CHAPTER 22
AMINES
SOLUTIONS TO TEXT PROBLEMS
22.1 (b) The amino and phenyl groups are both attached to C-1 of an ethyl group.
(c)
22.2 N,N-Dimethylcycloheptylamine may also be named as a dimethyl derivative of cycloheptanamine.
22.3 Three substituents are attached to the nitrogen atom; the amine is tertiary. In alphabetical order, the
substituents present on the aniline nucleus are ethyl, isopropyl, and methyl. Their positions are
specified as N-ethyl, 4-isopropyl, and N-methyl.
(CH
3
)
2
CH CH
2
CH
3
CH
3
N
N-Ethyl-4-isopropyl-N-methylaniline
N(CH
3
)
2
N,N-Dimethylcycloheptanamine
CHCH
2
NH
2
H
2
C
Allylamine, or
2-propen-1-amine
C
6
H
5
CHCH
3
NH
2
1-Phenylethylamine, or
1-phenylethanamine
604
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AMINES 605
22.4 The electron-donating amino group and the electron-withdrawing nitro group are directly conjugated
in p-nitroaniline. The planar geometry of p-nitroaniline suggests that the delocalized resonance form
shown is a major contributor to the structure of the compound.
22.5 The pK
b
of an amine is related to the equilibrium constant K
b
by
pK
b
H11005H11002log K
b
The pK
b
of quinine is therefore
pK
b
H11005H11002log (1 H11003 10
H110026
) H11005 6
the values of K
b
and pK
b
for an amine and K
a
and pK
a
of its conjugate acid are given by
K
a
H11003 K
b
H11005 1 H11003 10
H1100214
and
pK
a
H11001 pK
b
H11005 14
The values of K
a
and pK
a
for the conjugate acid of quinine are therefore
K
a
H11005H11005 H110051 H11003 10
H110028
and
pK
a
H11005 14 H11002 pK
b
H11005 14 H11002 6 H11005 8
22.6 The Henderson–Hasselbalch equation described in Section 19.4 can be applied to bases such as
amines, as well as carboxylic acids. The ratio [CH
3
NH
3
H11001
]H20862[CH
3
NH
2
] is given by
H11005
The ionization constant of methylammonium ion is given in the text as 2 H11003 10
H1100211
. At pH H11005 7 the
hydrogen ion concentration is 1 H11003 10
H110027
. Therefore
H11005
H11005 5 H11003 10
3
22.7 Nitrogen is attached directly to the aromatic ring in tetrahydroquinoline, making it an arylamine,
and the nitrogen lone pair is delocalized into the H9266 system of the aromatic ring. It is less basic than
tetrahydroisoquinoline, in which the nitrogen is insulated from the ring by an sp
3
-hybridized
carbon.
See Learning By Modeling for the calculated charges on nitrogen.
N
H
Tetrahydroquinoline
(an arylamine): less basic,
K
b
1.0 H11003 10
H110029
(pK
b
9.0)
NH
Tetrahydroisoquinoline
(an alkylamine): more basic,
K
b
2.5 H11003 10
H110025
(pK
b
4.6)
1 H11003 10
H110027
H5007H5007
2 H11003 10
H1100211
[CH
3
NH
3
H11001
]
H5007H5007
[CH
3
NH
2
]
[H
H11001
]
H5007
K
a
[CH
3
NH
3
H11001
]
H5007H5007
[CH
3
NH
2
]
1 H11003 10
H1100214
H5007H5007
1 H11003 10
H110026
10
H1100214
H5007
K
b
NH
2
H11001
NH
2
N
H11001
OO OO
N
H11001
H11002H11002H11002
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606 AMINES
22.8 (b) An acetyl group attached directly to nitrogen as in acetanilide delocalizes the nitrogen lone
pair into the carbonyl group. Amides are weaker bases than amines.
(c) An acetyl group in a position para to an amine function is conjugated to it and delocalizes the
nitrogen lone pair.
22.9 The reaction that leads to allylamine is nucleophilic substitution by ammonia on allyl chloride.
Allyl chloride is prepared by free-radical chlorination of propene (see text page 371).
22.10 (b) Isobutylamine is (CH
3
)
2
CHCH
2
NH
2
. It is a primary amine of the type RCH
2
NH
2
and can be
prepared from a primary alkyl halide by the Gabriel synthesis.
Isobutyl bromide
(CH
3
)
2
CHCH
2
Br H11001
N-Potassiophthalimide
NK
O
O
N-Isobutylphthalimide
NCH
2
CH(CH
3
)
2
O
O
Phthalhydrazide
H11001
Isobutylamine
(CH
3
)
2
CHCH
2
NH
2
H
2
NNH
2
O
O
NH
NH
H11001H11001Cl
2
Chlorine
H
2
C CHCH
3
Propene
H
2
C CHCH
2
Cl
Allyl chloride
HCl
Hydrogen
chloride
400H11034C
H11001H110012NH
3
Ammonia
H
2
C CHCH
2
Cl
Allyl chloride
H
2
C CHCH
2
NH
2
Allylamine
NH
4
Cl
Ammonium
chloride
O
CH
2
N
CH
3
H11002
H11001
O
CH
2
N
CH
3
O
N
H
CCH
3
O
N
H
CCH
3
H11002
H11001
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(c) Although tert-butylamine (CH
3
)
3
CNH
2
is a primary amine, it cannot be prepared by the
Gabriel method, because it would require an S
N
2 reaction on a tertiary alkyl halide in the first
step. Elimination occurs instead.
(d) The preparation of 2-phenylethylamine by the Gabriel synthesis has been described in the
chemical literature.
(e) The Gabriel synthesis leads to primary amines; N-methylbenzylamine is a secondary amine
and cannot be prepared by this method.
( f ) Aniline cannot be prepared by the Gabriel method. Aryl halides do not undergo nucleophilic
substitution under these conditions.
H11001
Br
Bromobenzene
no reaction
N-Potassiophthalimide
NK
O
O
CH
2
CH
3
H
N
N-Methylbenzylamine
(two carbon substituents on
nitrogen; a secondary amine)
2-Phenylethyl bromide
C
6
H
5
CH
2
CH
2
Br H11001
N-Potassiophthalimide
NK
O
O
N-(2-Phenylethyl)phthalimide
NCH
2
CH
2
C
6
H
5
O
O
Phthalhydrazide
H11001
2-Phenylethylamine
C
6
H
5
CH
2
CH
2
NH
2
H
2
NNH
2
O
O
NH
NH
tert-Butyl
bromide
(CH
3
)
2
CBr H11001H11001H11001
N-Potassiophthalimide
NK
O
O
KBr
Potassium
bromide
Phthalimide
NH
O
O
2-Methylpropene
(CH
3
)
2
CCH
2
AMINES 607
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22.11 For each part of this problem, keep in mind that aromatic amines are derived by reduction of the
corresponding aromatic nitro compound. Each synthesis should be approached from the standpoint
of how best to prepare the necessary nitroaromatic compound.
(b) The para isomer of isopropylaniline may be prepared by a procedure analogous to that used
for its ortho isomer in part (a).
After separating the ortho, para mixture by distillation, the nitro group of p-isopropyl-
nitrobenzene is reduced to yield the desired p-isopropylaniline.
(c) The target compound is the reduction product of 1-isopropyl-2,4-dinitrobenzene.
This reduction is carried out in the same way as reduction of an arene that contains only a
single nitro group. In this case hydrogenation over a nickel catalyst gave the desired product
in 90% yield.
The starting dinitro compound is prepared by nitration of isopropylbenzene.
80H11034C
HNO
3
, H
2
SO
4
Isopropylbenzene
CH(CH
3
)
2
1-Isopropyl-2,4-
dinitrobenzene
(43%)
CH(CH
3
)
2
NO
2
NO
2
reduce
CH(CH
3
)
2
NH
2
NH
2
4-Isopropyl-1,3-
benzenediamine
CH(CH
3
)
2
NO
2
NO
2
1-Isopropyl-2,4-
dinitrobenzene
H
2
, Ni; or
1. Fe, HCl; 2. HO
H11002
or
1. Sn, HCl; 2. HO
H11002
CH(CH
3
)
2
NH
2
CH(CH
3
)
2
NO
2
H11001
(CH
3
)
2
CHCl
AlCl
3
HNO
3
H
2
SO
4
CH(CH
3
)
2
NO
2
p-Isopropylnitro-
benzene
CH(CH
3
)
2
NO
2
o-Isopropylnitro-
benzene
CH(CH
3
)
2
IsopropybenzeneBenzene
NH
2
Ar NO
2
Ar HAr
(Ar H11005 substituted aromatic ring)
608 AMINES
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(d ) The conversion of p-chloronitrobenzene to p-chloroaniline was cited as an example in the text
to illustrate reduction of aromatic nitro compounds to arylamines. p-Chloronitrobenzene is
prepared by nitration of chlorobenzene.
The para isomer accounts for 69% of the product in this reaction (30% is ortho, 1% meta).
Separation of p-chloronitrobenzene and its reduction completes the synthesis.
Chlorination of nitrobenzene would not be a suitable route to the required intermediate,
because it would produce mainly m-chloronitrobenzene.
(e) The synthesis of m-aminoacetophenone may be carried out by the scheme shown:
The acetyl group is attached to the ring by Friedel–Crafts acylation. It is a meta director, and
its nitration gives the proper orientation of substituents. The order of the first two steps cannot
be reversed, because Friedel–Crafts acylation of nitrobenzene is not possible (Section 12.16).
Once prepared, m-nitroacetophenone can be reduced to m-nitroaniline by any of a number of
reagents. Indeed, all three reducing combinations described in the text have been employed
for this transformation.
Yield
Reducing agent (%)
m-Nitroacetophenone H
2
, Pt 94
↓
Fe, HCl 84
m-Aminoacetophenone Sn, HCl 82
22.12 (b) Dibenzylamine is a secondary amine and can be prepared by reductive amination of benz-
aldehyde with benzylamine.
H
2
, Ni
C
6
H
5
CH
O
C
6
H
5
CH
2
NH
2
C
6
H
5
CH
2
NHCH
2
C
6
H
5
H11001
Dibenzylamine Benzylamine Benzaldehyde
Benzene
HNO
3
H
2
SO
4
reduce
AlCl
3
CH
3
CCl
O
Acetophenone
CCH
3
O
m-Nitroacetophenone
NO
2
CCH
3
O
m-Aminoacetophenone
NH
2
CCH
3
O
p-Chloroaniline
Cl
NH
2
p-Chloronitrobenzene
Cl
NO
2
1. Fe, HCl; 2. HO
H11002
or
1. Sn, HCl; 2. HO
H11002
H
2
, catalyst; or
Benzene Chlorobenzene
Cl
o-Chloronitrobenzene
NO
2
Cl
p-Chloronitrobenzene
Cl
NO
2
Cl
2
FeCl
3
HNO
3
H
2
SO
4
H11001
AMINES 609
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(c) N,N-Dimethylbenzylamine is a tertiary amine. Its preparation from benzaldehyde requires
dimethylamine, a secondary amine.
(d ) The preparation of N-butylpiperidine by reductive amination is described in the text in
Section 22.11. An analogous procedure is used to prepare N-benzylpiperidine.
22.13 (b) First identify the available H9252 hydrogens. Elimination must involve a proton from the carbon
atom adjacent to the one that bears the nitrogen.
It is a proton from one of the methyl groups, rather than one from the more sterically hindered
methylene, that is lost on elimination.
(c) The base may abstract a proton from either of two H9252 carbons. Deprotonation of the H9252 methyl
carbon yields ethylene.
Deprotonation of the H9252 methylene carbon yields 1-butene.
1-Butene
H
2
C CHCH
2
CH
3
H11001
N-Ethyl-N,N-dimethylbutylammonium
hydroxide
heat
(H11002H
2
O)
N,N-Dimethylethylamine
H11001
N
CH
3
CH
3
H
CH
2
CHCH
2
CH
3
OH
H11002
CH
3
CH
2
N(CH
3
)
2
CH
3
CH
2
Ethylene
H
2
CCH
2
H11001
N-Ethyl-N,N-dimethylbutylammonium hydroxide
H11001
NCH
2
CH
2
CH
2
CH
3
CH
3
CH
3
CH
2
CH
2
HOH
H11002 heat
(H11002H
2
O)
N,N-Dimethylbutylamine
(CH
3
)
2
NCH
2
CH
2
CH
2
CH
3
OH
(1,1,3,3-Tetramethylbutyl)-
trimethylammonium
hydroxide
(CH
3
)
3
CCH
2
CH
2
CH
N(CH
3
)
3
CH
3
H11002
2,4,4-Trimethyl-1-pentene
(only alkene formed,
70% isolated yield)
(CH
3
)
3
CCH
2
CCH
2
CH
3
H11001
Trimethylamine
(CH
3
)
3
N
H11001
(CH
3
)
3
CCH
2
CH
3
C
H11001
N(CH
3
)
3
CH
3
A methylene group
H9252H9252
H9252
Two equivalent methyl groups
H
2
, Ni
H11001C
6
H
5
CH
O
Benzaldehyde Piperidine
H
N
N-Benzylpiperidine
C
6
H
5
CH
2
N
H
2
, Ni
C
6
H
5
CH
O
(CH
3
)
2
NH C
6
H
5
CH
2
N(CH
3
)
2
H11001
N,N-Dimethylbenzylamine Dimethylamine Benzaldehyde
610 AMINES
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The preferred order of proton removal in Hofmann elimination reactions is H9252 CH
3
H11022
H9252 CH
2
H11022 H9252CH. Ethylene is the major alkene formed, the observed ratio of ethylene to 1-butene
being 98 : 2.
