Chapter 23
Amines
Review of Concepts
Fill in the blanks below. To verify that your answers are correct, look in your textbook at
the end of Chapter 23. Each of the sentences below appears verbatim in the section
entitled Review of Concepts and Vocabulary.
• Pyridine is a stronger base than pyrrole, because the lone pair in pyrrole
participates in ____________.
• An amine moiety exists primarily as ______________________________ at
physiological pH.
• The azide synthesis involves treating an ________________ with sodium azide,
followed by _______________.
• The __________ synthesis generates primary amines upon treatment of
• Primary amines react with a nitrosonium ion to yield a ______________ salt in a
process called diazotization.
• Sandmeyer reactions utilize copper salts (CuX), enabling the installation of a
halogen or a ________ group.
• In the Schiemann reaction, an aryl diazonium salt is converted into a
604
CHAPTER 23
Review of Skills
Fill in the blanks and empty boxes below. To verify that your answers are correct, look
in your textbook at the end of Chapter 23. The answers appear in the section entitled
SkillBuilder Review.
23.1 Naming an Amine
23.2 Preparing a Primary Amine via the Gabriel Reaction
N
O
O
H
NH
2
IDENTIFY REAGENTS THAT WILL ACHIEVE THE FOLLOWING TRANSFORMATION:
1)
2)
3)
23.3 Preparing an Amine via a Reductive Amination
CHAPTER 23
605
23.4 Synthesis Strategies
AZIDE
SYNTHESIS
REDUCTIVE
AMINATION
IDENTIFY THE REACTANT THAT IS USED AS A NUCLEOPHILE IN EACH OF THE THREE PROCESSES SHOWN BELOW:
23.5 Predicting the Product of a Hofmann Elimination
PREDICT THE MAJOR PRODUCT OF THE FOLLOWING REACTION:
NH
2
1) excess CH
3
I
2) Ag
2
O, H
2
O, heat
23.6 Determining the Reactants for Preparing an Azo Dye
606
CHAPTER 23
Review of Reactions
Identify the reagents necessary to achieve each of the following transformations. To
verify that your answers are correct, look in your textbook at the end of Chapter 23. The
answers appear in the section entitled Review of Reactions.
Preparation of Amines
NO
2
NH
2
O
NH
2
CHAPTER 23
607
Reactions of Amines
H
NR
H
N
O
R
H
Reactions of Aryldiazonium Salts
N N
IClBr CN
608
CHAPTER 23
Reactions of Nitrogen Heterocycles
Solutions
23.1.
a) 3,3-dimethyl-1-butanamine
23.2.
NH
2
23.3.
H
H
23.4.
N NH
2
N
H
Increasing boiling point
CHAPTER 23
609
23.5.
23.6.
a)
N
b)
N
c)
N
O
d)
N
23.7.
N
N
Br
N
H
O
Increasing Basicity
23.8. In the reactant, the lone pair of the amino group is delocalized via resonance. In
23.9.
H
NH
3
OH
23.10.
a)
Br
1) NaCN
2) LAH
3) H
2
O
610
CHAPTER 23
b)
NH
2
1) NaCN
2) LAH
3) H
2
O
Br
c)
NH
2
1) NaCN
2) LAH
3) H
2
O
Br
23.11. This compound cannot be prepared from an alkyl halide or a carboxylic acid,
using the methods described in this section, because there are two methyl groups at the
alpha position (the carbon atom connected to the amino group). These two methyl
groups cannot be installed with either of the synthetic methods above, because both
methods produce an amine with two alpha protons.
23.12.
a)
NH
2
N
O
O
H
1) KOH
3) H
3
O
+
2)
Br
CHAPTER 23
611
c)
NH
2
N
O
O
H
1) KOH
3) H
3
O
+
2)
Br
23.13.
a)
NH
2
1) Br
2
,
hv
2)
t
-BuOK
3) HBr, ROOR
Br
1)
N
O
O
2) H
3
O
+
K
b)
23.14.
a)
[ H
+
] , NaBH
3
CN
H
O
H
2
N
612
CHAPTER 23
b)
[ H
+
] , NaBH
3
CN
O
H
N
H
c)
[ H
+
] , NaBH
3
CN
O
H
N
H
d)
N
[ H
+
] , NaBH
3
CN
HN
O
H
O
CHAPTER 23
613
e)
[ H
+
] , NaBH
3
CN
NH
O
23.15.
23.16. The last step of reductive amination is the reduction of a C=N bond. That step
introduces a proton on the alpha position (the carbon atom that is connected to the
nitrogen atom in the product):
C
NR
C
NRH
H
23.17.
