Amino compounds by dr. pramod r. padole

For B. Sc. & M.Sc. students

1. AMINO COMPOUNDS:

2. Do you know?
AMINO COMPOUNDS :

3. Contents
•Introduction
•Basicity and effect of substituents
•Methods of preparation of Aniline from
Nitrobenzene
•Reactions of amines
with :
acetyl and benzoyl chlorides,
Br
2(aq) and Br
2(CS
2),
Carbylamine reaction,
Alkylation

4. 4

5. 5

6. Introduction:
(Amines are responsible for the foul smell of dead fish.)
Amines are stronger bases and better nucleophiles
than other neutral organic compounds.
Defination:
Hydrocarbon derivatives of ammonia(NH
3) are
called as amino compounds(amines).
Or
Amines (R-NH
2) are organic nitrogen compounds, formed
by replacing one or more hydrogen atoms of ammonia
(NH
3) with same number of alkyl groups.
When the hydrocarbon group of an amine is alkyl, cycloallyl
or allyl, it is said to be an aliphatic amine;
while if the hydrocarbon group is aryl, the amine is called an
aromatic amine.

7. Classification of Amines:
7
Amines are also classified as
primary(1
0
),
secondary(2
0
) and
Tertiary(3
0
)
according to number of alkyl group attached to the nitrogen atom.NR
H
H
NR
H
R' NR
R''
R'
N
H
3C
H
H
sp
3
hybridised
N
H
3C
H
CH
3
sp
3
hybridised
N
H
3C
CH
3
CH
3
sp
3
hybridised
Methylamine Dimethylamine Trimethylamine
Primary
(1
o
)
Secondary
(2
o
)
Tertiary
(3
o
)

8. 23-8
Structure & Classification
Amines are further divided into aliphatic,
aromatic, and heterocyclic amines:
•Aliphaticamine:An amine in which nitrogen is bonded
onlyto alkyl groups.
•Aromatic amine:An amine in which nitrogen is bonded
to one or more aryl groups.Aniline
(a 1° aromatic amine)
N-Methylaniline
(a 2° aromatic amine)
Benzyldimethylamine
(a 3° aliphatic amine)
NH
2 N-H CH
2-N-CH
3
CH
3
CH
3
::
:

9. Do you know?
Basic Character of Amines:

10. Structure and bonding
Carbon
Four valence electrons
Four equi. sp
3
hybridized orbitals
Each orbital contains one electron
Nitrogen
Five valence electrons
Four equi. sp
3
hybridized orbitals
Three orbitals contain one electron each
Fourth orbital contains two electrons

11. Basic Character of Amines:
11
Amines are basic in nature.
Their basic character is due to presence of lone pair
(unshared pair) of electrons on nitrogen atom due to which
they behave as Lewis base.
Basic strength of the amine depends upon the availability of
lone pair of electronson nitrogen atom. This lone pair of electrons
is available more easily, and then more is the basicity.
Any factor that increases the electron density on the
N atom increases an amine’s basicity.
Any factor that decreases the electron density on N
decreases an amine’s basicity.
Q.1) Discuss / Explain the basic character of Amines. (W-15 & S-16, 2-4 Mark)

12. Basic Character of Amines:
12NR
H
H
Primary Amine
(1
o
)
..
H+
H+
RN
H
H
H
HN
H
H
HNH
H
H
Ammonia
..
More Available
Less Available
RN
H
H
H[
[
OR
HN
H
H
H[
[
OR NH
2 NH
2
....
CH
3 NO
2
electrons
less available
for bonding
electrons
more available
for bonding
(D)
(W)
p-nitroaniline is much weaker base than aniline
Any factor that increases the electron density on the N atom increasesan amine’s basicity.
Any factor that decreases the electron density on N decreasesan amine’s basicity.

