same
reason described in the previous problem, a methyl group destabilizes the AG° and pA^, making this phenol less acidic.
negatively charged intermediate, thus raising
Review Unit
6:
Conjugation and Aromaticity
Major Topics Covered (with vocabulary): Conjugated dienes: derealization 1 ,4-addition allylic position thermodynamic control kinetic control vulcanization Diels-Alder cycloaddition dienophile endo product exo product s-cis conformation Ultraviolet spectroscopy:
highest occupied molecular orbital absorptivity
(HOMO)
lowest unoccupied molecular orbital
Aromaticity: aromatic arene phenyl group benzyl group Hiickel 4/i + 2 rule antiaromatic heterocycle
(LUMO)
molar
ortho, meta, para substitution degenerate polycyclic aromatic compound ring current
Chemistry of aromatic compounds: sulfonation F-TEDA-BF4 Friedel-Crafts alkylation Friedel-Crafts acylation ortho- and para-directing activator ortho- and para-
electrophilic aromatic substitution
polyalkylation
directing deactivator meta-directing deactivator
aromatic substitution
inductive effect
resonance effect
nucleophilic
Meisenheimer complex benzyne benzylic position
Types of Problems: After studying these chapters, you should be able to:
-
Predict the products of electrophilic addition to conjugated molecules.
Understand the concept of kinetic vs. thermodynamic control of reactions. Recognize diene polymers, and draw a representative segment of a diene polymer. Predict the products of Diels-Alder reactions, and identify compounds that are good dienophiles and good dienes. Calculate the energy required for UV absorption, and use molar absorptivity to calculate concentration.
-
Predict
-
Name
-
if
and where a compound absorbs
in the ultraviolet region.
and draw substituted benzenes. structures and molecular
Draw resonance
orbital diagrams for benzene and other cyclic conjugated molecules. Use Huckel's rule to predict aromaticity. Draw orbital pictures of cyclic conjugated molecules. data to deduce the structures of aromatic compounds. Use NMR, IR and
UV
Predict the products of electrophilic aromatic substitution reactions.
Formulate the mechanisms of electrophilic aromatic substitution reactions. Understand the activating and directing effects of substituents on aromatic rings, and use inductive and resonance arguments to predict orientation and reactivity. Predict the products of other reactions of aromatic compounds. Synthesize substituted benzenes.
Review Unit 6
Points to *
It's
401
Remember:
not always easy to recognize Diels-Alder products, especially
bond of the
product has been hydrogenated.
if
the carbon-carbon double
no hydrogenation has taken place, look for a double bond in a six-membered ring and at least one electron-withdrawing group across the ring from the double bond. When a bicyclic product has been formed, it has probably resulted from a Diels-Alder reaction in which the diene is cyclic.
*
To be
initial
If
aromatic, a molecule must be planar, cyclic, conjugated, and
electrons in
its
it
must have 4n + 2
n system.
*
The carbocation
*
Nucleophilic aromatic substitution reactions and substitution reactions proceeding through benzyne intermediates take place by different routes. In the first reaction, the substitution takes place by an addition, followed by an elimination. In the second case, the substitution involves an elimination, followed by an addition. Virtually all substitutions are equivalent to an addition and an elimination (in either order).
*
Activating groups achieve their effects by making an aromatic ring more electron-rich and reactive toward electrophiles. Ortho and para directing groups achieve their effects by
intermediate of electrophilic aromatic substitution loses a proton to yield the aromatic product. In all cases, a base is involved with proton removal, but the nature of the base varies with the type of substitution reaction. Although this book shows the loss of the proton, it often doesn't show the base responsible for proton removal. This doesn't imply that the proton flies off, unassisted; it just means that the base involved has not been identified in the problem.
from ortho or para addition of an electrophile to the aromatic ring. The intermediate resulting from addition to a ring with an ortho or para director usually has one resonance form that is especially stable. The intermediate resulting from addition to a ring with a meta director usually has a resonance form that is especially unfavorable when addition occurs ortho or para to the functional group. Meta substitution results because it is less unfavorable than ortho or para substitution. stabilizing the positive charge that results
Self-test:
a-Farnesene a-Farnesene (A), an important biological intermediate in the synthesis of many natural products, has double bonds that are both conjugated and unconjugated. Show the products you would expect from conjugate addition of HBr; of Eto. What products would you expect from ozonolysis of A? Give one or more distinctive absorptions that you might see in the IR spectrum of
A and distinguishing features of the H NMR of A. Would you expect A to be UV-active? ]
Review Unit 6
402
C
B
D
Paroxypropione Describe the n orbitals in the ring of B. Might this ring be described as aromatic? Paroxypropione (C) is a hormone inhibitor. Predict the products of reaction of C with: (a) Br2 , FeBr 3 (b) CH3CI, A1C1 3 (c) KMn0 4 , H3O; (d) H 2 Pd/C. If the product of (d) is treated with the reagents in (a) or (b), does the orientation of substitution change? What significant information can you obtain from the IR spectrum of C? Name D. Plan a synthesis of D from benzene. Describe the *H of D (include spin-spin splitting). Where might D show an absorption in a UV spectrum? ;
,
;
NMR
Multiple choice: 1
.
What (a)
2.
are the hybridizations of the carbons in
sp 2 , sp 2 , sp 2 , sp 2
more (c)
which the
sp 2 sp 2 sp 2 sp 3 ,
,
,
1
,2-butadiene, starting with (c)
sp 2 sp, sp 2 sp 3 ,
,
CI?
(d) sp, sp,
sp 2 sp 3 ,
product is formed at lower temperature, and the formed at higher temperature: AGras ° > AGi s ° and AGms * > AGis* (b) AG ms ° > AGi s ° and AGi s * > AGms * ° ° * < and AG > * AGms (d) AGms ° < AGi s ° and AGi s * > AGms * AGi s AGi s ms
In a reaction in (a)
(b)
less stable (Is)
stable product (ms) is
Note: In this problem, a large value for AG° means a large ne gative value. 3
.
Which of the following combinations
is
most
likely to
undergo a successful Diels-Alder
reaction?
(a)
4
.
(b)
Which of the following
(d)
(c)
groups,
when bonded
to the terminal carbon of a conjugated k
system, probably affects the value of Xmax the least?
(a)-NH 2 (b)-Cl (c)-OH (d)-CH 3 5
.
value of Xmax for an unsubstituted diene is approximately 220 nm, and each additional double bond increases the value of Xmax by 30 nm, what is the minimum number of double bonds present in a compound that absorbs in the visible range of the electromagnetic spectrum? (a) 6 (b)7 (c)8 (d)9 If the
Review Unit 6
6
.
7
.
Which of the following compounds
9
.
.
10
(b) 11
(c)
12
(d)
14
Which of the following functional groups isn't (a)-N0 2 (b)-CONHCH 3 (c)-N(CH 3 ) 3 + Which of the following compounds substitution reaction that (a)
10.
aromatic?
How many benzene isomers of CvH6Br2 can be drawn? (a)
8
is
m-Cresol
we have
can't
a meta-directing deactivator? (d)
-NHCOCH3
be synthesized by an electrophilic aromatic
studied?
(b) p-Chloroaniline (c) 2,4-Toluenedisulfonic acid
compounds can you reduce Which compound is it?
In only one of the following
reducing the side chain. (a)
p-Bromoanisole
(d)
Phenylacetylene
(b)
(d)
m-Bromotoluene
the aromatic ring without also
Acetophenone (methyl phenyl ketone)
(c)
Styrene
403
Chapter 17 - Alcohols and Phenols
Chapter Outline
I.
Naming
alcohols and phenols (Section 17.1). A. Alcohols are classified as primary, secondary or organic groups bonded to the -OH carbon. B Rules for naming simple alcohols. 1 The longest chain containing the -OH group
tertiary,
depending on the number of
.
.
name replaces 2
.
3
.
is
the parent chain, and the parent
-e with -ol.
Numbering begins at the end of the chain nearer the -OH group. The substituents are numbered according to their position on the chain and
cited in
alphabetical order.
II.
C. Phenols are named according to rules discussed in Section 15.1 for aromatic compounds. Properties of alcohols and phenols (Section 17.2). A. Hydrogen-bonding of alcohols and phenols. 1 Alcohols have sp hybridization and a nearly tetrahedral bond angle. 2. Alcohols and phenols have elevated boiling points, relative to hydrocarbons, due to hydrogen-bonding a. In hydrogen-bonding, an -OH hydrogen is attracted to a lone pair of electrons on another molecule, resulting in a weak electrostatic force that holds the .
molecules together.
These weak forces must be overcome in boiling. Acidity and basicity of alcohols and phenols. 1 Alcohols and phenols are weakly acidic as well as weakly basic. 2 Alcohols and phenols can be reversibly protonated to form oxonium ions. Alcohols and phenols dissociate to a slight extent to form alkoxide ions and 3 phenoxide ions.
b
B
.
.
.
.
4.
Acidity of alcohols. a.
b. c.
Alcohols are similar in acidity to water. Alkyl substituents decrease acidity by preventing solvation of the alkoxide ion. Electron- withdrawing substituents increase acidity by delocalizing negative charge.
d.
5
.
Alcohols don't react with weak bases, but they do react with alkali metals and
strong bases. Acidity of phenols. Phenols are a million times a.
more
acidic than alcohols
and are soluble
in dilute
NaOH. b. c.
III.
Phenol acidity is due to resonance stabilization of the phenoxide anion. Electron-withdrawing substituents increase phenol acidity, and electrondonating substituents decrease phenol acidity.
Alcohols (Sections 17.3-17.8). A. Preparation of alcohols (Sections 17.3-17.5). 1
.
Familiar methods (Section 17.3). a. Hydration of alkenes. i. Hydroboration/oxidation yields non-Markovnikov products. ii. Oxymercuration/reduction yields Markovnikov products. b. 1,2-diols can be prepared by OSO4 hydroxylation, followed by reduction. This reaction occurs with syn stereochemistry. i. Ring-opening of epoxides produces 1,2-diols with anti stereochemistry. ii.
Alcohols and Phenols
2.
405
Reduction of carbonyl compounds (Section 17.4). Aldehydes are reduced to primary alcohols. a. b. Ketones are reduced to secondary alcohols. Either NaBH4(milder) or LiAlH4(more reactive) can be used to reduce 1 aldehydes and ketones. c. Carboxylic acids and esters are reduced to primary alcohols with LLA.1H4. i. These reactions occur by addition of hydride to the positively polarized carbon of a carbonyl group. ii. Water adds to the alkoxide intermediate during workup to yield alcohol .
3
.
product. Reaction of carbonyl compounds with Grignard reagents (Section 17.5). a. RMgX adds to carbonyl compounds to give alcohol products. i. Reaction of RMgX with formaldehyde yields primary alcohols. ii. Reaction of RMgX with aldehydes yields secondary alcohols. iii. Reaction of RMgX with ketones yields tertiary alcohols. iv. Reaction of RMgX with esters yields tertiary alcohols with at least two identical R groups bonded to the alcohol carbon. v No reaction occurs with carboxylic acids because the acidic hydrogen quenches the Grignard reagent. b Limitations of the Grignard reaction. Grignard reagents can't be prepared from reagents containing other reactive i. .
.
functional groups.
Grignard reagents can't be prepared from compounds having acidic hydrogens. Grignard reagents behave as carbon anions and add to the carbonyl carbon, A proton from water is added to the alkoxide intermediate to produce the i. ii.
c.
alcohol.
B. Reactions of alcohols (Sections 17.6-17.8). 1 Conversion to alkyl halides (Section 17.6). a. Tertiary alcohols (ROH) are converted to RX by treatment with HX. The reaction occurs by an SnI mechanism. i. b. Primary alcohols are converted by the reagents PBr3 and SOCI2. i. The reaction occurs by an Sn2 mechanism. 2 Conversion into tosylates. a. Reaction with /?-toluenesulfonyl chloride converts alcohols to tosylates. b Only the O-H bond is broken. .
.
.
Tosylates behave as halides in substitution reactions. d Sn2 reactions involving tosylates proceed with inversion of configuration. Dehydration to yield alkenes. a. Tertiary alcohols can undergo acid-catalyzed dehydration with warm aqueous c.
.
3
.
H2SO4. i. ii.
iii.
Zaitsev products are usually formed. The severe conditions needed for dehydration of secondary and primary alcohols restrict this method to tertiary alcohols. Tertiary alcohols react fastest because the intermediate carbocation formed in this
El reaction
is
more
stable.
Secondary alcohols are dehydrated with POCI3 in pyridine. This reaction occurs by an E2 mechanism. i. ii. Pyridine serves both as a base and as a solvent. Conversion into esters. Oxidation of alcohols (Section 17.7). a. Primary alcohols can be oxidized to aldehydes or carboxylic b Secondary alcohols can be oxidized to ketones. b.
4
.
5
.
.
acids.
Width: 612 Height: 792
406
Chapter 17
c.
Tertiary alcohols aren't oxidized.
d.
Oxidation to ketones and carboxylic acids can be carried out with KMnC>4, Cr0 3 , or Na 2 Cr2 7 Oxidation of a primary alcohol to an aldehyde is achieved with the Dess-Martin .
e.
periodinane.
The Dess-Martin periodinane is also used on sensitive alcohols. Oxidation occurs by a mechanism closely related to an E2 mechanism. Protection of alcohols (Section 17.8). a. It is sometimes necessary to protect an alcohol when it interferes with a reaction involving a functional group in another part of a molecule. b The following reaction sequence may be applied: i.
f 5.
.
.
Protect the alcohol.
i.
Carry out the reaction.
ii. iii.
A
c.
Remove the protecting group. trimethylsilyl (TMS) ether can be used for protection.
i.
ii. iii.
TMS ether formation occurs by an Sn2 route. TMS ethers are quite unreactive. TMS ethers can be cleaved by aqueous acid or by F~ to regenerate the alcohol.
IV. Phenols (Sections 17.9-17.10). A. Preparation and uses of phenols (Section 17.9).
Phenols can be prepared by treating chlorobenzene with NaOH. Phenols can also be prepared from isopropylbenzene (cumene). a. Cumene reacts with 2 by a radical mechanism to form cumene hydroperoxide. b Treatment of the hydroperoxide with acid gives phenol and acetone. The mechanism involves protonation, rearrangement, loss of water, i. readdition of water to form a hemiacetal, and breakdown to acetone and phenol. 3 Chlorinated phenols, such as 2,4-D, are formed by chlorinating phenol. 4. BHT is prepared by Friedel-Crafts alkylation of /7-cresol with 2-methylpropene. B. Reactions of phenols (Section 17.10). 1 Phenols undergo electrophilic aromatic substitution reactions (Chapter 16). a. The -OH group is a o,/?-director. 2. Strong oxidizing agents convert phenols to quinones. Reaction with Fremy's salt to form a quinone occurs by a radical mechanism. a. b The redox reaction quinone -* hydroquinone occurs readily. c Ubiquinones are an important class of biochemical oxidizing agents that function as a quinone/hydroquinone redox system. V. Spectroscopy of alcohols and phenols (Section 17.1 1). A. IR spectroscopy. -1 1 Both alcohols and phenols show -OH stretches in the region 3300-3600 cm -1 a. Unassociated alcohols show a peak at 3600 cm -1 b Associated alcohols show a broader peak at 3300-3400 cm -1 2. Alcohols show a C-0 stretch near 1050 cm -1 3 Phenols show aromatic bands at 1500-1600 cm -1 4. Phenol shows monosubstituted aromatic bands at 690 and 760 cm 1
.
2.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
B
.
NMR spectroscopy. In C NMR spectroscopy, carbons bonded to -OH groups absorb in the range 1
1
.
2.
50-80 6. *H NMR. a. Hydrogens on carbons bearing -OH groups absorb in the range 3.5-4.5 The hydroxyl hydrogen doesn't split these signals. i. b.
D2
exchange can be used
to locate the
O-H
signal.
6.
407
Alcohols and Phenols
Spin-spin splitting occurs between protons on the oxygen-bearing carbon and neighboring -H. d Phenols show aromatic ring absorptions, as well as an O-H absorption in the range 3-8 6. C. Mass Spectrometry. Alcohols undergo alpha cleavage to give a neutral radical and an oxygen-containing 1 c.
.
.
cation.
2
.
Alcohols also undergo dehydration to give an alkene radical cation.
Solutions to Problems
17.1
The
parent chain must contain the hydroxyl group, and the hydroxyl group(s) should receive the lowest possible number.
(a)
OH
OH
I
(b)
(c)
OH
HO
I
CH3CHCH2CHCHCH3
^^^s^^^* CH2CH2CCH3
CHo
CH<
CH 3
CH 3 5 -Methy 1-2 ,4-hexanediol
(d)
2-Methy l-4-phenyl-2-butanol (e)
4,4-Dimethylcyclohexanol
(f)
HoC
--Br
or
H
(1
5,25)-2-Bromocyclopentanol
4-Bromo-3-methylphenol 2-Cyclopenten- 1 -ol
17.2 (a)
HoC \
/
(c)
CHoOH 2
c=c
/
H
\
CH2CH2
(Z)-2-Ethyl-2-buten- 1 -ol
3-Cyclohexen- 1 -ol
rrafls-3-Chlorocycloheptanol
and enantiomer
(<*)
OH
(e)
(f)
I
CH3CHCH2CH2CH2OH
H3C
^^^ 1,4-Pentanediol
CH 2CH 2 OH
CH3
2,6-Dimethylphenol
^^ o-(2-Hydroxyethyl)phenol
408
Chapter 17
17.3
In general, the boiling points of a series of isomers decrease with branching. The more nearly spherical a compound becomes, the less surface area it has relative to a straight
A
chain compound of the same molecular weight and functional group type. smaller surface area allows fewer van der Waals interactions, the weak forces that cause covalent molecules to be attracted to each other. In addition, branching in alcohols makes it more difficult for hydroxyl groups to approach each other to form hydrogen bonds. given volume of 2-methyl-2-propanol therefore contains fewer hydrogen bonds than the same volume of 1-butanol, and less energy is needed to break them in boiling.
A
17.4 Most acidic
Least acidic (a)
HC=CH <
(CH 3 2 CHOH < )
alkyne
hindered
< (CF 3) 2 CHOH
CH 3 OH
alcohol with electron-
alcohol
withdrawing groups
alcohol
<
(b) /7-Methylphenol
Phenol
/?-(Trifluoromethyl)phenol
phenol with electrondonating groups (c)
phenol with electronwithdrawing groups
<
Benzyl alcohol
17.5
<
Phenol phenol
alcohol
/?-Hydroxybenzoic acid carboxylic acid
We saw in Chapter
16 that a nitro group is electron- withdrawing. Since electronwithdrawing groups stabilize anions, p-nitrobenzyl alcohol is more acidic than benzyl alcohol. The methoxyl group, which is electron-donating, destabilizes an alkoxide ion, making p-methoxybenzyl alcohol less acidic than benzyl alcohol.
17.6 (a)
CH3CH2 /
CHod
1
c=c\ /
.
2.
CHo3
BH 3 THF ,
H2
2
,
I
CH3CH2CHCHCH3
OH
CHo
H
OH 2-Methyl-3-pentanol
In a hydroboration/oxidation reaction, the hydroxyl group
is
bonded
to the less substituted
H3C 1
Hg(OAc) 2 H 2 Q ,
2.
NaBH 4
2-Methyl-4-phenyl-2-butanol
Markovnikov product results from oxymercuration/reduction. (c)
H \
HO
H /
1
\
2.
Q— Q
/
C4H9
C4H9
.
Os0 4 NaHS0 3 H 2
\
OH /
,-C— C-, u A
,
=
k
C4H9 C4H9 meso-5 ,6-Decanediol
Hydroxy lation
results in a diol with
syn stereochemistry.
HO
P4.H0
\
ijvr
,C— H'A C4H9
\
OH
Alcohols and Phenols
409
17.7 (a)
O
O
II
II
OH 1.NaBH 4
CH 3 CCH 2 CH 2 COCH3
NaBH4 reduces (
H3
2.
O II
I
CH 3 CHCH 2 CH 2 COCH3
+
aldehydes and ketones without interfering with other functional groups.
b)
OH 1
II
II
LiAIH 4
.
~~
CH 3 CCH 2 CH 2 COCH 3
d.
H3
/-«+
I
* CH 3 CHCH 2 CH 2 CH 2 OH
LiAlH4, a stronger reducing agent, reduces both ketones and
esters,
(c)
)H
1
UAIH4
2.
H3 +
LiAlH4 reduces carbonyl functional groups without reducing double bonds.
17.8 f?
OH ^
'
Benzyl alcohol may be the reduction product of an aldehyde, a carboxylic NaBH4 may be used to reduce the aldehyde.
acid, or
an
ester.
(b)
C"
[^jf
CH 3
1
.
LiAIH,
+
2H3
Reduction of a ketone yields the secondary alcohol.
NaBH4 may
also be used here
and
in
(c).
OH
(c)
LiAIHj^ + 2 -H 3 1
(
r
T—
d)
1
(CH 3 2 CHCHO or )
(CH 3) 2 CHC0 2 H or
(CH 3 2 CHC0 2 R
.
UAIH4 (CH 3
)
3
)
2 CHCH 2 OH
410
Chapter 17
17.9
All of the products have an ketone carbon.
17.10
type of alcohol. If the alcohol is primary, it can only be synthesized from formaldehyde plus the appropriate Grignard reagent. If the alcohol is secondary, it is synthesized from an aldehyde and a Grignard reagent. (Usually, there are two combinations of aldehyde and Grignard reagent). A tertiary alcohol is synthesized from a ketone and a Grignard reagent. If all three groups on the tertiary alcohol are different, there are often three different combinations of ketone and Grignard reagent. If two of the groups on the alcohol carbon are the same, the alcohol may also be synthesized from an ester and two equivalents of Grignard reagent.
-OH and a methyl group bonded to what was formerly a
First, identify the
(a)
2-Methyl-2-propanol
is
a tertiary alcohol.
To
synthesize a tertiary alcohol, start with a
ketone.
OH
O II
CHoCCHo d d
1
.
CH 3 MgBr_
2.H 3
t +
^
'
CHoCCHo d d |
CH 3 two or more alkyl groups bonded to the carbon bearing the -OH group are the same, an alcohol can be synthesized from an ester and a Grignard reagent.
If
OH
O II
CH 3 COR
1
.
2
CH 3 MgBr
2.H 3
t +
^
CHoCCHo d 6 j
|
CH 3 2-Methyl-2-propanol
Alcohols and Phenols
(b) Since 1-methylcyclohexanol
is
a tertiary alcohol,
start
411
with a ketone.
.0 1
2.
CH 3 MgBr H3 + -Methylcyclohexanol
1
(c)
3-Methyl-3-pentanol is a tertiary alcohol. When two of the three groups bonded to the alcohol carbon are the same, either a ketone or an ester can be used as a starting material.
II
1
CH3CH2CCH2CH3
.
£
CH 3 MgBr |_|
0+
OH
or
I
CH3CH2CCH2CH3 1
CH 3 CH 2 CCH 3
.
CH 3 CH 2 MgBr^
~l
CH 3
2 H ^ 0+
3-Methyl-3-pentanol
or
O 1
CH3COR (d)
.
2.
2
CH 3 CH 2 MgBr +
H3
Three possible combinations of ketone plus Grignard reagent can be used to synthesize this tertiary alcohol.
CH 2 CH 3
i
1
2.
CHa MgBr H3
+
>
HO
or
CH
1
.
2.
or
CH 3 CH 2 MgB H3 +
II
1
2.
C 6 H 5 MgB H3
+
v
U
CH2CH3
2-Phenyl-2-butanol
o
CH3CH2CCH3
CHo6
\ /
r
412
Chapter 17
(e)
Formaldehyde must be used
to synthesize this
primary alcohol,
^s^CH2OH f?
C 6 H 5 MgBr
2.
H3
H
1
H
+
Benzyl alcohol (f)
As
in (e), use
formaldehyde
B H'
17.11
1
.
primary alcohol.
(CH 3 2 CHCH 2 CH 2 MgBr )
2.H 3
H
to synthesize a
(CH 3 2 CHCH2CH2CH 2 OH
+
)
4-Methyl- 1 -pentanol
of the alcohol. This alcohol, 1-ethylcyclohexanol, is a tertiary alcohol that can be synthesized from a ketone. Only one combination of ketone and Grignard reagent is possible. First, interpret the structure
OH 1
2.
17.12
CH 3 CH 2 MgBr H3
aCHoCHo
+
Recall from Chapter 1 1 that -OH is a very poor leaving group in reactions run under Sn2 conditions. toluenesulfonate, however, is a very good leaving group, and reaction of the toluenesulfonate of the alcohol with ~CN proceeds readily under Sn2 conditions to give the desired product with inversion of configuration at the chirality center.
A
Alcohols and Phenols
413
17.13 (a)
OH
——
CH3CH2
POCI3
CH 3 CH2 CHCH(CH 3 2 )
K
^-^3
/C =C^
,0=0
+
CHg
H
^CH(CHg)2
H
H3C minor
major
^CH(CH3 )2 + /
C=C\ H
H minor
The major product has
the
more
substituted double bond.
POCI3 pyridine
" CH 3 3-Methylcyclohexene In
E2
elimination, dehydration proceeds
have an
most readily when the two groups to be eliminated compound, the only hydrogen with the proper
anti periplanar relationship. In this
stereochemical relationship to the is formed.
-OH group is
at
C6. Thus, the non-Zaitsev product 3-
methylcyclohexene
POCIg
_
pyridine
1
-Methylcyclohexene
Here, the hydrogen at C2 is trans to the hydroxy 1 group, and dehydration yields the Zaitsev product, 1 -methylcyclohexene.
414
Chapter 17
(d)
H3
f CH3CH
c=c\ / H3C
CH3
CH0CHCCH0CH0
...
+ H
2-Ethyl-3-methyl- 1 -butene
minor
major
PQCI3
=cN
CH3CH2
2,3-Dimethyl-2-pentene
OH
HoC
Jri
/C
_
'
pyridine
I
CHo
HoC
H3p
CHgCH /
CH3CH
^CHg
c=c\
H3 C
H
C=C
+ /
\
H3C
H
CH3
(E)-3 ,4-Dimethyl-2-pentene
(Z)-3,4-Dimethyl-2-pentene
minor
minor
Four different products (including E,Z isomers) can result from dehydration of 2,3dimethyl-2-pentanol. The major product has the most substituted double bond, according to Zaitsev's rule.
(e)
OH
CH 3 CH 2
POCI3
CH3CH2CH2CCH3
_
pyridine
/
C=C\ /
+
rj
CH3CH2CH2C — CH2
CH 3
H
CHo
CHo
CH 3
2-Methyl-2-pentene
2-Methyl- 1 -pentene
minor
major
17.14 Aldehydes
are synthesized from oxidation of primary alcohols, and ketones are synthesized from oxidation of secondary alcohols.
CH-
(b)
CHo3
CHo3 I
CH 3 CHCH 2 OH
(c)
Periodinane
CH2CI2
OH H
HoO +
I
CH3CHCHO
Alcohols and Phenols
415
17.15
Cr02
Starting material (a)
H3
+ Product
Periodinane Product
CH3CH 2 CH 2 CH 2 CH 2 CHO
CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 OH CH3CH 2 CH 2 CH 2 CH 2 C0 2 H
(b)
.CH<
C H OH
ds-3-Methylcyclohexanol (S)- 1 -Cyclopentylethanol
OH
N0 2
H 3C
4-Methyl-3-nitrophenol
17.21 The
reduction product
is
a racemic mixture. Reaction of the (S) enantiomer
is
(a)
(5>5-Methyl-2-hexanol
Dess-Martin periodinane
Na*"
"O
H
H
CI
(iii) I
shown,
Alcohols and Phenols
417
418
Chapter 17
Dess-Martin periodinane
> CH2CI2
^
CH 3
CH 3
O I.UAIH4 f*-
II
I
CH3CHCH2CH2COCH3
*
CH 3
1
CH3CHCH2CH2COCH3 ^
.
2
+
HOCH3
3
O II
I
I
CH3CHCH2CH2CH2OH
—
f
OH
CH 3
CHoCHoMgBr 2
'
'
»
CH3CHCH2CH2CCH2CH3
3
+
HOCH3
^
CH2CH3
17.24 VF
CH 3
^CHo6 .C.S / CH /S C 2
H3C H
H3C H
H3C H 1
.
\f
CHoCHoMgBr
2 -H 3
+
*~
.C.
if
„
^ CHo
/^C.S.CHo S
CH3 CH 2
CHoCHp's
HO
CH 2 CH 3
(3/e,45)-3,4-Di-
methyl-3-hexanol
The product
+
CH 3 CH 2 OH (35,45)-3,4-DImethyl-3-hexanol
is a mixture of the (3R,4S) and (35,45) diastereomers. The diastereomers are formed in unequal amounts, and the product mixture is optically active. We can't predict which diastereomer will predominate.
Alcohols and Phenols
419
Additional Problems
Naming Alcohols 17.25 (a)
(b)
CHo3 I
HOCH 2 CH 2 CHCH 2 OH
CH3CHCHCH2CH3 CH2CH2CH3
HO 2-Methyl- 1 ,4-butanediol
(
d)
& OH
H
3-Ethyl-2-hexanol
cis- 1
,3-Cyclobutanediol
m
(e)
N=C ds-3-Phenylcyclopentanol
ds-2-Methyl-4-cyclohepten- 1 -ol
2-Bromo-4-cyanophenol or
3-Bromo-4-hydroxybenzonitrile
17.26 None
of these alcohols has multiple bonds or rings.
OH CH3CH2CH2CH2CH2OH 1-Pentanol
J
CH 3 CH2CHCH2CH 3
2-Pentanol
3-Pentanol
I
CH 3 CH2CCH 3 CH 3
CH 3 2-Methyl- 1-butanol
I
CH 3 CH2CH2CHCH 3
OH CH 3 CH 2 CHCH 2 OH
OH
2-Methyl-2-butanol
OH I
CH 3 CHCHCH 3 *
I
CHo 3-Methyl-2-butanol
CHo3 I
HOCH 2 CH 2 CHCH 3 CH 3 3-Methyl- 1 -butanol
CH 3 CCH 2 OH
CH 3 2,2-Dimethyl- 1 -propanol
2-Pentanol, 2-methyl- 1-butanol and 3-methyl-2-butanol have chiral carbons (starred).
420
Chapter 17
17.27 Primary
alcohols react with CrOs in aqueous acid to form carboxylic acids, secondary alcohols yield ketones, and tertiary alcohols are unreactive to oxidation. Of the eight alcohols in the previous problem, only 2-methyl-2-butanol
is
unreactive to
CrO, +
CH 3 CH2CH2CH2C02H
3j +
CH 3 CH2CH2CCH 3
3] +
CH 3 CH2CCH2CH 3
CH3CH2CH2CH2CH2OH H3
OH I
CrQ
CH3CH2CH2CHCH3 H3
H
?
CrQ
CH3CH2CHCH2CH3 H3
CH 3 CH 2 CHCH 2 OH CH 3
O03j H3 +
OH CH3CHCHCH3 CH 3
H3 CrQ
HOCH 2 CH 2 CHCH 3
CH 3
O CrQ
I
CH 3 CH 2 CHC02 H
II
3] +
CH 3 CCHCH 3
CH 3 3)
HoO+
H0 2 CCH 2 CHCH 3
CH 3
fH 3 CH 3CCH 2 OH
CH 3 CHo3 CrQ
I
3) +
HoO
CH 3 CC0 2 H
CH 3
CH 3 17.28
Bombykol
(
0E, 1 2Z)- 1 0, 1 2-Hexadecadien- 1 -ol
17.29
Carvacrol
5-Isopropyl-2-methylphenol
C1O3
oxidation.
Alcohols and Phenols
Alcohols
Synthesizing
17.30
some of these problems, different combinations of Grignard reagent and carbonyl compound are possible. Remember that aqueous acid is added to the initial Grignard In
adduct to yield the alcohol.
(a)
CH3CHO
H
CH 3 CH 2 MgBr
+
CHgCHCH 2 CHo
or
i
CH 3 CH 2 CHO
+
CH 3 MgBr
2-Butanol
(b)
OH CH 3 CH 2 CHO
CHgCH 2 CHCH 2 CHo
CH 3 CH 2 MgBr
+
(
3-Pentanol
(c)
CHo3
CHo3 I
I
^Cx
H2 C
+
MgBr
CH 2
H2C
CH 2 OH
2-Methyl-2-propen- 1 -ol
MgBr
HO or
MgBr
OR +
2
Triphenylmethanol
422
Chapter 17
(e)
CCH 3
CH 3 MgBr
or
=\
HO \ /
ff
COR
^
+
+
CHo6
2CH 3 MgBr
//
or n
MgBr
(f)
+
CH3CCH3
2-Phenyl-2-propanol
CH 2 OH
MgBr
+
CH 2
17.31 Carbonyl precursor( s)
Alcohol (a)
CH 3
CHo
CHo
I
I
CH3CH2CH2CH2CCHO
CH3CH2CH2CH2CCH2OH
CH 3
CH3CH2CH2CH2CCO2H
CH 3
CHg CHo3 I
CH3CH2CH2CH2CCO2R
CH 3 (b)
OH I
(CH 3 3 CCHCH 3
(CH 3 ) 3 CCCH 3
)
o
(c)
CHCH2CH 3
o II
CCH2CH 3
17.32 Grignard Reagent + (a)
Carbonyl Compound
Product (after dilute acid workup)
o II
CH 3 MgBr
+
CH 3 CCH 3
o 2
CH 3 MgBr
CH 3 CH 2 MgBr
+
(CH 3 ) 3 COH
CH 3 COR
OH CH2CH 3 +
Alcohols and Phenols
Carbonyl Compound
Grignard Reagent + (c)
Product {after dilute acid workup)
o
-o
II
CH 3 CH 2 MgBr
CH 3 CH 2 C
+
or
CH 3 CH 2 MgBr
2
O ROC"
+
OH
\
/
I
CH3CH2CCH2CH3
or
MqBr d \— MgBr
CH 3 CH 2 CH 2 MgBr
+
CH3CH2CCH2CH3
+
+
CH 3 CH 2 CH 2 C
y
— MgBr
,
OH
v
—f
I
CH 3 CH2CH2CCH 3
y _
or
(?
7
(f
O
or
CH 3 MgBr
CH 3 C
+
CH 3 CH 2 CH 2 CCH 3
^^^CH 2 MgBr
(e)
XJ
+
CH 2 CH 2 OH
H2C=0 HoC
(0
O II
CH 3 MgBr
+
CH2CCH 3
or
O
OH
II
2
CH 3 MgBr
+
CH 2 COR
CH 3
or ft
CH 2 MgBr
CH2CCH 3
CH 3 CCH 3
423
424
Chapter 17
17.33
All of these syntheses involve a Grignard reaction at some step. Both the carbonyl compound and the Grignard reagent must be prepared from alcohols.
(a)
Mg
PBiv
CH 3 CH 2 OH
CH 3 CH 2 Br
»
CH 3 CH 2 MgBr
ether
.OH
OH CrO,
CHo3
CH<
CHr.
PBro
I
Periodinane
ch 2 c 2 '
Mg
'
3* CH 3 CH 2 CH 2 CHCH 2 Br
CH 3 CH 2 CH 2 CHCH 2OH
CH 3° H
CH 3 CH 2 MgB r
2.H 3 +
+
H3
(b)
1.
I
CH 3 CH 2 CH 2 CHCH 2 MgBr
ether
H * c=0
9 H3 CH 3 CH 2 CH 2 CHCH 2 MgBr
CH< +
ether - " ~+ 2. H 3
1
H 2 C=0
.
I
» CH 3CH 2 CH 2CHCH 2CH 2 OH
(c)
MgBr Br<
Mg
FeBrc
ether
OH CrO,
CH 3 CH 2 CH 2 CHCH 3
H3
CH 3 CH 2 CH 2 CCH 3
+
HO \ /
O
MgBr
1
CH 3 CH 2 CH 2 CCH 3
(
d)
CH 3
22
CHoCHCHoCHoOH 3
CH 3 I
ether
2-H 3
CH 3 ^
.
Periodinane *-
CH 2 CI 2
CH 3 CHCH 2 CH
CH 2 CH 2 CH 3
+
O ^
^ ^
CHoCHCHoCH 2 3
O
CH 3
II
1.
+
CH 3 CH 2 MgBr from
(a)
CHo3
2
ether
H 3°
+
I
OH I
CH 3 CHCH 2 CHCH 2 CH3
Alcohols and Phenols
Reactions of Alcohols
17.34 ( a)
CH3CH2CH2CH2CH2OH (b)
CH3CH2CH2CH2CH2OH
PBr 3
CHgCH^CH^CH^CH^Br
soci,
CH3CH2CH2CH2CH2CI
(c)
CrOc
CH3CH2CH2CH2CH2OH
+
H3 (d)
CH3CH2CH2CH2CH2OH
Periodinane
CH3CH2CH2CH2CO2H
CH3CH2CH2CH2CHO
CH2CI2
17.35 (a)
aCH
^\^CH=CH 2
2 CH 2 OH
POCI3 pyridine
2-Phenylethanol
(b)
Styrene
^^CH CHO
^\^CH CH OH 2
2
2
Periodinane
CH2CI2 Phenylacetaldehyde
(c)
^^CH C0
CH 2 CH 2 OH
2
CrQ
2H
3|
HoO + Phenylacetic acid
(d)
CH 2 CH 2 OH
KMnO/
HoO Benzoic acid
(e)
^v^CH=CH
CH2CH3
2 He
Pd from
(a)
Ethylbenzene
425
Width: 612 Height: 792
426
Chapter 17
(f)
.^/CHO
CH=CH 2 2. Zn,
from
H3 +
L^JJ Benzaldehyde
(a)
(g)
CH— CHo
>^s.
^,
CHCHq
Hg(OAc) 2 H 2 Q ,
2.
from (h)
NaBH 4 1-Phenylethanol
(a)
^s^ C
^\^CH CH OH 2
2
H 2 CH 2 Br
PBrc
1
-Bromo-2-phenylethane
17.36 (a)
OH
^^CHCH
/\
3
XCHo
CrO<
H3
+
Acetophenone
1-Phenylethanol (b)
OH I
XHCHo
r
KMnO,
.CH 2 OH
.C0 2 H 1
if
HoO
.
LiAIHx
2-H3
+"
r
ii
^Jl Benzyl alcohol
(c)
aC0 from
2H
FeBo m-Bromobenzoic acid
(b)
(d)
o
aCCH
HO
(a)
CHod
\ /
3 1
.
2.
from
.C0 2 H
Br
CH 3 MgBr H3
^CH<
+
2-Phenyl-2-propanol
Alcohols and Phenols
427
17.37
from
Remember
(c)
that hydroboration
added have a
(a)
(b)
cis relationship.
proceeds with syn stereochemistry, and the
-H
and
-OH
428
Chapter 17
(c)
CH 3
H 2 S04
OH
O
(d)
CH 3
Na 2 Cr2 7
no reaction
OH Tertiary alcohols aren't oxidized
by sodium dichromate.
Mechanisms 17.39 HgC^ ^CHg
HgC^ ^CHg
CH<
HoC d
CH<
C
H3 C
/ \
H
HOS020— H>
L.
:OH
C
H
P H2
'
CH 3 HoO +
+
H3 C I
CH3
Step
1: Protonation.
Step 2: Loss of Step 3: Alkyl
H 2 0.
shift to
Step 4: Loss of
H3
form the + .
tertiary carbocation.
+
Alcohols and Phenols
17.40 This mechanism
consists of the same steps as are seen in two different cycloalkenes.
Problem
17.39.
alkyl shifts result in
two
different
alkyl shifts
Step 1: Sn2 reaction of Grignard reagent. Step 2: Protonation of alkoxide oxygen. The methyl group and the hydroxy 1 group have a trans relationship.
Two
different
429
430
Chapter 17
17.42 CH 3
1.
CH 3
OH 2
HO-
H
CH 3 2. J
CH 3
CH 3
HO^
H
CH 3
H3 C
CHg
HgC
Step 1: Protonation. Step 2: Addition of H 2 0. + Step 3: Loss of H .
17.43
O H 3 C^
HOH
HO H
^C^
1
CH 3
C
.
NaBH
2 H3
+
v
\!
4.
H sC
.
H H
^
"CH3
S
"C^CH
+
H H
Reaction of 2-butanone with
3
H H
NaBH4 produces
a racemic mixture of (/?)-2-butanol
and
2-butanol.
Spectroscopy
17.44 2.32 6
—
HoC
2.43 6
7.10 6,7.17 6
4.50 6
/?-Methylbenzyl alcohol
17.45 (a)
OH
a=
c
(b)
0.93 6
OH
b
a=
1.42 6
b =
2.43 6
c =
4.80 6
d =
7.32 6
I
CH3 CH2CHCH2CH 3 a
b
d
b
3-Pentanol
a
b =
1.42 5
c=
1.83 6
d =
3.41 6
I
CHCH 3 c
1-Phenylethanol
a
(S)-
Alcohols and Phenols
17.46
3
.
C8H18O2 has no double bonds or rings, based on degree of unsaturation. 1 The IR band at 3350 cm" shows the presence of a hydroxyl group. The compound is symmetrical (simple NMR).
4
.
There
1
.
2.
is
no
431
splitting.
CHo3
CHo3 I
~*r-
1.24 6
I
HOCCH 2 CH 2 COH
1.95 5
1
N
CHo3
CH n3 ^
1.56 6
2,5-Dimethyl-2,5-hexanediol
17.47 e
d
CHo6
H \
/
C= C
3-Methyl-3-buten-3-ol
CH 2 CH 2 OH
H
a
b
c
f
The peak absorbing
at 1.76 6 (3
the allylic region of the spectrum,
H)
is
due to the d protons. This peak, which occurs
The peak absorbing at 2.13 6 (1 H) is due to the -OH proton a. The peak absorbing at 2.30 6 (2 H) is due to protons c. The peak splitting
in
is unsplit.
is
a
triplet
because of
by the adjacent b protons.
The peak absorbing at 3.72 6 (2 H) is due to the b protons. The adjacent oxygen causes the peak to be downfield, and the adjacent -CH2- group splits the peak into a triplet.
The peaks
17.48
(a)
at
4.79 6 and 4.85 6 (2 H) are due to protons e and
C 5 H, 2 0, C 4 H 8
2,
f.
C3H4O3
The H NMR data show that the compound has twelve protons. -1 shows that the compound is an alcohol. (c) The IR absorption at 3600 cm (d) The compound contains five carbons, two of which are identical. (e) C5H12O is the molecular formula of the compound. (b)
(f) , (g)
b
a = 0.9 6
H
?
b
=i. 08
CH 3 CCH 2 CH 3 c
d
I
CH 3
1
9 §
2-Methyl-2-butanol
a d =
1.4 5
c
17.49 c
OH =
1.41 5
= 2.24 6 =
5.00 6
=
6.97 5
432
Chapter 17
General Problems
17.50
In these present.
compounds you want to reduce some, but not all, of the To do this, choose the correct reducing agent.
functional groups
H2 with
a palladium catalyst hydrogenates carbon-carbon double bonds without affecting carbonyl double bonds.
LiAlH4 reduces carbonyl groups without affecting carbon-carbon double bonds.
17.51
433
Alcohols and Phenols
17.52 Remember that electron- withdrawing groups acidity. Electron-donating
stabilize phenoxide anions and increase groups decrease phenol acidity.
Most acidic
Least acidic
CH 3 electron-
electron-
electron-
donating
withdrawing by inductive
by resonance
group
withdrawing
effect
17.53 +
^^CH -CI
H I
c
2
H
-:o-
^:Base
ci"
1.
C,
ry i :0
UH
+
O 2.
C
76^
H
1:
Sn2
f H
:Base
Step
H
substitution.
Step 2: E2 elimination.
17.54 H 3 QH
H 3 CH
CH3
:OH /
C~ C
\
H 3 C-7
CCH 3
HgC
OH3
Pinacol
H3 +
+
Pinacolone
Step 1: Protonation. Step 2: Loss of H 2 Step 3: Alkyl shift. + Step 4: Loss of H .
CHr
we
Alcohols and Phenols
437
17.62 The hydroxyl group is axial in the cis isomer, which is expected to oxidize faster than the trans isomer. (Remember that the bulky tert-butyl group is always equatorial in the more stable isomer.)
OH CrO<
(CH 3 3C-
+
H3
)
ds-4-tert-Butylcyclohexanol
O
faster
(CH 3 3 C )
OH
CrQ
(CH 3 3C)
3> +
H3
slower
Jrafts-4-tert-Butylcyclohexanol
17.63 1
2.
UAIH4
H3
+
PBr
'
Mg, ether
1
.Cyclohexanone
2.
H3
MgBr
+
B icy clohexy lidene 17.64 An
alcohol adds to an aldehyde by a mechanism that we will study in a later chapter. The hydroxyl group of the addition intermediate undergoes oxidation (as shown in Section 17.7), and an ester is formed.
:o9 y
OH
\\
OH
I
I
HCCHo
CHoCHoO^ 3 2
S
CH 3
v CH 3 CHoCHoO^ 3 2
H
CH 3 CH 2 OH "0 Cr 3
H.
Aj O
? CH 3 CH 2 0^
O E2
CrO>' H
C
or enantiomer
CH 3 CH 2 0" £
^tT^lBese CH.
elimination
H:Base +
CH 3 CH 2 + Cr0 3
CH 3
2-
438
Chapter 17
17.65
(a)
H 3 + (b) PBr3 (c) Mg, ether, then CH2 (d) Dess-Martin periodinane, (e) C 6H 5 CH 2MgBr, then H 3 CT (f) POCl 3 pyridine
NaBH4
CH 2 C12
,
then
,
17.66
2.
NAD
OH UDP
UDP
UDP-Glucose
UDP-Galactose Step
1:
Base deprotonates the C4 hydroxyl group while
UDP
NAD+ oxidizes the alcohol to a
ketone.
Step
2:
When
the ketone
the starred carbon
is
is
reduced by the NADH formed in Step and UDP-glucose is formed.
1,
the configuration at
inverted,
17.67 (a)
OH
c
I
CHCH^CHg d
1
b
a
-Phenyl- 1 -propanol
a =
0.88 6
a = 2.60
b =
1.8 6
b =
3.76 6
c =
2.32 6
c =
4.53 6
d =
4.54 6
d =
6.85 6
e =
7.24 6
e =
7.23 6
p-Methoxybenzyl alcohol
6
Alcohols and Phenols
439
17.68 Structural formula:
CgHioO
contains 4 multiple bonds and/or rings. 1
3500 cm" indicates a hydroxyl group. The absorptions at -1 1500 cm" and 1600 cm" are due to an aromatic ring. The absorption at 830 cm shows
Infrared:
The broad band
1
at
1
that the ring is p-disubstituted. 1
Compound A
is
probably a phenol.
H
NMR: The triplet at 1.16 6 (3 H) is coupled with the quartet at 2.55 6 (2 H). These two absorptions are due to an ethyl group. The peaks at 6.74 6-7.02 6 (4 H) are due to aromatic ring protons. The symmetrical splitting pattern of these peaks indicate that the aromatic ring is /?-disubstituted. The
singlet absorption at 5.50 6 (1
H)
is
due to an
-OH proton.
Compound A
p-Ethylphenol
17.69 +.
Step Step
1:
2:
The nucleophile CN adds to the positively polarized carbonyl carbon. The tetrahedral intermediate is protonated to give the addition product.
17.70
r
The
reaction
is
an Sn2 displacement of iodide by phenoxide ion.
Chapter 18 - Ethers and Epoxides; Thiols and Sulfides
Chapter Outline I.
Acyclic ethers (Sections 18.1-18.4).
A. Naming ethers (Section 18.1). Ethers with no other functional groups are named by 1 substituents and adding the word "ether". .
citing the
two organic
When other functional groups are present, the ether is an alkoxy substituent. Properties of ethers. Ethers have the same geometry as water and alcohols. 1 2 Ethers have a small dipole moment that causes a slight boiling point elevation. Ethers can react slowly with oxygen to give explosive peroxides. 3 2.
B
.
.
.
.
C. Synthesis of ethers (Section 18.2). Symmetrical ethers can be synthesized by acid-catalyzed dehydration of alcohols, 1 This method is used only with primary alcohols. i. .
2.
Williamson ether synthesis. a. Metal alkoxides react with primary alkyl halides and tosylates to form ethers. b. The alkoxides are prepared by reacting an alcohol with a strong base, such as
NaH. c.
i. Reaction of the free alcohol with the halide can also be achieved with Ag20. The reaction occurs via an Sn2 mechanism. The halide component must be primary. i. ii. In cases where one ether component is hindered, the ether should be synthesized from the alkoxide of the more hindered reagent and the halide of
the less hindered reagent. 3
.
Alkoxymercuration of alkenes. Ethers can be formed from the reaction of alcohols with alkenes. a.
b
.
c.
The reaction is carried out in the presence of mercuric trifluoroacetate. The mechanism is similar to that for hydration of alkenes.
i. NaBH4 is used for demercuration of the intermediate. d Many different types of ethers can be prepared by this method. D. Reactions of ethers (Sections 18.3-18.4). 1 Ethers are relatively unreactive and often used as solvents. 2. Acidic cleavage (Section 18.3). Strong acids can be used to cleave ethers. a. Cleavage can occur by Sn2 or SnI routes. b i. Primary and secondary alcohols react by an Sn2 mechanism, .
.
.
halide attacks the ether at the less hindered
in
which the
site.
route selectively produces one halide and one alcohol. and allylic ethers react by either an SnI or an El route. Claisen rearrangement (Section 18.4). (a). This
ii.
3
.
a.
b
.
c.
II.
Tertiary, benzylic,
The Claisen rearrangement is specific to allyl aryl ethers or aryl vinyl The result of Claisen rearrangement is an o-allyl phenol. The reaction takes place in a single step by a pericyclic mechanism,
ethers.
i. Inversion of the allyl group is evidence for this mechanism. Cyclic ethers (Sections 18.5-18.7). A. Epoxides (oxiranes) (Sections 18.5-18.6). 1 The three-membered ring of epoxides gives them unique chemical reactivity (Section 18.5). 2 The nonsystematic name -ene oxide describes the method of formation. 3 The systematic prefix epoxy- describes the location of the epoxide ring. .
.
.
Ethers and Epoxides; Thiols and Sulfides
4.
441
Preparation of epoxides. Epoxides can be prepared by reaction of an alkene with a peroxyacid RCO3H. a. i. The reaction occurs in one step with syn stereochemistry. b. Epoxides are formed when halohydrins are treated with base. i This reaction is an intramolecular Williamson ether synthesis. Ring-opening reactions of epoxides (Section 18.6). a. Acid-catalyzed ring opening. Acid-catalyzed ring opening produces 1,2 diols. i. ii. Ring opening takes place by back-side attack of a nucleophile on the protonated epoxide ring. (a) trans-l, 2-diol is formed from an epoxycycloalkane. (b) If is used, the product is a trans halohydrin. iii. When both epoxide carbons are primary or secondary, attack occurs primarily at the less hindered site. iv. When one epoxide carbon is tertiary, attack occurs at the more highly .
5
.
.
.
A
HX
substituted v.
site.
The mechanism is midway between Sn2 and SnI routes. (a). The reaction occurs by back-side attack (Sn2), but positive charge stabilized
b.
by a
is
tertiary carbocation-like transition state (SnI).
Base-catalyzed ring-opening. Base-catalyzed ring opening occurs because of the reactivity of the strained i. ii.
iii.
epoxide ring. Ring-opening takes place by an Sn2 mechanism, in which the nucleophile attacks the less hindered epoxide carbon. Other nucleophiles can bring about ring opening. (a) Epoxides react with Grignard reagents to form a product with two more carbons than the starting alkyl halide. (b) Epoxide rings also react with amines in a ring-opening reaction. .
.
B. Crown ethers (Section 1
.
2
.
3
.
18.7).
Crown ethers are large cyclic ethers. Crown ethers are named as x-crown-y, where x =
the ring size and y
= # of
oxygens.
Crown a.
b. c.
d.
ethers are able to solvate metal cations.
crown ethers solvate different cations. Complexes of crown ethers with ionic salts are soluble in organic solvents. This solubility allows many reactions to be carried out under aprotic conditions. The reactivity of many anions in Sn2 reactions is enhanced by crown ethers. Different sized
IV. Thiols and sulfides (Section 18.8).
A.
Naming
thiols
and
sulfides.
Thiols (sulfur analogs of alcohols) are named by the same system as alcohols, with the suffix -thiol replacing -ol. a. The -SH group is a mercapto- group. 2. Sulfides (sulfur analogs of ethers) are named by the same system as ethers, with sulfide replacing ether. a. The -SR group is an alkylthio- group. B. Thiols. 1 Thiols stink! 2 Thiols may be prepared by Sn2 displacement with a sulfur nucleophile. 1
.
.
.
The
a.
3
.
reaction
may proceed
to
form
sulfides.
Better yields occur when thiourea is used. Thiols can be oxidized by Br2 or I2 to yield disulfides, RSSR. a. The reaction can be reversed by treatment with zinc and acid.
b
b
.
.
The
thiol-disulfide interconversion is an important biochemical interconversion.
Chapter 18
442
C. Sulfides. 1 Treatment of a thiol with base yields a halide to form a sulfide. .
2 3
.
.
4.
III.
thiolate anion,
which can
react with an alkyl
Thiolate anions are excellent nucleophiles. Dialkyl sulfides can react with alkyl halides to form trialkylsulfonium salts, which are also good alkylating agents. a. Many biochemical reactions use trialkylsulfonium groups as alkylating agents. Sulfides are easily oxidized to sulfoxides (R2SO) and sulfones (R2SO2). a. Dimethyl sulfoxide is used as a polar aprotic solvent.
Spectroscopy of ethers (Section 18.9). A. IR spectroscopy. 1
.
Ethers are difficult to identify by IR spectroscopy because -1 occur at 1050-1 150 cm where ethers absorb. spectroscopy. spectroscopy.
many
other absorptions
,
B
.
NMR H NMR 1
.
a.
b.
Hydrogens on a carbon next to an ether oxygen absorb downfield (3.4-4.5 6). Hydrogens on a carbon next to an epoxide oxygen absorb at a slightly higher field (2.5-3.5 6).
13
2.
C
a.
NMR spectroscopy. Ether carbons absorb downfield (50-80
6).
Solutions to Problems
18.1
Ethers can be
bonded
to
named
either as alkoxy-substituted
compounds or by
citing the
two groups
oxygen, followed by the word "ether".
Diisopropyl ether
p-Bromoanisole
Propoxycyclopentane
or
or
Cyclopentyl propyl ether
p-Bromomethoxybenzene
® '3
1-Methoxycyclohexene
18.2
Ethyl isobutyl ether
H 2 C= CHCH 2 OCH= CH 2 Allyl vinyl ether
first step of the dehydration mechanism is protonation of an alcohol. Water is then displaced by another molecule of alcohol to form an ether. If two different alcohols are present, either one can be protonated and either one can displace water, yielding a mixture of products. If this procedure were used with ethanol and 1-propanol, the products would be diethyl
The
and dipropyl ether. If there were equimolar amounts of the were of equal reactivity, the product ratio would be diethyl ether
ether, ethyl propyl ether,
alcohols, and if they
ethyl propyl ether
:
dipropyl ether
=
1:2: 1.
:
Ethers and Epoxides; Thiols and Sulfides
18.3
443
Remember that the halide in the Williamson ether synthesis should be primary or methyl, in order to avoid competing elimination reactions. The alkoxide anions shown are formed by
treating the corresponding alcohols with
NaH.
(a)
CH 3 CH 2 CH 2 0~
+
CH 3 Br
+
CH 3 0~
or
CH 3 CH 2 CH 2 Br
CH 3 CH 2 CH 2 OCH 3
+
Br"
+
Br"
Methyl propyl ether
(b)
OCH 3
CH 3 Br
Methyl phenyl ether (Anisole)
(c)
CHo3
CH.
I
CH 3 CHO"
CH 2 Br
CH 3 CHOCH 2
+
Benzyl isopropyl ether
(d)
CH 3
CHo3 I
CH 3 CCH 2 0" CHo
CH 3 CH 2 Br
CH 3
CH3CHCH3
»
primary
secondary
aryl halide
halide
halide
(not reactive)
(b)
CH 3 CH 2 Br
>
CH3CH 2 CI
»
+ NaBr
Tetrahydrofuran
Ho
The compounds most reactive in the Williamson ether any Sn2 reaction (review Chapter 1 1 if necessary).
CH 3 CH 2 Br
CHp
on 2 HoC^ /,CH O
+
"O Na +
18.6
3
\
/
/
Br
)
must be used.
HpC NaH
I
(CH 3
.sec-Butyl tert-butyl ether
synthesis
CH2 \
6
2
NaBH 4
The Williamson
/
—
CH0CHCH0CH0
(CFoCOp)pHg,
CH 3 CH=CHI
better
poorer
vinyhc
leaving group
leaving group
(not reactive)
most reactive
in
Width: 612 Height: 792
446
18.7
Chapter 18 (a) First, notice the substitution pattern of the ether. Bonded to the ether oxygen are a primary alkyl group and a tertiary alkyl group. When one group is tertiary, cleavage occurs by an SnI or El route to give either an alkene or a tertiary halide and a primary alcohol.
CH-
0^
HBr
tertiary
Br
+
CH 3 OH
methyl -
problem, the groups are primary and secondary alkyl groups. Br attacks at the less hindered primary group, and oxygen remains with the secondary group, to give a secondary alcohol. (b) In this
CH 3
CH3 HBr
ft I
CH 3 CH 2 CH— O— CH 2 CH 2CH 3
I
^
CH 3 CH 2 CHOH
Br— CH2CH 2CH3
+
primary
secondary
18.8 HoC \ /
s CH 3
H3C\
CHod
i=
II
CHo
H 3 C^
H— +
HX
The
first step of acid-catalyzed ether cleavage is protonation of the ether oxygen to give an intermediate oxonium ion, which collapses to form an alcohol and a tertiary carbocation. The carbocation then loses a proton to form an alkene, 2-methylpropene. This is an example of El elimination. The acid used for cleavage is often trifluoroacetic acid.
18.9
H— Kb
(
^
\._
R— X
+
HO— R'
H
:x:
HX first protonates the oxygen atom, and halide then brings about a nucleophilic displacement to form an alcohol and an organic halide. The better the nucleophile, the more effective the displacement. Since I~ and Br~ are better nucleophiles than Cl~, ether cleavage proceeds more smoothly with HI or HBr than with HC1.
Ethers and Epoxides; Thiols and Sulfides
447
18.10 Draw
the ether with the groups involved in the rearrangement positioned as they will appear in the product. Six bonds will either be broken or formed in the product they are shown as dashed lines in the transition state. Redraw the bonds to arrive at the intermediate enone, which rearranges to the more stable phenol. ;
2-Butenyl phenyl ether
o-(l-Methylallyl) phenol
intermediate
transition state
18.11 Epoxidation by use of ra-chloroperoxybenzoic acid (RCO3H) is a syn addition of oxygen to a double bond. The original bond stereochemistry is retained, and the product is a meso compound.
,H :c=cl r
H. l
3
C
RCOoH
^CHo
°
\o
/
/ic-c^ Hn M V "if CH 3
H3C
ds-2-Butene
c/s-2,3-Epoxybutane
In the epoxide product, as in the alkene starting material, the methyl groups are cis.
•"k
_
,-
CH 3
RCO3H
O R H3 C
trans-2-Butene
/
^
O Ft
S
S
H
H
CH 3
trans-2 ,3 -Epoxy butane
Reaction of rra/w-2-butene with m-chloroperoxybenzoic acid yields trans-2,3epoxybutane. A mixture of enantiomers is formed because the peroxyacid can attack either the top or bottom of the double bond.
448
Chapter 18
18.12 As discussed
epoxide ring opening occurs primarily at the one of the epoxide carbons is tertiary. In both parts of this problem, one epoxide carbon is tertiary. in this section, acid-catalyzed
more hindered carbon
if
OH CI
HCI
CH,
HgC
ether
major
tertiary
(b)
CHg
O
CI
CH 2 OH
HCI ether
major
tertiary
18.13 Notice
the relationship of the hydroxyl groups in the two diols. In diol (a), the two hydroxyls are cis, and in (b) they are trans. Since ring-opening of epoxides forms trans 1,2-diols, only diol (b) can be formed by this route. The cis- 1,2-diol in (a), results from treatment of 1-methylcyclohexene with OSO4. The enantiomers of the diols are also formed.
1
Os0 4 pyridine NaHS0 3 H 2
.
,
2.
1
,
(b)
RCO3H
OH
.CHg
aCHc OH
+
—
HoO
H
18.14
(a)
Attack of the basic nucleophile occurs at the less substituted epoxide carbon.
O A HoC —
/
CH 2 CH 3
OH N£|8 OH
HoC-cr" 18
CH<
Ho
18
OH
CH2CH3
HoO
W OCH CCH CH .18
2
CH 3
2
3
Ethers and Epoxides; Thiols and Sulfides
Under
(b)
of the carbons
HoC—
substituted epoxide carbon
is tertiary.
18
CH 2 CH 3 /
/ \
more
acidic conditions, ring-opening occurs at the
when one
449
18
H3
+
OH
HOCH 2 (j)CH2CH3
CH,
CHo
Addition of a Grignard reagent takes place at the less substituted epoxide carbon.
(c)
,CH2CH3
H3C
s />
CH 3
H
1.
2.
-=\
I
+
H3
CHo OH
3 \>— CH-CCH
C 6 H 5 MgBr
I
2 CH 3
CH,
18.15 .
/
\
^q.
;
%Q"
.
o>
.X.
o
|
-or
^.o'"
o 15-Crown-5 Bases on ionic
radii, the
12-Crown-4 ion-to-oxygen distance in 15-crown-5
is
about
40%
longer than
the ion-to-oxygen distance in 12-crown-4.
18.16
Thiols are
named by the same rules as alcohols, with the suffix -ol replaced by the suffix named by the same rules as ethers, with "sulfide" replacing "ether".
thiol. Sulfides are
(a)
CHo3
(b)
I
CHo3 SH I
(c)
CHo3
I
I
CH 3 CCH2CHCH2CHCH 3
CH 3 CH2CHSH
CH 3
(d)
CHo3 I
CH 3 CHSCH2CH 3
2-Cyclopentene- 1 -thiol
2,2,6-Trimethyl4-heptanethiol
2-Butanethiol
(e)
m
SCH<
o
a.
SCH 3
Ethyl isopropyl sulfide
o-(Dimethylthio)benzene
SCH2CH 3
3-(Ethylthio)cyclohexanone
450
Chapter 18
18.17 Thiourea
is
used to prepare thiols from alkyl halides.
O LLWJHa o t| ^fr 2.H
II
CH 3 CH = CHCOCH3
„ CH
H
3
CH= CHCH 2 OH
PBr3
CH 3 CH= CHCH 2 Br
3
Methyl 2-butenoate
1
.
(H 2 N) 2 C=S
2."OH, H 2
H 2 C-~ CHCH
HBr
CH 2
CH 3 CH=CHCH 2 Br
1
.
(H 2 N) 2 C=S
_ 2. "OH, H 2
CH 3 CH= CHCH 2 SH 2-Butene-l -thiol
1,3-Butadiene
18.18
O / \
CH 3 CH 2 C-C
rH H
H
d,e
a=
1.0 5
b =
1.5 6
c =
2.9 6
c
1,2-Epoxybutane
d,e
= 2.5
6,
2.75
Visualizing Chemistry
18.19 (a)
(b)
OCH 2 CH 3
H 3 C^/
C
Cn7CH
f=\ HoC
3
H Br
2s-2-(oBromophenyl)-2,3-epoxybutane
cis- 1 -Ethoxy-3-methylcyclohexane (c)
C i*.
H SH (5)- 1 -Cyclopentylethanethiol
18.20 Ring-opening state.
occurs
at the tertiary
carbon to give carbocation-like stability to the transition the C-OH bond, as it would in an Sn2 reaction.