H
O
1) Br
2
, hv
2) NaOEt
1) O
3
2) DMS
614
CHAPTER 23
23.18.
a)
N
H
H
O
H
O
NH
3
[ H
+
] , NaBH
3
CN
NH
2
[ H
+
] , NaBH
3
CN
d)
N
H
H
O
H
O
NH
3
[ H
+
] , NaBH
3
CN
NH
2
[ H
+
] , NaBH
3
CN
CHAPTER 23
615
N
H
O
[ H
+
] , NaBH
3
CN
23.19. The first alkyl group is installed via a Gabriel synthesis, and the remaining alkyl
groups are installed via reductive amination processes. For most of the following
syntheses, there is a choice regarding which group to attach via the initial Gabriel
synthesis. In such cases, the least sterically hindered group is chosen (the group
whose installation involves the least hindered alkyl halide):
a)
N
O
O
1) Br
H
2
N
H
ON
H
K[ H
+
] , NaBH
3
CN
2) H
3
O
+
616
CHAPTER 23
d)
Br
H
2
N
H
ON
H
N
O
O
1)
K[ H
+
] , NaBH
3
CN
2) H
3
O
+
N
f)
O
NEtNH
2
O
N
H
Br
1)
K[ H
+
] , NaBH
3
CN
2) H
3
O
+
23.20. The first alkyl group is installed via an azide synthesis, and the remaining alkyl
groups are installed via reductive amination processes. For most of the following
syntheses, there is a choice regarding which group to attach via the initial azide
synthesis. In such cases, the least sterically hindered group is chosen (the group
whose installation involves the least hindered alkyl halide):
CHAPTER 23
617
b)
NH
2
Br
1) NaN
3
2) H
2
, Pt
d)
N
H
H
O
[ H
+
] , NaBH
3
CN
Br H
2
N
1) NaN
3
2) H
2
, Pt
618
CHAPTER 23
23.21.
H
O O
H
O
H
O
N
NH
3
[ H
+
] , xs NaBH
3
CN
1) O
3
2) DMS
23.22.
NH
2
O
HN
O
HN
O
NH
2
23.23.
NH
2
O
HN
O
HN
O
Cl
NH
2
Cl
NO
2
CHAPTER 23
619
23.24.
1) excess NH
3
2) LAH
3) H
2
O
23.25.
a)
1) excess CH
3
I
2) Ag
2
O, H
2
O, heat
NH
2
c)
1) excess CH
3
I
2) Ag
2
O, H
2
O, heat
N
23.27.
1) excess CH
3
I
2) Ag
2
O, H
2
O
NH
2
620
CHAPTER 23
23.28.
N
N
1) excess CH
3
I
2) Ag
2
O, H
2
O, heat
+ +
N
23.29.
23.30.
a)
CNNH
2
1)
Cl
O
2) (CH
3
)
2
CHCl,
3) H
3
O
+
NH
2
1) NaNO
2
, HCl
2) CuCN
AlCl
3
c)
Cl
O
AlCl
3
O
HNO
3
H
2
SO
4
O
NO
2
CHAPTER 23
621
d)
e)
Cl NO
2
Cl F
1) Cl
2
, AlCl
3
2) HNO
3
, H
2
SO
4
1) Fe, H
3
O
+
2) NaNO
2
, HCl
3) HBF
4
23.31.
a)
HO
3
S N
N NH
2
SO
3
H
1) NaNO
2
, HCl
2)
HO
3
S NH
2
SO
3
HNH
2
b)
622
CHAPTER 23
23.32.
NH
2
Cl
O
AlCl
3
Cl
2) HNO
3
, H
2
SO
4
1) Fuming H
2
SO
4
NO
2
NH
2
1)
23.33.
CHAPTER 23
623
23.34. Attack at either C2 or C4 generates an intermediate that exhibits a resonance
structure with a nitrogen atom that lacks an octet (highlighted below). Attack at
C3 generates a more stable intermediate:
N
C2
ATTACK
NH
E
NH
E
NH
E
E
23.35.
N
H
HNO
3
, H
2
SO
4
0 C
N
H
NO
2
23.36.
a) The second compound will have an N-H stretching signal between 3300 and 3500 cm
-1
.
The first compound will not have such a signal.
23.37.
a) The
1
H NMR spectrum of the first compound will have a singlet resulting from the N-
methyl group.
1
H NMR spectrum of the second compound will not have any singlets.
624
CHAPTER 23
23.38.
23.39.
a) The lone pair that is farthest away from the rings is the most basic, because its lone
pair is localized. The lone pair of the other nitrogen atom is delocalized via resonance.
23.40.
N O
O N
H
N
23.41.
23.42.
N
NH
2
CHAPTER 23
625
NH
N
O
N
H
23.44.
a) two b) two c) one
23.45.
a) 2,2,3,3-tetramethyl-1-hexanamine
23.46.
