1390 Nitrogen Containing Compounds
(a)
(c)
(d)
(e)
If both assertion and reason are true and the reason is the correct
explanation of the assertion.
If both assertion and reason are true but reason is not the correct
explanation of the assertion.
If assertion is true but reason is false.
If the assertion and reason both are false.
If assertion is false but reason is true.
1.
Assertion
:
Reason
:
2.
Assertion
Reason
:
:
3.
Assertion
:
Reason
:
Assertion
:
Reason
:
Assertion
Reason
Assertion
:
:
:
Reason
:
(b)
Benzene diazonium chloride does not give tests
for nitrogen:
N 2 gas lose takes place during heating
or hydroxyazo compounds is called coupling
reaction.
Condensation of diazonium salt with phenol is
carried out in weakly acidic medium.
Carbylamine reaction involves the reaction
between 1° amine and chloroform in basic
medium.
In carbylamine reaction,  NH 2 group is
converted into –NC group.
Me3 N reacts with BF3 whereas Ph3 N does
not.
The electron pair on nitrogen atom in Ph3 N is
delocalised in the benzene ring and is not
available to boron in BF3
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
:
17.
Assertion
Reason
:
:
p-Anisidine is weaker base than aniline.
OCH 3 group in anisidine exerts –R effect.
18.
Assertion
:
Reason
:
Lower aldehydes and ketones are soluble in water
but the solubility decreases as the molecular
mass increases.
Distinction between aldehydes and ketones can
be made by Tollen’s test.
Assertion
:
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
:
15.
16.
[AIIMS 1999]
4.
5.
6.
7.
Assertion
:
Amines are basic in nature.
Presence of lone pair of electron on nitrogen
atom.
[AIIMS 1999]
Methyl isocyanide reacts with ozone to form
methyl isocyanate.
Methyl isocyanate was responsible for Bhopal
tragedy.
Alkyl cyanide can be prepared by carbylamine
reaction
Ethyl amine when heated with chloroform in
presence of alcoholic KOH, cyanide is formed.
CN  ion is an ambident nucleophile.
Nucleophiles are electron rich species.
Sulphanilic acid exists as dipolar ion whereas paminobenzoic acid does not.
Carboxyl group being more acidic than
SO 3 H group can easily transfer a H  to
the amino group.
Nitrating mixture used for carrying our nitration
of benzene consists of conc. HNO 3  conc.
:
In presence of H 2 SO 4 , HNO 3 acts as a base
and produces NO 2
8.
Assertion
:
Reason
:
9.
Assertion
:
10.
Reason
Assertion
:
:
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
:
Assertion
Reason
Assertion
:
:
:
11.
12.
13.
14.
20.
21.
H 2 SO 4 .
Reason
[AIIMS 1999]
19.