Bromine approaches 180° from 6+
O R i_N R C H-J° C-C 6 H 5 H3 C
CH 3
OH
Hj H3C
r—rP b+
H5 f C"C CH 6
(
:B*r:
3
H ' H3 C
S\ Br
Ethers and Epoxides; Thiols and Sulfides
18.21 The Grignard
451
reagent attacks the epoxide at the less hindered carbon in an Sn2 reaction. to the tertiary carbon.
The oxygen remains bonded secondary
1
.
2-
18.22
CH 3 MgBr, H3
ether
+
A molecular model shows that approach to the upper face of the double bond is hindered by a methyl group. Reaction with RCO3H occurs at the lower face of the double bond to produce epoxide A.
and H2O, the intermediate bromonium ion also forms at the lower water yields a bromohydrin which, when treated with base, forms
In the reaction of Br2 face. Reaction with
epoxide B.
452
Chapter 18
Additional Problems
Naming Ethers 18.23 (a)
CH2CH3
CH3CH2OCHCH2CH3 Ethyl 1-ethylpropyl ether
Di(/?-chlorophenyl) ether
O
(d)
(c)
3,4-Dimethoxybenzoic acid
(e)
OCH,
Cyclopentyloxycyclohexane
CH2CH
— CH2
4-Allyl-2-methoxyphenol
18.24 (b)
(a)
aOCH
(c)
3
OCH< Cyclohexyl isopropyl sulfide
(d)
o-Dimethoxybenzene
(e)
O
CHr.
1,2-Epoxycyclopentane
(f)
CHr
CH 3 CH— O
SH
—<^j N0 2
2-Methyltetrahydrofuran
Cyclopropyl isopropyl ether
o-Nitrobenzenethiol
or
Isopropoxycyclopropane
(g)
(h)
CHo I
CHo
I
I
,3
CH0CH0CHCHCHSCHCH0 3
2,
CH3 2-(Isopropylthio)-3,4-
dimethylhexane
(i)
OCHo3 CH3CCH3
OCHo 2,2-Dimethoxypropane
a
SCHr SCH,
l,l-(Dimethylthio)-
cyclohexane
Ethers and Epoxides; Thiols and Sulfides
Synthesizing Ethers
(b)
CH 3 CH — CH2
1
C 6 H 5 OH, Hg(OCOCF 3 )2^
2.
NaBH 4
CH(CH 3
(c)
H 3C ^ r
H
.-
H
)
2
O
RCOoH
/ \
H 3 C"i
XH 3
C" C
C"H CHo
H
[RCO3H = raeta-Chloroperoxybenzoic acid] (d) 1.
OH
Hg(OCOCF 3 )2, (CH 3
)
2
C(CH 3
C=CH2
2.
)
3
O
5
NaBH 4
H
(e)
"OCH.
RC0 3 H
O
H
OH 1
HOCH 3
H
' .
NaH
2.CH 3 I
H
OCH 3
OCH3
+ enantiomer (f)
+ enantiomer
H
H
OH 1.BD<
-OCH-
'
1
NaH 1
2.H 2
2.
2 ,"OH
CH31
D H
+ enantiomer
-OCHj
H
+ enantiomer
453
454
Chapter 18
18.26 (a)
0"Na+
OH
OCH,
/\
XHCHo CH,*
CHCHo
fY
f^Y
Na H>
XHCHo NaBr
Methyl 1-phenylethyl ether (b)
O
OH
/\
aCH—
I
CHCH 3 POCI<
CH 2 RCOo,H
C
^
T
CH— CH2
pyridine
Phenylepoxyethane
Styrene
(c)
OC(CH 3
a
CHCH 3 1
Hg(OCOCF3 )2, (CH 3) 2 C= CH 2_
2.
NaBH 4
aCHCH
)
3
3
tert-Butyl 1-phenylethyl ether
SH
Br
^^CHCH
I
^\.CHCH
3 1.
(H 2 N) 2 C=S
2.
"OH, H 2 1
U
3
-Phenylethanethiol
18.27 CHo
CH<
CH 3 C-r- CH 2
^
^*H T A
U
CHo3
CHo3
I
CH3 CCH 3 •:o I
H
I
CH 3 CCH 3
&
H^ C. :
CH 3 CCH 3
R
Base
Step 1: Protonation. Step 2: Attack of alcohol oxygen on carbocation. Step 3: Loss of proton. Notice that
(Problem
this reaction is the reverse
18.8).
of acid-catalyzed cleavage of a
tertiary ether
Ethers and Epoxides; Thiols and Sulfides
455
18.28
n.
r
H 1
c/s-2-Chlorocyclohexanol
enol
,2-Epoxycyclohexane
Cyclohexanone
-OH and -CI are in the trans orientation that allows epoxide formation to occur as described in Section 18.5. Epoxidation can't occur for the cis isomer, however. Instead, the base ~OH brings about E2 elimination, producing an enol, which tautomerizes to cyclohexanone. In the trans isomer, the
Reactions of Ethers and Epoxides
18.29
The enol tautomerizes
to
an aldehyde.
Width: 612 Height: 792
456
Chapter 18
(d)
LIT
(CH 3
)
3
»
CCH 2 OCH 2 CH3
(CH 3 3 CCH 2 OH
+
)
^ q
CH 3 CH 2 I
18.30 (a)
CHoCHo
Hg(OCOCF3 ) 2 CH 3 CH 2 OH
1
,
2.
(b)
NaBH 4
OH
OCHo H HoCi
H PBr3
HI
HoCi
H2
-Br
HoC
CH 3 I
+
H
H
H
H
(C)
O
.O
RCQ 3 H
'OH
H2
T C(CH 3
)
C(CH 3
3
)
C(CH 3
3
)
C(CH 3
3
)
3
(d) He
CH 3 CH 2 CH 2 CH 2 C — CH
CH 3 CH 2 CH 2 CH 2 CH^- CH 2
Lindlar catalyst
1
.
2.
CH3CH2CH2CH2CH2CH2OCH3
^
BH 3 THF H 2 2 "OH ,
,
CH3CH2CH2CH2CH2CH2OH CH"~l"
3
(e)
_ CH0CH0CH0CH0CH — CHo from(d)
1.
Hg(OCOCF3 ~
2.
NaBH 4
18.31 O. HI
HoO
ICH 2 CH 2 CH 2 CH 2 OH
)
2
,
CH 3 OH^
^
OCH 3 I
CH0CH0CH0CH0CHCH0
Ethers and Epoxides; Thiols and Sulfides
18.32
c«-5,6-Epoxydecane
C
H'j°
C~ H (C 4 H 9
(C4H9)
)
protonation of
epoxide oxygen
tt
+
QH
a attack of
"
H2
b
"Vl ft" H
(C4H9)
attack at carbon a
(
V/
attack of
C 4 H 9>
H2
attack at carbon
b
H 2 b: H
OH
+/ HCT*)
V H
(C 4 H 9 ),
H 2 or
Vc
* H
C—
/
HO
loss of proton
tt
H
H (C4 H 9 )
+
1
HoO"
Y<
\R_ />(C4 Hg)
/
(C 4 H 9 )"^r
OH
The product of acid hydrolysis of ds-5,6-epoxydecane S,S diols.
\^~^ H
H2
H (C 4 H 9 k \S
^OH
(C 4 H 9 )
(C 4 H 9 )
It
H >(C4h9)
HO
HO is
R
\
OH
a racemic mixture of R,R and
457
458
Chapter 18
18.33
r H-OH
2 trans-5 ,6-Epoxy decane
ft C
C*(C 4 H 9 H
(C 4 H 9 )
)
protonation of
epoxide oxygen
it
OH
+ attack of
-
H2
attack at carbon a
a
r-r
(C4H9)
V/
b
attack of
H2
attack at carbon
H
b
Hob:
OH
H (C4Hg)^*»
_
V^^Hg)
+/ H^T*)
H 2 or
(C4H9)
_ I^H
/
^(pH
(C4 H 9 )
H
H
H2
H
V
It
It loss of proton
HO
OH
H (C 4 Hg).\S
\R
/
F ^ Cv"(C4Hg)
HO
+
H3
H
.-C—
+
?4 H 9 _'^H
(C4 H 9 )
is a meso diastereomer of the products formed in the previous problem.
The product of acid hydrolysis of frans-5,6-epoxy decane
OH compound
18.34
OH H
HoO +
Z-^^H
H
C(CH3
O )
3
cw-3-tert-Butyl- 1 ,2-epoxycyclohexane
The hydroxyl groups
in the
product have a trans-diaxial relationship.
that is a
Ethers and Epoxides; Thiols and Sulfides
459
18.35 (a)(b)
Hj
PH2CH3
HO
O R r-r R
,C— C^ CH3 N
w CH 2 CH 3
H'i H3 C
(2#,3#)-2,3-Epoxy-
OH
(2/?,35)-3-Methyl2,3-pentanediol
3-methylpentane
Reaction with aqueous acid causes ring opening to occur at C3 because the positive charge of the transition state is more stabilized at the tertiary carbon. Ring opening produces a diol in which the hydroxyl groups have a trans-diaxial relationship. (c)
Since ring opening occurs exclusively at C3, the product is the 2R,3S isomer and opening occurred equally at either carbon, the product would be a mixture of chiral enantiomers).
is
chiral. (If ring
(d)
The product
is
optically active because only
one enantiomer
is
produced.
18.36 H
H3O+ * 2.
1.
H
Step 1: Attack of the hydride nucleophile. Step 2: Protonation of the alkoxide anion.
The
reaction
is
an Sn2 epoxide cleavage with ":H" as the nucleophile. The exact nature of
the attacking nucleophile
is
not clear.
18.37
OH
:0: H
H3
+
H H
H
D
D
+ enantiomer Deuterium and
-OH have a trans-diaxial relationship in the product.
460
Chapter 18
Spectroscopy
18.38
M + =116 corresponds to a sulfide of molecular formula C6H12S, indicating one degree of The IR absorption
unsaturation.
at
890
cm
1
is
due
to a
a
a=
1.74 5
CHo3
b =
2.11 6
c =
2.27 5
d =
2.57 6
R2C=CH2
group.
I
H2 C
— CCH 2CH 2 SCHg
e
c
d
b
e = 4.73 6
2-Methyl-4(methylthio)- 1 -butene
18.39 Chemical
Peak
H 3 C— H
Anethole
Multiplicity
shift
Split by:
a
1.84 6
doublet
b
3.76 6
singlet
c
6.09 6
two
d
6.36 5
doublet
e
6.82
doublet
f
doublet
e
f
6,
7.23 6
quartets
c
a,d
c
18.40 (a)
(b)
b
H3
a =
0.99 6
CH0CH0CSH d
b =
1.34 6
a
c:
d
I
CH 3
c,d
OCH 2 CH 2 CH 2 Br
cab
= 1.61 5
b
General Problems
18.41
(b)
CHo3
CHo3 1.
I
CH3CHCH 2 CH 2 CH 2 Br
2.
(c)
Brc
(H 2 N) 2 C=S
"OH,H 2
I
CH3CHCH 2 CH 2 CH 2 SH
~0
a=
2.31 6
b =
3.58 6
c =
4.08 6
d =
6.90-7.25 6
Ethers and Epoxides; Thiols and Sulfides
461
(d)
CH 2 CH 3
h2
2
,
CH 2 CH3
H2
18.42 OCH-
NaBH 4 HoO + OCH<
Anethole
c H3 C
CHgCH 2
H
CHgCH 2
two functional groups - an ether and a hydrocarbon side chain with a double bond. The ether is synthesized first - by a Williamson ether synthesis from phenol and CH3I. The hydrocarbon side chain results from a Friedel-Crafts acylation of the ether. Reduction of the ketone, bromination and dehydrohalogenation are used to introduce the
The anethole
ring has
double bond.
18.43 + O" Na
NaH +
(b)
He
CH 2 Br CH3CI
(PhC0 2
AlCIo
(c)
-
M„ + 0"Na
CH 2 Br
)
2
CH 2 0. +
from
(a)
from
(b)
Benzyl phenyl ether
NaBr
462
Chapter 18
18.44 CHp
HpC H2C
HO
HO
3.
CHp
HpC H 2C
HpC
\^CH 3
1
N
/ CV UH 3 O
The
+
H2n
I CHp
H 2 C \../G-
..
+ H3
CH 3
HO^
H— OSOoH
Step Step Step
C\
H 2C
,(X
+ /-Tch
i
1.
/
k
H2C
CHo
/
\*CH<
/
CHp vhGCHq
HpC
CHp
HpC
\^CH3
/
+
ch 3 :OH 2
Protonation of the tertiary hydroxyl group. Loss of water to form a tertiary carbocation. 3: Nucleophilic attack on the carbocation by the second hydroxyl group. 1:
2:
tertiary
carbocation
hydroxyl group is
more
is
more
likely to
be eliminated because the resulting
stable.
18.45
CH 3I
This reaction
is
an Sn2 displacement and can't occur
aprotic solvent that increases the rate of an
Sn2
at
DMF
an aryl carbon. is a polar by making anions more
reaction
nucleophilic.
18.46
CH 3 CH 2
i
0(CH 2 CH 3
)
2
+
0(CH 2 CH 3
)
2
CH 3 CH 2>V
CH3 CH 2 >^
:OH I.
o:
:Base 2.
Step 1: Attack of the alcohol on the triethyloxonium cation, with loss of diethyl Step 2: Loss of proton.
ether.
Trialkyloxonium salts are more reactive alkylating agents than alkyl iodides because a neutral ether is an even better leaving group than an iodide ion.
Ethers and Epoxides; Thiols and Sulfides
463
18.47
0"Na +
OH
o-
CH 2
O
OH H 2 C=
2 NaH
CHCH 2 CI
AICk +2NaBr
+ 2H,
Safrole
18.48 The mechanism
of Grignard addition to oxetane is the same as the mechanism of Grignard addition to epoxides, described in Section 18.6. The reaction proceeds at a reduced rate because oxetane is less reactive than ethylene oxide. The four-membered ring oxetane is less strained, and therefore more stable, than the three-membered ethylene oxide ring.
18.49
HoO + 2 HBr
BBr 3 forms
Br
a Lewis acid complex with
to
acts as a nucleo-
phile in an
form
CH 3 Br
+
S N 2 reaction
CH 3 Br.
B(OH) 3
Water cleaves the Lewis acid complex.
the ether.
Step Step Step
BBr3 forms
a Lewis acid complex with the ether. Br~ acts as a nucleophile in an Sn2 reaction to form CH3Br. 3: Water cleaves the Lewis acid complex. 1:
2:
18.50
—
—— =
1.06 g vanillin ; * ;
,
, n„
_ 6.97 x 1A 10 J3
mol
6.81 x 10-3
mol Agl
,
vanillin
152 g/mol 1.60
gAgl
=
234.8 g/mol 6.81 x 10" 3
mol
3 6.81 x 10~
Agl
mol
6.81 x 10" 3
CH 3 I
I-
mol
6.81 x 10" 3
mol
-OCH3
Thus, 6.97 x 10 mol of vanillin contain 6.81 x 10 mol of methoxyl groups. Since the ratio of moles vanillin to moles methoxyl is approximately 1:1, each vanillin contains one ~
methoxyl group.
CH3O
CHO
Vanillin
464
Chapter 18
18.51
Disparlure, C19H38O, contains one degree of unsaturation, at 2.8 5 identifies as
an epoxide
which the
!
H NMR absorption
ring.
KMnO/
HO Undecanoic acid
6-Methylheptanoic acid
18.52 NaNH
HC=C
7:C=C +
Br
C=C H2
,
Lindlar catalyst
I
RCO3H
+ Br"
Ethers and Epoxides; Thiols and Sulfides
465
18.53 CX+
HoO
H— OH 2 /lOH
:0 / \
Ph 2 C—
PhoC— CH
CH 2
H
Step 1: Protonation. Step 2: Epoxide opening. Step 3: Hydride shift. Step 4: Loss of proton. Reaction occurs by this route because of the
stability
of the intermediate tertiary
carbocation.
18.54 Use
the aldehyde-forming reaction
shown
in the previous
problem.
o-Hydroxyphenylacetaldehyde
18.55
H
CH 3MgBr, ether; ~OH, H 2 0. (a)
(b)
H 2 S0 4 H2 0; ,
(c)
NaH, then
CH 3 I;
(d)
m-ClC 6 H 4 C0 3 H;
(e)
Width: 612 Height: 792
466
Chapter 18
18.56
an Sn2 reaction because the rate depends on the concentrations of both reagents, reaction is a Williamson ether synthesis, in which an alkoxide displaces a halogen. In this reaction, KOH is used to form the phenoxide anion. (a)
This
is
Sn2
(b) This
18.57 The
reaction
complex
is
is
a nucleophilic aromatic substitution. The intermediate Meisenheimer by the -NO2 group.
stabilized
Step 1: Addition of phenoxide. Step 2: Elimination of fluoride.
18.58 (a)
b
OCH<
CHoCH 3 a
|c
OCHo3 b
(b)
a=
1.27 6
b =
3.31 5
c =
4.57 6
CH= CHOCH3
a=
3.71 6
b =
5.17 5
c =
6.08 6
d =
7.1-7.6 5
Ethers and Epoxides; Thiols and Sulfides
18.59 (a)
pCH 2 CH3
O CH 3 CH 2OH H
+
OH
OCH 2 CH 3
CH 3 CH 2OH H
catalyst
+
OCH 2 CH 3
catalyst
hemiacetal
acetal
+
OCH 2 CH 3
r*;ocH 2 cHo
+H2 °
- rn. \^ 2
I.
HOCH 2 CH 3
-
3
-
tl
OCH 2 CH 3
OCH 2 CH 3
OCH 2 CH 3
f
J^OCH 2CH3
4. acetal
+ H3
+
OH,
Step 1: Protonation. Step 2: Loss of water. Step 3: Addition of ethanol. Step 4: Loss of proton.
18.60
O / \
CH 2 Br Br"
Step Step
1:
2:
Addition of hydride to the ketone. Displacement of bromide by the alkoxide anion.
The intermediate resulting from addition of H: is similar to the intermediate in a Williamson ether synthesis. Intramolecular reaction occurs to form the epoxide.
468
Chapter 18
Review Unit
Major Topics Covered The
and Related Compounds
7: Alcohols, Ethers,
(with vocabulary):
-OH group:
alcohol
phenol
glycol
wood
alcohol
hydrogen bonding
alkoxide ion
phenoxide ion
acidity constant
Alcohols:
Grignard Reagent Dess-Martin periodinane Phenols: cumene hydroperoxide
quinone
Acyclic ethers: Williamson ether synthesis Cyclic ethers: oxirane
epoxide
tosylate
hydroquinone
protecting group
TMS
ether
ubiquinone
Claisen rearrangement
vicinal glycol
Thiols and sulfides: Thiol sulfide mercapto group sulfoxide sulfone
peroxyacid
crown ether
alkylthio group
disulfide
18-crown-6
thiolate ion
trialkylsulfonium salt
Types of Problems: After studying these chapters, you should be able to:
-
Name and draw
structures of alcohols, phenols, ethers, thiols and sulfides. Explain the properties and acidity of alcohols and phenols. Prepare all of the types of compounds studied. Predict the products of reactions involving alcohols, phenols and ethers. Formulate mechanisms of reactions involving alcohols, phenols and ethers. Identify alcohols, phenols and ethers by spectroscopic techniques.
Points to
Remember:
great biochemical importance of hydroxy 1 groups is due to two factors:(l) Hydroxy 1 groups make biomolecules more soluble because they can hydrogen-bond with water. (2) Hydroxyl groups can be oxidized to aldehydes, ketones and carboxylic acids. The presence of a hydroxyl group in a biological molecule means that all functional groups derived from alcohols can be easily introduced.
*
The
*
Carbon-carbon bond-forming reactions are always more skeleton.
The
difficult to learn than functional
group
often difficult to recognize the components that form a carbon product of a Grignard reaction contains a hydroxyl group bonded to at least one
transformations because
it
is
two or three ). When looking at a product that might have been formed by a Grignard reaction, remember that a tertiary alcohol results from the addition of a Grignard reagent to either a ketone or an ester (the alcohol formed from the ester has two identical -R groups), a secondary alcohol results from addition of a Grignard reagent to an aldehyde, and a alkyl group (usually
470
Review Unit 7 primary alcohol results from addition of a Grignard reagent to formaldehyde or to ethylene Remember that any molecule taking part in a Grignard reaction must not contain functional groups that might also react with the Grignard reagent. oxide.
Ethers are quite unreactive, relative to many other functional groups we study, and are often used as solvents for that reason. Concentrated halogen acids can cleave ethers to alcohols and halides. Remember that the halide bonds to the less substituted alkyl group when the ethers are primary or secondary alkyl ethers.
Epoxide rings can be opened by both acid and base. In basic ring-opening of an unsymmetrical epoxide (and in ring-opening using a Grignard reagent), attack occurs at the less substituted carbon of the epoxide ring. In acidic ring opening, the position of attack depends on the substitution pattern of the epoxide. When one of the epoxide carbons is tertiary, attack occurs at the more substituted carbon, but when the epoxide carbons are both primary or secondary, attack occurs at the less substituted carbon. *
The most useful spectroscopic data for these compounds: (1) A broad IR absorption in the range 3300 cm_1 -3600 cm -1 shows the presence of the -OH group of an alcohol or a phenol. (2) Hydrogens bonded to the -O-C- carbon of an alcohol or ether absorb in the range 3.5-4.5 6 in an H NMR spectrum or in the range 50-80 8 in a 13 C NMR spectrum. !
Self-Test: *2^
H3C-7 HCI
10
(d)
(c)
H 3 C-> H3 C
H3C-7 H3 C
OH
HCI
Ethers are stable to all of the following reagents except: (a) nucleophiles (b) bases (c) strong acids (d) dilute acids
~OH
Preview of Carbonyl Compounds
Chapter Outline
The carbonyl
I.
group.
A. Kinds of carbonyl compounds. 1
.
2.
All carbonyl compounds contain an acyl group (R-C=0). The groups bonded to the acyl group can be of two types: a.
b.
B
.
Groups that can't act as leaving groups. Examples: aldehydes and ketones. i. Groups that can act as leaving groups. ii. Examples: carboxylic acids, esters, amides, acid
halides, lactones, acid
anhydrides, lactams. Nature of the carbonyl group. 1 The carbonyl carbon is s/? 2 -hybridized. .
A k bond is formed between carbon and oxygen. Carbonyl compounds are planar about the double bond. 2 The carbon-oxygen bond is polar. The carbonyl carbon acts as an electrophile. a b. The carbonyl oxygen acts as a nucleophile. Reactions of carbonyl compounds. A. Nucleophilic addition reactions of aldehydes and ketones. 1 A nucleophile adds to the carbonyl carbon. a.
b.
.
.
II.
.
2
.
The resulting tetrahedral intermediate has two fates: a The negatively charged oxygen can be protonated to form an b Loss of water leads to formation of a C=Nu double bond. .
alcohol.
.
B
.
Nucleophilic acyl substitution reactions. nucleophile adds to the carbonyl carbon. 1 2 The resulting tetrahedral intermediate expels a leaving group to form a .
A
.
new
carbonyl
compound. 3
C
.
.
This type of reaction takes place with carbonyl compounds other than aldehydes
and ketones. Alpha substitution 1
.
reactions.
Reaction can occur at the position next to the carbonyl carbon ( a position). a. This type of reaction is possible because of the acidity of alpha hydrogens. b Reaction with a strong base forms an enolate anion, which behaves as a .
nucleophile.
2 All carbonyl compounds can undergo a substitution reactions. D. Carbonyl condensation reactions. 1 Carbonyl condensation reactions occur when two carbonyl compounds react with each other. 2. The enolate of one carbonyl compound adds to the carbonyl group of a second .
.
compound. Solutions to Problems 1
.
According to the electrostatic potential maps, the carbonyl carbon of acetyl chloride is more electrophilic and the oxygen of acetone is more nucleophilic. This makes sense, because acetyl chloride has two electron-withdrawing groups that make its carbonyl carbon electron-poor and thus electrophilic. Because acetyl chloride has two electron- withdrawing groups, neither group is as nucleophilic as the carbonyl oxygen of acetone.
Preview of Carbonyl Compounds
The
reaction of cyanide ion with acetone
is
OH
HoC
Cyanide anion adds
+
CN H3 C
to the positively polarized carbonyl
intermediate. This intermediate (a)
H3
I
H 3C j
HqC*
is
CN
carbon to form a tetrahedral
protonated to yield acetone cyanohydrin.
is a nucleophilic acyl substitution. Ammonia adds to acetyl chloride, and eliminated, resulting in formation of an amide.
This reaction
chloride
is
(b) In this nucleophilic addition reaction, addition
of the nucleophile
is
followed by loss of
water. (c)
473
a nucleophilic addition reaction.
O"
"CN
1
Two
molecules of cyclopentanone react in
this
carbonyl condensation.
Chapter 19 - Aldehydes and Ketones: Nucleophilic Addition Reactions
Chapter Outline I.
General information about aldehydes and ketones (Sections 19.1-19.3). A. Naming aldehydes and ketones (Section 19.1). 1
.
Naming a.
b
.
c.
d 2
.
.
aldehydes.
Aldehydes are named by replacing the -e of the corresponding alkane with -al. The parent chain must contain the -CHO group. The aldehyde carbon is always carbon 1. When the -CHO group is attached to a ring, the suffix -carbaldehyde is used.
Naming
ketones.
Ketones are named by replacing the -e of the corresponding alkane with -one. b Numbering starts at the end of the carbon chain nearer to the carbonyl carbon. c. The word acyl is used when a RCO- group is a substituent. B. Preparation of aldehydes and ketones (Section 19.2). a.
.
Preparation of aldehydes. a. Oxidation of primary alcohols with Dess-Martin periodinane. b Partial reduction of carboxylic acid derivatives. 2 Preparation of ketones. a. Oxidation of secondary alcohols. b. Ozonolysis of alkenes with at least one disubstituted unsaturated carbon. c. Friedel-Crafts acylation of aromatic compounds. d. Preparation from carboxylic acid derivatives. C. Oxidation of aldehydes and ketones (Section 19.3). 1 Aldehydes can be oxidized to carboxylic acids by many reagents. a. Cr03 is used for normal aldehydes. b Oxidation occurs through intermediate 1 1 -diols. 2 Ketones are generally inert to oxidation, but can be oxidized to carboxylic acids with strong oxidizing agents. Nucleophilic addition reactions of aldehydes and ketones (Sections 19.4-19.13). A. Characteristics of nucleophilic addition reactions (Section 19.4). Mechanism of nucleophilic addition reactions. 1 a. nucleophile attacks the electrophilic carbonyl carbon from a direction 105° 1
.
.
.
.
,
.
.
II.
.
A
b
.
c.
d
.
opposite to the carbonyl oxygen. 2 3 rehybridizes from sp to sp and a tetrahedral alkoxide intermediate is produced. The attacking nucleophile may be neutral or negatively charged. i. Neutral nucleophiles usually have a hydrogen atom that can be eliminated. The tetrahedral intermediate has two fates: i. The intermediate can be protonated to give an alcohol. ii. The carbonyl oxygen can be eliminated as -OH to give a product with a
The carbonyl group
C=Nu 2.
,
double bond.
Relative reactivity of aldehydes and ketones. Aldehydes are usually more reactive than ketones in nucleophilic addition a. reactions for two reasons: nucleophile can approach the carbonyl group of an aldehyde more readily i because only one alkyl group is in the way.
A
Aldehydes and Ketones: Nucleophilic Addition Reactions
475
Aldehyde carbonyl groups are more strongly polarized and electrophilic because they are less stabilized by the inductive effect of alkyl groups, Aromatic aldehydes are less reactive than aliphatic aldehydes because the electron-donating aromatic ring makes the carbonyl carbon less electrophilic. ii.
b
.
B. Nucleophilic addition reactions (Section 19.5-19.13). 1. Hydration (Section 19.5). a. Water adds to aldehydes and ketones to give 1,1-diols (often referred to as gemdiols or hydrates).
b
.
c.
d
.
The reaction compound. Reaction
is
is
reversible, but generally the equilibrium favors the carbonyl
slow
is catalyzed by both aqueous acid and base. an addition of -OH, followed by protonation
in pure water, but
i.
The base-catalyzed
ii.
of the tetrahedral intermediate by water. In the acid-catalyzed reaction, the carbonyl oxygen neutral water adds to the carbonyl carbon.
The i.
catalysts
Base Acid
reaction
have different
is
is
protonated, and
effects.
catalysis converts water to a better nucleophile.
ii. catalysis makes the carbonyl carbon a better electrophile. Reactions of carbonyl groups with H-Y, where Y is electronegative, are reversible; the equilibrium favors the aldehyde or ketone. Cyanohydrin formation (Section 19.6). a. HCN adds to aldehydes and ketones to give cyanohydrins. i. The reaction is base-catalyzed and proceeds through a tetrahedral
e.
2.
intermediate.
Equilibrium favors the cyanohydrin adduct. b is one of the very few protic acids that add to a carbonyl group c. Cyanohydrin formation is useful for the transformations that the -CN group can undergo. The -CN group can be reduced, to form an amine. i. ii. The -CN group can be hydrolyzed, to produce a carboxylic acid. Addition of hydride and Grignard reagents (Section 19.7). ii.
.
3
.
HCN
a.
Hydride addition. i. LiAlHi and NaBH4 act as if they are H:~ donors and add to carbonyl compounds to form tetrahedral alkoxide intermediates. ii. In a separate step, water is added to protonate the intermediate, yielding an
b
Addition of Grignard reagents. i. Mg 2+ complexes with oxygen, making the carbonyl group more
alcohol. .
electrophilic. ii.
iii.
R:~ adds to the carbonyl carbon to form a tetrahedral intermediate. in a separate step to protonate the intermediate, yielding an
Water is added alcohol.
Grignard reactions are irreversible because R:~ is not a leaving group. Addition of amines (Section 19.8). a. Amines add to aldehydes and ketones to form imines and enamines. b An imine (R2C=NR) is formed when a primary amine adds to an aldehyde or iv.
4.
.
ketone. i.
The process
ii.
A proton transfer converts the initial adduct to a carbinolamine.
iii.
Acid-catalyzed elimination of water yields an imine.
is
acid-catalyzed.
Width: 612 Height: 792
476
Chapter 19
iv.
The
reaction rate
enough
maximum occurs
at
pH = 4.5. At this pH,
to catalyze elimination of water, but
[H
low enough so
+ ]
is
that the
high
amine
is
nucleophilic.
v
.
Some imine derivatives
Enamines (R2N=CR-CR2)
c.
and ketones, produced when aldehydes and ketones react with
are useful for characterizing aldehydes
are
secondary amines. i
5
.
.
The mechanism is similar to that of imine formation, except the a carbon is lost in the dehydration step.
Addition of hydrazine: the Wolff-Kishner reaction (Section 19.9). a. Hydrazine reacts with aldehydes and ketones in the presence of
a proton from
KOH to form
alkanes. i.
b 6.
The intermediate hydrazone undergoes base-catalyzed bond of N2 and protonation to form the alkane.
migration, loss
The Wolff-Kishner reduction can
also be used to convert an acylbenzene to an alkylbenzene. Addition of alcohols: acetal formation (Section 19.10). a. In the presence of an acid catalyst, two equivalents of an alcohol can add to an aldehyde or ketone to produce an acetal. The initial intermediate, a hemiacetal (hydroxy ether), is formed when the i. first equivalent of alcohol is added. Protonation of -OH, loss of water, with formation of an oxonium ion, and ii. addition of a second molecule of yields the acetal. b Since the reaction is reversible, changing the reaction conditions can drive the .
ROH
.
reaction in either direction.
Because acetals are inert to many reagents, they can be used as protecting groups in syntheses. i. Diols are often used as protecting groups, forming cyclic acetals.
c.
7
.
The Wittig reaction (Section 19.1 1). a. The Wittig reaction converts an aldehyde b
.
or ketone to an alkene. Steps in the Wittig reaction: i. An alkyl halide reacts with triphenylphosphine to form an ii. iii.
alkyltriphenylphosphonium salt. Butyllithium converts the salt to an ylide (phosphorane). The ylide adds to an aldehyde or ketone to form a dipolar betaine. (a). In
c.
some
iv.
The
i.
The Wittig
cases, the addition is a one-step cycloaddition.
betaine forms a four-membered ring intermediate (oxaphosphatane), which decomposes to form the alkene and triphenylphosphine oxide. Uses of the Wittig reaction. reaction can be used to produce mono-, di-, and trisubstituted
alkenes, but steric hindrance keeps tetrasubstituted alkenes ii.
8.
The Wittig
from forming.
reaction produces pure alkenes of known stereochemistry
(excluding E,Z isomers). Biological reductions (Section 19.12). a.
b
.
The Cannizzaro
reaction is unique in that the tetrahedral intermediate of addition of a nucleophile to an aldehyde can expel a leaving group. Steps in the Cannizzaro reaction. i.
HCT
adds to an aldehyde with no a hydrogens to form a tetrahedral
intermediate. ii.
c.
FT
is
The
expelled and adds to another molecule of aldehyde.
which one molecule of aldehyde is oxidized and a second molecule is reduced. The Cannizzaro reaction isn't synthetically useful, but it resembles the mode of action of the enzyme cofactor NADH. iii.
result is a disproportionation reaction, in
Aldehydes and Ketones: Nucleophilic Addition Reactions
9
.
Conjugate addition to ^-unsaturated aldehydes and ketones (Section
477
19. 13).
Steps in conjugate addition.
a.
Because the double bond of an a,/3-unsaturated aldehyde/ketone is conjugated with the carbonyl group, addition can occur at the y3 position, which is an electrophilic site.
i.
Protonation of the a carbon of the enolate intermediate results in a product having a carbonyl group and a nucleophile with a 1,3 relationship. Conjugate addition of amines. ii.
b
.
Primary and secondary amines add to a,/3-unsaturated aldehydes and
i.
ketones. ii. The conjugate addition product Conjugate addition of water.
c.
is
often formed exclusively.
Water can add
to yield ^-hydroxy aldehydes and ketones. Conjugate addition of water also occurs in living systems. d Conjugate addition of organocopper reagents. Conjugate addition of organocopper reagents (R^CuLi) alkylates the double i. bond of a,/3-unsaturated ketones. ii. This type of addition doesn't occur with other organometallic reagents. iii. Primary, secondary, tertiary, aryl, and alkenyl groups can be added. iv. The mechanism may involve conjugate addition of the diorganocopper anion, followed by transfer of an -R group. Spectroscopy of aldehydes and ketones (Section 19.14). A. IR spectroscopy. The C=0 absorption of aldehydes and ketones occurs in the range 1660-1770 1 i.
ii.
.
III.
.
cm
-1
.
The exact position of absorption can be used
a.
to distinguish
between an aldehyde
and a ketone. b
.
The
position of absorption also gives information about other structural
and angle strain. absorption values are constant from one compound to another. c. The -1 Aldehydes also show absorptions in the range 2720-2820 cm spectroscopy. spectroscopy. features, such as unsaturation
2
B
.
.
.
NMR H NMR 1
.
a.
b.
2.
13
a.
b. c.
Aldehyde protons absorb near 10 6, and show spin-spin coupling with protons on the adjacent carbon. Hydrogens on the carbon next to a carbonyl group absorb near 2.0-2.3 6. Methyl ketone protons absorb at 2. 1 6. i.
C NMR
spectroscopy. carbonyl-group carbons absorb in the range 190-215 The These absorptions characterize aldehydes and ketones. Unsaturation lowers the value of 6.
6.
C. Mass spectrometry. 1
.
Some aliphatic a.
aldehydes and ketones undergo McLafferty rearrangement.
A hydrogen on the y carbon is transferred to the carbonyl oxygen, the bond between the a carbon and the p carbon produced. The remaining cation radical
is
broken, and a neutral alkene fragment
is
is detected. b Alpha cleavage. a. The bond between the carbonyl group and the a carbon is cleaved. b. The products are a neutral radical and an acyl cation, which is detected. .
2
.
478
Chapter 19
Solutions to Problems
19.1
Remember that the principal
chain must contain the aldehyde or ketone group and that an aldehyde group occurs only at the end of a chain. The aldehyde carbon is carbon 1 in an acyclic compound, and the suffix -carbaldehyde is used when the aldehyde group is attached to a ring.
(b)
(a)
^\.CH CH CHO 2
2
f?
CH3CCH 2 CH 2 CH 2 CCH 2 CH3
CHgCH2CCH(CHg)2 2-Methyl-3-pentanone
3-Phenylpropanal
2,6-Octanedione
(e)
CH 3 CH= CHCH 2 CH 2 CHO
4-Hexenal
/ra«s-2-Methylcyclo-
ds-2,5-Dimethylcyclohexanone
hexanecarbaldehyde
19.2 (b)
(c)
o
CI
CH 2 CHO
I
CHgCHCh^CCHg 4-Chloro-2-pentanone
(e)
CHO
ds-3-tert-Butylcyclo-
(f)
CHo3
CHo3
I
CHoCHCI 3
I
H 2 C=CCH 2 CHO
3-Methyl-3-butenal
hexanecarbaldehyde
19.3
Phenylacetaldehyde
2-(l-Chloroethyl)-5methylheptanal
We have seen the first two methods of aldehyde preparation in earlier chapters. (a)
Dess-Martin
CH3CH 2 CH 2 CH 2 CH 2 OH 1-Pentanol
periodinane
CH 2 CI 2
^
I
CH3 CH 2 CHCH 2 CH 2 CHCHO
CH3CH 2 CH 2 CH 2 CHO
Aldehydes and Ketones: Nucleophilic Addition Reactions
(b)
Or
1.
CH3CH2CH2CH2CH — CH2
~^
+
Zn H
* CH3CH2CH2CH2CHO
+
CH2O
1-Hexene (c)
CH3CH2CH2CH2CO2CH3 2.
(d)
_
1
CH3CH2CH2CH — CH2
.
2.
1-Pentene
19.4
THF
BH3,
H2
CH3CH2CH2CH2CHO
+
H3
CH3CH2CH2CH2CH2OH
z
2 ,"OH
(a)
All of these methods are familiar, (a)
O
HoO +
CH3CH2O— CCH2OH3
II
CH3CH2CH2CCH2CH3
HgS0 4 (b)
n
CH3COCI
Br
CH3
Br2 *»
AICk
FeBro
O^
HO.
(c)
^CH<
CHCH 3
MgBr
Br
Mg
1
CH3CHO
.
2.
H3
CrO<
l^^J
+
(d)
H3
+
^
H 1
BH3,
V CH3
THF
CrO
:
2.
H2
2
,
"OH
HoO
!^H OH
O
19.5
^HCo; H :o:
:CN
HO
CN
CN
6+
+
"OH
2.
1.
cyanohydrin
Step
1:
Cyanide anion adds
to the positively polarized carbonyl
carbon to form a
tetrahedral intermediate.
Step
2:
This intermediate
is
protonated to yield the cyanohydrin.
479
480
Chapter 19
19.6
CH 3 electron-withdrawing
electron-donating
The electron-withdrawing nitro group makes the aldehyde carbon of /7-nitrobenzaldehyde more electron-poor (more electrophilic) and more reactive toward nucleophiles than the aldehyde carbon of p-methoxybenzaldehyde.
19.7 :o:s-
OH
:o: I
CI 3 C'^
H
CIqC
C
H
-oh 2 '
OH
H
:OH,
Chloral hydrate
19.8
Q
:o:
:o:
II
I
R
R
OH OH
OH,
R
R :OH,
The above mechanism
is
mechanisms we have we can write the above mechanism in reverse to
similar to other nucleophilic addition
studied. Since all steps are reversible,
show how labeled oxygen
is
incorporated into an aldehyde or ketone.
+.
OH C r ~> R
O*
C OH.
I
Sh
This exchange
II
is
H2
+
R J * A O'. R \j-
R
R
very slow in water but proceeds more rapidly
when
either acid or base is
present.
19.9
O
NC
NC
O"
CH 3
H3C
CH 3 +
"CN
H3 C
HCN
OH CH 3
CH 3 +
"CN
2,2,6-Trimethylcyclohexanone is an equilibrium process. Because addition of ~CN to 2,2,6trimethylcyclohexanone is sterically hindered by the three methyl groups, the equilibrium lies toward the side of the unreacted ketone.
Cyanohydrin formation
Aldehydes and Ketones: Nucleophilic Addition Reactions
481
of a ketone or aldehyde with a primary amine yields an imine, in which C=0 has been replaced by C=NR. Reaction of a ketone or aldehyde with a secondary amine yields an enamine, in which C=0 has been replaced by C-NR2, and the double bond has moved.
19.10 Reaction
CH 2 CH 3
O H
+
CH 3 CH 2 NH 2
1
mine
:
N(CH 2 CH 3 2 )
H
+
+
(CH 3 CH 2 2 NH
H2
)
enamine
19.11
carbinolamine
Step 1: Protonation of nitrogen. Step 2: Addition of water. Step 3: Loss of proton. Step 4: Proton transfer. Step 5: Loss of amine.
19.12 The
structure
is
an enamine, which
the amine and ketone
is prepared from a ketone and a secondary amine. Find components and draw the reaction.
from diethylamine
from cyclopentanone
482
Chapter 19
19.13
O
CH 3
CHo
CHgC — CHCCHg
(b)
l_
H 2 NNHj
CH3C — CHCH2CH3
KOH
H2
(a)
H2
(c)
Pd/C
Pd/C
O
CHo3
(c)
II
I
CH3CHCH2CCH3
19.14 Formation
CHo3
H 2 NNH;
I
CH3CHCH2CH2CH3
KOH
of the hemiacetal
is
the
first step,
(a)
H— CI
CH0CH02
C"OH V11
/
:OH
.
OH
OH
^
^
R
*\
^
+ HoO 3
OH+
:OH 2
I
I
D -- C<+- ^ H •
CH 2 CH 2 OH
CH 2 CH 2 OH hemiacetal
Step 1: Protonation of oxygen. Step 2: Addition of -OH. Step 3: Loss of proton. (b)
:OH
H
^
CI
R^
R'
..
+^
9
,CH 2
I
o:
O:
1.
R'
R'
CH 2 CH 2 OH
CH,
2.
:0-
3.
-c
I
CH 2 CH 2 OH
H H
HH
V
R
.CH 2
I
H3
+
+
0^ c' i\
H H acetal
Step 1: Protonation. Step 2: Loss of H 2 0. Step 3: Addition of -OH. Step 4: Loss of proton. Protonation of the hemiacetal hydroxyl group is followed by loss of water. Attack by the second hydroxyl group of ethylene glycol forms the cyclic acetal ring.
Aldehydes and Ketones: Nucleophilic Addition Reactions
19.15 Locate acetal.
by
=0
the two identical -OR groups to identify the alcohol that was used to form the (The illustrated acetal was formed from methanol.) Replace these two -OR groups to find the carbonyl compound.
2
19.16 Locate
483
CH 3 OH
+
bond that is formed by the Wittig reaction. The simpler or less component comes from the ylide, and the more substituted component comes
the double
substituted
from the aldehyde or ketone. Triphenylphosphine oxide
is
a byproduct of
all
these
reactions.
(a)
from ylide
CH 3
cr
+
+
(Ph) 3 P
-
— CHCH3
from ketone
from aldehyde
(c) ?l
CH3CCH3 from ketone
from
ylide
+
(Ph) 3
P— CHCH 2CH2 CH3
484
Chapter 19
(d)
H
^\zC^'
O
.CH 3
II
from
CH3CCH3
CHo ketone from ylide
(e)
CHo
CH, (Ph) 3
P-CH
O
from ketone
The Z isomer
is
from
ylide
also produced,
(f)
OVCH
2
from ylide
.0 (Ph) 3
from ketone
19.17
CHO
p-Ionylideneacetaldehyde
+
P-CH 2
Aldehydes and Ketones: Nucleophilic Addition Reactions
19.18 (To:
:OH
OH
H
H
C _
/ \
H
:0:
3.
ti
O C
"OH
|^J^ CH
H 3 Q+
OC
2 OH
.h c
/ \
HO
H
Step 1: Addition of "OH. Step 2: Expulsion, addition of ~H. Step 3: Proton transfer. Step 4: Protonation. This
is
an intramolecular Cannizzaro reaction.
19.19 Addition of the pro-R hydrogen
of
NADH takes place at the Re face of pyruvate.
Re
face
t
NADH H3 C
\_/ C02
"
\ pro-R
I
HO
H
^ \ r
H 3C
_
co 2
+ C //
(^-Lactate
— NHo •
NAD+
H
C-NH 2 7/
Width: 612 Height: 792
486
Chapter 19
19.20 The -OH group adds
to the
Re
face at carbon 2, and
-H + adds to
the
Re
face at carbon 3,
to yield (2tf,3S)-isocitrate.
CO. "0 C2
2^ COH 2C
H
'
Re facet
H2
"OH
ofC2 \
Re
f
"0 2 C^
face
sj* HO
of C3
^CO H
(2/?,35)-Isocitrate
19.21 The product
is
formed by
1,4 addition of
CN, followed by
H— CN
protonation.
_
O
:o:
:CN nucleophilic addition
CN
rV ^^^CN
+ ~CN
CN protonation
Aldehydes and Ketones: Nucleophilic Addition Reactions
19.22 To choose the reactants
that
form a conjugate addition product, follow these
steps:
Give to the aldehyde or ketone carbon the number "1", and count two carbons away from the carbonyl carbon. The double bond in the a, -unsaturated starting material connected the carbons numbered "2" and "3". (2) The grouping bonded to the "3" carbon (circled here) came from the alkyllithium (1)
(3
reagent.
(a)
o
_ H 2 C — CHCCHg II
o 1
2
.
-
Li(CH 3 CH 2 CH2)2Cu^ *^ +
H 3°
II
\
CH3CH2CH2PH2CH2CCH3 N*,
3
2
2-Heptanone
1
487
488
Chapter 19
19.23
2-Cyclohexenone
3 -Methylcyclohexanone
l-Methyl-2cyclohexen-l-ol
2-Cyclohexenone is a cyclic (^-unsaturated ketone whose carbonyl IR absorption occurs -1 at 1685 cm If direct addition product A is formed, the carbonyl absorption vanishes and -1 a hydroxyl absorption appears at 3300 cm If conjugate addition produces B, the 1 carbonyl absorption shifts to 1715 cm" where 6-membered-ring saturated ketones .
.
,
absorb.
19.24 Find (a)
the type of aldehyde or ketone and check Table 19.2 for absorptions.
H 2C=CHCH 2 COCH 3 absorbs at
1715 cm"
1 .
(4-Penten-2-one
is
not an aJ3-un saturated
ketone.) (b)
CH3CH=CHCOCH3
is
an ^-unsaturated ketone and absorbs
(d)
(c)
-1
at
1685
cm
.
(e)
(c)
2,2-Dimethylcyclopentanone, a five-membered-ring ketone, absorbs
(d)
m-Chlorobenzaldehyde shows an absorption cm" and 2820 cm"1
at
1750
cm
.
1
at
1705 cm" and two absorptions
at
2720
1
.
(e)
3-Cyclohexenone absorbs
(f)
CH 3 CH 2CH2 CH=CHCHO
at
is
1715 cm"
1 .
an ^-unsaturated aldehyde and absorbs
at
1705
-1
cm
.
Aldehydes and Ketones: Nucleophilic Addition Reactions
19.25
489
In mass spectra, only charged particles are detected. The McLafferty rearrangement produces an uncharged alkene (not detected) and an oxygen-containing fragment, which is a cation radical and is detected. Alpha cleavage produces a neutral radical (not detected) and an oxygen-containing cation, which is detected. Since alpha cleavage occurs primarily on the more substituted side of the aldehyde or ketone, only this cleavage is shown.
(a)
H 3 (X
H I
^
H
HoC«
OH
CH McLafferty
H
C
H
CH 2
rearrangement
CH,
I
II
C
CHo3
/ \ I
CH 3
H
CH 3 m/z = 72
H 3 C,
o
c C
H
\
/ \
CHc
cleavage
CHod
III
c
H-
I
r
H'
CH 3
H
O
H 3 C.
Alpha
H
CH 3
3-Methyl-2-hexanone
CH 3 m/z = 43
m/z = 114
H HoCX H-
I
*
H
HoC»
PL
o
II
rearrangement
A H H
h3 c
OH
CH
McLafferty
I
CH
CHs
HoC
CH 3
H2 C
m/z = 58
H HoC^
I
M
O
C H-
H3 C
Alpha cleavage
I
C^n ^ch-
CHc
H
O III
H-
H 3C
/ \
H
H 3 C,
c '
^CHo
I
CH 3_ m/z = 43
4-Methyl-2-hexanone
m/z = 114
Both isomers exhibit peaks
at
m/z = 43 due to a-cleavage. The products of McLafferty
rearrangement, however, occur at different values of m/z and can be used to identify each isomer.
490
Chapter 19
(b)
H H3C
HoC.
FCH
McLafferty
H
rearrangement
HC
c.
2
H
v_»n2 -»n2 CHoCH
HoC 2^
CHoCHo
v
m/z = 72
H
H H 3 C^L,H or
H 3 C.
I
o
H-
H.
cleavage
C
H
/ \
H
S,
if
CHoCHo
s ^
O
CHc
Alpha
? CH2CH2
CHc
H m/z = 57 m/z = 114
3-Heptanone
H
OH
CHc'
McLafferty
II I
rearrangement
CH0CH0CH0 d d
C
'
H
CHc
H2
*
/ \
C
CH2CH2CH2
H
H
m/z = 86
H
H^I^H
O
C H
C
H
CH 2 CH 2 CHo
n
/ \
H
I
H-
cleavage
'
+
III
11
1
-1
O
CHo3
Alpha
C /
H
^CH 2
-
I
CH2CH2CH3 _
H
m/z = 71 m/z = 114
4-Heptanone
The isomers can be distinguished on the basis of both a-cleavage products (m/z = 57 vs m/z = 71) and McLafferty rearrangement products (m/z = 72 vs m/z = 86).
(c)
-1
HoC.
I
X
5
H
HoC
O
C
H
OH
CH McLafferty
H
/ \
H
+
rearrangement
II
CHc
I
C^
H
I
CH 3
2-Methylpentanal
m/z = 100
CH 3 m/z = 58
Aldehydes and Ketones: Nucleophilic Addition Reactions
H
O
HoC. Alpha
H
C
H
H
\
O
CH 2
cleavage
III
c
/ \
CH 3
H
I
H
CH 3
2-Methylpentanal
m/z = 100
m/z = 29
H HoC.
I
.H
H 3 C,
o
McLafferty
rearrangement
C
u r
H
CH
OH
II
I
CH
H
H
HoC
HoC'
H m/z = 44
H
O
C H-
II
I
-a v,a
H3 C
Alpha cleavage
O
CH.
III
H.
H3C
/ \
H
H 3 C,
C CHo
H
I
H
m/z = 29
3-Methylpentanal
m/z = 100
The fragments from McLafferty rearrangement, which occur serve to distinguish the two isomers.
at different values
-1
of m/z,
19.26 IR: The only important IR absorption for the compound is seen at 1750 cm where 5membered ring ketones absorb. Mass spectrum: The products of alpha cleavage, which occurs in the ring, have the ,
same mass
m/z
as the molecular ion.
=140
m/z - 84
491
492
Chapter 19
Visualizing Chemistry
19.27
It
helps to
Look
know
that all of these substances
were prepared from aldehydes or ketones.
for familiar groupings of atoms to identify the starting materials.
Notice that the substance pictured is a cyclic acetal. The starting materials were a diol (because cyclic acetals are prepared from diols) and an aldehyde (because an -H is bonded to the acetal carbon). Replace the two -OR groups with =0 to identify the aldehyde starting (a)
material (acetaldehyde).
acetal
(b) We know that the product is an imine because it contains a carbon-nitrogen double bond. The carbon that is part of the C=N bond came from a ketone, and the nitrogen came from a primary amine.
(c) The product is an enamine, formed from a ketone and a secondary amine. Nitrogen bonded to the carbon that once bore the carbonyl oxygen.
is
H— enamine (d) The secondary alcohol product might have been formed by either of two routes reduction of a ketone or by Grignard addition to an aldehyde.
- by
Aldehydes and Ketones: Nucleophilic Addition Reactions
493
19.28 The
intermediate results from the addition of an amine to a ketone. The product is an enamine because the amine nitrogen in the carbinolamine intermediate comes from a
secondary amine.
C
OH
o
N— H
+
N—
II
CH(CH 3
\
H3 C
C
/
CH(CH 3) 2
)
2
CH 3
C(CH 3 2 )
//
N—
3-Methyl-2-butanone
19.29
The nitrogen atom
(a)
is
sp -hybridized, and the geometry
(b)
A p orbital holds the lone-pair electrons of nitrogen.
(c)
The p
orbital holding the lone-pair electrons
is
of nitrogen
planar.
is
aligned for overlap with the
n electrons of the enamine double bond. With this geometry, the nitrogen lone-pair electrons can be conjugated with the double bond, thus lowering energy.
CHo _-.
I
ft H3 C
H3C
1/
/N^ ^CH2 CHo
Additional Problems
Naming Aldehydes and Ketones 19.30 (a)
(b)
O
X
II
CH 3 CCH 2 Br
HoC Bromoacetone (d)
CHO
(c)
CHO s OH
CH 3 CH 2 CH 2 CH 2 CCH(CH 3 2-Methyl-3-heptanone
(5')-2-Hydroxypropanal
(f)
(e)
CH 3
HO^l^H
CH 3 C^— CHCCH 3
(CH 3) 3 CCC(CH 3 3 )
»
O
II
s
VT ^OH
O
2,2,4,4-Tetramethyl-
4-Methyl-3-penten-2-one
3-pentanone
;
CH 2OH (2S,3/?)-2,3,4-Tri-
hydroxybutanal (h)
(g)
(i)
O
OHCCH 2 CH 2 CHO
Butanedial
CH=CHCHO 3 -Pheny 1-2-propenal
6,6-Dimethyl-2,4cyclohexadienone
)
2
494
Chapter 19
O
(j)
OoN-
p-Nitroacetophenone
19.31 Only 2-methylbutanal
is
chiral.
CH<
CH 3 CH 2 CHCHO
CH 3 CH 2 CH 2 CH 2 CHO
2-Methylbutanal
Pentanal
(CH 3 3 CCHO
CH3 CHCH 2 CHO
)
2,2-Dimethylpropanal
3-Methylbutanal
O
O ft
CH 3 CH 2 CH 2 CCH 3
CH 3 CHCCH 3
CH 3 CH 2 CCH 2 CH 3
CH 3 3-Pentanone
2-Pentanone
3-Methyl-2-butanone
19.32 (b)
(a)
QHO
H^C-OH i
CH 2 OH
3-Methyl-3-cyclo-
hexenone
(/?)-2,3-Dihydroxypropanal (D-Glyceraldehyde)
(d)
(e)
I
CHO
O II
CH 3 CHCH 2 CH
OHC
CHo 2-Methy 1- 3 -pentanone
19.33
(a)
The
a,B-unsaturated ketone
2-cyclohexenone
(f)
OH CH 3 CHCCH 2 CH 3
5-Isopropyl-2-methyl-
3 -Hydroxy butanal
C6HsO contains one
p-Benzenedicarbaldehyde ring. Possible structures include:
HoC
CH 3
H 3C
Cyclobutenones and cyclopropenones are also possible.
Aldehydes and Ketones: Nucleophilic Addition Reactions
(b)
O
O
II
II
CH 3 C — CCH 3
and many other
495
structures.
(c) f?
C
H3C
^CH3
CH 2 CHg
CH.
CHo
HoC
(d)
CHO
s^^J
and many other
structures.
Reactions of Aldehydes and Ketones
19.34
Reactions of phenylacetaldehyde:
(b)
(a)
(c)
OH N
O
aCH (d)
II
II
OH
(e)
OCHo3
(f)
I
I
CH 2 CHCHg
CH 2 CHg
CH 2 CHOCH 3
(h)
(g)
CH 2 CH
CH 2 COH
2 CH 2 OH
OH I
CH 2 CHCN
CH 2 CH^- CH2
Reactions of acetophenone:
(c)
(b)
(a)
H
OH
OH
f
\ /
no reaction
Width: 612 Height: 792
496
Chapter 19
HO
(d)
CHo6
(e)
HoCO
\ /
a
G,
CH,
(g)
(f)
OCHod
\ /
CH2CH3 CH<
CH 2 II
19.35 Remember:
RCH 2
—
(C 6 H 5 ) 3 P:
Triphenyl
alkyl halide
(C 6H 5 ) 3 PCH 2 R
X"
phosphonium
salt
phosphine (C 6 H 5 ) 3 PCH 2 R
X
phosphonium
salt
(C 6 H 5 ) 3
P-CHR
+
-
CH3CH2CH2CH2"
:+ Li 1
/
o=c
+
\ aldehyde or ketone
Aldehyde/ketone
Alkyl halide (a)
CH 2 Br
O
O
P-CHR
ylide
Butyllithium
ylide
CH 2 Br
(C 6 H 5 ) 3
R s
/
C=C \ h' alkene
Product
Aldehydes and Ketones: Nucleophilic Addition Reactions
19.36 Remember from Chapter
497
17:
Primary alcohols are formed from formaldehyde + Grignard reagent. Secondary alcohols are formed from an aldehyde + Grignard reagent. Tertiary alcohols are formed from a ketone (or an ester) + Grignard reagent.
Aldehyde/ Ketone (a)
Grignard
Product (after acidic workup)
reagent
O II
CH 3 CH
OH
CH 3 CH 2 CH 2 MgBr
O
I
or
CH 3 CH 2 CH 2 CHCH 3
CH 3 CH 2 CH 2 CH
CH 3 MgBr
CH 2
CH 3 CH 2 CH 2 MgBr
(b)
CH 3 CH 2 CH 2 CH 2 OH
OH C 6 H 5 MgBr
C6 H 5
C 6 H 5 MgBr or 2
19.37
C 6 H 5 MgBr
In general, ketones are less reactive than aldehydes for both steric (excess crowding) and electronic reasons. If the keto aldehyde in this problem were reduced with one equivalent
NaBH4,
the aldehyde functional group would be reduced in preference to the ketone. For the same reason, reaction of the keto aldehyde with one equivalent of ethylene glycol selectively forms the acetal of the aldehyde functional group. The ketone can then be reduced with NaBH4 and the acetal protecting group can be removed.
of
O
O 1
II
.
equiv ^
1
CH3CCH2CH2CH2CHO
NaBH 4
^
II
CH3CCH2CH2CH2CH2OH
3
o HOCH0CH0OH 2 2
o-
—
'
-
/
nLJ ^' CH OH 3 CCH 2 CH 2 CH 2 C^H '
'
acid catalyst
11. 2.
NaBH,
H3
O'
+
OH I
CH3 CHCH 2 CH 2 CH 2 CHO
P H2 |
CHc
498
19.38
Chapter 19
Aldehydes and Ketones: Nucleophilic Addition Reactions
499
19.39
CHO
^s^CH
CH 2OH
1.
NaBH 4
2.
H3
2 Br
PBr,
+
.
aCHoCHoOH
CH 2 CHO Periodinane
CH2CI2
1.
CH 2
2.
H3
Mg, ether
+
The product resembles
the starting material in having an aldehyde group, but a -CHfebetween the aldehyde and the aromatic ring. The product aldehyde results from oxidation of an alcohol that is the product of a Grignard reaction between formaldehyde and benzylmagnesium bromide. The Grignard reagent is formed from benzyl bromide, which results from treatment of benzyl alcohol with PBr3. Reduction of benzaldehyde
group
lies
yields the alcohol.
An
alternate route:
H
I (Ph) 3
a*0
P-CH 2
'
a 1
i
CH 2 CHO Periodinane
.
2.
I
BH 3 THF H 2 2 ,"OH ,
aCH
2 CH 2 OH
CH 2 CI 2 from hydroboration of a double bond that introduced by a Wittig reaction between benzaldehyde and methylenetriphenylphosphorane. In this scheme, the intermediate alcohol results
(b)
1
p
|
.
CH 3 MgBr
2.H 3 +
CHr
is
500
Chapter 19
When you
see a secondary amine and a double bond, you should recognize an enamine. is formed from the amine and acetophenone. Acetophenone, in turn, results
The enamine
from reaction of benzaldehyde with methylmagnesium bromide, followed by oxidation. (c)
aCH from
The
CH-P(Ph) 3
2 Br 1
(Ph) 3 P:
2.
BuLi
Cyclopentanone
(a)
trisubstituted
double bond suggests a Wittig reaction. Reaction of cyclopentanone with
the Wittig reagent formed from benzyl bromide (formed from benzaldehyde in (a)) yields the desired product.
19.40 H
OH
NHCH 3
C6H 5
(f)
no reaction
(h) I O.
\ .0
Aldehydes and Ketones: Nucleophilic Addition Reactions
501
19.41
HO
CH 3 H3 +
CH 3 MgBr
1
2.
+
H3
1 -Methylcy clohexene
The methyl group
introduced by a Grignard reaction with methylmagnesium bromide. Dehydration of the resulting tertiary alcohol produces 1-methylcyclohexene. (b)
X
is
HQ C 6 H 5 MgB
1.
2.
H3
^6 H 5
C6H5
y
1 -p6 H 5
1.BH 3
H 3 Ot
r
+
2.
H2
OH 2
OH
,
and enantiomer I
1
Cr0 3
+ HoO 3*
6 H5
O
2-Phenylcyclohexanone Reaction with phenylmagnesium bromide yields a tertiary alcohol that can be dehydrated. resulting double bond can be treated with BH3 to give an alcohol that can be oxidized to produce the desired ketone.
The
(c)
H 1
2.
OH
H
NaBH,
H3
1
+
.
2.
Os0 4 NaHS0 3
OH
,
H2 cis- 1 ,2-Cyclohexanediol
Reduction, dehydration and hydroxylation yield the desired product, (d)
MgBr
)H
1.PBr 3 2.
from
Mg, ether
(c)
A Grignard reaction forms
1
.
2-
Cyclohexanone^ *
H3 + 1
-Cyclohexylcyclohexanol
1 -cyclohexylcyclohexanol.
502
Chapter 19
Spectroscopy
19.42 Use Table
19.2 if
you need
help.
Only carbonyl absorptions
Due to:
Absorption:
1750 1685 1720
(a)
(b)
are noted.
cm-l cm-l cm-l
5-membered
ring ketone
a^-unsaturated ketone
5-membered
ring
and
aromatic ketone -l
1750 cm -1 1705 cm 2720 1715 cm" 1
(c)
(d)
,
5-membered
cm
-1 ,
2820
ring ketone aromatic aldehyde aliphatic ketone
-1
cm
Compounds in parts (b)-(d) also show aromatic ring IR -1 cm - 1600 cm" and in the range 690-900 cm" 1
absorptions in the range 1450
1
.
19.43
HoO +
or
C 6R n H5
6 n5
B 3-Hydroxy-3-phenylcyclohexanone
Compound A is a cyclic, nonconjugated enone whose carbonyl infrared absorption occur at 1715 cm" Compound B is an a,8-unsaturated, cyclic ketone; additional conjugation with the phenyl ring should lower its IR absorption below 1685 cm" Because the actual IR absorption occurs at 1670 cm" B is the correct structure.
should
.
1
.
1
,
19.44
(b)
NMR
that the unknown is a ketone, and indicates that the carbonyl C flanked by a secondary carbon and a tertiary carbon. unknown is an aldehyde and contains an isopropyl group.
