Amines
Chapter 23
23-1
Structure & Classification
 Amines
are classified as:
• 1°, 2°, or , 3° amines: Amines in which there are 1, 2, or
3 alkyl or aryl groups.
:
:
CH3 - NH 2
CH3 - NH
CH3
CH3
CH3 - N:
CH3
Me th yl ami n e Di me th yl am in e Tri me th yl ami n e
(a 1° am in e )
(a 2° am in e )
(a 3° am in e )
23-2
Structure & Classification
 Amines
are further divided into aliphatic,
aromatic, and heterocyclic amines:
• Aliphatic amine: An amine in which nitrogen is bonded
only to alkyl groups.
• Aromatic amine: An amine in which nitrogen is bonded
to one or more aryl groups.
N -H
CH3
CH2 - N- CH3
:
N H2
:
:
CH3
An i li n e
N-Me th yl an i li n e
Be n z yl dim eth ylami n e
(a 1° aromati c ami n e)(a 2° aromati c ami n e) (a 3° al iph ati c ami n e)
23-3
Structure & Classification
• Heterocyclic amine: An amine in which nitrogen is one
of the atoms of a ring.
N
N
H
H
Pyrrolidine Piperid ine
(h eterocyclic alip hatic amines)
N
N
H
Pyrrole
Pyridine
(h eterocyclic aromatic amin es)
23-4
Nomenclature
 Aliphatic
amines: replace the suffix -e of the
parent alkane by -amine.
NH2
H2 N
NH2
2-Propanamine
(S)-1-Phen ylethanamine
NH2
1,6-Hexanediamine
23-6
Nomenclature
 The
IUPAC system retains the name aniline.
N H2
N H2
N H2
N H2
OCH3
N O2
CH3
An i li ne 4-Ni troan il in e 4-Meth ylan il i ne 3-Meth oxyan i li n e
(p-Ni troan i li n e ) (p-Tolu i di n e )
(m-An i si di n e )
23-7
Nomenclature
 Among
the various functional groups, -NH2 is
one of the lowest in order of precedence.
H2 N
OH
2-A min oeth anol
OH
NH2
(S )-2-Amin o-3-methyl1-bu tanol
Amine vs
alcohol
H2 N
COOH
4-A minoben zoic acid
Amine vs
acid
23-8
Nomenclature
 Common
names for most aliphatic amines are
derived by listing the alkyl groups bonded to
nitrogen in one word ending with the suffix
-amine.
H
CH3 NH2
NH2
N
Et 3 N
Methylamine tert -Butylamine Dicyclopentylamine Triethylamine
23-9
Nomenclature
 When
four groups are bonded to nitrogen, the
compound is named as a salt of the
corresponding amine.
23-10
Chirality of Amines
• Consider the unshared pair of electrons on nitrogen as
a fourth group, then the arrangement of groups around
N is approximately tetrahedral.
• An amine with three different groups bonded to N is
chiral and exists as a pair of enantiomers and, in
principle, can be resolved.
23-11
Chirality of Amines
• In practice, however, they cannot be resolved because
they undergo inversion, which converts one
enantiomer to the other.
23-12
Chirality of Amines
• Pyramidal inversion is not possible with quaternary
ammonium ions, and their salts can be resolved.
ClN
Et
Me
S Enan tiomer
ClN
Et
Me
R Enan tiomer
23-13
Physical Properties
 Amines
are polar compounds, and both 1° and 2°
amines form intermolecular hydrogen bonds.
• N-H- - -N hydrogen bonds are weaker than O-H- - -O
hydrogen bonds because the difference in
electronegativity between N and H (3.0 - 2.1 =0.9) is
less than that between O and H (3.5 - 2.1 = 1.4).
Using bp as an indication of H bonding
CH3 CH3
MW (g/m ol )
bp (°C )
30.1
-88.6
CH3 NH 2
31.1
-6.3
Increasing strength
CH3 OH
32.0
65.0
23-14
Basicity
 All
amines are weak bases, and aqueous
solutions of amines are basic.
H
CH3 - N
H
+ CH3 - N- H O-H
+ H- O-H
H
Me thyl ami ne
H
Me thyl am monium hydroxide
• It is common to discuss their basicity by reference to
the acid ionization constant of the conjugate acid.