Aniline hydrogen sulphate on heating forms a
mixture of ortho and para aminobenzene
sulphonic acids.
The sulphonic acid group is electron
withdrawing.
[AIIMS 1996]
p  O 2 N  C 6 H 5 COCH 3 is prepared by
Friedel Crafts acylation of nitrobenzene.
Nitrobenzene easily undergoes electrophilic
substitution reaction.
[AIIMS 2005]
Alkyl isocyanides in acidified water give alkyl
formamides.
In isocyanides, carbon first acts as a nuclephile
and then as an electrophile.
ions.
[AIIMS 2005]
In order to convert R–Cl to pure R–NH , Gabriel
pthalimide synthesis can be used.
With proper choice of alkyl halides, pthalimide
synthesis can be used to prepare 1°, 2° or 3°
amines.
Ammonolysis of alkyl halides involves the
reaction between alkyl halides and alcoholic
ammonia.
Reaction can be used to prepare only 2° amines.
Nitroalkanes, but not nitroarenes can be distilled
at normal atmospheric pressure.
Nitroalkanes are sparingly soluble in water while
nitroarenes are insoluble.
In Hofmann bromide reaction, the amine formed
has one carbon atom less than the parent 1°
amide.
N-methyl acetamide undergoes Hofmann
bromamide reaction.
Nitrobenzene does not undergo Friedel Craft
alkylation.
Nitrobenzene is used as solvent in laboratory and
industry.
Ammonia is less basic than water.
Nitrogen is less electronegative than oxygen.
The reaction between a diazo salt and an
aromatic amine or a phenol, giving an aminoazo
2
Introduction of Nitrogen Containing Compounds
1
a
2
c
3
a
4
d
5
c
6
d
7
c
8
d
9
c
10
a
11
a
12
b
13
b
14
a
15
b
Preparation of Nitrogen Containing Compounds
1
c
2
d
3
b
4
b
5
b
6
b
7
c
8
a
9
c
10
a
11
d
12
d
13
b
14
b
15
d
16
c
17
a
18
b
19
a
20
c
21
b
22
a
23
a
24
a
25
b
26
b
27
c
28
d
29
c
30
a
Nitrogen Containing Compounds
1391
31
a
32
a
33
a
34
d
35
c
11
a
12
b
13
c
14
a
15
a
36
c
37
b
38
a
39
d
40
b,c
16
c
17
b
18
c
19
d
20
a
41
d
42
b
43
c
44
c
45
c
21
c
22
d
23
a
46
b
47
b
48
a
49
c
50
a
51
c
52
a
53
c
54
a
55
b
Properties of Nitrogen Containing Compounds
1
d
2
d
3
b
4
b
5
a
6
c
7
c
8
d
9
b
10
c
11
c
12
b
13
b
14
c
15
c
16
a
17
d
18
a
19
b
20
c
21
a
22
d
23
b
24
c
25
a
26
b
27
b
28
c
29
b
30
b
31
d
32
a
33
c
34
a
35
b
36
c
37
b
38
b
39
d
40
c
41
a
42
c
43
b
44
c
45
d
46
d
47
d
48
b
49
d
50
b
51
d
52
d
53
c
54
c
55
c
56
d
57
d
58
c
59
c
60
c
61
b
62
a
63
c
64
a
65
b
66
c
67
e
68
c
69
a
70
c
71
d
72
c
73
a
74
b
75
b
76
a
77
a
78
b
79
c
80
b
81
b
82
d
83
a
84
b
85
d
86
b
87
b
88
b
89
b
90
b
91
c
92
d
93
d
94
c
95
c
96
b
97
c
98
c
99
b
100
c
101
d
102
a
103
c
104
b
105
a
106
a
107
a
108
d
109
b
110
c
111
c
112
a
113
c
114
a
115
d
116
c
117
d
118
b
119
a
120
b
121
b
122
d
123
b
124
d
125
d
126
b
127
d
128
b
129
c
130
d
131
a
132
d
133
b
134
a
135
b
136
c
137
c
138
b
139
b
140
c
141
d
142
b
143
a
Tests for Nitrogen Containing Compounds
1
c
2
b
3
b
4
a
5
b
6
a
7
b
8
a
9
d
10
d
11
a
Critical Thinking Questions
1
c
2
c
3
a
4
c
5
b
6
b
7
b
8
a
9
bc
10
a
Nitrogen Containing Compounds
Assertion & Reason
1
a
2
a
3
b
4
d
5
b
6
c
7
a
8
c
9
c
10
b
11
c
12
b
13
e
14
c
15
a
16
a
17
d
18
b
19
d
20
d
21
a
1393
NaOH / Br
4.
2
(b) CH 3  CO  NH 2 
 CH 3  NH 2
Hofmann' s bromamide
(1c)
(2c)
5.
Na  C 2 H 5 OH
(b) CH 3 C  N  4[H ] 

 CH 3 CH 2 NH 2
6.
(b) CH 3  CH 2  CO  NH 2  Br2  4 KOH 
Reduction
Propionami de
CH 3 CH 2 NH 2  K 2 CO 3  2 KBr  2 H 2 O
7.
(c) C 2 H 5 I  NH 3  HI  C 2 H 5  NH 2
C 2 H 5 OH  NH 3  H 2 O  C 2 H 5  NH 2
9.
Sn / HCl

(c) CH 3  CH 2  NO 2  6 [H ] 
Nitro ethane
CH 3  CH 2  NH 2  2 H 2O
Ethyl amine
11.
NH
CH I
3
3
(d) CH 3 I 
 CH 3 NH 2 
 (CH 3 )2 NH

Methyl amine
Dimethyl amine
Introduction of Nitrogen Containing Compounds
4.
CH 3 I

 (CH 3 )3 N
Trimetyhlamine
(d) C 3 H 9 N can form all the 3 amines.
CH 3 CH 2 CH 2  NH 2 , CH 3  CH 2  NH  CH 3
1
CH 3
o
N

o
amine

|
2 amine
12.
(d)
CH 3
+
CH 3  CO
O
CH 3  CO
Aniline
CH 3
o
3 amine
5.
NH  CO  CH 3
NH 2
(c) (CH 3 ) 2 C  O  H .CH 2  COCH 3  NH 3

(CH 3 )2 C CH 2 COCH 3
16.
Acetanilide

Reduction

 CH 3  NH  CH 3
(c) CH 3  N  C  4 [H ] 
2
17.
|
o
amine
(a) CH 3 NO 2  3Cl 2  3 NaOH 
Nitromethane
NH 2
CCl 3  NO 2  3 NaCl  3 H 2 O.
diacetonea mine
8.