13. Basic Character of Amines:
13
Thus amines react with acids; to form salts.
Amine salts on treatment with a base like NaOH or KOH (OH
-
), regenerate
the parent amine.RNH
2 + H
2O + XRNH
3X + OH
..
salt from base
(NaOH / KOH)
Amine RNH
2
H X RNH
3 X
NH
2
H Cl
NH
3Cl
+
Salt
+
Aniline
Quaternary Ammonium Chloride
e.g.
..
Amine Acid
..

14. Calculation of Equilibrium constant or
Dissociation constant orBasicityconstant (K
b):
14
Aqueous solution of ammonia and amines are alkaline in nature
because of the formation of hydroxide ions (OH
-
).
Basic character or Base Strength of amines can be explained
in terms of their K
band pK
bvalues.
Basicityconstantis defined
as the concentration of the
products of ionization in
moles per liter divided by
the concentration of the
un-ionisedbase,
Stronger bases will have higher (larger) values of
basicityconstant (K
b) or smaller value of pK
b.RNH
2+H
2O
RNH
3]
RNH
2]
[OH]
[H
2O]
RNH
3
+OH
K =
pK
b=- log K
b
[
[
RNH
3]
RNH
2]
[OH]
[H
2O]K x =
[
[
RNH
3]
RNH
2]
[OH]
K
b =
[
[
Basicity constant =
K
b
Basicity
pK
bBasicity
..
..
..
..

15. Relation between Basicity& K
b
15
Stronger bases will have higher (larger) values of
basicity constant (K
b) or smaller value of pK
b.NH
3C
H
HNH
3C
CH
3
CH
3
Methyl Amine
(1
o
)
Trimethyl Amine
(3
o
)
NH
H
H
Ammonia
.. ..
Amines are more basic than ammonia because the electron release (donating) by alkyl groups
increases lone-pair availability.
NH
3C
H
CH
3
Dimethyl Amine
(2
o
)
....
K
b =6.7 x 10
-5
54 x 10
-5
37 x 10
-5 1.8 x 10
-5
Low value High value

16. Structure -Basicityrelationship of amines:
16
Basicity of amines is related to their structure.
Basic character of an amine depends upon the formation of the
cation by accepting a proton from the acid.
The more stable the cation is relative to the amine, more basic is
the amine.NR
H
H
Primary Amine
(1
o
)
..
H+
H+
RN
H
H
H
HN
H
H
HNH
H
H
Ammonia
..
More Available
Less Available
RN
H
H
H[
[
OR
HN
H
H
H[
[
OR

17. LOGO
Alkanamines
Versus
Ammonia:
Q.1) Why alkyl amines are more basic than ammonia? (W-12, 1 Mark)
Q.2) Explain why dimethylamine [(CH
3)
2NH] is more basic than methyl amine
[CH
3NH
2] whereas trimethylamine [(CH
3)
3N] is less basic than dimethylamine
[(CH
3)
2NH]? (W-12, 2 Mark)
Q.3) Which is the strongest base. (S-13, ½ Mark)
(a) Methyl amine (b) Dimethylamine (c) Trimethylamine (d) Ammonia
Q.4) Discuss the relative basic character of ammonia and aliphatic amines.
(W-18, 4 Mark)

18. Alkanamines VersusAmmonia:
Alkyl amines (Aliphatic amines) aremore basic
thanammonia:
Aliphatic amines are stronger basesthan
ammonia due to +I effect of alkyl groups,
increasing thehigh electron densityon the
nitrogen atom.
18NR
H
H
Amine
NH
H
H
Ammonia
.. ..
More Available Less Available
Amines are more basic than ammonia because the electron release (donating) by alkyl groups
increases lone-pair availability.