IR shows
(a)
group
is
The The IR absorption shows
that this compound is an a,p-unsaturated ketone, and the 13 spectrum indicates 3 molecular formula shows 3 degrees of unsaturation. The C 2 s/? -hybridized carbons and 3 secondary carbons. (c)
NMR
O
(a)
(b)
(c)
II
(CHg^CHCCH^CHg
(CH 3 2 CHCH 2 CHO )
A O
A has 4 degrees of unsaturation and is a five-membered ring ketone. The ^C NMR spectrum has only three peaks and indicates that A is very symmetrical. 1
19.45 Compound
O
O
Compound A
Aldehydes and Ketones: Nucleophilic Addition Reactions
19.46 As
always, calculate the degree of unsaturation assign the principal functional groups. (b)
(a)
p ell
a
first,
503
then use the available IR data to
Q a
b
CH 3 CHCCH 3
c
||
b
(CH 3 3 CCH2CCH 3 )
CI
a=
1.62 6
a =
1.02 6
b =
2.33 6
b =
2.12 6
c = 4.32 6
c =
2.33 6
General Problems
19.47 (a)
+ (Ph) 3 P: +
BrCH 2 OCH 3
+
BuLi
(Ph) 3 PCH 2
OCH 3
Br"
(Ph) 3
-
P— CHOCH 3 +
LiBr
(b)
H-OHj
..+
C/°
H— .C
CH 3
H d2
H
CH HoV^ oV° C p -h 3 +
2.
3.
M HO H3
+
H 2o:
+
5.
6.
HOCH 3 Steps 1,4: Protonation. Step 2: Addition of water. Steps 3,6: Deprotonation. Step 4: Loss of CH 3 OH.
+
4
H
504
Chapter 19
19.48 4-Hydroxybutanal forms
a cyclic hemiacetal
when
the hydroxyl
oxygen adds
to the
aldehyde group.
/**H—
HO
X."
H2 C
H 2 C— /
,
•:o:
H2 C
1.
^OH
R
H
A
\
H 2 C.
II
I
H
HH+H 3
H H
hemiacetal
Step 1: Protonation. Step 2: Addition of -OH. Step 3: Loss of proton. Methanol reacts with the cyclic hemiacetal to form 2-methoxytetrahydrofuran.
H 2 C— Q:
\„oh
7
HyA
H 2 C,
A
\\4
A
"
HH
+
H2
\< OHc
/
H2 C L
H 2 C— <^>
HoC— o:-)
^>
H2C
A
H
HH
).
HH
HOCHo 3.
tl
H 2 C— O: /
HoO + +
\
HoC. /\
H H
H 2 C— O: H
OCH<
/
\
H
H 2 C,
A £pCH 3 HH H
a cychc acetal
Hob:-*
Step 1: Protonation. Step 2: Loss of water. Step 3: Addition of methanol. Step 4: Loss of proton.
2-Methoxytetrahydrofuran is a cyclic acetal. The hydroxyl oxygen of 4-hydroxybutanal reacts with the aldehyde to form the cyclic ether linkage.
+
Aldehydes and Ketones: Nucleophilic Addition Reactions
505
19.49 H0:-n
H
H—
HO: ^6 H 5
intermediate,
which loses Br~
0=C\
CeHsCHB^
on
+
+ HoO *
Br
C6 H 5
6 n5
+ Br"
substitution of hydroxide ion
Sn2
s- C\J
yields an unstable bromoalcohol
to give benzaldehyde.
19.50 :CN
H3c
CN
CN
HCN
I
^c==0
H 3 C'y
CHgCHj
C ^0-
CHgCH 2
C
^OH
:CN
OH
:CN
CH3CH2
H3C CH 3 CH 2<
^
CH3CH2
>s
H 3 C'y
HCN CHgCHp^
-
I
T
CN
^:CN
CN
Attack can occur with equal probability on either side of the planar carbonyl group to yield a racemic product mixture that is optically inactive.
19.51 H
:0:
I
I I
CL
H2 H
H
/J SCoA
H 2 o:-^
I
:o: 1
^SCo A
+
1.
H
H
H2 +
2:
Conjugate addition of water. Proton shift.
HO
C
:o: II
C /\
H H
:o:
H 1:
v
I _
Step Step
H
II
SCoA
SCoA
Width: 612 Height: 792
506
Chapter 19
19.52
Hv
(PLP)
R'O"
V
~C0 2 ~ ~T
V( pLp
(PLP)
)
N
COc
H H
Step 1: Elimination to form the unsaturated imine. Step 2: Conjugate addition of cysteine to the imine.
19.53
(a)
Grignard addition to a conjugated ketone yields the 1,2 product, not the 1,4 product. to an alcohol. The correct scheme:
LiAlH4 reduces a ketone
1
.
2-
H 2 NNH;
Li(CH 3 ) 2 Cu^
H3
+
KOH
(b) Oxidation of an alcohol with acidic Cr03 converts primary alcohols to carboxylic acids, not to aldehydes. The correct scheme:
C6 H 5 CH=CHCH 2 OH
Periodinane
^
^
» C 6 H 5 CH=CHCHO
Treatment of a cyanohydrin with correct scheme: (c)
H30 + produces
OH HCN^
CHgCCHg
I
CH3CCH3
CHoOH ^j-** C H CH=CHCH(OCH — 6 5
a carboxylic acid, not an amine.
OH 1.LiAIH 4
~ _ CH3CCH3
~ ~
I
2
CN
3)2
CH 2 NH 2
The
Aldehydes and Ketones: Nucleophilic Addition Reactions
507
19.54 CH2
CH3CCH2CH2COCH2
HOCH 2 CH 2 OH ^
,0 N fp CH3CCH2CH2COCH2
+
1
.
2. 1.
H2C /
CH2
2.
DIBAH + H3
H2C
\
Periodinane
H3
+
CH2 \
/
CH 3 CCH 2 CH 2 CHO
UAIH4
CH 3 CCH2CH2CH20H
CH2CI2
(C 6 H 5 ) 3 P-C(CH 3 )2
H2C
^^2
O
4
,0 CH 3 CCH 2 CH 2 CH=C(CH 3 2
H HoO 3
)
*
•
CH 3 CCH 2 CH 2 CH— C(CH 3
)
2
6-Methyl-5-hepten-2-one
The
reaction sequence involves protecting the ketone, converting the ester to an aldehyde, using a Wittig reaction to introduce a substituted double bond, and deprotecting the ketone.
19.55 The same
series of steps
used
to
form an
^OH
acetal is followed in this
mechanism.
HO^SCH 2CH 3
3.
hemithioacetal
Step
1: Protonation.
Step 2: Addition of RSH.
Step 3: Loss of proton.
+
CH 3 CH 2 S^)
HO: SCH2CH3
H 2°
HSCH2CH 3 1.
hemithioacetal
3 IT tl
CH 3 CH2S» «SCH2CH 3
CH 3 CH2S»
-
H-*^-
:a"
SCH2CH 3
HA + thioacetal
Step 1: Protonation. Step 2: Loss of water. Step 4: Loss of proton.
Step 3: Addition of RSH.
508
Chapter 19
19.56 Even though the product looks
unusual, this reaction
is
made up
of steps with which you
are familiar.
:q: I
— S(CH
CH 2
:o— CH 2
3) 2
CH2S(CHg)2 + + :S(CH 3 ) 2 2.
1.
Step 1: Addition of ylide. Step 2: Sn2 displacement of dimethyl sulfide by O".
19.57 :o:
:OH
II
=^
c 1.
H3C Step Step
1:
H 3 C 6r
CN
/C^
2.
+
CH 3
H3C
Deprotonation by OH. of ~CN.
2: Elimination
This sequence
Ok
is
the reverse of the
:o:
:CN
mechanism shown
in Section 19.6.
HO
CN
CN
HoO + 3.
4.
Step 3: Nucleophilic addition of cyanide Step 4: Protonation of tetrahedral intermediate. This step
is
a nucleophilic addition of cyanide.
HO"
:cn
Aldehydes and Ketones: Nucleophilic Addition Reactions
509
19.58
O CCH2CH3
1?
CH 3 CH 2 CCI
/ \
AICI 3
1
UAIH4
2.
H3
+
OH
\
CHCH^CHg
MgBr 1
2.
Br2
,
FeBr^
(
H
H2CHQ
2 ^ Q+
Mg, ether
OH
^" pheny
1' 1
-p r°p ano1
MgBr
P
CHCH2CH3 PBr 3>
VSs
r"^^
N^'
aCHCH2CH
CHCH2CH3
3
Mg, ether
•J
MgBr
aCHCHoCHo 1
.
C=0 2-
(CH 3 2 NCH2CH 2
(CH 3
)
ether
H3
)
+
CH2CH 3
HO— C— C—
2 NCH 2 CH 2
CH2CH 3
c=c
(CH 3 )2NCH 2 CH 2
When you
Tamoxifen
see a product that contains a double bond, and you also know that one of the is a ketone, it is tempting to use a Wittig reaction for synthesis. In this
starting materials
case, however, the tetrasubstituted double bond can't be formed by a Wittig reaction because of steric hindrance. The coupling step is achieved by a Grignard reaction between the illustrated ketone and a Grignard reagent, followed by dehydration. The Grignard reagent is synthesized from 1 -phenyl- 1-propanol. which can be prepared from benzene by either of
two
routes.
510
Chapter 19
19.59 H\
H—
u
:0^ II
t.
H
HoC
H3 C
^
q=c
^^
2
V
CH 3
^ CH 3
^ :0:
H
:OH
c
-
3
-
tl
H
HA +
,:a-
CH 3
1
:o
C
H
^
H (
,
0+
O:
C
C
o:
I
H 3 C/
.
I
^o"
C
^H
5.
H 3 C^/ H
-o" "
^H
4
"
CHo
CHo
H
Paraldehyde
Step 1: Protonation makes the carbonyl carbon more electrophilic. Steps 2,3,4: Three successive additions of the carbonyl oxygen of acetaldehyde to the electrophilic carbonyl carbon, followed by loss of a proton (Step 5), give the cyclic product.
19.60 :o:
\/
"
II
Al
CH3CCH3
C(CH 3 2 )
,_
r
2.
Aluminum, a Lewis acid, complexes with the carbonyl oxygen. 2: Complexation with aluminum makes the carbonyl group more electrophilic and facilitates hydride transfer from isopropoxide. Step Step
1:
Step 3: Treatment of the reaction mixture with aqueous acid cleaves the aluminum-oxygen bond and produces cyclohexanol.
Both the Meerwein-Ponndorf-Verley reaction and the Cannizzaro reaction are hydride transfers in which a carbonyl group is reduced by an alkoxide group, which is oxidized. Note that each aluminum triisopropoxide molecule is capable of reducing three ketone molecules.
Aldehydes and Ketones: Nucleophilic Addition Reactions
19.61
(a)
511
Nucleophilic addition of one nitrogen of hydrazine to one of the carbonyl groups, followed by elimination of water, produces a hydrazone.
HoC 3
^
T
ff
/
C
^ C>^
ft
if
-«
CHo3
•
C^.
HoC
1
1.
9f* ^ c«
T
53
R
HH
+^h
H3 C
H H H 2N~NH 2
/
HH
H 2 G + h 2 n:
2.
H 2 N— NH 2
_
3.
tl
:a ("OH.
.£>H-A :OH
I?
f?
I
c H3 C
CH 3
C
H 3C
5.
>A CH3
ff
H H
HH
N
4.
— NH 2
H3 C
\ CH3 HN— NH2
& |_j
|_|
H
HoN \:
n: C.\
HoC 3
C
y ^ v.
+
HA
CH
/\
hydrazone
H H manner, the other nitrogen of hydrazine can add to the other carbonyl group of 2,4-pentanedione to form the pyrazole.
(b) In a similar
H— A
LA
:
^
u H"
NH 2 \
^:q:
H 2 0+
HO + x/ CHo3
A HH
HoC
1
A HH
CHg
2
.
CH.
JP-H-, « tl'
\
r :N— n:
y C\ / C v
£C
CH 3
5.
H 3C
H H
CH 3 /\
4.
HoC
HH
C
CH,
H H
:n— n: w
/
H 3C
C H H
H
If
H3C
1
/\
1^
n:
H3C
A"
\
n:
C
HoC 3
:
C
+ HA
3,5-Dimethylpyrazole
CH 3
H
The
driving force behind this reaction is the formation of an aromatic ring. The reactions in both parts of this problem are nucleophilic addition of a primary amine (Step 2), followed by elimination of water to yield an imine or enamine (Step 5). All of the other steps are protonations (Steps 1,4) and deprotonations (Steps 3,6).
512
Chapter 19
19.62 The same sequence of steps used in the previous problem leads to the formation of 3,5dimethylisoxazole when hydroxylamine is the reagent. Loss of a proton in the last step of (b) results in a ring that is aromatic. (a)
H-A B C
HoC
CHo3
M
l
HoC
H3 C
A
H H
HH
£ H H
\ CH3 H 2N"
0H
H 2 N— OH
H2
+
HQ
C
^
A"
^
f?
H3 C
:
f?
CHo
5
H H
H3 C
C H H
A
:6h
CH 3
IN— OH I
H tl
HO. n:
HoC 3
+ HA
C
CHo3
H H
oxime
A
4.
H3 C
C H H
h—
\ CH 3 HN— OH
Aldehydes and Ketones: Nucleophilic Addition Reactions
513
19.63
Go:
R u
:p:
H
1,1
rotate
'
H •
R (Ph) 3
j
I
H
(Ph) 3 P
w H
R'
^C-rC"
180° (Ph) 3 P
:p:
/
c=c
/
3.
2.
H
\
H
R
\ R'
+
P:^ (Ph) 3
P=0
Step 1: Addition of the phosphine nucleophile. Step 2: Rotation of C-C bond. Step 3: Elimination of triphenylphosphine oxide. final step is the same as the last step in a Wittig reaction. converts a trans alkene to a cis alkene.
The
The same
series
of steps
19.64
HO— O— H
+
:pH
Hydrogen peroxide and hydroxide
:p— OH react to
+
H 2p:
form water and hydroperoxide anion.
"OH
Conjugate addition of hydroperoxide anion (Step 1) is followed by elimination of hydroxide ion, with formation of the epoxide ring (Step 2).
514
Chapter 19
19.65
NAD +
:
Base
NADH
^
HUNH2 "0
'
2C
+
c
+ oxidation of
CO,
the
amine by
"0
2C
CO.
NAD+
Glutamate
H 2 0:-
—
imine
addition of
water to form a tetrahedral
J
intermediate
:NHc
"
C0 2
"OoC
proton
HO-
shift
"0
2C
HoO +
CO,
loss of
ammonia :
I
"0
Base
Q+
:
2C
deprotonation
CO,
+
NH 3
COc
"OoC
a-Ketoglutarate
of Compound A shows that the molecular formula of A is C5H10O (one degree of unsaturation), and the IR absorption shows that is an aldehyde. The uncomplicated is that of 2,2-dimethylpropanal.
19.66 The molecular weight
A
H NMR
(CH 3
)
3 CCH
T 1.2 6
9.7 5
Compound A
19.67 The IR of Compound B shows
H NMR
a ketone absorption. The splitting pattern of the spectrum indicates an isopropyl group and indicates that the compound is a methyl ketone.
2.4 5
O
CHgCHCCHg
2.1 6
CHo t 1.2 6
Compounds
Aldehydes and Ketones: Nucleophilic Addition Reactions
515
H NMR
at the spectrum, we know that the compound of formula C9H10O has 5 degrees of unsaturation, and we know from the IR spectrum that the unknown is an aromatic ketone. The splitting pattern in the spectrum shows an ethyl group next to a ketone, according to chemical shift values.
19.68 Before looking
H NMR
19.69 The IR two
absorption
is that
of an aldehyde that
isn't
conjugated with the aromatic ring. The
H NMR spectrum are due to two adjacent methylene groups.
triplets in the
O
aCh^Ch^CH
19.70 (a)
a =
1.44 6
b = 4.08 5 c,d
= 6.98
5,
7.81 6
e = I
9.87 6
a=
a
b
OCH 2 CH 3
b,c
1.86 6
=
6.00
d =
6,
6.31 6
9.57 6
(b)
O II
(CH 30)2CHCH 2 CCH3 d
c
b
a=
2.18 5
b = 2.47 6
b =
2.74 6
c =
3.66 6
c
d =
7.28 5
d =
= 3.37 6 4.79 6
a
Width: 612 Height: 792
516
Chapter 19
19.72
CH3O..O
+
CH 3 NH 2
CH 3 NH 2 H3 C
Q
!
CU.'
CH 3 NH\^ C
P
—x
HoC
H3 C
^
CH3O
CH 3°\
CH 3<\
CHoO
O
CH3O
o:
3.
t(
CH3OH
H3C
CH3OH
N
O
H 3 C.
H3 C
t C.
2H
CHo
3-Methyl-2-butenoic acid
1,3-Cyclopentadienecarboxylic acid
OH I
o H3C H (5)-3-Cyclopentyl-2methylpropanoic acid
20.18 (b)
C0 2 H
C0 2 H
(CH 3 2 N )
(a) /7-Bromobenzoic acid is more acidic than benzoic acid because the electron-withdrawing bromine stabilizes the carboxylate anion. (b) This /^-substituted aminobenzoic acid is less acidic than benzoic acid because the electron-donating group destabilizes the carboxylate anion.
20.19 H H HoC 3 CHo 3
H H HoC CHo3
v
V
V
HO
C
C0 2 H
HO
HH
HI
C
K
Br
H H
be used to synthesize the above carboxylic acid because the tertiary halide precursor (shown on the right) doesn't undergo Sn2 substitution with cyanide. Grignard carboxylation also can't be used because the acidic hydroxyl hydrogen interferes Nitrile hydrolysis can't
with formation of the Grignard reagent. If the hydroxyl group Grignard carboxylation can take place.
20.20 The electrostatic potential maps show
is
protected, however,
that the aromatic ring of anisole
(red) than the aromatic ring of thioanisole, indicating that the
is
more electron-rich is more
methoxyl group
strongly electron-donating than the methylthio group. Since electron-donating groups
decrease acidity, /?-(methylthio)benzoic acid
methoxybenzoic
acid.
is likely to
be a stronger acid than p-
Width: 612 Height: 792
526
Chapter 20
Additional Problems
Naming Carboxylic Acids and
Nitriles
20.21 (a)
COoH 2 I
COoH 2
(b)
CHo3 I
I
CH 3 CC0 2 H
CHgCHCH 2 CH 2 CHCH3
CH 3 2,2-Dimethylpropanoic acid
2,5-Dimethylhexanedioic acid
(d)
(c)
C0 2 H
C0 2 H
m-Cyanobenzoic acid (e)
CHo
(£)-2-Cyclodecenecarboxylic acid
®
CH 2 C02 H CH3CH 2 CH 2 CHCH 2 CH3
CHoCCN 3 |
CH 3 3-Ethylhexanoic acid
2,2-Dimethylpropanenitrile
(g)
(h)
Br
CN
I
BrCH 2 CHCH 2 CH 2 C02H 4,5-Dibromopentanoic acid
2-Cyclopentenecarbonitrile
20.22 (a)
H
(b)
H0 2 CCH 2 CH 2 CH 2 CH 2 CH 2 C0 2 H
\/T^C02 H
Heptanedioic acid
H cis- 1 ,2-Cyclohexanedicarboxylic acid
(d)
(c)
CH 3 C= CCH= CHC0 2 H 2-Hexen-4-ynoic acid
CHgCH 2
(pH 2 CH 2 CHg
CH3CH 2 CH 2 CH 2 CHCH 2 CHC0 2 H 4-Ethyl-2-propyloctanoic acid
Carboxylic Acids and Nitriles
(e)
(f)
C0 2 H
(C 6 H 5 ) 3 CC02 H Triphenylacetic acid
C0 2 H 3-Chlorophthalic acid
(h)
(g)
CN
m-Benzoylbenzonitrile
2-Cyclobutenecarbonitrile
20.23 (a)
CH 3 CH2CH2CHCH2CO2H
CH2CH2CH2CHCO2H
CH2CH2CH2CH2CH2CO2H
2-Methylpentanoic acid
Hexanoic acid
3-Methylpentanoic acid
CH2OH3
CHo3
THF
CH2CO2H
OH H3
H3
+
+
CHgC^C^C-~ CH2CH2OH OH
CH,
CH 3 CH 2 CH 2 C= CHC0 2 H 3-Methyl-2-hexenoic acid
As
in all of these
more complex syntheses, other routes to the target compound are was chosen because the Grignard reaction introduces a double bond
possible. This route
without removing functionality at carbon 3. Dehydration occurs in the desired direction produce a double bond conjugated with the carboxylic acid carbonyl group.
to
Spectroscopy
20.43 The peak
at 1.08 6 is
acid group.
due
to a terr-butyl group,
The compound
is
and the peak
at 11.2 6 is
due
to a carboxylic
3,3-dimethylbutanoic acid, (CH3)3CCH2CC>2H.
Carboxylic Acids and Nitriles
20.44
U NMR
Either C or carboxylic acids.
*H
535
NMR can be used to distinguish among these three isomeric
Compound
Number of !H NMR
Splitting
C NMR
absorptions
absorptions
signals
Number of 13
]
CH 3 (CH 2 3 C02 H )
of
H NMR
1 triplet,
peak area
3, 1.0 5
1 triplet,
peak area
2, 2.4 5
2 multiplets, peak area 4, 1.5 5
(CH3) 2 CHCH 2 C02H
peak area
12.0 6
1
singlet,
1
doublet, peak area 6, 1.0 6
1
doublet, peak area 2, 2.4 6
1
multiplet,
1
singlet,
peak area
1,
1
singlet,
peak area
9, 1.3 6
1
singlet,
peak area
1,
(CH 3 3 CC0 2 H
1,
peak area
1,
1.6 5
12.0 5
)
20.45
In
all
12.1 5
of these pairs, different numbers of peaks occur in the spectra of each isomer,
Use
(a), (b)
either
!
H NMR or C NMR to distinguish between the isomers. 1
Number of
Compound
13
(a)
C NMR
Number of ]
H NMR
absorptions
absorptions
3
3
C0 2 H
H0 2 C H02 C
C0 2 H
(b)
H0 2 CCH 2 CH 2 C0 2 H CH 3 CH(C0 2 H) 2 (c)
Use *H
NMR to distinguish between these two compounds. The carboxylic acid proton
of CH3CH2CH2CO2H absorbs near 12 6, and the aldehyde proton of HOCH2CH2CH2CHO absorbs near 10 5 and is split into a triplet.
Width: 612 Height: 792
536
Chapter 20
H13 NMR and 13 C NMR spectra. (CH3)2C=CHCH2C0 2H shows six absorptions in its C NMR and five in its H NMR spectrum; one of the H NMR signals occurs in the vinylic region (4.5 - 6.5 C NMR spectrum of the unsaturated acid also shows two 6) of the spectrum. The (d)
Cyclopentanecarboxylic acid shows four absorptions in both
its
!
absorptions in the
C=C bond region
(100-150
20.46 The compound has one degree
6).
of unsaturation, which absorption seen in the IR spectrum.
a
b
c
d
CH3CH2OCH2CO2H
a =
1.26 5
b =
3.64 5
c _
4 14 §
d =
11.125
is
due to the carboxylic acid
General Problems a tertiary alkyl halide and ~CN is a base. Instead of the desired Sn2 reaction of cyanide with a halide, E2 elimination occurs and yields 2-methyl-2pentene.
20.47 2-Chloro-2-methylpentane is
CH3CH2CH2CCH3
20.48
Ibuprofen
Carboxylic Acids and Nitriles
20.49
(a)
Use CO2 instead of NaCN to form the carboxylic acid, or eliminate scheme and form the acid by nitrile hydrolysis.
537
Mg from this
reaction (b)
Reduction of a carboxylic acid with LiAU-L* yields an alcohol, not an alkyl group.
(c) Acidic hydrolysis of the nitrile will also dehydrate the tertiary alcohol. hydrolysis to form the carboxylic acid.
Use basic
20.50
OH Step
1:
Protonation of acetal oxygen.
Step
2:
Loss of cyanohydrin.
Step 3: Addition of water, followed by deprotonation.
":CN
Deprotonation of the cyanohydrin hydroxyl group butanone.
is
followed by loss of
CN, forming
2-
538
Chapter 20
20.51 H 2 0:
f-H— OH2
:OH 2
R— C= NH
R— c=Nr
H
H3 +
3
-
ll
:6h 2
:0
:o:
NH 2
R
^
NH
R
2.
4.
NH 2
R
^NH 2
R
amide
Step 1: Protonation. Step 3: Proton transfer.
The
first
Step 2: Addition of water. Step 4: Deprotonation.
equivalent of water adds to a nitrile to produce an amide.
.s
H — OH 2
H
H :0'
II
R^j
NH 2
R
FT
NH 2
2.
H 2 o:
NH/
+
O
OH
R Step Step Step
^
n
I :nh 3
Proton transfer. 5: Deprotonation
R^ "bH Step Step
/
HO:
Addition of water.
4:
Loss of ammonia
to the
R
4.
2:
The second equivalent of water adds
amide
""-NH 2
3
-
H
C
5.
1: Protonation.
3:
^h
/
H2 +
TWa ^
to yield a carboxylic acid, plus
ammonium ion. 20.52
C0 2 H HNQ 3
KMnO,
H 2 S04
HoO
C0 2 H 1. Fe,
1
2.
H 3 Q+
"OH
PABA
Notice that the order of the reactions is very important. If toluene is oxidized first, the nitro group will be introduced in the meta position. If the nitro group is reduced first, oxidation to the carboxylic acid will reoxidize the -NH2 group.
Carboxylic Acids and Nitriles
539
20.53 Br c
FeBrv
— 1
^
^_
C H 2 OH
.
CH 2 Q
MgBr
Periodinane
CH2CI2
\
//
2.
H3
+
Fenclorac to this compound are possible. The illustrated route was chosen because it introduced the potential benzylic functional group and the potential carboxylic acid in one step. Notice that the aldehyde functional group and the cyclohexyl group both serve to direct the aromatic chlorination to the correct position. Also, reaction of the hydroxy acid + with SOCl 2 converts -OH to -CI and -C0 2 to -COC1. Treatment with 3 regenerates the carboxylic acid.
Other routes
H
H
20.54 Substituent
pKa
-PCI 2
3.59
-OS02CH 3
3.84
-CH= CHCN
4.03
-HgCH 3
4.10
-H
4.19
-Si(CH 3 3 )
4.27
Acidity
Most acidic
*E.A.S. reactivity
Least reactive
(most deactivating)
Least acidic
Most reactive (least deactivating)
*Electrophilic aromatic substitution
Recall from Section 20.4 that substituents that increase acidity also decrease reactivity in electrophilic aromatic substitution reactions. Of the above substituents, only -Si(CH3)3 is
an activator.
540
Chapter 20
Again, other routes to this compound are possible. The above route was chosen because it has relatively few steps and because the Grignard reagent can be prepared without competing reactions. Notice that nitrile hydrolysis is not a possible route to this compound because the halide precursor is tertiary and doesn't undergo Sn2 substitution.
The product results from two Grignard to this compound.
reactions.
As
in (a), nitrile hydrolysis is not a route
20.56 As we have
seen throughout this book, the influence of substituents on reactions can be resonance effects. For m-hydroxybenzoic acid, the negative charge of the carboxylate anion is stabilized by the electron-withdrawing inductive effect of -OH, making this isomer more acidic. For p-hydroxybenzoic acid, the negative charge of the anion is destabilized by the electron-donating resonance effect of -OH that acts over the
due
to inductive effects and/or
k electron system of the ring but
is
not important for m-substituents.
Carboxylic Acids and Nitriles
541
O"
H2
+
and other resonance forms
20.57
C0 2 H
CHO (a)
H3
BH 3 + );
THF, then H 2 2 OH"; (b) PBr3 (c) Mg, then C0 2 + (d) LiAlH 4 then H 3 (e) Dess-Martin periodinane, ,
,
,
,
;
;
then
+
H3
CH2 C12
;
XN, then H2 NNH2 ,KOH
(or
(f)
20.58 :OH
:o: :.OH •
H 3C
Br
HoC
h3C
\^ Nucleophilic addition observed product.
(1), alkyl shift (2),
3
C0 2 H
T)
and displacement of bromide
+
Br"
(3) lead to the
20.59 +
g
*\t~>
"/ C<^ C\ /CH2 OP2 O
i
+
P04 3
"
Hoc
II
:
C0 2
C
C
H H
H H
35
3-Phosphomevalonate 5 -diphosphate
*
^C^ /CH 2OP2 f
It
H
H H
35
Isopentenyl diphosphate
542
Chapter 20
20.60
A compound with the formula C4H7N has two degrees of unsaturation.The IR absorption -1 at
2250
cm a
identifies this b
c
CH 3 CH 2 CH 2 C=N
20.61 Both compounds nonequivalent).
compound
as a nitrile.
a=
1-06 6
b =
1.68 5
c =
2.31 6
contain four different kinds of protons (the H2C= protons are acid proton absorptions are easy to identify; the other three
The carboxylic
absorptions in each spectrum are
more complex.
It is possible to assign the spectra by studying the methyl group absorptions. The methyl group peak of crotonic acid is split into a doublet by the geminal (CH3CH=) proton, while the methyl group absorption of methacrylic acid is a singlet. The first spectrum (a) is that of crotonic acid, and the second spectrum (b) is that of methacrylic
acid.
20.62
(a)
From
the formula,
we know
compound has 2 degrees of unsaturation, one of 13 group that absorbs at 183.0 6. The C NMR spectrum
that the
which is due to the carboxylic acid also shows that no other sp carbons are present in the sample and indicates that the other degree of unsaturation is due to a ring, which is shown to be a cyclohexane ring by symmetry and by the types of carbons in the structure. (b) The compound has 5 degrees of unsaturation, and is a methyl-substituted benzoic The symmetry shown by the aromatic absorptions identifies the compound as p-
acid.
methylbenzoic acid.
20.63
O II
HgC Step
1:
Deprotonation
Step 2: Decarboxylation
+
"OH
CH3
Step 3: Protonation.
This reaction proceeds because of the loss of CO2 and the stability of the enolate anion.
Carboxylic Acids and Nitriles
20.64
The following
steps take place in the Ritter reaction:
Step 1: Protonation of the alkene double bond; Step 2: Attack of the nitrogen lone pair electrons on the carbocation; Step 3: Attack of water on the nitrile carbon; Step 4: Deprotonation; Step 5: Tautomerization to the ketone.
543
Chapter 21 - Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
Chapter Outline I.
Introduction to carboxylic acid derivatives (Sections 21.1-21.2). A. Naming carboxylic acid derivatives (Section 21.1). 1
.
Acid
b
2
.
halides.
acyl group is named first, followed by the halide. For acyclic compounds, the -ic acid or -oic acid of the carboxylic acid name is replaced by -oyl, followed by the name of the halide. i. There are eight exceptions, in which -yl is used c. For cyclic compounds, the -carboxylic acid ending is replaced by -carbonyl, followed by the name of the halide. Acid anhydrides. a. Symmetrical anhydrides are named by replacing acid by anhydride. b Unsymmetrical anhydrides are named by citing the two acids alphabetically, followed by anhydride. a.
The
.
.
3.
Esters. a.
named by
Esters are
first
identifying the alkyl group and then the carboxylic
acid group, replacing -oic acid
by
-ate.
4.
Amides.
5
Amides with an unsubstituted -NH2 group are named by replacing -oic acid or —ic acid by -amide or by replacing -carboxylic acid with -carboxamide. b If nitrogen is substituted, the nitrogen substituents are named in alphabetical order, and an N- is put before each. Thioesters. a.
.
.
a.
Thioesters are
named
like esters, using the prefix thio- before the
name of the
ester derivative of the carboxylic acid.
6.
Acyl phosphates. a. Acyl phosphates are named by
citing the acyl group and adding the word phosphate. B. Nucleophilic acyl substitution reactions (Section 21.2). 1 Mechanism of nucleophilic acyl substitution reactions. a. nucleophile adds to the polar carbonyl group. b The tetrahedral intermediate eliminates one of the two substituents originally .
A
.
bonded c.
to
it,
resulting in a net substitution reaction.
Reactions of carboxylic acid derivatives take this course because one of the groups bonded to the carbonyl carbon is a good leaving group.
The addition step is usually rate-limiting. Relative reactivity of carboxylic acid derivatives. d.
2
.
a.
Both
steric
and electronic factors determine relative reactivity. group decreases reactivity.
i.
Steric hindrance in the acyl
ii.
More
polarized acid derivatives are
more
reactive than less polarized
derivatives. hi.
The
effect of substituents
on
reactivity is similar to their effect
on
electrophilic aromatic substitution reactions.
b.
It is i.
3
.
possible to convert
more
reactive derivatives into less reactive derivatives.
In order of decreasing reactivity: acid chlorides thioesters > esters > amides.
> acid
Only esters, amides, and carboxylic acids are found ii. Kinds of reactions of carboxylic acid derivatives: a.
anhydrides
in nature.
Hydrolysis: reaction with water to yield a carboxylic acid.
>
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
b
.
c.
545
Alcoholysis: reaction with an alcohol to yield an ester. Aminolysis: reaction with ammonia or an amine to yield an amide.
Reduction. i. Reaction with a hydride reducing agent yields an aldehyde or an alcohol, ii Amides are reduced to yield amines. Reaction with an organometallic reagent to yield a ketone or alcohol. e. Reactions of carboxylic acids and their derivatives (Section 21.3-21.9). A. Nucleophilic acyl substitution reactions of carboxylic acids (Section 21.3). Carboxylic acids can be converted to acid chlorides by reaction with SOCI2. 1 a. The reaction proceeds through a chlorosulfite intermediate. 2 Acid anhydrides are usually formed by heating the corresponding carboxylic acid to remove 1 equivalent of water. d.
II.
.
.
3
.
Conversion to esters. a. Conversion can be effected by the Sn2 reaction of a carboxylate and an alkyl halide.
b
.
Esters can be produced an alcohol.
by
the acid-catalyzed reaction of a carboxylic acid and
This reaction is known as a Fischer esteriflcation. Mineral acid makes the acyl carbon more reactive toward the alcohol. iii. All steps are reversible. iv. The reaction can be driven to completion by removing water or by using a large excess of alcohol. v Isotopic labelling studies have confirmed the mechanism. 4. Conversion to amides. a. Amides are difficult to form from carboxylic acids because amines convert carboxylic acids to carboxylate salts that no longer have electrophilic carbons. b The reagent DCC (dicyclohexylcarbodiirnide) can be used; it is used in the laboratory to form peptide bonds. 5 Reduction of carboxylic acids. Reduction to alcohols can be achieved by use of L1AIH4. a. b BH3 in THF easily reduces carboxylic acids to alcohols. B. Chemistry of carboxylic acid halides (Section 21.4). Carboxylic acid halides are prepared by reacting carboxylic acids with either SOCI2 1 or PBr3 to form the corresponding acyl halide. 2. Acyl halides are very reactive. a. Most reactions occur by nucleophilic acyl substitution mechanisms. 3. Hydrolysis. Acyl halides react with water to form carboxylic acids. a. b The reaction mixture usually contains a base to scavenge the HC1 produced. i.
ii.
.
.
.
.
.
.
3
.
4.
Anhydride formation. Acid halides react with carboxylate ions a.
to
form anhydrides.
Alcoholysis. Acyl halides react with alcohols to form esters. a. b. Base is usually added to scavenge the HC1 produced.
Primary alcohols are more reactive than secondary or tertiary alcohols, It's often possible to esterify a less hindered alcohol selectively. i. Aminolysis. Acid chlorides react with ammonia and amines to give amides. a. b Either two equivalents of ammonia/amine must be used, or NaOH must be present, in order to scavenge HC1. Reduction. a. LiAlH4 reduces acid halides to alcohols. i. The reaction is a substitution of H~ for CF that proceeds through an c.
5
.
.
6.
intermediate aldehyde, which
is
then reduced.
Width: 612 Height: 792
546
Chapter 21 Reaction with organometallic reagents. Reaction with Grignard reagents yields tertiary alcohols and proceeds through a. an intermediate ketone. b Reaction with diorganocopper (Gilman) reagents yields ketones. i. Reaction occurs by a radical mechanism. ii. This reaction doesn't occur with other carboxylic acid derivatives. C. Chemistry of carboxylic acid anhydrides (Section 21.5). 1 Acid anhydrides can be prepared by reaction of carboxylate anions with acid
7
.
.
.
chlorides.
2
.
a. Both symmetrical and unsymmetrical anhydrides can be prepared by this Acid anhydrides react more slowly than acid chlorides. a. Acid anhydrides undergo most of the same reactions as acid chlorides.
route.
Acetic anhydride is often used to prepare acetate esters. In reactions of acid anhydrides, half of the molecule is unused, making anhydrides inefficient to use. D. Chemistry of esters (Section 21.6). Esters can be prepared by: 1 a. Sn2 reaction of a carboxylate anion with an alkyl halide. b. Fischer esterification. c. Reaction of an acid chloride with an alcohol, in the presence of base. 2 Esters are less reactive than acid halides and anhydrides but undergo the same types b. c.
.
.
of reactions. 3.
Hydrolysis. a. Basic hydrolysis (saponification) occurs through a nucleophilic acyl substitution
mechanism. i. Loss of alkoxide ion yields a carboxylic acid which
is
deprotonated to give a
carboxylate anion. Isotope-labelling studies confirm this mechanism. Acidic hydrolysis can occur by more than one mechanism. i. The usual route is by the reverse of Fischer esterification. Aminolysis. a. Esters can be converted to amides by heating with ammonia/amines, but it's easier to start with an acid chloride. Reduction. a. LiAlFLt reduces esters to primary alcohols by a route similar to that described for acid chlorides. at -78 °C is used, reduction yields an aldehyde. b If DIB Reaction with Grignard reagents. a. Esters react twice with Grignard reagents to produce tertiary alcohols containing ii.
b.
4.
5
.
AH
.
6
.
two
identical substituents.
E. Chemistry of amides (Section 21.7). Amides are prepared by the reaction of acid chlorides with ammonia/amines. 1 2. Hydrolysis. a. Hydrolysis occurs under more severe conditions than needed for hydrolysis of .
other acid derivatives.
b
.
c.
3
.
Acid hydrolysis occurs by addition of water to a protonated amide, followed by loss of ammonia or an amine. Basic hydrolysis occurs by attack of HO~, followed by loss of ~NH2-
Reduction. a.
LiAlH4 reduces amides
to amines.
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
547
F. Thiol esters and acyl phosphates (Section 21.8). 1 Nature uses thiol esters and acyl phosphates in nucleophilic acyl substitution reactions because they are intermediate in reactivity between acid anhydrides and .
esters.
Acetyl CoA is used as an acylating agent Polyamides and polyesters (Section 21.9). A. Formation of polyesters and polyamides. When a diamine and a diacid chloride react, a polyamide is formed. 1 2 When a diacid and a diol react, a polyester is formed. 3 These polymers are called step-growth polymers because each bond 2
III.
.
.
.
.
B
is
formed
independently of the others. Types of polymers.
.
1
.
2
.
Nylons are the most
common polyamides.
The most common polyester, Dacron,
is
formed from dimethylterephthalate and
ethylene glycol.
Biodegradable polymers are usually polyesters of naturally-occurring hydroxy carboxylic acids. IV. Spectroscopy of carboxylic acid derivatives and nitriles (Section 21.10). 3
.
A. Infrared spectroscopy. All of these compounds have characteristic carbonyl absorptions that help identify 1 them; these are listed in Table 21.3. B spectroscopy is of limited usefulness in distinguishing carboxylic acid .
.
NMR
derivatives. 1
.
Hydrogens next
to carbonyl groups absorb at
around
but this absorption can't be used to distinguish 2.
among
!
H NMR
spectrum, carboxylic acid derivatives.
2. 1
6 in a
Carbonyl carbons absorb in the range 160-180 5, but, again, be used to distinguish among carboxylic acid derivatives.
this absorption can't
Solutions to Problems
21.1
lists the suffixes for naming carboxylic acid derivatives. the functional group is part of a ring are in parentheses.
Table 21.1
when
(b)
(a)
The
suffixes used
(c)
P
4-Methylpentanoyl
Cyclohexylacetamide
chloride
Benzoic anhydride
Isopropyl cyclopentanecarboxylate
Isopropyl 2-methylpropanoate
Cyclopentyl 2-methylpropanoate
548
Chapter 21
(h)
(g)
(i)
O SCH 2 CHg
HgC^ HoC.
H 2 C=CHCH 2 CH 2 CNHCH 3
/
HO N-Methyl-4-pentenamide
21.2 (a)
O O
^ As
OP03
C V
2-
C=C \
/
H
(/?)-2-Hydroxypropanoyl phosphate
(b)
Ethyl 2,3-dimethyl2-butenethioate
(C)
CH 3 CH 2 CH 2 CNCH 2 CH 3
O
CH 3
CH 3 CHCH2 CHCCI CH-
Phenyl benzoate
Af-Ethyl-/V-methylbutanamide
2,4-Dimethylpentanoyl chloride
(d)
(e)
CH-
OCH.
O
O
II
II
CHgCH 2 CCH 2 COCH 2 CH 3
Methyl 1-methylcyclo-
SCH.
Ethyl-3-oxopentanoate
hexanecarboxylate
Methyl /7-bromobenzenethioate (h)
(g)
O
O
II
II
h
/^LrfCOBr
HC^ ^CCH 2CHg H Formic propanoic
cw-2-Methylcyclopentanecarbonyl bromide
anhydride
21.3
:OCH 3 c
:o:
:o: II
c CI
rocH 3
OCH3 +
addition of methoxide form a tetrahedral
to
intermediate
elimination of Cl~
cr
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
21.4 Use Figure
2 1 .2
if
you need
help.
Most reactive (
a)
549
Least reactive
O ?
II
>
CH 3 CCI (b)
CH3COCH3
O
>
CH3CNH2
o
O
II
CH3 COCH(CF3 The most reactive
2
)
is to
be the leaving group.
Identify the nucleophile (boxed)
group by the nucleophile (a)
CH 3 COCH 2 CH 3
>
acyl derivatives contain strongly electron-withdrawing groups in the part
of the structure that
21.5
CH 3 COCH 2 CCI 3
>
and the leaving group
(circled),
O
O
+
Na !"OH
II
HOCH3
c.
h3c
(b)
0"Na
H3 C
.;och 3
;
o
O 2
NHo'
;
NH 4+
I
HoC
(c)
O /''0 II
(d)
/
\
II
+
Na
CI"
O
!_"OCH3j
1?
II
CH 3 /'
\0
HoC
and replace the leaving
in the product.
HoC
OCH-
+
Na ~
O
O HCI
II
.C.
II
.C^
HOCHpCHoCHoCHo
+
OH
H3 C
1-Butanol
Acetic acid
+
OCH2CH2CH2CH3
H3 C
H2
Butyl acetate
(b)
HCI
?\
+
HOCH3
OH
CH3CH2CH2
ft
5==
(C>
H
V/
ff
CH3
H
S/
H
HCI
2-Propanol
Cyclopentane-
H2
Methyl butanoate
Methanol
Butanoic acid
+
^
OCH3
CH3CH2CH2
CH 3
Isopropyl cyclopentanecarboxylate
carboxylic acid
21.8
Under Fischer esterification conditions, many hydroxycarboxylic acids can form intramolecular esters (lactones).
B
8 \
2
C
C
^OH
HCI^
H2°^
C
I
C
C
/\
/\
HH
HH
5-Hydroxypentanoic acid
21.9
I
+
HoO
HpC^
^.CH20H
H2C,.
a lactone
Pyridine neutralizes the HCI byproduct by forming pyridinium chloride. This neutralization removes from the product mixture acid that might cause side reactions. As mentioned previously, positioning the reacting groups so that they face each other makes it easier to predict the products. < a)
O
O Pyridine
II
+
CH3CH2
CH3CH2
CI
Propanoyl chloride
II
HOCH3 Methanol
OCH3
Methyl propanoate
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
551
(b)
H3C
./Cn H3C
CI
Acetyl chloride
(c)
Pyridine
HOCH 2 CH 3 Ethanol
Ethyl acetate
O Pyridine
II
HOCH 2 CH 3
cr
CI
Benzoyl chloride
21 1 .
OCH2CH3
OCH 2 CH 3
Ethyl benzoate
Ethanol
As
explained in the text, only simple, low boiling alcohols are convenient to use in the Fischer esterification reaction. Thus, reaction of cyclohexanol with benzoyl chloride is the preferred method for preparing cyclohexyl benzoate.
'V
O
Pyridine
CI
HO
+
Cyclohexanol
Benzoyl chloride
Cyclohexyl benzoate
21.11 :OH
nucleophilic
addition of
morpholine
OCH-
deprotonation
by hydroxide
O ch3
°yyH' OCH,J Trimetozine
+
1
NaCI
elimination or chloride
OCH-
+
HoO
Chapter 21
552
21 12 .
An extra equivalent of base and
reactions. In (a)
(a)
(b),
must be added to neutralize the acid produced in these two equivalents of the amine may be used in place of NaOH.
O CH 3 CH 2
f >a
NaOH
H 2 NCH 3
+
(b)
NaCI
O
O HN(CH 2 CH 3
NaOH )
II
2
a^N(CH CH
CI
2
+
Benzoyl chloride (c)
+
N-Methylpropanamide
Methylamine
Propanoyl chloride
H2
+
NHCH 3
CH3 CH 2
CI
H2
+
3) 2
NaCI
AfAf-Diethylbenzamide
Diethylamine
O II
/C^ CH 3 CH 2
2
+
n
NH 3
CH 3 CH 2
CI
orNH4+^CI
NH 2
Propanamide
Propanoyl chloride
21.13 Two combinations
+
of acid chloride and organocopper reagent are possible,
(a)
^r\.C^
+
[(CH 3 ) 2 CH] 2 CuLi
c.
CH(CH 3 ) 2
of
/C^
•
CH(CH 3
Cr
)
2
(b)
B
H 2 C=CH
/C^ CI
+
(CH 3 CH 2 CH 2 2 CuLi )
_ — CH
or
H2 C
O (H 2
C=CH) 2 CuLi
ft
^C^
+ CI
CH 2 CH 2 CH 3
/C^ CH 2 CH 2 CH 3
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
553
21.14
Acetaminophen Step 1: Nucleophilic addition of /7-hydroxy aniline. Step 2: Deprotonation by hydroxide. Step 3: Loss of acetate ion.
21.15
Phthalic anhydride
The second half of the anhydride becomes a carboxylic
21.16
acid.
Acidic hydrolysis of an ester is a reversible reaction because the products are an alcohol and a carboxylic acid. Basic hydrolysis of an ester is irreversible because its products are an alcohol and a carboxylate anion, which has a negative charge and does not react with nucleophiles.
554
Chapter 21
21.17
J
.
2.
UAIH4
H3
A
(a)
1
DIBAH
2.
H3 +
O
+
21.18 Lithium aluminum hydride
9 CH 2 OH
reduces an ester to form two alcohols,
H
HoC
3f
ff
CH 3 CH 2 CH 2 CHCOCH3
1
.
UAIH4
H3Q
+
1
CH3CH2CH2CHCH2OH
Methanol
2-Methyl- 1 -pentanol
1
2.
H3
HO^^/n.
CH 2 OH
UAIH4
CH 3 OH
+
+
Benzyl alcohol
Phenol
21.19 Remember that Grignard
reagents can only be used with esters to form a tertiary alcohol two substituents, which come from the Grignard reagent, and work backward to select the ester (the alkyl group of the ester is unimportant).
two
that has
identical substituents. Identify these
Grignard Reagent
Tertiary Alcohol
+
Ester
(a)
OR 2
CH 3 MgBr
O H3 C
MgBr
OR
O
(c) »
x
CH 3 CH2 XyCh^Ch^
i
2
CH 3 CH2CH2CH2 OH
CH 3 CH 2 MgBr
+
CH 3 CH 2 CH 2 CH 2
OR
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
555
21.20
O (a)
Cv
NHCH 2 CH 3
j^^J^
H3
CH 2 OH
+
heat
Af-Ethylbenzamide
(c) 1
.
2.
UAIH4
H2
aCH
2 NHCH 2 CH 3
21.21
MgBr 1
.
2-
CQ 2 H3
,
ether
+
a
OH
SOCIc
CH 2 N(CH 3
)
J. 2.
2
LiAIH 4
H2
N(CH 3 2 )
(CH 3 2 NH
CI
)
NaOH
The product is a ./V-disubstituted amine, which can be formed by reduction of an amide. The amide results from treatment of an acid chloride with the appropriate amine. The acid chloride is the product of the reaction of SOCI2 with a carboxylic acid that is formed by carboxylation of the Grignard reagent synthesized from the starting material.
Width: 612 Height: 792
556
Chapter 21
21.22 Even though the entire molecule in this
problem only with the
of coenzyme A is biologically important, we are concerned group. The remainder of the structure is represented here
-SH
as "R".
Ho:
O
\ii
ii
HoC f
O
(
:
O
O".
II
O
I
— Adenosine
H3 C
O"
O
/
s
1.
O
I
— Adenosine
o-
N R
RS— H^lBase J*-
t
1? n
S
H3C Acetyl
o
O
l_
— Adenosine
CoA
1: Nucleophilic addition of the -SR group of adenylate to form a tetrahedral intermediate.
Step
CoA
(after deprotonation) to acetyl
Step 2: Loss of adenosine monophosphate.
2 1 .23 In each example, if n molecules of one component react with n molecules of the other component, a polymer with n repeating units is formed, and 2n small molecules are formed as byproducts; these are shown in each reaction. (a)
BrCH 2 CH 2 CH 2 Br +
HOCH 2 CH 2 CH 2 OH
/ \ -^-CH 2 CH 2 CH 2 OCH 2 CH 2 CH 2 0-^-
Base
+
<
b>
2n HBr
HOCH 2 CH 2 OH
+
H0 2 C(CH 2 6 C02 H
II
f
-V-
)
\
catalyst
)
o
/
H 2 N(CH 2 6 NH 2 )
+
CIC(CH 2 4 CCI )
m
2n H 2
(c) II
II
OCH 2 CH 2 0— C(CH 2 6 C-/+
II
0\
O
/
H 2 S0 4
o
-V-HNC(CH 2 6 NH )
+
\
— C(CH2 4C—h II
II
[
2n
)
HCI
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
557
21.24
o
O II
HOC
\_J^
C0H
+
1,4-Benzenedicarboxylic acid
21.25 Use Table
21.3
if
you need
21.26
cm
JR 1735
1,4-Benzenediamine
Functional group present Saturated ester or 6-membered ring lactone Saturated acid chloride
2500 - 3300 cm" 1 and 1710
(d) 1715
|
^/
help.
Absorption -l (a) 1735 cm -l (b) 1810 cm (c)
H2N
-1
cm -l
-l
Carboxylic acid Saturated ketone or 6-membered ring ketone
cm-1
corresponds to a saturated ester. and twelve hydrogens can be arranged in a number of ways to produce a structure for this compound. For example: (a)
The remaining
five carbons
CH3CH2COCH2CH2CH3 The
structural
m
o
formula indicates that
this
or
compound
11
CH 3 CN(CH 3 (c)
)
2
O
O
II
CH 3 CH = CHCCI
II
or
HCOCH2CH2CH2CH2CH3
H 2 C=C(CH 3 )CCI
can't
be a lactone.
Chapter 21
558
Visualizing Chemistry
21.27 (a)
(b)
H H
HH
fH 3
V
C
CH,
II
H 3 CH
iV
O
J\
II
HCH3 HH
O
3-Methylbutyl benzoate
Af,Af-Dimethyl-3-methylbutanamide
21.28 (a)
CI
*
pyridine
O
oBromobenzoic
\
2-Propanol
acid
=\\
m
CH 3 CHCHc^ /
/
SOCIc
CHr
Br
OH
OCHCHo /
* O
//
Isopropyl
o-bromobenzoate This compound can also be synthesized by Fischer esterification of o-bromobenzoic acid with 2-propanol and an acid catalyst. (b)
O //
1
.
socio
*-
CH 2 C \
2.
OH
2NHod
Cyclopentylacetic acid
I
y— ch c 2
NH 2 Cyclopentylacetamide
21.29
O
O
O" I
+
2 NH-
H 2 C=CHCHCH 2
H 2 C=CHCHCH^
CI
|
CH 3
NH 3
\ CI
H 2 C=CHCHCH 2
CH 3
CHo
+
starting material is 3-methyl-4-pentenoyl chloride, as indicated
tetrahedral intermediate.
which eliminates Cl~
by the -CI
Ammonia adds to give the observed tetrahedral
to yield the
NH 4+ cr
3-Methyl-4-pentenamide
3-Methyl-4-pentenoyl chloride
The
NH 2
above amide.
in the
intermediate,
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
559
21.30 According to
the electrostatic potential maps, the carbonyl carbon of acetyl azide is more electron-poor (less red) and therefore more reactive in nucleophilic acyl substitution reactions.
Resonance donation of nitrogen lone-pair electrons amide than in an acyl azide.
to the carbonyl
group
is
greater in an
:o:
:o:
^
II
^
NH 2
H 3C
I
NH 2
H3C
Acetamide
Additional Problems
Naming Carboxylic Acid
Derivatives
21.31 (a)
(b)
(c)
.
Cyclohexyl cyclohexanecarboxylate
Ethyl 2-cyclobutenecarboxylate
Succinic anhydride
can be drawn for each part of this problem,
structures
(a)
O
OCH 2 CH 3
o
21.33 Many
.
CH3CH2
.0 //
O
^CHg
//
N
C \
/
c=c\
C=0
H
CI
/
H?C=C
CHoC 2
\
CI
/
CH2CH3
CI
Cyclopentanecarbonyl chloride
(£)-2-Methyl-2pentenoyl chloride
3-Ethyl-3-butenoyl chloride
(b)
,0 ff
CHgCh^Ch^C^z CCH2C
h^C^-
CHCH^- CHC^
cr 1
-Cyclohexenecarboxamide
N(CH 3 ) 2
3-Heptynamide
N,N-Dimethyl2 ,4-pentadienamide
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
Nucleophilic Acyl Substitution Reactions
21.34 (a)
O II
CH 3 CH 2 CCI
(b)
CH3CH
ether
O II
CH 3 CH 2 CCI
(C)
<^!i
1
UAIH4
2.
H3 +
1
2
CH 3 CH 2 CH 2 OH
O II
OH .
CH 3 MgBr^
I
^
™
CH3CH2CCI
CHgCh^C^CHg^
(d) II
CH 3 CH 2 CCI
®
H3 +
II
*
CH 3 CH 2 COH
OO
O II
CH 3 CH 2 CCI
CHoC0 2"Na +
—
^
-
+
I'
*~
I'
CH 3 CH 2 C— OCCH 3
HCI
561
562
Chapter 21
21.35 The
reagents in parts
and
(a), (e),
(g) don't react with
methyl propanoate.
(b)
LLiAIH,
II
~
-
CH3CH2COCH3
3
(c)
CH 3 CH 2 CH 2OH
q+
O
+
CH3OH
OH 1.2 CHoMgBr
"
CH 3 CH 2 COCH 3
+
» CHgCHgCCCH^g + CH3OH
3
(d)
o
o
+
—— H3
II
II
-
CH 3 CH 2 COCH 3
CH 3 CH 2 COH
+
CH3OH
m
O CH 3 CH 2 COCH 3
21.36 The
reagents in parts
CH3CH2C
H2N
+
(a), (e), (f),
and
(g) don't react with
(b)
CH 3 CH2 CNH 2
(c)
g
H^
4 *
*
CH3CH2CH2NH2
O
O II
CH 3 CH 2 CNH 2
1
.
—
CHoMgBr
"
CH 3 CH 2 CNH 2
+
CH 4
3
(d)
O "
CH 3 CH 2 CNH 2
O
—
H3 -
H
+
II
*~
CH 3 CH 2 COH
+
NH 4 +
propanamide.
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
21.37 Dimethyl carbonate is a diester. Use your knowledge of the Grignard way through this problem.
reaction to
work your
Triphenylmethanol
The
overall reaction consists of three additions of eliminations of methoxide and one protonation.
563
phenylmagnesium bromide, two
564
Chapter 21
21.38 (a)
BH
1
CH3CH2CH2CO2H
* CH3CH2CH2CH2OH
+ 3
*
^ CH3CH2CH2CH2OH
Periodinane
CH 3 CH 2 CH 2 CHO
CH2C 2 '
from
(a)
(C)
PBr3
CHgCh^Ch^Ch^OH from
NaCN
K+ ~OC(CH 3 )3^
_
Ch^Ch^CH^- CH2
(c)
W
H 3Q
CH3CH2CH2CH2CN from
CH3CH2CH2CH2CN
(c)
CHgCh^Ch^Ch^Br from
Ch^Ch^Ch^Ch^Br
(a)
^nunununuD CH 3 CH2CH 2 CH2Br from
,
+
CH3CH2CH2CH2CO2H
(d)
|
CH 3 NH 2
CH3CH2CH2CH2CONHCH3 *
SOCI 2
CH3CH2CH2CH2COCI
(g)
CH3CH2CH2CH2CN
1 •
2
from(d) (h)
CH 3 MgBr
^
Ch^Ch^ChkCh^CCH;
'
^
socio
CH3CH2CH2CO2H
CH 3 CH 2 CH 2 COCI C6 H 6 I
AICI3I
aCH2CH
2 CH 2 CH3
^
^
2 NHo
SOCIo
CH3CH2CH2CO2H
||
CCH 2 CH 2 CH 3
CH 3 CH 2 CH 2 COCI
CH3CH2CH2CONH2 I
S0 2 +
2 HCI
SOCI 2
+ CH 3 CH 2 CH 2 CN
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
21.39 (a)
HC
OCH 2 CH 3
1
.
2.
(6)
CH 3 '
+
O II
H 2 C=CHCHCH 2 COCH 3
CH 2 CH 3
,
CH 2 CH 3
CH 3 CH 2 MgBr H3
\
+ CH 3 CH 2 OH
CH 3 LLiAIH, "
3
'
H 2 C= CHCHCH 2 CH 2 OH
+ CH 3 OH
565
Width: 612 Height: 792
566
Chapter 21
21.40 The reactivity
in saponification reactions is influenced by steric factors. Branching and alkyl portions of an ester hinders attack of the hydroxide nucleophile. This effect is less dramatic in the alkyl portion of the ester than in the acyl portion because alkyl branching is one atom farther away from the site of attack, but it is still significant.
of esters
in both the acyl
Most reactive
O
Least reactive
O
O CH 3
O
CH3COCH3 > CH 3 COCH 2 CH 3 > CH3COCHCH3 > CH3COCCH3 CH,
21.41
2,4,6-Trimethylbenzoic acid 2,4,6-Trimethylbenzoic acid has two methyl groups ortho to the carboxylic acid functional group. These bulky methyl groups block the approach of the alcohol and prevent esterification from occurring under Fischer esterification conditions. possible route to the
A
methyl
ester:
0H
1
.
2.
NaOH
OCH<
CH3I
This route succeeds because reaction occurs farther away from the site of steric hindrance. It is also possible to form the acid chloride of 2,4,6-trimethylbenzoic acid and react it with
methanol and pyridine.
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
Reaction of an ester with Grignard reagent produces a tertiary alcohol, not a ketone. (d)
O CN
.CCH, 1
2.
CH 3 MgBr H3
+
567
Chapter 21
43
Step
1:
The carboxylic
Step
2:
The carboxylate oxygen adds
Step
3:
The amine
acid protonates to
DCC.
DCC to form a reactive intermediate.
nitrogen adds to the carbonyl group to yield a tetrahedral intermediate.
Step 4: The intermediate loses dicyclohexylurea
to
produce the lactam.
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
Step
569
1: Protonation.
Step 2: Addition of methanol.
Step 3: Proton
transfer.
Step
4:
Loss of ethanol.
Step
5: Deprotonation.
In acidic methanol, the ethyl ester reacts by a nucleophilic acyl substitution mechanism to yield a methyl ester. The equilibrium favors the methyl ester because of the large excess of
methanol present.
21.45
2D
f?
C (CH 3 3 CO
(CH 3
)
)
3 CO'
(CH 3 3 CO^ )
addition
elimination
of azide
of chloride
+
^N3 ci~
This reaction is a typical nucleophilic acyl substitution reaction, with azide as the nucleophile and chloride as the leaving group.
570
Chapter 21
Step-Growth Polymers
21.46 Step
1:
Water opens the caprolactam ring
H
form the amino acid intermediate.
to
..
NH'
addition L
proton
rmg
of water
shift
opening
C II
o Step
2:
Reaction of the intermediate with a second molecule of caprolactam forms a dimer.
O
.0
H 2 N(CH 2 5 C
+N(CH2) 5C
)
OH addition of
OH
amino
group to carbonyl group
OH II
proton
1to H H o ^N^H^sC
o //
I
shift
H 2 N(CH 2 5 C-N(CH 2 ) 5 C
OH
)
nng
OH
opening
and beyond: Reaction of the dimer with caprolactam. This process repeats itself many, many times until the polymer stops growing. Remember that each new bond is formed in a discrete step. Heat forces the equilibrium in the direction of polymer formation. Steps 3
O HN(CH 2
)
5
O
C— N(CH 2 5 CH)
nylon 6
n
I
H
21.47 Look
for the
O
monomer
units,
which are difunctional compounds,
in the polymer.
O
HOC(CH 2 6 COH )
-^-C(CH 2
)
6
+
H2N
C— NH
(
CH,
}
— CH — 2
NH,
(
J
—
NH-}-
Qiana
+ 2n H 2
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
21.48 Hydroxide opens
the lactone ring,
571
and the resulting anion can add to a second lactone
molecule to produce a polyester.
HOCCH0CH0O:
|
|
••
i
o
o II
° <& HOCCH 2 CH 2 0^\_p
II
HOCCH 2 CH 2 0— CCH 2 CH 2
0~
repeat
many
.
times
O
O
CCH 2 CH 2 0— CCH 2 CH 2 021.49 The polyimide pictured
is
a step-growth polymer of a benzene tetracarboxylic acid and an
aromatic diamine.
O
o
II
II
HOC
HOC
YT
COH HoN
COH
II
II
o
o
1
NHc
,4-Benzenediamine
9
^
a
^c
1,2,4,5-Benzenetetracarboxylic acid
%
if
o
o a polyimide
+
2n H 2
Spectroscopy 2 1 .50 In some of these others, either
!
pairs,
IR spectroscopy alone can
differentiate
between the isomers. For
H NMR or a combination of H NMR and IR data is necessary.
(a)
O
O II
CH 3 CH 2 CNHCH 3 Af-Methylpropanarnide IR:
1680
cm"
(Af-substituted ]
H NMR:
1
amide)
one methyl group one ethyl group
CH 3 CN(CH 3 2 )
AfiV-Dimethylacetamide
1650
cm"
1
(Af,Af-disubstituted
amide)
three methyl groups
572
Chapter 21
if
(b)
HOCH 2 CH 2 CH 2 CH 2 C= N
CNH 2 Cyclobutanecarboxamide
5-Hydroxypentanenitrile
3300-3400 cm" 1
IR:
1690
(hydroxyl)
cm"
1
(amide)
2250 cm- 1 (nitrile)
(c)
O
O II
II
CICH 2 CH 2 CH 2 COH
CH 3 OCH 2 CH 2 CCI
4-Chloropentanoic acid
3-Methoxypropanoyl chloride
2500-3300 cm" 1
IR:
1810
(hydroxyl)
1710
cm"
cm"
1
(carboxylic acid chloride)
1
(carboxylic acid)
(d)
O
O II
CH3CH 2 COCH 2 CHg
CHgCOCh^Ch^CHg
Ethyl propanoate ]
H NMR:
Propyl acetate
two triplets two quartets
one
singlet
one
triplet
one quartet one multiplet
21.51 The IR spectrum CI
indicates that this
O II
I
CH3CHCOCH3 a
b
c
compound has
a=
1.69 6
b =
3.79 6
c =
4.41 6
a carbonyl group.
21.52 (a)
(b)
O
I?
CHgCH 2 CH 2 CCI a
b
II
N=CCH 2 COCH 2 CH 3 b
c
c
a=
1.00 5
a=
1.32 6
b =
1.75 6
b =
3.51 6
c =
2.86 6
c=
4.27 5
a
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
573
General Problems 2 1 .53
A negatively charged tetrahedral intermediate is formed when the nucleophile ~OH attacks the carbonyl carbon of an ester.