22.14 (b) The pattern of substituents in 2,4-dinitroaniline suggests that they can be introduced by dini-
tration. Since nitration of aniline itself is not practical, the amino group must be protected by
conversion to its N-acetyl derivative.
Hydrolysis of the amide bond in 2,4-dinitroacetanilide furnishes the desired 2,4-dinitroaniline.
(c) Retrosynthetically, p-aminoacetanilide may be derived from p-nitroacetanilide.
This suggests the sequence
HNO
3
H
2
SO
4
CH
3
CNH
O
Acetanilide
H
2
N
Aniline
1. Fe, HCl; 2. HO
H11002
or
1. Sn, HCl; 2. HO
H11002
or
H
2
, Pt
CH
3
CNH NO
2
O
p-Nitroacetanilide
(separate from ortho isomer)
CH
3
CNH NH
2
O
p-Aminoacetanilide
CH
3
COCCH
3
OO
CH
3
CNH
O
NH
2
p-Aminoacetanilide
CH
3
CNH
O
NO
2
p-Nitroacetanilide
2,4-Dinitroaniline
NH
2
NO
2
NO
2
2,4-Dinitroacetanilide
NHCCH
3
NO
2
NO
2
O
H
2
O, HO
H11002
, or
1. H
2
O, H
H11001
2. HO
H11002
NH
2
Aniline Acetanilide
NHCCH
3
O
2,4-Dinitroacetanilide
NHCCH
3
NO
2
NO
2
O
HNO
3
H
2
SO
4
CH
3
CCl
O
or
CH
3
COCCH
3
O O
AMINES 611
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22.15 The principal resonance forms of N-nitrosodimethylamine are
All atoms (except hydrogen) have octets of electrons in each of these structures. Other resonance
forms are less stable because they do not have a full complement of electrons around each atom.
22.16 Deamination of 1,1-dimethylpropylamine gives products that result from 1,1-dimethylpropyl
cation. Because 2,2-dimethylpropylamine gives the same products, it is likely that 1,1-dimethyl-
propyl cation is formed from 2,2-dimethylpropylamine by way of its diazonium ion. A carbocation
rearrangement is indicated.
Once formed, 1,1-dimethylpropyl cation loses a proton to form an alkene or is captured by water to
give an alcohol.
22.17 Phenols may be prepared by diazotization of the corresponding aniline derivative. The problem
simplifies itself, therefore, to the preparation of m-bromoaniline. Recognizing that arylamines are
ultimately derived from nitroarenes, we derive the retrosynthetic sequence of intermediates:
The desired reaction sequence is straightforward, using reactions that were discussed previously in
the text.
Br
OH
Br
NH
2
Br
NO
2
NO
2
HNO
3
H
2
SO
4
Br
2
Fe
1. Fe, HCl
2. NaOH 2. H
2
O, heat
1. NaNO
2
,
H
2
SO
4
H
2
O, 0–5H11034C
OH
Br Br
NH
2
Br
NO
2
NO
2
m-Bromophenol m-Bromoaniline m-Bromonitrobenzene Nitrobenzene
H11002H
H11001
H
2
O
CH
3
CCH
2
CH
3
CH
3
H11001
1,1-Dimethylpropyl
cation
(CH
3
)
2
CCH
2
CH
3
OH
2-Methyl-2-butanol
H11001H
2
C CCH
2
CH
3
CH
3
2-Methyl-1-butene
(CH
3
)
2
C CHCH
3
2-Methyl-2-butene
HONO
H11002N
2
CH
3
CCH
2
N
H11001
CH
3
CH
3
N
2,2-Dimethylpropyldiazonium
ion
CH
3
CCH
2
CH
3
CH
3
H11001
1,1-Dimethylpropyl
cation
CH
3
CCH
2
NH
2
CH
3
CH
3
2,2-Dimethylpropylamine
NN
O
H
3
C
H
3
C
NN
O
H
3
C
H
3
C
H11002
H11001
612 AMINES
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22.18 The key to this problem is to recognize that the iodine substituent in m-bromoiodobenzene is derived
from an arylamine by diazotization.
The preparation of m-bromoaniline from benzene has been described in Problem 22.17. All that
remains is to write the equation for its conversion to m-bromoiodobenzene.
22.19 The final step in the preparation of ethyl m-fluorophenyl ketone is shown in the text example im-
mediately preceding this problem, therefore all that is necessary is to describe the preparation of
m-aminophenyl ethyl ketone.
Recalling that arylamines are normally prepared by reduction of nitroarenes, we see that ethyl
m-nitrophenyl ketone is a pivotal synthetic intermediate. It is prepared by nitration of ethyl
phenyl ketone, which is analogous to nitration of acetophenone, shown in Section 12.16.
The preparation of ethyl phenyl ketone by Friedel–Crafts acylation of benzene is shown in Sec-
tion 12.7.
Reversing the order of introduction of the nitro and acyl groups is incorrect. It is possible to nitrate
ethyl phenyl ketone but not possible to carry out a Friedel–Crafts acylation on nitrobenzene, owing
to the strong deactivating influence of the nitro group.
22.20 Direct nitration of the prescribed starting material cumene (isopropylbenzene) is not suitable,
because isopropyl is an ortho, para-directing substituent and will give the target molecule
NO
2
CCH
2
CH
3
O
Ethyl m-nitrophenyl
ketone
CCH
2
CH
3
O
Ethyl phenyl ketone
F
CCH
2
CH
3
NH
2
NO
2
Ethyl m-nitrophenyl
ketone
Ethyl m-fluorophenyl
ketone
m-Aminophenyl ethyl
ketone
CCH
2
CH
3
O
CCH
2
CH
3
O O
I
Br
m-Bromoiodobenzene
Br
NH
2
m-Bromoaniline
1. NaNO
2
,
HCl, H
2
O
2. KI
I
Br Br
NH
2
m-Bromoiodobenzene m-Bromoaniline
AMINES 613
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m-nitrocumene as only a minor component of the nitration product. However, the conversion of
4-isopropyl-2-nitroaniline to m-isopropylnitrobenzene, which was used to illustrate reductive
deamination of arylamines in the text, establishes the last step in the synthesis.
Our task simplifies itself to the preparation of 4-isopropyl-2-nitroaniline from cumene. The follow-
ing procedure is a straightforward extension of the reactions and principles developed in this
chapter.
Reductive deamination of 4-isopropyl-2-nitroaniline by diazotization in the presence of ethanol or
hypophosphorous acid yields m-nitrocumene and completes the synthesis.
22.21 Amines may be primary, secondary, or tertiary. The C
4
H
11
N primary amines, compounds of the type
C
4
H
9
NH
2
, and their systematic names are
tert-Butylamine
(2-methyl-2-propanamine)
(CH
3
)
3
CNH
2
sec-Butylamine
(2-butanamine)
CH
3
CHCH
2
CH
3
NH
2
Butylamine
(1-butanamine)
CH
3
CH
2
CH
2
CH
2
NH
2
Isobutylamine
(2-methyl-1-propanamine)
(CH
3
)
2
CHCH
2
NH
2
HNO
3
H
2
SO
4
HO
H11002
, H
2
O, or
1. H
2
O, H
H11001
2. HO
H11002
CH(CH
3
)
2
CH(CH
3
)
2
NHCCH
3
NO
2
CH(CH
3
)
2
NO
2
NH
2
O
NHCCH
3
O
p-Isopropylacetanilide 4-Isopropyl-2-nitroacetanilide 4-Isopropyl-2-nitroaniline
CH(CH
3
)
2
Cumene
CH(CH
3
)
2
NO
2
p-Nitrocumene
CH(CH
3
)
2
NH
2
p-Isopropylaniline
CH(CH
3
)
2
NHCCH
3
O
p-Isopropylacetanilide
HNO
3
H
2
SO
4
1. Fe, HCl
2. HO
H11002
CH
3
CCl
O
CH(CH
3
)
2
Cumene
CH(CH
3
)
2
NH
2
NO
2
4-Isopropyl-2-
nitroaniline
CH(CH
3
)
2
NO
2
m-Nitrocumene
614 AMINES
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Secondary amines have the general formula R
2
NH. Those of molecular formula C
4
H
11
N are
There is only one tertiary amine (R
3
N) of molecular formula C
4
H
11
N:
22.22 (a) The name 2-ethyl-1-butanamine designates a four-carbon chain terminating in an amino
group and bearing an ethyl group at C-2.
(b) The prefix N- in N-ethyl-1-butanamine identifies the ethyl group as a substituent on nitrogen
in a secondary amine.
(c) Dibenzylamine is a secondary amine. It bears two benzyl groups on nitrogen.
(d) Tribenzylamine is a tertiary amine.
(e) Tetraethylammonium hydroxide contains a quaternary ammonium ion.
( f ) This compound is a secondary amine; it bears an allyl substituent on the nitrogen of cyclo-
hexylamine.
N
CH
2
CH CH
2
H
N-Allylcyclohexylamine
Tetraethylammonium
hydroxide
(CH
3
CH
2
)
4
N HO
H11002
H11001
Tribenzylamine
(C
6
H
5
CH
2
)
3
N
Dibenzylamine
C
6
H
5
CH
2
NCH
2
C
6
H
5
H
N-Ethyl-1-butanamine
CH
3
CH
2
CH
2
CH
2
NCH
2
CH
3
H
2-Ethyl-1-butanamine
CH
3
CH
2
CHCH
2
NH
2
CH
2
CH
3
N,N-Dimethylethylamine
(N,N-dimethylethanamine)
(CH
3
)
2
NCH
2
CH
3
Diethylamine
(N-ethylethanamine)
(CH
3
CH
2
)
2
NH
N-Methylpropylamine
(N-methyl-1-propanamine)
CH
3
NCH
2
CH
2
CH
3
H
N-Methylisopropylamine
(N-methyl-2-propanamine)
CH
3
NCH(CH
3
)
2
H
AMINES 615
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(g) Piperidine is a cyclic secondary amine that contains nitrogen in a six-membered ring.
N-Allylpiperidine is a tertiary amine.
(h) The compound is the benzyl ester of 2-aminopropanoic acid.
(i) The parent compound is cyclohexanone. The substituent (CH
3
)
2
N— group is attached to C-4.
( j) The suffix -diamine reveals the presence of two amino groups, one at either end of a three-
carbon chain that bears two methyl groups at C-2.
22.23 (a) A phenyl group and an amino group are trans to each other on a three-membered ring in this
compound.
(b) This compound is a tertiary amine. It bears a benzyl group, a methyl group, and a 2-propynyl
group on nitrogen.
N-Benzyl-N-methyl-2-propynylamine
(pargyline)
C
6
H
5
CH
2
CH
2
C CH
CH
3
N
trans-2-Phenylcyclopropylamine
(tranylcypromine)
C
6
H
5
H
H NH
2
H
2
NCH
2
CCH
2
NH
2
CH
3
CH
3
2,2-Dimethyl-1,3-
propanediamine
4-(N,N-Dimethylamino)-
cyclohexanone
(CH
3
)
2
N
O
H
CH
3
CHCOCH
2
C
6
H
5
NH
2
O
Benzyl 2-aminopropanoate
CH
2
N-Allylpiperidine
NCH
2
CH
616 AMINES
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(c) The amino group is at C-2 of a three-carbon chain that bears a phenyl substituent at its
terminus.
(d) Phenylephrine is named systematically as an ethanol derivative.
22.24 (a) There are five isomers of C
7
H
9
N that contain a benzene ring.
(b) Benzylamine is the strongest base because its amine group is bonded to an sp
3
-hybridized
carbon. Benzylamine is a typical alkylamine, with a K
b
of 2 H11003 10
H110025
. All the other isomers are
arylamines, with K
b
values in the 10
H1100210
range.
(c) The formation of N-nitrosoamines on reaction with sodium nitrite and hydrochloric acid is a
characteristic reaction of secondary amines. The only C
7
H
9
N isomer in this problem that is a
secondary amine is N-methylaniline.
(d) Ring nitrosation is a characteristic reaction of tertiary arylamines.
None of the C
7
H
9
N isomers in this problem is a tertiary amine; hence none will undergo ring
nitrosation.
HCl, H
2
O
NaNO
2
Tertiary arylamine
NR
2
p-Nitroso-N,N-dialkylaniline
NR
2
ON
N-Methylaniline
C
6
H
5
NHCH
3
N-Methyl-N-
nitrosoaniline
C
6
H
5
NCH
3
NO
HCl, H
2
O
NaNO
2
o-Methylaniline
CH
3
NH
2
m-Methylaniline
CH
3
NH
2
p-Methylaniline
CH
3
NH
2
Benzylamine
C
6
H
5
CH
2
NH
2
N-Methylaniline
C
6
H
5
NHCH
3
1-(m-Hydroxyphenyl)-
2-(methylamino)ethanol
CHCH
2
N
HO
OH
H
CH
3
C
6
H
5
CH
2
CHCH
3
NH
2
1-Phenyl-2-propanamine
(amphetamine)
AMINES 617
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22.25 (a) Basicity decreases in proceeding across a row in the periodic table. The increased nuclear
charge as one progresses from carbon to nitrogen to oxygen to fluorine causes the electrons to
be bound more strongly to the atom and thus less readily shared.
(b) The strongest base in this group is amide ion, H
2
N
–
, and the weakest base is water, H
2
O.