19. Alkylamines are stronger bases than ammonia ?
Due to the electron donating (releasing) nature of alkyl group,
it (R) pushes electrons towards nitrogen and thus makes the
unshared electron pair (lone pair electron) more available for
sharing with the proton of the acid.
Moreover, the substituted ammonium ion formed from the
amine gets stabilizeddue to dispersal of the positive charge by
the +I effect of the alkyl group.
19NR
H
H
Primary Amine
(1
o
)
..
H+
H+
RN
H
H
H
HN
H
H
HNH
H
H
Ammonia
..
More Available
Less Available
RN
H
H
H[
[
OR
HN
H
H
H[
[
OR

20. tertiary amine > secondary amine> primary amine > NH
3
Thus, the basic nature of aliphatic amines should increase with increase
in the number of alkyl groups.
This trend is followed in the gaseous phase.
The order of basicity of amines in the gaseous phase follows the expected order:
tertiary amine > secondary amine> primary amine > NH
3.
20NR
H
HNR
R''
R'
Primary Amine
(1
o
)
Tertiary Amine
(3
o
)
NH
H
H
Ammonia
.. ..
Amines are more basic than ammonia because the electron release (donating) by alkyl groups
increases lone-pair availability.
NR
H
R'
Secondary Amine
(2
o
)
....

21. [email protected]
Order of Basicity:
Case-I
2
o
> 1
o
Secondary
amines (R-NH
2)
are more basic
than Primary
amine
Order
of
Basicity:
Case-II
2
o
> 1
o
> 3
o
Tertiary amines
are less basic
(weaker base)
than Secondary
and Primary
amine
2
o
> 1
o 2
o
> 1
o
> 3
o

22. Case-I): 2
o
> 1
o
Secondary amines (R-NH
2) are more basicthan Primary
amine.
These are because of theloan pair of electrons onnitrogen
atomeasily available fordonation due to the presence of
two electrons donating (releasing) alkyl group.
22NR
H
H
Primary Amine
(1
o
)
..
NR
H
R'
Secondary Amine
(2
o
)
..
>
Most basic
Least basic

23. Case-II): 2
o
> 1
o
> 3
o
Tertiary aminesare less basic (weaker base) than
Secondary and Primary amine
23Least basic
Most basic
NR
H
H NR
R''
R'
Primary Amine
(1
o
)
Tertiary Amine
(3
o
)
NH
H
H
Ammonia
.. ..
NR
H
R'
Secondary Amine
(2
o
)
.. ..
> > > NH
3C
H
H NH
3C
CH
3
CH
3
Methyl Amine
(1
o
)
Trimethyl Amine
(3
o
)
NH
H
H
Ammonia
.. ..
NH
3C
H
CH
3
Dimethyl Amine
(2
o
)
.. ..
K
b 6.7 x 10
-5
54 x 10
-5
37 x 10
-5 1.8 x 10
-5
> > >

24. 2
o
> 1
o
> 3
o
Why?
3
o
amine is less basic
than
2
o
amine & 1
o
amine

25. Do you know?
Tertiary aminesare less basic (weaker base)
than
Secondary and Primary amine

26. Tertiary aminesare less basic(weaker base) than
Secondary and Primary amine
Stability of
Ammonium ion:
depends upon the
solvationenergy
Moreis
the solvationenergy;
stableare
the ammonium ions.
Stericeffect:
3
o
amines,
There are three bulky alkyl groups
(overcrowding of the three alkyl
groups attached to nitrogen)
These causes sterichindrance due
to this, a protection of nitrogen
of tertiary amine
become relatively difficult.
3
o
amine
Less basic
than1
o
& 2
o

27. LOGO
More is the solvation energy; stable are the ammonium ions.
In the aqueous solution, the substituted ammonium cationgets
stabilized not only by electron releasing effect of the alkyl group (+I) but
also by solvationwith water molecules.
The greater the size of the ion, lesserwill be the solvationand the less
stabilized is the ion.
The orders of stability of ions are as follows:RN
H
H
H
O H
RN
H
H
R'
RN
H
R'
R''
1
0
2
0
3
0
H
O H
H
O H
H
O H
H
O H
H
O H
H
Greater is the stability of the substituted ammonium cation,
stronger should be the corresponding amine as a base.