An
electron-withdrawing substituent can stabilize this
negatively charged tetrahedral intermediate and increase the rate of reaction. (Contrast this effect with substituent effects in electrophilic aromatic substitution, in which positive charge developed in the intermediate is stabilized by electron-donating substituents.) Substituents that are deactivating in electrophilic aromatic substitution are activating in ester
The substituents strongly electron-donating.
hydrolysis, as the observed reactivity order shows. electron- withdrawing;
-NH2 is
Most reactive
Y
= -N0 2 >
-CN and -CHO are Least reactive
-ON >-CHO
> -Br > -H > -CH 3 > -OCH 3 > -NH 2
21.54 :o:
CH 2 0"^ \" R / SCoA CHOH
CH 2 OH .C^ " CoAS R -H +
—
I
CHOH
?
I
j
CH 2 0*
W
CHOH
C>
R
+
"SCoA
2.
2-
2-
CH 2 OP0 3
:
I
CH 2 OP0 3
2-
CH 2 OP03
'
;
Glycerol 3-phosphate
1-Acylglycerol
3-phosphate
Addition of -OH to the fatty acyl
CoA
(Step
1),
followed by loss of -SCoA from the
tetrahedral intermediate (Step 2), produces 1-acylglycerol 3-phosphate.
21.55 :o:
OH I
I
FT^rOH H2 HoO:
The
tetrahedral intermediate
:
OH HO* T
T can eliminate any one of the three -OH groups to reform
either the original carboxylic acid or labeled carboxylic acid. Further reaction of water with
mono-labeled carboxylic acid leads to the doubly labeled product.
574
Chapter 21
21.56 18
O 18
CHoC d — OH
1.
2.
BH H3
18
V** +
CHoCHoOH d
CH 3 CH 2 COCI
tL
Pyridine
I?
18
CH 3 CH 2 C— OCH 2 CH 3
+
H2
Ethyl propanoate 1
Remember that
the
Q
O label appears in both oxygens of the acetic acid starting material.
21.57
:o:
F3 C" addition of carboxylic
x
O
:o:
- H+
«+ R 0C0CF3 loss
•*r^>
F3 C
of proton
II
y<"
^°COCF3
R
elimination of
acid
trifluoroacetate
FoC
(b)
The electron-withdrawing
?\
if
"OCOCF 3
o
fluorine atoms polarize the carbonyl group,
R
making
it
more
reactive toward nucleophiles. (c)
Because trifluoroacetate
is
a better leaving group than other carboxylate anions, the
reaction proceeds as indicated.
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
21.58 Formation
575
of the dipeptide:
I
HoN
R
H
H
. ISL
+
.
CO,
C
H3 N
6
H
R
+
i r H Step
1:
r H
The carboxylate group from one amino
acid adds to
DCC to form a reactive
intermediate.
Step
2:
The amino group of the second amino
acid adds to the carbonyl group to yield a
tetrahedral intermediate.
Step 3: The intermediate loses dicyclohexylurea to produce the amide. Proton transfers occur in steps
1
and
3.
Width: 612 Height: 792
576
Chapter 21
Formation of the 2,5-diketopiperazine:
Addition of carboxylate to
DCC.
Step
1:
Step
2: Intramolecular nucleophilic attack
acylating agent.
Step 3: Loss of dicyclohexylurea. Proton transfers occur in steps
1
and
2.
of the amino terminal end of the amide on
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
577
21.59
1 HoC" C\ 2
;o: 7
I
H2
HoC*" C\
C
H2
1.
:0
a
?+ H"T NH U
C
O
II
II
H 2 C-" C v
,°
1
O
H 2C
2.
II
H 2 C-*" C\
O
O +
HoC-^'
3.
4
.
HoC"
c
H 2 C^
C
£)NH 2 HO:
ho: VNH^3
OH
^NH 2
:OH +
3
+ H 0^ 3
.. + + OH
H2
by OH
HO:
GoH
ho: :oh
;
\ /
H 2 C~" Cx
HoC
C H2 C H2
NH 9.
H2 C
^NH
NH
HoC
8.
I
H 2 C^ /
7.
C
s o
O
H 2 C- C s
\.
:
C
II
fi
o
o
^OH
HoC
+
NH, 6.
I
.NH 2
HoC
A summary of steps: Step 1: Protonation Steps 3,5,7,9: Proton transfers Step 6: Nucleophilic addition of -NH2
Step Step Step
Nucleophilic addition of
2:
This reaction requires high temperatures because the intermediate amide nucleophile and the carboxylic acid carbonyl group is unreactive.
21.60
NH3
Ring opening 8: Loss of H2O
4:
is
a poor
This synthesis requires a nucleophilic aromatic substitution reaction, explained in Section 16.7.
NHCOCH 3
NHc
(CH 3
)
3 CO-
1.SnCI 2 2.
Nucleophilic aromatic
OC(CH 3
,
H 3 Q^
*
I
NaOH
"OH
Reduction ) 3 of nitro
substitution
CH 3 COC
group
Aminolysis
0(CH 3) 3
0(CH 3 ) 3 Butacetin
The amide can be formed by the reaction of acetyl chloride with the appropriate amine, which is produced by reduction of the nitro group of the starting material. A nucleophilic aromatic substitution of -F by -OC(CH3)3 can take place because the ring has an electronwithdrawing nitro group para to the to acetylate the amine.
site
of substitution. Acetic anhydride can also be used
578
Chapter 21
21.61
soci.
Formation of
1
acid chloride
.
SnCI 2 H 3
+
,
2.
"OH
Reduction of nitro group
Phenyl 4-aminosalicylate
21.62 HoC
HoC
1
%y I
C0 2 H
HoC
MgBr -
CQ 2 +
III
2.
H3
Grignard
SOCI 2
carboxylation
formation of acid chloride
O
COCI
HN(CH 2 CH 3
)
2
NaOH Af,Af-Diethyl-m-toluamide
'XX
formation of amide
Grignard carboxylation yields m-methylbenzoic acid, which can be converted to an acid chloride and treated with diethylamine to produce the amide.
21.63
C0 2 H H3
+
C0 2 H NBS
excess
NHo*
(PhC0 2 ); CH<
C0 2 H
CH 2 Br
H
CH 2 NH 2
Tranexamic acid
Using a rhodium catalyst, the aromatic ring is hydrogenated to form the cis-substituted cyclohexane, which is converted to the trans isomer by heating to 300°. The nitrile starting material is hydrolyzed to form a carboxylic acid, and the methyl group is brominated and treated with ammonia to form the amine.
579
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
21.64(a) +
H 2 C— n=n:
*-
H 2 C=N=N:
Resonance forms show that the carbon of diazomethane can occur to form a methyldiazonium ion. M
^C^..^H
^..- +
basic,
f?
h 2 c— n=n:
C£)
R
is
+
R
ch 3
and reaction with an acid
— n=n:
O:
(b)
O If
R'
II
/C^
CHo-j-N=N: 3
U
"of
An Sn2 reaction takes place in which to
form the methyl
21.65 Both
+
OCH 3
R
the carboxylate ion displaces
N2
as the leaving
group
ester.
steps involve nucleophilic acyl substitutions.
Formation of acyl phosphate: :0:
CoAS
Acyl phosphate Step Step
1:
2:
Reaction of the phosphate oxygen with the carbonyl carbon of succinyl Co A. Loss of ~SCoA from the tetrahedral intermediate, yielding acyl phosphate.
Conversion of acyl phosphate to succinate:
-q
9
o-
=.97 :0
— POPO— Guanosine I
IT
\fl POPO- Guanosine II
"OPOPOPO- Guanosine
I
0"0"
GDP
I
I
0"0"
0~0"0"
GTP
1: Reaction of the diphosphate oxygen of GDP with the phosphorus of the acyl phosphate to produce an intermediate similar to the intermediates formed in nucleophilic acyl substitutions of carboxylic acid derivatives. Step 2: Loss of phosphate to form GTP and succinate.
Step
580
Chapter 21
2 1 .66 In
all of these reactions, a nucleophile adds to either carbon or phosphorus to form an intermediate that expels a leaving group to give the desired product.
Formation of 1,3-bisphosphoglycerate:
:|5 / II
\^0— ADP
9f /\^_0-ADP
S.°"
/
'o-
-.or
o.
+
H— C— OH I
"
CH 2 OP0 32-
CH 2 OP0 32
CH 2 OP03 2 1,3
Route
enzyme-bound
to
Bisphosphoglycerate
thioester:
Enz
O
Base vwv
est;
P04 3"
Enz
.TV?
H
o~
_.\(\
H— C— OH
/
+
O"
O^
I
2-
CH 2 OP0 3 2
CH 2 OP03
CH 2 OP0 3
':
-.S^-Enz
H— C— OH
H— C— OH I
Enzyme-bound
Reduction
OADP
H— C— OH
H— C— OH
"
thioester
to glyceraldehyde 3-phosphate:
NAD j
")/^S-t-Enz
:o
S-t-E nz
H
o.
+ "S-t-Enz +
H
— C— OH I
CH 2 OP0 3 2
H— C— OH "
H— C— OH "
CH 2 OP0 3 2
I
CH 2 OP0 32
NAD + "
Glyceraldehyde 3-phosphate
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
581
21.67
(a)
^°^ uF
C °2
Enz
Enz-Nu
:B
'•Aw /"V,
r
(b)
N
O
O
A— H
\
(c)
I
P°2
H
N—
N
(
Nu Enz
H3 C
\0
S=0
^N. H
o
CO.
In (c), the imine rearranges to an a,(3-unsaturated ester, to
which the nucleophile adds
give the trapped p-lactamate.
21.68 :o:
:o:
Go: /
Ph
Ph Benzil
Ph
nucleophihc addition of hydroxide
c— a \ y\phOH
\
HoO 2
* ,
,
phenyl
,C-C Ph^/ Ph
shift
HO +
^C—
Ph^/ Ph
o //
o //
\
HoO +
N
OH
,
proton transfer
HO \
O //
.0— c OH
protonation
Benzylic acid
Ph^7 Ph
*
O
to
582
Chapter 21
21.69
OH r
HOCH 2 CHCH 2 OH Dimethyl terephthalate
O
.
C
o
.
—%
II
Glycerol
o
— COCH CHCH OC II
(\
II
/)
/
2
O ff
\
COCH 2 CHCH 2 0-
2
?
o=c
OCH 2 CHCH 2 OC
COCHoCHCHoOC 2 2
o
II
II
O
O
The product of the reaction of dimethyl terephthalate with cross-linking and is more rigid than Dacron.
glycerol has a high degree of
21.70 (a)
(b)
O
O II
II
CH 3 COCH(CH 3 b
)
CH=CHCOCH 2 CH 3
2
e
a
c
b
c
a =
1.22 5
a=
1.32 5
b =
2.01 6
b =
4.24 5
c =
4.99 6
c =
6.41 6
d =
7.36, 7.49 6
e =
7.68 5
21.71 (a)
(b)
O II
CICH 2 CH 2 COCH 2 CH 3 c
a
b
a=
da
Q
Q
ii
n
CH 3 CH 2 OCCH 2 COCH 2 CH 3 a
c
b
1.26 5
a=
1.27 6
b =
2.77 6
b =
3.34 6
c=
3.76 5
c =
4.20 6
d = 4.19 6
c
a
Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions
21.72 Addition
21.73
This
is
583
of the triamine causes formation of cross-links between prepolymer chains.
a nucleophilic acyl substitution reaction whose mechanism
is
similar to others
we
have studied.
o
:o:
II
HCL
.a Fry
R'
addition
OH
R elimination
OH
R
O"
acid-base reaction
I3C:
can act as a leaving group because the electron- withdrawing iodine atoms stabilize the
carbanion.
Review Unit
8:
Carbonyl Compounds
1.
Reaction at the Carbonyl Group
Major Topics Covered
(with vocabulary);
Aldehydes and ketones: -carbaldehyde acyl group
acetyl group
formyl group
benzoyl group
hydrate
Reactions of aldehydes and ketones: nucleophilic addition reaction gem diol cyanohydrin imine enamine carbinolamine 2,4-dinitrophenylhydrazone Wolff-Kishner reaction acetal hemiacetal Wittig reaction
betaine
Cannizzaro reaction
conjugate addition
Carboxylic acids and their derivatives: carboxylation carboxylic acid derivative
a,p-unsaturated carbonyl
acid halide
acid anhydride
ylide
compound
amide
ester
nitrile
-carbon itrile Reactions of carboxylic acids and their derivatives: nucleophilic acyl substitution hydrolysis alcoholysis aminolysis Fischer esterification reaction lactone saponification DIB lactam thiol ester acyl phosphate polyamide polyester step-growth polymer chain-growth polymer nylon
AH
Types of Problems: After studying these chapters you should be able to: - Name and draw aldehydes, ketones, carboxylic acids and their derivatives. - Prepare all of these compounds.
-
-
Explain the reactivity difference between aldehydes and ketones and between carboxylic acids and all their derivatives. Calculate dissociation constants of carboxylic acids, and predict the relative acidities of substituted carboxylic acids. Formulate mechanisms for reactions related to the reactions we have studied. Predict the products of the reactions for all functional groups we have studied. Use spectroscopic techniques to identify these compounds. Draw representative segments of step-growth polymers.
Points to *
Remember:
In all of these reactions, a nucleophile adds to a positively polarized carbonyl carbon to form a tetrahedral intermediate. There are three possible fates for the tetrahedral intermediate: (1) The
intermediate can be protonated, as occurs in Grignard reactions, reductions, and cyanohydrin The intermediate can lose water (or "OH), as happens in imine and enamine The intermediate can lose a leaving group, as occurs in most reactions of
formation. (2) formation. (3)
carboxylic acid derivatives. *
Many of the reactions in these three chapters require acid or base catalysis. An acid catalyst, protonates the carbonyl oxygen, making the carbonyl carbon more reactive toward nucleophiles, and/or protonates the tetrahedral intermediate, making loss of a leaving group easier. base catalyst deprotonates the nucleophile, making it more nucleophilic. The pH
A
optimum
for these reactions
is
a compromise between the two needs.
Review Unit 8
585
Here are a few reminders for drawing the mechanisms of nucleophilic addition and substitution reactions. (1) When a reaction is acid-catalyzed, none of the intermediates are negatively charged, although, occasionally, a few may be neutral. Check your mechanisms for charge balance. (2) Make sure you have drawn arrows correctly. The point of the arrow shows the new location of the electron pair at the base of the arrow. (3) In a polar reaction, two arrows never point at each other. If you find two arrows pointing at each other, redraw the mechanism. Reactions of acyl halides are almost always carried out with an equivalent of base present. The base is used to scavenge the protons produced when a nucleophile adds to an acyl halide. If base were not present, hydrogen ions would protonate the nucleophile and make it unreactive.
The products of acidic cleavage of an amide
are a carboxylic acid and a protonated amine. products of basic cleavage of an amide are a carboxylate anion and an amine.
The
In some of the mechanisms shown in the answers, a series of protonations and deprotonations occur. These steps convert the initial tetrahedral intermediate into an intermediate that more
These deprotonations may be brought about by the solvent, by the conjugate base of the catalyst, by other molecules of the carbonyl compound or may occur intramolecularly. When a "proton transfer" is shown as part of a mechanism, the base that removes the proton has often not been shown. However, it is implied that the proton transfer is assisted by a base: the proton doesn't fly off the intermediate unassisted. easily loses a leaving group.
The most useful spectroscopic information
for identifying carbonyl compounds comes from IR spectroscopy and 13 C spectroscopy. Carbonyl groups have distinctive identifying absorptions in their infrared spectra. 13 C is also useful for identifying aldehydes, ketones, and nitriles, although other groups are harder to distinguish. The *H absorptions of aldehydes and carboxylic acids are also significant. Look at mass spectra for McLafferty rearrangements and alpha-cleavage reactions of aldehydes and ketones.
NMR
NMR
NMR
Self-Test:
A with:
LiAlH4, then H30+ (b) CgHsMgBr, then H 3 0+; (c) (CH3 ) 2NH, H 3 0+; (d) CH3 OH, H+ catalyst (e) (C 6H5 ) 3P+-CH2-; (f) 1 equiv. CH 3 CH2NH2, H30+ How would you reduce A to yield a saturated hydrocarbon? Where would you expect the carbonyl absorption of A to occur in its IR spectrum? Predict the products of the reaction of
(a)
;
.
Predict the products of B with the reagents (a) - (d) above. What product(s) would be formed if B was treated with Br2, FeBr 3 ? Where do the carbonyl absorptions occur in the IR spectrum of B? Describe the 13 C NMR spectrum of B.
Width: 612 Height: 792
Review Unit 8
586
COCH 3 CH 3 CH 2
CH3CHCCHOH
C 6 H 5 CH 2 CH 2 — N
OH C
NHCO(pHCH 2 CH 3
O
D
Kethoxal
I
OCH3
CH 3
Julocrotine
(antiviral)
E
Xanthoxylin
Kethoxal (C) exists in solution as an equilibrium mixture. With what compound Why does the equilibrium lie on the side of kethoxal?
is it in
equilibrium.
Identify the carboxylic acid derivatives present in D.
with
(a)
-OH,
H2
Name E.
(b)
LiAlH 4 then ,
Show
the products of treatment of
D
H 20.
Describe the IR spectrum and
l
H NMR spectrum of E.
Multiple Choice: 1
.
In which of the following nucleophilic addition reactions does the equilibrium
of the products? (a) Propanal + HCN (b) Acetone + H 2 (d) 2,2,4,4-Tetramethyl-3-pentanone + 2
.
3
.
(c)
Acetaldehyde +
lie
on the side
HBr
HCN
alcohol can be formed by three different combinations of carbonyl compound + Grignard reagent? (a) 2-Butanol (b) 3-Methyl-3-hexanol (c) Triphenylmethanol (d) 1-Phenylethanol
Which
A nitrile can be converted to all of the following except: (a) an aldehyde (b) an amide (c) an amine (d) A nitrile can be converted to all of the above compounds.
4.
Which of the following p(a)
5
.
CH COC 6 H5 C0 2 H 3
substituted benzoic acids
(b)
C^OC^CO^
is
(c)
the least acidic?
BrC^CC^H
A carboxylic acid can be reduced by all of the following except: (a) UA1H4, then H 3 0+ (b) BH 3 THF, then H 3 0+ (c) NaBH 4 ,
,
(d)
then
NCC6 H5 C0 2 H
H 30+
(d) All
of these
reagents can reduce a carboxylic acid. 6.
Which of the
following carboxylic acids can be formed by both Grignard carboxylation and hydrolysis? (a) Phenylacetic acid (b) Benzoic acid (c) Trimethylacetic acid (d) 3-Butynoic acid
by
7
.
nitrile
Acid anhydrides are used mainly (a) synthesizing
(d)
8.
carboxylic acids
for:
(b)
forming alcohols
(c)
forming aldehydes
A ketone is formed from an acid halide by reaction with: (a) DIB AH (b) L1AIH4 (c) RMgBr (d) (CH 3 CH 2 )CuLi
introducing acetyl groups
Review Unit 8
9
.
From which
carboxylic acid derivative can you form a ketone as the product of a Grignard
reaction? (a) acid chloride
10.
An
(b) ester
infrared absorption at 1650
aromatic acid chloride (d) aromatic ester (a)
(c) nitrile
cm-1
(d)
amide
indicates the presence of:
(b) Af,7V-disubstituted
amide
(c) a,p-unsaturated
ketone
587
Chapter 22 - Carbonyl Alpha-Substitution Reactions
Chapter Outline I.
Keto-enol tautomerism (Section 22.1). A. Nature of tautomerism. 1 Carbonyl compounds with hydrogens bonded their corresponding enols. .
2.
This rapid equilibration
is
called tautomerism,
to their
a carbons
equilibrate with
and the individual isomers are
tautomers. 3
.
4
.
Unlike resonance forms, tautomers are isomers. Despite the fact that very little of the enol isomer is present enols are very important because they are reactive.
at
room
temperature,
B Mechanism of tautomerism. .
1
2
form an
.
In acid-catalyzed enolization, the carbonyl carbon is protonated to
.
intermediate that can lose a hydrogen from its a carbon to yield a neutral enol. In base-catalyzed enol formation, an acid-base reaction occurs between a base and
an a hydrogen. a.
b II.
.
The
resultant enolate ion is protonated to yield an enol. Protonation can occur either on carbon or on oxygen.
c. Only hydrogens on the a positions of carbonyl compounds are acidic. Enols (Sections 22.2-22.4). A. Reactivity of enols (Section 22.2). 1 The electron-rich double bonds of enols cause them to behave as nucleophiles. a. The electron-donating enol -OH groups make enols more reactive than alkenes. 2 When an enol reacts with an electrophile, the initial adduct loses -H from oxygen to give an a-substituted carbonyl compound. B. Reactions of enols (Sections 22.3-22.4). Alpha halogenation of aldehydes and ketones (Section 22.3). 1 a. Aldehydes and ketones can be halogenated at their a positions by reaction of X2 .
.
.
in acidic solution.
b
.
c.
d 2
.
.
The reaction proceeds by acid-catalyzed formation of an enol intermediate. Halogen isn't involved in the rate-limiting step: the rate doesn't depend on the + identity of the halogen, but only on [carbonyl] and [H ]. a-Bromo ketones are useful in syntheses because they can be dehydrobrominated by base treatment to form a,/3-unsaturated ketones.
Alpha-bromination of carboxylic acids (Section 22.4). In the Hell-Volhard-Zelinskii (HVZ) reaction, a mixture of Br2 and PBr 3 can a. be used to brominate carboxylic acids in the a position. b The initially formed acid bromide reacts with Br2 to form an a-bromo acid bromide, which is hydrolyzed by water to give the a-bromo carboxylic acid. The reaction proceeds through an acid bromide enol. c .
.
HI.
Enolates (Sections 22.5-22.7). A. Enolate ion formation (Section 22.5). 1 Hydrogens alpha to a carbonyl group are weakly acidic. a. This acidity is due to overlap of a filled p orbital with the carbonyl group p orbitals, allowing the carbonyl group to stabilize the negative charge by resonance. b. The two resonance forms aren't equivalent: the form with the negative charge on oxygen is of lower energy. .
Carbonyl Alpha-Substitution Reactions
589
Strong bases are needed for enolate ion formation. a. Alkoxide ions are often too weak to use in enolate formation. b Lithium diisopropylamide can be used to form the enolates of many different carbonyl compounds. 3 When a hydrogen is flanked by two carbonyl groups, it is much more acidic, a. Both carbonyl groups can stabilize the negative charge. Reactivity of enolate ions (Section 22.6). Enolates are more useful than enols for two reasons: 1 a. Unlike enols, stable solutions of enolates are easily prepared. b Enolates are more reactive than enols because they are more nucleophilic. 2 Enolates can react either at carbon or at oxygen. 2.
.
.
B
.
.
.
.
Reaction at carbon yields an a-substituted carbonyl compound. Reaction at oxygen yields an enol derivative. C. Reactions of enolate ions (Sections 22.6-22.7). a.
b
1
.
.
Base-promoted a-halogenation. a. Base-promoted halogenation of aldehydes and ketones proceeds readily because each halogen added makes the carbonyl compound more reactive. b Consequently, polyhalogenated compounds are usually produced. c. This reaction is only useful with methyl ketones, which form HCX3 when .
d.
reacted with halogens. This reaction is known as the haloform reaction. i. ii.
2.
The HCX3 is a solid that can be identified. The last step of the reaction involves a carbanion leaving group.
Alkylation reactions of enolates (Section 22.7). a.
General features. i. ii.
b
.
Alkylations are useful because they form a new C-C bond. Alkylations have the same limitations as Sn2 reactions; the alkyl groups must be methyl, primary, allylic or benzylic.
The malonic ester synthesis. The malonic ester synthesis i.
used for preparing a carboxylic acid from a by two carbon atoms. useful because its enolate is easily prepared by reaction is
halide while lengthening the chain ii.
iii.
Diethyl malonate is with sodium ethoxide. Since diethyl malonate has two acidic hydrogens, two alkylations can take place.
Heating in aqueous HC1 causes hydrolysis and decarboxylation of the alkylated malonate to yield a substituted monocarboxylic acid.. (a). Decarboxylations are common only to /3-keto acids and malonic acids. v. Cycloalkanecarboxylic acids can also be prepared. iv.
c.
The acetoacetic ester synthesis. i. The acetoacetic ester synthesis ii.
iii.
is used for converting an alkyl halide to a methyl ketone, while lengthening the carbon chain by 3 atoms. As with malonic ester, acetoacetic ester has two acidic hydrogens which are flanked by a ketone and an ester, and two alkylations can take place. Heating in aqueous HC1 hydrolyzes the ester and decarboxylates the acid to
yield the ketone.
Most /3-keto esters can undergo this type of reaction. Direct alkylation of ketones, esters, and nitriles. i. in a nonprotic solvent can be used to convert the above
iv.
d
.
LDA
compounds
to
their enolates. ii.
Alkylation of an unsymmetrical ketone leads to a mixture of products, but the major product is alkylated at the less hindered position.
590
Chapter 22
Solutions to Problems
22.1-22.2
Acidic hydrogens in the keto form of each of these compounds are bold. these hydrogens is removed by base when an enolate is formed.
One of
Number of Keto Form (a)
Enol Form
O
(b)
r
ft
SCHg
SCHg
H
H
H
(c) ft
H
OCH 2 CH 3
/ \
H
OCH 2 CH 3
C
H
H
(d)
H ft
HoC.
HoC.
C
H
/ \
H
In (d) and
C
H
I
H
(f),
cis
H
and trans enolates are possible.
Acidic Hydrogens
Carbonyl Alpha-Substitution Reactions
591
22.3
The
first
two monoenols
equivalent;
equivalent;
more
less stable
are
stable
more
stable because the enol double
bond
conjugated with
is
the carbonyl group.
22.4
T
deuteration
loss of proton
h
h
of carbonyl
HO 2 HOD
+
enol
at alpha position
oxygen
0+ * D- OD 2 +
:
ff
{?
C
D
C
CH3 deuteration
enol
CH<
/ \ -
H
H
H
of enol double bond
22.5
D3
+
H
loss of deuterium
on carbonyl oxygen
Alpha-bromination, followed by dehydration using pyridine, yields the enone below.
o CH3CH2
II
II
II
C
^ CH 3
/\
H H
Br2
CH3CO0H
*
CH0CH0
C
^CH3
/\
H Br
pyridine '
heat
^
,
CH0CH0
C I
H l-Penten-3-one
H
592
Chapter 22
22.6
OH
O
I
PBr<
H,
C
OH
C H
Br
Br
/ \
/ \
CH(CH 3 )CH 2 CH 3
H
CH(CH 3 )CH 2 CH 3
CH(CH 3 )CH 2 CH 3
enolization
formation of acid bromide
1 alphasubstitution
Br,
O f
CH 3 OH
Br^ C
OCH-
CH(CH 3 )CH 2 CH 3
with methanol L
Methyl 2-bromo-3-methylpentanoate
The mechanism of the ester-forming
step
is
A H CH(CH
Br
reaction
/ \
H
II
Br.
3 )CH 2 CH 3
a nucleophilic acyl substitution, which was
described in Chapter 21.
(To:
:0:
v
ll
Br
-c^
Br.
Br.
C
Br
/\
H 2 CH 3
addition
v Base C"
of methanol
22.7
Hydrogens a
to
one carbonyl group (or
carbonyl groups are protons.
(a)
O
much more
loss of
elimination
proton
of bromide
nitrile) are
weakly
acidic.
+ Br"
Hydrogens a
to
two
acidic, but they are not as acidic as carboxylic acid
(b)
(c)
O
II
ft
II
(CH 3 3 C
H
/ \
H
weakly acidic
o
c
)
/ \
H-S J
C
H H
OCH-
/\
PCH 3
H
H
weakly acidic
(e)
CH 3 CH 2
most
H
ac idic
weakly acidic
m ft
/j-CSN H H weakly acidic
* V
H
I
H
CH 3
weakly acidic
Carbonyl Alpha-Substitution Reactions
22.8
weakly acidic because the bonds of the nitrile group.
Nitriles are
the
jz
H 2 C— C=n:
22.9
Halogenation
-«
anion can be stabilized by resonance involving
H 2 C=C=n:
*-
in acid
nitrile
medium
is
acid-catalyzed because hydrogen ions are regenerated:
O
O H
II
x2
C
H
593
+
W'
H
+
1
Na
2.
CH 3 CH 2 CH 2 Br
CH 2 (C0 2 Et) 2
~OEt
CH 3 CH 2 CH 2
— CH(C02 +
1
Na
.
HoO
CHC0 2 H
CH 3 CH 2 CH 2
CH 3CH 2 CH 2
heat
"OEt
CH 3 Br
1 2. +
NaBr
+
Et) 2
— C(C02
CH 3
Et)
NaBr
+
2
CHo
2-Methylpentanoic acid +
C0 2
+ 2
EtOH
(c)
fromhaUde((CH 3 2 CHCH^H;CH 2 C02 H ) from malonic )
CH 2 (C0 2 Et) 2
Na+
1
0Et
^
:
2.
(CH 3
)
2 CHCH 2 Br
(CH 3
)
2 CHCH 2
— CH(C0 H3
|
(CH 3 ) 2 CHCH 2
ester
+ ,
+
2 Et) 2
NaBr
heat
— CH2 C02 H
+
C0 2
+ 2
EtOH
4-Methylpentanoic acid
22.11
Since malonic ester has only two acidic hydrogen atoms,
it
can be alkylated only two
times. Formation of trialkylated acetic acids is thus not possible.
22.12 from malonic
ester
C0 2 H
from halides +
1
CH 2 (C0 2 Et);
2.
Na
~OEt
CH 3 CHCH 2 Br
CH 3
CH 3
CHoCHCHo— CHCOoH I
CH 3
2,4-Dimethylpentanoic acid +
C0 2
+ 2
EtOH
+
1
.
1 2.
heat
CH 3
— CH(C02 Na
Et)
+
2
NaBr
"OEt
|
+
- HqO
I
CH 3 CHCH 2
CH 3 CHCH 2 CH 3
CH 3 Br
— C(C0 2 ^^ 3
Et) 2
+
NaBr
Carbonyl Alpha-Substitution Reactions
595
22.13 As
in the malonic ester synthesis, you should identify the structural fragments of the target compound. The acetoacetic ester synthesis converts an alkyl halide to a methyl ketone ("substituted acetone"). The methyl ketone component comes from acetoacetic ester; the other component comes from a halide.
from halide
N
../T%
(
(CH 3 )2CHCH 2 t\CH 2 CCH 3 / from acetoacetic ester
n
1
CHd2 CCHo6
Na
.
(CHo) dV2 CHCH d2
(CH 3 2 CHCH 2 Br
2.
\
^
"OEt
— CHCCh \
)
C02 Et
C0 2 Et
|
H3 +
,
heat
O (CH 3 2 CHCH 2 )
— CH2 CCH
3
C0 2
+
EtOH
+
5 -Methy 1-2-hexanone (b) II
.'
from halide
\
(c 6 H 5 CH 2 CH 2 t\CH 2 CCH 3 / from acetoacetic ester O
f?
1
CHoCCHo 2 6
.
Na+ "OEt
"
2.C 6 H 5 CH 2 CH 2 Br
I
CHCCHo6 + CfiHcCHpCHo— b 5 2 2
NaBr
|
C0 2 Et
C0 2 Et |
H3
+ ,
heat
O II
C 6 H 5 CH 2 CH 2
— CH2 CCH3 +
C0 2
EtOH
+
5-Phenyl-2-pentanone
22.14 The acetoacetic ester synthesis can only be used for certain products: (1)
(2) (3)
Three carbons must originate from acetoacetic ester. In other words, compounds of the type RCOCH3 can't be synthesized by the reaction of RX with acetoacetic ester. Alkyl halides must be primary or methyl. The acetoacetic ester synthesis can't be used to prepare compounds that are trisubstituted at the
a position. (b)
(a)
H
(c)
H
\ /
r^V Phenylacetone (a)
CH3
i^V
Cv
Acetophenone
H3
ch3
V
H 3C
c^
OH 3
3,3-Dimethylbutan-2-one
Phenylacetone can't be produced by an acetoacetic ester synthesis because bromobenzene, the necessary halide, does not enter into Sn2 reactions. [See above.]
ch 3
(2)
Width: 612 Height: 792
596
Chapter 22
(b)
Acetophenone
(c)
3,3-Dimethyl-2-butanone can 't be prepared because
can't
be produced by an acetoacetic ester synthesis. [See it is
(1) above.]
trisubstituted at the
a
position.
[See (3) above.]
22.15 f?
CH 2 CCH 3
+ 1.2 Na ~OEt
BrCH 2 CH 2 CH 2 CH 2 Br
2.
C0 2 Et
'
H c
+ 2 NaBr
C0 2 Et H3 +
,
heat
J
C0 2
+
+
EtOH
HoC
22.16
Direct alkylation
is
used to introduce substituents a to an
ester,
the target molecule to identify these substituents. Alkylation starting material with
(a)
LDA,
is
ketone or nitrile. Look at achieved by treating the
followed by a primary halide.
o
O 1.
II
CH2 CCH 3
2.
LDA
II
\
CH3I
CHCCHo3
#
I
CH 3
3-Phenyl-2-butanone Alkylation occurs at the carbon next to the phenyl group because the phenyl group can help stabilize the enolate anion intermediate. (b)
_
CH 3 CH 2 CH 2 CH 2 C= N
1
.
2.
LDA
CH 3 CH 2 I
CH 3 CH 2 CH 2 CHC= N
CH 2 CH 3 2-Ethylpentanenitrile
(c) 1.
O
LDA
2.ICH 2 CH=CH 2
nr
CH 2 CH — CH 2
v.
O 2-Allylcyclohexanone
(d)
H3 Q .
excess
LDA
2.
excess
CH 3 I
1
tf
Cl
/
CH<
2,2,6,6-Tetramethylcyclohexanone
Carbonyl Alpha-Substitution Reactions
597
(e) ff
CH 2 CH 3
1.
2.
LDA CH 3I
CHCHo3
a"
I
CHo
Isopropyl phenyl ketone (f)
CH 3 O
O
CHo3
II
I
CH3CHCH0COCH3
1.
LDA
2.
CH 3 CH 2 I
CH3CHCHCOCH3 CH2CH3
Methyl 2-ethyl-3-methylbutanoate
Visualizing Chemistry
22.17
Check to
is a methyl ketone or a substituted carboxylic acid. a methyl ketone, and the reaction is an acetoacetic ester synthesis.) Next, identify the halide or halides that react with acetoacetic ester. (The halide is 1-bromo3-methyl-2-butene.) Formulate the reaction, remembering to include a decarboxylation
(a)
see
if
the target molecule
(The target molecule
is
step.
^CH 3
from halide
^
)
2
+
f?
1
CH2CCH3
.
Na
11
C= CHCH 2,'K( CH2CCHy
•
"OEt
from acetoacetic
(CH 3 2 C=
2.(CH 3 ) 2 C=CHCH 2 Br
CHCH 2
)
ester
— CHCCH
C0 2 Et
3
+
NaBr
C0 2 Et HoO
I
+ ,
heat
9
\
co 2
EtOH + (CH 3 2 C=CHCH 2
+
)
— CH 2CCH 3
6-Methyl-5-hepten-2-one (b)
This product
is
formed from the reaction of malonic
ester with both benzyl
bromomethane. t
C 6 H 5 CH2rCHC0 2 H
J
from malonic
ester
H>'
fromhafides
'^9.
+
1
CH 2 (C0 2 Et) 2
2.
Na "OEt C 6 H 5 CH 2 Br
C 6 H 5 CH 2
— CH(C02 1
,
CfiHcCHo 2 6 5
CHCOoH 2 |
HoP
+
heat
C 6 H 5 CH 2
CH 3 2-Methyl-3-phenylpropanoic acid
.
2. (
Na
+
Et)
2
CH 3 Br
— C(C0 2 C0 2
NaBr
"OEt
Et) 2
CH 3 +
+
+ 2
EtOH
+
NaBr
bromide and
598
Chapter 22
22.18 H
H
O
HO.
Ordinarily, p-diketones are acidic because they can form enolates that can be stabilized by derealization over both carbonyl groups. In this case, loss of the proton at the bridgehead carbon doesn't occur because the strained ring system doesn't allow formation of the bridgehead double bond. Instead, enolization takes place in the opposite direction, and the
diketone resembles acetone, rather than a /3-diketone, in
it
pKa and degree of dissociation.
22.19
O Enolization can occur on only one side of the carbonyl group because of the two methyl groups on the other side. The circled axial hydrogen is more acidic because the p orbital that remains after
its
removal
is
aligned for
optimum overlap with
the
% electrons of the
carbonyl oxygen.
Additional Problems Acidity of Carbonyl
22.20
Compounds
Acidic hydrogens are bold. The most acidic hydrogens are the two between the carbonyl groups in (b) and the hydroxy 1 hydrogen in (c). The hydrogens in (c) that are bonded to the methyl group are acidic (draw resonance forms to prove it). (b)
(a)
(c)
H
H
CH3CH2CHCCH;
CH 3 H
(d)
(e)
H
(f)
Carbonyl Alpha-Substitution Reactions
22.21 Check your answer by
using the
i?
(CH 3 CH 2 2 NH < CH3CCH3 < )
22.22 (a)
:o:
:o: •
C<^
in
Table 22. 1 »-
Least Acidic
HoC
pKa s
"y C\ c I
H
599
Most Acidic
!?!?{?{?
CH 3 CH 2 OH < CH 3 CCH 2 CCH 3 < CH 3 CH 2 COH < CCI3COH
600
Chapter 22
;0:
.0:
rc \
OCH:0:
22.23
Enolization at the y position produces a conjugated enolate anion that derealization of the negative charge over the n system of five atoms.
22.24 The
illustrated
compound, l-phenyl-2-propenone, doesn't
with base because the hydrogen on the a carbon
is
i l-Phenyl-2-propenone
H
stabilized
yield an anion
when
by
treated
vinylic and isn't acidic (check Table
22.1 for acidity constants).
|
is
Carbonyl Alpha-Substitution Reactions
a-Substitution Reactions
22.25 (a)
C0 2 H
C0 2 H heat
j^^^
C0 2
+
OH 3
(b)
O
+
1
Na
.
CH 3 I
2.
(c)
~OEt
PBro
CH 3 CH 2 CH 2 C0 2 H
CHoCHoCHCOBr 2 3
Br2
2
CH 3 CH 2 CHC0 2 H \
|
A
Br
(d)
—H
B
Br
O II
a"CH
"OH, H 2 Q
3
O" + HCI3
h
22.26 +
(a)
CH 2 (C0 2 Et) 2
1
2.
Na
~OEt
CH 3 CH 2 CH 2 Br
CH 3 CH 2 CH 2
|
EtOH
CH 3 CH 2 CH 2 CH 2 C0 2 Et
H
Ethyl pentanoate
^ CH (C0 ™ 2
2 Et) 2
2.
+
H3
,
(CH 3 2 CHBr
+
+ 2
(CH 3) 2 CH— CH(C0 2 Et) 2
EtOH
+
)
EtOH
)
H
+
NaBr
heat
C0 2
H3
+ ,
heat
(CH 3 2 CHCH 2 C0 2 H )
catalyst
+
Ethyl 3-methylbutanoate
Some elimination product will
2
CH 3 CH 2 CH 2 CH 2 C0 2 H
Na+ "OEt
(CH 3 2 CHCH 2 C0 2 Et
Et)
catalyst
+
1
.
+
— CH(C02
also be formed.
C0 2
+ 2
EtOH
NaBr
601
602
Chapter 22
(c)
Na
1
CH 2 (C0 2 Et) 2
+
~OEt
CH 3 CH 2 CH(C0 2 Et) 2
1
2.
Na+ ~OEt
1
CH 3 CH 2 Br
2.
NaBr
+
CH 3 Br
CHo3 I
CH 3 CH 2 C(C0 2 Et) 2 NaBr
EtOH
I
CH 3 CH 2 CHC0 2 Et
H
+
H3
|
CH<
CHo3
CH 3 CH 2 CHC0 2 H
C0 2
+
+ 2
+ ,
heat
EtOH
catalyst
Ethyl 2-methylbutanoate
The malonic
(d)
acid synthesis can't be used to synthesize carboxylic acids that are
trisubstituted at the alpha position.
22.27 Look back
to Problem 22.14, which describes compounds that can be prepared by an acetoacetic ester synthesis. Neither (a) or (c) are products of an acetoacetic ester synthesis
because the halide component that would be needed for each synthesis doesn't undergo Sn2 reactions. Compound (b) can be prepared by the reaction of acetoacetic ester with 1 ,5dibromopentane.
22.28 Two
alkylations are needed
two
the target molecule has
if
alkyl groups «to the carbonyl
group, (a)
+
f?
^
1
CH 2 CCH 3
2.
2 Na 2
O H3 Q+
f?
OEt
CH 3 CH 2 Br
(CH 3 CH 2 2 CCCH 3 )
C0 2 Et
II
(CH 3 CH 2 2 CHCCH 3 )
heat
C0 2 Et
3-Ethyl-2-pentanone +
+ 2 NaBr
C0 2
(b)
P
+
— Na
II
1
CHoCCHo d 6
2.
|
p "
~OEt
CH 3 CH 2 CH 2 Br
CHoCHoCHoCHCCHo 6 d 6 *
11
Na
.
HoC 3 I
Q
CH 3 CH 2 CH 2 CH — CCH 3 3-Methyl-2-hexanone +
HoC 3
II
EtOH
*
HoO^
^
+
~OEt
CH 3 Br
2.
C0 2
NaBr
C02 Et
C0 2 Et
+
+
\
I
O II
CH 3 CH 2 CH 2 C— CCH 3
C0 2 Et
+
NaBr
+
EtOH
Carbonyl Alpha-Substitution Reactions
22.29 Use a malonic derivative.
ester synthesis if the product
Use an
you want
is
603
an a-substituted carboxylic acid or you want is an a-substituted
acetoacetic acid synthesis if the product
methyl ketone. (a)
1.2 Na+ ~OEt
CH 2 (C0 2 Et) 2
2. 2
CH 3 C(C0 2 Et) 2
CH 3 Br
+ 2 NaBr
CHo
(b)
O II
+
1.2 Na
f?
CH 2 CCH 3
CCH 3
"OEt
BrCH 2 (CH 2 4 CH 2 Br
2.
+
NaBr
+
C0 2
C0 2 Et
)
C0 2 Et H3
+ ,
heat
O II
CCH 3
EtOH
+
H
(c)
+
1.2 Na
CH 2 (C0 2 Et) 2
C0 2 Et
~OEt
BrCH 2 CH 2 CH 2 Br
2.
0<:C0 |
H3
+ 2 NaBr
2 Et + ,
heat
/\.C02 H
+
C0 2
+ 2
EtOH
(d)
+
f?
1
CH 2 CCH 3
.
2.
Na
"OEt
H 2 C=CHCH 2 Br
H 2 C=CHCH 2 CHCCH 3
C0 2 Et
+
NaBr
C0 2 Et H3
+ ,
heat
J
H 2 C— CHCH 2 CH 2 CCH 3
+
qq
+
22.30 The haloform reaction (Problem 22.25d) is an alpha-substitution reaction by
which a
is
CH 3 CCH 3
(b)
C 6 H 5 CCH 3
is
Negative haloform reaction:
Positive haloform reaction:
(a)
in
trihalogenated at the alpha position, and the trihalomethyl group -OH. It is a test for methyl ketones.
methyl ketone displaced
EtOH
(c)
CH 3 CH 2 CHO
(d)
CH 3 C0 2 H
(e)
CH 3 CN
604
22.31
Chapter 22
with PBr 3 to form
First, treat geraniol
(CH 3 )2C=CHCH2CH2C(CH3)=CHCH 2 Br
(geranyl bromide).
(a)
CH 3 C0 2 Et
"
2 Geranyl*
C= CHCH2CH2C(CH 3)= CHCH 2CH 2C0 2 Et
(CH 3) 2
b ro m d e
Ethyl geranylacetate
i
Alternatively:
+
1
CH 2 (C0 2 Et) 2
Na
.
~OEt
—
-r-z c.
oeranyi
.
(CH 3 2 C=CHCH 2 CH 2 C(CH 3 )=CHCH 2 CH 2 (C0 2 Et) 2 )
bromide
C0 2
+ 2
I
H3 +
,
heat
EtOH + (CH 3 2 C=CHCH 2 CH 2 C(CH 3 )=CHCH 2 CH 2 C0 2 H )
SOCI 2 EtOH, pyridine
(CH 3 2 C= CHCH 2 CH 2 C(CH 3 )= CHCH 2 CH 2 C0 2 Et )
Ethyl geranylacetate
(b)
O U
CH 2 CCH 3
g
C0 2 Et
Qerany?^
(CH 3 ) 2
C= CHCH2CH2C(CH3)= CHCH 2CHCCH3
bromide
I
H
+ ,
C0 2 Et
heat
1
t
C0 2
EtOH
+
+
(CH 3 2 C=CHCH 2 CH2 C(CH 3 )=CHCH 2 CH 2 CCH 3 )
Geranylacetone
22.32 Dialkylation of diethylmalonate:
Et0 2
C^^C02 Et
.C0 2 Et
Et0 2 C. +
1
.
2.
Na "OEt
H 2 C=CHCH 2 Br
Et0 2 C. +
\
1
\
.
Na
.C0 2 Et
"OEt
2.(CH 3 2 CHBr )
V
^
Carbonyl Alpha-Substitution Reactions
605
Nucleophilic acyl substitution:
This series of steps
is
repeated to form the 6-membered ring.
General Problems hydrogen atoms is treated with NaOD in D2O, all by deuterium atoms. For each proton (atomic weight 1) lost, a deuteron (atomic weight 2) is added. Since the molecular weight of cyclohexanone increases by four after NaOD/D20 treatment (from 98 to 102), cyclohexanone contains four acidic hydrogen atoms.
22.33 When
a
compound containing
acidic
acidic hydrogens are gradually replaced
22.34
Reaction of (/?)-2-methylcyclohexanone with aqueous base is shown below. Reaction with aqueous acid proceeds by a related mechanism through an enol, rather than an enolate ion, intermediate.
(K)-2-Methylcyclohexanone its chirality when the enolate ion double bond is formed. Protonation occurs with equal probability from either side of the planar s^-hybridized carbon 2, resulting in a racemic product.
Carbon 2 loses
Width: 612 Height: 792
606
Chapter 22
22.35
(S)-3-Methylcyclohexanone
(5)-3-Methylcyclohexanone isn't racemized by base because involved in the enolization reaction.
22.36 The Hell-Volhard-Zelinskii
its
chirality center is not
reaction involves formation of an intermediate acid bromide
enol, with loss of stereochemical configuration at the chirality center. Bromination of (R)-
2-phenylpropanoic acid can occur from either face of the enol double bond, producing racemic 2-bromo-2-phenylpropanoic acid. If the molecule had a chirality center that didn't take part in enolization (Problem 22.35), the product would be optically active.
22.37
(a)
Na + ~OCH 3
,
then
CH3
dimethylcyclohexanone
I;
(b)
may
H3
+ ,
heat; (c)
LDA,
then
CH 3 I (some 2,2-
also be formed).
22.38
carbonyl oxygen
protonation at y position
of a proton
loss of proton
on oxygen
Carbonyl Alpha-Substitution Reactions
607
22.39 :Q:
:o:
:0:
:o:
H
H
H-[OH
I
abstraction
+ protonation
H
H
H
of a proton
The
"OH
aty position
enolate of 3-cyclohexenone can be protonated at three different positions. Protonation
at the y position yields the a,6-unsaturated ketone.
22.40
O
:o:
CH<
H
H
Vu nr*
abstraction
H
ho:
H
of Y proton
H
'3 is used for oxidizing alcohols to carbonyl compounds, ecgonine has the structure shown above. Again, the stereochemistry is unspecified.
reaction with hydroxide.
The complete
reaction sequence:
Cocaine
Ecgonine
Carbonyl Alpha-Substitution Reactions
22.54 Laurene
613
differs in stereochemical configuration from the observed product at the carbon a methylene group. Since this position is a to the carbonyl group in the precursor to laurene, enolization and isomerization must have occurred during the reaction. Isomerization of the ketone precursor is brought about by a reversible reaction with the basic Wittig reagent, which yields an equilibrium mixture of two diastereomeric ketones. One of the ketone isomers then reacts preferentially with the Wittig reagent to give only the observed product.
to the
22.55 The key
step
is
an intramolecular alkylation reaction of the ketone a-carbon, with the
tosylate in the second ring serving as the leaving group.
614
Chapter 22
22.56
O
H
if
I
O Na + ~QEt_
r.
II
CH 3 CNHCC0 2 Et
CH 3 CNHCC0 2 Et formation of enolate
C0 2 Et
C0 2 Et
J
CH 3 I
H 2 NCHCOH
+
H 63 C
ll|
\
O II
>
CH 3 C-|-NHC— C-| OEt '"fy ou 2 t
decarboxylation
C0 2
CH 3 C0 2 H
+
Acid hydrolyzes both
shown
heat
amide hydrolysis
Alanine
EtOH
,
ester hydrolysis
CH 3
+ 2
O
+
H3
f?
alkylation
in Figure 21.8
ester bonds, as well as the amide bond, by mechanisms that were and Section 21.6. Decarboxylation of the /S-keto acid produces
alanine.
22.57 1
CH 2 (C0 2 Et) 2
.
2.
Na+ "OEt
(CH 3 2 CHCH 2 CH(C0 2 Et) 2
(CH 3 2 CHCH 2 Br
)
)
|
Br 1
(CH 3 2 CHCH 2 CHC02 H )
*
1
.
2.
Br d 9 PBro ,
H2
^
H3 +
,
heat
(CH 3 2 CHCH 2 CH 2 C02 H )
+
C0 2
+ 2
EtOH
NH. (CH 3 2 CHCH 2 CHC02 H Leucine )
A malonic ester synthesis is used to form 4-methylpentanoic acid. Hell-Volhard-Zelinskii bromination of the acid, followed by reaction with ammonia, yields leucine. The reaction is an Sn2 displacement of bromide by ammonia.
last
Carbonyl Alpha-Substitution Reactions
615
22.58
Step Step
1:
Protonation.
3:
Deprotonation.
22.59 This sequence resembles
Step 2: Hydride shift. Step 4: Enolization to form the aromatic the
one shown
in
Problem 22.32.
Sodium Pentothal The
series
of steps
is
repeated to form the 6-membered ring.
ring.
Width: 612 Height: 792
Chapter 23 - Carbonyl Condensation Reactions
Chapter Outline
I.
The
aldol reaction (Sections 23.1-23.6).
A. Characteristics of the aldol reaction (Sections 23.1-23.2). 1
.
The
aldol condensation
is
a base-catalyzed dimerization of
two aldehydes or
ketones.
2
.
3
.
The reaction can occur between two components that have a hydrogens. One component (the nucleophilic donor) is converted to its enolate and undergoes an a-substitution reaction.
4.
5
.
6
.
The other component
(the electrophilic acceptor) undergoes nucleophilic addition. For simple aldehydes, the equilibrium favors the products, but for other aldehydes and ketones, the equilibrium favors the reactants. Carbonyl condensation reactions require only a catalytic amount of base (Section
23.2). Alpha-substitution reactions, on the other hand, use one equivalent of base. a. B. Dehydration of aldol products (Section 23.3).
Aldol products are easily dehydrated to yield a,/?-unsaturated aldehydes and ketones. a. Dehydration is catalyzed by both acid and base. b. Reaction conditions for dehydration are only slightly more severe than for condensation. Often, dehydration products are isolated directly from condensation reactions. c. Conjugated enones are more stable than nonconjugated enones. 2 3 Removal of the water byproduct drives the aldol equilibrium towards product formation. C. Aldol products (Sections 23.4-23.5). 1 Using aldol reactions in synthesis (Section 23.4). a. Obvious aldol products are: 1
.
.
.
.
i.
a,/S-Unsaturated aldehydes/ketones.
^-Hydroxy aldehydes/ketones. it's possible to work backwards from a compound that doesn't seem resemble an aldol product and recognize aldol components. ii.
b 2
.
.
Often,
Mixed a.
b
.
to
aldol reactions (23.5).
two similar aldehydes/ketones react under aldol conditions, 4 products may be formed - two self-condensation products and two mixed products. A single product can be formed from two different components i. If one carbonyl component has no a-hydrogens. ii. If one carbonyl compound is much more acidic than the other. If
:
D. Intramolecular aldol condensations (Section 23.6). 1 Treatment of certain dicarbonyl compounds with base can lead to cyclic products. 2 A mixture of cyclic products may result, but the more strain-free ring usually .
.
predominates. II.
The Claisen condensation
(Sections 23.7-23.9).
A. Features of the Claisen condensation (Section 23.7). Treatment of an ester with 1 equivalent of base yields a /3-keto ester. 1 2 The reaction is reversible and has a mechanism similar to that of the aldol reaction. .
.
Carbonyl Condensation Reactions 3
.
4
.
5
.
617
A major difference from the aldol condensation is the expulsion of an alkoxide ion from the tetrahedral intermediate of the initial Claisen adduct. Because the product is often acidic, one equivalent of base is needed; addition of this amount of base drives the reaction to completion. Addition of acid yields the final product. Claisen condensations (Section 23.8). Mixed Claisen condensations of two different esters can succeed
B Mixed 1
.
2.
if
one component
has no a hydrogens. Mixed Claisen condensations between a ketone and an ester with no a hydrogens are also successful.
C. Intramolecular Claisen condensations: the Dieckmann cyclization (Section 23.9). The Dieckmann cyclization is used to form cyclic /3-keto esters. 1 1,6-Diesters form 5-membered rings. a. b 1 ,7-Diesters form 6-membered rings. 2. The mechanism is similar to the Claisen condensation mechanism. 3 The product /3-keto esters can be further alkylated. a. This is a good route to 2-substituted cyclopentanones and cyclohexanones. Other carbonyl condensation reactions (Sections 23.10-23.13). A. The Michael reaction (Section 23.10). The Michael reaction is the conjugate addition of an enolate to an a,/3-unsaturated 1 carbonyl compound. a. The highest-yielding reactions occur between stable enolates and unhindered a,/3-unsaturated carbonyl compounds. 2. Stable enolates are Michael donors, and «,y5-unsaturated compounds are Michael .
.
.
III.
.
acceptors.
B. The Stork 1
2
.
.
reaction (Section 23.1 1). ketone that has been converted to an enamine can act as a Michael donor in a A reaction known as the Stork reaction. The sequence of reactions in the Stork reaction: a. Enamine formation from a ketone.
b
.
c.
Michael-type addition to an a,/3-unsaturated carbonyl compound.
Enamine hydrolysis back
to a ketone.
This sequence is equivalent to the Michael addition of a ketone to an a,/3-unsaturated carbonyl compound and yields a 1,5 diketone product.. C. The Robinson annulation reaction (Section 23.12). The Robinson annulation reaction combines a Michael reaction with an 1 intramolecular aldol condensation to synthesize substituted ring systems. 3
.
.
The components
are a nucleophilic donor, such as a /3-keto ester, and an a,punsaturated ketone acceptor. 3 The intermediate 1 ,5-diketone undergoes an intramolecular aldol condensation to yield a cyclohexenone. D. Biological carbonyl condensation reactions (Section 23.13). 1 Many biomolecules are synthesized by carbonyl condensation reactions. 2 The enzyme aldolase catalyzes the addition of a ketone enolate to an aldehyde, a. This mixed aldol reaction is successful because of the selectivity of enzyme
2
.
.
.
.
catalysis.
3
.
Acetyl Co A is the major building block for the synthesis of biomolecules. Acetyl Co can act as an electrophilic acceptor by being attacked at its carbonyl a. group.
A
b
.
Acetyl
CoA
can act as a nucleophilic donor by loss of its acidic a hydrogen.
618
Chapter 23
Solutions to Problems
23.1
( 1)
Form the
enolate of one molecule of the carbonyl compound.
f?
CH 3 CH 2 CgCH
+ :OH
:
CH3CCH3
+
1
(3) Reprotonate the enolate anion.
O HO"
+
CH2CCH3
Problem
23.1.
619
Chapter 23
As in Problem 23.1, align the two carbonyl compounds so that the location of the new bond is apparent. After drawing the addition product, form the conjugated enone product by dehydration. In parts (b) and (c), a mixture of E,Z isomers may be formed.
heat |J
(C)
O
OH M
(CH 3 ) 2 CHCH 2 CH
II
+
CH 2 CH CH(CH 3) 2
NaOH, «
*
l|
I
(CH 3 2 CHCH 2 )
Et0H
C— CHCH H
CH(CH 3
)
2
heat
ti
O
= CCH II
(CH 3
)
2 CHCH 2 CH
+
CH(CH 3) 2
H2
Carbonyl Condensation Reactions
23.4
621
Including double bond isomers, 4 products can be formed. The major product is formed by reaction of the enolate formed by abstraction of a proton at position "a" because position "b" has more steric hindrance.
major
(less hindered)
minor (more hindered)
23.5 (a)
OH O I
CH3CH2CH2C
II
CH
2-Hydroxy-2-methylpentanal
CHa This is not an aldol product. The hydroxyl group in an aldol product must be fi, not the carbonyl group. (b)
H3C Jj)
CH 3 CH 2 C= C— CCH 2 CH 3
5-Ethyl-4-methyl-4-hepten-3-one
CH2CH3 This product results from the aldol self-condensation of 3-pentanone, followed by dehydration.
a, to
Chapter 23
O
O
II
2
CH 3 CH
NaOH, EtOH
1
2.
heat
NaBH 4 + H ^+
1.
II
= CHCH CHoCH 6
n
,
2.
CH 3 CH=CHCH 2 OH
,
3
|h 2 Pd/C ,
CH 3 CH2CH 2 CH 2 OH
1-Butanol
O //
H—
H— NaOH
N
EtOH
H
c—
O-C\
H H
H
H
H
OH \ /
H
heat .0
Cyclopropylacetaldehyde
H— C— I
H
H
(a)
OH
O II
O II
I
CH
•
NaOH,
Or-?
CHCH 2 CCH 3 heat
=
CH0CCH03
+ HoO
EtOH [
4-Phenyl-3-buten-2-one This mixed aldol will succeed because one of the components, benzaldehyde, is a good acceptor of nucleophiles, yet has no a-hydrogen atoms. Although it is possible for acetone to
undergo self-condensation, the mixed aldol reaction
(b)
o
a
CHo
II
is
much more
O
O
II
I
II
aC=CHCCH
CCH3
3
1
NaOH, EtOH
2.
heat
favorable.
)
+
CHq6
O
I
O
II
CH3CCH3 +
^nn
(CH 3 2 C=CHC Vs
O
U
II
.C=CHC.
(CH 3 2 C=CHCCH 3 + )
h,o
^s.
kJ
Four products result from the aldol condensation of acetone and acetophenone. The two upper compounds are mixed aldol products, and the bottom two are self-condensation products.
Carbonyl Condensation Reactions
(c)
HoC 1
NaOH, EtOH
2.
heat
CH 3 CH 2 CHO
623
CHO
CH 3 CH 2 CH=CCHO CHo
As
in (b), a mixture of products is formed because both carbonyl partners contain ahydrogen atoms. The upper two products result from mixed aldol condensations; the lower two are self-condensation products.
23.9
O
o
it-
ii
CH3CCHCCH3 A
NaOH
O
o
//
CH3CCH2CCH3
O
NaOH
O
O
II
II."
|
CH3CCH2CCH2
is in
I
C=CH HoC
B 2,4-Pentanedione
H 2 C—
equilibrium with two enolate ions after treatment with base. Enolate
A is stable and unreactive, while enolate B can undergo internal aldol condensation to form a cyclobutenone product. But, because the aldol reaction is reversible and the cyclobutenone product is highly strained, there is little of this product present when equilibrium is reached. At equilibrium, only the stable, diketone enolate ion A is present.
23.10
This intramolecular aldol condensation gives a product with a seven-membered ring fused to a
flve-membered
To
ring.
O NaOH,
EtOH
EtOH
H2
+
heat
624
Chapter 23
23.11 As
in the aldol condensation, writing the
makes
it
two Claisen components
in the correct orientation
easier to predict the product.
(a) I?
ff
(CH 3 2 CHCH 2 COEt
CH 2 COEt
+
)
1
.
Na+ "OEt
H^ Q
2
+
ff
I?
(CH 3 2 CHCH 2 C— CHC( CHCOEt )
I
CH(CH 3 ) 2
CH(CH 3
EtOH
+
)
2
O
(b)
•I
II
CH 2 COEt
+
CH 2 COEt
Na+ "OEt» - + 2 H 3°
1.
ff
ff
CH 2 C— CHCOEt
-
I
EtOH
+
(c)
+
ff
1
CH 2 COEt
+
CH 2 COEt I
.
d.
Na "OEt u H 3 n+ U
ff
ff
CH 2 C— CHCOEt
+ EtOH
23.12
O
c
I
CH 3 3.
C0 2 CH 3
OCH 3
Step 1: Conjugate addition of Step 3: Loss of methoxide.
H3C
.
Step 2: Claisen condensation.
Carbonyl Condensation Reactions
23.62
C0 2 Et
C0 2 Et
C02 Et
OEt 1.
C0 2 Et EtOH
"OEt +
o Step Is Enolate formation. Step 2: Intramolecular aldol condensation. Step 3: Retro-aldol condensation. Step 4: Protonation.
23.63
Step 1: Deprotonation and retro aldol reaction. Step 2: Equilibration between two enolates. Step 3: Internal aldol condensation. Step 4: Protonation.
651
652
Chapter 23
23.64
H-A
(a)
ej)H2 H3C
H
!.
3
Z -
£ HN(CH 3
)
3
HN(CH 3
3 )
-
+
H3 c^Sr h
C n(CH :
2
3 )2
4.
2
ti
HoO
+
CH3 CH= N(CH 3
)
2
iminium ion
:
|j
CH=N(CH 3 2 )
L. \J
5 - Aminopy rimidine
24.4
are less basic than hydroxide but more basic than amides. The pKa values of the conjugate acids of the amines in (c) are shown. The larger the pKa the stronger the base.
Amines
,
More Basic
Less Basic
(a)
CH 3 CH 2 NH 2
CH 3 CH 2 CONH 2
(b)
NaOH
CH 3 NH 2
CH3NHCH3 pKa = 10.73
pyridine
(C)
pKa =
5.25
24.5
aCH pKa =
2 NrV
CH 3 CH 2 CH 2 NH 3 + pKa =
9.33
stronger acid (smaller
aCH pKb = 14-9.33 =
pKa)
10.71
weaker acid
(larger
p#a)
2 NH 2
CH 3 CH 2 CH 2 NH 2 pKb = 14-10.71 =
4.67
weaker base
3.29
stronger base
The
stronger base (propylamine) holds a proton more tightly than the weaker base (benzylamine). Thus, the propylammonium ion is less acidic (larger p^a ) than the benzylammonium ion (smaller pKa).
To
calculate
p#b Ka -Kh = :
10~ 14 ,
pKa + pKb =
14 and
pKb =
14
-
pKa
.
660
Chapter 24
24.6
The
basicity order of substituted arylamines is the
same
as their reactivity order in
electrophilic aromatic substitution reactions because, in both cases, electron- withdrawing
make
substituents
24.7
Use
the expressions
[RNH2 log
[RNH 3
=
j
[RNH3
24.8
=
1
pH = 7.3,
shown
more electron-poor and
destabilize a positive charge.
in Section 24.5.
pH-pK a
= 7.3-1.3 =
6.0
]
[RNH2
At
the site of reaction
antilog (6.0)
= 10 6 [RNH 2 ] = 10 6 [RNH 3 :
+ ]
]
virtually
100% of the pyrimidine molecules
are in the neutral form.