Ammonia is a weaker base than hydroxide ion; the equilibrium lies to the left.
The correct order is
(c) These anions can be ranked according to their basicity by considering the respective acidities
of their conjugate acids.
Base Conjugate acid K
a
of conjugate acid
H
2
N
H11002
H
3
N10
H1100236
HO
H11002
H
2
O1
H1100216
7.2 H11003 10
H1100210
2.5 H11003 10
1
The order of basicities is the opposite of the order of acidities of their conjugate acids.
(d) A carbonyl group attached to nitrogen stabilizes its negative charge. The strongest base is the
anion that has no carbonyl groups on nitrogen; the weakest base is phthalimide anion, which
has two carbonyl groups.
H11022
Strongest base
N
H11002
H11022N
O
H11002
Weakest base
N
O
O
H11002
H11022 NO
3
H11002
Weakest
base
Strongest
base
H
2
N
H11002
H11022HO
H11002
H11022CN
H11002
HON
O
O
H11002
H11001
H11002
ON
O
O
H11002
H11001
HC NCN
H11002
NH
3
H11022
Strongest
base
H
2
N
H11002
H11022HO
H11002
H11022 H
2
O
Weakest
base
H11001
Weaker
acid
H
2
O
Weaker
base
NH
3
H11001
Stronger
base
OH
H11002
Stronger
acid
NH
4
H11001
10
H1100260
Strongest
base
H
3
C
H11002
H11022H11022H11022
10
H1100236
H
2
N
H11002
10
H1100216
HO
H11002
3.5 H11003 10
H110024
Weakest
base
F
H11002
of conjugate
acid
K
a
618 AMINES
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22.26 (a) An alkyl substituent on nitrogen is electron-releasing and base-strengthening; thus methyl-
amine is a stronger base than ammonia. An aryl substituent is electron-withdrawing and base-
weakening, and so aniline is a weaker base than ammonia.
(b) An acetyl group is an electron-withdrawing and base-weakening substituent, especially when
bonded directly to nitrogen. Amides are weaker bases than amines, and thus acetanilide is a
weaker base than aniline. Alkyl groups are electron-releasing; N-methylaniline is a slightly
stronger base than aniline.
(c) Chlorine substituents are slightly electron-withdrawing, and methyl groups are slightly
electron-releasing. 2,4-Dimethylaniline is therefore a stronger base than 2,4-dichloroaniline.
Nitro groups are strongly electron-withdrawing, their base-weakening effect being especially
pronounced when a nitro group is ortho or para to an amino group because the two groups are
then directly conjugated.
(d) Nitro groups are more electron-withdrawing than chlorine, and the base-weakening effect of
a nitro substituent is greater when it is ortho or para to an amino group than when it is meta
to it.
H11022H11022
NO
2
Cl
NH
2
4-Chloro-2-nitroaniline,
weakest base:
K
b
1 H11003 10
H1100215
pK
b
15.0
3,4-Dichloroaniline,
strongest base:
K
b
H11015 10
H1100211
pK
b
H11015
11
NH
2
Cl
Cl
4-Chloro-3-nitroaniline:
K
b
8 H11003 10
H1100213
pK
b
12.1
NH
2
NO
2
Cl
H11022H11022
CH
3
CH
3
NH
2
2,4-Dimethylaniline,
strongest base:
K
b
8 H11003 10
H1100210
pK
b
9.1
NH
2
Cl
Cl
2,4-Dichloroaniline:
K
b
1 H11003 10
H1100212
pK
b
12.0
NO
2
NO
2
NH
2
2,4-Dinitroaniline,
weakest base:
K
b
3 H11003 10
H1100219
pK
b
18.5
H11022
N-methylaniline,
strongest base:
K
b
8 H11003 10
H1100210
pK
b
9.1
C
6
H
5
NHCH
3
H11022
Aniline:
K
b
3.8 H11003 10
H1100210
pK
b
9.4
C
6
H
5
NH
2
Acetanilide,
weakest base:
K
b
1 H11003 10
H1100215
pK
b
15.0
C
6
H
5
NHCCH
3
O
H11022
Methylamine,
strongest base:
K
b
4.4 H11003 10
H110024
pK
b
3.4
CH
3
NH
2
H11022
Ammonia:
K
b
1.8 H11003 10
H110025
pK
b
4.7
NH
3
Aniline,
weakest base:
K
b
3.8 H11003 10
H1100210
pK
b
9.4
C
6
H
5
NH
2
AMINES 619
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(e) According to the principle applied in part (a) (alkyl groups increase basicity, aryl groups
decrease it), the order of decreasing basicity is as shown:
22.27 Nitrogen
H17034
a
is the most basic and the most nucleophilic of the three nitrogen atoms of physostigmine
and is the one that reacts with methyl iodide.
The nitrogen that reacts is the one that is a tertiary alkylamine. Of the other two nitrogens,
H17034
b is
attached to an aromatic ring and is much less basic and less nucleophilic. The third nitrogen,
H17034
c
, is
an amide nitrogen; amides are less nucleophilic than amines.
22.28 (a) Looking at the problem retrosynthetically, it can be seen that a variety of procedures are avail-
able for preparing ethylamine from ethanol. The methods by which a primary amine may be
prepared include
NCH
2
CH
3
O
O
Gabriel synthesis
Reduction of
an azide
CH
3
CH
2
N
3
CH
3
CH
2
NH
2
Reductive
amination
CH
3
CH
O
Reduction of
an amide
CH
3
CNH
2
O
Methyl
iodide
CH
3
II
H11002
H11001
Physostigmine
CH
3
CH
3
N
N
CH
3
N
OCNHCH
3
O
b
a
c
“Physostigmine methiodide”
H11001
H
3
C
CH
3
N
OCNHCH
3
O
H11022H11022(CH
3
)
2
NH
Dimethylamine,
strongest base:
K
b
5.1 H11003 10
H110024
pK
b
3.3
C
6
H
5
NHCH
3
N-Methylaniline:
K
b
8 H11003 10
H1100210
pK
b
9.1
(C
6
H
5
)
2
NH
Diphenylamine,
weakest base:
K
b
6 H11003 10
H1100214
pK
b
13.2
620 AMINES
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Two of these methods, the Gabriel synthesis and the preparation and reduction of the
corresponding azide, begin with ethyl bromide.
To use reductive amination, we must begin with oxidation of ethanol to acetaldehyde.
Another possibility is reduction of acetamide. This requires an initial oxidation of ethanol
to acetic acid.
(b) Acylation of ethylamine with acetyl chloride, prepared in part (a), gives the desired amide.
Excess ethylamine can be allowed to react with the hydrogen chloride formed in the acylation
reaction. Alternatively, equimolar amounts of acyl chloride and amine can be used in the pres-
ence of aqueous hydroxide as the base.
(c) Reduction of the N-ethylacetamide prepared in part (b) yields diethylamine.
N-Ethylacetamide
CH
3
CNHCH
2
CH
3
O
Diethylamine
CH
3
CH
2
NHCH
2
CH
3
1. LiAlH
4
2. H
2
O
Ethylamine
2CH
3
CH
2
NH
2
H11001
Acetyl
chloride
CH
3
CCl
O
H11001
N-Ethylacetamide
CH
3
CNHCH
2
CH
3
O
Ethylammonium
chloride
CH
3
CH
2
NH
3
Cl
H11002
H11001
Acetic acid
CH
3
CO
2
H
Ethylamine
CH
3
CH
2
NH
2
K
2
Cr
2
O
7
, H
2
SO
4
H
2
O, heat
Ethanol
CH
3
CH
2
OH
Acetamide
CH
3
CNH
2
O
1. SOCl
2
2. NH
3
1. LiAlH
4
2. H
2
O
Ethylamine
CH
3
CH
2
NH
2
NH
3
, H
2
, Ni
Acetaldehyde
CH
3
CH
O
Ethanol
CH
3
CH
2
OH
PCC or PDC
CH
2
Cl
2
Acetaldehyde
CH
3
CH
O
Ethyl azide
CH
3
CH
2
N
3
Ethylamine
CH
3
CH
2
NH
2
Ethyl
bromide
CH
3
CH
2
Br
NaN
3
1. LiAlH
4
2. H
2
O
H11001 N
H11002
K
H11001
O
O
N-Potassiophthalimide
NCH
2
CH
3
O
O
N-Ethylphthalimide
CH
3
CH
2
NH
2
Ethylamine
CH
3
CH
2
Br
Ethyl
bromide
H
2
NNH
2
Ethanol
CH
3
CH
2
OH
Ethyl bromide
CH
3
CH
2
Br
PBr
3
or HBr
AMINES 621
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Diethylamine can also be prepared by reductive amination of acetaldehyde [from part (a)]
with ethylamine.
(d) The preparation of N,N-diethylacetamide is a standard acylation reaction. The reactants,
acetyl chloride and diethylamine, have been prepared in previous parts of this problem.
(e) Triethylamine arises by reduction of N,N-diethylacetamide or by reductive amination.
( f ) Quaternary ammonium halides are formed by reaction of alkyl halides and tertiary amines.
22.29 (a) In this problem a primary alkanamine must be prepared with a carbon chain extended by one
carbon. This can be accomplished by way of a nitrile.
The desired reaction sequence is therefore
1-Butanol
CH
3
CH
2
CH
2
CH
2
OH
Butyl bromide
CH
3
CH
2
CH
2
CH
2
Br
Pentanenitrile
CH
3
CH
2
CH
2
CH
2
CN
1-Pentanamine
CH
3
CH
2
CH
2
CH
2
CH
2
NH
2
1. LiAlH
4
2. H
2
O
PBr
3
or
HBr
NaCN
RCH
2
NH
2
RCN RBr ROH
(R )H11005 CH
3
CH
2
CH
2
CH
2
Triethylamine
(CH
3
CH
2
)
3
NH11001
Ethyl bromide
CH
3
CH
2
Br
Tetraethylammonium
bromide
(CH
3
CH
2
)
4
N Br
H11002
H11001
Diethylamine
(CH
3
CH
2
)
2
NHH11001
Acetaldehyde
CH
3
CH
O
Triethylamine
(CH
3
CH
2
)
3
N
H
2
, Ni
or
NaBH
3
CN
N,N-Diethylacetamide
O
CH
3
CN(CH
2
CH
3
)
2
Triethylamine
(CH
3
CH
2
)
3
N
1. LiAlH
4
2. H
2
O
Diethylamine
(CH
3
CH
2
)
2
NHH11001
Acetyl
chloride
CH
3
CCl
O
N,N-Diethylacetamide
O
CH
3
CN(CH
2
CH
3
)
2
HO
H11002
Ethylamine
CH
3
CH
2
NH
2
H11001
Acetaldehyde
CH
3
CH
O
Diethylamine
CH
3
CH
2
NHCH
2
CH
3
H
2
, Ni
or NaBH
3
CN
622 AMINES
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(b) The carbon chain of tert-butyl chloride cannot be extended by a nucleophilic substitution
reaction; the S
N
2 reaction that would be required on the tertiary halide would not work. The
sequence employed in part (a) is therefore not effective in this case. The best route is car-
boxylation of the Grignard reagent and subsequent conversion of the corresponding amide to
the desired primary amine product.
The reaction sequence to be used is
Once the carboxylic acid has been obtained, it is converted to the desired amine by reduction
of the corresponding amide.
(c) Oxidation of cyclohexanol to cyclohexanone gives a compound suitable for reductive
amination.
(d) The desired product is the reduction product of the cyanohydrin of acetone.
The cyanohydrin is made from acetone in the usual way. Acetone is available by oxidation of
isopropyl alcohol.
Acetone
CH
3
CCH
3
O
Acetone
cyanohydrin
CH
3
CCH
3
OH
CN
KCN
H
2
SO
4
Isopropyl
alcohol
CH
3
CHCH
3
OH
K
2
Cr
2
O
7
, H
2
SO
4
H
2
O
Acetone
cyanohydrin
CH
3
CCH
3
OH
CN
1-Amino-2-methyl-
2-propanol
CH
3
CCH
3
OH
CH
2
NH
2
1. LiAlH
4
2. H
2
O
Cyclohexanol
OH
N-Methylcyclohexylamine
NHCH
3
Cyclohexanone
O
K
2
Cr
2
O
7
H
2
SO
4
, H
2
O
CH
3
NH
2
,
H
2
, Ni
or
CH
3
NH
2
,
NaBH
3
CN
(CH
3
)
3
CCO
2
H
2,2-Dimethylpropanoic
acid
(CH
3
)
3
CCH
2
NH
2
2,2-Dimethyl-1-
propanamine
2,2-Dimethylpropanamide
(CH
3
)
3
CCNH
2
O
1. SOCl
2
2. NH
3
1. LiAlH
4
2. H
2
O
1. Mg, diethyl ether
2. CO
2
3. H
3
O
H11001
(CH
3
)
3
CCl
tert-Butyl
chloride
(CH
3
)
3
CCO
2
H
2,2-Dimethylpropanoic
acid
(CH
3
)
3
CCH
2
NH
2
(CH
3
)
3
CCO
2
H(CH
3
)
3
CCNH
2
O
(CH
3
)
3
CCl
AMINES 623
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(e) The target amino alcohol is the product of nucleophilic ring opening of 1,2-epoxypropane by
ammonia. Ammonia attacks the less hindered carbon of the epoxide function.
The necessary epoxide is formed by epoxidation of propene.
( f ) The reaction sequence is the same as in part (e) except that dimethylamine is used as the nucleo-
phile instead of ammonia.
