Chapter 24.
Amines and Heterocycles
Why this Chapter?
Amines and carbonyl compounds are the most
abundant and have rich chemistry
In addition to proteins and nucleic acids, a majority of
pharmaceutical agents contain amine functional
groups
Amines – Organic Nitrogen Compounds



Organic derivatives of ammonia, NH3,
Nitrogen atom with a lone pair of electrons, making
amines both basic and nucleophilic
Occur in plants and animals
2
24.1 Naming Amines
Alkyl-substituted (alkylamines) or aryl-substituted
(arylamines)
 Classified:
◦ 1° (RNH2), methyl (CH3NH2),
◦ 2° (R2NH),
◦ 3° (R3N)

Quaternary Ammonium Ions


A nitrogen atom with four attached groups is positively
charged
Compounds are quaternary ammonium salts
4
IUPAC Names – Simple Amines

For simple amines, the suffix -amine is used as the parent
name of the alkyl substituent
tert-Butylamine

Cyclohexylamine
Aniline
The suffix -amine can be used in place of the final -e in the parent
name
1,4-Butanediamine
4,4-Dimethylcyclohexanamine
5
IUPAC Names – Amines With More Than
One Functional Group

Consider the NH2 as an amino substituent on the parent
molecule
4-Amino-2-butanone
2-Aminobutanoic acid
2,4-Diaminobenzoic acid
6
IUPAC Names – Multiple Alkyl Groups

Symmetrical secondary and tertiary amines are named by
adding the prefix di- or tri- to the alkyl group
Triethylamine
Diphenylamine
7
IUPAC Names – Multiple, Different Alkyl
Groups
Named as N-substituted primary amines
 Largest alkyl group is the parent name, and other alkyl
groups are considered N-substituents

N,N-Dimethylpropylamine
N-Ethyl-N-methylcyclohexylamine
8
Common Names of Heterocyclic Amines


If the nitrogen atom occurs as part of a ring, the
compound is designated as being heterocyclic
Each ring system has its own parent name
9
Learning Check:

Name the following:
Solution:

Name the following:
N-methylethanamine
N-Ethyl-N-methylcyclohexylamine
(or Ethyl methyl amine)
Tricyclohexylamine
1,3-Butanediamine
N-methylpyrrolidine
Diisopropylamine
24.2 Properties of Amines

Bonding to N is similar to that in ammonia
◦ N is sp3-hybridized
◦ C–N–C bond angles are close to 109° tetrahedral value
12
Chirality Is Possible (But Not Observed)


An amine with three different substituents on nitrogen is
chiral (in principle but not in practice): the lone pair of
electrons is the fourth substituent
Most amines that have 3 different substituents on N are
not resolved because the molecules interconvert by
pyramidal inversion
13
Amines Form H-Bonds
Amines with fewer than five carbons are water-soluble
 Primary and secondary amines form hydrogen bonds,
increasing their boiling points

14
24.3 Basicity of Amines
The lone pair of electrons on nitrogen makes amines basic
and nucleophilic
 They react with acids to form acid–base salts and they react
with electrophiles

15
Relative Basicity




Amines are stronger bases than alcohols, ethers, or water
Amines establish an equilibrium with water in which the
amine becomes protonated and hydroxide is produced
The most convenient way to measure the basicity of an
amine (RNH2) is to look at the acidity of the
corresponding ammonium ion (RNH3+)
High pKa → weaker acid and stronger conjugate base.
16
The most convenient
way to measure the
basicity of an amine
(RNH2) is to look at the
acidity of the
corresponding
ammonium ion (RNH3+)
Weak Acid
H
Strong Base
CH3
N H
H
Most simple alkylammmonium
ions have pKa's of 10 to 11
High pKa →
weaker acid and
stronger
conjugate base.
Arylamines and heterocyclic
aromatic amines are
considerably less basic than
alkylamines (conjugate acid
pKa 5 or less)
Strong Acid
Weak Base
N
H
H
Relative Basicity
Weaker Acid (pKa ~60)
Stronger base
Weak Acid (pKa ~40)
Strong base
Purification of Amines:

Amines can be separated from a mixture by reaction with
acids to form soluble amine salts
Basicity of Amides