Amide
reduction can be used to synthesize most amines, but nitrile reduction can be used compounds in (b) and (d) can be synthesized only by amide reduction.
to synthesize only primary amines. Thus, the
Amine
Nitrile
Precursor
Amide Precursor
(a)
CH 3 CH 2 CH 2 NH 2
CH 3 CH 2 C= N
CH 3 CH 2 CNH 2
(b)
CH 3 CH2 CNHCH 2 CH2 CH 3
(CH 3 CH 2 CH2)2NH (c)
O
C=N
CH2 NH 2 (d)
o
The compounds
"12
o II
NHCH 2 CH 3 in parts (b)
NHCCH 3 and
(d) can't
be prepared by reduction of a
nitrile.
Amines and Heterocycles
661
24.9
Step 1: Addition of hydroxide. Step 2: Ring opening. Step 3: Proton transfer. Step 4: Addition of hydroxide. Step 5: Elimination of amine. Step 6: Proton transfer.
24.10 The upper reaction
is
the azide synthesis, and the lower reaction
OH
is
the Gabriel synthesis.
662
Chapter 24
24.11 Look
at the target molecule to find the groups bonded to nitrogen. One group comes from the aldehyde/ketone precursor, and the other group comes from the amine precursor. In
most
cases,
two combinations of amine and aldehyde/ketone
Amine
Amine Precursor
are possible.
Carbonyl Precursor
O
CH 3
(a)
II
CH 3 CH 2 NHCHCH 3
CH 3 CCH 3
CH 3 CH 2 NH. or
CHo3 I
CH 3 CHO
H 2 NCHCH 3
— NHCH CH
,
Pt »»
HoN-
1 J
CH 3
CNH-^^-S02 -NH-<^ N
(CH 3 CO) 2
HgN-/""^— S0 2 - NH— ^
jj
N 4.
Sulfathiazole
j)
Amines and Heterocycles
24.17
665
nitrated and the nitro group is ultimately reduced, but important in arriving at the correct product. In (a), nitrobenzene is immediately reduced and alkylated. In (c), chlorination occurs before reduction so that chlorine can be introduced in the m-position. In (b) and (d), nitrobenzene is reduced and then acetylated in order to overcome amine basicity and to control reactivity. In both cases, the acetyl group is removed in the last step.
In
all
of these reactions, benzene
the timing of the reduction step
is
is
Either method of nitro group reduction (SnCl2, H2) can be used in problem; both methods are shown.
Mono- and
trialkylated anilines are also formed.
all
parts of this
Width: 612 Height: 792
666
Chapter 24
24.18
Br
Br
+
"
N 2 HS0 4
/7-Bromobenzoic acid
The
shown above
one of several ways to synthesize p- bromobenzoic acid and is definitely ngj the simplest way. (The simplest route is Friedel-Crafts alkylation -» bromination -* oxidation). The illustrated synthesis shows the use of the diazonium replacement reaction that substitutes bromine for a nitro group. Oxidation of the methyl group yields the substituted benzoic acid. route
is
m-Bromobenzoic acid Again, this isn't the easiest route to this compound. In this case, nitration is followed by bromination, then by diazotization, treatment with CuCN, and hydrolysis of the nitrile.
Amines and Heterocycles
from
(b)
667
ra-Bromochlorobenzene
/?-(Af,Af-Dimethylamino)azobenzene
Coupling takes place between Af,N-dimethylaniline and a benzenediazonium the desired product.
salt to yield
668
Chapter 24
24.20
Thiazole
Thiazole contains six ^electrons. Each carbon contributes one electron, nitrogen
two electrons
contributes one electron, and sulfur contributes sulfur
and nitrogen have lone electron pairs
in sp
2
to the ring
orbitals that
he
n system. Both
in the plane of the ring.
24.21 log
[RNH 2| [RNH 3 [RNH 2
[RNH 3 [RNH 3
]
pH-pK a
= 7.37-6.00 =
1.37
]
]
=
+
=
antilog(1.37)
23.4:
[RNH 2 ] = 23.4[RNH 3
]
]
+ 23.4 [RNH 3
[RNH 3
=
]
+ ]
= 24.4 [RNH 3
]
= 100%
=100% -24.4 =4.1%
[RNH 2 ] = 100% -
4.1
= 95.9%
4.1% of histidine molecules have in the protonated
form
at
the imidazole nitrogen physiological pH.
24.22 Attack
at
C2:
unfavorable
J
Amines and Heterocycles
E
H
E
H
E
669
H
unfavorable
Reaction
at
C3
is
favored over reaction
at
C2 or C4. The positive charge
of the cationic
intermediate of reaction at C3 is delocahzed over three carbon atoms, rather than over carbons and the electronegative pyridine nitrogen as occurs in reaction at C2 or C4..
two
24.23
CH 2 CH 2 N(CH 3
)
2
Af,Af-Dimethyltryptarnine
atom of Af,Af-dimethyltryptarnine is more basic than the ring lone electron pair is more available for donation to a Lewis The aromatic nitrogen electron lone pair is part of the ring k electron system.
The
side chain nitrogen
nitrogen atom because
its
acid.
670
Chapter 24
Attack at C3:
Positive charge can be stabilized by the nitrogen lone-pair electrons in reaction at both C2 and C3. In reaction at C2, however, stabilization by nitrogen destroys the aromaticity of
the fused benzene ring. Reaction at C3 is therefore favored, even though the cationic intermediate has fewer resonance forms, because the aromaticity of the six-membered-ring is
preserved.
24.25
NH 2 NHi
(CH 3 3 CCCH 3 )
NaBH 3 CN
A
(CH 3
)
3 CCHCH 3
B -1
The IR spectrum (pair of bands at around 3300 cm ) shows that B is a primary amine, and the *H NMR spectrum shows a 9-proton singlet, a one-proton quartet, and a 3-proton doublet.
An
absorption due to the amine protons
is
not visible.
Visualizing Chemistry
24.26 (b)
(a)
H 63 C
(c)
H
\ /
HoC 3
=v /
N
CH(CH 3
HoN
I
H A^-Methylisopropylamine
secondary amine
trans-(2-Methy\cyclopentyl)amine
N-Isopropylaniline
primary amine
secondary amine
)
2
Amines and Heterocycles
671
24.27
(1
S,2S)-( 1 ,2-Diphenylpropy l)amine
Hofmann be 180°
elimination
apart.
is
(Z)- 1 ,2-Diphenyl- 1 -propene
an E2 elimination,
in
The product that results from
which the two groups to be eliminated must geometry is the Z isomer.
this elimination
24.29 most basic
/
H
The
indicated nitrogen
is
Lewis
acid.
it is more electron-rich. The electrons of the n system and are less available for donation to a
most basic because
other nitrogens are part of the fused-ring
672
Chapter 24
Additional Problems
Naming Amines 24.30 (b)
(a)
Br
(c)
[^^— CH2CH2 NH 2
v
1^^— NHCH2CH
3
Br
2,4-Dibromoaniline
(2-Cyclopentylethyl)amine
(d)
A^Ethylcyclopentylamine
(e)
(f)
CH,
N— CH2 CH2 CH 3
N
H 2 NCH 2 CH 2 CH 2 CN
CH 3 A/,Af-Dimethylcyclopentylamine
4-Aminobutanenitiile
Af-Propylpyrrolidine
24.31 (a)
(b)
N(CH 3
)
NHCH 3
CH 2 NH 2
3
(Cyclohexylmethyl)amine
Af-Dimethylaniline
«
(c)
CH 3
(
A^-Methylcyclohexylamine
e> (H 3 C) 2 NCH 2 CH 2 C0 2 H
NH 2 (2-Methylcyclohexyl)amine
3-(N, Af-dimethylaminopropanoic acid
24.32 (a)
fi
— tertiary
secondary amine (b)
amine
NHCH 3
^
*\ ||
secondary amine
N
—
•*
H \
H
Lysergic acid diethylamide
secondary amine
Amines and Heterocycles
Amine
673
Basicity
24.33 The
pyrrole anion, C4H4N:",
is
as the cyclopentadienyl anion.
a 6 n electron species that has the same electronic structure Both of these anions possess the aromatic stability of 6 x
electron systems.
24.34
The The
"a" nitrogen is
most basic because
"c" nitrogen is the least basic
part of the ring
its
electron pair
is
most available
to
Lewis
acids.
because the lone-pair electrons of the pyrrole nitrogen are
k electron system.
24.35
inductive effect of the electron-withdrawing nitro group makes the amine nitrogens of both m-nitroaniline and p-nitroaniline less electron-rich and less basic than aniline.
The
When nitro
the nitro group is para to the amino group, conjugation of the amino group with the group can also occur. /7-Nitroaniline is thus even less basic than m-nitroariiline.
674
Chapter 24
Synthesis of Amines
24.36 (a)
NaN iai \^ 3
PBr<, rpi3
CH2CH2CH2CH2OH
i
Ch^Ch^Ch^Ch^Br
^
CH2CH2CH2CH2N3 1
.
2.
UAIH4
H2
CH2CH2CH2CH2NH2 Butylamine
(b)
CH2CH2CH2CH2OH
^
Cr0 3 H3
SOCIc
CH2CH2CH2CO2H
CH 2 CH 2 CH 2 COCI CH2CH2CH2CH2NH2 from
(a)
NaOH ?
CH3CH2CH2CH2NHCH2CH2CH2CH3
^
1
.
"
L1AIH4
2 |_|
CH3CH2CH2CNHCH2CH2CH2CH3
Q
Dibutylamine
(c)
CrQ 3 H 3 Q ,
CH2CH2CH2CH2OH
Propylamine
^
CH2CH2CH2CO2H
1
— — CH2CH2CH2CONH2-*— Br2
j CH2CH2CH2NH2''
+
,
NaOH
777;
I
-v
2U
\ H 2 0,
S0C 2 '
2 NHo
JSJV
CH2CH 2 CH 2 CON 3
heat
CH 2 CH 2 CH 2 COCI
(d)
CH2CH2CH2CH2OH
PBr 3 ^
^
NaCN_
^
CH2CH2CH2CH2Br
CH2CH2CH2CH2CN 1
.
2.
UAIH4
H2
CH2CH2CH2CH2CH2NH2 Pentylamine (e )
^
Periodinane
CH2CH2CH2CH2OH
CH
» C|
NaBhU
CH 2 CH 2 CH 2 CHO
CH 2 CH2CH2CH 2 N(CH 3 )2
MA^-Dimethylbutylamine
W
excess
CH 2 CH 2 CH2NH 2 from ( c )
QH 3
+
* CH 2 CH 2 CH 2 N(CH 3 )r
1
.
Ag P 0, H P
yf^j '
"
CH 3 CH=CH 2 Propene
+ (CH 3 ) 3 N
Amines and Heterocycles
675
24.37 (a)
CH3CH2CH2CH2CO2H
CH3CH2CH2CH2COCI
2 NHt **
CH3CH2CH2CH2CONH2 Pentanamide
(b)
from
CH3CH2CH2CH2NH2
H2
(a)
^
1
Butylamine
LiAIH 4
.
CH3CH2CH2CH2CONH2 from
NaOH
Br ?)
CH3CH2CH2CH2CONH2
CH3CH2CH2CH2CH2NH2
2
(a)
Pentylamine
1
Br2 PBr3
2.
H2
,
CH3CH2CH2CH2CO2H
**"
CH3CH2CH2CHCO2H
2-Bromopentanoic acid
(e)
LBH3 ~ /-^+^
CH3CH2CH2CH2CO2H
2.
ChtaChtaChtaChtaChtaOH
H3O
I
NaCN
CH3CH2CH2CH2CH2CN
PBr 3
CH3CH2CH2CH2CH2Br
Hexanenitrile (f) 1
CH3CH2CH2CH2CH2CN ~
.
2.
from
UAIH4
H2
^
CH3CH2CH2CH2CH2CH2NH2 Hexylamine
(e)
24.38 (a)
N0 2 HNOi
1
,
H 2 S0 4
(b)
2.
CONH-
Hair
SnCI 2 H 3 Q+
NaOH,.h 2 o
Y
T
NH 2
NHc Br2
,
NaOH
H2
(c)
COCI
CONH-
NHc
Width: 612 Height: 792
676
Chapter 24
24.39
First,
shown
synthesize aniline from benzene, as
in
Problem 24.38
(a).
"
HS0 4
N2
H 2 NH 2
CuCN
HNO 2_ H 2 S0 4
1
2.
LiAH H2
24.40 (a)
CH 3 CH 2 CH 2 CH 2 CONH 2
1
UAIH4
2.
H2
CHgCH 2 CH 2 CH 2 CH 2 NHj
(b) 1
.
UAIH4
CHgCH 2 CH 2 CH2 CN ~ ~ ~ *
(c)
CH0CH0CHi=r.M. — CHo2
32
1
.
2.
BHo,
CH3CH 2 CH 2 CH 2 CH 2 NH 2
THF
PBi>
CH0CH0CH0CH0OH 2 3
H 2 2 rOH
2
2
CH3CH 2 CH 2 CH 2 Br NaCN |
CH3CH 2 CH 2 CH 2 CH 2 NH 2
(d)
CH 3 CH 2 CH 2 CH 2 CH 2 CONH 2
Br 2
,
J
.
UAIH4
—
—
H~~0
Cri3CH 2 CH 2 CH 2 CN
NaOH CHgCH 2 CH 2 CH 2 CH 2 NHj
H2
(e)
PBrv
CH3CH 2 CH 2 CH 2OH
CH 3 CH 2 CH 2 CH 2 Br
NaCN
CH 3 CH 2 CH 2 CH 2 CN 1
I
.
\ 2.
UAIH4
H2
CHgCH 2 CH 2 CH 2 CH 2 NH 2 (f)
CHoCH 2 CH 2 CH 2 CH — CHCH 2 CH 2 CH 2 CHg CH3CH 2 CH 2 CH 2 CHO
NH-
NaBK
1.
-~
0<
~
2. Zn,
H3O
+
2
CH3CH 2 CH 2 CH 2 CH 2 NH 2
CHgCH 2 CH 2 CH 2 CHO ^
'
Amines and Heterocycles
(g)
SOCI^
CH 3 CH2CH 2 CH 2 C02H
CH 3 CH2CH 2 CH 2 COCI 1 2 1
CH 3 CH 2 CH 2 CH 2 CH 2 NH 2
^
.
LiAIH 4 - -4
NH 3
CH 3 CH 2 CH 2 CH 2 CONH 2
24.41
OH
OH
O II
I
I
CH— CCH 3 +
H 2 NCH 3
I
CH— CHCH3 NaBH 4 Ephedrine
CH 3 OH
Reactions of Amines
24.42
(b)
24.43
NHCHo3
677
678
Chapter 24
In (e), the
amine
reacts with the Grignard reagent
and inactivates
it.
Amines and Heterocycles
24.45 Hydrogens that can be eliminated
are starred. In cases where more than one alkene can form, the alkene with the less substituted double bond is the major product.. 1
.
CH 3I
Ag 2 Q, H 2 Q
3.
heat
Amine (a)
excess
2.
Alkene
Amine
*
r> (b)
O
NHCH 3
H 2C
CHo3
N(CH 3 3 )
— CHCH2 CH2CH 2CH3
N(CH 3
2
)
I
major
HNCHCH 2 CH2 CH 2 CH 3
CH 3 CH^— CHCH 2 CH 2 CH 3 minor
(c)
CHo3
CHo3 r
I
_
CH 3 CHCH — CHCH 2 CH3
CH3 CHCHCH 2 CH 2 CH 3
N(CH 3 3 )
major
NHo CHo3 r
CH 3 C — CHCH 2 CH 2 CH 3 minor
24.46 (a)
HNO'
H 2 Pt
H 2 S0 4
EtOH*
,
(b)
CH 2 NH, HNO, H 2 S0 4
CH 3 from
(a)
1
LiAIH 4
2.
H2
679
680
Chapter 24
Amines and Heterocycles
(d)
HoC
— CHo
H2C
NaOH
CH,
Br
HoO
Br
equiv
+
2
H2
+
2
NaBr
N I
NH 2 1
681
CH-
I
CHo
Spectroscopy
24.48 The HNMRofthe amine shows !
5 peaks.
at
3.40 5
is
Two
two are due
electronegative element (oxygen),
due to an ethyl group bonded to an 4 aromatic ring hydrogens, and the peak
are
to
due to 2 amine hydrogens.
HCOCH 3 "OH
HoO
OCH 2 CH 3
OCH0CH3 +
"
CH 3 C0 2
CgH^NO
C 10 H 13 NO 2 Phenacetin
/7-Ethoxyaniline
C 6 H 7 NO /7-Aminophenol
24.49 (b)
(a)
HOCH2CH2CH2NH2 e
d
a
c
b
a=
1.68 6
b =
2.69 5
c=
2.89 5
d = 3.72 5
(CH 3 0)2CHCH 2 NH2 c
d
b
a
a=
1.29 6
b =
2.78 6
c=
3.39 6
d = 4.31 6
682
Chapter 24
General Problems
24.50
CHCHg
CH 3
a
b,
A
c
aCHCH NCH
CH_CH2
2
£
3
^j^ NH
(a) 3 NaBH4 (b) excess CH 3 I; (c) Ag 2 0, H 2 0, heat; (d) RCO3H (e) (CH 3) 2 NH. Step (e) is an Sn2 ring opening of the epoxide by nucleophilic substitution of the amine at the primary carbon. ,
;
24.51
NO (J Oxazole
is
an aromatic 6
electron are in
p
jt
electron heterocycle.
orbitals that are part
from each carbon.
Oxazole
Two oxygen electrons
of the n electron system of the
and one nitrogen along with one
ring,
2
An oxygen
lone pair and a nitrogen lone pair are in sp orbitals that lie in the plane of the ring. Since the nitrogen lone pair is available for donation to acids, oxazole is more basic than pyrrole. electron
24.52 H
:o'
:o: II
R'
A^-Protonation
(no resonance stabilization)
I
R'
R'
NH 2
O-Protonation (resonance stabilization)
Protonation occurs on oxygen because an O-protonated amide
is
stabilized
by resonance.
Amines and Heterocycles
683
24.53 HpC
HpC \
/
C—
R— Cw
R'
CHp
CHp
HpC
R— C ^C— R'
Ik
\
i
//
O
o
,:0:
^ H—
nh 3
Ho 3.
tl
HpC
— CH
HoC—
CHp
HpC
R
"9
C
rC— R*
:0:
~, R'
\
I
r-<5
O
:kH
H
HoO
+
~xP
O H ? 0> NH 2
L.NH2
«« HoC— CH
R— C^ C— /II
HoC— CH 2
W
/
(
R'
/
Qokt:nh 2
R
w-
H 2 C— CH
w
;c
H.q:
c
R
R-
*
+
CX
H 2 (X^
.
V C-R"
IJJ
H H
H 9.
* H
rU C— CH
HC— CH
w
w
//
R—
C—
N N
tl
R— C.
R'
+
10.
.. '
Steps Steps Steps Steps Steps
R'
N
I
H
.C—
H
+
H 2°
1,6: Protonation. 2,7: Nucleophilic addition.
3,8: Proton transfer. 4,9: Loss of water. 5,10: Loss of proton.
The mechanism
consists of the nucleophilic addition of ammonia, first to one of the ketones, and then to the other, with loss of two equivalents of water.
.
684
Chapter 24
24.54
HQ
CHo
H2
qp/CH3 c/i.
CCH2.
3.
V. NHoOH
H 2C
NHOH
HoC \
/CCHg
,CCHo
CCH-
o
o
o 4.
tl
CH 3
CH<
CH 3 ^:b
CHo3
H
I
N
HoC
^
+
CH<
HO:
HoC
\
\p:OH
OH /
f
HoC
u
H
v NOH
5.
H 2 C'
\N0H
CCHo6
.CCHo
//
CH 3
O
+
H2
tl
10.
9.
HoC
CH 3
CHo
CH<
/%
HC
I
\\
C O
H 2 COCH 3
r H2
+
I
CH 2 CHCH 3
+
Icr03 H 3
CHo3
|
OH
H 2 C— CHCH 3
3
NaBH„
Cyclopentamine
/
[
— CH CCH 2
3
The
last step of the synthesis is a reductive amination of a ketone that is formed by oxidation of the corresponding alcohol. The alcohol results from the Grignard reaction between cyclopentylmagnesium bromide and propylene oxide.
24.60 (a)
COCH 2 CH 2 N(CH 3
CH 3 (CH 2 3 NH )
CH 3 CH 2 CH 2 CHO );
NaBH,
O
COCH 2 CH 2 N(CH 3 2
OoN
)
1
(b)
.
2.
H 2>
Pt
e^qh
II
» H 2N
COCH 2 CH 2 N(CH 3
SOCI 2
(c)
HOCH 2 CH 2 N(CH 3
)
2
)i
Benzene
CH 3 CI
,
Pyridine
AICI 3
O II
OoN\\
The
/
COH-«
KMn0 ± 4 H2
HNO'
OoN d
H 2 S0 4
\
CH<
//
synthesis in (a) is achieved by a reductive amination reaction. Reactions in (b) include formation of an acid chloride, esteriflcation, and reduction of the nitro group.
688
Chapter 24
24.61
Atropine
Tropidene
Tropine
ch 2OH
q
H0 2CCHC 6 H 5 CH 2 OH
Tropic acid
We know the location of the -OH group of tropine because optically inactive alcohol. This hydroxyl group results that is
composed of tropine and
it is stated that tropine is an from basic hydrolysis of the ester
tropic acid.
24.62
1
.
CH 3 I
1
.
CH3I
2.
Ag 2 Q, H 2 Q
2.
Ag 2 Q, H 2Q
3.
heat
3.
heat
Tropidene Tropilidene results from two cycles of
24.63 The formula C9H17N
N(CH 3 Tropilidene
Hofmann
elimination on tropidene.
indicates two degrees of unsaturation in the product. Both are probably due to rings since the product results from catalytic reduction.
Step 1: Reduction of nitrile Step 2: Nucleophilic addition. Step 3: Dehydration. Step 4: Reduction of double bond.
)
3
Amines and Heterocycles
689
24.64 The molecular formula indicates that coniine has one double bond or ring, and the Hofmann elimination product shows that the nitrogen atom is part of a ring. 1
.
excess
CH 3 I
2.
Ag 2 Q, H 2 Q
3.
heat
(CH 3 2 N )
5-(N, A^Dimethylamino)- 1 -octene
Coniine
24.65
C0 2 Et
HC0 2 Et
CH 2 CH0CH0CH9C 3 2 2jf
O
+
1
CHo2
.
2.
II
CHCN
Na
"OEt, +
H3
EtOH
CH 2
CHoCHoCHoC 3 2 ZJj
|
CH 2 CN
O
t.
H3
H
H
H H 2 ,Pt
CH 2
4.
O H2
H
Step
1:
Michael addition.
Step 2: Ester hydrolysis; decarboxylation.
Step
3: Reduction.
Step 4: Cyclization.
Step 5: Reduction.
3.
CH 3 CH 2 CH 2 C
O
CH 2
N-^
Coniine
,
heat
H
\ /
\ /
CHoCHpCHoC
+
+
h"
C0 2
CH 2 CH 2 CN + EtOH
690
Chapter 24
24.66
OH
OH
OH
OH
Tyramine
When you
see
-CH2NH2,
substitution of a benzylic
think of the reduction of a
nitrile.
The
nitrile
comes from
bromide by ~CN.
24.67
The
reactive intermediate is benzyne, which undergoes a Diels-Alder reaction with cyclopentadiene to yield the observed product.
Amines and Heterocycles
691
24.68
CH 3 I
1
2.
Ag 2 Q, H 2 Q
2.
Ag 2 Q, H 2 Q
3.
heat
3.
heat
1
I
N j
.
N(CH 3
Two
successive cycles of
Hofmann
CH3I
.
N(CH 3
)
3
Cyclooctatriene )
2
elimination lead to formation of cyclooctatriene.
NBS EtOH Cyclooctatetraene Allylic bromination followed
by elimination
yield cyclooctatetraene.
24.69 H HC)
-V
H
f?
C
"nh2
..
c
H fX
.^c
+ +
Br2
,
NH 3
NaOH
H2
Hofmann rearrangement
(the mechanism is shown in Section 24.6) of an a-hydroxy amide produces a carbinolamine intermediate that expels ammonia to give an aldehyde.
692
Chapter 24
24.70
OCH 3
eeri-
^rj:NH2 CH3
I-
er NHCH 3 /\
O
HH 3.
I -
..
:o:
H+ :OCH 3
:0:
OCHo
r-X.
NH-CH3
N- CH 3 5.
O Step Step Step Step Step
1:
+
4.
o
CH3OH
Conjugate addition of amine.
2: Proton transfer. 3: Nucleophilic addition of amine. 4: Proton transfer. 5: Elimination of methanol.
24.71 Br
C0 2 CH 3
CO0CH0
HoC 2
1.
OH Et 3 N tl
Y
'
Br
2.
:
CO0CH3
I"
N
+
C0 2 CH 3 "*
+ OH
3.
C0 2CH 3 -
Step 1: Sn2 displacement of Br by amine. Step 2: Deprotonation. Step 3: Conjugate addition of alcohol. Step 4: Proton transfer.
°H
+ Et 3 NH
n
Amines and Heterocycles
24.72
(5)-Norcoclaurine
Step
1:
Nucleophilic addition of the amine to the protonated aldehyde.
Step 2: Proton
transfer.
Step 3: Loss of water. Step 4: Electrophilic aromatic substitution.
Step
5:
Loss of proton.
693
694
24.73
Chapter 24
Amines and Heterocycles
Steps
1-3: El
Steps
4-6: Sn2
elimination (protonation, loss of
HOCH3,
deprotonation).
substitution (protonation, substitution, deprotonation).
Step 7: El elimination of carbamate. Steps 8-9: conjugate addition of
DNA (addition, deprotonation).
Notice that five of the nine steps are either protonations or deprotonations.
695
Width: 612 Height: 792
696
Chapter 24
24.74 (a)
N— CH 2
N(CH 3 2
HoC
)
-
b
Ch^CHg b
dore
a
a=
2.25 6
a=
1.14 6
b =
2.89 6
b =
3.40 6
c =
6.66 6, 7.03 6
c =
4.47 6
d =
6.65 6
d ore
e = 7.24 6
24.75
\||
I
.
-NHg
R
H
R
NH 2 I
^C R ^C0 2 H H
:OH
:0:
Go:
^ CC-NH 3
H3 +
*4.
NH 2
\
H
H
NHo
Cnh 2
I
«-
R
CCN H
A
.C^
C
3.
R
:CN
H
Step 1: Addition of NH 3 Step 2: Elimination of ~OH. Step 3: Addition of ~CN. Step 4: Hydrolysis of nitrile. .
The mechanism of acid-catalyzed
nitrile
hydrolysis
is
shown
in
Problem 20.51.
Amines and Heterocycles
24.76
Step
1:
Conjugate addition of hydrazine.
Step 2: Proton Step
transfer.
3: Nucleophilic acyl substitution,
Step 4: Proton
transfer.
forming the cyclic amide.
697
Review Unit
9:
Carbonyl Compounds
Reaction at the
II
a Carbon; Amines
Major Topics Covered (with vocabulary): Carbonyl a-substitution reactions: tautomerism tautomer enolate ion Hell-Volhard-Zelinskii reaction (3-diketone p-keto eater malonic ester synthesis acetoacetic eater synthesis LDA
a-substitution reaction
Carbonyl condensation reactions: carbonyl condensation reactions aldol reaction enone mixed aldol reaction Claisen condensation reaction Dieckmann cyclization Michael reaction Michael acceptor Michael donor Stork enamine reaction Robinson annulation reaction
Amines: primary, secondary, tertiary amine quaternary ammonium salt arylamine heterocyclic amine pyramidal inversion K\> azide synthesis Gabriel amine synthesis reductive amination Hofmann rearrangement Curtius rearrangement Hofmann elimination reaction
arenediazonium salt diazotization Sandmeyer reaction azo compound diazonium coupling reaction pyrrole thiophene furan pyridine fused-ring heterocycle pyrimidine
purine
nitrogen rule
Types of Problems: After studying these chapters, you should be able to:
-
Draw keto-enol
tautomers of carbonyl compounds, identify acidic hydrogens, and draw the resonance forms of enolates. Formulate the mechanisms of acid- and base-catalyzed enolization and of other a-substitution reactions.
-
Predict the products of a-substitution reactions.
-
Use
-
Predict the products of carbonyl condensation reactions. Formulate the mechanisms of carbonyl condensation reactions.
-
Name
-
Predict the basicity of alkylamines, arylamines and heterocyclic amines. Synthesize alkylamines and arylamines by several routes.
-
-
a-substitution reactions in synthesis.
Use carbonyl condensation
reactions in synthesis.
and draw amines, and classify amines as primary, secondary, tertiary arylamines, or heterocyclic amines.
,
quaternary,
Predict the products of reactions involving alkylamines and arylamines.
Use diazonium
involving arylamines, including diazo coupling reactions. and explain their acid-base properties. Explain orientation and reactivity in heterocyclic reactions, and predict the products of reactions involving heterocycles. Propose mechanisms for reactions involving alkylamines, arylamines, and heterocycles. Identify amines by spectroscopic techniques.
Draw
salts in reactions
orbital pictures of heterocycles
Review Unit 9
Points to *
699
Remember:
unusual to think of a carbonyl compound as an acid, but the protons a to a carbonyl group can be removed by a strong base. Protons a to two carbonyl groups are even more acidic: in some cases, acidity approaches that of phenols. This acidity is the basis for a-substitution reactions of compounds having carbonyl groups. Abstraction by base of an a proton produces a resonance-stabilized enolate anion that can be used in alkylations involving alkyl hahdes and It is
tosylates.
*
Alkylation of an unsymmetrical LDA-generated enolate generally occurs at the less hindered carbon.
*
When you
need
to synthesize a (3-hydroxy
a
ketone or aldehyde or an a,p-unsaturated ketone or
aldehyde, use an aldol reaction. When you need to synthesize a p-diketone or p-keto ester, use a Claisen reaction. When you need to synthesize a 1,5-dicarbonyl compound, use a Michael reaction. The Robinson annulation is used to synthesize polycyclic molecules by a combination of a Michael reaction with an aldol condensation.
many of the mechanisms in this group of chapters, the steps involving proton transfer are not explicitly shown. The proton transfers occur between the proton and the conjugate base with the most favorable pK of those present in the solution. These steps have been omitted at times to simplify the mechanisms.
*
In
*
In the Claisen condensation, the enolate of the p-dicarbonyl
compound
is
treated with
H3O"
to
yield the neutral product. *
For an amine, the larger the value of pKa of its
ammonium ion,
the stronger the base.
The
smaller the value of pA'b of the amine, the stronger the base. *
The Sandmeyer
reaction allows the synthesis of substituted benzenes that can't be formed by These reactions succeed because N2 is a very good
electrophilic aromatic substitution reactions.
leaving group.
Self-Test:
O
o
CH3CH2 H
(CH 3 )2CHCH 2 CH2
||
A
into
how
they?
B
Pentymal
Dypnone
(a sedative)
(sunscreen)
The six-membered
What are
O
ring in
A is formed by the cyclization of two difunctional compounds.
What type of reaction
occurs to form the ring? The two alkyl groups are introduced to cyclization. What type of reaction is occurring, and type of reaction occurs in the formation of Dypnone (B)? Why might B
one of the difunctional compounds prior is it
carried out?
What
be effective as a sunscreen?
Review Unit 9
700
o
CHoCHNCH.
CH
;
C Benzphetamine
Butralin (an herbicide)
(an appetite suppressant)
What type of amine is C? Do you expect it to be more or less basic than ammonia? Than What product do you expect from Hofmann elimination of C? What significant absorptions might be seen in the IR spectrum of C? What information can be obtained from the aniline?
mass spectrum? Plan a synthesis of D from benzene.
Multiple Choice: 1
.
Which of the following compounds has (a)
2
.
2-Pentanone
(b)
3-Pentanone
(c)
four acidic hydrogens? Acetophenone (d) Phenylacetone
In which of the following reactions is an enol, rather than an enolate, the reacting species? alkylation (d) (a) acetoacetic acid synthesis (b) malonic ester synthesis (c) Hell-Volhard-Zelinskii reaction
LDA
3
4
.
.
Cyclobutanecarboxylic acid is probably the product of a: (a) malonic ester synthesis (b) acetoacetic ester synthesis (d) Hell-Volhard-Zelinskii reaction
An LDA (a)
5
.
6
.
alkylation can be used to alkylate
aldehydes
(b) ketones
(c) esters
all
(c)
LDA alkylation
of the following, except:
(d) nitriles
you want to carry out a carbonyl condensation, and you don't want to form a-substitution product, you should: (a) lower the temperature (b) use one equivalent of base (c) use a catalytic amount of base (d) use a polar aprotic solvent If
Which reaction forms a cyclohexenone? (a) Dieckmann cyclization (b) Michael reaction
(c)
Claisen condensation
(d) intramolecular aldol condensation
7
.
All of the following molecules are good Michael donors except: (a) Ethyl acetoacetate (b) Nitroethylene (c) Malonic ester (d) Ethyl 2-oxocyclohexanecarboxylate
8
.
The ammonium ion of which of the following amines has (a)
9
.
10.
Methylamine
(b)
Trimethylamine
(c)
Aniline
(d)
the smallest value of
pKa ?
p-Bromoaniline
All of the following methods of amine synthesis are limited to primary amines, except: (a) Curtius rearrangement (b) reductive animation (c) Hofmann rearrangement (d) azide synthesis
To form an azo compound, an aryldiazonium (a)
CuCN
(b)
benzene
(c) nitrobenzene
(d)
should react with: phenol
salt
Chapter 25 - Bio molecules: Carbohydrates
Chapter Outline I.
Classification of carbohydrates (Section 25.1).
A. Simple vs. complex: Simple carbohydrates (monosaccharides) can't be hydrolyzed to smaller units. 1 2. Complex carbohydrates are made up of two or more simple sugars linked together. .
a.
b
.
A disaccharide is composed of two monosaccharides. A polysaccharide is composed of three or more monosaccharides.
B. Aldoses vs. ketoses: 1 A monosaccharide with an aldehyde carbonyl group is an aldose. 2 A monosaccharide with a ketone carbonyl group is a ketose. C. Tri-, tetr- pent-, etc. indicate the number of carbons in the monosaccharide. Monosaccharides (Sections 25.2-25.7). A. Configurations of monosaccharides (Section 25.2-25.4). .
.
,
II.
1
.
Fischer projections (Section 25.2). a. Each chirality center of a monosaccharide
is
represented by a pair of crossed
lines.
The horizontal line represents bonds coming out of the page. The vertical line represents bonds going into the page. Allowed manipulations of Fischer projections:
i.
ii.
b
.
i.
A Fischer projection can be rotated on the page by
1
80°, but not
by 90° or
270°. ii.
Holding one group steady, the other three groups can be rotated clockwise or counterclockwise.
c.
d.
Rules for assigning R,S configurations. i. Assign priorities to the substituents in the usual way (Section 5.5). ii. Perform one of the two allowed motions to place the lowest priority group at the top of the Fischer projection. iii. Determine the direction of rotation of the arrow that travels from group 1 to group 2 to group 3, and assign R or S configuration. Carbohydrates with more than one chirality center are shown by stacking the centers on top of each other. i The carbonyl carbon is placed at or near the top of the Fischer projection. .
2.
D,L sugars (Section 25.3). a. (/?)-Glyceraldehyde is also
b
.
In
D
sugars, the
known
as D-glyceraldehyde.
-OH group farthest from the carbonyl group points to the right
in a Fischer projection. i.
3
.
Most naturally-occurring sugars are D sugars. L sugars, the -OH group farthest from the carbonyl group points
c.
In
d
a Fischer projection. D,L designations refer only to the configuration farthest from the carbonyl
.
to the left in
carbon and are unrelated to the direction of rotation of plane-polarized Configurations of aldoses (Section 25.4). a. There are 4 aldotetroses - D and L erythrose and threose. b. c.
light.
There are 4 D,L pairs of aldopentoses: ribose, arabinose, xylose and lyxose. There are 8 D,L pairs of aldohexoses allose, altrose, glucose, mannose, gulose, idose, galactose, and talose. :
702
Chapter 25
d
.
A scheme for drawing and memorizing the D-aldohexoses: i
.
ii.
Draw Draw
all
the
-OH groups at C5 pointing to the right. first four -OH groups at C4 pointing to the right and the second
four pointing to the left. Alternate -OH groups at C3: two right, two left, two right, two left. iv. Alternate -OH groups at C2: right, left, etc. v Use the mnemonic " All .altruists .gladly make .gym in .gallon Xanks " to assign hi.
.
names. B. Cyclic structures of monosaccharides (Section 25.5). Hemiacetal formation. 1 a. Monosaccharides are in equilibrium with their internal hemiacetals. i. Glucose exists primarily as a six-membered pyranose ring, formed between the -OH group at C5 and the aldehyde group at CI Fructose exists primarily as a five-membered furanose ring. ii. b Structure of pyranose rings. Pyranose rings have a chair-like geometry. i. .
.
ii. iii.
iv.
The hemiacetal oxygen is at the right rear for D-sugars. group on the right in a Fischer projection is on the bottom face pyranose ring, and an -OH group on the left is on the top face. For D sugars, the -CH2OH group is on the top.
An -OH
Mutarotation. a. When a monosaccharide cyclizes, a new chirality center The two diastereomers are anomers. i.
2.
is
in a
generated.
The form with the anomeric -OH group trans to the -CH2OH group is the a anomer (minor anomer). iii. The form with the anomeric -OH group cis to the -CH2OH group is the p anomer (major anomer). When a solution of either pure anomer is dissolved in water, the optical rotation ii.
b
.
of the solution reaches a constant value. This process is called mutarotation. i. ii. Mutarotation is due to the reversible opening and recyclizing of the hemiacetal ring and is catalyzed by both acid and base. C. Reactions of monosaccharides (Section 25.6). 1
.
Ester and ether formation.
by treatment with an acid anhydride or acid chloride. Ethers are formed by treatment with methyl iodide and Ag20. c Ester and ether derivatives are crystalline and easy to purify. Glycoside formation. a. Treatment of a hemiacetal with an alcohol and an acid catalyst yields an acetal. Acetals aren't in equilibrium with an open-chain form and do exhibit i. mutarotation. Aqueous acid reconverts the acetal to a monosaccharide. ii. b. These acetals, called glycosides, occur in nature. c. Glycosides are named by first citing the alkyl group and then replacing the -ose suffix of the sugar with -oside. d. The laboratory synthesis of glycosides is achieved by the Koenigs-Knorr a.
b
Esterification occurs
.
.
2
.
reaction. i.
ii.
iii.
Treatment of the acetylpyranose with HBr, followed by treatment with the appropriate alcohol and Ag20, gives the acetylglycoside. Both anomers give the same product. The reaction involves neighboring-group participation by acetate.
Biomolecules: Carbohydrates
3
.
703
Phosphorylation.
Monosaccharides can be phosphorylated by ATP to form a glycosyl monophosphate. b The resulting glycosyl monophosphate can react with a second nucleoside triphosphate to produce a glycosyl diphosphate. This product can react with a lipid or a protein to form a glycoconjugate. c. 4. Reduction of monosaccharides. a. Reaction of a monosaccharide with NaBFLt yields an alditol (a poly alcohol). Oxidation of monosaccharides. 5 Several mild reagents can oxidize the carbonyl group to a carboxylic acid a. a.
.
.
(aldonic acid).
Br2
i.
In the laboratory, aqueous
ii.
Historically, Tollens reagent, Fehling's reagent
is
used to oxidize aldoses (not ketoses). and Benedict's reagent have
served as tests for reducing sugars. All aldoses and some ketoses are reducing sugars, but glycosides are nonreducing. b. The more powerful oxidizing agent, dilute HNO3, oxidizes aldoses to dicarboxylic acids (aldaric acids). Enzymes can oxidize the -CH2OH of a monosaccharide (with oxidizing the c. aldehyde) to form a uronic acid. 6. Chain-lengthening: the Kiliani-Fischer synthesis. a. In the Kiliani-Fischer synthesis, an aldehyde group becomes C2 of a chainlengthened monosaccharide and the added carbon is the new CI. b The reaction involves cyanohydrin formation, reduction and hydrolysis. The products are two diastereomeric aldoses that differ in configuration at C2. c. 7 Chain-shortening: the Wohl degradation. a. The Won! degradation shortens an aldose by one carbon. b The reaction involves treatment of the aldose with hydroxylamine, dehydration and loss of from the resulting cyanohydrin. D. Eight essential monosaccharides (Section 25.7). 1 Glucose, galactose, mannose and xylose are aldoses. 2 Fucose is a deoxy sugar. 3 N-Acetylglucosamine and ^-acetylgalactosamine are amino sugars. 4. A^-Acetylneuraminic acid is the parent compound of the sialic acids. All of the essential monosaccharides arise from glucose. 5 HI. Other carbohydrates (Sections 25.8-25.1 1). A. Disaccharides (Section 25.8). 1 Cellobiose and maltose. ii.
.
.
.
HCN
.
.
.
.
.
Cellobiose and maltose contain a l-» 4-glycosidic acetal bond between two glucose monosaccharide units.
a.
Maltose consists of two glucopyranose units joined by a l-» 4-a-glycosidic bond. c. Cellobiose consists of two glucopyranose units joined by a l-» 4-(3-glycosidic bond. d. Both maltose and cellobiose are reducing sugars and exhibit mutarotation. Humans can't digest cellobiose but can digest maltose. e.
b
2
.
.
Lactose. a.
Lactose consists of a unit of galactose joined by a (3-glycosidic bond between
CI and C4 of a glucose b.
Lactose
is
unit.
a reducing sugar found in milk.
704
Chapter 25 Sucrose. Sucrose is a disaccharide that yields glucose and fructose on hydrolysis. a. Sucrose is called "invert sugar" because the sign of rotation changes when i. sucrose is hydrolyzed. ii. Sucrose is one of the most abundant pure organic chemicals in the world. b. The two monosaccharides are joined by a glycosidic link between CI of glucose and C2 of fructose. c. Sucrose isn't a reducing sugar and doesn't exhibit mutarotation. B. Polysaccharides and their synthesis (Section 25.9). Polysaccharides have a reducing end and undergo mutarotation, but aren't 1 considered to be reducing sugars because of their size. 2 Important polysaccharides. 3.
.
.
a.
b.
Cellulose. i.
Cellulose consists of thousands of D-glucose units linked by l-> 4-p-
ii.
glycosidic bonds. In nature, cellulose
is
used as structural material.
Starch. i.
Starch consists of thousands of D-glucose units linked by 1-* 4-a-glycosidic
bonds. ii.
Starch can be separated into amylose (water-soluble) and amylopectin (water-insoluble) fractions. (a).
iii.
Amylopectin contains l-» 6-a-glycosidic branches. is digested in the mouth by glycosidase enzymes, which only cleave
Starch
a-glycosidic bonds. c.
3
.
An a.
b
.
c.
Glycogen. Glycogen i.
is
an energy-storage polysaccharide.
Glycogen contains both 1^-4- and l-> 6-links. ii. outline of the glycan assembly method of polysaccharide synthesis. A glycal (a monosaccharide with a C1-C2 double bond) is protected at C6 by formation of a silyl ether and at C3-C4 by formation of a cyclic carbonate ester. The protected glycal is epoxidized. Treatment of the glycal epoxide (in the presence of ZnCl2) with a second glycal having a free C6 hydroxy 1 group forms a disaccharide. The process can be repeated.
d Other important carbohydrates (Section 25.10). C. Deoxy sugars have an -OH group missing and are components of nucleic 1 2 In amino sugars, an -OH is replaced by a -NH2. Amino sugars are found in chitin and in antibiotics. a. surface carbohydrates and influenza viruses (Section 25.11). D. Cell .
.
acids.
.
1
.
Polysaccharides are involved in cell surface recognition. a. Polysaccharide markers on the surface of influenza viruses are variants of two types of glycoproteins -hemagglutinin (H - Type 5 or Type 1), and
neuraminidase (N - Type
b
.
Infection occurs
by c.
when
1).
a virus binds to a receptor on a target cell and
is
engulfed
the cell.
New viral particles are produced, pass out of the cell,
and are held
to surface
receptors.
d
.
A neuraminidase enzyme cleaves the receptor-virus bond, allowing the virus to invade a
2
.
new
cell.
Antiviral vaccines block the neuraminidase enzyme, limiting the spread of the virus.
Biomolecules: Carbohydrates
705
Solutions to Problems
25.1 (b)
(a)
(d)
(c)
?
r° H^C-*OH
HO^Cj^H
H^C-OH
r° HO*- C-" H
I
CH 2 OH
I
H^C—H H^C-OH
I
H»-C-OH
I
Hy>0
CH 2 OH
CH 2 OH
HO"-
I
H
I
H^
I
H^C- OH
CH 2 OH
CH 2 OH
I
CH 2 OH Threose
an
25.2
Ribulose
2-Deoxyribose
Tagatose
a ketopentose
aldotetrose
an aldopentose
a ketohexose
Horizontal bonds of Fischer projections point out of the page, and vertical bonds point into the page. (a)
C0 2 H
(b)
C0 2 H
90 2 H H
H2 N
-H
H 2 N-
NH,
HoC
CH<
CHc
CHO
CHO OH
CHO R
i
'
i
H
OH
H-
CH 3
i
(c)
CH<
CH 3
CH<
CH 3
decide
.it CH2CHg :ho CH
CH 2 CHg
two Fischer projections
if
CH-
H j
CH 2 CH3 To
HO
'^CHO
H
-CHO
H"
25.3
OH
are identical, use the
superimpose two groups of each projection. after rotation, the projections represent the
If the
two allowable
rotations to
remaining groups are also superimposed
same enantiomer.
in both A and B, keep it steady, and rotate the other groups are superimposed, the two projections are identical. If only two groups are superimposed, the projections are enantiomers. Thus, A (a)
Since
-H
is in
three groups. is
If,
the
same position
after rotation, all
identical to B.
A HO
B
CHO r I
Ml
OH H
=
HOCH 2
1
steady
CH 2 OH
CHO
H
Width: 612 Height: 792
706
Chapter 25
B
OH
HOCH
H
r
H
CHO
D
CH 2 OH
HO-
CHO
steady
B
OH
CH 2OH 180°
CHO
OHC-
»
OH
HOCH,
H
CHO
CH 2 OH
B
Projections A,
25.4
+
OH
and
C
are identical, and
D is their enantiomer.
Rotate the structure 1 80° around the horizontal axis to arrive at a drawing having the hydrogen at the rear. Assign the R,S configuration as usual, and draw the Fischer projection
H
CI
T
HOCH 2
— C~CH
HOCHo*^ 3
'
^ CH<
R
H
25.5
CH<
HOCHc CI
the skeleton of the Fischer projection and add the -CHO and -CH2OH groups to the top and bottom, respectively. Look at each carbon from the direction in which the -H and -OH point out of the page, and draw what you see on the Fischer projection.
Draw
View C3 from
-OH
this side;
on the
is
right.
\
CHO
HO H
M
1
.a
HO^
A
CHO
H*-l— OH
i\
HOH
H H
I
CH 2 OH
t View C2 from
-OH
is
on the
H»-j— OH
CHO R OH R OH
this side; right.
CH 2OH
Hi onio ecu Its: I
25.6
The hydroxy! group bonded the right in a
D
S
HO-
.
H
H
CH 2 OH
(c)
—
R
C=0
H
)
CH 2 OH
sugar.
—
S
HO-
L
CHO R OH
(b)
HO-
R
OH
H-
CH 2 OH
OH
CH 2 OH
D-Xylose
L-Erythrose
D-Xylulose
25.7
CHO R OH
H-
HO-
H
HO-
H
L-(+)-Arabinose
CH 2 OH 25.8
CHO
(a)
HO-
HO-
H
H-
CHO
(b)
CHO
(c)
H
HO-
OH
H-
OH
HO-
H
H-
OH
HO-
H
HO-
HO-
CH 2 OH
HO-
CH 2OH
CH 2 OH L-Xylose
25.9
An
L-Galactose
aldoheptose has 5 chirality centers. Thus, there are 2
L-Allose 5
= 32
aldoheptoses
aldoheptoses and 16 L aldoheptoses.
25.10 See Problem 25.5 View C3 from
-OH
is
for the
method of solution.
this side;
on the
right.
CHO
OH OH
— R OH — R OH — R OH
CH 2 OH
CH 2 OH
1
Hdta
OHC
2
.a
C
3
C
H OH
H OH
t
View C2,C4 from
-OH
is
CH 2 OH
l\
l\
t
4
on the
this side;
right.
707
carbon farthest from the carbonyl group points to
sugar, and points to the left in an
CHO
(a)
to the chiral
Carbohydrates
-OH
D-Ribose
- 16 D
708
Chapter 25
25.11 The
steps for drawing a furanose are similar to the steps for drawing a pyranose. Ring formation occurs between the -OH group at C4 and the carbonyl carbon. groups on the right in a Fischer projection is on the bottom face of the ring.
A
CHO R OH R OH R OH
— — —
CH 2 OH
HOCH O
OH D-Ribose
OH
(Furanose form)
of fructose results from ring formation between the -OH group at C5 and the C2. In the a anomer, the anomeric -OH group is trans to the C6 -CH2OH group, and in the /j anomer the two groups are cis. In the pyranose form, cyclization occurs between the -OH group at C6 and the ketone. The more stable chair conformations are
25.12 The furanose ketone
at
shown.
a-D-Fructopyranose
a-D-Fructofuranose
/3-D-Fructopyranose
/3-D-Fructofuranose
Biomolecules: Carbohydrates
709
25.13 There
are two ways to draw these anomers: (1) Following the steps in Worked Example draw the Fischer projection, lay it on its side, form the pyranose ring, and convert it to a chair, remembering that the anomeric -OH group is cis to the C6 group; (2) Draw p-Dglucopyranose, and exchange the hydroxyl groups that differ between glucose and the other two hexoses.
25.3,
a
/3-D-Mannopyranose
/3-D-Galactopyranose
p-D-Galactopyranose and (3-D-mannopyranose each have one hydroxyl group position and are therefore of similar stability.
25.14
In the previous problem
we drew /?-D-galactopyranose. In this problem, invert the D enantiomer and perform a ring-flip to
configuration at each chirality center of the the structure of the
in the axial
arrive at
L enantiomer.
/3-D-Galactopyranose
/3-L-Galactopyranose
All substituents, except for the
-OH at C4,
are equatorial in the
more
stable conformation
of /3-L-galactopyranose.
25.15 From the model, we can
form of a D-hexose. It is an a-anomer because the anomeric hydroxyl group is trans to the group at C6. Comparing the model with a-D-glucopyranose, we see that all groups have the same axial/equatorial relationship, except for the hydroxyl group at C3, which is axial in the model and equatorial in a-D-glucopyranose. The monosaccharide is a-D-allopyranose. Use see that the monosaccharide
Figure 25.3 as a reference.
OH
HOCH 2
OH OH
OH
OH
a-D-Allopyranose
OH CH 2 OH D-Allose
is
the pyranose
710
Chapter 25
25.16
CH 3 OCH 2
CH 3I,Ag 2
(a)
OCH,
HOCH
OCH 3 OH
OCH3
OH
/?-D-Ribofuranose
\
(b)
(c^CO^O, pyridine
CHoCO 3 H
25.17
CHO H-
OH
H-
H
LNaBH^
H-
OH
2.H 2
H-
OH
HO-
OH
OH
H-
-OH
H-
H
HO-
H
1.
NaBH,
HO-
-H
H-
OH
HO-
H
2.
H2
HO-
-H
H-
OH
H-
HO-
CH 2 OH
CH 2 OH
D-Glucose
D-Glucitol
Reaction of D-galactose with
CH 2 OH
CHO
CH 2 OH
NaBH4 yields
OH
CH 2 OH
CH 2 OH D-Galactose
an
alditol that
-OH
H-
Galactitol
has a plane of symmetry and
is
a
meso compound.
25.18
CHO H-
HO-
9H 2 OH
OH
H-
H
1
H-
OH
2.H 2
H-
OH CH 2 OH
D-Glucose
.
NaBH 4
HO-
-OH
HO-
H
-H
HO-
H
H-
-OH
H-
H-
-OH
HO-
CH 2 OH D-Glucitol
CHO
CH 2 OH
1
.
HO-
H
NaBH 4 HO-
H
OH 2.H 2 H
CH 2 OH
H-
HO-
OH H
CH 2 OH L-Gulose
Reaction of an aldose with NaBH4 produces a polyol (alditol). Because an alditol has the same functional group at both ends, two different aldoses can yield the same alditol. Here, L-gulose and D-glucose form the same alditol (rotate the Fischer projection of L-gulitol 180° to see the identity).
Biomolecules: Carbohydrates
711
25.19
CHO
C0 2 H
OH
H-
HO-
H
H-
OH
H-
OH
-OH
H-
OH
-H
H-
OH
H-
-OH
H-
OH
H-
-OH
H-
OH
H-
HO-
HNO,
dil.
heat
CH 2 OH
C0 2 H
D-Glucose
Glucaric acid
25.20
dil.
HNO;
H-
OH
H-
OH
H-
OH
H-
OH
heat
CH 2 OH
C0 2 H
D-Allose
Allaric
acid
symmetry and no symmetry plane.
Allaric acid has a plane of
acid has
C0 2 H
CHO
is
an optically inactive meso compound. Glucaric
D-Allose and D-galactose yield meso aldaric acids. All other D-hexoses produce optically active aldaric acids on oxidation because they lack a plane of symmetry.
25.21 The
products of Kiliani-Fischer reaction of D-ribose have the same configuration at C3,
C4 and C5
Use Figure 25.3
as D-ribose.
as a reference.
CHO CHO 1.
HCN
OH
2.
H 2 Pd
H-
OH
3.
H3
H-
OH
H-
,
catalyst
+
CHO
H-
OH
HO-
H-
OH
H-
OH
H-
OH
H-
OH
H-
OH
H-
OH
H
CH 2 OH
CH 2 OH D-Ribose
D-Allose
CH 2 OH D-Altrose
25.22 The aldopentose, L-xylose has the same configuration as the configuration C5 of L-idose and L-gulose. Use Figure 25.3 as a reference.
CHO CHO
H1
HOH-
HO-
.
H
OH H
CH 2 OH L-Xylose
CHO
OH
HO-
H
H
HO-
H
HCN
2.
H 2 Pd
3.
H3
,
+
HOcatalyst
H-
HO-
OH H
CH 2 OH L-Idose
H-
HO-
OH H
CH 2 OH L-Gulose
at
C3,
C4 and
712
Chapter 25
25.23 The aldopentoses have
the
CHO HO-
H
1
H
or
OH
H"
C3 and C4
as D-threose.
H 2 NOH
.
HOH-
CH 2 OH
H
OH
3.
CH 3 C0 2"Na+ + Na ~OCH 3
CH 2 OH
HO-
H
OH
H-
CH 2OH
D-Lyxose
D-Xylose
CHO
(CH^CO)pO,
2.
HO
at
CHO
OH
H-
same configurations
D-Threose
25.24 COc
c=o I
HoC
—
> Base
COc "
C=0
C0 2
c=o
C=0
CH2 H-
"
C0 2
H .C "^ O:
CH 2 H-
-OH
CH3CONH
CH3CONH-
H
CH3CONH-
-H
HO
HO-
H
HO-
-H
CH 2 OH A^-Acetylmannosamine
H-
OH
H-
-OH
H-
OH
H-
-OH
CH 2 OH
CH 2 OH N-Acetyl-D-neuraminic acid
Hi onio ecu Its: I
Carbohydrates
713
25.25
CH 2 OH
CH 2 OH 2.H 2
OH CH 2 OH
HO Cellobiose
(b)
CH3COCI
CH 2OH
CH 2 OH
^AjtOH
O
o
H
HoO*
OH
A cO
(c)
Ac0
pyridine
OAc OAc
OAc
Ac = CH 3 C
Visualizing Chemistry
25.26
Convert the model to a Fischer projection, remembering that the aldehyde group is on and that the groups bonded to the carbons below point out of the page. The model represents a D-aldose because the -OH group at the chiral carbon farthest from the aldehyde points to the right. (a)
top, pointing into the page,
CHO
HO H
HO^Jc^CHO^ A\
HO-
A\
H-
HO H
H H
H
OH CH 2 OH
D-Threose
Break the hemiacetal bond and uncoil the aldohexose. Notice that all hydroxyl groups point to the right in the Fischer projection. The model represents the (3 anomer of D(b)
allopyranose.
CHO
CH 2OH
-OH -OH
OH
H
-OH
/3-D-Allopyranose
-OH CH 2 OH
714
Chapter 25
25.27 The
hints in the previous
problem also apply
™
_
(a)
u HOH
^
here.
Molecular models are helpful,
(b)
CHO
HO^ C
HO-
CHO
i\
ohcVg
H
CH 2OH
H H
CHO
HO H
is
X^oh
,'k
OH OH
,'k
HOH HH
CH 2 OH
L-Glyceraldehyde
D-Erythrose structure represents an a anomer because the anomeric -OH group and the -CH2OH group are trans. The compound is a-L-mannopyranose because the -OH group at C2 is the only non-anomeric axial hydroxyl group.
25.28 The
trans
*OH
OH a-L-Mannopyranose
25.29 (a)
HOH
HO H
OHC,
C ,'k
H
CHO
CHO
CHO
Vf
vr
C
.c x
OH
,'k
OH HO H
H-
-OH
HO-
H
H-
H-
-OH
HO-
H
HO-
c"
H H
OH H
HO-
-H
H-
OH
H-
OH
HO-
-H
H-
OH
H-
OH
CH 2 OH L-Mannose
The model
CH 2 OH
CH 2 OH
D-Mannose
D-Glucose
(enantiomer)
(diastereomer)
represents an L-aldohexose because the hydroxyl group on the chiral carbon from the aldehyde group points to the left. (c) This is tricky! The furanose ring of an aldohexose is formed by connecting the -OH group at C4 to the aldehyde carbon. The best way to draw the anomer is to lie L-mannose on its side and form the ring. All substituents point down in the furanose, and the anomeric -OH and the -CH(OH)CH2 OH group are cis. (b)
farthest
HOCH
p-L-Mannofuranose
Hi onio ecu Its: I
Carbohydrates
Additional Problems Carbohydrate Structure
25.30 (b)
(a)
(c)
(pH 2 OH
CHO
CH 2 OH
r° CH OH
OH
H-
C=0
2
H—
a ketotriose
OH
H-
HO-
-OH
H-
CH 2 OH
HO-
a ketopentose
H
OH H
OH
H"
CH 2 OH an aldoheptose
25.31 (a)
(b)
CH 2 OH
(c)
CHoOH 2
(d)
CHO
CHO
I
C=0
C=0
-OH
H—
CH 2 OH
HO—
H-
-OH -H
H-
H
H-
NH 2
H-
OH
H-
OH
H
H-
OH
HO-
CH 2 OH
OH
H-
CH 2OH
CH 2 OH a ketopentose
a ketotetrose
25.32 D-Ribose and L-xylose
a deoxyaldohexose
a five-carbon amino sugar
are diastereomers and differ in all physical properties (or if they
have identical physical properties
in
any category,
it is
a coincidence).
25.33-25.34 Ascorbic acid has an L configuration because the hydroxyl group center points to the
HO.
HO
C=0 H-
HO-
R
left.
o -H
CH 2 OH L- Ascorbic acid
OH
^ ^
H
Mannose
c
\\~\
HO— C r ~^~H— OH
HO— H^jw
25.64 O-
C0 2 H H-
HO-
OH H
H-
OH
H-
OH
HOCH 2 H
OH
H"
HO-
H
O OH
OH
H-
enohzation
H-
lactone
CH 2 OH
CH 2 OH
formation
D-Gluconic acid
+
H2
-H
HO-
-H
H-
-OH
H-
-OH
enediol
OH enohzation
pyridine
jj
HOCH 2 OH
O
HO
HO
CH 2 OH
H
formation
D-Mannonic acid
double bond.
v
OH
lactone
Isomerization at
pyridine
\l tt
HOCH 2 HO'^V^----'
C0 2 H HO-
O
CH 2 OH +
H2
C2 occurs because the enediol
can be reprotonated on either side of the
728
Chapter 25
25.65 There
are eight diastereomeric cyclitols.
H
HQ
HO
HO
OH HO
HO
HO^\ ^- ^*\^OH
OH
,
-i
OH
OH
OH
I
OH HO
QH OhJ
OH HO HO
y^^\^OH
HO^^-*^\^ OH
OHJ
HO^^*^^X
OH
1
OH
OH
^4 OH
OH
25.66
CHO
C0 2 H
OH
H-
dil.
D-Ribose
OH
H-
H-
-OH
H-
-OH
H-
-OH
HNO<
heat
OH
Hh
C0 2 H
CH 2 OH A 1.
HCN
2.
H 2 Pd
3.
-H
H-
"OH
H-
-OH -OH C0 2 H E
,
H3
CHO
H
H
OH
H-
OH
H
OH
H-
OH
H
OH
H-
OH
H
OH
HOdil.
HN0 3
heat
catalyst
+
CHO
C0 2 H HO-
B
CH 2 OH C D-Altrose
CH 2 OH D
C0 2 H
-OH dil.
HNO<
heat
H-
-OH
H-
-OH -OH C0 2 H F
D-Allose
Because A is oxidized to an optically inactive aldaric acid, the possible structures are Dribose and D-xylose. Chain extension of D-xylose, however, produces two hexoses that,
when
oxidized, yield optically active aldaric acids.
Biomolecules: Carbohydrates
729
25.67
.0
(a)
HN ^NNHPh C
CH 2 OH
C
C=0
OH or
H
HOH-
HO-
HO-
H
C= NNHPh
H
2
H 2 NNHPh
OH
H-
OH
OH
H-
OH
H-
OH
-OH
CH 2 OH
CH 2OH
D-Fructose
D-Mannose
CH 2 OH An osazone
(c)
^N— NHPh
H.
.
NHt- NHPh
NH + I
HO-
-OH
H-
C H-
-H
HO-
H-
CH 2 OH
H^
H
OH
D-Glucose (b)
HOor
OH
C=0
o-^h
H
HO-
H
HO-
H-
OH
H-
OH
H-
OH
H-
OH
(d)
H
^NH PhNH — NH 2
0=0
HO-
H
I
-H
-OH H-
CH 2OH enol
CH 2 OH phenylhydrazone
H 2 NPh
-OH
CH 2 OH keto imine
H
H
I
H
,q PhNHNmJ-c— NH, i
PhNHN— C— NHo2
..
I
PhNHN=C
I '
C=0
H
HO-
-
C=
c=o
H
HO-
+ NHo
-H
HO-
H-
OH
H-
OH
H-
OH
H-
-OH
H-
OH
H-
OH
H-
OH
H-
-OH
CH 2 OH
CH 2 OH
CH 2 OH
V
V
PhNHN=C
PhNHN=C
CH 2 OH
H
H
PhNHN=C
PhNH— Kh^C=0: PhNHNH 2 -C-0:~ H
HO-
H
HO-
H-
OH
H-
OH
H-
OH
H-
CH 2 OH
I
PhNHNH— C— OH HO
+
PhNHN=C
—— |
HoO
PhNHN=C _^ HO-
H
-OH
H-
OH
OH
-OH
H-
OH
CH 2 OH
CH 2 OH
H-
CH 2 OH
two nucleophilic addition reactions take place to yield imine products. The mechanism has been worked out in greater detail in Section 19.8, but the essential In these last steps,
steps are additions of phenylhydrazine, first to the imine, then to the ketone. Proton transfers are followed
by eliminations,
first
of ammonia, then of H2O.
730
Chapter 25
25.68
(a)
more
less stable
stable
p-D-Idopyranose
OH
OH more
less stable
stable
a-D-Idopyranose (b)
is more stable than (3-D-idopyranose because only one group is axial more stable chair conformation, whereas f3-D-idopyranose has two axial groups in more stable conformation.
a-D-Idopyranose in its its
(c)
1,6-Anhydro-D-idopyranose is formed from the p anomer because the axial hydroxyl groups on carbons 1 and 6 are close enough for the five-membered ring to form. (d)
The hydroxyl groups stable conformation
at
carbons
and
1
and 6 of D-glucopyranose are equatorial
are too far apart for a ring to form.