CH3 NH3 + + H2 O
CH3 NH2 + H3 O+
+
Ka =
[ CH3 NH2 ] [H3 O ]
+
[CH3 NH3 ]
= 2.29 x 10-1 1
pK a = 10.64
23-15
Basicity
• Using values of pKa, we can compare the acidities of
amine conjugate acids with other acids.
CH3 NH2 + CH3 COOH
pK a 4.76
(s tronger
acid)
+
-
CH3 NH3 + CH3 COO
pK a 10.64
(w eak er
acid )
pK eq = -5.88
Ke q = 7.6 x 105
23-16
Basicity-Aliphatic Amines
 Aliphatic
Amines
• note that pKa + pKb = 14
Ami n e
S tru ctu re
Amm on ia
N H3
Prim ary Am in e s
me th yl ami n e
CH3 N H2
e th yl am in e
CH3 CH 2 NH 2
cycl ohe xylam in e C6 H1 1 NH 2
pK a
pK b
9.26
4.74
10.64
10.81
10.66
3.36
3.19
3.34
S e con dary Ami n es
dim e th ylam in e ( CH3 ) 2 N H
10.73
die th yl ami n e ( CH3 CH2 ) 2 NH 10.98
Te rtiary Am in e s
tri me th yl amin e ( CH3 ) 3 N
9.81
tri e th ylam in e ( CH3 CH2 ) 3 N
10.75
Stronger bases
3.27
3.02
4.19
3.25
23-17
Basicity-Aromatic Amines
Amine
Structure
pK a of Conju gate A cid
Aromatic A min es
An iline
4-Methylanilin e
NH2
NH2
CH3
4-Ch loroaniline
Cl
4-N itroanilin e
O2 N
4.63
NH2
NH2
5.08
Weaker
bases
4.15
1.0
Heterocyclic Aromatic A min es
Pyrid ine
N
5.25
Intermediate
N
Imidazole
6.95
N
H
23-18
Basicity-Aromatic Amines
• Aromatic amines are considerably weaker bases than
aliphatic amines.
NH2 + H2 O
Cycloh exylamine
NH2 + H2 O
A niline
+
-
NH3 OH
pK a = 10.66
Cyclohexylammonium
hydroxid e
NH3 + OH-
pK a = 4.63
A niliniu m h yd roxide
23-19
Basicity-Aromatic Amines
 Aromatic
amines are weaker bases than aliphatic
amines because of two factors:
• Resonance stabilization of the free base, which is lost
on protonation.
H
N H
H .
.
H + H
N
H
un hybridized 2p orb ital of N
..
.
.
H .
H
H+ H
N
.
N
H
H + H
N
H
H
nitrogen is sp 2 h yb rid ized
23-20
Basicity-Aromatic Amines
• The greater electron-withdrawing inductive effect of
the sp2-hybridized carbon of an aromatic amine
compared with that of the sp3-hybridized carbon of an
aliphatic amine.
And note the effect of substituents
 Electron-releasing
groups, such as alkyl groups,
increase the basicity of aromatic amines.
 Electron-withdrawing groups, such as halogens,
the nitro group, and a carbonyl group decrease
the basicity of aromatic amines by a combination
of resonance and inductive effects.
23-21
Example: Basicity-Aromatic Amines
3-nitroaniline is a stronger base than 4-Nitroaniline.
O2 N
NH2
O2 N
4-N itroaniline
pK a 1.0
de localiz ation of th e n itrog en
lon e p air on to th e oxyg en atom s
of th e n itro grou p
3-N itroanilin e
pK a 2.47
-
O
+N
-
O
NH 2
NH2
O
+N
NH2
+
-O
Cannot do this kind of resonance in 3 nitroaniline
23-22
Basicity-Aromatic Amines
 Heterocyclic
aromatic amines are weaker bases
than heterocyclic aliphatic amines.
N
N
H
Piperidine
p Ka 10.75
N
Pyridin e
pK a 5.25
N
H
Imid azole
pK a 6.95
23-23
Basicity-Aromatic Amines
• In pyridine, the unshared pair of electrons on N is not
part of the aromatic sextet.
.
H
.
H
H
.
.
H
.