Amides (RCONH2) in general are not proton acceptors except in very
strong acid
•The C=O group is strongly electron-withdrawing, making the N a very
weak base
•Addition of a proton occurs on O but this destroys the double bond character
of C=O as a requirement of stabilization by N
20
24.4 Basicity of Substituted Arylamines

The N lone-pair electrons in arylamines are delocalized by
interaction with the aromatic ring  electron system and are
less able to accept H+ than are alkylamines
Weak Base
Weak Nucleophile
Electrons tied up in resonance aren’t as available for bonding.
21
Substituted
Arylamines

Electron-donating
substituents (such as
CH3, NH2, OCH3)
increase the basicity
of the
corresponding
arylamine

Electronwithdrawing
substituents (such
as Cl, NO2, CN)
decrease
arylamine basicity
22
Learning Check:
Without looking at a table,
 Name the following compounds then
 Rank the structures in each set in order of decreasing basicity (#1 = strongest)
A.
NH2
NH2
H
O2N
NH2
C
Br
O
B.
NH2
NH2
NH2
F
F
C
CH3
F
F
CH2
Solution:
Without looking at a table,
 Name the following compounds then
 Rank the structures in each set in order of decreasing basicity (#1 = strongest)
A.
NH2
NH2
H
O2N
3
C
B.
Br
2
O
p-Nitroaniline
NH2
p-Aminobenzaldehyde
NH2
1
p-Bromoaniline
NH2
NH2
F
F
C
F
3
p-(Trifluoromethyl)aniline
CH3
1
p-Methylaniline
F
CH2
2
p-(Fluoromethyl)aniline
24.5 Biological Amines

In what form do amines exist at a physiological pH of 7.3?
H
CH3
N
H
+
CH3
H2O
N
H
H
H
+
H3O+
pKa = -1.7
pKa = 10.64
[HA]
[A-]
-
pH = pKa + log [A ]
[HA]
At pH = 10.64 we expect the amount of [HA] and [A-] to be the same.
As the solution becomes more acidic (Below a pH of 10.64) we expect
methylamine to be protonated. (Higher concentration of HA)
As the solution becomes more basic (pH above 10.64) we expect the acidic
proton to be removed. (Higher concentration of A-)
25
Biological Amines

The amino acid, alanine, at pH = 7.3
pKa = 9.69
pKa = 2.34
A pH below pKa atom is protonated.
At pH above pKa proton is removed.
24.6 Synthesis of Amines

Reductions of Nitriles, Amides, & Nitro’s
◦ Reactions we’ve already seen:
Synthesis of Amines
Arylamines are prepared from nitration of an aromatic
compound and reduction of the nitro group
 Reduction by catalytic hydrogenation over platinum is
suitable if no other groups can be reduced


Iron, zinc, tin, and tin(II) chloride are effective in acidic solution
28
SN2 Reactions of Alkyl Halides

Ammonia and other amines are good nucleophiles
Remember: Nucleophilicity parallels Bacisity
29
Uncontrolled Multiple Alkylation

Primary, secondary, and tertiary amines all have similar
reactivity, the initially formed monoalkylated substance
undergoes further reaction to yield a mixture of products

Need better method to avoid mixture of products
30
Selective Preparation of Primary Amines:
the Azide Synthesis



Azide ion, N3 displaces a halide ion from a primary or
secondary alkyl halide to give an alkyl azide, RN3
Alkyl azides are not nucleophilic (but they are explosive)
Reduction gives the primary amine
31
Gabriel Synthesis of Primary Amines


A phthalimide alkylation for preparing a primary amine
from an alkyl halide
The N-H in imides (CONHCO) can be removed by
KOH followed by alkylation and hydrolysis
32
Reductive Amination of Aldehydes & Ketones

Treatment of an aldehyde or ketone with ammonia or an amine in the
presence of a reducing agent
Forms an imine that is then reduced
33
Reductive Amination Is Versatile

Ammonia, primary amines, and secondary amines yield
primary, secondary, and tertiary amines, respectively
34
Mechanism of Reductive Amination
35
Reducing Step


Sodium cyanoborohydride, NaBH3CN, reduces C=N but
not C=O
Stable in water
36
Learning Check:

Prepare using a reductive amination.
Solution:

Prepare using a reductive amination.
Hofmann and Curtius Rearrangements

Carboxylic acid derivatives can be converted into primary
amines with loss of one carbon atom by both the
Hofmann rearrangement and the Curtius rearrangement
39
Hofmann
Rearrangement

Amide reacts with Br2 and base

Rearranges to lose carbonyl
carbon making chain 1 carbon
shorter

Gives high yields of arylamines
and alkylamines
40
Hofmann example:
Curtius Rearrangement


Heating an acyl azide prepared from an acid chloride
Migration of R from C=O to the neighboring nitrogen
with simultaneous loss of a leaving group
42
Curtius Example:
Learning Check:


Prepare o-methylbenzylamine from a carboxyic acid
using a Hofmann rearrangement.
Prepare o-methylbenzylamine from a carboxyic acid
using a Curtius rearrangement.
Solution:

Prepare o-methylbenzylamine from a carboxyic acid
using a Hofmann rearrangement.
Hofmann
Curtius

Prepare o-methylbenzylamine from a carboxyic acid
using a Curtius rearrangement.
24.7 Reactions of Amines

We’ve already seen: Alkylation and acylation
46
Hofmann Elimination


Converts amines into alkenes
NH2 is very a poor leaving group so it converted to an
alkylammonium ion, which is a good leaving group
47
Silver Oxide Is Used for the Elimination
Step

Ag2O in water forms Ag(OH)2 which exchanges hydroxide
ion for iodide ion in the quaternary ammonium salt, thus
providing the base necessary to cause elimination
48
Orientation in Hofmann Elimination

We would expect that the more highly substituted
alkene product predominates in the E2 reaction of an
alkyl halide (Zaitsev's rule)

However, the less highly substituted alkene
predominates in the Hofmann elimination due to the
large size of the trialkylamine leaving group

The base must abstract a hydrogen from the most
sterically accessible, least hindered position
49
Steric Effects Control the Orientation
50
24.8 Reactions of Arylamines


Amino substituents are strongly activating, ortho- and
para-directing groups in electrophilic aromatic substitution
reactions
Reactions are controlled by conversion to amide
51
Arylamines Are Not Useful for FriedelCrafts Reactions


The amino group forms a Lewis acid–base complex with
the AlCl3 catalyst, preventing further reaction
Therefore we use the corresponding amide
52
Diazonium Salts: The Sandmeyer
Reaction

Primary arylamines react with HNO2, yielding stable
arenediazonium salts
53
Uses of Arenediazonium Salts

The N2 group can be replaced by a nucleophile
54
Preparation of Aryl Halides


Reaction of an arenediazonium salt with CuCl or CuBr
gives aryl halides (Sandmeyer Reaction)
Aryl iodides form from reaction with NaI without a
copper(I) salt
55
Aryl Nitriles and Carboxylic Acids

An arenediazonium salt and CuCN yield the nitrile, ArCN,
which can be hydrolyzed to ArCOOH
56
Formation of Phenols (ArOH)

From reaction of the arenediazonium salt with copper(I)
oxide in an aqueous solution of copper(II) nitrate
57
Reduction to a Hydrocarbon

By treatment of a diazonium salt with hypophosphorous
acid, H3PO2
58
Mechanism of Diazonium
Replacement

Through radical (rather than polar or ionic) pathways
59
Diazonium Coupling Reactions

Arenediazonium salts undergo a coupling reaction with
activated aromatic rings, such as phenols and arylamines,
to yield brightly colored azo compounds, ArN=NAr
60
How Diazonium Coupling Occurs


The electrophilic diazonium ion reacts with the electronrich ring of a phenol or arylamine
Usually occurs at the para position but goes ortho if para
is blocked
61
Azo Dyes

Azo-coupled products have extended  conjugation that
lead to low energy electronic transitions that occur in
visible light (dyes)
62
24.9 Heterocycles

A heterocycle is a cyclic compound that
contains atoms of two or more elements in
its ring, usually C along with N, O, or S
63
Pyrole and Imidazole

Pyrole is an amine and a conjugated diene,
however its chemical properties are not
consistent with either of structural features
64
Chemistry of Pyrole
Electrophilic substitution reactions occur at C2 b/c it is
position next to the N
 A more stable intermediate cation having 3 resonance
forms
 At C3, only 2 resonance forms