(g) The key to performing this synthesis is recognition of the starting material as an acetal of
acetophenone. Acetals may be hydrolyzed to carbonyl compounds.
Once acetophenone has been obtained, it may be converted to the required product by reduc-
tive amination.
22.30 (a) The reaction of alkyl halides with N-potassiophthalimide (the first step in the Gabriel synthe-
sis of amines) is a nucleophilic substitution reaction. Alkyl bromides are more reactive than
alkyl fluorides; that is, bromide is a better leaving group than fluoride.
H11001 FCH
2
CH
2
Br
1-Bromo-2-
fluoroethane
N-Potassiophthalimide
NK
O
O
2-Phthalimidoethyl fluoride
NCH
2
CH
2
F
O
O
H11001
Acetophenone
C
6
H
5
CCH
3
O
Piperidine
N
H
N-(1-Phenylethyl)-
piperidine
N
C
6
H
5
CHCH
3
NaBH
3
CN
or H
2
, Ni
1,2-Ethanediol
HOCH
2
CH
2
OHH11001
Acetophenone
C
6
H
5
CCH
3
O
2-Methyl-2-phenyl-
1,3-dioxolane
CH
3
C
6
H
5
OO
H
3
O
H11001
H11001
Dimethylamine
(CH
3
)
2
NH
1-(N,N-Dimethylamino)-
2-propanol
CH
3
CHCH
2
N(CH
3
)
2
OH
1,2-Epoxypropane
[prepared as in part (e)]
CH
3
CH CH
2
O
Propene
CH
3
CH CH
2
Isopropyl
alcohol
CH
3
CHCH
3
OH
H
2
SO
4
heat
CH
3
COOH
1,2-Epoxypropane
CH
3
CH CH
2
O
O
1,2-Epoxypropane
CH
3
CH CH
2
1-Amino-2-propanol
CH
3
CHCH
2
NH
2
OH
NH
3
O
624 AMINES
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(b) In this example one bromine is attached to a primary and the other to a secondary carbon.
Phthalimide anion is a good nucleophile and reacts with alkyl halides by the S
N
2 mechanism.
It attacks the less hindered primary carbon.
(c) Both bromines are bonded to primary carbons, but branching at the adjacent carbon hinders
nucleophilic attack at one of them.
22.31 (a) Amines are basic and are protonated by hydrogen halides.
(b) Equimolar amounts of benzylamine and sulfuric acid yield benzylammonium hydrogen
sulfate as the product.
(c) Acetic acid transfers a proton to benzylamine.
Benzylamine Acetic acid Benzylammonium acetate
C
6
H
5
CH
2
NH
2
H11001 CH
3
COH
O
C
6
H
5
CH
2
NH
3
H11002
OCCH
3
H11001
O
Benzylamine Sulfuric acid
C
6
H
5
CH
2
NH
2
HOSO
2
OHH11001
Benzylammonium hydrogen
sulfate
C
6
H
5
CH
2
NH
3
H11002
OSO
2
OH
H11001
Benzylamine
C
6
H
5
CH
2
NH
2
HBrH11001
Benzylammonium
bromide
C
6
H
5
CH
2
NH
3
Br
H11002
H11001
1,4-Dibromo-2,2-dimethylbutane N-4-Bromo-3,3-dimethylphthalimide
(only product, 53% yield)
NCH
2
CH
2
CCH
2
Br
O
O
CH
3
CH
3
H11002
OO
N
CH
2
Br CH
2
C Br
More
crowded
Less
crowded
CH
2
H
3
C
CH
3
1,4-Dibromopentane
N-4-Bromopentylphthalimide
(only product, 67% yield)
NCH
2
CH
2
CH
2
CHCH
3
O
O
Br
H11002
OO
N
CH
2
Br
Br
CH
2
CH
CH
2
CH
3
More crowded
Less crowded
AMINES 625
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(d) Acetyl chloride reacts with benzylamine to form an amide.
(e) Acetic anhydride also gives an amide with benzylamine.
( f ) Primary amines react with ketones to give imines.
(g) These reaction conditions lead to reduction of the imine formed in part ( f ). The overall reac-
tion is reductive amination.
(h) Amines are nucleophilic and bring about the opening of epoxide rings.
(i) In these nucleophilic ring-opening reactions the amine attacks the less sterically hindered car-
bon of the ring.
( j) With excess methyl iodide, amines are converted to quaternary ammonium iodides.
C
6
H
5
CH
2
NH
2
H11001
Benzylamine
C
6
H
5
CH
2
N(CH
3
)
3
I
H11002
Benzyltrimethylammonium
iodide
Methyl
iodide
3CH
3
I
H11001
C
6
H
5
CH
2
NH
2
C
6
H
5
CH
2
NHCH
2
CHCH
3
H11001
Benzylamine 1-(N-Benzylamino)-2-propanol1,2-Epoxypropane
H
2
C CHCH
3
O
OH
C
6
H
5
CH
2
NH
2
C
6
H
5
CH
2
NHCH
2
CH
2
OHH11001
Benzylamine 2-(N-Benzylamino)ethanolEthylene oxide
H
2
C CH
2
O
C
6
H
5
CH
2
NH
2
(CH
3
)
2
CHNHCH
2
C
6
H
5
CH
3
CCH
3
O
H11001
Benzylamine Acetone N-Isopropylbenzylamine
H
2
, Ni
C
6
H
5
CH
2
NH
2
(CH
3
)
2
CCH
3
CCH
3
O
H11001
Benzylamine Acetone N-Isopropylidenebenzylamine
NCH
2
C
6
H
5
2C
6
H
5
CH
2
NH
2
CH
3
CNHCH
2
C
6
H
5
CH
3
COCCH
3
O O O
H11001H11001
Benzylamine Acetic anhydride N-Benzylacetamide
O
C
6
H
5
CH
2
NH
3
H11002
OCCH
3
Benzylammonium acetate
H11001
2C
6
H
5
CH
2
NH
2
CH
3
CNHCH
2
C
6
H
5
CH
3
CCl
O O
H11001H11001
Benzylamine Acetyl
chloride
N-Benzylacetamide
C
6
H
5
CH
2
NH
3
Cl
H11002
H11001
Benzylammonium
chloride
626 AMINES
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(k) Nitrous acid forms from sodium nitrite in dilute hydrochloric acid. Nitrosation of benzylamine
in water gives benzyl alcohol via a diazonium ion intermediate.
Benzyl chloride will also be formed by attack of chloride on the diazonium ion.
22.32 (a) Aniline is a weak base and yields a salt on reaction with hydrogen bromide.
(b) Aniline acts as a nucleophile toward methyl iodide. With excess methyl iodide, a quaternary
ammonium salt is formed.
(c) Aniline is a primary amine and undergoes nucleophilic addition to aldehydes and ketones to
form imines.
(d) When an imine is formed in the presence of hydrogen and a suitable catalyst, reductive ami-
nation occurs to give an amine.
(e) Aniline undergoes N-acylation on treatment with carboxylic acid anhydrides.
( f ) Acyl chlorides bring about N-acylation of arylamines.
(g) Nitrosation of primary arylamines yields aryl diazonium salts.
C
6
H
5
NH
2
C
6
H
5
N HSO
4
H11002
NaNO
2
, H
2
SO
4
H
2
O, 0–5H11034C
H11001
N
Aniline Benzenediazonium
hydrogen sulfate
2C
6
H
5
NH
2
H11001H11001C
6
H
5
CCl
O
Benzoyl
chloride
C
6
H
5
NHCC
6
H
5
O
Benzanilide Anilinium
chloride
C
6
H
5
NH
3
Cl
H11002
H11001
Aniline
2C
6
H
5
NH
2
C
6
H
5
NHCCH
3
CH
3
COCCH
3
O O O
H11001H11001
Acetic anhydride Acetanilide Anilinium acetate
C
6
H
5
NH
3
H11002
OCCH
3
O
H11001
Aniline
C
6
H
5
NH
2
CH
3
CH C
6
H
5
NHCH
2
CH
3
H11001
Aniline Acetaldehyde N-Ethylaniline
O
H
2
, Ni
C
6
H
5
NH
2
CH
3
CH C
6
H
5
NH11001H11001
Aniline Acetaldehyde N-Phenylacetaldimine Water
CHCH
3
H
2
O
O
C
6
H
5
NH
2
H11001
Aniline
3CH
3
I
Methyl
iodide
C
6
H
5
N(CH
3
)
3
I
H11002
H11001
N,N,N-Trimethylanilinium
iodide
C
6
H
5
NH
2
HBrH11001
Aniline Hydrogen
bromide
C
6
H
5
NH
3
Br
H11002
H11001
Anilinium
bromide
C
6
H
5
CH
2
NH
2
C
6
H
5
CH
2
OH
Benzylamine Benzyl alcohol
NaNO
2
, HCl
H
2
O
H11002N
2
H
2
O
Benzyldiazonium
ion
C
6
H
5
CH
2
N N
H11001
AMINES 627
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The replacement reactions that can be achieved by using diazonium salts are illustrated in
parts (h) through (n). In all cases molecular nitrogen is lost from the ring carbon to which it
was attached and is replaced by another substituent.
(o) The nitrogens of an aryl diazonium salt are retained on reaction with the electron-rich ring of
a phenol. Azo coupling occurs.
(p) Azo coupling occurs when aryl diazonium salts react with N,N-dialkylarylamines.
22.33 (a) Amides are reduced to amines by lithium aluminum hydride.
N-Ethylaniline
C
6
H
5
NHCH
2
CH
3
1. LiAlH
4
, diethyl ether
2. H
2
O
Acetanilide
C
6
H
5
NHCCH
3
O
Benzenediazonium
hydrogen sulfate
HSO
4
H11002
C
6
H
5
N
H11001
N
p-(Azophenyl)-N,N-dimethylanilineN,N-Dimethylaniline
C
6
H
5
N(CH
3
)
2
H11001 N(CH
3
)
2
C
6
H
5
NN
Benzenediazonium
hydrogen sulfate
HSO
4
H11002
C
6
H
5
N
H11001
N
p-(Azophenyl)phenolPhenol
C
6
H
5
OHH11001 OHC
6
H
5
NN
Benzenediazonium
hydrogen sulfate
C
6
H
5
N
H11001
N HSO
4
H11002
Phenol
C
6
H
5
OH(h)
H
H11001
, H
2
O
heat
Chlorobenzene
C
6
H
5
Cl(i)
CuCl
Bromobenzene
C
6
H
5
Br(j)
CuBr
Benzene
C
6
H
6
(l)
H
3
PO
2
Iodobenzene
C
6
H
5
I(m)
KI
Fluorobenzene
C
6
H
5
F(n)
1. HBF
4
2. heat
Benzonitrile
C
6
H
5
CN(k)
CuCN
628 AMINES
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(b) Acetanilide is a reactive substrate toward electrophilic aromatic substitution. An acetamido
group is ortho, para-directing.
(c) Sulfonation of the ring occurs.
(d) Bromination of the ring takes place.
(e) Acetanilide undergoes Friedel–Crafts alkylation readily.
( f ) Friedel–Crafts acylation also is easily carried out.
Acetanilide Acetyl chloride
H11001H11001 ortho isomerC
6
H
5
NHCCH
3
O
CH
3
CCl
O
AlCl
3
p-Acetamidoacetophenone
O
NHCCH
3
CCH
3
O
Acetanilide tert-Butyl
chloride
H11001H11001 ortho isomerC
6
H
5
NHCCH
3
O
(CH
3
)
3
CCl
AlCl
3
p-tert-Butyl-
acetanilide
O
NHCCH
3
C(CH
3
)
3
Acetanilide p-Bromoacetanilide
H11001C
6
H
5
NHCCH
3
ortho isomer
O
Br
2
acetic acid
O
NHCCH
3
Br
Acetanilide p-Acetamidobenzenesulfonic acid
H11001C
6
H
5
NHCCH
3
O
SO
3
H
2
SO
4
O
NHCCH
3
SO
3
H
ortho isomer
Acetanilide o-Nitroacetanilide p-Nitroacetanilide
H11001C
6
H
5
NHCCH
3
O
HNO
3
H
2
SO
4
O
NHCCH
3
NO
2
O
NHCCH
3
NO
2
AMINES 629
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(g) Acetanilide is an amide and can be hydrolyzed when heated with aqueous acid. Under acidic
conditions the aniline that is formed exists in its protonated form as the anilinium cation.
(h) Amides are hydrolyzed in base.
22.34 (a) The reaction illustrates the preparation of a secondary amine by reductive amination.
(b) Amides are reduced to amines by lithium aluminum hydride.
(c) Treatment of alcohols with p-toluenesulfonyl chloride converts them to p-toluenesulfonate
esters.
p-Toluenesulfonate is an excellent leaving group in nucleophilic substitution reactions.
Dimethylamine is the nucleophile.
(d) Amines are sufficiently nucleophilic to react with epoxides. Attack occurs at the less substi-
tuted carbon of the epoxide.