.
N
H
:
an sp 2 hybrid orbital; the electron
pair in this orbital is not a
part of the aromatic s extet
nitrogen is sp 2 h yb ridized
• Pyridine is a weaker base than heterocyclic aliphatic
amines because the free electron pair on N lies in an
sp2 hybrid orbital (33% s character) and is held more
tightly to the nucleus than the free electron pair on N
in an sp3 hybrid orbital (25% s character).
23-24
Basicity-Aromatic Amines
 Imidazole
Which N lone pair is protonated? The
one which is not part of the aromatic system.
Th is e l ectron pai r
i s n ot a part of th e
arom atic s e xte t
:
H
N
N+
+ H2 O
N
H
Imidaz ole
:
:
Th is e l ectron
pair i s a part
of th e aromati c
s exte t
Aromati ci ty is
mai n tain e d wh e n
i midaz ole i s proton ated
+ OH-
N
H
Imidaz oli u m i on
23-25
Basicity-Guanidine
 Guanidine
is the strongest base among neutral
organic compounds.
NH
H2 N C NH2
Guanidine
NH2
+
H2 O
+
H2 N C NH2
+
OH
Guanidinium ion
-
pKa = 13.6
• Its basicity is due to the delocalization of the positive
charge over the three nitrogen atoms.
NH2
H2 N
NH2
C
NH2
:
C
NH2
:
:
:
H2 N
+
:
NH2
:
+
H2 N
C
+
NH2
Thre e e quivale nt contributing s tructure s
23-26
Reaction with Acids
 All
amines, whether soluble or insoluble in water,
react quantitatively with strong acids to form
water-soluble salts.
OH
HO
OH
NH2
+ HCl
HO
(R)-N orepinep hrine
(on ly s ligh tly s olu ble in w ater)
H2 O
HO
NH3 + Cl-
HO
(R)-N orepinep hrine h yd roch loride
(a w ater-s olu ble salt)
23-27
Reaction with acids
 Separation
and purification of an amine and a
neutral compound.
23-28
Preparation
 We
have already covered these methods
• nucleophilic ring opening of epoxides by ammonia and
amines.
• addition of nitrogen nucleophiles to aldehydes and
ketones to form imines
• reduction of imines to amines
• reduction of amides to amines by LiAlH4
• reduction of nitriles to a 1° amine
• nitration of arenes followed by reduction of the NO2
group to a 1° amine
23-29
Preparation
 Alkylation
of ammonia and amines by SN2
substitution.
SN 2
+
N
H
CH3 Br
3
CH3 NH 3 + Br Me th yl am mon i um
bromi de
• Unfortunately, such alkylations give mixtures of
products through a series of proton transfer and
nucleophilic substitution reactions.
CH3 Br + NH3
+
-
+
-
CH3 NH3 Br + (CH3 ) 2 NH2 Br
+
-
+
+ (CH3 ) 3 NH Br + (CH3 ) 4 N Br
polyalkylations
23-30
-
Preparation via Azides
 Alkylation
Ph CH2 Cl
Benzyl chloride
K N3
-
RN 3
R
N
+
N
:
N: -
Az ide i on
(a good n u cle oph il e )
+
:
N
+
N
:
-:
:
N3
-
of azide ion.
N: -
An al k yl az i de
1. LiAlH4
Ph CH2 N3
2. H2 O
Benzyl azide
Ph CH2 NH2
Benzylamine
Overall
Alkyl Halide  Alkyl amine
23-31
Example: Preparation via Azides
• Alkylation of azide ion.
A rCO 3 H
Cycl oh e xe n e
O
1. K
+
N3 -
2 . H2 O
1,2-Epoxycycl oh e xan e
OH
1 . LiA lH4
2 . H2 O
N3
trans- 2-Az i docycl oh e xan ol
(racem ic)
OH
N H2
trans- 2-Am in ocycl oh e xan ol
(racem ic)
Note retention of
configuration, trans  trans
23-32
Reaction with HNO2
 Nitrous
acid, a weak acid, is most commonly
prepared by treating NaNO2 with aqueous H2SO4
or HCl.
HNO2 + H2 O
 In
H3 O+ + NO2 -
pK a = 3.37
its reactions with amines, nitrous acid:
• Participates in proton-transfer reactions.