25.69
HO D-Ribofuranose
is
OH
the sugar present in acetyl
CoA.
in the
most
Hi onio ecu Its: I
25.70 Cleavage of fructose
Carbohydrates
731
1,6-bisphosphate occurs by a retro aldol reaction.
CH 2 OP0 3 2 CH 2 OP0 3 2-
"
CH 2OP03 2
C— 6:^ " HO— C— H LH "jjA J
C=Q
—
C=0
11
Ho{-C— H
+
— C— O — H
,D
H
"
I
CH 2 OH Dihydroxyacetone 3-phosphate
— C= O
H
I
H— C— OH CH 2 OP0 3 2 Fructose
25.71
1
I
H— C— OH "
^~
I
CH 2 OP0 3 2
,6-bisphosphate
Glyceraldehyde 3-phosphate
+
Oxidation by NADP elimination, and conjugate reduction by NADPH give the + observed product. Notice that there is no net consumption of NADP The mechanism of + NADP oxidations and reductions has been shown many times in this book and also appears in part (c). (a)
,
.
HOCH 2 OH
HO
HO<.
NADP + ADPH/H
O
HO.
+
O
OGDP
CH 2
OH Q
OGDP
HO
GDP-D-Mannose
HoC
NADP +
OGDP
NADPH/H" i
CH 2 OH
O HO
O
OGDP
Chapter 25
(b)
Two epimerizations, both a to the carbonyl
group, cause a change in stereochemistry.
HO (c)
Reduction
at
C4 by
NADPH forms GDP-L-fucose.
GDP-L-Fucose
Chapter 26 - Biomolecules: Amino Acids, Peptides and Proteins
Chapter Outline I.
Amino
acids (Sections 26.1-26.3).
A. Structure of amino acids (Section 1
.
26.1).
Amino acids exist in solution as a.
zwitterions.
Zwitterions are internal salts and have
many of the properties
associated with
salts. i. ii.
hi.
b
.
Zwitterions can act either as acids or as bases. i. The -CO2 group acts as a base.
The ammonium group
ii.
2
.
3
.
4
.
5
.
They have large dipole moments. They are soluble in water. They are crystalline and high-melting.
All but one (proline) of the 20 common amino acids are primary amines. All of the amino acids are represented by both a three-letter code and a one-letter code. See Table 26. 1 All a.
b
.
amino acids except glycine are chiral. Only one enantiomer (L) of each pair is naturally-occurring. In Fischer projections, the carboxylic acid is at the top, and the amino group points to the
c
6
.
acts an acid.
All natural amino acids are a-amino acids: the amino group and the carboxylic acid group are bonded to the same carbon.
.
left.
a- Amino acids are referred to as L-amino acids.
Side chains can be neutral, acidic, or basic. Fifteen of the 20 amino acids are neutral. b. Two (aspartic acid and glutamic acid) are acidic. i. At pH = 7.3, their side chains exist as carboxylate ions. a.
c.
Three
At
i.
(lysine, arginine
pH =
and
histidine) are basic.
7.3, the side chains of lysine
and arginine exist as
ammonium
ions.
not quite basic enough to be protonated at pH = 7.3. hi. The double-bonded nitrogen in the histidine ring is basic. d. Cysteine and tyrosine are weakly acidic but are classified as neutral. 7 Humans are able to synthesize only 1 1 of the 20 amino acids. a. These are nonessential amino acids. b The 9 essential amino acids must be supplied in the diet. B. The Henderson-Hasselbalch equation and isoelectric points (Section 26.2). Histidine
ii.
is
.
.
1
.
The Henderson-Hasselbalch a.
b
.
equation.
and pKa we can calculate the percentages of protonated, neutral and deprotonated forms of an amino acid. If we do these calculations at several pH values, we can construct a titration If
we know
the values of
pH
,
curve for each amino acid. 2
.
The
isoelectric point (pi) is the
pH at which
an amino acid exists as a neutral,
dipolar zwitterion. a.
pi i. ii.
hi.
is
related to side chain structure.
The 15 amino acids that are neutral have pi near neutrality. The two acidic amino acids have pi at a lower pH. The 3 basic amino acids have pi at a higher pH.
734
Chapter 26
For neutral amino acids, pi is the average of the two pKa values. For acidic amino acids, pi is the average of the two lowest pKa values. d For basic amino acids, pi is the average of the two highest pKa values. e. Proteins have an overall pi. 3 Electrophoresis allows the separation of amino acids by differences in their pi. a. A buffered solution of amino acids is placed on a paper or gel. b. Electrodes are connected to the solution, and current is applied. Negatively charged amino acids migrate to the positive electrode, and positively c. charged amino acids migrate to the negative electrode. d. Amino acids can be separated because the extent of migration depends on pi. C. Synthesis of a-amino acids (Section 26.3). 1 The Hell-Volhard-Zelinskii method and the phthalimide method. An a-bromo acid is produced from a carboxylic acid by a-bromination. a. b Displacement of -Br by ammonia gives the a-amino acid. 2. The amidomalonate synthesis. b. c.
.
.
.
.
a.
An alkyl halide
reacts with the anion of diethyl
amidomalonate.
Hydrolysis of the adduct yields the a-amino acid. Reductive animation. a. Reductive animation of an a-keto carboxylic acid gives an a-amino acid. b This method is related to the biosynthetic pathway for synthesis of amino acids. 4 All of the methods listed above produce a racemic mixture of amino acids. D. Enantioselective synthesis of amino acids. Resolution of racemic mixtures: 1 a. The mixture can react with a chiral reagent, followed by separation of the diastereomers and reconversion to amino acids. b Enzymes selectively catalyze reactions that form one of the enantiomers, but not
b
3
.
.
.
.
.
.
2
.
the other. Enantioselective synthesis. a.
Enantioselective hydrogenation of Z-enamido acids produces chiral a-amino
b
The most effective
acids. .
catalysts are
complexes of rhodium
(I),
cyclooctadiene and a
chiral diphosphine. II.
Peptides (Sections 26.4-26.8). A. Peptide structure (Section 26.4). 1 Peptide bonds. a. A peptide is an amino acid polymer in which the amine group of one amino acid forms an amide bond with the carboxylic acid group of a second amino acid. b The sequence of -N-CH-CO- is known as the backbone of the peptide or .
.
protein.
Rotation about the amide bond is restricted. The N-terminal amino acid of the polypeptide is always drawn on the left. The C-terrninal amino acid of the polypeptide is always drawn on the right. Peptide structure is described by using three-letter codes, or one-letter codes, for the individual amino acids, starting with the N-terminal amino acid on the left. c.
2
.
3
.
4.
5
.
Disulfide bonds. a.
b
.
Two cysteines
can form a disulfide bond
Disulfide bonds can link polypeptide chain.
(
-S-S-).
two polypeptides or introduce a loop within a
Biomolecules:
B
.
Amino
Acids, Peptides and Proteins
735
Structure determination of peptides (Sections 26.5-26.6). 1 Amino acid analysis (Section 26.5). .
a.
b. c.
Amino acid analysis provides the identity and amount of each amino acid present in a protein or peptide. First, all disulfide bonds are reduced and all peptide bonds are hydrolyzed. The mixture is placed on a chromatography column, and the residues are eluted. i. As each amino acid elutes, it undergoes reaction with ninhydrin, which produces a purple color that
is
detected and measured
spectrophotometrically d.
2
.
Alternatively, the mixture can be analyzed
Amino
by HPLC.
reproducible on properly maintained equipment; residues always elute at the same time, and only small sample sizes are needed. The Edman degradation (peptide sequencing) (Section 26.6). a. The Edman degradation removes one amino acid at a time from the -NH2 end of a peptide. i. The peptide is treated with phenylisothiocyanate (PITC), which reacts with the arnmo-terrninal residue. The PITC derivative is split from the peptide. ii. iii. The residue undergoes acid-catalyzed rearrangement to a PTH, which is identified chromatographically. iv. The shortened chain undergoes another round of Edman degradation. b Since the Edman degradation can only be used on peptides containing fewer than 50 amino acids, a protein must be cleaved into smaller fragments. Partial acid hydrolysis is unselective and therefore is of limited usefulness. i. The enzyme trypsin cleaves proteins at the carboxyl side of Arg and Lys ii. e.
acid analysis
is
.
residues. iii.
c.
The enzyme chymotrypsin
cleaves proteins at the carboxyl side of Phe, Tyr
and Trp residues. The complete amino acid sequence of a protein results from determining the individual sequences of peptides and overlapping them.
C. Synthesis of peptides (Sections 26.7-26.8). Laboratory synthesis of peptides (Section 26.7). 1 a. Groups that are not involved in peptide bond formation are protected. Carboxyl groups are often protected as methyl or benzyl esters. i. ii. Amino groups are protected as Boc or Fmoc derivatives. b The peptide bond is formed by coupling with DCC (dicyclohexylcarbodiimide). c. The protecting groups are removed. i. Boc groups are removed by brief treatment with trifluoroacetic acid. Fmoc groups are removed by treatment with base. ii. iii. Esters are removed by mild hydrolysis or by hydrogenolysis (benzyl). 2 Automated peptide synthesis - Merrifield solid-phase method (Section 26.8). a. The carboxyl group of a Boc-protected amino acid is attached to a polystyrene resin by formation of an ester bond. b The resin is washed with trifluoroacetic acid, and the Boc group is removed. A second Boc-protected amino acid is coupled to the first, and the resin is c. .
.
.
.
washed. d.
The cycle
(deprotecting, coupling, washing)
is
repeated as
many
times as
needed. e.
Finally, treatment with
anhydrous
HF removes the final Boc group and frees
the polypeptide. f
.
Robotic peptide synthesizers have improved yield and preparation time.
Width: 612 Height: 792
736
Chapter 26
HI. Proteins (Section 26.9). A. Classification of proteins.
Fibrous proteins consist of long, filamentous polypeptide chains. 1 Globular proteins are compact and roughly spherical. 2 Protein structure. 1 Levels of protein structure. a. Primary structure refers to the amino acid sequence of a protein. b Secondary structure refers to the organization of segments of the peptide backbone into a regular pattern, such as a helix or sheet. c. Tertiary structure describes the overall three-dimensional shape of a protein. d Quaternary structure describes how protein subunits aggregate into a larger .
.
B
.
.
.
.
structure.
2.
Examples of structural a.
features.
a-Helix. i.
An a-helix is
a right-handed coil; each turn of the coil contains 3.6 amino
acids. ii.
b.
structure is stabilized
and
C=0 groups four residues
by hydrogen bonds between amide
N-H groups
away.
/3-Pleated sheet. i.
ii.
c.
The
In a /^-pleated sheet, hydrogen bonds occur between residues in adjacent chains. In a /3-pleated sheet, the peptide chain
is
extended, rather than coiled.
Tertiary structure. i.
The nonpolar amino
acid side chains congregate in the center of a protein to
avoid water. ii.
hi.
The polar
side chain residues are on the surface, where they can take part hydrogen bonding and salt bridge formation. Other important features of tertiary structure are disulfide bridges and hydrogen bonds between amino acid side chains.
in
Denaturation of proteins. a. Modest changes in temperature and pH can disrupt a protein's tertiary structure. i. This process is known as denaturation. Denaturation doesn't affect protein primary structure. ii. b. Denaturation affects both physical and catalytic properties of proteins. c. Occasionally, spontaneous renaturation can occur. C. Enzymes (Sections 26.10-26.11). 1 Description of enzymes and cofactors (Section 26. 10). a. An enzyme is a substance (usually protein) that catalyzes a biochemical reaction. b. An enzyme is specific and usually catalyzes the reaction of only one substrate, i. Some enzymes, such as papain, can operate on a range of substrates. 3
.
.
c.
How enzymes function. i.
Enzymes form an enzyme-substrate complex, within which
the conversion
to product takes place. ii.
Enzymes
accelerate the rate of reaction
by lowering the energy of the
transition state.
The rate constant for the conversion of E S to E + P is the turnover number. Enzymes are grouped into 6 classes according to the reactions they catalyze. hi.
d.
i. ii. iii.
iv.
Oxidoreductases catalyze oxidations and reductions. Transferases catalyze the transfer of a group from one substrate to another. Hydrolases catalyze hydrolysis reactions. Lyases catalyze the addition or loss of a small molecule to or from a substrate.
v.
Isomerases catalyze isomerizations.
Amino
Biomolecules:
Acids, Peptides and Proteins
737
Ligases catalyze bond formation between two molecules, often coupled with hydrolysis of ATP The name of an enzyme has two parts, ending with -ase. vi.
e.
The first part identifies the substrate. The second part identifies the enzyme's class. Most enzymes are globular proteins, and many consist of a protein portion i.
ii.
f
.
(apoenzyme) and a cofactor. Cofactors may be small organic molecules (coenzymes) or inorganic ii. Many coenzymes are derived from vitamins.
i.
2
.
How enzymes work - citrate a.
.
c.
d e.
1).
Citrate synthase catalyzes the aldol-like addition of acetyl
produce
b
synthase (Section 26. 1
Co A to oxaloacetate to
citrate.
Functional groups in a cleft of the enzyme bind oxaloacetate. Functional groups in a second cleft bind acetyl CoA. The two reactants are now in close proximity. i. Two enzyme amino acid residues generate the enol of acetyl CoA. The enol undergoes nucleophilic addition to the ketone carbonyl group of oxaloacetate.
f
Two enzyme amino
.
g
.
acid residues deprotonate the enol and protonate the
carbonyl oxygen. Water hydrolyzes the thiol
ester, releasing citrate
and CoA.
Solutions to Problems
26.1
Amino Acids with aromatic rings: Phe, Tyr, Trp, His. Amino acids containing sulfur: Cys, Met. Amino acids that are alcohols: Ser, Thr (Tyr is a phenol.) Amino acids having hydrocarbon side chains: Ala, He, Leu,
Val, Phe.
26.2
C0 2 H H2N
H
H
H0 2 C
NH 2 R
R
A Fischer projection of the a-carbon of an L-amino acid is pictured above. For most L-amino acids:
Group
ions.
Priority
-NH 2 -C02 H
For cysteine:
Group
Priority
2
-NH 2 -CH 2 SH
-R
3
-CO?H
3
-H
4
-H
4
1
1
2
4
R
2
738
Chapter 26
26.3
—
—c
zo 2
(
c
— R NH — R OH
s •1
J
2
R
s )
i
CH 3
Diastereomers of Threonine
L-Threonine
26.4
CH 3
3H 3
(
On the low pH
net positive charge at
26.5
and on the high pH hemoglobin (pi = 6.8) has a
(acidic) side of pi, a protein has a net positive charge,
(basic) side of pi, a protein has a net negative charge. Thus,
pH = 5.3
and a net negative charge
at
pH = 7.3.
This method of amino acid synthesis is simple and uses methods we have already studied. The phthalimide synthesis can also be used to introduce the amino group. Remember that only racemic amino acids are produced by this method. (a)
NH 2
Br
Br2 PBr3
1.
C 6 H 5 CH 2 CH 2 C02H
,
CgHgCH 2 CHC0 2 H
H Q
,,.
NH 3
I
I
C 6 H 5 CH 2 CHC0 2
•xcess*
3-Phenylpropanoic acid
Phenylalanine
(b)
NH 2
Br 1
(CH 3 2 CHCH 2 C0 2 H )
PBro J
Br?,
.
(CH 3 ) 2 CHCHC0 2 H
H q
2
Nhk
I '
I
(CH 3
excess
3-Methylbutanoic acid
)
2 CHCHC02
~
Valine
26.6
O^O-B C02 Et H
+
1
— C — C02 Et ^N^ C
~OEt
RX
^CH 3
|
N
heat
>V CH3
*
II
+
H qO
— C-r COoEt
H-
o
+ 2 EtOH
I
R 2.
H
Na
OH
I
+
C0 2
+
CH 3 C0 2 H
H
II
O
In the amidomalonate synthesis,
RCH(NH3 +)C02H. Choose an
shown above, an
alkyl halide
Amino Acid
Halide
NH 3 + I
"
(CH 3 2 CHCH 2 CHC0 2 )
Leucine
RX is converted to
alkyl halide that completes the structure of the target
acid.
(a)
I
R— C— u H— N—
(CH 3 2 CHCH 2 Br )
amino
Amino
Biomolecules:
Amino Acid
Acids, Peptides and Proteins
739
Halide
(b)
N
N
"
CH 2 CHC0 2
CH 2 Br H
H Histidine (C)
NHo3
+
I
~
CH 2 Br
CH 2 CHC0 2
Tryptophan (d)
NHo
+
1
CH3SCH 2 CH 2 CHC0 2
CH3SCH 2 CH 2 Br
Methionine
26.7
The precursor to an amino acid prepared by enantioselective hydrogenation has bond conjugated with a carboxylic acid carbonyl group.
O HoC
OH ^ NHCOCH3
(3
H3C H
[Rh(DiPAMP)(COD)]
1.
hfe,
2
NaOH, H 2
.
"
BF4
\F
II
HoC,
C
C H NH 2
H 3C H
Leucine
26.8
Val-Tyr-Gly (VYG) Val-Gly-Tyr (VGY)
26.9
The N-terminal group
Tyr-Gly-Val Tyr-Val-Gly is
on the
H C ,2 ~CH 2
u r 2
right,
c %.^iJk^^ 9 \ 9
N-terminal
$,
"1
+
II
O ^C*"*Q "
HoN. 3
'
M
\
H;.
C0 2
V
M
C-terminal
H
"CH(CH3) 2
CH3SCH 2 CH 2 H Met
amide bonds
—
ProP
is
HH w
II
I
Gly-Val-Tyr (GVY) Gly-Tyr-Val (GYV)
and the C-terminal group
O
H
V
(YGV) (YVG)
Val
Gly
v—
G
Z double
O
H H +
a
on the
left.
O
740
Chapter 26
26.10
C0 2H H 3 NCHCH 2 SH
The cysteine
H 3 NCHCH 2 S— CH 2 C0 2 H
ICH 2 C02 H
sulfur
is
a
good nucleophile, and iodide
is
+
T~
a good leaving group.
26.11 One product of the
reaction of an amino acid with ninhydrin is the extensively conjugated purple ninhydrin product. The other major product is the aldehyde derived from the side chain of the amino acid. When valine reacts, the resulting aldehyde is 2-methylpropanal. The other products are carbon dioxide and water. The identity of the aldehyde is determined by the amino acid side chain.
T3
OH +
(CH 3 2 CHCHC0 2 )
OH
O O
O
II
(CH 3 2 CHCH + )
26.12
C0 2
+ 3
H2
Trypsin cleaves peptide bonds at the carboxyl (right) side of lysine and arginine. Chymotrypsin cleaves peptide bonds at the carboxyl side of phenylalanine, tyrosine and tryptophan.
—
Trypsin
Asp-Arg
*-
+
Val-Tyr-Ile-His-Pro-Phe
Asp-Arg-Val-Tyr-Ile-His-Pro-Phe Chymotrypsin
Asp-Arg-Val-Tyr +
26.13 The part of the PTH
Ile-His-Pro-Phe
derivative that lies to the right of the indicated dotted lines comes from Complete the structure to identify the amino acid, which in this
the N-terminal residue.
problem
is
methionine.
O
C 6 H 5, (I
/
"0
CH0CH9SCH3
t
N
2
^H
CCH— CH 2CH 2SCH 3 +
H
+
C 6 H 5 N=C=S
NH 3 Methionine
26.14 The N-terminal residue of angiotensin II is C«H 6 n 5.
aspartic acid.
-CH2CO2H to arrive O
derivative in Figure 26.4 with
N— X
/
c
,.CH 2 C0 2 H H
H
Replace the
-R
group of the
at the correct structure.
PTH
Amino
Biomolecules:
26.15 Line up the (a)
Acids, Peptides and Proteins
741
fragments so that the amino acids overlap.
Arg-Pro Pro-Leu-Gly
(b)
V-M-W W-N-V V-L
Gly-Ile-Val
The complete sequence:
The complete sequence:
V-M-W-N-V-L
Arg-Pro-Leu-Gly-Ile-Val
26.16
:<£) \\y
(CH 3
)
3
:o:
:o:
ii
CO— C— O— C— OC(CH3 3 5=*
)
addition of
H 2 NCHRC0 2 H
(CH 3 ) 3 CO— C—
C02
amino acid nitrogen
NHCHRC0 2 ~ .
Et 3
N:v^
)
loss of
x
)
deprotonation
H
:o: r :6: (CH 3 3 CO- C- O 3- C-r OC(CH 3 3
-*
HOC(CH 3) 3 .
+
..
II
C- O- C- OC(CH 3 3
HI^HRC0 2 H
nucleophilic
(
+
(CH 3 3 CO-
)
:o: ..
I
)
y HNCHRC0 H i
+
2
C02and
Et 3
fer?-butoxide
26.17 CH 2 CH(CH 3 +
.
Protect the
R = CH 2 CH(CH 3) 2
R
2
+1
Leu = H 3 NCHC0 2~
1
)
1
f
"
H 3 NCHC0 2
=
amino group of leucine.
o o
O
(CH 3 ) 3 COCOCOC(CH 3 3 + )
H 2 NCHC0 2
EtoN
"
II
~ (CH 3 3 COCNHCHC0 2 +
2. Protect the carboxylic acid group of alanine.
+
Ala
+
CH 3 OH
H CaaySs
*
I
)
Leu
H 2 NCHC0 2 H
R
H 2 NCHC0 2 CH 3
C0 2
+
HOC(CH 3
)
3
NH +-
742
Chapter 26
3
.
Couple the protected amino acids with DCC.
9
O
II
II
(CH 3 3 COCNHCHC0 2 ~
+
N=C=N
H 2 NCHCOCH 3
)
R
CHo
O
I
II
O (CH 3
)
3
ff f COCNHCHC— NHCHCOCH3
CH 3
R 4.
Remove the
leucine protecting group.
{?
(CH 3
)
3
NH-C-NH
ff
CF 3 CQ2 H
f?
COCNHCHC— NHCHCOCH3
ff
CH 3
R
R (CH 3 )2C
5
.
CH<
— CH2
ff
f?
R
1
NaOH, H 2 (3
2.
H3 +
H 3 NCHC
— NHCHCOH
(b) (c)
+
CH 3 OH
CH 3
(CH 3 )2CHCH2
CH<
Leu(a)
COc
+
Remove the alanine protecting group.
H 3 NCHC— NHCHCOCH 3
26.18
ff
H 3 NCHC— NHCHCOCH 3
-Ala
Pyruvate decarboxylase is a lyase. Chymotrypsin is a hydrolase. Alcohol dehydrogenase is an oxidoreductase.
Visualizing Chemistry
26.19 (a)
(b)
-
H
H
(c)
^C02
H 3 N+
H 3 N+ H
A "OoC
H3N
CO.
j H
N
Isoleucine
26.20
c
/
II
N-terminal
H 3 N.
N •
H
HCHoSH Cys-
c—
Lys-
-K —
C
a\
||
n u
CH 2 C0 2
II
I
C
~
O
H
H
I
N \
H CHo3 H Ala
-
A-
Asp -D
CO.
C-terminal
H
C \
II
H
Glutamine
Histidine
H
O
H
O
Biomolecules:
Amino
Acids, Peptides and Proteins
743
26.21 H H
w .a H 3 C^
s C
cho H
H3 N s +
26.22
It's
C0 2
possible to identify this representation of valine as the
configuration at the chirality center.
The configuration
is
D enantiomer by
noting the
R, and thus the structure
is
D-
valine.
~0 2
C^R^ CH(CH 3 2 )
D-Valine (/?)-Valine
H3N H
26.23
After identifying the amino acid residues, notice that the tetrapeptide has been drawn with the amino terminal residue on the right. To name the sequence correctly, the amino terminal residue must be cited
first.
Thus, the tetrapeptide should be named Ser-Leu-Phe-Ala.
Additional Problems
Amino Acid 26.24 Both
Structures and Chirality
(/?)-serine
on the
and
(/?)-alanine are
D-amino
acids. In a
right.
COc H-
-NH3+
H-
-NH 3+
CH 2 OH (tf)-Serine
(R)- Alanine
D-amino
acid, the
-NH2 group
is
744
Chapter 26
26.25
+
H
HoN-
NHo
"OoC
+
R
CH 2 Br L-Bromoalanine (/?)-Bromoalanine
This L "amino acid" also has an R configuration because the in priority than the -CO2H group.
-CH2Br "side chain"
26.26 +
H
H 2 N-
(5)-Proline
7
H 2C
CH 2
26.27 (a)
CH 2 CHC0 2
-
(b)
(c)
CHo3 I
"
CH 3 CH 2 CHCHC0 2 +
Tryptophan (Trp)
Histidine (His)
NHo
Isoleucine (lie)
HSCH0CHCO0" 2 2,
NHo
Cysteine (Cys)
is
higher
Amino
Biomolecules:
Acids, Peptides and Proteins
745
26.28
C—
O"
C—
O"
-H+ K,a2
H 2 A+ deprotonated
protonated
At pH =
2.50:
log
[h 2 a +]
= p H - pKa] = 2.50 -
1.99
At pH = 2.50, approximately three times form as exist in the protonated form.
=
= Q.51;J^h +
H 2A
as
many
3. 24
]
proline molecules exist in the neutral
At pH = 9.70: [A"]
log
[HA] At 26 29 .
(a)
(b)
= pH -
pH = 9.70,
p/ra 2
= 9.70 - 10.60 = -0.90;J^r = 0.126 [HA]
the ratio of deprotonated proline to neutral proline
Val-Leu-Ser
V-L-S
Ser-Val-Leu
S-V-L
Val-Ser-Leu
V-S-L
Leu-Val-Ser
L-V-S
Ser-Leu-Val
S-L-V
Leu-Ser-Val
L-S-V
is
approximately
Ser-Leu-Leu-Pro
S-L-L-P
Leu-Leu-Ser-Pro
L-L-S-P
Ser-Leu-Pro-Leu
S-L-P-L
Leu-Leu-Pro-Ser
L-L-P-S
Ser-Pro-Leu-Leu
S-P-L-L
Leu-Ser-Leu-Pro
L-S-L-P
Pro-Leu-Leu-Ser
P-L-L-S
Leu-Ser-Pro-Leu
L-S-P-L
Pro-Leu-Ser-Leu
P-L-S-L
Leu-Pro-Leu-Ser
L-P-L-S
Pro-Ser-Leu-Leu
P-S-L-L
Leu-Pro-Ser-Leu
L-P-S-L
26.30 Aldehydes and ketones can undergo
1:8.
nucleophilic addition reactions. In particular,
aldehydes and ketones can react with amines to form imines and enamines, reactions that might compete with formation of amide bonds between amino acids. Because of this reactivity, aldehydes and ketones are unlikely to be found in amino acid side chains.
Width: 612 Height: 792
746
Chapter 26
Amino Acid
Synthesis and Reactions
26.31 The diethylamidomalonate ethoxide.
anion is formed by treating diethylamidomalonate with sodium Choose the appropriate halide based on the amino acid side chain.
C0 2 Et
(a)
C0 2 Et
—
(CHo)oCHCHoBr
":C— C02 Et
.N
I
(CH 3 2 CHCH 2 )
C— C02 Et
CH3
H II
O
o HoO +
CH 3 C0 2 H
+
C0 2
+
EtOH
2
(CH 3 2 CHCH 2 CHC0 2 H
+
+
(b)
Leucine
)
NHo
C0 2 Et
COoEt 2 I
CH 2 -C— C0 2 Et
CH 2 Br
~:C— C02 Et
,CH3 II
O CH 2 CHC0 2 H CH 3 C0 2 H
+
C0 2
+
2
EtOH
+
nrv N
26.32 (a) I?
CH 3 SCH 2 CH 2 CC0 2 H
NH 2
NH-
NaBH,
CH 3 SCH 2 CH 2 CHC0 2 H Methionine
W
HoC NH 2 3
HoC O 3 |
II
CH 3 CH 2 CHCC0 2 H
NH.
NaBH,
|
I
CH 3 CH 2 CHCHC0 2 H Isoleucine
NH 3
Tryptophan u H
Amino
Biomolecules:
Acids, Peptides and Proteins
747
26.33 (a)
O H
C— OH
n \.
1
NCOCH3
(b)
y
2
.
H2, [Rh(DiPAMP)(COD)]
a
~
O"
BF4
NH
NaOH, H 2
Proline
O
CHo
HoC 3
.
+
II
C
y C\
CH 3
1
OH
.
H2, [Rh(DiPAMP)(COD)]
2.
I
+
V/
~
BF4
H3C
NaOH, H 2
fj>
y ^\ y
O
C
NHCOCH3
_
Valine
H NH 2
26.34
9 .
H
Cv (
H
C02 Et
C0 2 Et I
:c— C02 Et
"OCH 2
— C— C0 2
CH 3
N
H 3 0!
Et
HOCH 2 CHC0 2 H +
NH 3 Serine
x CHd
y
ff
I?
O
O
+ +
CH 3 C0 2 H
C0 2
+
2 EtOH
26.35 (a) ~
(CH 3 2 CHCHC0 2 )
+
CH0CH0OH H
+
**
(
CH 3)2 CH HC02 CH 2 CH 3 p
catalyst
NH 3
NH<
Valine
(b)
O O II
(CH 3 2 CHCHC0 2 )
+ NH
II
(CH 3 3 COCOCOC(CH 3 )
Et3 N
)
3
(CH 3 2 CHCHC0 2~ )
(CH 3 2 CHCHC0 2 ~ )
NH,
HOC(CH 3
NHCOC(CH 3 3 )
O (c)
+
KOH, HoO
^—
(CH3 ) 2 CHCHC02"K NH'
+
+ H2
)
+ C0 2
3
748
Chapter 26
(
d) (CH 3 2 CHCHC0 2"
1
-
.
CHoCOCI, -
)
+
NHo3
pyridine
**
~
(CH 3 2 CHCHC0 2 )
g "
NHCCHo3 „
O
26.36
O
+
H2
Step 1: Dehydration. Step 2: Nucleophilic addition of the amino group of the amino Step 3: Proton transfer. Step 4: Loss of water. (b)
Decarboxylation produces a different imine.
o
acid.
R
Biomolecules:
Amino
Acids, Peptides and Proteins
Step 1: Addition of water. Step 2: Proton transfer. Step 3: Bond cleavage to yield an aldehyde and an amine, (d)
Step 1: Addition of the amine to a carbonyl carbon of a second ninny drin molecule. Step 2: Proton shift. Step 3: Loss of water to form the purple anion. Notice that the amino nitrogen
is all
that remains of the original
amino
acid.
749
750
Chapter 26
It is
also possible to
draw many other resonance forms
aromatic 6-membered rings.
that involve the
n electrons of the
Biomolecules:
Peptides and
Amino
Acids, Peptides and Proteins
751
Enzymes
26.38 (a)
N— CH 2 CH 2 SCHg HsN
C
>- °V
-
HSCH 2 H
N
°f ^ f co f OH
-
CHgCH^COg"
Cys-
Met
Glu-
His-
C—
H—
•E
—
M
(b)
H H
1
y
P
V
O
.
ft
^
H 2CH 2 C02
j»
"
/
H
~
CH 2 C0 2
H 3 C OH
H CH 2 CH 2 C0 2
6
E Pro-
-
H
N
+
N /\
H H
3^ / CH 2 CH3
N\ /C^ X CV /N^ /C^
H
H
"
2
Glu
—
E Pro-
Thr
Asp-
lie
26.39 The tripeptide
Leu
is cyclic.
Leu
/ \ Ala
Phe
—\
/
Phe
Ala
Glu
752
Chapter 26
26.40 Step
1:
Valine II
is
protected as
its
Boc
+
O
—^ Et 3 N
II
(CH3) 3 COCOCOC(CH 3 )3
derivative.
Val
II
(CH 3 3 COC— Val— )
OH
(Boc— Val-OH) Step 2: Boc-Val bonds to the polymer in an S N2 reaction.
Boc
— Val— OH
Step
+
CICH 2
The polymer
3:
Base
^Polymer)
is first
washed, then
is
»
Boc
,
treated with
— Val— OCH
2
—( Polymer
CF3C02H to cleave the Boc
group.
— Val— OCH2 —/Polymer) ^/ ""Sl
Boc
1
.
f
Step 4:
2.
S*
wash
"X,
Val— OCH 2 —f Polymer!
CFgC0 2 H
A Boc-protected Ala is coupled to the polymer-bound valine by reaction with
DCC. The polymer is washed. Boc— Ala
+
/* Val— OCH 2 —(Polymer)
Step 5: The polymer
is
treated with
S
6:
.
DCC
Boc- Ala— Val— OCH 2
— (^Polymer
CF3 C02H to remove Boc.
—^—^
\
CFoCOoH *
Boc-Ala— Val— OCH 2 — (Polymer) Step
1
-
Ala— Val— OCH 2 ~( Polymer
A Boc-protected Phe is coupled to the polymer by reaction with DCC. The
polymer
is
Boc-Phe
washed. +
X
Ala— Val— OCH 2 —f Polymer
1
.
2.
_
DCC wash
Boc— Phe -Ala— Val— OCH 2— (Polymer Step
7:
Treatment with anhydrous HF removes the Boc group and cleaves the ester the peptide and the polymer.
bond between
Boc— Phe—Ala— Val— OCH 2— Phe-Ala-Val
+
f
(CH 3 2 C=CH 2 )
"\
HF
(Polymer) +
C0 2
+
HOCH 2
—
(Polymer
Amino
Biomolecules:
Acids, Peptides and Proteins
753
26.41 PITC Peptide (a)
Phenylthiohydantoin
C6H5 S
Ile-Leu-Pro-Phe
— P—
I— L
Shortened Peptide
/0
H
N-C
9 3 \ ^--CHCHoCHo
/
n
S *
">
N
Leu-Pro-Phe
L
— P—
H
I
H
(b)
O
H 5, C6 n fi
Asp-Thr-Ser-Gly-Ala
— T— S— G—
D
N—
Thr-Ser-Gly-Ala V
/
,-CH 2 C0 2 H
c
T— S— G—
H H
26.42 Phe|-Leu— Met— Lys-j-Tyr|-Asp— Gly— Gly— Arg-j-Val— lie— Pro— Tyr Cleaved by trypsin = Cleaved by chymotrypsin = ++++
26.43
(a)
(b) (c)
Hydrolases catalyze the cleavage of bonds by addition of water (hydrolysis). Lyases catalyze the elimination of a small molecule (H2O, CO2) from a molecule. Transferases catalyze the transfer of a functional group between substrates.
26.44 Amino
acids with polar side chains are likely to be found on the outside of a globular
where they can form hydrogen bonds with water and with each other. Amino acids with nonpolar side chains are found on the inside of a globular protein, where they can avoid water. Thus, aspartic acid (b) and lysine (d) are found on the outside of a
protein,
globular protein, and valine (a) and phenylalanine (c) are likely to be found on the inside.
Refer to Table 26.1.
754
Chapter 26
26.45
Leuprolide
Glu E
His-
—
H
—
Trp-
Ser
Tyr
Leu-
w-
S
Y—
L
The N-terminal glutamic
(a)
—
acid
is
a cyclic lactam.
Leu-L
The C-terminal
Pro-NHEt P NHEt
Arg-
R
—
proline
is
an Af-ethyl
amide. (b)
One of the
(c)
See above.
leucines (indicated above) has
(d) The charge
on a peptide
side chain that
is
neutral
charged
is
due
D stereochemistry. According to Table 26. 1 the only Thus, leuprolide has a charge of +1 at
to the side chains.
at neutral
pH is arginine.
,
pH.
General Problems
26.46
A proline residue in a polypeptide chain interrupts a-helix formation because the amide nitrogen of proline has no hydrogen that can contribute to the hydrogen-bonded structure
of an a-helix.
26.47 Formation of cation:
H 3 C— O-1 CH 2
—
SnCI 4
CI
Electrophilic aromatic substitution:
rv
V
H 2 C=OCH 3 -
CH 2 OCH 3
"
SnCI 5 Protonation of the ether oxygen, followed by displacement of methanol by Cl~.
h— a
9
\
CH20CH2
V
•
//
\
CH 2 CI
CHpT.OCHo H :ci:
+
HOCH3
Biomolecules:
Amino
Acids, Peptides and Proteins
755
Step 1: NaOH brings about ehmination of the carboxylated peptide. Step 2: Loss of C0 2 .
The Fmoc group which
The
is
first
is
acidic because the
resonance-stabilized and
step
is
is
Fmoc
anion
is
similar to the cyclopentadienyl anion,
aromatic.
a substitution similar to the nucleophilic acyl substitution reactions that
studied in Chapter 21
we
Width: 612 Height: 792
756
Chapter 26
O
(b)
f NHCHCH — C-^NHCHC Ik
I
-|-NHCH— C=NHCHC-|-
-|-
I
p
H2 C
R'
2
X
\0\J
H2C
H 2 C^..
^S-C=N:
+
R'
H 3 C— S— C=N:
CHo Sn2 displacement of sulfide results in formation of a 5-membered ring containing an iminium group. Internal
(c)
H 2or>
-4-
o
;
4 NHCH — C— NHCHC
NHCH— C= NHCHC-f-
I
H 2 C^
of water
H2C
NHCH— C H2C
R
R'
proton transfer
O
r :OH
^1 * lis -4- NHCH— C— NHpCHC-4 s_
.,
H 2 NCHC-
O
N
I
O
H2 C
O ..
I
H 2 C^
addition
R'
,>+
O
OH;
H 2 Cv
1
breaking of peptide bond
H 2C
O
R'
H2 C
lactone
bond
In this sequence of steps, water adds to the imine double bond, and the peptide
is
cleaved.
(d)
H2 I
b:^h
on^OH r£OH
—5* NHCH
HoO +
..
—C
\k
>
—J-
\
„UH ^OH
NHCH— C
^ -4-
\>^OH
_t -S-NHCH— C
NHCH— C
*
H 2C
\
H2C
/
H2 C
O addition
of water
\
H 2C
/
+
HO:-)
O proton transfer
\
OH
\
rin
H 2C
\
I
HoC
*OH
H2 C
opemng
tt
O +
'I
CH 2 OH
>
H 3 NCHC-^-
+
O
-|-NHCH — C^ I
H2 C
R from
Water opens the lactone ring
to give the product
shown.
\
(c)
OH CH 2 OH
Amino
Biomolecules:
Acids, Peptides and Proteins
757
26.50
Qo:
:0:
C x o:
:0: ii
,-C^/R R
C0 2
4
H
«
N
R
N
R'
OH
1
N
R
+
/ \
l
H(OH
+
I
H
H
C0 2 + "OH
R'
26.51
^
II
NHR
H2N
+
H+
NH
NH,
NH,
NHR
H2 N
The protonated guanidino group can be
26.52
NHR
H2N
stabilized
100 g of cytochrome c contains 0.43 g
™Ql
^
0.43 g Fe x i 55.8 g Fe
Assuming
that
NH,
NH,
H2 N
NHR
H2 N
+
NHR
by resonance.
iron, or
0.0077 mol Fe:
= 0.0077 mol Fe
each mole of protein contains
1
mol
Fe, then
mol Fe = mol
protein.
100 g Cytochrome c _ 13,000 g Cytochrome c " 0.0077 mol Fe 1 mol Fe
Cytochrome c has a minimum molecular weight of 13,000
26.53
!
g/ mol.
H NMR shows that the two methyl groups of Af,Af-dimethylformamide are nonequivalent
room temperature. If rotation around the CO-N bond were unrestricted, the methyl ! absorptions would coalesce into a groups would be interconvertible, and their H at
NMR
single signal.
(a)
O
H3C
c
(b)
H d3 C
(b) rotate
H 3C
H
\
»
,0 //
N— C
H3C
(a)
H
The presence of two methyl absorptions shows
that there is a barrier to rotation
around the
CO-N bond. This barrier is due to the partial double-bond character of the CO-N bond, indicated
by
the
two resonance forms below. Rotation
to interconvert the
two methyl
slow at room temperature, but heating to 180° supplies enough energy rapid rotation and to cause the two absorptions to merge. groups
is
NMR
HcC
qfy
HoC
/
/
\
N=C
N-*-C
H3 C
H
:o:
\+
H3 C
H
to allow
as
758
Chapter 26
26.54 Gly Gly-Asp-Phe-Pro Phe-Pro-Val Val-Pro-Leu
The complete sequence: Gly-Gly-Asp-Phe-Pro-Val-Pro-Leu
26.55 Cjly
He yai
?
lu
f I
i
I
I
Gm^ys-Cys-Tbj-Ser-ne-Cys-Ser-Leu-Tyr|Gln-Leu-Glu-Asn-T S
>
—
rps-Leu^ys^ly-Ser-ms-Leu-Val^lu-Ala-Ixu-Tyr|Leu-Val-^ys '
Glu
Gly
r
i
^
sn
Glu
w phe
Jal
Arg s
i
s
s
Tm*]-Lys-Pro-TliiiTyiiphetphe-Gly Cleaved by Trypsin = Cleaved by Chymotrypsin =
26.56
Ser-Ile-Arg-Val-Val-Pro-Tyr-Leu-Arg
26.57 Reduced oxytocin:
Cys-Tyr-Ile-Gln-Asn-Cys-Pro-Leu-Gly-NH 2
Oxidized oxytocin:
Cys-Tyr-Ile-Gln-Asn-Cys-Pro-Leu-Gly-NH 2
S
S
The C-terminal end of oxytocin information given.
is
actually an amide, but this can't
be determined from the
Amino
Biomolecules:
Acids, Peptides and Proteins
26.58 (b)
(a)
O
O .C^
H 2 N^ j
N
C
CH 2 ^
o
C
.OCHo3
HofM O M.
>
\
H
H
\
^ OCHg
I
CH 2H
o
co 2
Aspartame (nonzwitterionic form) 7.3,
^
^
~
C0 2 H
At pH =
C
Aspartame
at
aspartame exists in the zwitterionic form, as
pH = does
it
5.9, 7.3
at
pH =
5.9.
26.59
^
S-C*
H 2 0:-n /
VV
c
r
H
r
C6 H 5 H
C 6 H5 H
I OH /^OH
H
3 :
H C H A
c-. :s
H
r
H
R
4.
C6 H5
2
H
I
H 2b:^ H
\:N-CrOH QPH
?5o„
I
C6 H5 H
H
4
.
^
fi
\
C6
\
2.
C6 H 5 H
C 6 H5 >>+
V,oh'
:S—
1.
?
H2 .
+
//+
HoO +
H 5. C6 n
.0
ft
:N— \
/
7.
H 1
H
H I
H
Step 1: Protonation Step 3: Proton transfer. Step 5: Bond rotation, addition of amine. Step 7: Loss of water
^ H
H
Step 2: Addition of water Step 4: Ring opening. Step 6: Proton transfer. Step 8: Deprotonation.
759
760
Chapter 26
26.60
26.61
4-Methylideneimidazol-5-one (MIO)
Step 1: Nucleophilic addition of the amino group of the amino Step 2: Proton transfer. Step 3: Elimination. Step 4: Elimination.
acid.
Biomolecules:
Step
1:
Step
2:
Amino
Acids, Peptides and Proteins
Nucleophilic addition of the amine to a-ketoglutarate.
Loss of water.
Review Unit Carbohydrates,
10: Bio molecules I
Amino
-
Acids, Peptides
Major Topics Covered (with vocabulary): Monosaccharides: carbohydrate monosaccharide furanose
aldose
ketose
anomer anomeric center a anomer
Fischer projection
D,L sugars
pyranose
p anomer mutarotation glycoside Koenigs-
Knorr reaction aldonic acid alditol reducing sugar aldaric acid Kiliani-Fischer synthesis degradation rucose glucosamine galactosamine neuraminic acid
Wohl
Other sugars: disaccharide
1,4'
cellobiose
link
maltose
lactose
sucrose
polysaccharide
cellulose
amylose amylopectin glycogen glycal assembly method deoxy sugar amino sugar cell-surface carbohydrate
hemagglutinin
neuraminidase
Amino acids: amino acid zwitterion amphoteric electrophoresis resolution
a-amino acid side chain isoelectric point (pi) amidomalonate synthesis reductive amination
Henderson-Hasselbalch equation
enantioselective synthesis
Peptides:
residue backbone Af-terminal amino acid C-terminal amino acid disulfide link amino acid analysis Edman degradation phenylthiohydantoin trypsin chymotrypsin peptide synthesis protection Boc derivative Fmoc derivative Merrifield solid-phase technique
DCC
Proteins:
simple protein
conjugated protein
quaternary structure
a-helix
primary structure
p-pleated sheet
apoenzyme holoenzyme coenzyme vitamin oxidoreductase
transferase
secondary structure
tertiary structure
bridge prosthetic group isomerase hydrolase ligase salt
enzyme
cofactor
lyase
denaturation
Types of Problems: After studying these chapters, you should be able to:
-
Classify carbohydrates as aldoses, ketoses,
D
or L sugars, monosaccharides, or
polysaccharides.
-
Draw monosaccharides
-
Identify the
-
as Fischer projections or chair conformations.
Predict the products of reactions of monosaccharides and disaccharides. Deduce the structures of monosaccharides and disaccharides.
Formulate mechanisms for reactions involving carbohydrates.
common amino acids and draw them with correct stereochemistry in dipolar form. Explain the acid-base behavior of amino acids. Synthesize amino acids. Draw the structure of simple peptides. Deduce the structure of peptides and proteins. Outline the synthesis of peptides. Explain the classification of proteins and the levels of structure of proteins. Draw structures of reaction products of amino acids and peptides.
Review Unit 10
Remember:
Points to
Aldohexoses, ketohexoses and aldopentoses can forms. *
763
all
exist in both pyranose
forms and furanose
A reaction that produces the same functional group at both ends of a monosaccharide halves the number of possible stereoisomers of the monosaccharide.
*
The
*
At physiological pH,
*
reaction conditions that form a glycoside are different from those even though both reactions, technically, form -OR bonds.
that
form a polyether,
the side chains of the amino acids aspartic acid and glutamic acid exist as anions, and the side chains of the amino acids lysine and arginine exist as cations. The imidazole ring of histidine exists as a mixture of protonated and neutral forms.
Since the amide backbone of a protein is neutral and uncharged, the isoelectric point of a is determined by the relative numbers of acidic and basic amino acid residues present in the peptide. protein or peptide
*
In the Merrifield technique of protein synthesis, a protecting group
carboxyl group because
it is
isn't
needed for the
attached to the polymer support.
Self-Test:
OH CH
O
NHo3
+
I
OH
CH 3
H3NCH2CH2CH2CHCO2"
OH
A
C
Digitalose
Ornithine
(hydrolysis product of digitoxigenin, a heart
medication) Digitalose (A)
is
related to
which D-aldohexose? Provide a name for A, including the
configuration at the anomeric carbon. Predict the products of the reaction of + catalyst; (b) I, Ag 0.
H
CH 3
Vicianose (B) of B with CH3I and
is
(a)
CH3OH,
2
a disaccharide associated with a natural product found in seeds. Treatment followed by hydrolysis, gives 2,3,4-tri-O-methyl-D-glucose and
Ag 2 0,
2,3,4-tri-O-methyl-L-arabinose.
Ornithine (C)
A with:
is
What
is
the structure of
B? Is B
a reducing sugar?
a nonstandard amino acid that occurs in metabolic processes.
Which amino
acid does it most closely resemble? Estimate pKa values and pi for ornithine, and draw the major form present at pH = 2, pH = 6, and pH = 1 1. If ornithine were a component of proteins, how
would
it
affect the tertiary structure
of a protein?
Review Unit 10
764
Tyr-Gly-Gly-Phe-Leu-Arg-Arg-Ile-Arg-Pro-Lys-Leu-Lys-Trp-Asp-Asn-Gln Porcine Dynorphin (D)
Dynorphin (D)is a neuropeptide. Indicate the /^-terminal end and the C-terminal end. Show the products of cleavage with: (a) trypsin; (b) chymotrypsin. Show the Af-phenylthiohydantoin that results from treatment of with phenyl isothiocyanate. Do you expect to be an acidic, a neutral
D
D
or a basic peptide? Kallidin (E) is a decapeptide that serves as a vasodilator. The composition of E is Arg2 Gly Lys Phe2 Pro3 Ser. The C-terminal residue is Arg. Partial acid hydrolysis yields the following
fragments:
What
is
Pro-Gly-Phe, Lys-Arg-Pro, Pro-Phe-Arg, Pro-Pro-Gly, Phe-Ser-Pro
the structure of E.
Multiple choice: 1
.
The enantiomer of a-D-glucopyranose (a)
2
.
3
.
p-D-Glucopyranose
(b)
is:
a-L-Glucopyranose
(c)
p-L-Glucopyranose
(d)
none of these
All of the following reagents convert an aldose to an aldonic acid except: (a) dilute HNO3 (b) Fehling's reagent (c) Benedict's reagent (d) aqueous Br2
Which two
aldoses yield D-lyxose after
Wohl degradation?
D-Glucose and D-Mannose (b) D-Erythrose and D-Threose D- Altrose (d) D-Galactose and D-Talose (a)
(c)
D-Galactose and
4.
All of the following disaccharides are reducing sugars except: (a) Cellobiose (b) Sucrose (c) Maltose (d) Lactose
5
.
Which of the following polysaccharides contains p-glycosidic bonds? (a) Amylose (b) Amylopectin (c) Cellulose (d) Glycogen
6
.
To
find the p/ of an acidic average of the
(a) find the
pKa values
(c) find the
amino acid: two lowest pKa values
average of
all
pKa values
(b) find the average of the (d) use the value
of the
two highest
pKa of the side
chain.
7
.
Which of the following techniques can
synthesize a single enantiomer of an amino acid? Hell-Volhard-Zelinskii reaction (b) reductive amination (c) amidomalonate synthesis (d) hydrogenation of a Z enamido acid (a)
8
.
9
.
The purple product
that results from the reaction of ninhydrin with an amino acid contains which group of the amino acid? (a) the amino group (b) the amino nitrogen (c) the carboxylic acid group (d) the side chain
Which of the following
reagents is not used in peptide synthesis? Phenylthiohydantoin (b) Di-terf-butyl dicarbonate (c) Benzyl alcohol (d) Dicyclohexylcarbodiimide
(a)
10.
Which
element is not present in myoglobin? group (b) regions of a-helix (c) hydrophobic regions
structural
(a) a prosthetic
structure
(d) quaternary
Chapter 27 - Bio molecules: Lipids
Chapter Outline I.
Esters (Sections 27.1-27.3).
A. Waxes, 1
.
and oils (Section 27.1). are esters of long-chain carboxylic acids with long-chain alcohols.
fats
Waxes
2.
Fats and oils are triacylglycerols. Hydrolysis of a fat yields glycerol and three fatty acids. a. b The fatty acids need not be the same.
3
Fatty acids. a. Fatty acids are even-numbered, unbranched long-chain (C12-C20) carboxylic
.
.
acids.
b.
The most abundant
saturated fatty acids are palmitic (Ci6) and stearic (Ci8)
acids. c.
The most abundant unsaturated
d
Cig). Linoleic and arachidonic acids are polyunsaturated fatty acids. i. Unsaturated fatty acids are lower-melting than saturated fatty acids because the double bonds keep molecules from packing closely. The C=C bonds can be catalytically hydrogenated to produce higher-melting
.
e.
fatty acids are oleic
and
linoleic acids (both
fats.
Occasionally, cis-trans bond isomerization takes place. i. (Section 27.2). Soap is a mixture of the sodium and potassium salts of fatty acids produced by hydrolysis (saponification) of animal fat. Soap acts as a cleanser because the two ends of a soap molecule are different. a. The hydrophilic carboxylate end dissolves in water. b. The hydrophobic hydrocarbon tails solubilize greasy dirt. c. In water, the hydrocarbon tails aggregate into spherical clusters (micelles), in which greasy dirt can accumulate in the interior. 2+ 2+ or Ca cations. Soaps can form scum when a fatty acid anion encounters a. This problem is circumvented by detergents, which don't form insoluble metal
B Soap .
1
.
2
.
3
.
Mg
salts.
C. Phospholipids (Section 27.3). Glycerophospholipids. a. Glycerophospholipids consist of glycerol, two fatty acids (at CI and C2 of glycerol), and a phosphate group bonded to an amino alcohol at C3 of glycerol. 2. Sphingomyelins. a. Sphingomyelins have sphingosine or a related dihydroxy amine as their backbone. b They are abundant in brain and nerve tissue. Phospholipids comprise the major lipids in cell membranes. 3 a. The phospholipid molecules are organized into a lipid bilayer, which has polar groups on the inside and outside, and nonpolar tails in the middle. Prostaglandins and other eicosanoids (Section 27.4). 1
.
.
.
II.
A. Prostaglandins. 1
.
2
.
3
.
Prostaglandins are C20 lipids that contain a C5 ring and two side chains. Prostaglandins are present in small amounts in all body tissues and fluids. Prostaglandins have many effects: they lower blood pressure, affect blood platelet aggregation, affect kidney function and stimulate uterine contractions.
Width: 612 Height: 792
766
Chapter 27
B. Eicosanoids. 1
.
Prostaglandins, thromboxanes, and leukotrienes
make up
the eicosanoid class of
compounds. 2 3
.
.
Eicosanoids are named by their ring system, substitution pattern and number of double bonds. Eicosanoids are biosynthesized from arachidonic acid, which is synthesized from linoleic acid. a.
The transformation from arachidonic
b.
(COX) enzyme. One form of the
acid
is
catalyzed by the cyclooxygenase
COX enzyme catalyzes the usual functions, and a second form produces additional prostaglandin as a result of inflammation.
IE. Terpenoids (Section 27.5).
A. Facts about terpenoids. Terpenoids occur as essential oils in lipid extractions of plants. 1 2. Terpenoids are small organic molecules with diverse structures. .
3
.
All terpenoids are structurally related.
a. Terpenoids arise from head-to-tail bonding of isopentenyl diphosphate units. b Carbon 1 is the head, and carbon 4 is the tail. 4. Terpenoids are classified by the number of five-carbon multiples they contain. Monoterpenoids are synthesized from two five-carbon units. a. b. Sesquiterpenoids are synthesized from three five-carbon units. c Larger terpenoids occur in both animals and plants. Biosynthesis of terpenoids. 1 Nature uses the isoprene equivalent isopentenyl diphosphate (IPP) to synthesize .
.
B
.
.
terpenoids. a.
IPP
is
biosynthesized by two routes that depend on the organism and the
structure of the terpenoid. i.
The mevalonate pathway produces sesquiterpenoids and most animals and
ii.
triterpenoids in
plants.
The 1-deoxy xylulose 5-phosphate pathway gives monoterpenoids, diterpenoids, and tetraterpenoids.
2.
The mevalonate pathway.
CoA undergoes Claisen condensation to form acetoacetyl CoA. Another acetyl CoA undergoes an aldol-like addition to acetoacetyl CoA Acetyl
a.
b
.
to give
(35)-3-hydroxy-3-methylglutaryl CoA (HMG-CoA). CoA is reduced by NADPH, yielding (7?)-mevalonate. c. d. Phosphorylation and decarboxylation convert (/?)-mevalonate to IPP. Conversion of IPP to terpenoids. a. IPP is isomerized to dimethylallyl diphosphate (DMAPP) by a carbocation
HMG
3
.
pathway.
b
.
c.
The C=C bond of IPP
displaces the
PPO~ group
of dimethallyl diphosphate, to
form geranyl diphosphate (GPP), the precursor to all monoterpenoids. Geranyl diphosphate reacts with IPP to yield farnesyl diphosphate (FPP), the
precursor to sesquiterpenoids. GPP is isomerized and cyclizes on the way to yielding many monoterpenoids. IV. Steroids (Sections 27.6-27.7). A. Steroids are derived from the triterpenoid lanosterol. 1 Steroids have a tetracyclic fused ring system, whose rings are designated A, B, C, and D. 2. The three six-membered rings adopt chair geometry and do not undergo ring-flips. B Stereochemistry of steroids (Section 27.6). Two cyclohexane rings can be joined either cis or trans. 1 a. In a trans-fused ring, the groups at the ring junction are trans. b. In cis-fused rings, the groups at the ring junction are cis. d.
.
.
.
Biomolecules: Lipids
A
Cis ring fusions usually occur between rings and B. In both kinds of ring fusions, the angular methyl groups usually protrude above the c.
2
.
3
.
4
.
rings.
A-B trans fusions are more common. Substituents can be either axial or equatorial. Steroids with a.
Equatorial substituents are
more favorable
for steric reasons.
C. Types of steroid hormones. 1 Sex hormones. a. Androgens (testosterone, androsterone) are male sex hormones. b Estrogens (estrone, estradiol) and progestins are female sex hormones. 2. Adrenocortical hormones. + + Mineralocorticoids (aldosterone) regulate cellular Na and K balance. a. b Glucocorticoids (hydrocortisone) regulate glucose metabolism and control .
.
.
inflammation. Synthetic steroids. a. Oral contraceptives and anabolic steroids are examples of synthetic steroids. D. Biosynthesis of steroids (Section 27.7). All steroids are biosynthesized from lanosterol. 1 Lanosterol is formed from squalene, which is the product of dimerization of 2 farnesyl diphosphate (FPP). 3 Squalene is first epoxidized to form 2,3-oxidosqualene. 3
.
.
.
4.
Nine additional
5
The first several steps are cyclization reactions. b The last steps are hydride and methyl shifts involving Other enzymes convert lanosterol to cholesterol.
needed
steps are
to
form
lanosterol.
a.
carbocations.
.
.
Solutions to Problems
27.1
CH 3 (CH 2 from
)
Carnaubawax
18 C^
C20 acid
0(CH 2 from
)
3 i CH 3
C 32
alcohol
27.2
O
O
II
II
CH 2 OC(CH 2
)
14 CH 3
CH 2 OC(CH 2
)
7
CH= CH(CH 2 7CH 3 )
(cis)
O
O
II
II
CHOC(CH 2
)
CHOC(CH 2 7 CH= CH(CH 2 7 CH 3
14 CH 3
)
)
(cis)
O I
767
ff
CH 2 OC(CH 2
II
)
CH 2 OC(CH 2 7 CH = CH(CH 2 7 CH 3
14 CH 3
)
Glyceryl tripalmitate
Glyceryl tripalmitate
is
)
Glyceryl trioleate higher melting because
it is
saturated.
(cis)
768
Chapter 27
27.3
CH 3 (CH2)7CH=CH(CH 2 )7CO" Mg 2+ "OC(CH 2 )7CH=CH(CH 2 )7CH3 Magnesium oleate
The double bonds
are cis.
27.4
O IP
CH 2 OC(CH 2
)
CH 2 OH
14 CH 3
+
Na ~OC(CH 2
O
)
14 CH 3
Sodium palmitate
NaOH HoO
II
CHOC(CH 2
)
7
CH= CH(CH2 7 CH 3 )
O
+
CHOH
+
2 Na
f
CH 2 OC(CH 2 7 CH= CH(CH 2 7 CH 3 )
CH 2 OH
)
Glyceryl dioleate monopalmitate (cis double bonds)
"OC(CH 2 7 CH )
= CH(CH 2 7 CH 3
Sodium oleate
)
cis
Glycerol
27.5
C0 2 H Prostaglandin E2
h
27.6
6h
h H
The pro-S hydrogen
OH
(blue) ends
up
cis to the
methyl group, and the pro-R hydrogen (red)
ends up trans.
O pro-R
pro -S i
9 H 3 C O— H ,CH 2 OPP ^C. .C. C
^C^ ^CH 2 OPP
C
A \
H
T pro -S
A
P
t \
H H
?H3
/ \
|
H
pro -R
H
H
H
Biomolecules: Lipids
27.7
769
As described in Worked Example 27. 1 draw the diphosphate precursor so that it resembles the product. Often, the precursor is linalyl diphosphate, which results from isomerization of geranyl diphosphate (the mechanism is shown in Figure 27.10). In (a), it's not easy to see the relationship, but once you've arrived at the product, rotate the ,
structure.
770
Chapter 27
27.8
Both ring systems are trans-fused, and both hydrogens at the ring junctions are axial. Refer back to Chapter 4 if you have trouble remembering the relationships of substituents on a cyclohexane ring.
equatorial
axial
27.9
Draw
the three-dimensional structure and note the relationship of the hydroxyl group to
groups whose orientation
OH-*—
equatorial
is
known.
Biomolecules: Lipids
771
27.10
1 /\ H HgC CH3
Cholesterol
Lanosterol 1
.
Two
methyl groups
2.
One methyl group
3
4.
C5-C6 C8-C9
5
Double bond
.
.
Cholesterol at
at
1
.
Two
2
.
One hydrogen
C4.
CI 4.
single bond. double bond.
at
at
C4.
C 1 4.
4.
C5-C6 double bond C8-C9 single bond.
5
Saturated side chain.
3.
in side chain
hydrogens
.
Visualizing Chemistry
27.11
C0 2 H
Cholic acid
equatorial
Cholic acid
have a
is
an
A-B
cis steroid
because the groups
at the fusion
of ring
A and ring B
cis relationship.
27.12
:
Draw
Base
Helminthogermacrene
farnesyl diphosphate in the configuration that resembles the product, then draw its isomer (the mechanism for the formation of the isomer is shown in Problem 27.7).
allylic
In this reaction, a cyclization, followed by loss of a proton to form the double bond, gives helminthogermacrene.
772
Chapter 27
26.13
C0 2 H Linoleic acid
A polyunsaturated fat such as linoleic acid is more likely to be found in peanut oil. Additional Problems Fats, Oils,
and Related Lipids
27.14
C0 2 H Eicosa-5,8,1 1,14,17-pentaenoic acid (all cis)
27.15
CH 2 OC(CH 2 )i6CH3
CH 2 OC(CH 2
)
16 CH 3
o
I
II
I
I
pHOC(CH 2
)
16 CH 3
f?
CHOC(CH 2
or
O
)
7
CH 2 OC(CH 2
)
)
)
(cis)
p
CH 2 OC(CH 2 7 CH= CH(CH 2 7 CH 3
CH= CH(CH2 7CH 3
)
16 CH 3
(cis)
optically active
1
.
2.
CH 2 OH
optically inactive
"OH, H 2 + H3
O
•"
II
II
I I
CHOH
+
HOC(CH 2 7 CH=CH(CH 2 7 CH 3 )
)
(cis)
+
2
HOC(CH 2
)
16 CH 3
I
Oleic acid
CH 2 OH Four
bonded
different groups are
Stearic acid
to the central glycerol
fat.
27.16
CH 3 (CH 2
)
14 C
Cetyl palmitate
OCH 2 (CH 2
)
-|
4 CH 3
carbon atom in the optically active
Biomolecules: Lipids
27.17 O II
CH 2OC(CH2)7CH=CH(CH 2 )7CH3
(cis)
O II
CHOC(CH 2 I
)
7
CH= CH(CH2 7 CH 3
(cis)
)
{?
CH 2 OC(CH 2 7 CH= CH(CH 2 7 CH 3 )
(cis)
)
Glyceryl trioleate (a)
O CH 2OC(CH 2 I
Glyceryl
Brc
trioleate
CH 2 CI 2
I
)
7 CH(Br)CH(Br)(CH 2 ) 7 CH 3
Q II
CHOC(CH 2
7 CH(Br)CH(Br)(CH 2 ) 7 CH 3
)
O II
CH 2 OC(CH 2 (b)
)
7 CH(Br)CH(Br)(CH 2 ) 7 CH 3
O II
CH 2 OC(CH 2 H 2 /Pd
Glyceryl
I
)
16 CH 3
ff
I
CH 2 OC(CH 2
)
(c)
Glyceryl
16 CH 3
CHoOH 2
NaOH
I
CHOH
HoO
trioleate
16 CH 3
n
CHOC(CH 2
trioleate
)
+
3
+
"0 C(CH 2 2
Na
)
7
CH= CH(CH 2 7CH 3 )
CH 2OH (d)
o
O
CH 2 OC(CH 2 7 CH )
o
Glyceryl
1.
trioleate
2. Zn,
o
o CH 3 C0 2 H
CHOC(CH 2 7 CH )
O
O +
3
HC(CH 2 7 CH 3 )
O
CH 2 OC(CH 2 7CH )
(e)
CH 2 OH Glyceryl trioleate
1.