(d) Allyl isocyanide. CH 2  CH  CH 2  N  C
9.
(c)
(chloropicrin)
18.
(b)
NH 2
NH 2
+ CH 3 COOH
Aceticanhydride
1o
Reduction
[H]
R  CH 2  NH 2
Hydrolysis
R  COOH  NH 3
1 o amine
R C  N
Nitrile
NH 2

o
12.
13.
15.
1 amine
(b) CH 3 CH 2  O  N  O is a nitrite derivative, hence it
is not a nitro derivative.
(b) CH 3 CN is called acetonitrile....
19.
(a) C 2 H 5 OH  NH 3  C 2 H 5 NH 2  H 2 O .
20.
(c) R  CN  H 2 O 2 RCOOH  NH 3
H O/H

It yield amine when reduced as –
R  CN  H 2  R  CH 2  NH 2
(b) Four 1 o amines are possible
CH 3 CH 2 CH 2 CH 2 NH 2 , (CH 3 )2 CHCH 2 NH 2 ,
CH 3 CH (NH 2 )CH 2 CH 3 , (CH 3 )3 CNH 2
acid
H3O
alumina
21.
PO
4H
2 5
(b) CH 3 CONH 2 
 CH 3 CN 
 CH 3 CH 2 NH 2
O
Preparation of Nitrogen Containing Compounds
1.
(c) Hofmann’s bromamide reaction
CH 3  CO  NH 2  Br2  4 KOH 
||
22.
(a) CH 3  CH 2  N  O  3 H 2  CH 3 CH 2 NH 2  2 H 2 O .
23.
2
 CH 3 OH  N 2  H 2 O
(a) CH 3 NH 2 
NO
Methyl amine
H 2O
Acetamide
CH 3 NH 2  K 2 CO 3  2 KBr  2 H 2 O
24.
(a) R  NH 2  CHCl 3  3 KOH  R  NC  3 KCl  3 HO
1 o  amine
Methyl amine
2.
NaOBr
(d) CH 3 CONH 2 

 CH 3 NH 2 .
3.
(b) CH 3 CONH 2  Br2  4 NaOH 
NH 2
25.
(b)
NaNO
Methyl amine
/ HCl
2

Acetamide
CH 3 NH 2  Na 2 CO 3  2 NaBr  2 H 2 O
N 2Cl
0
Aniline
o
o
5 C
+ 2 H 2O .
Benzene diazonium chloride
1394 Nitrogen Containing Compounds
26.
SOCl
2
(b) CH 3 CH 2 COOH 

 CH 3 CH 2 COCl  SO 2  HCl
51.
CH 3 CH 2 COCl  NH 3  CH 3 CH 2 CONH 2  HCl
CH 3 CH 2 CONH 2  Br2 / NaOH  CH 3 CH 2 NH 2  CO 2
Ethyl amine
27.
PCl5
(c) Secondary amines gives oily nitrosomine with nitrous
acid.
(CH 3 CH 2 )2 NH  HONO  (CH 3 CH 2 )2 N . NO  H 2 O
oily
52.
NH 3
(c) CH 3 COOH  CH 3 COCl  CH 3 CONH 2
(a) When aniline is treated with HNO 2 at 0–5°C then
diazonium salt is formed and by the coupling of
diazonium salt and phenol azo dyes are prepared.
NaOBr


 CH 3 NH 2
NH 2
Br
28.
(d)
Br
Br
Br
NaNO 2  HCl

 

0 5 C
NaNO 2
Br2



HCl
Benzene
diazonium chloride
Aniline
Br
Br
Br
Boiling



Br
N   NCl 
C 2 H 5 OH
29.
N   NCl 
NH 2
N 2 Cl
NH 2
Br
LiAlH4
 CH 3  NH  CH 3
(c) CH 3  N  C 
 CHCl 3  3 KOH 
 3 KCl  3 H 2 O .
(a)
35.
(c) Methyl amine is the strongest base.
36.
2
(c) C6 H 5 NO 2  6 H 

 C6 H 5 NH 2  2 H 2 O
38.
39.
HO
54.
(a) p-nitrobenzene from p-nitroaniline.
HCl , 0  5 C
(a) C2 H 5 NH 2  CS 2  HgCl2  C2 H 5 NCS  2 HCl  HgS .
NaNO 3
N2+Cl–
NH2
(d) C6 H 5 NH 2  C6 H 5 N 2 Cl 
H 2O
C6 H 5 OH  N 2  HCl
KI


NaNO 2

o 
Aniline
p-nitroaniline
55.
NO2
NO2
NO2
HCl
41.
N
N
N
Azo dye
pt / H
Nitrobenzene
weakly alkaline,OH 
Phenol
Na  ROH
(a) CH 3 CONH 2 