28. LOGOOrder of basicityof aliphatic amines
Thus, the order of basicity of aliphatic amines: Solvation energy primary > secondary > tertiary
Inductive effect tertiary > secondary > primary
Steric Hindrance or effect tertiary > secondary > primary Least basic
Most basic
NR
H
H NR
R''
R'
Primary Amine
(1
o
)
Tertiary Amine
(3
o
)
NH
H
H
Ammonia
.. ..
NR
H
R'
Secondary Amine
(2
o
)
.. ..
> > >

29. 29

30. LOGOBy Dr. P. R. Padole
Arylamines Vs Ammonia:
Q.1) Aniline is a weakerbase than methyl amine. (S-12, S-14 & W-16, ½ Mark)
Q.2 Explain: Aniline is a weaker base than aliphatic amines. (S-12, 2 Mark)
Q.3) Explain why, aromatic primary amine is weaker base than aliphatic amine. (W-14, 2 Mark)
Q.4) Explain why: Methyl amine is much stronger base than aniline? (W-16, 2 Mark)
Q.5) Explain why: p-nitroaniline is much weaker base than aniline? (W-16, 2 Mark)
Q.6) Aniline is ________ in nature. (W-17, ½ Mark)
(a) Acidic (b) Basic (c) Neutral (d) Amphoteric
Q.7) The nature of Aniline is: (W-19, ½ Mark)
(a) Acidic (b) Basic (c) Phenolic (d) Neutral

31. LOGO
Arylamines Versus Ammonia:
Aromatic aminesare weaker basesthan ammonia
due to the electron withdrawing nature of the aryl group
(benzene ring) by resonance effect (+ R-effect).
31NH
2
Aniline
..
+H
(Weak Base)
(from Acid)
lone pair of electrons present on N
is Less available or
not easily available for protonation
NH
3
Ammonium cation
(Salt)
So, it is weaker than
aliphatic amine
Aromatic Amine (Stronger Base)
(Salt)
NR
H
H
Aliphatic Amine
..
H
+
+
RN
H
H
H
More easily Available

32. LOGO
Arylamines Versus Ammonia:
In aniline (aromatic amine), -NH
2group is attached
directly to the benzene ring because of that the
lone pair of electrons present on nitrogen atom is
involved in the conjugation with the π-bondof the
benzene ring by + R-effect and thus making it
less available for protonation.
32NH
2
Aniline
.. NH
2
Aniline
..
Not easily available for protonation
H+ HN
H
H
HNH
H
H
Ammonia
..
More Available
(Weak base)
(Strong base)
<
lone pair electrons
less available on N atom
lp electrons
easily involved
in the
delocalization

33. LOGO
Arylamines Versus Ammonia:
Hence aromatic amines are less basic than aliphatic
amines which show no resonance.
If we write different resonating structures of aniline, we will find that
aniline is a resonance hybrid of the following five structures.
On the other hand, anilinium ion obtained by accepting a proton can
have only two resonating structures (Kekule structure).
33NH
2 NH
2
NH
2
I II III
NH
2
IV
V
Electron density on N is less
H N H
[
[
Resonance hybrid



 NH
2
Aniline
..
+H
(Weak Base)
(from Acid)
Aromatic Amine
NH
3
I
NH
3
II
[
[
Two resonating structures

34. LOGO
Arylamines VersusAmmonia:
We know that greater the number of resonating structures, greater is
the stability. Thus aniline(more resonating structures) is more stable
than aniliniumion. Hence, the proton acceptability or the basic nature
of aniline or other aromatic amines would be less than that of ammonia.
Hence aromatic amines are less basic than ammonia
which shows no resonance.
In case of substituted aniline, it is observed that electron donating (releasing) groups (D)
like –OCH
3, –CH
3increase basic strength
whereas electron withdrawing groups (W) like –NO
2, –SO
3H, –COOH, –X
decreasethe basic strength (basicity) of aromatic amines.
34NH
2 NH
2
....
CH
3 NO
2
electrons
less available
for bonding
electrons
more available
for bonding
(D)
(W)
p-nitroaniline is much weaker base than aniline

