Online Test — Amines
20 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 20
Which of the following is a secondary amine?
CH3NH2
(CH3)2NH
(CH3)3N
C6H5NH2
Explanation: (CH3)2NH has two carbon groups on nitrogen, so it is a secondary amine.
Question 2 of 20
The hybridisation of nitrogen in methanamine is:
sp
sp2
sp3
sp3d
Explanation: Nitrogen in amines is sp3 hybridised and pyramidal, with a lone pair.
Question 3 of 20
The Gabriel phthalimide synthesis is best for preparing:
pure tertiary amines
aromatic amines
pure primary aliphatic amines
quaternary salts
Explanation: It gives uncontaminated 1° aliphatic amines and fails for aromatic amines like aniline.
Question 4 of 20
Hofmann bromamide degradation of CH3CH2CONH2 gives:
propan-1-amine
ethanamine
propan-2-amine
methanamine
Explanation: The amide loses its carbonyl carbon as carbonate, so propanamide (3 C) gives ethanamine (2 C).
Question 5 of 20
Reduction of nitrobenzene with Sn and HCl gives:
nitrosobenzene
aniline
phenol
azobenzene
Explanation: Sn/HCl reduces –NO2 to –NH2, giving aniline.
Question 6 of 20
In the gas phase the order of basicity of methylamines is:
NH3 > CH3NH2 > (CH3)2NH
(CH3)3N > (CH3)2NH > CH3NH2
CH3NH2 > (CH3)2NH > (CH3)3N
all are equal
Explanation: In the gas phase only +I operates, so more alkyl groups means a stronger base: 3° > 2° > 1°.
Question 7 of 20
Aniline is a weaker base than ammonia because:
of the +I effect of phenyl
the lone pair is delocalised into the ring
of intramolecular H-bonding
of its low molar mass
Explanation: Resonance delocalisation of the nitrogen lone pair into the benzene ring lowers its availability for protonation.
Question 8 of 20
Which test is positive only for primary amines?
Hinsberg test
carbylamine test
azo coupling
Lucas test
Explanation: The carbylamine (isocyanide) test with CHCl3 + alc. KOH is given only by primary amines.
Question 9 of 20
In the Hinsberg test, a tertiary amine:
gives a KOH-soluble product
gives a KOH-insoluble product
does not react
gives an azo dye
Explanation: A 3° amine has no N–H to react with benzenesulphonyl chloride, so it does not react.
Question 10 of 20
A 2° amine reacting with nitrous acid gives:
a diazonium salt
an alcohol and N2
a yellow N-nitrosamine
an isocyanide
Explanation: Secondary amines give yellow oily N-nitrosamines with HNO2.
Question 11 of 20
Bromination of aniline with bromine water gives mainly:
o-bromoaniline
m-bromoaniline
p-bromoaniline
2,4,6-tribromoaniline
Explanation: The strongly activating –NH2 group leads to substitution at all three o/p positions, giving 2,4,6-tribromoaniline.
Question 12 of 20
Aniline does not undergo Friedel–Crafts reaction because:
it is too volatile
AlCl3 binds the basic nitrogen, deactivating the ring
the ring is deactivated by –NH2
it is insoluble
Explanation: The Lewis acid AlCl3 coordinates to the basic N, making it positive and deactivating the ring.
Question 13 of 20
Diazotisation of aniline must be done at:
0–5°C
25°C
50°C
100°C
Explanation: Above 5°C the diazonium salt decomposes to phenol, so 0–5°C is required.
Question 14 of 20
The Sandmeyer reaction with CuBr converts a diazonium salt to:
a phenol
an aryl bromide
benzene
an azo dye
Explanation: CuBr/HBr replaces –N2+ with –Br, giving an aryl bromide.
Question 15 of 20
Treatment of benzenediazonium chloride with H3PO2 gives:
phenol
chlorobenzene
benzene
iodobenzene
Explanation: Hypophosphorous acid replaces –N2+ by –H (deamination), giving benzene.
Question 16 of 20
Azo coupling of a diazonium salt with phenol occurs at the:
meta position
para position
ipso position
carbonyl
Explanation: The weak diazonium electrophile couples at the para position of the activated phenol ring.
Question 17 of 20
The reagent used in the Balz–Schiemann reaction is:
CuCN
KI
HBF4
CuCl
Explanation: Diazonium fluoroborate (from HBF4) is heated to give the aryl fluoride.
Question 18 of 20
Which amine cannot be acylated?
a primary amine
a secondary amine
aniline
a tertiary amine
Explanation: Tertiary amines have no N–H, so they cannot form amides (cannot be acylated).
Question 19 of 20
Reduction of CH3CH2CN with LiAlH4 gives:
ethanamine
propan-1-amine
N-methylethanamine
propan-2-amine
Explanation: R–CN → R–CH2NH2; propanenitrile gives propan-1-amine.
Question 20 of 20
The correct aqueous basicity order for ethyl amines is:
(C2H5)3N > (C2H5)2NH > C2H5NH2
(C2H5)2NH > (C2H5)3N > C2H5NH2
C2H5NH2 > (C2H5)2NH > (C2H5)3N
NH3 > (C2H5)2NH
Explanation: For ethyl amines in water the experimental order is (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3.