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
23-1
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
23-2
-
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
23-3
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
23-4
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.
23-5
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.
23-6
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
23-7
Secondary Amines with HNO2
• 2° Aliphatic and aromatic amines react with NO+ to
give N-nitrosamines.
N-H + HNO2
Piperidin e
N-N=O + H2 O
N-N itrosopiperidin e
carcinogens
23-8
Amines with HNO2
 Reaction
of a 2° amine to give an N-nitrosamine.
• Step 1: Reaction of the 2° amine (a nucleophile) with
the nitrosyl cation (an electrophile).
• Step 2: Proton transfer.
•
•
+
•
•
••
+
N O
(1)
+
N
N
••
N
N=O
•
•
N=O
•
•
••
•
•
••
••
•
•
H
H
+
+ H O H
H
••
(2)
•
•
H O
H
23-9
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
23-10
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
23-11
1° RNH2 with HNO2
 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%)
23-12
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
23-13
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
23-14
Pinacol Rearrangement: an example of
stabilization of a carbocation by an adjacent
lone pair.
Overall:
23-15
Mechanism
Reversible
protonation.
Elimination
of water to
yield tertiary
carbocation.
This is a
protonated
ketone!
1,2
rearrangement
to yield
resonance
stabilized
cation.
Deprotonation.
23-16
1° ArNH2 with HNO2
-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
23-17
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
23-18
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
23-19
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
23-20
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
23-21
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
23-22
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
23-23
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.
23-24
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
23-25
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
23-26