35. LOGO
Preparation of Aniline fromNitrobenzene:
Reduction of Nitrobenzene:
Aniline

36. LOGO
Preparation of Aniline fromNitrobenzene:
Reduction of Nitro compounds:
When nitrobenzeneis reduced by the reducing
reagent like,
(i) H
2in the presence of Ni, Pd or Pt as catalyst,
(ii) Zn or Fe or Snand HCl;
(iii) Lithium Aluminum hydride (LiAlH
4); to form aniline.
36
Aniline
Q.1) Explain: Conversion of nitrobenzene to aniline using Ni catalyst. (S-12, 2 Mark)
Q.1) How will you prepare the Aniline from nitrobenzene? (W-14, 2 Mark)NO
2 NH
2(i) H
2 / Ni or Pd or Pt as catalyst
(ii)Zn/HCl or Fe/ HCl or Sn/HCl
(iii) LiAlH
4
Aniline
Nitrobenzene
+ 6 [H]
+ 2 H
2O
Reduction by
or
3 H
2

37. LOGOElectrophilic Substitution in the Ring:
The amino group (-NH
2) activates the benzene ring by
resonance mechanism (+R-effect) and the ortho–para
positionsare centers of high electron density.
37
Q.1) In aniline electophilicsubstitution occurs at: (S-15, ½ Mark)
(a) Ortho-para (b) Meta (c) All positions (d) None of these.NH
2 NH
2
NH
2
I II III
NH
2
IV
V
Electron density on N is less
H N H
[
[
Resonance hybrid





38. LOGO
Aniline

39. Chemical Reactions of Aniline:
AnilineAcetanilide
Benzanilide
Phenyl
isocyanide
N-methyl
aniline
Br
2
Aqueous Br
2Br
2 in CS
2
2,4,6-
tribromoaniline
Ortho & para-
bromoaniline
1
Reaction with
Acetyl chloride
Or Acetic anhydride
Reaction with
CHCl
3and
alco. KOH

40. LOGO
(i) Preparation of Acetanilidefrom Aniline:
Or
Action of Acetyl Chloride or Acetic Anhydride
on Aniline:
Or
Reaction with Acetyl chloride:
AcylationReaction
Q.1) Explain the Conversion of aniline to Acetanilide.(W-13, 2 Mark)
Q.2) How does aniline react with Carbonyl chloride?(W-16, 2 Mark)
Q.3) How will you prepare Acetanilide from Aniline?(S-17, 2 Mark)
Q.4) What happen when Aniline is treated with Acetyl chloride?(W-19, 2 Mark)

41. LOGO
Preparation of Acetanilide fromAniline:
Reaction with Acetyl chloride (Acylation Reaction):
Whenanilineis reacted with acetyl chloride (or acetic anhydride);
to form acetanilide.
41N
H
H
ClCCH
3
O
CCH
3
O
N
H
Aniline
+
Acetanilide
Acetyl chloride + HCl
ORN
H
H
C
O
CCH
3
O
N
H
Aniline
+
Acetanilide
Acetic anhydride
+ CH
3COOH
CH
3
C
O
CH
3
O
gacial
CH
3COOH

42. LOGO
Q.1) How will you convert Aniline to benzanilide? (S-12, S-13, W-13, S-14, W-14, S-15, W-15 & W-18, 2 M)
Q.2) Complete the following reaction. (W-16, 2 Mark)
Q.3) How will you prepare Benzanilidefrom Aniline? (S-17, 2 Mark)
(ii) Action of Benzoyl chloride:
OrPreparation of Benzanilide fromAniline:
Or Reaction with Benzoyl chloride:NH
2
COCl
NaOH
?