 CH 3 CH 2 OH  H 2 O

NH 2
N C
33.
coupling


+
sec. amine
32.
OH
I
p-nitroiodobenzene
alcohol
(b) C 2 H 5 Br  KCN 

 C 2 H 5 CN  KBr
H O
(d) 2C6 H 5 Cl  2 NH 3 2

in xylene 570 o K
chlorobenzene
Properties of Nitrogen Containing Compounds
C6 H 5 NH 2  Cu 2 Cl 2  H 2 O
(Aniline)
1.
44.
(c) C6 H 5 NH 2  HCl  NaNO 2  C6 H 5 N 2Cl
46.
Sn  HCl
(b) C6 H 5 NO 2  6 H 
 C6 H 5  NH 2  2 H 2 O
2.
o
NaNO 2  HCl 0 C
47.
(b) C6 H 5 NH 2 
 C6 H 5 N 2 Cl
48.
 CH 3 NH 2  2 H 2 O
(a) CH 3 NO 2  6 H 
3.
Sn
(d) Tertiary amine does not react with nitrous acid
because in it -H atom is absent.
(d) Due to +ve I.E. of alkyl group, N-atom of amines
acquires patrial –ve charge and thus electron pair is
easily donated.
(b) CH 3  C H  COOH
|
HCl
49.
NH 2
2
 C6 H 5 N 2 Cl  


(c) C6 H 5 NH 2 2
o
o
(X)
0 C 5 C
H 2O
NaNO
/ HCl
HNO
The compounds in which both amino ( NH 2 ) as well
as acidic (–COOH) group is present is called amino
acid.
C6 H 5 NO 2  N 2  HCl
(Y)
50.
(a) Halogen have –I and +M effect by which its electron
delocalized in benzene ring by resonance & due to its
–I effect its bonded with benzene ring and cannot be
substitute by CN  & show the inertness against
KCN while other option gives Aromatic nitrile
4.
ON O

O  N  O :





O  N  O : O  N  O :
+
+
+
Presence of  NO 2 group decreases electron density
at o- and p- positions. Hence, incoming electrophile
goes to m position. Therefore it is m-directing group.
ArN2  CuCN  ArCN  N 2  Cu 
P2 O5
ArCONH 2 
 ArCN
 H 2O
ArCONH 2  SOCl 2  ArCN  SO 2  2 HCl
(b)
6.
Hydrolysis

 R  COOH  NH 3
(c) R  C  N  2 H 2 O 
NH 2
Aniline
N  NCl
Benzene diazonium
chloride
1395
Nitrogen Containing Compounds
7.
(c)
23.
0o C
+ HNO 2  HCl 
+ 2 H 2O
Diazotization
24.
9.
(b) R  C  N  R  Mg  X  R  C  N  Mg  Br
Alkylcyanide
|
25.
(b)
CH NH
Red
(Litmus paper)
3
2



 Blue
This litmus paper test shows basic nature of amine.
(c) Presence of alkyl group increases electron density on
nitrogen atom due to +I effect. Thus basic nature
increases.
(a) Mustard oil reaction
HgCl
2
CH 3  CH 2  NH 2  CS 2 


R
Hydrolysis
Ethyl amine
CH 3  CH 2  N  C  S  H 2 S
Br
OH
R  CO  R  NH 3  Mg
Ketone
Ethyl isothiocyanate
NO 2
10.
(c)
+ 3 NHO 3 
fuming
NO 2
+ 3 H 2O
28.
(c)
NO 2
NO 2
sym-trinitro benzene
11.
KMnO
[O ]
3- nitro group can be introduced.
Aldimine
H 3O
NO 2
1, 3, 5-trinitrobenzene
4
(c) CH 3  CH 2  NH 2 
 CH 3  CH  NH
Ethylamine
Nitration


NO 2


 CH 3  CHO
34.
HNO
2
(a) R  CH 2  NO 2 

 R  C  NO 2
||
o
1 nitro
N  OH
Acetaldehyde
12.
13.
(b) Only primary aromatic amines
diazotisation.
(b) R  CH 2  NH 2  O  CH  R 
can
Nitroloic acid
undergo
NaOH


 R  C  NO 2
||
N  O  Na 
aldehyde
o
(Blood red)
1 amine
R  CH 2  N  CH  R  H 2O
Aldimine
14.
NaOH


 Blue colour.
NO2
conc. H 2 SO 4
+ HNO 3 

(c)
Nitrobenzene
PO
37.
2 5
 CH 3  C  N  H 2 O .
(b) CH 3 CONH 2 
38.
(b) CHCl 3  C2 H 5 NH 2  3 KOH 
Acetamide
NO2
conc.
Acetonitrile

C2 H 5 N  C  3 KCl  3 H 2 O
m-dinitrobenzene
 NO 2 group is meta directing group.
17.
|
N O
Acetic acid
NO2
16.
(R)2 C  NO 2
(b) (R)2 CH  NO 2 
(c) CH 3  CO  NH 2  HNO 2  CH 3 COOH  N 2   H 2O
Acetamide
15.
35.
HNO 2
Ethyl isocyanide
Sn / HCl
 ROH  NH 3  H 2 O
(a) R  O  N  O  6 [H ] 
Alkyl nitrite
Alcohol
(d) CH 3  CH 2  NH 2  HCl  CH 3 CH 2  NH 3 Cl 
39.
(d) CH 3  C  N  2 H 2 O  CH 3 COOH  NH 3
40.
(c) CH 3 CONH 2  Br2  4 KOH 
Methyl cyanide
Aceticacid
(2c )
Ethyl ammonium chloride
Amines are basic in nature they react with acid to form
salt.
18.