CHCH
2
NHCH(CH
3
)
2
OH
CH
3
O
OCH
3
1-(2,5-Dimethoxyphenyl)-2-
(isopropylamino)ethanol (67%)
H11001CH CH
2
O
CH
3
O
OCH
3
2-(2,5-Dimethoxyphenyl)oxirane
H
2
NCH(CH
3
)
2
Isopropylamine
N,N-Dimethyl-3-phenyl-
1-propanamine (86%)
3-Phenylpropyl p-toluenesulfonate Dimethyl-
amine
CH
3
(CH
3
)
2
NHC
6
H
5
CH
2
CH
2
CH
2
OSO
2
C
6
H
5
CH
2
CH
2
CH
2
N(CH
3
)
2
H11001
p-Toluenesulfonyl chloride 3-Phenylpropyl p-toluenesulfonate3-Phenyl-1-propanol
H11001
pyridine
C
6
H
5
CH
2
CH
2
CH
2
OH H
3
CSO
2
Cl CH
3
C
6
H
5
CH
2
CH
2
CH
2
OS
O
O
NCH
2
CH
3
O
NCH
2
CH
3
1. LiAlH
4
2. H
2
O, HO
H11002
6-Ethyl-6-
azabicyclo[3.2.1]octan-7-one
6-Ethyl-6-
azabicyclo[3.2.1]octane
Cyclohexylamine Dicyclohexylamine (70%)Cyclohexanone
H11001
H
2
, Ni
H
2
N NHO
Sodium
hydroxide
H11001H11001
Acetanilide
C
6
H
5
NHCCH
3
O
NaOH
Aniline
C
6
H
5
NH
2
Sodium acetate
O
Na
H11001 H11002
OCCH
3
H
2
O
Acetanilide Water
H11001H11001H11001C
6
H
5
NHCCH
3
O
H
2
O
Hydrogen
chloride
HCl
Anilinium
chloride
C
6
H
5
NH
3
Cl
H11002
H11001
Acetic acid
HOCCH
3
O
630 AMINES
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
(e) H9251-Halo ketones are reactive substrates in nucleophilic substitution reactions. Dibenzylamine
is the nucleophile.
Because the reaction liberates hydrogen chloride, it is carried out in the presence of added
base—in this case triethylamine—so as to avoid converting the dibenzylamine to its hydro-
chloride salt.
( f ) Quaternary ammonium hydroxides undergo Hofmann elimination when they are heated.
A point to be considered here concerns the regioselectivity of Hofmann eliminations: it is the
less hindered H9252 proton that is removed by the base giving the less substituted alkene.
.
(g) The combination of sodium nitrite and aqueous acid is a nitrosating agent. Secondary alkyl-
amines react with nitrosating agents to give N-nitroso amines as the isolated products.
22.35 (a) Catalytic hydrogenation reduces nitro groups to amino groups.
(b) Nitro groups are readily reduced by tin(II) chloride.
This reaction is the first step in a synthesis of the drug lidocaine.
1,3-Dimethyl-2-nitrobenzene
CH
3
NO
2
CH
3
2,6-Dimethylaniline
CH
3
NH
2
CH
3
1. SnCl
2
, HCl
2. HO
H11002
H
2
, Pt
ethanol
CH
3
CH
2
CH
3
CH
2
NO
2
1,2-Diethyl-4-nitrobenzene
CH
3
CH
2
CH
3
CH
2
NH
2
3,4-Diethylaniline (93–99%)
NaNO
2
HCl, H
2
O
(CH
3
)
2
CHNHCH(CH
3
)
2
Diisopropylamine
(CH
3
)
2
CHNCH(CH
3
)
2
N
O
N-Nitrosodiisopropylamine (91%)
Elimination to give does not occur
CH
3
CH
3
H
3
C
H11002H
2
O
H
3
C
CH
2
H
N(CH
3
)
3
H11001
OH
H11002
H11001
H
3
C
CH
2
CH
3
trans-1-Isopropenyl-4-
methylcyclohexane (98%)
(CH
3
)
3
N
Trimethylamine
H
3
C
1-(Dibenzylamino)-2-
propanone (87%)
CH
3
CCH
2
N(CH
2
C
6
H
5
)
2
O
H11001(C
6
H
5
CH
2
)
2
NH
Dibenzylamine 1-Chloro-2-propanone
CH
3
CCH
2
O
Cl
AMINES 631
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
(c) The amino group of arylamines is nucleophilic and undergoes acylation on reaction with
chloroacetyl chloride.
Chloroacetyl chloride is a difunctional compound—it is both an acyl chloride and an alkyl
chloride. Acyl chlorides react with nucleophiles faster than do alkyl chlorides, so that acyla-
tion of the amine nitrogen occurs rather than alkylation.
(d) The final step in the synthesis of lidocaine is displacement of the chloride by diethylamine
from the H9251-halo amide formed in part (c) in a nucleophilic substitution reaction.
The reaction is carried out with excess diethylamine, which acts as a base to neutralize the
hydrogen chloride formed.
(e) For use as an anesthetic, lidocaine is made available as its hydrochloride salt. Of the two
nitrogens in lidocaine, the amine nitrogen is more basic than the amide.
( f ) Lithium aluminum hydride reduction of amides is one of the best methods for the preparation
of amines, including arylamines.
(g) Arylamines react with aldehydes and ketones in the presence of hydrogen and nickel to give
the product of reductive amination.
Aniline
C
6
H
5
NH
2
N-Heptylaniline (65%)
C
6
H
5
NHCH
2
(CH
2
)
5
CH
3
H
2
, Ni
Heptanal
H11001 CH
3
(CH
2
)
5
CH
O
N-Phenylbutanamide
C
6
H
5
NHCCH
2
CH
2
CH
3
O
N-Butylaniline (92%)
C
6
H
5
NHCH
2
CH
2
CH
2
CH
3
1. LiAlH
4
2. H
2
O
Lidocaine
CH
3
NHCCH
2
N(CH
2
CH
3
)
2
CH
3
O
HCl
Lidocaine hydrochloride
CH
3
NHCCH
2
N(CH
2
CH
3
)
2
Cl
H11002
CH
3
OH
H11001
N-(Chloroacetyl)-
2,6-dimethylaniline
H11001
Diethylamine
(CH
3
CH
2
)
2
NH
CH
3
NHCCH
2
Cl
CH
3
O
Lidocaine
CH
3
NHCCH
2
N(CH
2
CH
3
)
2
CH
3
O
N-(Chloroacetyl)-
2,6-dimethylaniline
2,6-Dimethylaniline
CH
3
NH
2
CH
3
H11001
Chloroacetyl
chloride
ClCH
2
CCl
O
CH
3
NHCCH
2
Cl
CH
3
O
632 AMINES
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
(h) Acetanilide is a reactive substrate toward electrophilic aromatic substitution. On reaction with
chloroacetyl chloride, it undergoes Friedel–Crafts acylation, primarily at its para position.
Acylation, rather than alkylation, occurs. Acyl chlorides are more reactive than alkyl chlorides
toward electrophilic aromatic substitution reactions as a result of the more stable intermediate
(acylium ion) formed.
(i) Reduction with iron in hydrochloric acid is one of the most common methods for converting
nitroarenes to arylamines.
( j) Primary arylamines are converted to aryl diazonium salts on treatment with sodium nitrite in
aqueous acid. When the aqueous acidic solution containing the diazonium salt is heated, a
phenol is formed.
(k) This problem illustrates the conversion of an arylamine to an aryl chloride by the Sandmeyer
reaction.
(l) Diazotization of primary arylamines followed by treatment with copper(I) bromide converts
them to aryl bromides.
(m) Nitriles are formed when aryl diazonium salts react with copper(I) cyanide.
NaNO
2
, HCl
H
2
O
CuCN
(H11002N
2
)
NO
2
NH
2
o-Nitroaniline
NO
2
CN
o-Nitrobenzonitrile
(87%)
NO
2
NN
H11001
NaNO
2
, HBr
H
2
O
CuBr
(H11002N
2
)
NH
2
Br
m-Bromoaniline
Br
Br
m-Dibromobenzene
(80–87%)
NN
Br
H11001
NaNO
2
, H
2
SO
4
H
2
O
2,6-Dinitroaniline 2-Chloro-1,3-
dinitrobenzene (71–74%)
CuCl
(H11002N
2
)
NH
2
O
2
NNO
2
N N
H11001
O
2
NNO
2
Cl
O
2
NNO
2
4-Bromo-4H11032-hydroxybiphenyl (85%)
Br OH
H
2
O, heat
(H11002N
2
)
Br N N
H11001NaNO
2
, H
2
SO
4
H
2
O
4-Amino-4H11032-bromobiphenyl
Br NH
2
4-Bromo-4H11032-nitrobiphenyl
1. Fe, HCl
2. HO
H11002
Br NO
2
4-Amino-4H11032-bromobiphenyl (94%)
Br NH
2
Chloroacetyl
chloride
Acetanilide p-Acetamidophenacyl chloride
(79–83%)
H11001
AlCl
3
CH
3
CNH
O O
CCH
2
ClClCH
2
CCl
O
CH
3
CNH
O
AMINES 633
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
(n) An aryl diazonium salt is converted to an aryl iodide on reaction with potassium iodide.
(o) Aryl diazonium fluoroborates are converted to aryl fluorides when heated. Both diazonium
salt functions in the starting material undergo this reaction.
( p) Hypophosphorous acid (H
3
PO
2
) reduces aryl diazonium salts to arenes.
(q) Ethanol, like hypophosphorous acid, is an effective reagent for the reduction of aryl diazo-
nium salts.
(r) Diazotization of aniline followed by addition of a phenol yields a bright-red diazo-substituted
phenol. The diazonium ion acts as an electrophile toward the activated aromatic ring of the
phenol.
NaNO
2
, H
2
SO
4
H
2
O
CH
3
H
3
C
H
3
C
OH
CH
3
H
3
C
H
3
C
OH
NNC
6
H
5
2,3,6-Trimethyl-4-
(phenylazo)phenol (98%)
C
6
H
5
NH
2
Aniline Benzenediazonium
hydrogen sulfate
C
6
H
5
N HSO
4
H11002
N
H11001
NaNO
2
, HCl
H
2
O
CH
3
CH
2
OH
NH
2
I
CO
2
H
2-Amino-5-
iodobenzoic acid
I
CO
2
H
m-Iodobenzoic acid
(86–93%)
N
H11001
N
I
CO
2
H
NaNO
2
, H
2
SO
4
H
2
O, H
3
PO
2
NO
2
O
2
N
NO
2
NH
2
2,4,6-Trinitroaniline
NO
2
O
2
N
NO
2
1,3,5-Trinitrobenzene
(60–65%)
2BF
4
NN
H11001
NN
H11001
H11002
4,4H11032-Bis(diazonio)biphenyl fluoroborate
heat
F F
4,4H11032-Difluorobiphenyl (82%)
NaNO
2
, H
2
SO
4
H
2
O
KI
NO
2
II
NH
2
2,6-Diiodo-
4-nitroaniline
NO
2
II
I
1,2,3-Triiodo-
5-nitrobenzene
(94–95%)
NO
2
II
NN
H11001
634 AMINES
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(s) Nitrosation of N,N-dialkylarylamines takes place on the ring at the position para to the
dialkylamino group.
22.36 (a) 4-Methylpiperidine can participate in intermolecular hydrogen bonding in the liquid phase.
These hydrogen bonds must be broken in order for individual 4-methylpiperidine molecules
to escape into the gas phase. N-Methylpiperidine lacks a proton bonded to nitrogen and so
cannot engage in intermolecular hydrogen bonding. Less energy is required to transfer a mol-
ecule of N-methylpiperidine to the gaseous state, and therefore it has a lower boiling point
than 4-methylpiperidine.
(b) The two products are diastereomeric quaternary ammonium chlorides that differ in the con-
figuration at the nitrogen atom.
(c) Tetramethylammonium hydroxide cannot undergo Hofmann elimination. The only reaction
that can take place is nucleophilic substitution.
H11001
H
3
C
H
3
C
CH
3
N
Trimethylamine
CH
3
OH
MethanolTetramethylammonium
hydroxide
H
3
C
H
3
C
CH
3
NCH
3
H11002
OH
H11001
N
CH
3
C(CH
3
)
3
4-tert-Butyl-N-
methylpiperidine
H11001
H
3
C
N
H11001
C
6
H
5
CH
2
C(CH
3
)
3
H
Cl
H11002
C
6
H
5
CH
2
N
H11001
CH
3
C(CH
3
)
3
H
Cl
H11002
C
6
H
5
CH
2
Cl
N
CH
3
N-Methylpiperidine;
no hydrogen bonding possible
to other N-methylpiperidine molecules
CH
3
H
3
C
N
N
H
H
CH
3
N
H
CH
3
(CH
3
)
2
N
N,N-Dimethyl-m-toluidine
CH
3
(CH
3
)
2
NN
O
3-Methyl-4-nitroso-N,N-
dimethylaniline (83%)
NaNO
2
, HCl, H
2
O
AMINES 635
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(d) The key intermediate in the reaction of an amine with nitrous acid is the corresponding dia-
zonium ion.
Loss of nitrogen from this diazonium ion is accompanied by a hydride shift to form a sec-
ondary carbocation.
Capture of isopropyl cation by water yields the major product of the reaction, 2-propanol.
22.37 Alcohols are converted to p-toluenesulfonate esters by reaction with p-toluenesulfonyl chloride.
None of the bonds to the stereogenic center is affected in this reaction.
Displacement of the p-toluenesulfonate leaving group by sodium azide in an S
N
2 process and pro-
ceeds with inversion of configuration.
Reduction of the azide yields a primary amine. A nitrogen–nitrogen bond is cleaved; all the bonds to
the stereogenic center remain intact.