43. Preparation of Benzanilide from Aniline:
Reaction with Benzoyl chloride:
Whenanilineis reacted with benzoyl chloride in
presence of NaOH; to form benzanilide.
This reaction is called Benzoylation Reaction.
43N
H
H
ClCC
6H
5
O
CC
6H
5
O
N
H
Aniline
+
Benzanilide
Benzoyl chloride
+ HCl
NaOH

44. LOGO
Alkyl halides (Alkylation):
Reaction with
Preparation of N-methyl aniline:
Preparation of N-dimethyl aniline:
Preparation of Quaternary ammonium iodide:
Aniline

45. Reaction with Alkyl halides
(Alkylation Reaction):
When Anilineis heated with methyl iodide under
pressure; to form N-methyl aniline (2
0
amine),
which is then converted to dimethyl aniline (3
0
amine) and finally to quaternary ammonim iodide
(salt).
45
Q.1) How will you obtain N-methyl aniline form Aniline? (S-17 & W-19, 2 Mark)N
H H
N
CH
3 H
N
CH
3
CH
3CH
3N
CH
3 CH
3
Aniline
N-Methyl aniline
+ CH
3-I
Quaternary ammonium
Iodide
N-Dimethyl
aniline
I
- HI - HI
.. .. ..
+ CH
3-I + CH
3-I

46. Do you know?
Carbylamines Reaction

47. Carbylamines Reaction
(Isocyanides Test):
47
OrPreparation of Phenyl isocyanidefrom Aniline
by CarbylamineReaction:
Q.1) How will you convert Aniline to phenyl isocyanideby Carbylaminereaction? (S-12 & S-14, 2 Mark)
Q.2) How will you convert Aniline to phenyl isocyanide? (S-13 & S-15, 2 Mark)
Q.3) Explain the Carbylaminereaction of aniline. (W-13 & S-15, 2 Mark)
Q.4) How will you prepare phenyl isocyanidefrom aniline? (W-17, 2 Mark)
Q.5) What happen when Aniline is treated with Chloroform and alcoholic KOH? (W-19, 2 Mark)
Q.6) The reaction of aniline with chloroform and alcoholic caustic potash solution is
called as Carbylamines reaction
When aniline is reacted with chloroform (CHCl
3) and
alcoholic KOH solution (caustic potash solution); to form
phenyl isocyanide(phenyl carbylamines).Aniline
NH
2
+
Phenyl isocyanide
Alcohol
NC
CHCl
3+3 KOH +3 KCl+3 H
2
O

48. Reaction with Bromine (Br
2):
Reaction with
aqueous Bromine:
Or
Preparation of
2,4,6-tribromo
Aniline
from Aniline:
Bromine
Br
2
Reaction with
Bromine in CS
2:
Or
Preparation of
ortho and para-
bromo Aniline
from
Aniline:

49. Reaction with aqueous Bromine:
OrPreparation of 2,4,6-tribromo Aniline fromAniline:
When Anilineis reacted with aqueous solution of
Bromine (Br
2); to form 2,4,6-tribromoaniline.
Q.1) What happens when aniline is treated with aqueous Br
2? (S-17 & S-18, 2 Mark)
Q.2) How will you prepare 2,4,6-tribromoaniline from Aniline? (W-17, 2 Mark)
Q.3) Complete the following reaction and predict the product: (S-19, 2 Mark)NH
2
NH
2
Br Br
Br
Aniline
+3 Br
2 (aq)
2,4,6-Tribromoaniline (White ppt)
H
H
H
+ 3 HBr
Br----Br
Br----Br
Br----Br ?
NH
2
+ Br
2(aq)

50. Reaction with Bromine in CS
2:
OrPreparation of ortho and para-bromo Aniline from Aniline:
When Anilineis treated or reacted with Bromine (Br
2) in Carbon
disulphide (CS
2); to form mixture of ortho and para-bromo aniline
(2-bromoaniline and 4-bromoaniline).