• A source of the nitrosyl cation, NO+, a weak
electrophile.
23-33
Reaction with HNO2
 NO+
is formed in the following way.
• Step 1: Protonation of HONO.
• Step 2: Loss of H2O.
H
+
+ H O N O
(1)
+
H O N O
(2)
H
H O +
H
+
N O
+
N O
Th e n itros yl cation
• We study the reactions of HNO2 with 1°, 2°, and 3°
aliphatic and aromatic amines.
23-34
Tertiary Amines with HNO2
• 3° Aliphatic amines, whether water-soluble or waterinsoluble, are protonated to form water-soluble salts.
• 3° Aromatic amines: NO+ is a weak electrophile and
participates in Electrophilic Aromatic Substitution.
Me2 N
1. NaNO2 , HCl, 0-5°C
2. NaOH, H2 O
N,N-Dimethylaniline
Me2 N
N=O
N,N-Dimethyl-4-nitrosoaniline
23-35
Secondary Amines with HNO2
• 2° Aliphatic and aromatic amines react with NO+ to
give N-nitrosamines.
carcinogens
N-N=O + H2 O
N-H + HNO2
N-N itrosopiperidin e
Piperidin e
Mechanism:
•
•
+
•
•
••
+
N O
(1)
+
N
N
••
•
•
N=O
N=O
•
•
H O
H
N
•
•
H
••
•
•
••
••
•
•
H
+
+ H O H
H
••
(2)
23-36
RNH2 with HNO2
 1°
aliphatic amines give a mixture of
unrearranged and rearranged substitution and
elimination products, all of which are produced
by way of a diazonium ion and its loss of N2 to
give a carbocation.
 Diazonium ion: An RN2+ or ArN2+ ion
23-37
1° RNH2 with HNO2
 Formation
of a diazonium ion.
Step 1: Reaction of a 1° amine with the nitrosyl cation.
:
An N-n itros ami n e
: :
:
:
:
:
:
+
R-NH2 + : N O :
A 1° al iph ati c
am in e
:
k e to-e n ol
tau tome ri s m
H
R-N-N=O :
R-N=N-O-H
A di azotic acid
Step 2: Protonation followed by loss of water.
: :
:
:
R-N=N-O-H
H
+
••
R-N
A di azotic acid
••
N
••
A di azon i u m i on
+
R
+
N
••
N
••
••
R
+
N
H
+ - H2 O
O-H
N
A carbocation
23-38
1° RNH2 with HNO2 (Aliphatic)
 Aliphatic
diazonium ions are unstable and lose
N2 to give a carbocation which may:
1. Lose a proton to give an alkene.
2. React with a nucleophile to give a substitution
product.
3. Rearrange and then react by Steps 1 and/or 2.
Cl
(5.2%)
OH
N H2 N aN O2 , H Cl
0-5o C
OH
+
(25%)
(13.2%)
(25.9%)
+
(10.6%)
23-39
1° RNH2 with HNO2
reaction: Treatment of a aminoalcohol with HNO2 gives a ketone and N2.
 Tiffeneau-Demjanov
OH
 
CH2 NH2 + HNO2
A-aminoalcohol
O
+ H2 O + N2
Cycloheptan on e
23-40
Mechanism of Tiffeneau-Demjanov
• Reaction with NO+ gives a diazonium ion.
• Concerted loss of N2 and rearrangement followed by
proton transfer gives the ketone.
:
:
: O-H
: OH
CH2NH2
HNO 2
CH2
+
N N:
-N2
(A diazonium ion)
:
:
:
:O H
+ O H
+ CH2
CH2
A resonance-stabilized cation
proton transfer
to H2O
O
Cycloheptanone
Similar to pinacol rearrangement
23-41
Pinacol Rearrangement: an example of
stabilization of a carbocation by an adjacent
lone pair.
Overall:
23-42
Mechanism
Reversible
protonation.
Elimination
of water to
yield tertiary
carbocation.
This is a
protonated
ketone!
1,2
rearrangement
to yield
resonance
stabilized
cation.
Deprotonation.
23-43
1° Primary Amines with HNO2 (Aromatic)
-N2+ group of an arenediazonium salt can be
replaced in a regioselective manner by these
groups.