22.38 (a) The overall transformation can be expressed as RBr → RCH
2
NH
2
. In many cases this can be
carried out via a nitrile, as RBr → RCN → RCH
2
NH
2
. In this case, however, the substrate is
1-bromo-2,2-dimethylpropane, an alkyl halide that reacts very slowly in nucleophilic substi-
(R)-2-Octanamine
(compound C)
CH
3
(CH
2
)
5
CH
3
H
CH
2
N
(R)-1-Methylheptyl azide
(compound B)
CH
3
(CH
2
)
5
CH
3
H
CNN N
H11002 H11001 1. LiAlH
4
2. H
2
O, HO
H11002
(S)-1-Methylheptyl p-toluenesulfonate
(compound A)
C OSO
2
H
3
C
CH
3
(CH
2
)
5
H
CH
3
H11001 OSO
2
CH
3
NN N
H11002H11002H11001
(R)-1-Methylheptyl azide
(compound B)
CH
3
(CH
2
)
5
CH
3
H
CNN N
H11002 H11001 H11002
pyridine
H11001COH
H
3
C
CH
3
(CH
2
)
5
H
(S)-2-Octanol
CH
3
C OSO
2
H
3
C
CH
3
(CH
2
)
5
H
(S)-1-Methylheptyl p-toluenesulfonate
(compound A)
H
3
CSO
2
Cl
p-Toluenesulfonyl chloride
H11001 H
H11001
H11001CH
3
CHCH
3
O
H11001
HH
CH
3
CHCH
3
H11001
Isopropyl
cation
H
2
O
Water
CH
3
CHCH
3
OH
2-Propanol
H11001CH
3
CHCH
3
H11001
Isopropyl
cation
NN
NitrogenPropyldiazonium ion
CH
3
CHCH
2
H
N
H11001
N
NaNO
2
, HCl
H
2
O
CH
3
CH
2
CH
2
NH
2
1-Propanamine
CH
3
CH
2
CH
2
N
H11001
N
Propyldiazonium ion
636 AMINES
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tution processes. Carbon–carbon bond formation with 1-bromo-2,2-dimethylpropane can be
achieved more effectively by carboxylation of the corresponding Grignard reagent.
The carboxylic acid can then be converted to the desired amine by reduction of the derived
amide.
The yields listed in parentheses are those reported in the chemical literature for this synthesis.
(b) Consider the starting materials in relation to the desired product.
The synthetic tasks are to form the necessary carbon–nitrogen bond and to reduce the carbonyl
group to a methylene group. This has been accomplished by way of the amide as a key inter-
mediate.
A second approach utilizes reductive amination following conversion of the starting car-
boxylic acid to an aldehyde.
The reducing agent in the reductive amination process cannot be hydrogen, because that
would result in hydrogenation of the double bond. Sodium cyanoborohydride is required.
(c) It is stereochemistry that determines the choice of which synthetic method to employ in in-
troducing the amine group. The carbon–nitrogen bond must be formed with inversion of
10-Undecenal
H
2
C CH(CH
2
)
8
CH
O
NaBH
3
CN
H11001
Pyrrolidine N-(10-Undecenyl)pyrrolidine
H
2
C CH(CH
2
)
8
CH
2
N
N
H
10-Undecenoic acid
H
2
C CH(CH
2
)
8
COH
O
10-Undecenal
H
2
C CH(CH
2
)
8
CH
O
10-Undecen-1-ol
H
2
C CH(CH
2
)
8
CH
2
OH
PCC or PDC
CH
2
Cl
2
1. LiAlH
4
2. H
2
O
10-Undecenoic acid
H
2
C CH(CH
2
)
8
COH
O
N-(10-Undecenoyl)pyrrolidine (75%)
O
H
2
C CH(CH
2
)
8
C N
N-(10-Undecenyl)pyrrolidine (66%)
H
2
C CH(CH
2
)
8
CH
2
N
1. SOCl
2
2. pyrrolidine
1. LiAlH
4
2. H
2
O
H11001
Pyrrolidine
N
H
10-Undecenoic acid
H
2
C CH(CH
2
)
8
COH
O
N-(10-Undecenyl)pyrrolidine
H
2
C CH(CH
2
)
8
CH
2
N
3,3-Dimethylbutanoic
acid
(CH
3
)
3
CCH
2
CO
2
H
3,3-Dimethyl-1-butanamine
(57%)
(CH
3
)
3
CCH
2
CH
2
NH
2
1. SOCl
2
2. NH
3
1. LiAlH
4
2. H
2
O
3,3-Dimethylbutanamide
(51%)
(CH
3
)
3
CCH
2
CNH
2
O
1-Bromo-2,2-
dimethylpropane
(CH
3
)
3
CCH
2
Br
3,3-Dimethylbutanoic
acid (63%)
(CH
3
)
3
CCH
2
CO
2
H
1. Mg
2. CO
2
3. H
3
O
H11001
AMINES 637
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configuration at the alcohol carbon. Conversion of the alcohol to its p-toluenesulfonate ester en-
sures that the leaving group is introduced with exactly the same stereochemistry as the alcohol.
Once the leaving group has been introduced with the proper stereochemistry, it can be dis-
placed by a nitrogen nucleophile suitable for subsequent conversion to an amine.
(As actually reported, the azide was reduced by hydrogenation over a palladium catalyst, and
the amine was isolated as its hydrochloride salt in 66% yield.)
(d) Recognition that the primary amine is derivable from the corresponding nitrile by reduction,
and that the necessary tertiary amine function can be introduced by a nucleophilic substitution
reaction between the two given starting materials suggests the following synthesis.
Alkylation of N-methylbenzylamine with 4-bromobutanenitrile has been achieved in 92%
yield in the presence of potassium carbonate as a weak base to neutralize the hydrogen bro-
mide produced. The nitrile may be reduced with lithium aluminum hydride, as shown in the
equation, or by catalytic hydrogenation. Catalytic hydrogenation over platinum gave the de-
sired diamine, isolated as its hydrochloride salt, in 90% yield.
(e) The overall transformation may be viewed retrosynthetically as follows:
The sequence that presents itself begins with benzylic bromination with N-bromosuccinimide.
N-bromosuccinimide
benzoyl peroxide, CCl
4
heat
CH
3
NC
p-Cyanotoluene
CH
2
BrNC
p-Cyanobenzyl bromide
ArCH
2
N(CH
3
)
2
ArCH
2
Br ArCH
3
Ar H11005 NC
BrCH
2
CH
2
CH
2
CNC
6
H
5
CH
2
NH
CH
3
C
6
H
5
CH
2
NCH
2
CH
2
CH
2
CN
CH
3
C
6
H
5
CH
2
NCH
2
CH
2
CH
2
CH
2
NH
2
CH
3
H11001
N-Methylbenzylamine 4-Bromobutanenitrile N-Benzyl-N-methyl-1,4-butanediamine
2. H
2
O
1. LiAlH
4
C
6
H
5
CH
2
NCH
2
CH
2
CH
2
CH
2
NH
2
CH
3
C
6
H
5
CH
2
NCH
2
CH
2
CH
2
C
CH
3
N
NaN
3
1. LiAlH
4
2. H
2
O
CH
3
OSO
2
C
6
H
5
O
cis-2-Phenoxycyclopentyl
p-toluenesulfonate
trans-2-Phenoxycyclo-
pentyl azide (90%)
N
3
C
6
H
5
O
trans-2-Phenoxycyclo-
pentylamine
NH
2
C
6
H
5
O
pyridine
H11001
OHC
6
H
5
O
cis-2-Phenoxycyclo-
pentanol
CH
3
OSO
2
C
6
H
5
O
cis-2-Phenoxycyclopentyl
p-toluenesulfonate
H
3
CSO
2
Cl
p-Toluenesulfonyl
chloride
638 AMINES
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The reaction shown in the equation has been reported in the chemical literature and gave the
benzylic bromide in 60% yield.
Treatment of this bromide with dimethylamine gives the desired product. (The isolated
yield was 83% by this method.)
22.39 (a) This problem illustrates the application of the Sandmeyer reaction to the preparation of aryl
cyanides. Diazotization of p-nitroaniline followed by treatment with copper(I) cyanide con-
verts it to p-nitrobenzonitrile.
(b) An acceptable pathway becomes apparent when it is realized that the amino group in the prod-
uct is derived from the nitro group of the starting material. Two chlorines are introduced by
electrophilic aromatic substitution, the third by a Sandmeyer reaction.
Two of the required chlorine atoms can be introduced by chlorination of the starting material,
p-nitroaniline.
The third chlorine can be introduced via the Sandmeyer reaction. Reduction of the nitro group
completes the synthesis of 3,4,5-trichloroaniline.
The reduction step has been carried out by hydrogenation with a nickel catalyst in 70% yield.
reduce
NH
2
ClCl
NO
2
2,6-Dichloro-4-
nitroaniline
Cl
ClCl
NO
2
1,2,3-Trichloro-5-
nitrobenzene
Cl
ClCl
NH
2
3,4,5-Trichloroaniline
1. NaNO
2
, HCl, H
2
O
2. CuCl
Cl
2
NH
2
NO
2
p-Nitroaniline
NH
2
ClCl
NO
2
2,6-Dichloro-4-
nitroaniline
Cl
ClCl
NH
2
Cl
ClCl
NO
2
NH
2
ClCl
NO
2
NH
2
NO
2
NH
2
NO
2
p-Nitroaniline
CN
NO
2
p-Nitrobenzonitrile
1. NaNO
2
, HCl, H
2
O
2. CuCN
H11001CH
2
BrNC
p-Cyanobenzyl bromide
(CH
3
)
2
NH
Dimethylamine
CH
2
N(CH
3
)
2
NC
p-Cyano-N,N-dimethylbenzylamine
AMINES 639
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(c) The amino group that is present in the starting material facilitates the introduction of the
bromine substituents, and is then removed by reductive deamination.
Hypophosphorous acid has also been used successfully in the reductive deamination step.
(d) Reduction of the nitro group of the 1,3-dibromo-5-nitrobenzene prepared in the preceding part
of this problem gives the desired product. The customary reducing agents used for the reduc-
tion of nitroarenes would all be suitable.
(e) The synthetic objective is
This compound, known as acetaminophen and used as an analgesic to reduce fever and
relieve minor pain, may be prepared from p-nitroaniline by way of p-nitrophenol.
Any of the customary reducing agents suitable for converting aryl nitro groups to arylamines
(Fe, HCl; Sn, HCl; H
2
, Ni) may be used. Acetylation of p-aminophenol may be carried out
with acetyl chloride or acetic anhydride. The amino group of p-aminophenol is more nucle-
ophilic than the hydroxyl group and is acetylated preferentially.
22.40 (a) Replacement of an amino substituent by a bromine is readily achieved by the Sandmeyer
reaction.
(b) This conversion demonstrates the replacement of an amino substituent by fluorine via the
Schiemann reaction.
2. CuBr
1. NaNO
2
, HBr, H
2
O
o-Anisidine
OCH
3
NH
2
o-Bromoanisole (88–93%)
OCH
3
Br
2. heat
1. NaNO
2
, H
2
O, H
2
SO
4
p-Nitroaniline p-Nitrophenol
OH
NO
2
NH
2
NO
2
1. reduce
2. acetylate
p-Acetamidophenol
OH
HNCCH
3
O
p-Acetamidophenol
NHCCH
3
HO
O
BrBr
NO
2
1,3-Dibromo-5-nitrobenzene
[prepared from p-nitroaniline as in part (c)]
BrBr
NH
2
3,5-Dibromoaniline
(80%)
H
2
, Ni
NH
2
NO
2
p-Nitroaniline
NH
2
BrBr
NO
2
2,6-Dibromo-4-
nitroaniline
(95%)
BrBr
NO
2
1,3-Dibromo-5-
nitrobenzene
(70%)
1. NaNO
2
, H
2
O, H
H11001
2. ethanol
Br
2
acetic acid
640 AMINES
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(c) We can use the o-fluoroanisole prepared in part (b) to prepare 3-fluoro-4-methoxyaceto-
phenone by Friedel–Crafts acylation.
Remember from Section 12.16 that it is the more activating substituent that determines the
regioselectivity of electrophilic aromatic substitution when an arene bears two different sub-
stituents. Methoxy is a strongly activating substituent; fluorine is slightly deactivating.
Friedel–Crafts acylation takes place at the position para to the methoxy group.
(d) The o-fluoroanisole prepared in part (b) serves nicely as a precursor to 3-fluoro-4-methoxy-
benzonitrile via diazonium salt chemistry.
The desired sequence of reactions to carry out the synthesis is
OCH
3
F
o-Fluoroanisole
OCH
3
NH
2
o-Anisidine 2-Fluoro-4-
nitroanisole
(53%)
OCH
3
F
NO
2
4-Amino-2-
fluoroanisole
(85%)
OCH
3
F
NH
2
3-Fluoro-4-
methoxybenzonitrile (46%)
OCH
3
F
CN
as in part (b) HNO
3
H
2
, Pt
1. NaNO
2
, HCl, H
2
O
2. CuCN
OCH
3
F
CN
OCH
3
F
NH
2
OCH
3
F
NO
2
OCH
3
F
[from part (b)]
o-Anisidine
OCH
3
NH
2
o-Fluoroanisole
OCH
3
F
as in part (b)
3-Fluoro-4-methoxyacetophenone
(70–80%)
F
OCH
3
AlCl
3
CH
3
COCCH
3
O O
CCH
3
O
2. HBF
4
1. NaNO
2
, HCl, H
2
O
heat
o-Anisidine
OCH
3
NH
2
o-Fluoroanisole
(53%)
OCH
3
F
o-Methoxybenzenediazonium
fluoroborate (57%)
OCH
3
N N
H11001
BF
4
H11002
AMINES 641
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Conversion of o-fluoroanisole to 4-amino-2-fluoroanisole proceeds in the conventional way
by preparation and reduction of a nitro derivative. Once the necessary arylamine is at hand, it
is converted to the nitrile by a Sandmeyer reaction.