Q.1) How does aniline react with Carbon disulphide (CS
2)? (W-16, 2 Mark)
Q.2) What happens when aniline is treated with Br
2in Carbon disulphide (CS
2)?
(S-17 &S-18, 2 Mark)
Q.3) How will you prepare orthoand parabromoanilinefrom aniline? (W-17, 2 Mark)NH
2
NH
2
Br
NH
2
Br
Aniline
+
2-Bromoaniline
(o-bromo aniline)
in CS
2
+
4-Bromoaniline
(p-bromo aniline)
Br
2

51. LOGO
Hoffmann’s Exhaustive Methylation:
Hoffmann’s Degradation or
Hoffmann’s Rearrangement:
Q.1) Explain Hoffmann’s exhaustive methylation for opening of heterocyclic ring.
(S-14, 4 Mark)
Q.2) Explain Hoffmann’s exhaustive methylation with mechanism.
(S-15 & S-16, 4 Mark)
Q.3) Explain Hoffmann’s exhaustive methylation reaction. (S-19, 4 Mark)

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Hoffmann’s Exhaustive Methylation:
Hoffmann’s Degradation orHoffmann’s Rearrangement:
The complete process of converting an amine (1
0
, 2
0
or 3
0
),
to an unsaturated compoundby the pyrolyticdecomposition
(by heating) of the quaternary ammonium hydroxide
(obtained by treating amine with the excess of methyl
iodide and then hydrolyzing the resulting quaternary
ammonium iodide with moist silver oxide) is called the
Hoffmann’s Exhaustive Methylationor the Hoffmann’s
Degradationor Hoffmann’s rearrangement.
52C
H
C
NH
2
CH
3I
C
H
C
N(CH)
3 I
C
H
C
N(CH
3)
3OH
CC+ N(CH
3)
3
H
2O
(excess)
Moist
Ag
2O Heat
+
Amine
(1
o
, 2
o
or 3
o
)
unsaturated
compound
Hydrolysis Decomposition

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Mechanism of Hoffmann’s Exhaustive Methylation:
Step-1) Reaction with excess of Methyl iodide by SN
2
reaction:
The conversion of an amine(1
0
, 2
0
or 3
0
) to quaternary
ammonium iodide by treating the amine with excess of methyl
iodide is an SN
2
reaction, here the amineis the incoming
nucleophileand iodineis the leaving nucleophile.
Step-2) Reaction with moist silver oxide is a simple ion -exchange reaction:
The conversion of the quaternary ammonium iodide to quaternary
ammonium hydroxide by the action of moist silver oxide is a simple
ion-exchange reaction.
53Amine
(1
o
, 2
o
or 3
o
)
CCN
H H
H
CH
3I CCN
H H
H
CCN
H H
H
CH
3
CCN
H CH
3
CH
3
CH
3CH
3
:+
 + CH
3I
excess

I II
quaternary ammonium iodide
 CCN
H CH
3
CH
3
CH
3I
quaternary ammonium iodide
CCN
H CH
3
CH
3
CH
3+
Ion exchange reaction
+AgOH OH AgI
Moist
Ag
2O
quaternary ammonium hydroxide

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Mechanism of Hoffmann’s Exhaustive Methylation:
Step-3) Pyrolyticdecomposition; to form unsaturated compound:
 The pyrolyticdecomposition of the quaternary ammonium
hydroxide toan unsaturated compound , water and 3
0
amine is a
-elimination reaction which usually follows the E
2pathway;
and so the Hoffmann’s product are usually obtained in the pyrolytic
decomposition of quaternary ammonium hydroxide.
54CCN
H CH
3
CH
3
CH
3OH
quaternary ammonium hydroxide
CCN
H CH
3
CH
3
CH
3 CC N
CH
3
CH
3
CH
3+ +
HO
H
2O
Pyrolytic
decomposition
By E
2
elimination reaction
:
unsaturated
compound

55. 55