 The
H2 O
HBF4
HCl, CuCl
Ar-NH2
HNO2
0-5°C
Ar-N2
+
(-N2 )
HBr, CuBr
Ar-OH
Ar-F
Sch ieman n
reaction
Ar-Cl
Ar-Br
San dmeyer
reaction
KCN, CuCN Ar-CN
KI
Ar-I
H3 PO2
Ar-H
23-44
1° ArNH2 with HNO2
A
1° aromatic amine converted to a phenol.
OH
NH2
Br
Br
1 . HNO2
2 . H2 O, heat
CH3
2-Bromo-4methylanilin e
CH3
2-Bromo-4meth ylp henol
23-45
1° ArNH2 with HNO2
Problem: What reagents and experimental conditions will
bring about this conversion?
CH3
CH3
(1)
COOH
(2)
N O2
COOH
(3)
N O2
COOH
(4)
N H2
OH
23-46
1° ArNH2 with HNO2
 Problem:
Show how to bring about each
conversion.
CH3
Cl
(5)
CH3
CH3
NH2
(6)
CH3
C N
(8)
CH3
CH3
NH2
Cl
CH2 NH2
(7)
Cl
(9)
Cl
Cl
23-47
Hofmann Elimination
 Hofmann
elimination: Thermal decomposition of
a quaternary ammonium hydroxide to give an
alkene.
• Step 1: Formation of a 4° ammonium hydroxide.
CH3 I +
CH2 -N - CH 3
+
A g2 O
H2 O
CH3
(C ycl oh exylm e th yl)tri me th yl -Si lve r
oxide
amm oni u m iodide
CH3 OH+
CH2 -N - CH 3
+
A gI
CH3
(C ycl oh exylm e th yl)tri me th yl ammon i u m h ydroxi de
23-48
Hofmann Elimination
• Step 2: Thermal decomposition of the 4° ammonium
hydroxide.
CH3
OH+
CH2 -N - CH 3
160°
CH3
(C ycl oh exyl me th yl )tri me thyl amm on i u m hydroxi de
CH2 +
Me th yl en e cycl ohe xan e
( CH3 ) 3 N + H2 O
Trim eth ylami n e
23-49
Hofmann Elimination
 Hofmann
elimination is regioselective - the major
product is the least substituted alkene.
CH3
+
N(CH3 ) 3 OH
heat
-
CH2 + (CH3 ) 3 N + H2 O
rule: Any -elimination that occurs
preferentially to give the least substituted alkene
as the major product is said to follow Hofmann’s
rule.
 Hofmann’s
23-50
Hofmann Elimination
HO H
H H
C
H
CH 3 CH 2
C
+
N ( CH 3 ) 3
HOH
E2 reaction
(concerted
elimination)
H
CH 3 CH 2
H
C
C
H
N ( CH3 ) 3
• The regioselectivity of Hofmann elimination is
determined largely by steric factors, namely the bulk of
the -NR3+ group.
• Hydroxide ion preferentially approaches and removes
the least hindered hydrogen and, thus, gives the least
substituted alkene.
• Bulky bases such as (CH3)3CO-K+ give largely Hofmann
elimination with haloalkanes.
23-51
Cope Elimination
 Cope
elimination: Thermal decomposition of an
amine oxide.
Step 1: Oxidation of a 3° amine gives an amine oxide.
O
+
CH2 N-CH3 + H2 O2
CH3
-
CH2 N-CH3 + H2 O
CH3
An amin e oxid e
Step 2: If the amine oxide has at least one -hydrogen, it
undergoes thermal decomposition to give an alkene.
O
-
H
+
100-150°C
CH2 N-CH3
CH3
CH2
Methylenecyclohexane
+ (CH3 ) 2 NOH
N,N-Dimethylhydroxylamine
23-52
Cope Elimination
• Cope elimination shows syn stereoselectivity but little
or no regioselectivity.
• Mechanism: a cyclic flow of electrons in a sixmembered transition state.
C
-
CH3
C
heat
N+
:O
CH3
Transition s tate
:
H
C
C
H
an alken e
CH3
N
O
N,N-d imeth ylhydroxylamin e
CH3
23-53