(e) Diazotization followed by hydrolysis of the 4-amino-2-fluoroanisole prepared as an interme-
diate in part (d ) yields the desired phenol.
22.41 (a) The carboxyl group of p-aminobenzoic acid can be derived from the methyl group of
p-methylaniline by oxidation. First, however, the nitrogen must be acylated so as to protect the
ring from oxidation.
The sequence of reactions to be used is
(b) Attachment of fluoro and propanoyl groups to a benzene ring is required. The fluorine sub-
stituent can be introduced by way of the diazonium tetrafluoroborate, the propanoyl group by
way of a Friedel–Crafts acylation. Because the fluorine substituent is ortho, para-directing,
introducing it first gives the proper orientation of substituents.
F
CCH
2
CH
3
O
Ethyl p-fluorophenyl
ketone
F
Fluorobenzene
NH
2
Aniline
K
2
Cr
2
O
7
, H
2
SO
4
H
2
O, heat
1. HCl, H
2
O
2. neutralize
CH
3
COCCH
3
OO
CH
3
NH
2
p-Methylaniline
CH
3
HNCCH
3
O
p-Methylacetanilide
CO
2
H
HNCCH
3
O
p-Acetamido-
benzoic acid
CO
2
H
NH
2
p-Amino-
benzoic acid
O
CO
2
H
NHCCH
3
CH
3
NH
2
p-Methylaniline
CO
2
H
NH
2
p-Aminobenzoic
acid
OCH
3
NH
2
o-Anisidine
as in part (d)
3-Fluoro-4-
methoxyphenol
(70%)
OCH
3
F
OH
4-Amino-2-
fluoroanisole
OCH
3
F
NH
2
1. NaNO
2
, H
2
SO
4
, H
2
O
2. heat
642 AMINES
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Fluorobenzene is prepared from aniline by the Schiemann reaction, shown in Section 22.18.
Aniline is, of course, prepared from benzene via nitrobenzene. Friedel–Crafts acylation of
fluorobenzene has been carried out with the results shown and gives the required ethyl
p-fluorophenyl ketone as the major product.
(c) Our synthetic plan is based on the essential step of forming the fluorine derivative from an
amine by way of a diazonium salt.
The required substituted aniline is derived from m-xylene by a standard synthetic sequence.
(d) In this problem two nitrogen-containing groups of the starting material are each to be replaced
by a halogen substituent. The task is sufficiently straightforward that it may be confronted
directly.
Replace amino group by bromine:
NH
2
NO
2
CH
3
2-Methyl-4-nitro-1-
naphthylamine
Br
NO
2
CH
3
1-Bromo-2-methyl-4-
nitronaphthalene (82%)
1. NaNO
2
, HBr, H
2
O
2. CuBr
Br
2
H
3
C CH
3
m-Xylene
NO
2
H
3
C CH
3
1,3-Dimethyl-4-
nitrobenzene (98%)
NH
2
H
3
C CH
3
2,4-Dimethylaniline
HNO
3
H
2
SO
4
1. Fe, HCl
2. HO
H11002
Br
F
H
3
C CH
3
1-Bromo-2-fluoro-
3,5-dimethylbenzene (60%)
2-Bromo-4,6-dimethylaniline
Br
NH
2
H
3
C CH
3
1. NaNO
2
, HCl,
H
2
O, 0H11034C
2. HBF
4
3. heat
Br
F
H
3
C CH
3
1-Bromo-2-fluoro-
3,5-dimethylbenzene
Br
NH
2
H
3
C CH
3
NH
2
H
3
C CH
3
2,4-Dimethylaniline
H11001
F
CCH
2
CH
3
O
Ethyl p-fluorophenyl
ketone (86%)
F
Fluorobenzene Propanoyl
chloride
CH
3
CH
2
CCl
O
AlCl
3
AMINES 643
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Reduce nitro group to amine:
Replace amino group by fluorine:
(e) Bromination of the starting material will introduce the bromine substituent at the correct
position, that is, ortho to the tert-butyl group.
The desired product will be obtained if the nitro group can be removed. This is achieved by its
conversion to the corresponding amine, followed by reductive deamination.
( f ) The proper orientation of the chlorine substituent can be achieved only if it is introduced after
the nitro group is reduced.
The correct sequence of reactions to carry out this synthesis is shown.
C(CH
3
)
3
Cl
NO
2
C(CH
3
)
3
NH
2
Cl
C(CH
3
)
3
NH
2
C(CH
3
)
3
C(CH
3
)
3
Br
NO
2
2-Bromo-1-tert-
butyl-4-nitrobenzene
C(CH
3
)
3
Br
NH
2
3-Bromo-4-tert-
butylaniline
C(CH
3
)
3
Br
o-Bromo-tert-
butylbenzene
H
2
, Ni
(or other appropriate
reducing agent)
1. NaNO
2
, H
H11001
2. H
3
PO
2
C(CH
3
)
3
NO
2
p-tert-Butyl-
nitrobenzene
C(CH
3
)
3
Br
NO
2
2-Bromo-1-tert-
butyl-4-nitrobenzene
Br
2
, Fe
Br
NH
2
CH
3
4-Bromo-3-methyl-
1-naphthylamine
Br
F
CH
3
1-Bromo-4-fluoro-2-
methylnaphthalene
(64%)
1. NaNO
2
, HCl,
H
2
O, 0–5H11034C
2. HBF
4
3. heat
Br
NO
2
CH
3
1-Bromo-2-methyl-
4-nitronaphthalene
Br
NH
2
CH
3
4-Bromo-3-methyl-
1-naphthylamine
1. Fe, HCl
2. HO
H11002
644 AMINES
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(g) The orientation of substituents in the target molecule can be achieved by using an amino group
to control the regiochemistry of bromination, then removing it by reductive deamination.
The amino group is introduced in the standard fashion by nitration of an arene followed by
reduction.
This analysis leads to the synthesis shown.
HNO
3
H
2
, Ni
ethanol
1. NaNO
2
, H
H11001
, H
2
O
2. H
3
PO
2
CH
3
CH
2
CH
2
CH
3
m-Diethylbenzene
CH
3
CH
2
CH
2
CH
3
NO
2
2,4-Diethyl-1-nitrobenzene
(75–80%)
1-Bromo-3,5-
diethylbenzene (70%)
CH
3
CH
2
CH
2
CH
3
Br
Br
2
CH
2
CH
3
CH
3
CH
2
NH
2
2,4-Diethylaniline
(80–90%)
2-Bromo-4,6-
diethylaniline (40%)
CH
3
CH
2
CH
2
CH
3
NH
2
Br
CH
2
CH
3
CH
3
CH
2
NH
2
CH
3
CH
2
CH
2
CH
3
Br CH
3
CH
2
CH
2
CH
3
NH
2
Br
hydrolysis to
remove acetyl
group
Cl
2
C(CH
3
)
3
NHCCH
3
O
4-tert-Butyl-2-
chloroacetanilide
Cl
1. NaNO
2
, H
H11001
2. H
3
PO
2
NH
2
Cl
C(CH
3
)
3
4-tert-Butyl-2-
chloroaniline
C(CH
3
)
3
Cl
m-tert-Butyl-
chlorobenzene
C(CH
3
)
3
NO
2
p-tert-Butyl-
nitrobenzene
C(CH
3
)
3
NH
2
p-tert-Butyl-
aniline
C(CH
3
)
3
NHCCH
3
O
p-tert-Butyl-
acetanilide
H
2
, Ni acetic anhydride
AMINES 645
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(h) In this exercise the two nitrogen substituents are differentiated; one is an amino nitrogen, the
other an amide nitrogen. By keeping them differentiated they can be manipulated indepen-
dently. Remove one amino group completely before deprotecting the other.
Once the acetyl group has been removed by hydrolysis, the molecule is ready for introduction
of the iodo substituent by way of a diazonium salt.
(i) To convert the designated starting material to the indicated product, both the nitro group and the
ester function must be reduced and a carbon–nitrogen bond must be formed. Converting the
starting material to an amide gives the necessary carbon–nitrogen bond and has the advantage
that amides can be reduced to amines by lithium aluminum hydride. The amide can be formed
intramolecularly by reducing the nitro group to an amine, then heating to cause cyclization.
This synthesis is the one described in the chemical literature. Other routes are also possible,
but the one shown is short and efficient.
22.42 Weakly basic nucleophiles react with H9251,H9252-unsaturated carbonyl compounds by conjugate addition.
R
2
CCH
2
CRH11032
O
Y
CHCRH11032R
2
CH11001
O
HY
H
2
Ni
heat
CH
3
O
CH
3
O
CH
2
COCH
3
O
2
N
O
CH
3
O
CH
3
O
CH
2
COCH
3
H
2
N
O
CH
3
O
CH
3
O
O
NH
CH
3
O
CH
3
O
NH
1. LiAlH
4
2. H
2
O
1. HCl, H
2
O
heat
2. HO
H11002
1. NaNO
2
,
HCl, H
2
O
2. KI
Br
CF
3
NHCCH
3
O
2-Bromo-6-(trifluoromethyl)-
acetanilide
Br
CF
3
NH
2
2-Bromo-6-(trifluoromethyl)-
aniline (69%)
Br
CF
3
I
1-Bromo-2-iodo-3-
(trifluoromethyl)benzene
(87%)
1. NaNO
2
, H
H11001
, H
2
O
2. H
3
PO
2
4-Amino-2-bromo-6-
(trifluoromethyl)acetanilide
H
2
N Br
CF
3
NHCCH
3
O
2-Bromo-6-(trifluoromethyl)-
acetanilide (92%)
Br
CF
3
NHCCH
3
O
646 AMINES
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Ammonia and its derivatives are very prone to react in this way; thus conjugate addition provides a
method for the preparation of H9252-amino carbonyl compounds.
(a)
(b)
(c)
(d) The conjugate addition reaction that takes place in this case is an intramolecular one and
occurs in virtually 100% yield.
22.43 The first step in the synthesis is the conjugate addition of methylamine to ethyl acrylate. Two
sequential Michael addition reactions take place.
CHCOCH
2
CH
3
CH
3
NHCH
2
CH
2
COCH
2
CH
3
CH
3
N(CH
2
CH
2
CO
2
CH
2
CH
3
)
2
H
2
C CHCO
2
CH
2
CH
3
CH
3
NH
2
H
2
CH11001
O O
Methylamine Ethyl acrylate
(CH
2
)
4
CH
3
O
N
H
CH
2
CH
2NH
2
CH
2
CH
(CH
2
)
4
CH
3
O
Morpholine 3-Morpholino-1,3-diphenyl-
1-propanone (91%)
1,3-Diphenyl-2-
propen-1-one
HN OC
6
H
5
CCH C
6
H
5
CCH
2
CHC
6
H
5
O O
CHC
6
H
5
H11001
N
O
HN
N
OO H11001
2-Cyclohexenone Piperidine 3-Piperidinocyclo-
hexanone (45%)
(CH
3
)
2
C (CH
3
)
2
CCH
2
CCH
3
CHCCH
3
NH
3
H11001
O O
NH
2
4-Methyl-3-penten-2-one 4-Amino-4-methyl-2-
pentanone (63–70%)
Ammonia
AMINES 647
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Conversion of this intermediate to the desired N-methyl-4-piperidone requires a Dieckmann
cyclization followed by decarboxylation of the resulting H9252-keto ester.
Treatment of N-methyl-4-piperidone with the Grignard reagent derived from bromobenzene gives
a tertiary alcohol that can be dehydrated to an alkene. Hydrogenation of the alkene completes the
synthesis.
22.44 Sodium cyanide reacts with alkyl bromides by the S
N
2 mechanism. Reduction of the cyano group
with lithium aluminum hydride yields a primary amine. This reveals the structure of mescaline to be
2-(3,4,5-trimethoxyphenyl)ethylamine.
22.45 Reductive amination of a ketone with methylamine yields a secondary amine. Methamphetamine is
N-methyl-1-phenyl-2-propanamine.
22.46 There is no obvious reason why the dimethylamino group in 4-(N,N-dimethylamino)pyridine
should be appreciably more basic than it is in N,N-dimethylaniline; it is the ring nitrogen of
Benzyl methyl
ketone
Methylamine
H
2
, Ni
CH
3
NH
2
H11001
N-Methyl-1-phenyl-
2-propanamine
(methamphetamine)
CH
2
CHCH
3
NHCH
3
CH
2
CCH
3
O
CH
3
O
CH
3
O
CH
3
OCH
2
Br
3,4,5-Trimethoxybenzyl
bromide
2-(3,4,5-Trimethoxyphenyl)-
ethanenitrile
CH
3
O
CH
3
OCH
2
CN
CH
3
O
2-(3,4,5-Trimethoxyphenyl)ethylamine
(mescaline)
CH
3
O
CH
3
OCH
2
CH
2
NH
2
CH
3
O
NaCN
1. LiAlH
4
2. H
2
O
Phenylmagnesium
bromide
C
6
H
5
MgBrH11001
1. diethyl ether
2. H
3
O
H11001
C
6
H
5
CH
3
OH
heat
H
H11001
H
2
, Pt
N-Methyl-4-
piperidone
CH
3
O
N N
C
6
H
5
CH
3
N
N-Methyl-4-
phenylpiperidine
(compound A)
C
6
H
5
CH
3
N
CH
2
CH
2
N
CH
2
CH
3
CH
2
OCH
2
CH
3
COCH
3
CH
2
OC
O
N
CH
3
CH
2
OC
O
O
CH
3
1. NaOCH
2
CH
3
2. H
H11001
1. HO
H11002
, H
2
O
2. H
H11001
3. heat N
O
CH
3
N-Methyl-4-
piperidone
648 AMINES
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4-(N,N-dimethylamino)pyridine that is more basic. Note that protonation of the ring nitrogen per-
mits delocalization of the dimethylamino lone pair and dispersal of the positive charge.
22.47 The
1
H NMR spectrum of each isomer shows peaks corresponding to five aromatic protons, so com-
pounds A and B each contain a monosubstituted benzene ring. Only four compounds of molecular
formula C
8
H
11
N meet this requirement.
Neither
1
H NMR spectrum is consistent with N-methylbenzylamine, which would have two
singlets due to the methyl and methylene groups. Likewise, the spectra are not consistent with
N-ethylaniline, which would exhibit the characteristic triplet–quartet pattern of an ethyl group. Al-
though a quartet occurs in the spectrum of compound A, it corresponds to only one proton, not the
two that an ethyl group requires. The one-proton quartet in compound A arises from an H—C—CH
3
unit. Compound A is 1-phenylethylamine.
Compound B has an
1
H NMR spectrum that fits 2-phenylethylamine.
22.48 Only the unshared electron pair on nitrogen that is not part of the H9266 electron cloud of the aromatic
system will be available for protonation. Treatment of 5-methyl-H9253-carboline with acid will give the
salt shown.
22.49 Write the structural formulas for the two possible compounds given in the problem and consider
how their
13
C NMR spectra will differ from each other. Both will exhibit their CH
3
carbons at high
field signal, but they differ in the positions of their CH
2
and quaternary carbons. A carbon bonded to
5-Methyl-H9253-carboline
H11001
H
N
N
H
H11001
N
N
CH
3
CH
3
C
6
H
5
CH
2
CH
2
NH
2
Singlet (H9254 1.1 ppm)
Pair of triplets
at H9254 2.75 ppm and 2.95 ppm
C
6
H
5
C
CH
3
NH
2
Quartet (H9254 3.9 ppm)
Doublet (H9254 1.2 ppm)
Singlet (H9254 1.3 ppm)
H
C
6
H
5
CH
2
NHCH
3
N-Methylbenzylamine
C
6
H
5
NHCH
2
CH
3
N-Ethylaniline
C
6
H
5
CHCH
3
NH
2
1-Phenylethylamine
C
6
H
5
CH
2
CH
2
NH
2
2-Phenylethylamine
Most stable protonated form of
4-(N,N-dimethylamino)pyridine
H
H11001
N
N(CH
3
)
2
H
N
H11001
N(CH
3
)
2
AMINES 649
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nitrogen is more shielded than one bonded to oxygen, because nitrogen is less electronegative than
oxygen.
In one isomer the lowest field signal is a quaternary carbon; in the other it is a CH
2
group. The spec-
trum shown in Figure 22.10 shows the lowest field signal as a CH
2
group. The compound is there-
fore 2-amino-2-methyl-1-propanol, (CH
3
)
2
CCH
2
OH.
=
NH
2
This compound cannot be prepared by reaction of ammonia with an epoxide, because in basic so-
lution nucleophiles attack epoxides at the less hindered carbon, and therefore epoxide ring opening
will give 1-amino-2-methyl-2-propanol rather than 2-amino-2-methyl-1-propanol.
SELF-TEST
PART A
A-1. Give an acceptable name for each of the following. Identify each compound as a primary,
secondary, or tertiary amine.
(a)(c)
(b)
A-2. Provide the correct structure of the reagent omitted from each of the following reactions:
(a)
(b)
(c) C
6
H
5
CH
2
Br C
6
H
5
CH
2
NH
2
1. ?
2. H
2
NNH
2
H11001
O
O
NH
NH
C
6
H
5
CH
2
Br C
6
H
5
CH
2
CH
2
NH
2
1. ?
2. LiAlH
4
3. H
2
O
C
6
H
5
CH
2
Br C
6
H
5
CH
2
NH
2
1. ?
2. LiAlH
4
3. H
2
O
NHCH
3
NHCH
2
CH
2
CH
3
Br
CH
3
CH
2
CCH
3
CH
3
NH
2
(CH
3
)
2
C CH
2
NH
3
O
H11001
2,2-Dimethyloxirane Ammonia
(CH
3
)
2
CCH
2
NH
2
OH
1-Amino-2-methyl-2-propanol
1-Amino-2-methyl-2-propanol
(CH
3
)
2
CCH
2
NH
2
OH
Lower field
signal
Higher field
signal
2-Amino-2-methyl-1-propanol
(CH
3
)
2
CCH
2
OH
NH
2
Higher field
signal
Lower field
signal
650 AMINES
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A-3. Provide the missing component (reactant, reagent, or product) for each of the following:
(a)
(b)
(c)
(d)
(e)
( f )
(g)
A-4. Provide structures for compounds A through E in the following reaction sequences:
(a)
(b)
A-5. Give the series of reaction steps involved in the following synthetic conversions:
(a)
(b) m-Chloroaniline from benzene
(c) C
6
H
5
N N(CH
3
)
2
from anilineN
from benzene
C(CH
3
)
3
I
DEH11001 CH
3
CH
2
NH
2
NaBH
3
CN
CH
3
OH
NaNO
2
, HCl
H
2
O
O
C H
2
C
CH
3
CHCH
2
CH
2
NCH
2
CH
3
heat
B
Ag
2
O
H
2
O
A
CH
3
I
NHCH
2
CH
3
?
NaNO
2
, HCl, H
2
O
N(CH
3
)
2
?
NaNO
2
, HCl, H
2
O
?
CH
2
CH
3
NHCCH
3
O
HNO
3
H
2
SO
4
?
NH
2
H
3
C NHCCH
3
H
3
C
O
Product of part (a) toluene
?
Product of part (a)?
CuBr
NH
2
H
3
C?
NaNO
2
, HCl
H
2
O
AMINES 651
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A-6. p-Nitroaniline (A) is less basic than m-nitroaniline (B). Using resonance structures, explain
the reason for this difference.
A-7. Identify the strongest and weakest bases among the following:
A-8. Write the structures of the compounds A–D formed in the following reaction sequence:
PART B
B-1. Which of the following is a secondary amine?
(a) 2-Butanamine
(b) N-Ethyl-2-pentanamine
(c) N-Methylpiperidine
(d) N,N-Dimethylcyclohexylamine
B-2. Which of the following C
8
H
9
NO isomers is the weakest base?
(a) o-Aminoacetophenone
(b) m-Aminoacetophenone
(c) p-Aminoacetophenone
(d) Acetanilide
B-3. Rank the following compounds in order of increasing basicity (weakest → strongest):
(a)4H11021 2 H11021 1 H11021 3(c)4H11021 3 H11021 1 H11021 2
(b)4H11021 1 H11021 3 H11021 2(d)2H11021 1 H11021 3 H11021 4
NH
2
CH
2
NH
2
O
CNH
2
NO
2
NH
2
1234
(CH
3
)
3
CCl
AlCl
3
NHCCH
3
O
ABC D
H
2
O, HCl
heat
Cl
2
(2 mol)
1. NaNO
2
, HCl
2. CuBr
N
H
N
H
N
H
O
2
N
N H O
AB C D
NH
2
NO
2
A
NH
2
NO
2
B
652 AMINES
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B-4. Which of the following arylamines will not form a diazonium salt on reaction with sodium
nitrite in hydrochloric acid?
(a) m-Ethylaniline
(b) 4-Chloro-2-nitroaniline
(c) p-Aminoacetophenone
(d) N-Ethyl-2-methylaniline
B-5. The amines shown are isomers. Choose the one with the lowest boiling point.
(a)(b)(c)(d)
B-6. Which of the following is the strongest acid?
(a)(d)
(b)(e)
(c)
B-7. The reaction
gives as final product
(a) A primary amine
(b) A secondary amine
(c) A tertiary amine
(d) A quaternary ammonium salt
B-8. A substance is soluble in dilute aqueous HCl and has a single peak in the region
3200–3500 cm
–1
in its infrared spectrum. Which of the following best fits the data?
(a)(c)
(b d)
CO
2
HNH
2
NHCH
3
N(CH
3
)
2
NHCH
2
CH
3
CH
3
I (excess) ?H11001
HH
N
H11001
HH
N
H11001
N H
H
H11001
H11001
H
H
HHN
H
H
H N
NH
2
NHCH
3
CH
3
N
H CH
3
N
AMINES 653
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B-9. Identify product D in the following reaction sequence:
B-10. Which one of the following is the best catalyst for the reaction shown?
(a)(c)(e)
(b)(d)
B-11. What will be the major product of each of the two reactions shown?
(a)1x, 2x (b)1x, 2y (c)1y, 2x (d)1y, 2y
H11001
N(CH
3
)
3
H11002
OH
1. CH
3
CH
2
CHCH
3
Br
2. CH
3
CH
2
CHCH
3
CH
3
CH
2
ONaH11001
H11001CH
3
CH CHCH
3
x
CH
3
CH
2
CH CH
2
y
heat
heat
NH
3
Cl
H11002
H11001
NH
2
CH
2
Cl NHCCH
3
O
CH
2
N(CH
3
)
3
Cl
H11002
H11001
CH
3
(CH
2
)
8
CH
2
Br CH
3
(CH
2
)
8
CH
2
CN
KCN
benzene
CH
3
N(CH
3
)
2
CH
3
CH
3
CCH
2
CHN(CH
3
)
2
CH
3
O
CH
3
CH
3
CCH
2
CN(CH
3
)
2
CH
3
CH
3
CH
3
CCH
2
CN
CH
3
CH
3
CH
3
CCH
2
CH
2
N(CH
3
)
2
CH
3
CH
3
OH
CH
3
CCH
2
CHN(CH
3
)
2
(a)
(b)
(c)
(d)
(e)
CH
3
CH
3
CH
3
CCH
2
CH
2
OH A
K
2
Cr
2
O
7
, H
2
SO
4
H
2
O, heat
1. LiAlH
4
, diethyl ether
2. H
2
O
SOCl
2
BC D
(CH
3
)
2
NH
(2 mol)
654 AMINES
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B-12. Which sequence represents the best synthesis of 4-isopropylbenzonitrile?
(a) 1. Benzene H11001 (CH
3
)
2
CHCl, AlCl
3
; 2. Br
2
, FeBr
3
; 3. KCN
(b) 1. Benzene H11001 (CH
3
)
2
CHCl, AlCl
3
; 2. HNO
3
, H
2
SO
4
; 3. Fe, HCl; 4. NaOH;
5. NaNO
2
, HCl, H
2
O; 6. CuCN
(c) 1. Benzene H11001 (CH
3
)
2
CHCl, AlCl
3
; 2. HNO
3
, H
2
SO
4
; 3. Fe, HCl; 4. NaOH; 5. KCN
(d) 1. Benzene H11001 HNO
3
, H
2
SO
4
; 2. (CH
3
)
2
CHCl, AlCl
3
; 3. Fe, HCl; 4. NaOH; 5. NaNO
2
,
HCl, H
2
O; 6. CuCN
(e) 1. Benzene H11001 HNO
3
, H
2
SO
4
; 2. Fe, HCl; 3. NaOH; 4. NaNO
2
, HCl, H
2
O; 5. CuCN;
6. (CH
3
)
2
CHCl, AlCl
3
B-13. The major products from the following sequence of reactions are
(a) (CH
3
)
2
CHCH
2
NH
2
H11001 H
2
C?CH
2
(b) (CH
3
)
2
NCH
2
CH
3
H11001 H
2
C?C(CH
3
)
2
CH
3
?
(c) (CH
3
)
2
CHCH
2
NCH
2
CH
3
H11001 H
2
C?CH
2
(d) (CH
3
)
3
N
H11001
CH
2
CH
3
I
H11002
H11001 H
2
C?CH
2
(e) None of these combinations of products is correct.
B-14. Which compound yields an N-nitrosoamine after treatment with nitrous acid (NaNO
2
, HCl)?
(a)(d)
(b)
(e)
(c)
NHCH
3
CNH
2
O
N
NH
2
H
3
CCH
2
NH
2
(CH
3
)
2
CHCH
2
N(CH
2
CH
3
)
2
CH
3
I Ag
2
O
H
2
O
heat
?
(CH
3
)
2
CH CN
4-Isopropylbenzonitrile
AMINES 655
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| 课件名称： | 有机化学（英文原版） |
| 课件分类： | 化学 |
| 课件类型： | 电子图书 |
| 文件大小： | 28.97MB |
| 下载次数： | 7 |
| 评论次数： | 5 |
| 用户评分： | 8.2 |
- 47. 有机化学（英文原版）：SGChapt13
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- 50. 有机化学（英文原版）：SGChapt16.PDF
- 51. 有机化学（英文原版）：SGChapt17
- 52. 有机化学（英文原版）：SGChapt18
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- 60. 有机化学（英文原版）：SGChapt26
- 61. 有机化学（英文原版）：SGChapt27
- 62. 有机化学（英文原版）：SGTOC
- 63. 有机化学（英文原版）：toc
