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Class 12 Chemistry Chapter 13 Amines MCQs | Class 12 Chemistry Chapter 13 Amines Multiple Choice Questions (MCQs) and Answers

(1) Reduction of aromatic nitro-compounds using Sn and HCl gives
[A] aromatic primary amines
[B] aromatic secondary amines
[C] aromatic tertiary amines
[D] aromatic amides
Answer: aromatic primary amines
(2) When aniline is heated with cone. H2SO4 at 455-475 K, it forms
[A] aniline hydrogensulphate
[B] sulphanilic acid
[C] amino benzene sulphonic acid
[D] benzenesulphonic acid
Answer: sulphanilic acid
(3) Which of the following compounds cannot be identified by carbylamine test?
[A] CH3CH2NH2
[B] (CH3)2CHNH2
[C] C6H5NH2
[D] C6H5NHC6H5
Answer: C6H5NHC6H5
(4) What is obtained when benzoyl chloride reacts with aniline in the presence of sodium hydroxide?
[A] Benzoic acid
[B] Benzanilide
[C] Acetanilide
[D] Azobenzene
Answer: Benzanilide
(5) Which of the following compounds reacts with NaNO2 and HCl at 0-4°C to give alcohol/phenol?
[A] C6H5NH2
[B] C2H5NH2
[C] CH3NHCH3
[D] C6H5NHCH3
Answer: C2H5NH2
(6) Which of the following has highest pKb value?
[A] (CH3)3CNH2
[B] NH3
[C] (CH3)2MH
[D] CH3NH3
Answer: NH3
(7) Acetylation of a secondary amine in alkaline medium yields
[A] N, N-dialkyl acetamide
[B] N, N-dialkyl amine
[C] N, N-dialkyl amide
[D] acetyl dialkyl amine
Answer: N, N-dialkyl acetamide
(8) The amines are basic in nature, hence they form salts with hydrochloric acid. Which of the following will be insoluble in dil. HCl?
[A] C6H5NH3
[B] (C6H5)3N
[C] C2H5NH2
[D] CH3NHCH3
Answer: (C6H5)3N
(9) Primary, secondary and tertiary amines may be separated by using
[A] iodoform
[B] diethyloxalate
[C] benzene sulphonyl chloride
[D] acetyl chloride
Answer: benzene sulphonyl chloride
(10) An organic compound (X) was treated with sodium nitrite and HCl in ice cold conditions. Bubbles of nitrogen gas were seen coming out. The compound (X) may be
[A] a secondary aliphatic amine
[B] a primary aromatic amine
[C] a primary aliphatic amine
[D] a tertiary amine
Answer: a primary aliphatic amine
(11) Among the following:
I. CH3NH5
II. (CH3)2NH
III. (CH3)3N
IV. C6H5NH2
Which will give the positive carbylamine test?
[A] I and II
[B] I and IV
[C] II and IV
[D] II and III
Answer: I and IV
(12) Cyclohexyiamine is stronger base than aniline become
[A] in aniline electron pair is involved in conjugation
[B] in cyclohexyiamine electron pair is involve conjugation
[C] in aniline-NH, group is protonated
[D] in cyclohexyiamine nitrogen has a negative charge
Answer: in aniline electron pair is involved in conjugation
(13) The strongest base among the following is
[A] C6H5NH3
[B] p-NH2C6H4NH
[C] m-NO2C6H4NH2
[D] C6H5CH2NH2
Answer: C6H5CH2NH2
(14) Nitrogen atom of amino group is ………. hybridised
[A] sp
[B] sp2
[C] sp3
[D] sp3d
Answer: sp3
(15) C3H8N cannot represent
[A] 1° ammine
[B] 2° ammine
[C] 3° ammine
[D] quartemary ammonium salt
Answer: quartemary ammonium salt
(16) Identify the correct IUPAC name
[A] (CH3CH2)2NCH3 = N-Ethyl-N-methylethanamine
[B] (CH3)3CNH2 = 2-methylpropan-2-amine
[C] CH3NHCH (CH3)2 = N-Methylpropan-2-amine
[D] (CH3)2CHNH2 = 2, 2-Dimethyl-N-propanamine
Answer: (CH3CH2)2NCH3 = N-Ethyl-N-methylethanamine
(17) The most convenient method to prepare primary (i Amine) amine containing one carbon atom less is
[A] Gabriel phthalmidie synthesis
[B] Reductive amination of aldehydes
[C] Hofmann bromamide reaction
[D] Reduction of isonitriles
Answer: Hofmann bromamide reaction
(18) When excess of ethyl iodide is treated with ammonia, the product is
[A] ethylamine
[B] diethylamine
[C] triethylamine
[D] tetrathylammonium iodide
Answer: tetrathylammonium iodide
(19) Amides may be converted into amines by a reaction named after
[A] Hofmann Bromide
[B] Claisen
[C] Perkin
[D] Kekule
Answer: Hofmann Bromide
(20) Reduction of CH3CH2NC with hydrogen in presence of Ni or Pt as catalvst gives
[A] CH3CH2NH2
[B] CH3CH2NHCH3
[C] CH3CH2NHCH2CH3
[D] (CH3)3N
Answer: CH3CH2NHCH3

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