N N H
ethyldimethylamine diethylamine
NH
methylpropylamine
NH
isopropylmethylamine
23.47. None of these compounds are chiral.
N
N
N
626
CHAPTER 23
23.48.
a)
N
O
O
+
HN
O
O
+
H
B
a
s
e
A
c
i
d
23.49.
O
23.50.
a)
NH
2
OH
1) PBr
3
2) NaN
3
3) H
2
, Pt
c)
NH
2
OH
1) PBr
3
2) t–BuOK
CHAPTER 23
627
23.51.
a)
b)
NH
2
Br
1) NaCN
2) LAH
3) H
2
O
c)
d)
NH
2
CN
1) LAH
2) H
2
O
23.52. Aziridine has significant ring strain, which would increase significantly during
pyramidal inversion. This provides a significant energy barrier that prevents
pyramidal inversion at room temperature.
23.53.
N
O
H
N
H
628
CHAPTER 23
23.55. In acidic conditions, the amino group is protonated to give an ammonium ion.
The ammonium group is a powerful deactivator and meta-director.
23.56.
a) The presence of the nitro group in the para position helps stabilize the conjugate base
23.57.
O
23.58. Protonation of the oxygen atom gives a resonance stabilized cation (as seen in
23.59.
a)
NH
2
1) CH
3
Cl, AlCl
3
2) NBS, heat
1) NaN
3
2) H
2
, Pt
Br
Cl
NO
2
AlCl
3
COOH
1)
2) HNO
3
, H
2
SO
4
2) NaNO
2
, HCl
3) CuCN
1) Fe, H
3
O
+
4) H
3
O
+
CHAPTER 23
629
23.60.
OH
O
O
N
R
H
OOH
O
O
N
O
R
H
O
O
O
OH
OH
N
R
H
H
OH
+
23.61.
N
O
O
RRNH
2
+
H
2
N NH
2
NH
NH
O
O
23.62.
630
CHAPTER 23
d)
1) HNO
3
/ H
2
SO
4
2) Fe, H
3
O
+
3) NaNO
2
, HCl
4) CuCN
CN
23.63.
NaBH
3
CN
(CH
)
NH
[ H
+
]
1) excess Me
I
2) Ag
2
O, H
2
O, heat
23.64. The conjugate base of pyrrole is highly stabilized because it is an aromatic anion
and it is resonance stabilized, spreading the negative charge over all five atoms of
the ring:
23.65.
O
CHAPTER 23
631
23.66.
[ H
+
] , NaBH
3
CN
HN
O
23.67.
a)
NN
Cl
b)
N
N
O
O
23.69.
a) b)
632
CHAPTER 23
d)
1) xs LAH
2) H
2
O
N
O
N
23.71.
23.72.
NaBH
3
CN
NH
O
NH
2
[ H
2
SO
4
]
CHAPTER 23
633
b)
1) HNO
3
, H
2
SO
4
2) Fe, H
3
O
+
NH
2
xs Cl
2
NH
2
ClCl
Cl
1) NaNO
2
, HCl
ClCl
Cl
O NH
2
2) CuCN
3) H
3
O
+
d)
Cl
O
AlCl
3
NH
2
HCl
NaNO
2
1)
2) HNO
3
, H
2
SO
4
3) HCl, Zn[Hg], heat
1) Cl
2
, AlCl
3
634
CHAPTER 23
23.74. The IR data indicates that we are looking for structures that lack an N-H bond (i.e
tertiary amines):
N N N
N
23.75.
O N
H
N
23.76.
23.77. The compound is a tertiary amine with the appropriate symmetry that provides
for only three signals:
N N
or
23.78.
CHAPTER 23
635
23.79.
N
H
H
N
23.80.
1) CH
3
CH
2
COCl, AlCl
3
3) H
2
O
1) conc. H
2
SO
4
, heat
2) MCPBA
2) MeMgBr
O
HO
23.81.
1) Cl
2
, AlCl
3
2) NaOH, heat
3) Et
I
Cl
O
1) HNO
3
, H
2
SO
4
2) Fe, H
3
O
+
3)
O O
N
O
H
636
CHAPTER 23
23.83.
N
23.85.
CH
3
OH 2) DMS
H
O
H
O
NH
3
NaBH
3
CN
HN NH
Na, NH
3
1) O
3
[ H
+
]
23.86.
CHAPTER 23
637
23.87.
N
H
N
O H
N N
O
H
N
H
N
O
HOH
H
HOH
HOH
H
H Cl
O
N
O
H
NaCl
H Cl
Na
O
N
O
+
638
CHAPTER 23
23.88. Protonation of the nitrogen highlighted below results in a cation that is highly
resonance stabilized. Protonation of either of the other nitrogen atoms would not
result in a resonance stabilized cation: