Nitrogen containing
organic compounds
1
1) The number of functional isomers
possible with C3H9N is _____
a) 2
b) 4
c)3
d)5
2
CH3CH2CH2NH2 or (CH3)2CHNH2
CH3CH2NHCH3,(CH3)3N
Hence the answer is (c) 3
3
2) An organic compound (x) containing
nitrogen on reduction with Fe / HCl
forms an amine which reacts with
NaNO2/H+ at 0°C to liberate nitrogen gas.
The compound (x) is _____
4
N2 liberation is possible when 1°amine is
not aryl (eg.: aniline). So the compound
(x) on reduction that gives a 1° amine that
is not aryl is (b)
Ans: (b)
5
3)
1.Li AlH 4
CHCl3 +AlcKOH
Br2 /NaOH
o
P 
 Q 
R 

2
amine
2.H 2O
P and R are:
a) Amide and 1°amine
b)1°amine and nitrile
c) Amide and isonitrile
d) nitrite and isonitrile
6
If P is an amide we have
CHCl3
Br2 / NaOH
Re duction
amide 
1 a min e 
 isonitrile 
 2 a min e
(P)
(Q)
AlcKOH
Hence P is an amide and R is isonitrile
Ans: (C) Amide and isonitrile
7
4) The variation in base strength of 1°, 2°,
3° methylamines in aqueous medium is
not due to ______
a) inductive effect
b) solvation effect
c) steric hindrance
d) resonance effect
8
In methylamines there is no resonance
effect to influence the base strength
Ans: (d) Resonance effect
9
5) Conjugate acid of
NH2 – CH2 - CH2 - COOH is ______
a) NH2 - CH2 - CH2 - COOb) +NH3- CH2 – CH2 - COOH
c) –NH - CH2 - CH2 - COOH
d) +NH3 - CH2 - CH2 - COO10
Conjugate acid can be obtained when a
basic group accepts H+.
In H2N-CH2-CH2-COOH , H2N accepts
H+ to form its conjugate acid
Ans: b)
11
6) Increase in base strength of
i) benzylamine
ii) cyclohexylamine
iii) o-methoxyaniline is_______
a) iii < i < ii
c) i > ii > iii
b) iii > ii > i
d) ii < iii > i
12
Hence iii) is the weakest base and
ii) is the strongest
Ans: (a) iii < i < ii
13
7) Schotten Baumann reaction helps to
convert
a) 1° amine to 2°amine
b) 1°amine into N-substituted amide
c) amide into 1°amine
d) 1° amine into a quaternary salt
14
Benzoylation in presence of alkali is
Schotten Baumann reaction 1°amine react
to form N-substituted amide
Eg: C6H5COCl + C6H5NH2 
C6H5CONHC6H5
Ans: b) 1° amine into N-substituted amide
15
8) CH3NH2 does not react with
a) NaOH
b) HCl
c) CH3COCl
d) C6H5SO2Cl
16
CH3NH2 is a base, does not react with
NaOH
Ans: (a) NaOH
17
9) Ethanamine and benzenamine can be
best distinguished using______
a) CHCl3 + Alc. KOH
b) CH3I
c) dil. HCl
d) Bromine water
18
C2H5NH2 and
both react with the reagents in a, b & c
but only benzenamine readily
decolourises orange colour of bromine
water.
Ans: d) Bromine water
19
10) Gabriel phthalimide synthesis is
used to prepare
a) amide
b) 1° aliphatic amine
c) amine
d) 1° aromatic amine
20
Take care to mark the correct answer.
Ans: (b) 1° aliphatic amine
21
11) Reagent that need not be used to
convert benzene into
is____
a) KI
c) I2
b) HNO3
d) HNO2
22
Ans: (C) I2
23
12)Alkyl halide
A and B are:
a) position isomers
b) functional isomers
c) metamers
d) chain isomers
24
X is R-CN
R-CH2NH2 (A)
Y is R-NC
R-NH-CH3 (B)
A and B are functional isomers
Ans: (b) functional isomers
25
13) N-methylbenzenamine reacts with
ethyliodide to form a salt that______
a) is achiral
b) is chiral
c) exhibits metamerism
d) exhibits geometric isomerism
26
N attached to four different groups is
chiral. Therefore the salt is chiral
Ans: (b) is chiral
27
14) p-nitroaniline 

1. NaNO2 /HCl 0 C
2.KI
gives
a) p-diiodobenzene
b) p-iodobenzenamine
c) 2, 4-dinitrobenzene
d) p-iodonitrobenzene
28
+N Cl2
NH2
2 / HCl
NaNO



NO2
I
I
KI

NO2
Ans: (d) p-iodonitrobenzene
NO2
29
15) 0.3g of organic compound containing
kjeldhal 's
C,O,H,N,
ammonia


method
absorbed in 100ml of 0.1 M H2SO4.
Unreacted acid reacted with 20ml of
0.5 M NaOH. The percentage of nitrogen
in the compound could be____
a) 46
c) 36
b) 56
d) 66
30
H2SO4 +2NaOHNa2SO4+ 2H2O
2NH3 + H2SO4  (NH4)2SO4
Initial amount of H2SO4 is 10 millimole
NH3 consumes 5millimole H2SO4
 Amount NH3 =10millimole
-3
10 10 14 100
 46.6%
%N=
0.3
Ans: a) 46
31
16) Coupling reaction between
benzene diazoniumchloride and phenol
involves the electrophile_________
a) C6 H

5
c) C6 H5O
b) C6 H5 N
-
d) N

2

2
32
In
, the electrophile is
Ans: (b) C6 H5 N

2
33
17) Zwitter ion is not formed by
R
a) H2N-CH-COOH
b) HOOC
NH2
c) H2N
SO3H
d) H2N-CH2-CH2-COOH
34
Zwitter ion is formed when there is
transfer of H+ from acidic to basic group
within a molecule. For this acidic or
basic groups must be fairly strong.
Options a, c, d, have either a strong
acidic
or
strong
basic
group
In option (b) NH2 is very weak, so also
is COOH.
Ans: (b) HOOC
NH2
35
18) CCl2 is an electrophile for_______
a) Hoffmann’s bromamide reaction
b) Carbylamine reaction
c) Sandmeyer’s reaction
d) Hinsberg test
36
In carbylamine reaction CHCl3 reacts
with alc. KOH to form CCl2 a neutral
electrophile
CHCl3
CCl2(dichlorocarbene)
Ans: (a) Carbylamine reaction
37
19)
is nitrated. The nitro group goes to
position________
a) c
b) a
c) x
d) y
38
In the molecule NH group is ereleasing, ring activating, o,p directing,
but CO group is e- withdrawing, ring
deactivating, m – directing. Nitration
happens faster for the ring attached to
NH group. The NO2 group mainly goes
to para position (c).
Ans: (a)
c
39
n -FeCl3
20) X 
 Y 
 violet colour
in water warm
X & Y have functional groups as
a) amine and phenol
b) amide and phenol
c) diazonium and phenol
d) nitrile and phenol
40
Y must be phenol to give violet colour
with n-FeCl3. Phenol is obtained from X
by hydrolysis. X must be a benzene
diazonium salt.
Ans: (c) X - diazonium group
Y - phenol
41
21) Amines are not
a) Lewis bases
b) Bronsted bases
c) nucleophiles
d) Arrhenius bases
42
All types of amines (1°, 2°, 3°) have a
lone pair of electrons on nitrogen.
Hence by definition they fit into a, b, c
but not d.
Ans: (d) Arrhenius base.
43
22) Ethanamine is not obtained by the
reduction of
a) methylcyanide
b) nitroethane
c) acetamide
d) ethylcyanide
44
Re duction

a) Methylcyanide – CH3CN  CH 3CH 2 NH 2
b) Nitroethane- C2H5NO2 Re
duction

 CH 3CH 2 NH 2
c) Acetamide – CH3CONH2
Re
duction

 CH 3CH 2 NH 2
d) Ethylcyanide–C2H5CN Re
duction

 CH 3CH 2CH 2 NH 2
Ans: (d) ethylcyanide
45
23) Which one of the following is most
reactive towards electrophilic
substitution reaction?
a) aniline
b) Acetanilide
c) benzamide
d) nitrobenzene
46
b, c, d have electron withdrawing groups,
which is ring deactivating. (a) has
electron releasing, ring activating group
Ans: (a) aniline
47
24) A tribromo derivative is obtained
when X is treated with bromine water. X
could be_______
a) acetophenone
b) acetanilide
c) o-cresol
d) m-toluidine
48
To get a tribromo derivative, the
functional group must be strongly
electron releasing group.(c) and (d) has
it. But in (c) ortho position is blocked
by CH3 group but not in (d). All o & p
positions are free, hence m-toluidine on
bromination forms tribromo derivative.
Ans: (d) m-toluidine
49
25)
C6 H5 NH 2  A 
B
HNO2 0 C
CuCN
HCN

C
1.Li AlH 4
2.H 2O
C is --------a) C6H5COONH4
b) C6H5CH2NH2
c) C6H5NHCH3
d) C6H5COOH
50
Ans: b) C6H5CH2NH2
51
26. In aniline, the polar effects due to NH2
group are
a) -I, - R
b) +I, +R
c) -I, +R
d) +I, -R
52
NH2 is electron releasing due to
resonance
(+R)
and
electron
withdrawing
due
to
high
electronegativity of N (-I).
Ans: c) –I, +R
53
27. In detection of nitrogen in an
organic compound via sodium fusion
extract, nitrogen is converted into
a) NH3
b) NaCN
c) NaNO3
d) N2
54
When nitrogen containing organic
compound is fused with sodium, the
compound formed is NaCN.
Ans: b) NaCN
55
28. Which of the following statements
is correct?
a) Amides are more basic than aliphatic
amines
b) Electron
withdrawing
group
increases the base strength of aniline
c) Diphenyl amine is less basic than
aniline
d) Amines are stronger bases than
ammonia
56
Ans:
c) Diphenyl amine
is less basic than aniline
57
29. The name of
a) ethylmethyl propylamine
b) sec-butyl methylamine
c) ethylmethyl isopropylamine
d) ethyl propylamine
58
It is a tertiary amine with alkyl groups
methyl, ethyl and isopropyl.
Ans: c) ethylmethyl isopropylamine
59
30. Benzene is converted into
The best sequence of reactions to do
this is
a) Methylation, nitration, reduction
b) Nitration, reduction, methylation
c) Methylation, reduction, nitration
d) Reduction, nitration, methylation
60
Ans: a) Methylation, nitration, reduction
61
31. Which one of these is neither a test
nor a method to prepare amines?
a) Hoffmann reaction
b) Carbylamine
c) Methylation
d) Sulphonation
62
Ans:
d) Sulphonation
63
32. X with NaOBr forms a 1° amine. X
could be
a) CH3CONHCH3
b) CH3COONH4
c) CH3CONH2
d) CH3NC
64
NaOBr is obtained from Br2/NaOH
X must be an amide (unsubstituted
RCONH2)
Ans: c) CH3CONH2
65
33. ethanenitrile on reduction forms
a) ethanamine
b) ethanamide
c) propanamine
d) propanamide
66
a) CH3CN reduction

 CH3CH2NH2
ethanenitrile
ethanamine
Ans: a) ethanamine
67
34. For maximum activity of sulpha
drugs, the minimum standard feature of
the drug should be
a)
b)
c)
d)
68
Ans:
c)
69
35. An aromatic compound with the
formula C7H9N cannot be
a) 1° amine
b) 2° amine
c) toluidine
d) 3° amine
70
3° amine is not possible.
Ans: d) tertiary amine
71
36. Aniline can be obtained by the
reduction of _____
a) benzonitrile
b) benzamide
c) anilide
d) nitrobenzene
72
C6H5CN 
 C6H5CH2NH2
(benzonitrile)
C6H5CONH2 reduction

 C6H5CH2NH2
(benzamide)
reduction
R-CONH-Ar 
RCH2NHAr
(anilide)
reduction


 C6H5NH2
C6H5NO2
(nitrobenzene)
Ans: d) Nitrobenzene
reduction
73
37. p-hydroxy azobenzene is obtained
when
a) Phenol is coupled with azobenzene
b)p-hydroxyaniline is coupled with
benzene
c)Phenol
is
coupled
with
bezenediazonium ion
d) Aniline is coupled with phenol
74
is obtained when phenol couples with
a benzene diazonium salt or ion
Ans:
c)
Phenol is
bezenediazonium ion
coupled
with
75
38. 1,3,5-tribromobenzene can be
obtained from aniline by a series of
reactions in the order
a) bromination, reduction, diazotisation
b) bromination, diazotisation, boil with
ethanol
c) carbylamine reaction, diazotisation,
boil with Br2 water
d) bromination, heat with Zn dust, boil
with ethanol
76
Ans: b) bromination, diazotisation, boil in
C2H5OH
77
39. Enantiomers are possible for
a) ethylmethyamine
b) secondary butylamine
c) N, N-dimethyl aniline
d) trimethylammonium ion
78
Enantiomers are possible if the given
compound is chiral or has a chiral
carbon.
has a chiral carbon and hence is chiral.
Ans: b) sec-butylamine
79
40. Acetylation of aniline, makes it
a) less reactive towards ESR & more
reactive towards oxidation
b) more reactive towards both ESR and
oxidation
c) less reactive towards both ESR and
oxidation
d) more reactive towards ESR but less
reaction towards oxidation
80
Acetylation of aniline gives
which has electron withdrawing
group in the side chain
Ans:
c) less reactive towards both ESR and
oxidation
81
41. Among i) aniline ii) benzaldehyde
iii) acetanilide iv) toluene,
the least and the most reactive towards
Friedel Craft’s alkylation is
a) i and iv
b) iv and ii
c) iii and ii
d) ii and i
82
Ans:
a) i and iv
83
42. Which one of the following gives a
product with Hinsberg reagent which is
insoluble in an alkali?
a) (CH3)3N
b) CH3CH2NH2
c) C6H5NHCH3
d) (CH3)2NC6H5
84
2° amine with Hinsberg reagent gives a
product insoluble in an alkali.
Ans: c) C6H5NHCH3
85
43. 2° amine reacts with NaNO2 and HCl
at 0° C to
a) form a 2° alcohol
b) liberate nitrogen gas
c) form a diazonium salt
d) form N-nitrosoamine
86
2° amine reacts with HNO2 liberated to
form N-nitrosoamine
Ans: d) form N-nitrosoamine
87
44. Aniline reacts with concentrated
H2SO4 at 450 K to give
a) phenol
b) sulphanilic acid
c) aniline hydrogensulphate
d) benzene sulphonic acid
88
Ans:
b) sulphanilic acid
89
45. The base strengths of the following
compounds decreases as
a) o-toluidine > aniline > o-nitroaniline
b) o-toluidine > o-nitroaniline > aniline
c) Aniline > o-nitroaniline > o-toluidine
d) o-nitroaniline<aniline<o-toluidine
90
Ans:
a) o-toluidine > aniline > o-nitroaniline
91
46. In the gaseous state the base strengths
of the following amine increases as:
a) CH3NH2<(CH3)2NH<(CH3)3N
b) (CH3)3N<CH3NH2<(CH3)2NH
c) (CH3)3N<(CH3)2NH<CH3NH2
d) CH3NH2<(CH3)3N<(CH3)2NH
92
In gaseous state only +I effect decides base
strength of amines.
Ans:
a) CH3NH2<(CH3)2NH<(CH3)3N
93
47. Gabriel phthalimide synthesis does not
involve
a) Neutralisation
b) Nucleophilic substitution
c) Hydrolysis
d) Hydration
94
Ans:
d) Hydration
95
48. A positive result for carbylamine test
will be obtained when the amine is
a) Aliphatic 1° amine only
b) Aromatic 1° amine only
c) 1° amine
d) 2° aliphatic amine
96
Ans:
c) 1° amine
97
49. Aniline and benzylamine can be best
distinguished using
a) NaNO2/ HCl at 0°C + b-naphthol in
NaOH
b) Hinsberg reagent
c) CHCl3 + alc.KOH
d) CH3COCl
98
Both are 1° amines, but aniline is an aryl
amine
Ans:
a) NaNO2/ HCl at 0°C + b-naphthol in
NaOH
99
50. Amide can be converted into 1° amine
by _____ reaction
a) Sandmeyer
b) Carbylamine
c) Hoffman’s bromamide
d) Rosenmund’s
100
Ans:
c) Hoffman’s bromamide
101
51. Which of the following statements is
correct?
i) Nitration of aniline gives meta isomer.
ii)Acetylation of aniline decreases the
tendency of aniline to get oxidized.
iii) anilinium ion is a strong conjugate
base.
iv) Aniline is an antioxidant.
a) i, ii, iv b) ii, iii c) ii, iii, iv d) iii, iv
102
Ans:
a) i, ii, iv
103
52. Number of resonance structures
possible for
is
a) 2
b) 3
c) 4
d) 5
104
are the only two possible resonance
structures
Ans: a) 2
105
53.
reagent
is done using the
a) H4P2O7
b) H3PO2
c) H3PO3
d) H3PO4
106
Ans:
b) H3PO2
107
54. Compounds P and Q that yield an
isocyanide when treated with alc. AgCN
and CHCl3/ alkali respectively are:
a) chlorobenzene and acetone
b) ethyl bromide and acetamide
c) methyl iodide and ethyl amine
d) ethyl alcohol and ethyl bromide
108
Ans:
c) methyl iodide and ethyl amine
109
55. An optically active compound A
(C8H11N) dissolves in HCl and answers
carbylamine test. The compound could
be
a) C6H5CH2CH2NH2
b) CH3NHCH2CH2C6H5
c) C6H5CH(NH2)CH3
d) C6H4(NH2) (C2H5)
110
Ans:
c) C6H5CH(NH2)CH3
111
56.
X is
a)
KOH
b)
NaNO3/ HCl
c)
C2H5OH
d)
NaNO2/ HCl
112
Ans:
d) NaNO2/ HCl
113
57. The pKb value is lowest for
a) C6H5NH2
b) C6H5(NH)CH3
c) (CH3)2NH
d) C6H5NHC6H5
114
Lower the pKb value, stronger is the base.
Among the options dimethyl amine in (c)
is the strongest base.
Ans: c) (CH3)2NH
115
58.
Z is
a) m-chlorophenol
b) resorcinol
c) catechol
d) m-nitroaniline
116
Ans: b) resorcinol
117
59.
X, Y, Z are
a) CH3OH, CH3NH2, CH3OH
b) CH3Cl, CH3NH2, CH3NO2
c) Cl-CH2-COOH, NH2-CH2-COOH,
HO-CH2-COOH
d) CH3COCl, CH3CONH2, CH3OH
118
Ans:
c) Cl-CH2-COOH, NH2-CH2-COOH,
HO-CH2-COOH
119
60. A mixture of aniline and nitrobenzene
can be separated using
a) NaOH solution
b) HCl acid
c) alcohol
d) ether
120
Aniline is basic hence dissolves in HCl
acid. Nitrobenzene remains undissolved
forms a separate layer, hence can be
separated.
Ans: b) HCl acid
121
61.
X should be
a) o-nitrobenzenamine
b) o-nitrobenzoic acid
c) m-nitrobenzoic acid
d) p-nitrobenzoic acid
122
Terphthalic acid is
Ans: d) p-nitrobenzoic acid
123
62. The reagent that will not help to
distinguish benzylamine from Nmethylbenzenamine is
a) CHCl3/ alc.KOH
b) C6H5SO2Cl
c) HNO2
d) HCl acid
124
Both of these react differently with
reagents in a, b and c but both dissolve in
HCl acid to form aqueous solution.
Ans: d) HCl acid
125
63. 1.78 g of an amine reacts with nitrous
acid at 0°C to liberate 4.48 dm3 of N2 at
STP. The molar mass of the amine is
_____ g mol-1.
a) 78
b) 89
c) 68
d) 48
126
4.48
=
0.02
mol
nitrogen
at
STP
22.4
Mass of 0.02 mole of amine = 1.78g
Molar mass of the amine = 1.78
0.02
= 89 gmol-1
Ans: b) 89
127
64. Aniline can be purified by steam
distillation. Efficiency of steam
distillation can be increased by the
addition of
a) alcohol
b) HNO3
c) NaCl
d) benzene
128
When NaCl is dissolved, vapour pressure
of salt solution (water + salt) is lowered,
boiling point of water increases. Aniline
does not dissolve NaCl and hence gets
distilled more efficiently.
Ans: c) NaCl
129
65.
Y is
a) 2° amine
b) 3° amine
c) 1° amine
d) amide
130
Ans:
a) 2° amine
131