(a) (CH 3 ) 2 NH  CH 3 NH 2  (CH 3 ) 3 N  N H 3
2 o amine is
most basic
1 o amine
o
3 amine
20.
(b)
+ CHCl 3  3 KOH 
Aniline
(1c)
41.
(a) (C2 H 5 )2 NH  (aq.) HONO  (C2 H 5 )2 N  N  O  H 2 O
42.
Na  EtOH

 CH 3  CH 2  NH 2
(c) CH 3  C  N 
Diethyl nitrosoami ne
Reduction

N C
NH 2
19.
Ammonia is
least basic
CH 3 NH 2  K 2 CO 3  3 KBr  2 H 2 O
K Cr O7

 CH 3 CH 2  OH 2 2 
 CH 3 COOH
HNO 2
H 2 SO 4
+ 3 KCl  3 H 2O
Phenyl isocyanide
(c) Because in tertiary nitroalkanes
absent.
43.
HNO
Ethyl amine
  H atom is
21.
(a) Primary amine reacts with CHCl 3 and alc. KOH to
22.
form isocyanide while secondary and tertiary amines
do not react.
(d) Friedel-craft’s reaction is used for the preparation of
alkyl benzene or acetophenone. It is not a method to
prepare amine.
PCl
2
5

 C 2 H 5 OH 
 C 2 H 5 Cl
(b) C 2 H 5 NH 2 
Ethyl alcohol
NH
3

 C 2 H 5 NH 2
Ethyl amine
44.
(c) CH 3 CH 2  NH 2  HNO 2  CH 3 CH 2  OH  N 2  H 2 O
1
o
Alcohol
(CH 3 CH 2 )2 NH  HNO 2  (CH 3 CH 2 )2 N  N  O  H 2O
2o
45.
(d)
Nitroso amine

Hydrolysis
RN  C  2 H 2 O 

 RNH 2  HCOOH
Alkyl Isocyanide
o
1 amine
Formic acid
1396 Nitrogen Containing Compounds
46.
(d) CH 3 NH 2  2 HNO 2  CH 3  O  N  O  N 2  2 H 2 O
2CH 3 NH 2  2 HNO 2  CH 3  O  CH 3  2 N 2  3 H 2 O
NO
NO 2
47.
(d)
63.
67.
NHOH
2H


2H


Zn / NH 4 Cl
2 Zn / NH 4 Cl
(c) 3 o amine cannot be Acetylated because replacable Hatom is absent.
(e) Because
N-N dimethyl propanimine
CH 3
|
CH 3  N  CH 2  CH 2  CH 3
3 o amine
Phenyl hydroxyl amine
50.
+ Bromine water 
(d)
N-Methyl aniline 
NH 2
NH 2
51.
NH  CH 3
(b) Because the N atom in aniline has a lone pair to
donate and also due to +I effect of  NH 2 group.
Br
o
2 amine
Br
68.
aniline  1 o amine.
(c) Replacable H  is absent.
excess
52.
Br

(d) R  NH 2  CHCl 3  3 NaOH  RN  C  3 NaCl  3 H 2O
53.
The unpleasant smell is due to the formation of
isocyanide.
(c) RNH 2  NaNO 2  HCl  R  OH  NaCl  N 2  H 2 O
(c)
+ CH 3  Cl

58.
60.
61.
Cl 

72.
In methyl amine only one electron releasing group is
present but in dimethyl amine two electron releasing
groups are present which increase the basicity higher
in dienethyl amine.
(d) Nitro compounds are not explosive but stable
compound.
(c) CH 3  NH 2  HNO 2  CH 3 OH  N 2  H 2 O
(c) R 3 N  HONO  R 3 N . HONO called as Quaternary
ammonium salt.
N 2Cl
NO2
4 NO 2


cold
62.
H O/H

 CH 3 COOH  NH 3
(a) CH 3 CN 2
NH2
(b)
NH 2
+ NH 4 HS 
NO 2
NO 2
76.
(a) Because of presence of electron withdrawing group NO 2 .
77.
78.
(a) To supress the concentration of the aniline available
for coupling other rise coupling occurs.
(b) R  NH 2  HNO 2  R  OH  N 2  H 2 O .
79.
(c) C6 H5 NH 2  (CH 3 )3 N  CH 3 NH 2  (CH 3 )2 NH
80.
(b) CH 3 CN  CH 3 MgI  (CH 3 ) 2 CNMgI 2
alcohol
H O/H
CH 3 COCH 3  Mg
(B)
82.
86.
87.

 NH 3
(A)
4 NO
2

 R  OH  N 2  H 2 O
But R  NH 2 
3, 4-aryne
NO2
83.
cold
NH 3


 Br
(a) R 2 NH  RNH 2  R 3 N  NH 3 .
+ HNO 3 / H 2 SO 4 A black mass.
75.
NO2
NO2
NaNH 2




(c)
Br
73.
Nitration of aniline without protecting the amino
group is not possible because HNO 3 is a strong
oxidising agent which oxidises aniline.