LiAIH 4
2.
H3
+
CHOH I
CH 2 OH
+
3
HOCH 2 (CH 2 7CH=CH(CH 2 7 CH 3 )
)
773
774
Chapter 27
®
CH 2OH B£
SoSte
^
CH 3
3
HOC(CH 2 7 CH=CH(CH 2 7 CH 3
+ 3
? H0H
)
CH 2OH
)
CH 3
27.18 CH 3 (CH 2
)
7
CH= CH(CH 2 7 C02 H
(cis)
)
Oleic acid (a)
^ Oleic acid .
CH 3*OH
.
,
CH 3 (CH 2
HCI
^ Methyl
from
2 /Pd -
i.o 3
Oleic acid
2.
)
)
Methyl
(c)
CH= CH(CH2 7 C02CH 3
CH 3 (CH 2
(a)
stearate
OHC(CH 2 7 C0 2 H
+
)
)
n ^ -a 9-Oxononanoic acid
'[
XT Nonanal
Cr0 3
9-Oxononanoic acid
C0 2 CH 3
16
CHo(CH 3V 2 7 CHO z
Zn,CH 3 C0 c2 H
(d)
from
7
Methyl oleate
—H
oleate
)
H0 2 C(CH 2 7 C02 H )
Nonanedioic acid
(c)
(e)
Bf2
Oleic acid
»
CH 2 CI 2
CH 3 (CH 2 7 CH(Br)CH(Br)(CH 2 7C02 H 1 3 NaNH 2 NH 3 )
)
.
,
\
H3 +
2.
CH 3 (CH 2 7 C= C(CH 2 7 C0 2 H )
)
StearoUc acid
Three equivalents of the base are needed because one of them
is
neutralized
by the
CH 3 (CH 2
CHC0 2 H
carboxylic acid. (f)
H 2 /Pd Oleic acid
1
CH 3 (CH 2
)
15 CH 2
C0 2 H
.
—
Br2 PBr 3 *^ ,
2 H2<~
'
)
15
>
Stearic acid (g)
+
1
2
CH 3 (CH 2
)
16
.
C0 2 CH 3 2.
from
Na H3
2-Bromostearic acid
"
O "OCH-:
CH 3 (CH 2
+
)
16 CCH(CH 2) 15 CH 3 +
'I
HOCH 3
C0 2 CH 3
(b)
H3
+ ,
heat
O CH 3 (CH 2 1
)
16 CCH 2 (CH 2 ) 15 CH 3
+
C0 2
+
HOCH 3
8-Pentatriacontanone
This synthesis uses a Claisen condensation, followed by a /3-keto ester decarboxylation.
Biomolecules: Lipids
27.19
775
Fats and plasmalogens are both esters of a glycerol molecule that has carboxylic acid ester groups at C2 and C3. The third group bonded to glycerol, however, differs with the type
of
lipid: a fat
has a carboxylic acid ester at CI, and a plasmalogen has a vinyl ether in that
location.
27.20 O
?
CH 2 OH
+
O +
I
HOCR'
P
NaOH
CHOCR'
*
+
Na
II
"OCR" yet
o
HoO +
if
CHOH
CH 2 OH
CH 2 OCR"
HOCR"
+
CHOH
H2
O
+
+
Na
"OCR'
II
CH 2 OH
+
CH 2 OCH=CHR
HCCH 2 R
CH 2 OCH=CHR
Basic hydrolysis cleaves the carboxylic acid ester bonds but doesn't affect the ether bond. Acidic hydrolysis cleaves all three groups bonded to glycerol and produces an aldehyde
from the vinyl ether group.
27.21
L RC-|OCH
H
CH 2 0|-CR O
2
I
ll
|
?
if? ,M
HOCH 2 CHCH 2 OH + 2 Na3 pQ 4
3
I
l
R'C^-OCH I
ll
n
CHO"|-CR"
NaOH
O
o
I
^
l
-|U
s
II
s
I
CH 2 0|-P-|-0CH 2 CHCH 2 0|-P^-0CH 2
OH
O"
O II
II
+
RCO" Na+
+
+ + R"CO~ Na + R"'CO~ Na
I
+
+
R'CO" Na
H 2° (j>
0~
jj)
a cardiolipin
Saponification of a cardiohpin yields 4 different carboxylates, 3 equivalents of glycerol and
two equivalents of phosphate.
27.22
CH 3 (CH 2 7 C=C(CH 2 7 C0 2 H )
)
Stearohc acid Stearolic acid contains a triple acids.
^
CH 3 (CH 2 7 C0 2 H )
Nonanoic acid
+
H02 C(CH 2 7C0 2 H )
Nonanedioic acid
bond because the products of ozonolysis are carboxylic
Width: 612 Height: 792
776
Chapter 27
27.23 CH 3 (CH 2
7
)
NaNHc
C= CH
CH 3 (CH 2 7 C=C: > Na +J
I
I-r
"
_
NaCN
CH 3 (CH 2 7 C= C(CH 2 6 CH 2 CN )
CH 3 (CH 2
H3
)
)
7
—
CI
_ C(CH CH — CI C= 2 6 2 )
+
_ 7 C_ C(CH 2 6 CH 2 C0 2 H
)
CH2 (CH 2 5 CH 2
CH 3 (CH 2
)
|
u
NH 3
+
)
NHo
)
Stearolic acid
-
rather than CI
-
displaced is a better leaving group than chloride. I
,
by
,
is
acetylide because iodide
Terpenoids and Steroids
27.24-27.26 Remember that a compound with n chirality
centers can have a
maximum of 2 n
stereoisomers. Not all the possible stereoisomers of these compounds are found in nature or can be synthesized. Some stereoisomers have highly strained ring fusions; others contain 1 ,3-diaxial interactions.
Sabinene
Guaiol (8 possible stereoisomers)
If
carbon
1
(4 possible stereoisomers)
Cedrene (16 possible stereoisomers)
of each diphosphate were isotopically labeled, the labels would appear
circled positions of the terpenoids.
at the
Biomolecules: Lipids
777
27.27 ff
^C^ CoAS
Co AS
—
^C^
CHo
*
CH 3
*
*~
CoAS
CH 2
*
CH 3
*
HSCoA
+
ff
H2
HoC 3 \ VOH .C. .C.
H 3C
Q
HoC 3 t OH /
Q A
"O
CH 2
*
CH 2
*
dimethallyl diphosphate
IPP
CH 2
*
is
*
isomerized to
is
CH 3
*
CH 2
*
+ HSCoA SCoA
IPP
^C. H 2 C^*
.
CH 2
CHoOPP +
COo
*
DMAPP. CHo
I
I
^C. *
CH 2
*
converted to isopentenyl diphosphate (IPP) and
CHo H 2C
II
(DMAPP).
.CHoOPP CH 2 *
.c'
"O
Q
.C.
PH
H3 C
ft
*
+ "SCoA
mevalonate 5-diphosphate
First,
H
"O
*
(#)-Mevalonate
27.28
2 NADPH/H -<
.CH 2 OH
SCoA
f
.
CH 2
CHoOPP
.CHoOPP
.C^ H3C
*
*
CH
DMAPP
*
DMAPP and IPP couple to give geranyl diphosphate (GPP). GPP *
OPP
+
'
*
*
OPP
Qpp
A second molecule of IPP adds to GPP to give farnesyl diphosphate, the precursor to acadinene.
778
Chapter 27 Notice that the
C labels are located at two different positions:
-OPP was bonded; Now,
(2) at the
(1) at the
carbon to which
carbon bonded to the methyl group.
arrange farnesyl diphosphate to resemble the skeleton of a-cadinene. The first step sequence is formation of the allylic isomer of FPP; the mechanism was
in the reaction
shown
27.29
in
Problem
27.7.
Farnesyl diphosphate (from the previous problem) dimerizes to form squalene.
OPP
Farnesyl diphosphate
Biomolecules: Lipids
27.30 Squalene
is
779
converted to lanosterol by the series of steps pictured in Figure 27.14.
27.31
Draw
farnesyl diphosphate in the correct orientation in order to
make
this
problem much
displacement of ~OPP by the electrons of one double bond is followed by attack of the electrons of the second double bond on the resulting carbocation. Loss of a proton from the carbon next to the resulting carbocation produces the double bond. easier. Internal
780
Chapter 27
General Problems
27.32
(^OPP Farnesyl diphosphate
f
^
OPP
Isopentenyl diphosphate
OPP Geranylgeranyl diphosphate
The precursor to
flexibilene
is
formed from the reaction of farnesyl diphosphate and
isopentenyl diphosphate.
+
The precursor cyclizes by
the now-familiar
OPP
Flexibilene
mechanism to produce
flexibilene.
27.33
O
i^-Ionone
(3-Ionone
+
H3 +
Acid protonates a double bond, and the electrons of a second double bond attack the carbocation. Deprotonation yields p-ionone.
Biomolecules: Lipids
27.34
The two hydrocarbon
substituents are equatorial in the
most
stable chair conformation.
27.35
OH As
H
H
OH
H
always, use the stereochemistry of the groups at the ring junction to label the other and then esterify the appropriate -OH group.
substituents as equatorial or axial
781
782
Chapter 27
27.37
Linalyl
diphosphate
Isoborneol
The
initial addition is followed by a carbocation rearrangement to produce a secondary carbocation, which reacts with water to yield the secondary alcohol.
27.38
HoO
2.
i.
Isoborneol 3.
H2
p:^
4.
CHr.
Camphene Step 1: Protonation. Step 3: Carbocation rearrangement.
The key
step ring bonds.
is
Step 2: Loss of water. Step 4: Loss of proton.
the carbocation rearrangement,
which occurs by the migration of one of the
Biomolecules: Lipids
783
27.39
O
Digitoxigenin
The hydroxy 1 group
in ring
A is axial, and the hydroxy
C and axial to ring D. fusion and a C-D cis ring fusion. equatorial to ring
1 group at the ring C-D fusion is Notice that digitoxigenin has both an A-B cis ring
27.40
O
Lithium aluminum hydride reduces the lactone ring to a hydroxyl group because the second group is tertiary.
diol.
Periodinane oxidizes only one
784
Chapter 27
27.41
CH 3 (CH 2 5 CHO )
v
/C
H
'
1
=cx
Heptanal
2. Zn,
H
CH 3 C0 2 H
+
OHC(CH 2 9 C0 2 H
Vaccenic acid
)
1
CH2I2
1-Oxoundecanoic acid
CH 2 (CH 2 8 C0 2 H
CH 3 (CH 2 4 CH 2 )
)
(and enantiomer)
Zn/Cu
a
r
q
Lactobacillic acid
27.42 r
r
H
CH0CH0CH0CH0 3 2 2 2
C
C
1
C
1
1
CH 2 (CH2) 6 C0 2 H
H
H
(9Z,1 1£,13£)-9,1 1,13-Octadecatrienoic acid (Eleostearic acid)
1-
3 1 2. Zn, I
CH 3 CH 2 CH 2 CH 2 CHO
+
OHC— CHO
CH 3 C0 2 H +
OHC— CHO
The stereochemistry of the double bonds
27.43
can't
+
OHC(CH 2 7 C0 2 H )
be determined from the information given.
This mechanism also appears in Problem 27.32
Biomolecules: Lipids
785
27.44 CHc -OH --H
^
HO Estradiol
H3 C
Diethylstilbestrol
Estradiol and diethylstilbestrol resemble each other in having similar carbon skeletons, in
having a phenolic ring, and in being
diols.
27.45
1.
NaH
2.
CH 3 Br CH 3
CH3CH2COCI AICI3
CH3O
XT" I
1
.
2.
MgBr
J
Br
NaBH,
H3 +
I
^^Syy, CHCH2CH3
/'
S^*****S*S
OH I
OHCH2CH3
CHCH2CH3 PBr,
C I
CH2CH3
CH3O Diethylstilbestrol
The key reaction is a Grignard reaction between two molecules that are both synthesized from phenol. Phenol is first converted to anisole, in order to avoid problems with acidic hydrogens interfering with the Grignard reaction. Next, anisole undergoes Friedel-Crafts acylation with propanoyl chloride.The resulting ketone is one of the Grignard components. The other component is prepared by reduction, bromination and treatment with magnesium of a quantity of the ketone. After the Grignard reaction, HI serves to both dehydrate the alcohol and cleave the methyl ether groups.
Width: 612 Height: 792
786
27.46
Chapter 27
Biomolecules: Lipids
787
27.47
One equivalent of H2 hydrogenates the least substituted double bond. Dihydrocembrene has no ultraviolet absorption because it is not conjugated.
27.48
a-Fenchone
The mechanism follows the usual path: cyclization of linalyl diphosphate, followed by attack of the k electrons of the second double bond, produces an intermediate carbocation.
A carbocation rearrangement occurs, and the resulting carbocation reacts with water to form an alcohol
that is oxidized to give a-fenchone.
788
Chapter 27
27.49 r=0: 11/
CH 3 C— S— Protein
CHoC—S— Protein 3 1.
|
CHC— S— Protein
f :CHC— S— Protein
I
II
o2c o
I
~0 C 2
2.
+
O
:0:~
C0 2
:ov / ii
CH 3 C = CHC — S — P HA
O
O ii
II
I
CHoCy CHC— S— Protein
rote n i
O*Cf..
3.
O
4.
O"
- + -
S— Protein-
1
O
CH 3 CCH 2 C— S— Protein 3-Ketobutyryl-protein
Step 1: Attack of malonyl-protein anion on acetyl-protein (Claisen condensation). Step 2: Loss of S-protein. Step 3: Decarboxylation. Step 4: Protonation and tautomerization.
27.50
PPO
In this series of steps, dissociation of diphosphate ion allows bond isomerization to take making it possible for ring formation to occur. This mechanism is very similar to the
place,
mechanism shown
in Figure 27.10.
Biomolecules: Lipids
Chapter 28 - Biomolecules: Nucleic Acids
Chapter Outline I.
Nucleic acids (Sections 28.1-28.2). A. Nucleotides (Section 28.1). 1 Nucleotides are composed of a heterocyclic purine or pyrimidine base, an aldopentose, and a phosphate group. a. In RNA, the purines are adenine and guanine, the pyrimidines are uracil and cytosine, and the sugar is ribose. b. In DNA, thymine replaces uracil, and the sugar is 2'-deoxyribose. 2 Positions on the base receive non-prime superscripts, and positions on the sugar receive prime superscripts. 3 The heterocyclic base is bonded to C 1' of the sugar. 4 is vastly larger than RNA and is found in the cell nucleus. B Nucleic acids. 1 Nucleic acids are composed of nucleotides connected by a phosphodiester bond between the 5' ester of one nucleotide and the 3' hydroxy 1 group of another. a. One end of the nucleic acid polymer has a free hydroxy 1 group and is called the .
.
.
.
DNA
.
.
3'
end.
The other end has a free phosphate group and is called the 5' end. The structure of a nucleic acid depends on the order of bases. The sequence of bases is described by starting at the 5' end and listing b.
2
.
3
.
the bases
by
their one-letter abbreviations in order of occurrence. in DNA (Section 28.2). DNA consists of two polynucleotide strands coiled in a double helix.
C. Base-pairing 1
.
Adenine and thymine hydrogen-bond with each other, and cytosine and guanine hydrogen-bond with each other. Because the two DNA strands are complementary, the amount of A equals the amount of T, and the amount of C equals the amount of G. The double helix is 20 A wide, there are 10 bases in each turn, and each turn is 34 a.
2.
3
.
4
.
A in height. The double
helix has a major groove and a minor groove into which polycyclic aromatic molecules can intercalate. D. The "central dogma" of molecular genetics. The function of is to store genetic information and to pass it on to RNA, 1 which, in turn, uses it to make proteins. 2. Replication, transcription and translation are the three processes that are responsible for carrying out the central dogma. transfer of genetic information (Sections 28.3-28.5). A. Replication of (Section 28.3). Replication is the enzyme-catalyzed process whereby makes a copy of itself. 1 2. Replication is semiconservative: each new strand of consists of one old strand and one newly synthesized strand. 3 How replication occurs:
DNA
.
II.
The
DNA
DNA DNA
.
.
a.
The i.
b
.
DNA helix partially unwinds. is catalyzed by the enzyme helicase. nucleotides form base pairs with their complementary partners.
This process
New
Biomolecules: Nucleic Acids
c.
Formation of new bonds 5'
catalyzed by
DNA polymerase and takes place in the
-* 3' direction.
Bond formation occurs by
i.
d.
is
791
attack of the
3'
hydroxy 1 group on the
5'
triphosphate, with loss of a diphosphate leaving group. in the 5' -» 3' direction.
Both new chains are synthesized i.
One
ii.
The other
chain
synthesized continuously (the leading strand). is synthesized in small pieces, which are later joined by ligase enzymes (the lagging strand). (Section 28.4). B. Transcription - synthesis of 1 There are 3 main types of RNA: a. Messenger (mRNA) carries genetic information to ribosomes when protein synthesis takes place. (rRNA), complexed with protein, comprises the physical b. Ribosomal makeup of the ribosomes. c. Transfer (tRNA) brings amino acids to the ribosomes, where they are joined to make proteins. d. There are also small RNAs, which carry out a variety of cellular functions. 2. contains "promoter sites", which indicate where synthesis is to begin, and base sequences that indicate where synthesis stops. polymerase binds to the promoter sequence. a. 3 is synthesized in the nucleus by transcription of DNA. a. The partially unwinds, forming a "bubble". bases. b Ribonucleotides form base pairs with their complementary is
strand
DNA
RNA
.
RNA
RNA
RNA
DNA
mRNA
mRNA
.
RNA mRNA DNA
DNA
.
c.
d
.
e.
4.
Bond formation
occurs in the 5' -» 3' direction. Only one strand of (the antisense, or noncoding, strand) is transcribed. Thus, the synthesized is a copy of the sense (coding) strand with
DNA mRNA
U
replacing T. Synthesis of a.
b.
mRNA is not necessarily continuous. DNA DNA
Often, synthesis begins in a region of called an exon and is interrupted by seemingly noncoding region of called an intron. a In the final mRNA, the noncoding sections have been removed and the
remaining pieces have been spliced together by specific enzymes. Translation (Section 28.5). C. 1 Translation is the process in which proteins are synthesized at the ribosomes by using as a template. 2. The message delivered by is contained in "codons" - 3-base groupings that .
mRNA
mRNA
amino acid. Amino acids are coded by 61 of the possible 64 codons. a. b. The other 3 codons are "stop" codons. Each tRNA is responsible for bringing an amino acid to the growing protein chain. a. A tRNA has a cloverleaf- shaped secondary structure and consists of 70-100 are specific for an
3
.
ribonucleotides.
b Each tRNA contains an anticodon complementary to the mRNA codon. 4. The protein chain is synthesized by enzyme-catalyzed peptide bond formation. 5 A 3-base "stop" codon on mRNA signals when synthesis is complete. .
.
IE.
DNA technology (Sections 28.6-28.8). A. DNA sequencing (Section 28.6). Before sequencing, the DNA chain is cleaved at specific sites by restriction 1
.
endonucleases.
The
restriction endonuclease recognizes both a sequence on the sense strand and complement on the antisense strand. b. The DNA strand is cleaved by several different restriction endonucleases, to produce fragments that overlap those from a different cleavage. a.
its
792
Chapter 28
2.
3
.
Maxam-Gilbert DNA sequencing. a. This method uses chemical techniques. Sanger dideoxy
DNA sequencing.
The following mixture is assembled: i. The restriction fragment to be sequenced. A primer (a small piece of DNA whose sequence ii. on the 3' end of the fragment). iii. The 4 DNA nucleoside triphosphates.
a.
iv.
complementary
to that
Small amounts of the four dideoxynucleotide triphosphates, each of which is
b.
is
labeled with a different fluorescent dye.
DNA polymerase is added to the mixture, and a strand begins to grow from the end of the primer.
B.
c.
Whenever a dideoxynucleotide
d.
When reaction is complete,
e.
Because fragments of all possible lengths are represented, the sequence can be read by noting the color of fluorescence of each fragment.
is
incorporated, chain growth stops. by gel electrophoresis.
the fragments are separated
DNA synthesis (Section 28.7). DNA synthesis is based on principles similar to those for peptide synthesis. 1 .
2
.
The following steps are needed: The nucleosides are protected and bound a. i. ii.
to a silica support.
Adenine and cytosine bases are protected by benzoyl groups. Guanine is protected by an isobutyryl group.
Thymine isn't protected. The 5' -OH group is protected as a DMT ether. The DMT group is removed. The polymer-bound nucleoside is coupled with a protected nucleoside iii.
iv.
b
.
c.
containing a phosphoramidite group.
d
.
e.
f
.
i.
One of the phosphoramidite oxygens
ii.
Tetrazole catalyzes the coupling.
is
protected as a /3-cyano ether.
The phosphite is oxidized to a phosphate with I2. Steps b - d are repeated until the desired chain is synthesized. All protecting groups are removed and the bond to the support treatment with aqueous ammonia.
is
cleaved by
C. The polymerase chain reaction (Section 28.8). The polymerase chain reaction (PCR) can produce vast quantities of a 1 .
DNA
fragment.
2
.
3.
The key to PCR is Taq DNA polymerase, a heat-stable enzyme. a. Newer heat-stable DNA polymerase enzymes have become available. Steps in PCR: a. The following mixture is heated to 95 °C (a temperature at which DNA becomes single-stranded); i.
Taq polymerase.
ii.
Mg2+ ion.
iii.
iv.
b c
.
.
d.
The 4 deoxynucleotide triphosphates. A large excess of two oligonucleotide primers, each of which
is
complementary to the ends of the fragment to be synthesized. The temperature is lowered to 37 °C - 50 °C, causing the primers to hydrogenbond to the single-stranded DNA. After raising the temperature to 72 °C, Taq catalyzes the addition of further nucleotides, yielding two copies of the original DNA.
The process
is
repeated until the desired quantity of
DNA is produced.
Biomolecules: Nucleic Acids
793
Solutions to Problems
28.1 5'
end
2'-Deoxyadenosine 5'-phosphate (A)
H
N
NH 2
2'-Deoxyguanosine 5 '-phosphate (G) 3'
end
OH
28.2 5*
end H
O" N*
I
"O— P=0
I
Uridine 5'-phosphate (U)
I
OCH
"O— P=0
Adenosine 5'-phosphate (A) 3'
28.3
end
OH
OH
DNA (5' end) GGCTAATCCGT (3' end) is complementary to DNA (3' end) CCGATTAGGCA (5* end) Remember
that the
complementary strand has the
3'
end on the
left
and the
5'
end on the
right.
The complementary sequence can
DNA used
(5'
(3'
end) to
5',
also be written as:
ACGGATTAGCC (3' end). Be sure that you know which format is being or
5'
to
3').
Chapter 28
794
28.4
Adenine
O
H
28.5
28.6
DNA RNA RNA DNA
28.7-28
(5'
(3'
end) end)
end) end)
(5' (3'
GATTACCGTA
(3'
CUAAUGGCAU
(5'
UUCGCAGAGU
(3'
AAGCGTCTCA
(5'
end) end)
end) end)
is
complementary
to
antisense (noncoding) strand
Several different codons can code for the same amino acid (Table 28. 1). The 8 corresponding anticodon follows the slash mark after each codon. The codons are written with the 5' end on the left and the 3' end on the right, and the tRNA anticodons have the 3' end on .
mRNA
the left
and the
Amino
5'
end on the
Ala
acid:
Codon sequence/
tRNA
right.
anticodon:
Leu
Phe
UUU/AAA UUC/AAG
GCU/CGA GCC/CGG GCA/CGU GCG/CGC
UUA/AAU UUG/AAC CUU/GAA CUC/GAG CUA/GAU CUG/GAC
Tyr
UAU/AUA UAC/AUG
28.9-28.10 The
mRNA base sequence:
The amino The
(5'
end)
DNA sequence:
(3*
end)
(5'
end)
Leu— Met— Ala— Trp— Pro-(stop)
acid sequence:
(antisense strand)
CUU-AUG-GCU-UGG-CCC-UAA
(3'
end)
GAA-TAC-CGA-ACC-GGG-ATT
Biomolecules: Nucleic Acids
28.11 OCH.
OCH-
°-Q-HJ O
O
CH 2 Ohv
H
j
Polymer)-
CB
CO
/
+
CH 2 OH
Polymer- CO
H-rOH 2
+
O
Base
H
HoO
O
Base
H
OCH<
CH 3
OH
O u .11 polymer)- CCX^vA
HoO
CHoOH
v
O
H
H-CH20H ft
Jolymej)+
HoO +
O
N Base
H
Cleavage of DMT ethers proceeds by an cation
is
CO^vX
Sjsjl
mechanism and
is
N Base
rapid because the
DMT
unusually stable.
28.12
^-:nh 3 r
..^7
0= P— O— CH2 This
is
nitrile
H r-i
CHCN
0=P— O"
+
H 2 C=CHCN
+
NH 4H
an E2 elimination reaction, which proceeds easily because the hydrogen a to the group is acidic.
795
Width: 612 Height: 792
796
Chapter 28
Visualizing Chemistry
28.13
H Guanine (G)
DNA RNA All three bases are found in
Uracil (U)
RNA RNA, but only
Cytosine (C)
DNA RNA guanine and cytosine are found in
DNA.
28.14
The triphosphate made from 2',3'-dideoxythymidine 5' phosphate is labeled with a fluorescent dye and used in the Sanger method of DNA sequencing. Along with the restriction fragment to be sequenced, a DNA primer, and a mixture of the four dNTPs, small quantities of the four labeled dideoxyribonucleotide triphosphates are mixed together.
DNA polymerase is added, and a strand of DNA complementary to the restriction fragment is synthesized. Whenever a dideoxyribonucleotide is incorporated into the DNA chain, chain growth stops. The fragments are separated by electrophoresis, and each terminal dideoxynucleotide can be identified by the color of its fluorescence. By identifying these terminal dideoxy nucleotides, the sequence of the restriction fragment can be read.
28.15 According
map, the nitrogen at the 7 position of 9methylguanine is more electron-rich (red) and should be more nucleophilic. Thus 9methylguanine should be the better nucleophile. to the electrostatic potential
9-Methylguanine
9-Methyladenine
Biomolecules: Nucleic Acids
797
Additional Problems
DNA that codes for natriuretic peptide (32 amino acids) consists of 99 bases; 3 bases code for each of the 32 amino acids in the chain (96 bases), and a 3 -base "stop" codon is
28.16 The
also needed.
28.17 Position
9:
Horse amino acid = Gly mRNA codons
(5'
GGU GGC GGA GGG
Human amino acid = — > 3'):
Ser
UCU UCC UCA UCG AGU AGC
DNA bases (antisense strand 3' — > 5'): CCC AGA AGG AGT AGC T£A TCG
CCA CCG CCT
The underlined horse DNA base underlined) by only one base.
triplets differ
from
their
human
counterparts (also
Position 30:
Horse amino acid = Ala mRNA codons
Human amino acid = Thr
(5'
—>
GCU GCC GCA GCG
3'):
ACU ACC ACA ACG
DNA bases (antisense strand 3' — > 5'): CGA CGG CGT CGC TGA TGG TGT TGC Each of the above groups of DNA bases from horse insulin has a counterpart in human insulin that differs from it by only one base. It is possible that horse insulin DNA differs from human insulin DNA by only two bases out of 159!
28.18 The percent of A always
equals the percent of T, since A and T are complementary. The G equals the percent C for the same reason. Thus, sea urchin DNA contains about 32% each of A and T, and about 18% each of G and C.
percent
28.19 Even though doesn't stop.
DNA
UAA
shown contains the stretch of in sequence, protein synthesis are read as 3-base individual units from start to end, and, in this
The codons
mRNA sequence, the unit UAA is read as part of two codons, not as a single codon. 28.20
Restriction endonucleases cleave
DNA base sequences that are palindromes, meaning that
complement when both are read in the (5') to (3') sequence in (c), CTCGAG is recognized. The sequence in (a), GAATTC, is also a palindrome and is recognized by a restriction endonuclease. The sequence in (b) is not a palindrome and is not recognized. the sequence reads the
same
as the
direction. Thus, the
28 21—28 23
mRNA
codon
Amino
DNA tRNA The
:
(5'->3')
acid:
sequence: anticodon:
(3'->5*) (3'->5')
(a)
AAU
(b)
GAG
(c)
UCC
(d)
CAU
Asn
Glu
Ser
His
TTA
CTC
GTA
UUA
CUC
AGG AGG
DNA sequence of the antisense (noncoding) strand is shown.
GUA
798
Chapter 28
UAC is a codon for tyrosine. DNA chain.
28.24-28.25 strand of a
It
was transcribed from
ATG of the antisense
mRNA codon 5'
DNA
end
3'
end
O— P=0 O
u
N
CH 2
J
f
O
>
N
0=P— O" I
I
NH'
o
£-0 CH,
T
O
V
H'
I
^CHoO=P— 0" I
o
ISL
.
N
o
CH,
M
>
?
0=P— O" I
O
3'
28.26
Tyr—
UAC
UAU
—
end
5'
Gly
Gly
Phe
GGU GGC GGA GGG
GGU GGC GGA GGG
UUU
A total of2x4x4x2x 1 x3 = metenkephalin!
o
-Met
AUG
UUC
194 different
(stop)
is
coded by
UAA UAG UGA
mRNA sequences can code for
end
799
Biomolecules: Nucleic Acids
28.27 Angiotensin
II:
Asp
Arg
Tyr
Val
lie
His
Phe
Pro
(stop)
mRNA sequence: GAU CGU GUU UAU AUU CAU CCU UUU UAA (5'->3*) GAC CGC GUC UAC AUC CAC CCC UUC UAG AUA UGA CGA GUA CCA CGG GUG CCG AGA AGG As in the previous problem, many mRNA sequences (13,824) can code for angiotensin 28.28
DNA coding strand mRNA (5'->3'):
(5'->3'):
CTT- CGA-CCA- GAC — AGC — TTT
CUU-CGA-CCA-GAC-AGC-UUU
Amino acid sequence: The mRNA sequence is
Arg Pro Asp Phe Ser complement of the DNA noncoding (antisense)
Leu
the
DNA
coding (sense) strand. Thus, the the complement of the of the coding (sense) strand, with T replaced by U. is
strand,
which
mRNA sequence is a copy
DNA
28.29
mRNA sequence (5'->3'): CUA-GAC-CGU-UCC-AAG-UGA Amino
Leu
Acid:
28.30
DNA coding strand (5'->3'): mRNA sequence (5'->3'):
Asp
Arg
Ser
Lys
(stop)
Original Sequence
Miscopied Sequence
-CAA-CCG-GAT-CAA-CCG-GAU-
-CGA-CCG-GAT-CGA-CCG-GAU-
— —
Amino
— —
acid sequence: -Gln Pro Asp-Arg Pro Aspgene sequence were miscopied in the indicated way, a glutamine in the original protein would be replaced by an arginine in the mutated protein. If this
28.31
1
.
First, protect the nucleosides.
(a)
II.
Bases are protected by amide formation.
NH 2 C 6 H 5 COQ Adenine
pyridine
C 6 H 5 COCI Cytosine
pyridine
(CH 3 2 CHCOCI )
Guanine
pyridine
NHCOCH(CH 3 Thymine does not need
to
be protected.
);
800
Chapter 28
The
(b)
hydroxyl group
5'
is
protected as
Base
its
p-dimethoxytrityl
(DMT) Base
DMTOCH'
HOCHc 1
.
ether.
Base i
2.
DMTBr
OCCH 2 CH 2 CO
O
OCCHoCHoCO 2 2
O
2. Attach a protected 2-deoxycytidine nucleoside to the
II
H
o
o
polymer support.
Cytosine
DMTOCH 2
OCCH 2 CH 2 CNH(CH 2
OCCHoCHoCO 2 2 II
H
o
o
Silica)
—
3.
O
— CCH CH CNH(CH —
3 Si
—
2 ) 3 Si
2
(^Silica
O
Support
DMT ether. Cytosine
Cytosine
DMTOCHc
HOCHc CHCI 2 C0 2 H
CH 2 CI 2
O
^Silica
{?
2
Cleave the
)
Si(CH 2 ) 3 NH 2
?\
Let
Cytosine
DMTOCH2
(Support
O
(Support
Biomolecules: Nucleic Acids
4.
801
Couple protected 2'-deoxythymidine to the polymer-2'-deoxycytidine. (The nucleosides have a phosphoramidite group at the 3' position.)
Thymine
DMTOCH,
Thymine
DMTOCH' Tetrazole
O
o
P— N(/- Pro)2 NCCH 2 CH 2
P-OCH2
Cytosine
NCCH 2 CH 2
O
C Support
HOCH
5
.
6.
Oxidize the phosphite product to a phosphate
triester,
using iodine.
Repeat steps 3-5 with protected 2'-deoxyadenosine and protected 2'-deoxyguanosine.
802
Chapter 28 7
.
Cleave
all
protecting groups with aqueous
yield the desired sequence.
Guanine
Guanine
DMTOCH,
ammonia to
HOCH 2
o
?
0=P— O"
I
0=P— OCH 2 CH2 CN Adenine
Adenine
OCH 2
OCH,
HoO
O
0=P— o~
I
0=P— OCHoCHoCN Thymine
I
OCH 2
o
?
0=P— O"
I
0=P — OCH2 CH 2CN Cytosine
O
Thymine
I
OCH 2
(Support
Cytosine
OCHc
OH
Biomolecules: Nucleic Acids
28.32 Both of these cleavages occur by the now-familiar nucleophilic
acyl substitution route. nucleophile adds to the carbonyl group, a proton shifts location, and a second group is eliminated. Only the reacting parts of the structures are sown.
Deprotection
at 1
HoN
Deprotection at
HoN
2:
OH :0.
O.
C / \
V^r (P.
mi
N H3
t
:orNH2
803
A
804
Chapter 28
28.34
This reaction involves addition of a thiol residue of the enzyme to malonic semialdehyde, + yielding a hemithioacetal (Step 1). Oxidation by (step 2), followed by nucleophilic acyl substitution by Co A (Step 3), gives malonyl Co A.
NAD
Malonic semialdehyde
O
SCoA
Malonyl
28.35 The
steps: (1)
CoA
phosphorylation by
ATP;
(2) cyclization; (3) loss
of phosphate; (4)
tautomerization.
ADPO— P0 3^ H
ADP
1
I.
5-Phosphoribose
Formylglycinamidine ribonucleotide
i
:b
H
P;
:B
H
/A
4.
NH
NH,
5-Phospho-
5-Phospho-
5-Phospho-
ribose
ribose
ribose
Aminoimidazole ribonucleotide
Biomolecules: Nucleic Acids
28.36 The
Addition of water; (2) Proton
steps: (1)
shift; (3)
Elimination of
NH 3
805
.
..-r< Guanine
H
+ NH 3
Xanthine
28.37 Both
steps are nucleophilic acyl substitutions,
(a)
"0 C. 2
o2c,
Pi,
- ..H 2 N
HoN
'»
H+
°2 C H 2N
r-H HoN'
"
C0 2
2-,
°="°) H H
°*
O
N I
r-H C0 2
H Carbamoyl aspartate
"
Width: 612 Height: 792
806
Chapter 28
28.38 (a), (b)
The -CH2- group (d)
C2 of deoxyribose
The missing atoms
ring, the
(e)
at
is
missing from ganciclovir.
are part of the relatively inflexible deoxyribose ring. Without the
DNA chain is floppy and base pairing to form a double helix can't occur.
As mentioned
in (d), the inability to
form base
pairs stops the replication of
DNA.
Chapter 29 - The Organic Chemistry of Metabolic Pathways
Chapter Outline I.
Overview of metabolism and biochemical energy (Section 29.1). A. Metabolism. The reactions that take place in the cells of organisms 1 .
are collectively called
metabolism.
The
reactions that produce smaller molecules from larger molecules are called catabolism and produce energy. b The reactions that build larger molecules from smaller molecules are called anabolism and consume energy. Catabolism can be divided into four stages: a. In digestion, bonds in food are hydrolyzed to yield monosaccharides, fats, and a.
.
2
.
amino b
.
c.
acids.
These small molecules are degraded In the citric acid cycle, acetyl
to acetyl
CoA.
CoA is catabolized to CO2,
and energy
is
produced. d.
Energy from the is
citric
acid cycle enters the electron transport chain,
where
ATP
synthesized.
B Biochemical energy. .
1
2
.
ATP, a phosphoric acid anhydride, is the storehouse for biochemical energy. The breaking of a P-O bond of ATP can be coupled with an energetically
unfavorable reaction, so that the overall energy change is favorable. The resulting phosphates are much more reactive than the original compounds. Lipid metabolism (Sections 29.2-29.4). A. Catabolism of fats (Section 29.2-29.3). Triacylglycerols are first hydrolyzed in the stomach and small intestine to yield 1 3
II.
.
.
.
glycerol plus fatty acids (Section 29.2). a. The reaction is catalyzed by a lipase. i.
Aspartic acid, serine and histidine residues in the
enzyme bring about
reaction.
b
.
Glycerol i
.
ii.
2.
3
.
is
phosphorylated and oxidized and enters glycolysis.
The mechanism of oxidation involves a hydride The addition to NAD + is stereospecific.
^-Oxidation
transfer to
NAD+
.
mitochondria) (Section 29.3). by /3-oxidation, a 4-step spiral that results in the cleavage of an n-carbon fatty acid into nil molecules of acetyl CoA. (in the
a.
Fatty acids are degraded
b.
Before entering /^-oxidation, a
fatty acid is first
converted to
its
fatty-acyl
CoA.
Steps of p oxidation. Introduction of a double bond conjugated with the carbonyl group. a. i. The reaction is catalyzed by acyl CoA dehydrogenase. ii. The enzyme cofactor FAD is also involved and is reduced. iii.
The mechanism involves
abstraction of the
pro-R a and p hydrogens,
resulting in formation of a trans double bond.
b
.
c.
Conjugate addition of water to form an alcohol. i. The reaction is catalyzed by enoyl CoA hydratase. Alcohol oxidation. i. The reaction is catalyzed by L-3-hydroxyacyl CoA dehydrogenase. + + ii. The cofactor NAD is reduced to NADH/H at the same time. iii. Histidine deprotonates the hydroxyl group.
808
Chapter 29
d
.
Cleavage of acetyl i.
The
reaction,
CoA from the chain.
which
is
catalyzed by /3-keto thiolase,
is
a retro-Claisen
reaction.
Nucleophilic addition of coenzyme A to the keto group is followed by loss of acetyl CoA enolate, leaving behind a chain-shortened fatty-acyl CoA. ^-carbon fatty acid yields nil molecules of acetyl CoA after (rc/2-1) passages of ii.
3
.
An
/^-oxidation. a.
Since most fatty acids have an even number of carbons, no carbons are
left
over
after p- oxidation.
B
.
b. Those with an odd number of carbons require further steps for degradation. Biosynthesis of fatty acids (Section 29.4). 1
.
General principles. a.
b
.
c.
2
.
most cases, the pathway of synthesis isn't the exact reverse of degradation. i. If AG° is negative for one route, it must be positive for the exact reverse, which is thus energetically unfavorable. ii. The metabolic strategy is for one pathway to be related to its reverse but not to be identical. All common fatty acids have an even number of carbons because they are synthesized from acetyl CoA. In vertebrates, a large multienzyme synthase complex catalyzes all steps in the pathway. In
Synthetic pathway. Steps 1-2: Acyl transfers convert acetyl CoA to a. Acetyl CoA is converted to acetyl ACP. i. ii.
b
.
The
acetyl group of acetyl
more
reactive species.
ACP is transferred to the synthase enzyme.
Steps 3-4: Carboxylation and acyl transfer. i. Acetyl CoA reacts with bicarbonate to yield malonyl (a). The
coenzyme
biotin, a
CoA
CO2 carrier, transfers CO2
in
and ADP. a nucleophilic acyl
substitution reaction. ii.
At
CoA is converted to malonyl ACP.
both acetyl groups and malonyl groups are bound to the synthase enzyme. Step 5: Condensation. i. Claisen condensation forms acetoacetyl CoA from acetyl synthase and
iii.
c.
Malonyl
this point,
A
malonyl ACP.
The reaction proceeds by an initial decarboxylation of malonyl an enolate that adds to acetyl synthase to form acetoacetyl CoA. Steps 6-8: Reduction and dehydrogenation. The ketone group of acetoacetyl CoA is reduced by NADPH. i. ii.
d.
ACP to give
The ^-hydroxy thiol ester is dehydrated. The resulting double bond is hydrogenated by NADPH to yield butyryl ACP. The steps are repeated with butyryl synthase and malonyl ACP to give a sixii.
iii.
e.
carbon
unit.
up to palmitic acid (16 carbon atoms) are synthesized by this route, Elongation of palmitic acid and larger acids occurs with acetyl CoA units as the two-carbon donor, rather than ACP. HI. Carbohydrate metabolism (Sections 29.5-29.8). A. Catabolism of carbohydrates (Sections 29.5-29.7). 1. Glycolysis (Section 29.5). Glycolysis is a 10-step series of reactions that converts glucose to pyruvate. a. f
.
Fatty acids i.
The Organic Chemistry b.
of Metabolic Pathways
809
Steps 1-2: Phosphorylation and isomerization.
Glucose
i.
is
phosphorylated
at the 6-position
The enzyme hexokinase
by reaction with ATP.
involved. ii. Glucose 6-P is isomerized to fructose 6-P by glucose-6-P isomerase. Step 3: Fructose 6-P is phosphorylated to yield fructose 1,6-bisphosphate. (a).
c.
is
(a). ATP and phosphofructokinase are involved. Step 4: Cleavage. Fructose 1,6-bisphosphate is cleaved to glyceraldehyde 3-phosphate and i. dihydroxyacetone phosphate. (a). The reaction is a reverse aldol reaction catalyzed by aldolase. Step 5: Isomerization. e. Dihydroxyacetone phosphate is isomerized to glyceraldehyde 3-phosphate. i. ii. The net result is production of two glyceraldehyde 3 -phosphates, both of which pass through the rest of the pathway. Steps 6-7: Oxidation, phosphorylation, and dephosphorylation. f Glyceraldehyde 3-phosphate is both oxidized and phosphorylated to give i. 1 ,3-bisphosphoglycerate. + (a). Oxidation by occurs via a hemithioacetal to yield a product that forms the mixed anhydride. ii. The mixed anhydride reacts with ADP to form ATP and 3 -phosphogly cerate (a). The enzyme phosphogly cerate kinase is involved. g Step 8: Isomerization. 3 -Phosphogly cerate is isomerized to 2-phosphogly cerate by 1 phosphoglycerate mutase. h. Steps 9-10: Dehydration and dephosphorylation. 2-Phosphoglycerate is dehydrated by enolase to give phosphoenolpyruvate. i. Pyruvate kinase catalyzes the transfer of a phosphate group to ADP, with ii. formation of pyruvate. The conversion of pyruvate to acetyl Co A (Section 29.6).
d.
.
NAD
.
The conversion pyruvate
a.
-» acetyl
Co A
is
catalyzed by an
enzyme complex
called pyruvate dehydrogenase complex.
b
.
Step i.
1
:
Addition of thiamin.
A nucleophilic ylide group on thiamin diphosphate adds to the carbonyl group of pyruvate
to yield a tetrahedral intermediate. Decarboxylation. d 3: Reaction with lipoamide. The enamine product of decarboxylation reacts with lipoamide, displacing i. sulfur and opening the lipoamide ring. Step 4: Elimination of thiamin diphosphate ylide. e. f Step 5: Acyl transfer. i. Acetyl dihydrolipoamide reacts with coenzyme to give acetyl Co A. ii. The resulting dihydrolipoamide is reoxidized to lipoamide by FAD. + iii. FADH 2 is reoxidized to FAD by g Other fates of pyruvate. In the absence of oxygen, pyruvate is reduced to lactate. i. ii. In bacteria, pyruvate is fermented to ethanol. The citric acid cycle (conversion of acetyl Co to CO2) (Section 29.7). a. Characteristics of the citric acid cycle. The citric acid cycle is a closed loop of eight reactions. i ii. The intermediates are constantly regenerated. and FADH2 are available, which means iii. The cycle operates as long as c.
.
Step Step
2:
.
A
NAD
.
A
.
NAD
that
oxygen must also be
available.
.
810
Chapter 29
b.
Steps 1-2: Addition to oxaloacetate. Acetyl CoA adds to oxaloacetate to form citryl
i.
Co A, which
is
hydrolyzed
to
citrate.
c.
The
is catalyzed by citrate synthase. isomerized to isocitrate by aconitase. (a). The reaction is an Elcb dehydration, followed by conjugate addition of water. Steps 3-4: Oxidative decarboxylations. i. Isocitrate is oxidized by isocitrate dehydrogenase to give a ketone that loses
(a).
Citrate
ii.
reaction
is
CO2 to give
d.
e
.
a-ketoglutarate.
is transformed to succinyl CoA in a reaction catalyzed by a multienzyme dehydrogenase complex. Steps 5-6: Hydrolysis and dehydrogenation of succinyl CoA. Succinyl CoA is converted to an acyl phosphate, which transfers a i. phosphate group to GDP in a reaction catalyzed by succinyl CoA synthase. ii. Succinate is dehydrogenated by FAD and succinate dehydrogenase to give
a-Ketoglutarate
ii.
fumarate; the reaction is stereospecific. Steps 7-8: Regeneration of oxaloacetate. i. Fumarase catalyzes the addition of water to fumarate to produce (5)-malate. + ii. (5)-malate is oxidized by and malate dehydrogenase to complete the
NAD
cycle.
C. Carbohydrate biosynthesis: gluconeogenesis (Section 29.8). 1
.
2
.
3
.
Step 1 Carboxylation. a. Pyruvate is carboxylated to yield oxaloacetate in a reaction that uses biotin and :
ATP.
4.
Step 2: Decarboxylation and phosphorylation. a. Concurrent decarboxylation and phosphorylation of oxaloacetate produce phosphoenolpyruvate. Steps 3-4: Hydration and isomerization. Conjugate addition of water gives 2-phosphogly cerate. a. b. Isomerization produces 3 -phosphogly cerate. Steps 5-7: Phosphorylation, reduction and tautomerization. Reaction of 3 -phosphogly cerate with ATP yields an acyl phosphate. a. + b The acyl phosphate is reduced by NADPH/H to an aldehyde. c. The aldehyde tautomerizes to dihydroxy acetone phosphate. Step 8: Aldol reaction. a. Dihydroxy acetone phosphate and glyceraldehyde 3-phosphate join to form .
5
.
fructose 1,6-bisphosphate.
This reaction involves the imine of dihydroxyacetone phosphate, which forms an enamine that takes part in the condensation. Steps 9-11: Hydrolysis and isomerization. a. Fructose 1,6-bisphosphate is hydrolyzed to fructose 6-phosphate. b Fructose 6-phosphate isomerizes to glucose 6-phosphate. c. Glucose 6-phosphate is hydrolyzed to glucose. Several of these steps are the reverse of steps of glycolysis.
b 6
.
.
.
7
.
The Organic Chemistry
of Metabolic Pathways
811
IV. Protein metabolism (Section 29.9). Catabolism of proteins: Deamination. 1
.
The pathway to amino acid catabolism: a. The amino group is removed as ammonia by transamination. b The ammonia is converted to urea. c. What remains is converted to a compound that enters the citric i. Each carbon skeleton is degraded in a unique pathway. .
2.
acid cycle,
Transamination.
The -NH2 group of an amino
acid adds to the aldehyde group of pyridoxal phosphate to form an imine (Schiff base). b The imine tautomerizes to a different imine. c. The second imine is hydrolyzed to give an a-keto acid and an amino derivative of pyridoxal phosphate. d The pyridoxal derivative transfers its amino group to a-ketoglutarate, to regenerate pyridoxal phosphate and form glutamate. 3 Deamination. a. The glutamate from transamination undergoes oxidative deamination to yield ammonium ion and a-ketoglutarate. conclusions about biological chemistry (Section 29.10). 1 The mechanisms of biochemical reactions are almost identical to the mechanisms of a.
.
.
.
V.
Some
.
laboratory reactions.
2
.
Most metabolic pathways are linear. a. Linear pathways make sense when a multifunctional molecule b.
transformation. Cyclic pathways
is
undergoing
may be more energetically feasible when a molecule
Solutions to Problems
29.1
This reaction
is
a substitution at phosphorus, with
ADP as the leaving group.
is
small.
812
Chapter 29
29.2
O II
CH 3 CH2 - CH 2 CH 2 - CH 2 CH 2 - CH 2 CSCoA Caprylyl
CoA
Of |
(passage 4)
O
II
II
CH 3 CH 2 -CH 2 CH 2 -CH 2 CSCoA Hexanoyl
CoA
|
+
CH3CSC0A
(passage 5)
O
CH 3 CH 2 -CH 2 CSCoA
CoA
Butanoyl
O
1
+
CH3CSC0A r.
.
(passage 6)
O
I T
II
II
CH3CSC0A 29.3
+
CH3CSC0A
A fatty acid with n carbons yields nil acetyl CoA molecules after {nl2 -
1)
passages of the
p-oxidation pathway.
(a)
\
\
\
\
\
\
\
CH 3 CH 2 — CH 2 CH 2 — CH 2 CH 2 — CH 2 CH 2 — CH 2 CH 2 — CH2 CH 2 — CH 2 CH 2 — CH 2 C0 2 H
IP
oxidation
O II
8
CH3CSC0A
Seven passages of the /^-oxidation pathway are needed. (b)
O
CH 3 CH 2
— (CH CH — CH C0 H 2
2) 8
2
j /
oxidation^
2
1
Q
CH3 q SCoA
Nine passages of the ^-oxidation pathway are needed.
29.4
p-Hydroxybutyryl ACP resembles the ^-hydroxy ketones that were described in Chapter that dehydrate readily by an ElcB mechanism.
23 and
SACP
:0^ ^SACP
^SACP c
H-C-H^ H— C— OH
" B
I
h-c; H— C-r OH
CH II
CH
I
CH 3
CH 3
+ HoO
The Organic Chemistry
29.5
of Metabolic Pathways
813
A fatty acid synthesized from 13CH3CC>2H has an alternating labeled and unlabeled carbon The carboxylic
chain.
acid carbon
is
unlabeled.
CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CO2H
29.6
The
face in front of the plane of the page is the it occurs at the Si face.
Re
face.
Since addition occurs from behind
the plane of the page,
—
29.7
ATP is produced
29.8
Step 1 is a nucleophilic acyl substitution at phosphorus (phosphate transfer) by the group at C6 of glucose, with ADP as the leaving group.
in step 7 (1,3-bisphosphogly cerate > 3-phosphoglycerate) and in step 10 (phosphoenolpyruvate > pyruvate). Refer to Figure 29.7.
—
-OH
Step 2 is an isomerization, in which the pyranose ring of glucose 6-phosphate opens, tautomerism causes isomerization to fructose 6-phosphate, and a furanose ring is formed. Step 3
is
a substitution, similar to the
one
in step
1
,
involving the
-OH
group
at
C
1
of
fructose 6-phosphate (phosphate transfer).
Step 4
is
a retro-aldol reaction that cleaves fructose 1,6-bisphosphate to glyceraldehyde
3-phosphate and dihydroxyacetone phosphate.
Step 5
is
an isomerization of dihydroxyacetone phosphate by keto-enol tautomerization.
to glyceraldehyde
3-phosphate
that occurs
Step 6 begins with a nucleophilic addition reaction to the aldehyde group of glyceraldehyde 3-phosphate by a thiol group of an enzyme to form a hemithioacetal, which + is oxidized by NAD to an acyl thioester. Nucleophilic acyl substitution by phosphate yields the product 1,3-bisphosphoglycerate.
Step 7
is
a nucleophilic acyl substitution reaction at phosphorus, in
with 1,3-diphosphoglycerate, yielding
ATP and 3-phosphoglycerate
Step 8
is
an isomerization of 3-phosphoglycerate
Step 9
is
an ElcB elimination of
H2O
to
to
ADP reacts
(phosphate transfer).
2-phosphogly cerate.
form phosphoenolpyruvate.
is a substitution reaction at phosphorus that forms tautomerizes to pyruvate (phosphate transfer).
Step 10
which
ATP and enolpyruvate,
which
814
Chapter 29
29.9 1
CHO
1
—2«C— OH
2C=0 HO— — H H— C— OH H-C-OH
H
1
HO—3 C—
J H-^C— OH
Dihydroxyphosphate
2.5
(pHOH
1.6
CH 2 OP03 2-
2.5 1 -
Glyceraldehyde 3 -phosphate
and
I
CHO 2-
I
2.5
dehydrogenase
6CH 3
C=0
1.6
complex
2
C0 2
CH 3
Acetyl
-CH3
+
I
Pyruvate
1 and 6 of glucose end up as glucose end up as CO2.
CHO
SCoA Pyruvate
C=0
Carbons
3
Glyceraldehyde 3-phosphate
COo" 2 I
CHOH
6CH 2 OP03
Fructose 1,6-bisphosphate 3.4
2
CoA
groups of acetyl CoA. and carbons 3 and 4 of
Refer to Figure 29.12.
isocitrate are tricarboxylic acids.
29.11
hO/rC0 2
OH
A— H^
'
OoC
"OoC
COc
H
H
HO—
cw-Aconitate
29.12 The pro-R hydrogen
is
removed during dehydration, and
the reaction occurs with anti
pro-S Citrate
OH
(2^,35")-Isocitrate
geometry.
pro-R
2-
H— C— OH
+
'
CHO
Citrate
4
CH 2 OP0 3
acetone
I
29.10
3CH 2 OH
Aldolase
6CH 2 OP0 3
CH 2 OH
>
-phosphate^
2-
Glucose
3.4
CH 2 OP0 3 2~ Tnose isomerase
I
6
1
2C=0
(j)
H—JC— OH
2-
CH 2OP0 3
cw-Aconitate
The Organic Chemistry
29.13
of Metabolic Pathways
815
1,3-bisphosphoglycerate reacts with a cysteine residue of the enzyme in a nucleophilic acyl substitution reaction, with loss of phosphate. Then, reduction by in a second nucleophilic acyl substitution reaction yields glyceraldehyde 3-phosphate. First,
NADH
29.14
CH 3
CH 3
PMP
a-Ketoglutarate
a-Keto acid imine tautomer
Nucleophilic acyl substitution, followed by loss of water, forms the imine tautomer.
H 2 N-frEnzi 2-
3 PO.
\/
O"
H*
H
C0 2
"
I
rr
I
N
n
N
C0 2
H H
H 3 N-f(En£j
+ 2 "OoPO. 6
>
.c.
H
CO,
COc CHr.
CH.
a-Keto acid imine tautomer
a-Keto acid imine
A lysine residue deprotonates the carbon next to the ring, leading to tautomerization.
Width: 612 Height: 792
816
Chapter 29
PLP-glutamate imine
a-Keto acid imine Enzymatic protonation of the keto acid imine yields
CH 3
PLP imine
PLP glutamate
imine.
CH 3 Glutamate
Addition of the enzyme, followed by loss of glutamate, regenerates
PLP imine.
The Organic Chemistry
29.15
of Metabolic Pathways
817
Position leucine and a-ketoglutarate so that the groups to be exchanged are aligned. This arrangement makes it easy to predict the products of transamination reactions.
"
CH 3
C0 2
2.0 mol of acetyl CoA
fatty acid with n carbons yields n/2 moles of acetyl palmitic acid (C15H31CO2H),
1.0
acid x 8 1
(c)
is
is
its
that also has a negative AG°'.
One mole of glucose is catabolized mole of acetyl CoA. Thus, 1.0
energetically unfavorable. Since exact reverse has a positive AG°; synthesized by gluconeogenesis, an
favorable reaction
energetically favorable (negative AG°'),
energetically unfavorable. Instead, glucose
alternate
29.34
is
mol of acetyl CoA mol of palmitic acid
CoA
per mole of fatty acid. For
— > 8.0 mol of acetyl CoA
Maltose is a disaccharide that yields two moles of glucose on hydrolysis. Since each mole of glucose yields two moles of acetyl CoA, 1.0
mol of maltose
— > 2.0 mol of glucose — > 4.0 mol of acetyl CoA
The Organic Chemistry
29.35
Glucose
(a)
(b) Palmitic acid
amu
256.4
of Metabolic Pathways
(c)
amu
Maltose
amu
Molecular weight
180.2
Moles
0.5549 mol
0.3900 mol
0.2921 mol
2 x 0.5549 mol
8 x 0.3900 mol = 3.120 mol
4x0.2921 mol = 1.168 mol
2526 g
945.6 g
in
342.3
100.0 g
Moles of acetyl
CoA
=1.1 10
mol
produced
Grams acetyl
898.6 g
CoA
produced Palmitic acid
is
the most efficient precursor of acetyl
CoA on a weight basis.
29.36 Amino acid
a-Keto acid
(a)
"
"
CH 3 CHCHC0 2
CH 3 CHCC0 2
OH
OH
(b) "
~
CH 2 CHC0 2
^
O II
NH 3 + I
"
H 2 NCCH 2 CHC0 2
CH 2 CC0 2
P
P
II
II
"
H 2 NCCH 2 CC0 2
821
822
Chapter 29 in Section 29.1, formation of glucose 6-phosphate from glucose and ATP is ') The reverse reaction, transfer of a phosphate group energetically favorable (negative
29.37 As we saw
AG ADP from glucose 6-phosphate, is energetically unfavorable and doesn't occur spontaneously. Phosphate transfers to ADP from either 3-phosphoglyceroyl phosphate or
to
phosphoenolpyruvate have negative AG°' values and are energetically favorable reactions. In chemical terms, the leaving groups in the reactions of 3-phosphoglyceroyl phosphate (carboxylate) and phosphoenolpyruvate (enolate) are more stable anions than the leaving group in the reaction of glucose (alkoxide), so these reactions are more favorable.
...
RO— P>^
RO— FL— O: :Nu'
+
Y
:o— P-
Nu
I
:0:~
:0:
Glucose
:0:'
Phosphoenolpyruvate
3-Phosphoglyceroyl phosphate
CHO
RO" =
ro:
COo" 2
1
COo" 2 I
I
H-•C — OH
H— C— OH
1
C—
I
HO-C — H-C — OH
CH 2 OP0 3
O"
II
2-
CH 2
1
1
H-•C — OH 1
CH 2 0" 29.38
H^
CH0H
;3 Q
B:
.0— h^:b
H,
.0
rL
0H
-OH
OH
H
OH
-OH
OH
H-
OH
-OH
H—
o
2-
CH 2OP0 3
'
CH 2OP0 3
Ribulose 5-phosphate
The isomerization of ribulose 5-phosphate intermediate enolate.
2-
2-
CH 2 OP0 3
'
'
Ribose 5-phosphate to ribose
5-phosphate occurs by
way of an
The Organic Chemistry
29.39 This
is
a reverse aldol reaction, similar to step
c.
—
<
^
u —o H-C-OH
+
C
«
II
I
H+
H— C— O— H— C— OH I
CH 2 OP0 3
823
4 of glycolysis.
p4r4enz5 "
Hv.
£enz4 NH P
of Metabolic Pathways
H2
H— C— OH H— C=0 H— C— OH
H— J O— H— C— OH I
2-
CH 2OP0 3
I
2-
CH 2 OP0 3 2I
|H 2 (enz)|
NH 2
v° CH 2OH
steps in the conversion of a-ketoglutarate to succinyl CoA are similar to steps in the conversion of pyruvate to acetyl CoA shown in Figure 29.1 1, and the same coenzymes are + involved: lipoamide, thiamin diphosphate, acetyl CoA and
29.40 The
NAD
.
824
Chapter 29
Step 1: Nucleophilic addition of thiamin diphosphate ylid. Step 2: Decarboxylation. Step 3: Addition of double bond to lipoamide, with ring opening. Step 4: Elimination of thiamin diphosphate. Step 5: Nucleophilic addition of acetyl Co A to succinyl lipoamide and elimination of dihydrolipoamide to give succinyl CoA. Step 6: Reoxidation of dihydrolipoamide to lipoamide.
29.41 Addition of -OH: For carbon face to give an
Addition of
R
H+
face to give an
:
2, the top face configuration at carbon 2.
For carbon
3, the
S configuration
at
top face
carbon
3.
is
the
face,
and
H + adds
-OH
adds from
this
from the bottom, or Re, reaction occurs with anti geometry.
is Si,
The
Re
and
The Organic Chemistry
of Metabolic Pathways
825
General Problems
29.42 sn-Glycerol
3-phosphate
pro-S
CONH,
CONH 2 Dihydroxyacetone phosphate
pro-R
NAD" The
face above the plane of the ring
-H takes place from the Re face,
is
problem states that addition of hydrogen comes from ^-glycerol. The added
the Si face. Since the
the circled
hydrogen has pro-R stereochemistry.
29.43
2
Acetyl
O
HSCOA HSp
P CH3CSC0A
O
H2
HSCoA
CH 3 CCH 2 CSC0A
CoA
Acetoacetyl
o
o
II
II
CH 3 CCH 2 CO~
CoA CO-
V
NADH/H +
Acetoacetate
NAD +
O
OH
IT
I
CH 3 CHCH 2 CO
CH3CCH3 Acetone
3-Hydroxybutyrate ketone bodies
29.44 :0: \||
— Q QT + » H 3 C^ ^SCoA H 2 C^ ^SCoA
29.45
o
O
II
II
CH 3 CCH 2 CSCoA
HsC^/n^CHoCSCoA d CoAS-«J
Addition occurs from the Si face to form the
O
+
"SCoA
R enantiomer.
Re H 3
<
OH
vr
_r ,CH 2 COSACP
HoC
fl
11
C H H
Si "h:^
O SACP
Width: 612 Height: 792
826
Chapter 29
29 .4 6
If
dehydration removes the pro-R hydrogen and the resulting double bond must have taken place.
is trans,
as
indicated, anti elimination
29.47
Butyryl
The reduction
29.48
is
a syn addition.
(a) The first sequence of steps in this mechanism involves formation of the imine (Schiff base) of sedoheptulose 7-phosphate, followed by retro-aldol cleavage to form erythrose 4-phosphate and the enamine of dihydroxyacetone.
CH 2 OH (enz^NH2
C=^U>
HO— C— T H— C — OH H— C — OH H— C — OH
CH 2 OH
CH 2 OH
C-NH^enz;
)= Nh&enz 3 HoO
^
f-*L
I
H-C7O7H
B4enz
I
CH 2OP0 3
H
CHO
H— C— OH H— C— OH I
H— C— OH 2-
+
I
H— C— OH
I
CH 2 OP03
HO
2-
CH 2 OP0 3 Erythrose 4-phosphate
2-
The Organic Chemistry
of Metabolic Pathways
827
The enamine of dihydroxyacetone adds to glyceraldehyde 3-phosphate to yield fructose 6-phosphate. This reaction is almost identical to the reaction pictured for Step 8 of gluconeogenesis in Section 29.8. (b)
CH 2 OH
CH 2OH
CH 2 OH
C=0
acetyl Co A is the formation of acetoacetyl Co A. This reaction also occurs as the first step in fatty acid catabolism. Although we haven't studied the mechanism, it involves formation of a mixed anhydride. HSCoA, ATP
XX The final
step
AMP, PPj
O .
is
SCoA
a retro-Claisen reaction, whose mechanism
is
pictured in Section 29.3 as
Step 4 of ^-oxidation of fatty acids.
:oT)
O
£0:
O
:dp
II
-4i
MO in the
-fy-
Ground ground
Since
all orbitals
are occupied in
Excited state
HOMO
making n>\ the and and the other occupies y>2 j/n the excited state, there is no
state,
In the excited state, one electron occupies
HOMO.
state
,
Orbitals and Organic Chemistry: Pericyclic Reactions
839
For 1,3-butadiene: ^4*
LUMO
A
HOMO hv
s-
D C
Four 2p atomic
4- ^2
orbitals
*-* Ground In the ground state,
HOMO, and y4*
is
the HOMO, and LUMO.
ip2 is
the
V3
is
the
LUMO.