65
Polycyclic Heterocycles
66
24.10 Spectroscopy of Amines Infrared


Characteristic N–H stretching absorptions 3300 to
3500 cm1
Amine absorption bands are sharper and less intense
than hydroxyl bands
◦ Protonated amines show an ammonium band in the
range 2200 to 3000 cm1
67
Examples of Infrared Spectra
68
Nuclear Magnetic Resonance
Spectroscopy


N–H hydrogens appear as broad signals without clear-cut
coupling to neighboring C–H hydrogens
In D2O exchange of N–D for N–H occurs, and the N–H
signal disappears
69
Chemical Shift Effects


Hydrogens on C next to N and absorb at lower field than
alkane hydrogens
N-CH3 gives a sharp three-H singlet at  2.2 to  2.6
70
13C

NMR
Carbons next to amine N are slightly deshielded - about
20 ppm downfield from where they would absorb in an
alkane
71
Mass Spectrometry


A compound with an odd number of nitrogen atoms has
an odd-numbered molecular weight and a corresponding
parent ion
Alkylamines cleave at the C–C bond nearest the nitrogen
to yield an alkyl radical and a nitrogen-containing cation
72
Mass Spectrum of N-Ethylpropylamine

The two modes of a cleavage give fragment ions at m/z = 58
and m/z = 72.
73
Select the best classification for the following
molecule:
1.
2.
3.
4.
5.
1˚ aliphatic amine
2˚ aliphatic amine
3˚ aliphatic amine
aromatic amine
heterocyclic aromatic
amine
Which of the following is not represented in the
molecule shown?
1.
2.
3.
4.
5.
aliphatic amine
aromatic amine
heterocyclic aromatic
amine
secondary amine
a pyridine ring
Determine the IUPAC name for the following
molecule:
1.
2.
3.
4.
5.
ethylpropylphenylamine
N-phenyl-N-ethylpropanamine
N-ethyl-N-phenylpropanamine
N-ethyl-N-propylaniline
N-phenyl-N-ethylaniline
If the pKb of an amine is 9.54, what is the pKa of
its conjugate acid?
1.
2.
3.
4.
5.
6.46
4.46
-9.54
not enough
information to
determine
pKa is unrelated to
pKb.
The major alkaloid present in tobacco leaves is nicotine, whose
structure is shown below. Which will be the major ammonium
ion formed when nicotine is treated with one equivalent of a
strong acid?
N
CH3
N
nicotine
1.
2.
Arrange the above from
strongest to weakest base:
1.
2.
3.
4.
5.
A, B, C
B, C, A
C, A, B
A, C, B
B, A, C
Amine A is more basic than amine B.
1.
2.
True
False
The equilibrium lies to
the right in the above
reaction:
1.
2.
True
False
Which of the following statements is true
regarding the following two molecules?
1.
2.
3.
4.
5.
Both A and B are
aromatic.
Both A and B are aliphatic
amines.
A is more basic than B.
B is more basic than A.
Both A and B are planar
molecules.
Select the most acidic compound from the
choices provided.
1.
2.
3.
4.
5.
If a protonated amine with a pKa of 10 is placed in a
solution of pH 12, the predominant form of the amine
in solution will be the protonated form.
1.
2.
True
False
The following transformation can be
accomplished as shown.
1.
2.
True
False
Identify the major product for the following
reaction.
O
1) NaN3
2) heat
Cl
3) H2O
1.
2.
O
NH2
NH2
3.
O
OH
5.
4.
NH2
O
C
N
Identify the structure of A based on the products that
result from a Hofmann elimination.
N(CH )
NH2 + 3 2
1) CH3I (excess)
A
+
2) Ag2O, H2O, heat
+
N(CH3)2
1.
2.
N
N
3.
N
H
4.
N
H
CH3
5.
NH
NH2
N(CH3)2
Select the order of chemical shift (most downfield →
most upfield) for the carbon atoms in the following
molecule:
1.
2.
3.
4.
5.
b, c, d, a
a, d, b, c
a, b, d, c
c, b, d, a
c, d, b, a
What is the intermediate that results from the
treatment of aniline with HNO2 and H2SO4?
1.
2.
3.
4.
5.
a diazonium salt
an aryl radical
a quaternary
ammonium salt
an isocyanate
an aryl azide