CH 3
(d)
NH and CH 3  NH 2 .
CH 3
(b)
HNO
NO2
NH 2
57.
Because oxidation of
aniline occur in
absence of making
effect.
O
P - Benzoquinone
Quaternary ammonium salt
56.
O
3



(c)
N-phenyl
hydroxylamine
Nitro benzene
Nitro benzene
70.
+ CH 3 Cl .
N  (CH 3 )3
+ CHCl 3
(c)
2H


NHCH 3
N (CH 3 )2
55.
Zn / HCl


(a)
NH 2
NH 2
54.
69.
NHOH
NO
NO 2
OH
I
 C 6 H 5 NH  NHC 6 H 5  4 H 2 O
(d) C 6 H 5  NO 2 
10 [ H ]
Zn / NaOH
Hydrazo benzene
(a) RCOCl  2 Me 2 NH  RCON

Me
 Me 2  N H 2 Cl 
Me
Me = Methyl.
(b) Phenol react with aniline to give diazonium salt by
coupling but Methyl amine not react with phenol.
(b) C6 H 5 SO 2 Cl is called Hinsberg’s reagent they react
with sec amine to form a product in soluble in
Nitrogen Containing Compounds
88.
alkalies. This reaction used to separate 1 o , 2 o and
3 o amine from their mixture.
(b) A mixture of benzene and aniline can be separated by
dil. HCl.
NO2
NO2
89.
(b)
H 2 SO 4
 HNO 3 

1397
H 2O
NO2
m-dinitrobenzene
90.
NaNO 2  HCl
KCN
(b) C6 H 5 NH 2 
 C6 H 5 N 2 Cl 
 C6 H 5 CN
o
(A)
Aniline
(B)
0 C
NH 2
(C)
H3O
115. (d)
2



HCl
CH3NH2 + COCl2  [CH3NH – CO – Cl]
92.

CH3 – N = C = O

 HCl
methyl
(d) R  NC  2 Hisocyanate
2 O  RNH 2  HCOOH .
93.
(d) CH 3  NC  2 H 2 O  CH 3 NH 2  HCOOH
95.
LiAlH4
(c) CH 3 NC  4 H 

(CH 3 )2 NH .
ether
NH2
NH2
99.
(b)
121. (b) We know that
C6 H 5 NH 2  CHCl 3  3 KOH  C6 H 5 NC  3 KCl  3 H 2 O
2, 4, 6 tribromo
aniline
||
Aniline
H 2  Ni
102. (a) R  C  NH 2  R  CH 2  NH 2
HCl
104. (b) CH 3 CN  2 H 2 O 
 CH 3 COOH  NH 3
106. (a) CH 3 CH 2 NH 2  CH 3 COClCH 3 CH 2 NHCOCH 3  HCl
N Ethyl acetanilide
O
||
CH 3
CH 3
NH  CH 3 COOH 
N  C  CH 3
107. (a)
CH 3
CH 3
108. (d) Anilinium hydrogen chloride produces chloride ion
which gives white precipitate with AgNO 3 . In fact
anilium chloride is a part of aniline.
C  NH 2
CN
 H 2O
POCl 3





SO 3 H
O
NH2
[H ]


[H ]


Conc. H 2 SO 4
H 2 SO 4
OH
p-amino phenol
Phenyl hydroxyl
amine
H
N
C6 H 5 CH  O 
H
Benzonitrile
||
NH-OH
Nitrobenzene
113. (c)
O
||
(A)
110. (c)
C6 H 5 NH 2  (CH 3 CO )2 O  CH 3 CONHC 6 H 5  CH 3 COOH
Acetanilide
(antipyretic)
Conc. H SO
NO 2
phenyl isocyanide
 KOH
K2CO 3  C6 H 5 NH 2 


 C6 H 5 NCO  KBr  H 2O
2 KOH
124. (d)
2
4




Chloroform
Thus in this reaction phenyl isocyanide is produced.
this is called carbylamine reaction.
122. (d) Isocyanides on hydrolysis forms primary amines not
ammonia
123.(b) C6 H 5 CONH 2  Br2  KOH  C6 H 5 CONHBr  KBr  H 2 O
NO 2
NO 2
109. (b)
119. (a) Basicity of amines increase with increase in number
of CH 3 groups (or any group which cause +I
effect), due to increase in electron density on N atom.
As a rule, the basicity of t-amine should be more than
that of s-amine, but actually it is found to be lesser
than s-amines. This is due to stearic hinderence of
bulkier alkyl groups, which decreases the availability
of lone pair of electron on the N atom of the amino
group. Hence the correct order of basicity is :
(CH 3 )2 NH  (CH 3 )3 N  CH 3 NH 2
2 RNH 2  H 2 SO 4  [RNH 3 ]2 SO 42 
Br
O
CH 3
120. (b) Amines are basic in nature, hence form salts with
acid.
Br
Br
Aqueous
 3Br2 