Excited state
state
In the excited state, ^3
is
the
30.2 conrotatory
—
CH 3 CH 3
CH3 cis
product
octatriene
H
C
CH 3
H
trans product
HOMO (not formed) The symmetry of the octatriene HOMO predicts that ring closure occurs by a disrotatory path in the thermal reaction and that only cis product
30.3
Note: rra/is-3,4-dimethylcyclobutene argument.
is chiral;
is
formed.
the S,S enantiomer will be used for this
Path A:
H
H E,E
CH<
Conrotatory ring opening of rra/2s-3,4-dimethylcyclobutene can occur in either a clockwise or a counterclockwise manner. Clockwise opening (path A) yields the E,E isomer; counterclockwise opening (path B) yields the Z,Z isomer. Production of (2Z,4Z)-2,4hexadiene is disfavored because of steric strain between the methyl groups in the transition state leading to ring-opened product.
840
Chapter 30
30.4 conrotato
CH *"
H H*l H
(2£,4Z,6£> trans-5 ,6-Dimethyl-
2,4,6-Octatriene
Ground
HOMO
state
Excited state
HOMO
1
,3-cyclohexadiene
conrotatory
h^7v ch3
h (y ch 3
HoC
(2E,4Z,6Z)cis-5 ,6-Dimethyl-
2,4,6-Octatriene
Ground
state
HOMO
Excited state
HOMO
1
,3-cyclohexadiene
Photochemical electrocyclic reactions of 6 n electron systems always occur in a conrotatory manner.
30.5 (2£,4£)-2,4-Hexadiene
(2£,4Z)-2,4-Hexadiene
Diene
LUMO Alkene
HOMO
H
HoC
^
H
CH<:
XIX
HoC
,
H
The Diels-Alder reaction is a thermal [4 + 2] cycloaddition, which occurs with geometry. The stereochemistry of the diene is maintained in the product.
H
CH<
suprafacial
Orbitals and Organic Chemistry: Pericyclic Reactions
841
30.6
reaction of cyclopentadiene and cycloheptatrienone is a [6 + 4] cycloaddition. This thermal cycloaddition proceeds with suprafacial geometry since five electron pairs are
The
involved in the concerted process. The % electrons of the carbonyl group do not take part in the reaction.
30.7
This
[1
,7]
sigmatropic reaction proceeds with antarafacial geometry because four electron
pairs are involved in the rearrangement.
842
Chapter 30
30.8
Scrambling of the deuterium label of 1 -deuterioindene occurs by a
series of [ 1 ,5] sigmatropic rearrangements. This thermal reaction involves three electron pairs - one pair
of jz electrons from the six-membered ring, the n electrons from the five-membered ring, and two electrons from a carbon-deuterium (or hydrogen) single bond - and proceeds with suprafacial geometry. [1,5]
D
30.9
The Claisen rearrangement of an
unsubstituted allyl phenyl ether is a [3,3] sigmatropic rearrangement in which the allyl group usually ends up in the position ortho to oxygen. In this problem both ortho positions are occupied by methyl groups. The Claisen intermediate undergoes a second [3,3] rearrangement, and the final product is /7-allyl phenol.
Orbitals and Organic Chemistry: Pericyclic Reactions
CH.
HoC, '
843
[3,3,1
shift
Number of
30.10 Type of reaction
electron pairs
(a)
Thermal electrocyclic
(b)
Photochemical electrocyclic Photochemical cycloaddition
(c)
(d) (e)
Thermal cycloaddition Photochemical sigmatropic
four four four four four
Stereochemistry conrotatory disrotatory suprafacial antarafacial
suprafacial
rearrangement
Visualizing Chemistry
30.11 [3,33 *shift
CH= CH 2 This reaction product.
is
CH— CH 2
a [3,3] sigmatropic rearrangement that yields
1
,5-cyclodecadiene as a
30.12 [3,3 ] "shift"
The
13
C NMR spectrum of homotropilidene would show five peaks if rearrangement were
NMR
slow. In fact, rearrangement occurs at a rate that is too fast for to detect. The C spectrum taken at room temperature is an average of the two equilibrating forms, in
NMR
which positions
1 and 5 are equivalent, as are positions 2 and 4. Thus, only three distinct 13 types of carbons are visible in the spectrum of homotropilidene. C
NMR
844
Chapter 30
Additional Problems Electrocyclic Reactions
30.13
Rotation of the orbitals in the 6
According to the rules
in
;r electron system occurs in a disrotatory fashion. Table 30. 1, the reaction should be carried out under thermal
conditions.
Ground
state
HOMO
Excited state
For the hydrogens to be trans This can happen only if the
in the product, rotation
HOMO
must occur
in
a conrotatory manner.
HOMO has the symmetry pictured. For a 6 n electron system,
this
HOMO must arise from photochemical excitation of a m electron. To obtain a product
having the correct stereochemistry, the reaction must be carried out under photochemical conditions.
Orbitals and Organic Chemistry: Pericyclic Reactions
30.14 The
diene can cyclize by either of two conrotatory paths to form cyclobutenes
A and B.
845
Width: 612 Height: 792
846
30.15
Chapter 30
A photochemical electrocyclic reaction involving two electron pairs proceeds in a disrotatory
manner (Table
Ground
30.1).
HOMO
state
Excited state
HOMO
ti
H
H
The two hydrogen atoms
in the
four-membered ring are
cis to
'
>
each other in the cyclobutene
product.
30.16 The cyclononatriene thermal conditions.
is a 6 k electron system that cyclizes by a disrotatory route under The two hydrogens at the ring junction have a cis relationship.
heat
30.17 heat
^
} H3C
CH
conrotatory'
HoC
CH^
HoC
CH<
HoC
CH<:
hv disrotatory
(2£,4Z,6Z,8£)2,4,6,8-Decatetraene
Four electron pairs undergo reorganization in this electrocyclic reaction. The thermal reaction occurs with conrotatory motion to yield a pair of enantiomeric fra/2s-7,8-dimethyl1,3,5-cyclooctatrienes. The photochemical cyclization occurs with disrotatory motion to yield the cz's-7,8-dimethyl isomer.
Orbitals and Organic Chemistry: Pericyclic Reactions
847
30.18
30.19
Two electrocyclic reactions,
involving three electron pairs each, occur in this
The thermal reaction is a disrotatory process that yields two ds-fused sixrings. membered The photochemical reaction yields the trans-fused isomer. The two pairs of n electrons in the eight-membered ring do not take part in the electrocyclic reaction. isomerization.
Cycloaddition Reactions
30.20
is a reverse [4 + 2] cycloaddition. The reacting orbitals have the correct for the reaction to take place by a favorable suprafacial process.
This reaction
symmetry
848
Chapter 30
H
H
This [2 + 2] reverse cycloaddition antarafacial
is
not likely to occur as a concerted process because the
geometry required for the thermal reaction
is
not possible for a four ^-electron
system.
Formation of the bicyclic ring system occurs by a suprafacial [4 + 2] Diels-Alder cycloaddition process. Only one pair of n electrons from the alkyne is involved in the reaction; the carbonyl k electrons are not involved.
Loss of CO2
is
a reverse Diels-Alder [4
+
2] cycloaddition reaction.
Orbitals and Organic Chemistry: Pericyclic Reactions
849
30.22
The
first
reaction
is
a Diels-Alder [4
+
2] cycloaddition,
which proceeds with suprafacial
geometry.
The second
reaction
is
a reverse Diels-Alder [4
+
2] cycloaddition.
Sigmatropic Rearrangements
30.23
This thermal sigmatropic rearrangement involved in the reaction.
is
a suprafacial process since five electron pairs are
850
Chapter 30
30.24 The product of this
[3,3] sigmatropic
rearrangement
is
an enol that tautomerizes to a
ketone.
2
OH 30.25 [1,3]
heat 1
This reaction
is
2
Cyclopentene
Vinylcyclopropane
a [1,3] sigmatropic rearrangement.
30.26 CHo
HoC-C^)
[3,3]
10
1
^C^
2C
shift
3
^
H 3 C— C= C= CH— CH 2 CCH 3 2
1
heat
CH 2
3
3
2
3
III'
3C I
H
An
allene
is
formed by a
[3,3]
sigmatropic rearrangement.
:OH
CH<
HgC
C
— C — CH
H 3 C— C= C= CH-^CH^ CCH 3
CH2CCH3
A—H^ CHo3
— H tlT
O II
I
H 3 C— C= C— CH= CHCCH 3
O
H Acid catalyzes isomerization of the allene
to a conjugated
dienone via an intermediate enol.
30.27
H3 HoC [3,3] shift
heat
Karahanaenone
Karahanaenone
is
formed by a
[3,3] sigmatropic
rearrangement (Claisen rearrangement).
Orbitals and Organic Chemistry: Pericyclic Reactions
851
General Problems
30.28 Tables
30.1-30.3
may be helpful. The first step
is
always to find the number of electron
pairs involved in the reaction.
Number of Type of reaction
Stereochemistry
electron pairs
Photochemical [1,5] sigmatropic rearrangement (b) Thermal [4 + 6] cycloaddition
antarafacial
(a)
(c)
Thermal
(d)
Photochemical [2
suprafacial
[1,7] sigmatropic rearrangement
+
antarafacial
suprafacial
6]
cycloaddition
30.29 H
HA
D
Ad
HoC
D
HoC
D D
h-7 H Each of the two
D
electrocyclic reactions involves
two
pairs of electrons
D and proceeds
in a
conrotatory manner.
30.30 Ring opening of Dewar benzene
is a process involving two electron pairs and, according to Table 30.1, should occur by a conrotatory pathway. However, if you look back to other ring openings of cis-fused cyclobutenes, you will see that conrotatory ring opening produces a diene in which one of the double bonds is trans. Since a trans double bond in a six-membered ring is not likely to be formed, ring opening occurs by a different, higher energy, nonconcerted pathway.
30.31
Ring opening of the /rarcs-cyclobutene isomer proceeds by the expected conrotatory route to form the observed product. For the ds-cyclobutene isomer, the observed product can be formed by a four-electron pericyclic process only if the four-membered ring geometry is trans. Ring opening of the cis isomer by a concerted process would form a severely strained six-membered ring containing a trans double bond. Reaction of the cis isomer to yield the observed product occurs instead by a higher energy, nonconcerted path.
Both reactions are
[2
+
2]
photochemical electrocyclic reactions, which occur with
disrotatory motion.
30.33 [3,3] shift
[3,3] shift etc.
Bullvalene
Bullvalene can undergo [3,3] sigmatropic rearrangements in all directions. At 100 °C, the rate of rearrangement is fast enough to make all hydrogen atoms equivalent, and only one l spectrum. signal is seen in the
U NMR
Orbitals and Organic Chemistry: Pericyclic Reactions
853
30.34
The observed products
A
B
and result from a [1,5] sigmatropic hydrogen shift with suprafacial geometry, and they confirm the predictions of orbital symmetry. and
C
D are
not formed.
30.35
This
[2,3] sigmatropic
suprafacial geometry.
rearrangement involves three electron pairs and should occur with
854
Chapter 30
30.36
Concerted thermal ring opening of a cis-fused cyclobutene is conrotatory and yields a product having one cis and one trans double bond. The ten-membered ring product of reaction 2 is large enough to accommodate a trans double bond, but a seven-membered ring containing a trans double bond is highly strained. Opening of the cyclobutene ring in reaction 1 occurs by a higher energy nonconcerted process to yield a seven-membered ring having two cis double bonds.
30.37
Thermal ring opening of the methylcyclobutene ring can occur by allowed conrotatory paths to yield the observed product mixture.
either of two
symmetry-
Orbitals and Organic Chemistry: Pericyclic Reactions
30.38
The
first
reaction
is
an electrocyclic opening of a cyclobutene
Formation of estrone methyl ether occurs by a Diels-Alder
[4
ring.
+
2] cycloaddition.
30.39
Reaction
1:
Reverse Diels-Alder [4 + 2] cycloaddition;
Reaction 2:
Conrotatory electrocyclic opening of a cyclobutene ring;
Reaction 3:
Diels-Alder [4
+ 2]
cycloaddition.
855
Width: 612 Height: 792
856
Chapter 30
H
^Ch^CHg
H
H
^CH^CHg H3
^Ch^CHg
+
OHC
H0 2 C Coronafacic acid
Treatment with base enolizes the ketone and changes the ring junction from trans to cis. A cis ring fusion is less strained when a six-membered ring is fused to a five-membered ring.
30.40 CHo
HoC
CH 2 P,3]
3^ CcH HoC^
NH(CH 3
shift
)
2
*heat
CH2CH=CH2
2
3
CH2CH=CH2
H
30.41(a)
(b)
The rearrangement product absorbs
at a longer wavelength because extensive system of conjugated double bonds.
it
has a more
Chapter 31 - Synthetic Polymers
Chapter Outline I.
Chain-growth polymers (Sections 31.1-31.3). A. General features of chain-growth polymerization reactions (Section 31.1). 1
.
How polymerization occurs. a.
An initiator adds to
b
The reactive intermediate adds The process is repeated.
.
c.
2
.
a carbon-carbon double bond of a vinyl monomer. to a second molecule of monomer.
Types of polymerization. a.
A radical initiator leads to radical polymerization.
b
An
.
acid causes cationic polymerization.
Acid-catalyzed polymerization is effective only if the vinyl monomers contain electron-donating groups. Anionic polymerization can be brought about by anionic catalysts. i. Vinyl monomers in anionic catalysis must have electron-withdrawing groups. Polymerization occurs by conjugate nucleophilic addition to the monomer. ii. iii. Acrylonitrile, styrene and methyl methacrylate can be polymerized i.
c.
anionically.
"Super glue" is an example of an anionic polymer. Stereochemistry of polymerization (Section 3 1 .2). There are three possible stereochemical outcomes of polymerization of a substituted 1 iv.
B
.
.
vinyl a.
monomer.
If the substituents all lie
on the same
side of the
polymer backbone, the polymer
is isotactic.
along the backbone, the polymer is syndiotactic. randomly oriented, the polymer is atactic. The three types of polymers have different properties. Although polymerization using radical initiators can't be control stereochemically, Ziegler-Natta catalysts can yield polymers of desired stereochemical orientation.
b 2
.
3
.
.
If the substituents alternate
c.
If the substituents are
a.
Ziegler-Natta catalysts are organometallic-transition metal complexes. i. They are usually formed by treatment of an alkylaluminum with titanium tetrachloride.
Ziegler-Natta polymers have very little chain-branching. c. Ziegler-Natta catalysts are stereochemically controllable. d Polymerization occurs by coordination of the alkene monomer to the complex, followed by insertion into the polymer chain. 4 Common Ziegler-Natta polymers. a. Polyethylene produced by the Ziegler-Natta process (high-density polyethylene) is linear, dense, strong, and heat-resistant. b Other high-molecular- weight polyethylenes have specialty uses. Copolymers (Section 31.3).
b
.
.
.
.
C
1
.
2
.
3
.
Copolymers are formed when two different monomers polymerize together. The properties of copolymers are different from those of the corresponding monomers.
Types of copolymers. a.
Random copolymers.
b.
Alternating copolymers. B lock copolymers
c
.
858
Chapter 31
Block copolymers are formed when an excess of a second monomer is added to a still-active mix. Graft copolymers i. Graft copolymers are made by gamma irradiation of a completed homopolymer to generate a new radical initiation site for further growth of a i.
d
.
chain. II.
Step-growth polymers (Section 31.4). A. Step-growth polymer are formed by reactions in which each bond is formed independently of the others. B Most step-growth polymers result from reaction of two difunctional compounds. 1 Step-growth polymers can also result from polymerization of a single difunctional .
.
compound. C. Types of step-growth polymers. Polyamides and polyesters. 1 2 Polycarbonates (formed from carbonates and alcohols or phenols). .
.
3
.
Polyurethanes. a. urethane has a carbonyl group bonded to both an -NR2 group and an -OR group. b Most polyurethanes are formed from the reaction of a diisocyanate and a diol. Polyurethanes are used as spandex fibers and insulating foam. c. Foaming occurs when a small amount of water is added during i. polymerization, producing bubbles of CO2. Polyurethane foams often use a polyol, to increase the amount of crossii.
A
.
linking.
HI. Olefin metathesis polymerization (Section 31.5).
A. General features. In an olefin metathesis reaction, two olefins (alkenes) exchange substituents. 1 2 The catalysts contain a carbon-metal (usually ruthenium) double bond. a. They react reversibly with an alkene to form a 4-membered metallacyte. b The metallacyte opens to give a different catalyst and a different alkene. 3 The reaction is compatible with many olefin functional groups. 4 The double bonds allow for further manipulations. B Ring-opening metathesis polymerization (ROMP). 1 The monomer is a moderately strained cycloalkene. 2 The resulting polymer has double bonds spaced regularly along the chain. .
.
.
.
.
.
.
.
C
Acyclic diene metathesis (ADMET). 1 The monomer is a long-chain dialkene with double bonds at the end of the chain. 2 As the reaction progresses, gaseous ethylene escapes, driving the reaction toward product. IV. Polymer structure and physical properties (Section 31.6). A. Physical properties of polymers. 1 Because of their large size, polymers experience large van der Waals forces, a. These forces are strongest in linear polymers. 2 Many polymers have regions held together by van der Waals forces; these regions .
.
.
.
.
3
.
are
known
a.
Polymer
b
Tm is
.
as crystallites.
Some polymers have a.
by the substituents on the chains. which the crystalline regions of a polymer melt.
crystallinity is affected
the temperature at little
ordering but are hard at room temperature. soft at a temperature T (glass transition temperature). g
These polymers become
Synthetic Polymers
B
.
859
Polymers can be classified by physical behavior. 1
.
Thermoplastics. a. Thermoplastics have a high Tg and are hard at room temperature. b Because they become soft at higher temperatures, they can be molded. c. Plasticizers such as dialkyl phthalates are often added to thermoplastics to keep them from becoming brittle at room temperature. .
2.
Fibers. a.
b.
Fibers are produced by extrusion of a molten polymer. On cooling and drawing out, the crystallite regions orient along the axis of the fiber to
3
.
add
tensile strength.
Elastomers. a. Elastomers are amorphous polymers that can stretch and return to their original shape. b. These polymers have a low Tg and a small amount of cross-linking. c. The randomly coiled chains straighten out in the direction of the pull, but they return to their random orientation when stretching is done. Natural rubber is an elastomer, but gutta percha is highly crystalline. Thermosetting resins. Thermosetting resins become highly cross-linked and solidify when heated. a. b Bakelite, a phenolic resin formed from phenol and formaldehyde, is the most familiar example. d.
4.
.
Solutions to Problems
Most reactive
**»
H 2 C=CHC 6 H 5 > H 2 C=CHCH 3 > H 2C=CHCI
Least reactive
> H 2 C=CHC0 2 CH 3
The alkenes most reactive to cationic polymerization contain electron-donating functional groups that can stabilize the carbocation intermediate. The reactivity order of substituents cationic polymerization
is
similar to the reactivity order of substituted benzenes in
electrophilic aromatic substitution reactions.
31.2 Most reactive
^-
Least reactive
H 2 C=CHC=N > H 2 C=CHC 6 H 5 > H 2 C=CHCH 3 Anionic polymerization occurs most readily with alkenes having electron-withdrawing substituents.
31.3
The intermediate anion can be
stabilized
by resonance involving the phenyl
ring.
in
860
Chapter 31
31.4 n
* -^CH 2 -CCI 2
H 2C=CCI 2
'
'n
Vinylidene chloride doesn't polymerize in isotactic, syndiotactic or atactic forms because no asymmetric centers are formed during polymerization.
31.5
None
of the polypropylenes rotate plane-polarized light. If an optically inactive reagent and an achiral compound react, the product must be optically inactive. For every chirality center generated, an enantiomeric chirality center is also generated, and the resulting polymer mixture is optically inactive.
31.6 CHo
CHo
I
n
H 2 C=CCH=CH 2
+
H 2 C=C
n
CHo
/
/ -4-
i_i+
I
CHo
I
I
CH 2 C= CHCH 2 CH 2 C V In L. CHf
*~
CHg 2-Methylpropene
2-Methyl- 1 ,3butadiene
31.7
-^CH2CH = CHCH 2 -^
—»*
-^-CH 2 CH=CHCH-^
|
m
H 2 C=CHC 6 H 5
CH 2 CH = CHCH
polybutadiene chain
C 6 H 5 CH -j/
Irradiation homolytically cleaves an allyhc
energy.
The
polystyrene
C-H bond because it has the lowest bond
resulting radical adds to styrene to produce a polystyrene graft.
31.8
O n
cham
m
HOCH 2 CH 2 OH
+ n
,
—
HOC
/>|v»
N^N
H2N Melamine,
.NHCH 2 NH
f
repeat
times
Melmac
NH 2
.NH2
N
N
CH2 many
YT >y n
NH2
Synthetic Polymers
869
31.29 •o-J
V
\
H 2 C— CHCH 2 CI
CICH 2 CH— CH 2
:o:
CI—
f — CHoCHCHo\_/—
H 2 C—
\\
/
\
v
CHCH 2
repeat several times
A H C CHCH 2
A
OH 2
O-^x
^-OCH 2 CHCHp
^~9~-C-^
CH2 CHCH 2 -tO-^v I
OH
V
y
Cross-linking occurs
prepolymer.
CH 3
^— OCH CHCH2|-0-^v
=/
^
when
CH3
2
OH
-
n
'
S_ CH3
the triamine opens epoxide rings on
two
OCH2CHCH2NCH2CH2 ^
OH different chains of the
4
870
31.30
Chapter 31 (a) The diamine is formed by an electrophilic aromatic substitution reaction of formaldehyde with two equivalents of aniline.
(b)
The diamine reacts with two
equivalents of phosgene.
Synthetic Polymers
871
31.31
O
O
CNH
—
7-CH
('
//
\
II
NHC— OCH2 CH 2 0-
31.32
NH 2
H2N
+
O
O
II
II
CH 2
k r
31.33
Step
1:
Polystyrene and the phthalimide combine in an electrophilic aromatic substitution
reaction
-(CH 2 -CH^
-(CH 2 -CH^
-(-
C H 2 -Ch\
-(CH2
-CH^
CF3SO3H CHo2
— OH
v y° I
CH 2 -rOH 2 I
N
o
o
k|
CH 2? I
°yy°
o
Chapter 31 Step
2:
The phthalimide
is
cleaved in a series of steps that involve nucleophilic acyl
substitution reactions.
— CH^n
-^CH 2
Let
= R
OH;
R
HgN-NHg
R
r
R
H 2 N— NH 2
H.
H ,N
N
o
~ NH 2
O:
^
// attack of / nucleophile
x
L
%
A
proton shift
ring
tt opening +
NHR NH 2 NH
RNH H 2 N-NH
RNH 2 HN- NH '
O attack of nucleophile
ti
HN-NH 0==(
+
RNH 2
=
"(cH 2
— CH-^ n
7=0
CH 2 NH 2
\\
v
/ "'
Synthetic Polymers
873
31.34
2
1
CH 3 CH 2 CH 2 CHO
.
NaOH, EtOH :
2.
heat
'
CH 3 CH 2 CH 2 CH= CCHO I
I
7 or KMn04 from aldehydes by oxidation from alkyl halides by conversion into Grignard reagents followed by
(Sec. 8.8) (Sec. 16.9) (Sec. 19.3) (Sec. 20.5)
reaction with (Sec. 20.5, 20.7) (Sec. 21.4) (Sec. 21.5) (Sec. 21.6) (Sec. 21.7)
Cyanohydrins,
alkyl halides
CO2
from nitriles by acid or base hydrolysis from acid chlorides by reaction with aqueous base from acid anhydrides by reaction with aqueous base from esters by hydrolysis with aqueous base from amides by hydrolysis with aqueous base
RCH(OH)CN from aldehydes and ketones by reaction with
(Sec. 19.6)
HCN
Cycloalkanes from alkenes by addition of dichlorocarbene from alkenes by reaction with CH2I2 and Zn/Cu (Simmons-Smith
(Sec. 8.9) (Sec. 8.9)
reaction)
from arenes by hydrogenation
(Sec. 16.10)
Disulfides,
RS-SR' from
(Sec. 18.8)
Enamines,
thiols
by oxidation with bromine
RCH=CRNR 2 from ketones or aldehydes by reaction with secondary amines
(Sec. 19.8)
O Epoxides,
R 2C— CR 2
(Sec. 8.9, 18.5) (Sec. 18.5)
Esters,
from alkenes by treatment with a peroxyacid from halohydrins by treatment with base
RC0 2 R'
(Sec. 21.3) (Sec. 21.3)
from carboxylic acid salts by Sn2 reaction with primary alkyl halides from carboxylic acids by acid-catalyzed reaction with an alcohol (Fischer esteriflcation)
(Sec. 2 1 .4) (Sec. 21.5) (Sec. 22.7) (Sec. 22.7)
from acid chlorides by base-induced reaction with an alcohol from acid anhydrides by base-induced reaction with an alcohol from alkyl halides by alkylation with diethyl malonate from esters by treatment of their enolate ions with alkyl halides
Functional-Group Synthesis
880
Ethers,
R-O-R'
(Sec. 16.8)
from activated haloarenes by reaction with alkoxide ions from unactivated haloarenes by reaction with alkoxide ions via benzyne
(Sec. 18.2)
from primary alkyl halides by Sn2 reaction with alkoxide ions (William-
(Sec. 18.2)
from alkenes by alkoxymercuration/demercuration from alkenes by epoxidation with peroxyacids
(Sec. 16.7)
intermediates
son ether synthesis) (Sec. 18.5)
Halides, alkyl,
R 3 C -X from alkenes by electrophilic addition of HX from alkenes by addition of halogen from alkenes by electrophilic addition of hypohalous acid (HOX)
(Sec. 7.7) (Sec. 8.2) (Sec. 8.3) (Sec. 9.3) (Sec. 9.3)
to yield
halohydrins from alkynes by addition of halogen from alkynes by addition of from alkenes by ally lie bromination with Af-bromosuccinimide (NBS) from alcohols by reaction with from alcohols by reaction with SOCI2 from alcohols by reaction with PBr3 from alkyl tosylates by Sn2 reaction with halide ions from arenes by benzylic bromination with Af-bromosuccinimide (NBS) from ethers by cleavage with either
HX
(Sec. 10.3)
HX
(Sec. 10.5) (Sec. 10.5) (Sec. 10.5) (Sec. 11.2, 11.3) (Sec. 16.9)
HX
(Sec. 18.3)
from ketones by a-halogenation with bromine from carboxylic acids by a-halogenation with phosphorus and PBr3
(Sec. 22.3) (Sec. 22.4)
(Hell-Volhard-Zelinskii reaction)
Ar-X
Halides, aryl,
(Sec. 16.1, 16.2) (Sec. 24.8)
from arenes by electrophilic aromatic substitution with halogen from arenediazonium salts by reaction with cuprous halides (Sandmeyer reaction)
Halohydrins,
R 2 CXC(OH)R 2
(Sec. 8.3) (Sec. 18.6)
Imines.
R 2 C=NR'
(Sec. 19.8)
Ketones,
from alkenes by electrophilic addition of hypohalous acid (HOX) from epoxides by acid-induced ring opening with HX
from ketones or aldehydes by reaction with primary amines
R2 C =
(Sec. 8.8) (Sec. 8.7) (Sec. 9.4) (Sec. 9.4) (Sec. 16.3) (Sec. 17.7, 19.2) (Sec. 19.2, 21.4)
from alkenes by ozonolysis from 1,2-diols by cleavage reaction with sodium periodate from alkynes by mercuric-ion-catalyzed hydration from alkynes by hydroboration/oxidation from arenes by Friedel-Crafts acylation reaction with an acid chloride from secondary alcohols by oxidation from acid chlorides by reaction with lithium diorganocopper (Gilman) reagents
(Sec. 19.13) (Sec. 20.7) (Sec. 22.7) (Sec. 22.7)
from conjugated enones by addition of lithium diorganocopper reagents from nitriles by reaction with Grignard reagents from primary alkyl halides by alkylation with ethyl acetoacetate from ketones by alkylation of their enolate ions with primary alkyl halides
Functional-Group Synthesis
881
R-C-N
Nitriles, (Sec.
1
from primary alkyl halides by Sn2 reaction with cyanide ion from primary amides by dehydration with SOCI2 from nitriles by alkylation of their a-anions with primary alkyl halides from arenediazonium ions by treatment with CuCN
20.7)
1.3,
(Sec. 20.7) (Sec. 22.7) (Sec. 24.8)
Nitroarenes, Ar-N(>2 from arenes by (Sec. 16.2)
Organometallics, (Sec. 10.6)
electrophilic aromatic substitution with nitric/sulfuric acids
R-M formation of Grignard reagents from organohalides by treatment with
magnesium formation of organolithium reagents from organohalides by treatment with
(Sec. 10.7)
lithium (Sec. 10.7)
formation of lithium diorganocopper reagents (Gilman reagents) from organolithium reagents by treatment with cuprous halides
Ar-OH
Phenols,
from arenediazonium salts by reaction with CU2O and Cu(NC>3)2 from aryl halides by nucleophilic aromatic substitution with hydroxide ion
(Sec. 24.8) (Sec. 16.7)
Quinones (Sec. 17.10)
Sulfides,
from
[(KS0 3 ) 2NO]
thiols
by Sn2 reaction of thiolate ions with primary
alkyl halides
R-S0 2 -R' from
(Sec. 18.8)
Sulfoxides,
sulfides or sulfoxides
by oxidation with peroxyacids
R-SO-R'
(Sec. 18.8)
Thiols,
salt
R-S-R'
(Sec. 18.8)
Sulfones,
from phenols by oxidation with Fremy's
from
sulfides
by oxidation with H2O2
R-SH
(Sec. 11.3) (Sec. 18.8)
from primary alkyl halides by Sn2 reaction with hydrosulfide anion from primary alkyl halides by Sn2 reaction with thiourea, followed by hydrolysis
Functional-Group Reactions
The following groups are
table
summarizes the reactions of important functional groups. The functional followed by a reference to the appropriate text section.
listed alphabetically,
Acetal 1
.
Hydrolysis to yield a ketone or aldehyde plus alcohol (Sec. 19. 10)
Acid anhydride 1
.
2
.
3
.
4
.
Hydrolysis to yield a carboxylic acid (Sec. 21.5) Alcoholysis to yield an ester (Sec. 2 1 .5) Aminolysis to yield an amide (Sec. 2 1 .5) Reduction to yield a primary alcohol (Sec. 2 1 .5)
Acid chloride 1
.
2.
3
.
4
.
5 6 7
.
.
.
compound to yield an aryl ketone (Sec. 16.3) Hydrolysis to yield a carboxylic acid (Sec. 21 .4) Alcoholysis to yield an ester (Sec. 2 1 .4) Aminolysis to yield an amide (Sec. 2 1 .4) Reduction to yield a primary alcohol (Sec. 2 1 .4) Grignard reaction to yield a tertiary alcohol (Sec. 21 .4) Reaction with a lithium diorganocopper reagent to yield a ketone (Sec. 21.4) Friedel-Crafts reaction with an aromatic
Alcohol 1
.
2.
Acidity (Sec. 17.2) Oxidation (Sec. 17.7) a. Reaction of a primary alcohol to yield an aldehyde or acid b Reaction of a secondary alcohol to yield a ketone Reaction with a carboxylic acid to yield an ester (Sec. 21 .3) Reaction with an acid chloride to yield an ester (Sec. 21.4) Reaction with an acid anhydride to yield an ester (Sec. 21.5) Dehydration to yield an alkene (Sec. 17.6) Reaction with a primary alkyl halide to yield an ether (Sec. 18.2) Conversion into an alkyl halide (Sec. 17.6) a. Reaction of a tertiary alcohol with b. Reaction of a primary or secondary alcohol with SOCI2 Reaction of a primary or secondary alcohol with PBr3 c. .
3
.
4.
5
.
6.
7
.
8.
HX
Aldehyde Oxidation to yield a carboxylic acid (Sec. 19.3) 2. Nucleophilic addition reactions a. Reduction to yield a primary alcohol (Sees. 17.4, 19.7) b. Reaction with a Grignard reagent to yield a secondary alcohol (Sees. 17.5, 19.7) Grignard reaction of formaldehyde to yield a primary alcohol (Sec. 17.5) c. d. Reaction with HCN to yield a cyanohydrin (Sec. 19.6) Wolff-Kishner reaction with hydrazine to yield an alkane (Sec. 19.9) e. Reaction with an alcohol to yield an acetal (Sec. 19. 10) f g Wittig reaction to yield an alkene (Sec. 19. 1 1) h Reaction with an amine to yield an imine or enamine (Sec. 19.8) 3 Aldol reaction to yield a (3-hydroxy aldehyde (Sec. 23. 1) 4. Alpha bromination of an aldehyde (Sec. 22.3) 1
.
.
.
.
.
Functional-Group Reactions
Alkane 1
Radical halogenation to yield an alkyl halide (Sees. 6.3, 10.3)
.
Alkene 1
Electrophilic addition of
.
HX to yield an alkyl halide (Sees. 7.7-7.1 1)
Markovnikov regiochemistry 2
.
3
.
4.
5
.
6.
7
.
8
.
9. 10. 1 1
.
12.
is
observed.
Electrophilic addition of halogen to yield a
1
,2-dihalide (Sec. 8.2)
Oxymercuration/demercuration to yield an alcohol (Sec. 8.4) Markovnikov regiochemistry is observed, yielding the more highly substituted alcohol. Hydroboration/oxidation to yield an alcohol (Section 8.5) Hydrogenation to yield an alkane (Sec. 8.6) Hydroxylation to yield a 1,2-diol (Sec. 8.7) Oxidative cleavage to yield carbonyl compounds (Sec. 8.8) Simmons-Smith reaction with CH2I2 to yield a cyclopropane (Sec. 8.9) Reaction with dichlorocarbene to yield a dichlorocyclopropane (Sec. 8.9) Allylic bromination with NBS (Sec. 10.4) Alkoxymercuration to yield an ether (Sec. 1 8.2) Reaction with a peroxyacid to yield an epoxide (Sees. 8.7, 18.5)
Alkyne 1
Electrophilic addition of
.
3
.
Hydroboration/oxidation to yield an aldehyde (Sec. 9.4) Alkylation of an alkyne anion (Sec. 9.8)
4.
5
HX to yield a vinylic halide (Sec. 9.3)
Electrophilic addition of halogen to yield a dihalide (Sec. 9.3) Mercuric-sulfate-catalyzed hydration to yield a methyl ketone (Sec. 9.4)
2.
.
Reduction (Sec. 9.5) Hydrogenation over Lindlar catalyst to yield a a. b. Reduction with Li/NH3 to yield a trans alkene
6.
cis
alkene
Amide 1
Hydrolysis to yield a carboxylic acid (Sec. 21.7) Reduction with LiAlH4 to yield an amine (Sec. 21.7) Dehydration to yield a nitrile (Section 20.7)
.
2.
3
.
Amine 1.
2.
3
.
Basicity (Sec. 24.3) Sn2 alkylation of an alkyl halide to yield an amine (Sec. 24.6) Nucleophilic acyl substitution reactions
Reaction with an acid chloride to yield an amide (Sec. 21.4) b. Reaction with an acid anhydride to yield an amide (Sec. 21 .5) Hofmann elimination to yield an alkene (Sec. 24.7) Formation of an arenediazonium salt (Sec. 24.8) a.
4. 5.
Arene 1
2
.
.
3.
4.
Oxidation of an alkylbenzene side chain to yield a benzoic acid (Sec. 16.9) Catalytic reduction to yield a cyclohexane (Sec. 16. 10) Reduction of an aryl alkyl ketone to yield an arene (Sec. 16. 10) Electrophilic aromatic substitution (Sees. 16.1-16.3) a.
b. c. d.
e.
Halogenation (Sees. 16.1-16.2) Nitration (Sec. 16.2) Sulfonation (Sec. 16.2) Friedel-Crafts alkylation (Sec. 16.3) Aromatic ring must be at least as reactive as a halobenzene Friedel-Crafts acylation (Sec. 16.3)
883
Functional-Group Reactions
884
Arenediazonium 1
.
2
.
3.
4.
5
.
6.
Conversion Conversion Conversion Conversion Conversion Conversion
salt into
an aryl chloride (Sec. 24.8) bromide (Sec. 24.8) an aryl iodide (Sec. 24.8) an aryl cyanide (Sec. 24.8) a phenol (Sec. 24.8) an arene (Sec. 24.8)
into an aryl into
into into into
Carboxylic acid 1
.
2.
Acidity (Sees. 20.2-20.4)
Reduction to yield a primary alcohol (Sees. 17.4, 21.3) a. Reduction with LiAlrLt b. Reduction with BH3
Nucleophilic acyl substitution reactions (Sec. 21.3) a. Conversion into an acid chloride b Conversion into an acid anhydride Conversion into an ester c. (1) Fischer esterification (2) Sn2 reaction with an alkyl halide 4. Alpha bromination (Hell-Volhard-Zelinskii reaction) (Sec. 22.4) 3.
.
Diene 1
.
2.
Conjugate addition of HX and X2 (Sec. 14.2) Diels-Alder reaction (Sees. 14.4, 14.5, 30.5)
Epoxide 1.
2.
HX
Acid-catalyzed ring opening with to yield a halohydrin (Sec. 18.6) Ring opening with aqueous acid to yield a 1,2-diol (Sec. 18.6)
Ester 1.
2. 3.
4. 5.
6
.
Hydrolysis to yield a carboxylic acid (Sec. 21.6) Aminolysis to yield an amide (Sec. 21.6) Reduction to yield a primary alcohol (Sees. 17.4, 21.6) Partial reduction with DIB to yield an aldehyde (Sec. 21.6) Grignard reaction to yield a tertiary alcohol (Sees. 17.5, 21.6) Claisen condensation to yield a p-keto ester (Sec. 23.7)
AH
Ether 1.
2.
Acid-induced cleavage to yield an alcohol and an alkyl halide (Sec. 18.3) Claisen rearrangement of an allyl aryl ether to yield an o-allyl phenol (Sees. 18.4, 30.8)
Halide, alkyl 1
.
2. 3 .
4.
5
.
6
.
Reaction with magnesium to form a Grignard reagent (Sec. 10.6) Reduction to yield an alkane (Sec. 10.6) Coupling with a diorganocopper reagent to yield an alkane (Sec. 10.7) Reaction with an alcohol to yield an ether (Sec. 18.2) Nucleophilic substitution (Sn 1 or Sn2) (Sees. 11.1-11.5) Dehydrohalogenation to yield an alkene (El or E2) (Sees. 1 1 .7-1 1 10) .
Halohydrin 1.
Conversion into an epoxide (Sec. 18.5)
Functional-Group Reactions
885
Ketone 1
.
Nucleophilic addition reactions a. Reduction to yield a secondary alcohol (Sees. 17.4, 19.7) b. Reaction with a Grignard reagent to yield a tertiary alcohol (Sees. 17.5, 19.7) Wolff-Kishner reaction with hydrazine to yield an alkane (Sec. 19.9) c. to yield a cyanohydrin (Sec. 19.6) d. Reaction with e. Reaction with an alcohol to yield an acetal (Sec. 19.10) f. Wittig reaction to yield an alkene (Sec. 19. 1 1) g. Reaction with an amine to yield an imine or enamine (Sec. 19.8)
HCN
2
.
3
.
Aldol reaction to yield a (3-hydroxy ketone (Sec. 23. 1) Alpha bromination (Sec. 22.3)
Nitrile Hydrolysis to yield a carboxylic acid (Sees. 20.5, 20.7) 1. Reduction to yield a primary amine (Sec. 20.7) 2. 3. Reaction with a Grignard reagent to yield a ketone (Sec. 20.7)
Nitroarene 1
.
Reduction to yield an arylamine (Sees. 16.2, 24.6)
Organometallic reagent 1
.
2.
3
.
4.
5
.
6.
Reduction by treatment with acid to yield an alkane (Sec. 10.6) Nucleophilic addition to a carbonyl compound to yield an alcohol (Sees. 17.5, 19.7) Conjugate addition of a lithium diorganocopper to an a,p-unsaturated ketone (Sec. 19. 13) Coupling reaction of a lithium diorganocopper reagent with an alkyl halide to yield an alkane (Sec. 10.7) Coupling reaction of a lithium diorganocopper with an acid chloride to yield a ketone (Sec. 21.4) Reaction with carbon dioxide to yield a carboxylic acid (Sec. 20.5)
Phenol 1
.
2. 3.
4.
Acidity (Sec. 17.2) Reaction with an acid chloride to yield an ester (Sec. 21.4) Reaction with an alkyl halide to yield an ether (Sec. 18.2) Oxidation to yield a quinone (Sec. 17. 10)
Quinone 1
.
Reduction to yield a hydroquinone (Sec. 17.10)
Sulfide 1
2.
3
.
Reaction with an alkyl halide to yield a sulfonium Oxidation to yield a sulfoxide (Sec. 18.8) Oxidation to yield a sulfone (Sec. 18.8)
salt (Sec. 18.8)
Thiol 1.
2.
Reaction with an alkyl halide to yield a sulfide (Sec. 18.8) Oxidation to yield a disulfide (Sec. 18.8)
Width: 612 Height: 792
Reagents in Organic Chemistry
table summarizes the uses of some important reagents in organic chemistry. The reagents are listed alphabetically, followed by a brief description of the uses of each and references to the appropriate text sections.
The following
Acetic acid, CH3CO2H: Used as a solvent for the reduction of ozonides with zinc (Section 8.8) and the a-bromination of ketones and aldehydes with Br2 (Section 22.3).
Acetic anhydride, (CH3CO)20: Reacts with alcohols to yield acetate esters (Sections 21.5 and 25.6) and with amines to yield acetamides (Section 21.5).
Aluminum
chloride, AICI3:
acylation reactions of aromatic
Acts as a Lewis acid catalyst
compounds (Section
in Friedel-Crafts alkylation
and
16.3).
Ammonia, NH3: Used -
as a solvent for the reduction of alkynes by lithium metal to yield trans alkenes (Section 9.5). Reacts with acid chlorides and acid anhydrides to yield amides (Sections 21.4 and 21.5).
Borane, BH3: Adds
-
to alkenes, giving alkylboranes that
can be oxidized with alkaline H2O2 to
yield alcohols (Section 8.5). Adds to alkynes, giving vinylic organoboranes that can be oxidized with aldehydes (Section 9.4).
- Reduces
H2O2
to yield
carboxylic acids to yield primary alcohols (Section 21.3).
Bromine, Br2: Adds to alkenes, yielding 1 ,2-dibromides (Sections 8.2, 14.2). - Adds to alkynes yielding either 1 ,2-dibromoalkenes or 1,1,2,2-tetrabromoalkanes (Section 9.3).
- Reacts
with arenes in the presence of FeBr3 catalyst to yield bromoarenes (Section 16. 1).
- Reacts with ketones
in acetic acid solvent to yield a-bromo ketones (Section 22.3). carboxylic acids in the presence of PBr3 to yield a-bromo carboxylic acids (Hell-Volhard-Zelinskii reaction; Section 22.4). Oxidizes aldoses to yield aldonic acids (Section 25.6).
- Reacts with -
N-Bromosuccinimide (NBS), (Ct^CO^NBr:
Reacts with alkenes in the presence of aqueous dimethylsulfoxide to yield bromohydrins (Section 8.3). - Reacts with alkenes in the presence of light to yield allylic bromides (Section 10.3). - Reacts with alkylbenzenes in the presence of light to yield benzylic bromides; (Section 16.9).
Di-terf-butoxy dicarbonate, protected amino acids suitable Butyllithium, anions,
CH3CH2CH2CH2Li: A
which can be alkylated (Section
- Reacts with dialkylamines -
(I-BuOCOhO:
strong base; reacts with alkynes to yield acetylide 9.8).
to yield lithium dialkylamide bases such as
amide] (Section 22.5). Reacts with alkyltriphenylphosphonium (Section 19.11).
Reacts with amino acids to give r-Boc
for use in peptide synthesis (Section 26.7).
salts to yield
LDA [lithium diisopropyl-
alkylidenephosphoranes (Wittig reagents
Reagents
Carbon
dioxide,
CO2: Reacts with Grignard
887
reagents to yield carboxylic acids (Section 20.5).
Chlorine, CI2: Adds to alkenes to yield 1,2-dichlorides (Sections 8.2 and 14.2). - Reacts with alkanes in the presence of light to yield chloroalkanes by a radical chain reaction
pathway (Section
- Reacts with arenes
10.2).
in the presence of
FeCb
catalyst to yield chloroarenes (Section 16.2).
wi-Chloroperoxybenzoic acid, m-ClC6H4CC>3H:
Reacts with alkenes to yield epoxides
(Sections 8.7, 18.5).
Chlorotrimethylsilane,
(CH3)3SiCl:
Reacts with alcohols to add the trimethylsilyl
protecting group (Section 17.8).
Chromium
trioxide, CrC^: Oxidizes alcohols in aqueous acid to yield carbonyl-containing products. Primary alcohols yield carboxylic acids, and secondary alcohols yield ketones (Sections 17.7 and 19.3).
Cuprous bromide, CuBr:
Reacts with arenediazonium
salts to yield
bromoarenes (Sandmeyer
chloride, CuCl: Reacts with arenediazonium
salts to yield
chloroarenes (Sandmeyer
reaction; Section 24.8).
Cuprous
reaction; Section 24.8).
Cuprous cyanide, CuCN: Reacts with arenediazonium (Sandmeyer
salts to yield substituted benzonitriles
reaction; Section 24.8).
Cuprous
iodide, Cul: Reacts with organolithiums to yield lithium diorganocopper reagents (Gilman reagents; Section 10.7).
Cuprous
oxide,
CU2O: Reacts with arenediazonium
salts to yield
phenols (Section 24.8).
Dess-Martin periodinane, C7H4l02(OAc)4: Oxidizes primary alcohols
to
aldehydes
(Sections 17.7 and 19.2)
Dichloroacetic acid,
CI2CHCO2H:
Cleaves
DMT
protecting groups in
DNA
synthesis
(Section 28.7).
Dicyclohexylcarbodiimide (DCC), C2,
Heat
ft
RCHp-CHoCCHo + C0 2 + CHoOH
H 3 + heat ,
a catalyst used for the hydrogenation of carbon-carbon
Aldol condensation reaction (Section 23.1): the nucleophilic addition of an enol or enolate ion to a ketone or aldehyde, yielding a p-hydroxy ketone.
ft
2
NaOH
I
R-C-(pH
H ft
*
?
I
|
r_g_^_g-H R
Amidomalonate amino acid synthesis to the
(Section 26.3): a multistep reaction sequence, similar malonic ester synthesis, for converting a primary alkyl halide into an amino acid.
—
RCH 2 X + -:C(C0 2 Et) 2
2.
I
NHAc
Cannizzaro reaction (Section
? 2
is
C02
+ 2 EtOH
NH 2 test for
aldehydes, involving treatment with cupric ion
19.12): the disproportionation reaction that occurs
when
a
treated with base.
1.HO-
RoCCH 3
+
heat
Benedict's test (Section 25.6): a chemical in aqueous sodium citrate.
nonenolizable aldehyde
RCH 2-CHCOH
*~
HoO+ 3
ff
RoCCOH + R0CCH0OH 3 3 2
2.H 3 0+
Claisen condensation reaction (Section 23.7): a nucleophilic acyl substitution reaction that occurs when an ester enolate ion attacks the carbonyl group of a second ester molecule. The product is a (3-keto ester.
ff
2
R-CH 2-C-OCH 3
1.HO-
-
fP
f?
R_CH 2-C-(j)H-C-OCH3 R
+
CH 3 OH
894
Name
Reactions
Claisen rearrangement (Sections 18.4 and 30.8): the thermal of an allyl vinyl ether or an allyl phenyl ether.
[3.3]
sigmatropic rearrangement
H
Cope rearrangement (Section to a new 1,5-diene.
30.8): the thermal [3.3] sigmatropic rearrangement of a 1,5-diene
heat
Curtius rearrangement (Section 24.6): the thermal rearrangement of an acyl azide isocyanate, followed by hydrolysis to yield an amine.
to
an
O II
+
-
R—C— N=N=N
1.
heat
2.
H2
RNH 2
Diazonium coupling reaction diazonium
salt
0~
and a phenol or
n+=n
Dieckmann reaction
C02
+ N2
(Section 24.8): the coupling reaction between an aromatic aniline.
0^
+
+
h
—*
O~ ^O~ n=n
0h
(Section 23.9): the intramolecular Claisen condensation reaction of a 1,6-
or 1,7-diester, yielding a cyclic (3-keto ester.
C0 2 Et C0 2 Et 1.
NaOEt
2-
H3 +
+ EtOH
Diels-Alder cycloaddition reaction (Sections 14.4-14.5 and diene and a dienophile to yield a cyclohexene ring.
heat
30.5): the reaction
between a
Name
Reactions
895
Edman
degradation (Section 26.6): a method for cleaving the N-terminal amino acid from a peptide by treatment of the peptide with Af-phenylisothiocyanate.
O 'I
Ph-N=C=S
+
H 2 N-CH-C-NH-£ R
Fehling's test (Section 25.6): a chemical test for aldehydes, involving treatment with cupric ion in
aqueous sodium
tartrate.
Fischer
esterification reaction (Section 21.3): the acid-catalyzed reaction carboxylic acid and an alcohol, yielding the ester.
O
H+
II
R-C-OH
O ,
heat
II
R-C-OR' + H 2
R'-OH
+
between a
Friedel-Crafts reaction (Section
16.3): the alkylation or acylation of an aromatic ring treatment with an alkyl- or acyl chloride in the presence of a Lewis-acid catalyst.
by
R-CI, AICI 3 Alkylation
O 11
R-C-CI, AICI3
R Acylation
Gabriel amine synthesis (Section 24.6): a multistep sequence for converting a primary alkyl halide into a primary amine by alkylation with potassium phthalimide, followed by hydrolysis.
NTK +
+
RCH 2X
1.
mix
2.
NaOH
2
NH 2
Gilman reagent
(Section 10.7): a lithium dialkylcopper reagent, R^CuLi, prepared by treatment of a cuprous salt with an alkyllithium. Gilman reagents undergo a coupling reaction with alkyl
halides, a 1,4-addition reaction with a,|3-unsaturated ketones,
chlorides to yield ketones.
and a coupling reaction with acid
Width: 612 Height: 792
Name
896
Reactions
Glycal assembly method (Section 25.9): a method of polysaccharide synthesis in which a glycal is converted into its epoxide, which is then opened by reaction with an alcohol.
Grignard reaction
(Section 19.7): the nucleophilic addition reaction of an alkylmagnesium halide to a ketone, aldehyde, or ester carbonyl group.
OH
O R-Mg-X
+
1
IT
mix
I
10.6): an organomagnesium halide, RMgX, prepared by reaction between an organohalide and magnesium metal. Grignard reagents add to carbonyl compounds
Grignard reagent (Section to yield alcohols.
Hell-Volhard-Zelinskii reaction (Section 22.4): the a-bromination of treatment with bromine and phosphorus tribromide.
HO III
Br
— C— C— OH
1
.
,
2. I
Hofmann amine
—
Br 2 PBr 3
H2
a carboxylic acid
by
O
— C— C— OH I
II
|
elimination (Section 24.7): a method for effecting the elimination reaction of an The amine is first treated with excess iodomethane, and the resultant
to yield an alkene.
quaternary
R2N
ammonium salt is heated H
V
1.CHJ
C— C
^\
" 2.
Ag 2 0, H 2
with silver oxide.
\
/
C=C / \
+ R2 NCH 3
Name
Hofmann rearrangement
Reactions
897
(Section 24.6): the rearrangement of an TV-bromoamide to a primary
amine by treatment with aqueous base.
O Br,
I
II
R— C— NH.
R—C— NHBr
NaOH
RNH 2
+
C0 2
Kiliani-Fischer synthesis (Section 25.6): a multistep sequence for chain-lengthening an aldose into the next higher homolog.
CHO CHO
i.HCN
R
2.
H2
3.
H3 +
,
I
CH(OH)
Pd,
BaS0 4
|
R
Knowles amino acid synthesis synthesis
(Section 26.3): an enantioselective method of amino acid chiral hydrogenation catalyst.
by hydrogenation of a Z enamido acid with a
C0 2 H
C=C NHCOCH3
H2) [Rh(DiPAMP)(COD)]
1
2.
+
C0 2 H
"
BF4
NaOH, H 2
H2 N
H
Koenigs-Knorr reaction
(Section 25.6): a method for synthesizing a glycoside by reaction of a pyranosyl bromide with an alcohol and Ag20.
CH 2 OAc
;h 2
ROH, Ag 2
AcO Ac 2.
OAcI
oh
H0
~OH, H 2
Br
Malonic ester synthesis (Section
f?
R-CH 2-X
+
1
.
:pH-C-OCH 3 2.
C0 2 CH 3
sequence for converting an alkyl halide two carbon atoms to the chain.
22.7): a multistep
into a carboxylic acid with the addition of
heat
H3O+
P
heat
RCH 2-CH 2 COCH 3
+
C0 2
+
CH 3 OH
898
Name
Reactions
McLafferty rearrangement (Section 19.14): a mass spectral fragmentation pathway for carbonyl compounds having a hydrogen three carbon atoms away from the carbonyl carbon.
Meisenheimer complex
(Section 16.7): an intermediate formed in the nucleophilic aryl
substitution reaction of a base with a nitro-substituted aromatic ring.
Merrifield solid-phase peptide synthesis (Section 26.8): a rapid and efficient means of peptide synthesis in which the growing peptide chain is attached to an insoluble polymer support.
Michael reaction (Section 23.10): the 1,4-addition reaction of a stabilized enolate anion such that from a 1,3-diketone to an a,p-unsaturated carbonyl compound.
—
O
O
c \
-c*
2 H 29 C=CHCR
Base
II
CH 29
/
as
+
9 CH-CH 29 CH 29 CR
\
—
C \\
o
II
/
\\
o
Robinson annulation reaction
(Section 23.12): a multistep sequence for building a new initial Michael reaction of the
cyclohexenone ring onto a ketone. The sequence involves an ketone followed by an internal aldol cyclization.
Sandmeyer reaction halide
(Section 24.8): a method for converting an aryldiazonium salt into an aryl
by treatment with a cuprous
0~
N+=N
+
CuBr
Sanger dideoxy method (Section
halide.
—
Br
"
28.6): an
O"'
(or CI,
enzymatic method for
I)
DNA
sequencing.
Name
Reactions
899
Simmons-Smith reaction
(Section 8.9): a method for preparing a cyclopropane by treating an alkene with diiodomethane and zinc-copper.
CH 2 2
+
I
Stork enamine reaction (Section 23.11): a multistep sequence whereby a ketone is converted into an enamine by treatment with a secondary amine, and the enamine is then used in Michael reactions.
^—
O
NFL2
HNR2 Suzuki-Miyaura reaction
HpC=CHCOR'
1.
I
O
O
A^\A II
||
R
,
(Section 10.7): an organometallic coupling of an aromatic or vinyl
substituted boronic acid with an aryl or vinyl substituted organohalide in the presence of a base
and a palladium
catalyst.
+
Ar-B(OH) 2
I— Ar'
(or vinyl)
Pd(Pho) 4
Ar-Ar'
CaC0 3 THF
Tollens' test (Section 25.6): a chemical test for detecting aldehydes by treatment with ammoniacal silver nitrate. A positive test is signaled by formation of a silver mirror on the walls of the reaction vessel.
Walden Sn2
inversion (Section 11.1): the inversion of stereochemistry
at
a chirality center during an
reaction.
Nuf
+
v c—
N
/
\
Williamson ether synthesis (Section
18.2): a
primary alkyl halide with an alkoxide
Na+
R-0
+
R'CH 2 Br
^
method
for preparing an ether
by treatment of a
ion.
R-0-CH 2 R'
+
NaBr
1): a general method of alkene synthesis by treatment of a ketone or aldehyde with an alkylidenetriphenylphosphorane.
Wittig reaction (Section 19.1
° +
C ^R'
\ C=PPh«3 /
R \
/
C=C
r^ 1
\
+
PhoP=0
Name
900
Reactions
Wohl degradation
(Section 25.6): a multistep reaction sequence for degrading an aldose into the next lower homolog.
CHO j
CH(OH) |
R
—— — 1.
NH 2 OH -
2.
Ac 2
3.
NaOEt
CHO |
R
Wolff-Kishner reaction (Section
19.9): a method for converting a ketone or aldehyde into the corresponding hydrocarbon by treatment with hydrazine and strong base.
9
N 2H 4
,
KOH
C
Woodward-Hoffmann
R-CH 9 -R'
orbital symmetry rules (Section 30.9): a series of rules for predicting the stereochemistry of pericyclic reactions. Even-electron species react thermally through either antarafacial or conrotatory pathways, whereas odd-electron species react thermally through either suprafacial or disrotatory pathways.
Abbreviations
A
symbol for Angstrom unit (10
ADMET
acyclic diene metathesis, a
Ac-
Acetyl group,
Ar-
aryl
8
cm =
10~ 10
m)
method of polymerization
ft
group
at.
no.
atomic number
at.
wt.
atomic weight
[a]o
CH 3 C-
specific rotation
I?
Boc
(CH 3 ) 3 COC-
te?t-butoxycarbonyl group,
bp
boiling point
n-Bu
rc-butyl group,
sec-Bu
CH 3 CH 2 CH2 CH 2sec-butyl group, CH 3 CH 2 CH(CH 3 )-
t-Bu
terf-butyl group,
cm
centimeter
cm
1
(CH 3 ) 3 C-
wavenumber, or reciprocal centimeter
D
stereochemical designation of carbohydrates and amino acids
DCC
dicyclohexylcarbodiimide,
6
chemical
A
symbol for
AH dm
heat of reaction
DMF DMSO DNA DNP
dimethylformamide,
(E)
entgegen, stereochemical designation of double bond geometry
E act
activation energy
El
unimolecular elimination reaction
ElcB
unimolecular elimination that takes place through a carbanion intermediate
E2
bimolecular elimination reaction
Et
ethyl group,
g
gram
shift in
-N=C=N-C 6H
1 j
ppm downfield from TMS symbol for change
heat; also
decimeter (0.1
1
m)
dimethyl sulfoxide,
(CH 3 ) 2 NCHO
(CH 3 ) 2 SO
deoxyribonucleic acid dinitrophenyl group, as in
CH 3 CH2-
2,4-DNP (2,4-dinitrophenylhydrazone)
902
Abbreviations
kv
symbol for
Hz
Hertz, or cycles per second (s
i-
iso
IR
infrared
J
Joule
J
symbol for coupling constant
K Ka
Kelvin temperature
kJ
kilojoule
L
stereochemical designation of carbohydrates and amino acids
LAH
lithium
Me
methyl group,
mg
milligram (0.001 g)
MHz mL
meg ahertz(10
s
milliliter (0.001
L)
mm
millimeter (0.001
mp
melting point
Hg
microgram (10
m\x
millimicron (nanometer, 10
MW
molecular weight
n-
normal, straight-chain alkane or alkyl group
ng
nanogram (10
light )
acid dissociation constant
aluminum hydride, LiAlH 4
CH 3 )
m)
-6
g) -9
m)
-9
gram) -9
nm
nanometer (10
NMR
nuclear magnetic resonance
-OAc
acetate group,
-OCCH 3
Ph
phenyl group,
-C 6 H 5
pH pKa pm
measure of acidity of aqueous solution
ppm
parts per million
/i-Pr
/i-propyl group,
z-Pr
isopropyl group,
pro-R
designation of a prochirality center
pro-S
designation of a prochirality center
R-
symbol
meter)
measure of acid strength (= -log
Ka
)
-12
picometer(10
m)
CH 3CH 2CH 2(CH 3 ) 2 CH-
for a generalized alkyl group
Abbreviations
(
rectus, designation of chirality center
R)
Re
2
face
RNA ROMP
a face of a planar, sp -hybridized carbon atom ribonucleic acid
ring-opening metathesis polymerization
of chirality center
(S)
sinister, designation
sec-
secondary
Si face
a face of a planar, sp -hybridized carbon atom
SN 1
unimolecular substitution reaction
SN2
bimolecular substitution reaction
tert-
tertiary
THF TMS
tetrahydrofuran tetramethylsilane
Tos
tosylate group,
2
nmr
standard, (CH3)4Si
—r
— CH
y
3
UV
ultraviolet
X-
halogen group (-F, -CI, -Br, -I)
(Z)
zusammen, stereochemical designation of double bond geometry chemical reaction in direction indicated
m
reversible chemical reaction
resonance symbol
^
—>^
curved arrow indicating direction of electron flow is
equivalent to
>
greater than
<
less than
«
approximately equal to
indicates that the organic fragment
m
6+,
|
6-
shown
is
a part of a larger molecule
single
bond coming out of the plane of the paper
single
bond receding
partial
bond
partial
charge
into the plane of the paper
denoting the transition state
903
Infrared Absorption Frequencies
Functional
Alcohol
Group
Frequency (cm
-O-H
3300-3600
(s)
1 )
Text Section
17.11
\
-C-O-
1050
(s)
/
Aldehyde aliphatic
-CO-H
2720, 2820 (m)
\
1725
(s)
/
1705
(s)
C=0
aromatic
Alkane
19.14
12.8
\
-C-H \
2850-2960
(s)
/
800-1300 (m)
Alkene
12.8
3020-3100
Alkyne
(s)
H
1650-1670 (m)
RCB=CH2
910, 990 (m)
R2C=CH2
890 (m)
sC-H
3300
-OC-
2100-2260 (m)
12.8
(s)
Alkyl bromide
12.8
\
-C-Br
500-600
(s)
/ Alkyl chloride
12.8
\
— C-Cl /
600-800
(s)
Infrared Absorptions
Amine, primary
24.10 3400, 3500
(s)
H secondary
\
r Ammonium
-H
3350
(s)
24.10
salt
\ + -HST-H
2200-3000 (broad)
Ar-H
3030 (m)
Ar-R
690-710
(s)
730-770
(s)
o-disubstituted
735-770
(s)
m-disubstituted
690-710
(s)
810-850
(s)
810-840
(s)
Aromatic ring monosubstituted
p-disubstituted
Carboxylic acid
-O-H
2500-3300 (broad)
associated
\
1710
(s)
free
/
1760
(s)
G=0
15.8
Acid anhydride
20.8
21.10
\
c=o
1760, 1820
(s)
/
Acid chloride
21.10
\
1810
(s)
/
}
1770
(s)
aliphatic
\
1810
(s)
aromatic
!
1770
(s)
N-substituted
1680
(s)
N, N-disubstitu ted
1650
(s)
aliphatic
aromatic
O=0
Amide
21.10
Width: 612 Height: 792
906
Infrared Absorptions
21.10
Ester aliphatic
\
1735
(s)
aromatic
/
1720
(s)
C=0
Ether
18.9
/
1050-1150
Ketone
19.14
aliphatic
\
1715
(s)
aromatic
f
1690
(s)
6-memb. ring
1715
(s)
5-memb. ring
1750
(s)
C=0
20.8
Nitrile
aliphatic
-C=N
2250 (m) 2230 (m)
-O-H
3500
aromatic
Phenol
(s)
=
strong; (m)
(s)
= medium
intensity
(s)
17.11
NMR Chemical Shifts
Proton
Chemical Shift
Type of Proton
(8)
Text Section
R-CH 3 R-CH2-R R 3 C-H
0.7-1.3
13.9
1
.2-1 .4
13.9
i
.4_i
.7
13.9
AUyto
-C=C-j:~H
1.6-1.9
13.9
atocarbonyl
-C-j:-H
2.0-2.3
19.14
Benzylic
Ar-j:-H
2.3-3.0
15.8
Acetylenic
R-OC-H
2.5-2.7
13.9
Alkyl chloride
Cl-C-H
3.0-4.0
13.9
Alkyl bromide
Br-^-H
2.5-4.0
13.9
Alkyl iodide
I-