CH 3
CH 3
CH 3
Br
Hydrolysis
CH 3 COOH
(D)
(c) Methyl isocyanate is industrially prepared by the
action of methyl amine with phosgene.
NH 2

Br
H


Br
Ac O
2
C 6 H 5 COOH
91.
NHCOCH 3
NHCOCH 3
N
CH 3
CH 3
CH 3
CH 3
Anyhydrons
ZnCl 2
N
CH 3
CH 3
125. (d) (i) RCNH 2 + Br2 + KOH  RCONHBr
+ KBr + H2O
(ii) RCONHBr + KOH  RNCO + KBr + H2O
(iii) RNCO + 2KOH  RNH2 + K2CO3
RCONH2 + Br2 + 4KOH  RNH2 + 2KBr + K2CO3
+2H2O
126. (b) Aniline reacts with benzaldehyde and forms Schiff's
base (benzal aniline) or anils.

C6 H 5  NH 2  O  CHC 6 H 5 
C6 H 5 N  CHC 6 H 5
H 2O
Benzylidine aniline
1398 Nitrogen Containing Compounds
127. (d) CH 3 CONH 2  Br2  4 KOH 

2H O
Boiling H 2 O

 CH 3 CHO
2
Acetamide
Acetaldeyde
CH 3 NH 2  2 KBr  2 K 2 CO 3
CH 3
(Methyl amine)
N
128. (b)
138. (b)
is most basic
|
CH 3
H

130. (d)

CN
N 2 Cl
NH 2
NaNO 2


CuCN



HCl
DIAZOTISATION
2 

H / Ni
B
CH 2  NH 2
A
N
CH 2 OH
139. (b) N-alkyl formamides when dehydrated with POCl3 in
presence of pyridine give isocyanides.
140. (c) Pollutants which are formed by reaction amongst the
primary pollutants (persist in the environment in the
form they are passed into it) are called as secondary
pollutants. e.g. peroxyacyl nitrates (PAN) are formed
through reaction between nitrogen oxides and
hydrocarbons in the presence of sunlight.
NH2
HNO 2



NH2
Nitrous acid
NH 2 C
NO 2
D
141.
+
H 2 SO 4

131. (a)
NO2
NO2
Aniline
132. (d) C 6 H 5 NH 2 is least basic compound due to resonance
by which the Lone pair of nitrogen takes part in
resonance & due to unavailability of lone pair on N
Aniline become less basic. The Lone pair of N is
delocalized into benzene ring by resonance
..
+
+
+
NH2
NH2
NH2
NH2
OH
OH
OH
OH
.–.
.–.
Carbylamine
NaNO 2  HCl

 

( NH 3 ) , which is of deactivating nature and of mdirective nature.
NH2
N2Cl
ion
diazotizat
 


142. (b)
Br
Br
Br
NaNO 2  HCl / 0 5 o C
Br
Br
Br
3,4,5-Tribromoaniline
2
3 

H PO
Br
Br
Br
1,2,3-Tribromobenzene
143. (a) Basicity order is
C4 H5 NH 2  (CH 3 )3 N  CH 3 NH 2  (CH 3 )2 NH
(CH 3 )3 N is less basic due to steric effect while
0 5 C
C4 H 5 NH 2 is less basic due to resonance.
Benzene
diazonium chloride
OH
o-nitro aniline
(2%)
The reason for this is that, in acidic condition
protonation of  NH 2 group gives anilinium ion
Secondary amine
134. (a) Azo dye is prepared by the coupling of phenol and
diazonium chloride.
NH 2
N   NCl 
m-nitro
aniline (47%)
p-nitro aniline
(51%)
..
–
133. (b) Carbylamine (or isocyanides) give
secondary amine
on reduction.
Ni / H 2
RN 
 C 

 R  NH  CH 3
Tests for Nitrogen Containing Compounds

coupling
+
WeaklyalkalineMedium
1.
Phenol
OH
135. (b)
+
Weakly
acidic medium
NH2
NO2
HNO 3



(d)
NH2
(c) CH 3 CONH 2  Br2  4 NaOH 
Acetamide
CH 3 NH 2  Na 2 CO 3  2 NaBr  2 H 2 O
N
N
N
p-hydroxyazobenzene
Methyl amine
NH 2
C6 H 5 NH 2 2
 C6 H 5 N 2 Cl
NaNO / HCl
5.
(X )
(b)
H 2O / H 

 C6 H 5 CN 
 C6 H 5  COOH
Cu 2 (CN ) 2
(Y )
HCl
 HC  CH
137. (c) CH 3 CN  2 H 
Ether
CH 3
Benzoic acid ( Z )
Thus product Z is identified as C6 H 5  COOH
136. (c) This is Hofman-bromide reaction. In this reaction
one carbon less amines are formed from amides.
Br2 / KOH
CH 3 CONH 2 

 CH 3 NH 2
is a type of 1 o amine
CH 3 and hence gives +ve
carbyl test
6.
(a)
N =N
NH 2
Orange Colour
8.
(a) Diazo-coupling is useful to prepare some dyes.
N = N-Cl + H
Benzenediazonium
chloride
NH 2 
N=N
NH 2
p-amino azo benzene (yellow dye)
Nitrogen Containing Compounds
9.
11.
(bc) (1) With NaHCO 3 
NH 3 Cl 
(a) CHCl 3 gives carbylamine test.
But p-chloro aniline is basic not acidic it does not
liberate CO 2 .
NH 3 Cl 
Critical Thinking Questions
(2) With AgNO3 
(c) R1  H and R 2  R3  CH 3
R1
R2
N  R3 
H
CH 3
(c) CH 3 CH 2 NH 2  HNO 2  CH 3 CH 2 OH  N 2  H 2 O
Ethyl amine
Ethyl alcohol
p-chloro aniline does not contain ionic chlorine to it
does not give white ppt with AgNO 3
(a)
(a)
11.
(a) C 6 H 5  NH 2  CHCl 3  3 KOH 
C6 H 5 NC  3 KCl
Phenyl Isocyanide


HCl
H  Cl  N  N
dimethyl aniline
NN
(CH 3 )2 N
12.
13.
14.
(c) Hofmann degradation of amide
R  CoNH 2  Br2  4 KOH 
(CH 3 )2 NH  CH 3 NH 2  (CH 3 )3 N
K b  5.4  10 4
4.5  10 4 0.6  10 4
(c) (CH 3 )2 NCOCH 3  HCl / H 2 O

(CH 3 )2 NH  CH 3 COOH
(a) Order of basicity of amines
(i) 2 o  1o  3 o
(ii) R 2 NH  RNH 2  ArCH 2  NH 2  NH 3  ArNH  R 
ArNH 2  ArNH  Ar
NO 2
primary amine
NH 2
NO 2
HNO 3 / H 2 SO 4


6.
7.
Aniline
8.
16.
CHMe 2
R  NH 2  NH 3  R 2 NH  R3 N
CMe 3
NH 3  R  NH 2  R 2 NH  R3 N
(b) The nitrogroup is very firmly linked to the benzene
nucleus and does not undergo any displacement
reaction. Nitro group deactivates the benzene
nucleus.
(a)
+H2O
Anil or Schiff's base
OH
KOH
(c) C6 H 5 SO 2 Cl  RNH 2  RNHSO 2 C6 H 5 

soluble in KOH
17.
(b) When sulphur and nitrogen both are present in
organic compound during Lassaigne's Test, both
changes into "sodium thiocyanate". (NaSCN) which
gives a blood red colouration with Ferric ion.
3 NaCNS  FeCl 3  Fe(CNS )3  3 NaCl
Ferric sulpho cyanide
(Blood red colour)
H
18.
(c)
N
|
H
N=CH–CH3
Trace of an acid..
 CH 3  CHO 

OH
+
RNKSO 2 C6 H 5
R 2 NH  R  NH 2  NH 3  R3 N
NH 2
heat
 KOH (solid) 

(a)
NO 2
NO 2
Because OH  is nucleophile.

(b) The relative basic charecter of 1 o ,2 o and 3 o amines
also depends upon the nature of the alkyl group.
Relative basic strength
R
CH 3
R 2 NH  R  NH 2  R3 N  NH 3
C 2 H 5
15.
Sn / HCl


Nitrobenzene
 3 H 2O
(b)
R  NH 2  2 KBr  K 2 CO 3  2 H 2 O
(b)
+ 2 H 2O
Aniline
0 5C
 NaNO 2  2 HCl 


(CH 3 )2 N
5.
Sn / HCl
+ 6 [H ] 

2
10.
Nitrobenzene
2 H 2 O  NaCl
4.
NH
NO 2
N 2 Cl
NH 2
3.
 AgNO3 
NH 3 NO 2  AgCl  (White ppt)
N  CH 3
Sec. amine reacts with Nitrous acid to form nitroso
amine yellow liquid.
2.
+ CO 2  NaCl  H 2 O
Anilinium hydrochloride is an acid salt and liberates
CO 2 from NaHCO 3 .
C  3 KCl  3 H 2 O
alkyl isocyanide
1.
NH 2
 NaHCO 3 

RNH 2  CHCl 3  3 KOH (alc.) 
RN
1399
19.
does not have aromaticity by which the Lone
pair of electron of Nitrogen does not
delocalised in benzene ring so it will be strong
base on other hand rest 3 have aromaticity i.e.,
they follow the huckel rule so the electron pair
of Nitrogen delocalised in ring by resonance &
resulting they become less basic.
(d) Liebermann’s Nitroso reaction.