Class 12 Chemistry MCQs Amines for JEE/NEET

Here, you will get Class 12 Chemistry MCQ of Chapter 13 Amines for cracking JEE and NEET/AIIMS. By solving these MCQs of Class 12 Chemistry Chapter 13 Amines, you will get the confidence to crack JEE or NEET. Practice MCQ Questions for Class 12 Chemistry with Answers on a daily basis and score well in exams. 

Amines 12 MCQs Questions with Answers

C3H8N cannot represent
(a) 1° ammine
(b) 2° ammine
(c) 3° ammine
(d) quartemary ammonium salt

Answer: (d) quartemary ammonium salt

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: (c) Hofmann bromamide reaction

When excess of ethyl iodide is treated with ammonia, the product is
(a) ethylamine
(b) diethylamine
(c) triethylamine
(d) tetrathylammonium iodide

Answer: (d) tetrathylammonium iodide

Secondary amines can be prepared by
(a) reduction of nitro compounds
(b) oxidation of N-substituted amides
(c) reduction of isonitriles
(d) reduction of nitriles

Answer: (c) reduction of isonitriles

Benzoic acid is treated with SOCl2 and the product (X) formed is reacted with ammonia to give (Y). (Y) on reaction with Br2 and KOH gives (Z). (Z) in the reaction is
(a) aniline
(b) chlorobenzene
(c) benzamide
(d) benzoyl chloride

Answer: (a) aniline

Tertiary amines have lowest boiling points amongst isomeric amines because
(a) they have highest molecular mass
(b) they do not form hydrogen bonds
(c) they are more polar in nature
(d) they are most basic in nature

Answer: (b) they do not form hydrogen bonds

The major product of the reaction between /n-dinitro benzene with NH4HS is
(a) p-Dinitro benzene
(b) m-Diamino benzene
(c) m-nitroaniline
(d) p-Diamino benzene

Answer: (c) m-nitroaniline

An organic compound C7H9N forms clear solution when dissolved in KOH after reacting with C6H5SO2Cl, ‘A’ on diazotisation at 0°C and then reaction with β-naphthol gives orangish red dye. ‘A’ on electrophilic substitution gives single product. ‘A’ is
(a) 4-Methyl aniline
(b) 2-Methyl aniline
(c) 3-Methyl aniline
(d) N-Methyl aniline

Answer: (a) 4-Methyl aniline

The best reagent for converting 2-Phenyl propanamide into 2-phenyl propanamine is
(a) Br2/NaOH
(b) excess of H2
(c) I2/P4
(d) LiAlH4 in ether

Answer: (d) LiAlH4 in ether

Choose the amide which on reduction with LiAlH4 gives secondary amine
(a) Ethanamide
(b) N-Methyl ethanamide
(c) N, N-diethyl ethanamide
(d) Benzamid

Answer: (b) N-Methyl ethanamide

Oxidation of aniline with K2Cr2O7/H2SO4 gives
(a) phenylhydroxylamine
(b) p-benzoquinone
(c) nitrosobenzene
(d) nitrobenzene

Answer: (b) p-benzoquinone

Which of the following pair of species will yield carbylamine?
(a) CH3CH2Br and KCN
(b) CH3CH2Br and NH3 (excess)
(c) CH3CH2Br and AgCN
(d) CH3CH2NH2 and HCHO

Answer: (c) CH3CH2Br and AgCN

C6H5CONHCH3 can be converted into C6H5CH2NHCH3 by
(a) NaBH4
(b) H2-Pd/C
(c) LiAlH4
(d) Zn-Hg/HCl

Answer: (c) LiAlH4

C6H5N+2 Cl + CuCN → C6H5CN + N2 + CuCl. The above chemical reaction is associated with which of the following name
(a) Balz Schiemen
(b) Gattermann
(c) Shimonini
(d) Sandmeyer

Answer: (d) Sandmeyer

Aniline was diazotised and is subsequently reduced with tin (II) chloride and HCl. The product formed is
(a) Phenylhydrazine
(b) Aniline
(c) Phenyl aniline
(d) p-aminoazobenzene

Answer: (a) Phenylhydrazine

Arrange the following compounds in increasing order of basicity
CH3NH2, (CH3)2 NH, NH3, C6H5NH2
(a) C6H5NH2 < NH3 < (CH3)2NH < CH3NH2
(b) CH3NH2 < (CH3)2NH < NH3 < C6H5NH2
(c) C6H5NH2 <NH3 < CH3NH2<(CH3)2NH
(d) (CH3)2NH < CH3NH2 <NH3 < C6H5NH2

Answer: (c) C6H5NH2 <NH3 < CH3NH2<(CH3)2NH

Which of the following from isocyanide on reaction with CHCl and KOH?
(a) C6H5NHCH3
(b) CH3C6H4NH2
(c) C6H5NHC4H9.
(d) C6H5N (C2H5)2

Answer: (b) CH3C6H4NH2

Among the compounds C3H7NH2, CH3NH2, C2H5NH2, and C6H5NH2. Which is the least basic compound?
(a) CH3NH2
(b) C2H5NH2
(c) C3H7NH2
(d) C6H5NH2

Answer: (d) C6H5NH2

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: (a) aromatic primary amines

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: (b) sulphanilic acid

Which of the following has highest pKb value?
(a) (CH3)3CNH2
(b) NH3
(c) (CH3)2MH
(d) CH3NH3

Answer: (b) NH3

Primary, secondary and tertiary amines may be separated by using
(a) iodoform
(b) diethyloxalate
(c) benzene sulphonyl chloride
(d) acetyl chloride

Answer: (c) benzene sulphonyl chloride

In order to prepare a 1° amine from an alkyl halide with simultaneous addition of one CH2 group in the carbon chain, the reagent used as source of nitrogen is ____________
(a) Sodium amide, NaNH2
(b) Sodium azide, NaN3
(c) Potassium cyanide, KCN
(d) Potassium phthalimide, C6H4(CO)2NK+

Answer: (c) Potassium cyanide, KCN

4-Nitrotoluene is treated with bromine to get P. ‘P’ is reduced with Sn/HCl to get compound ‘Q\ ‘Q’ is diazotised and the product is treated with phosphinic acid to get compound ‘R’. ‘R’ is oxidised with alkaline KMn04 to get ‘S’. Compound ‘S’ is
(a) 2-Bromo-4-hydroxy benzoic acid
(b) 2-Bromo benzoic acid
(c) 3-Bromo benzoic acid
(d) 4-Bromo benzoic acid

Answer: (b) 2-Bromo benzoic acid

Amongst the given set of reactants, the most appropriate for preparing 2° amine is ____________
(a) 2° R—Br + NH3
(b) 2° R—Br + NaCN followed by H2/Pt
(c) 1° R—NH2 + RCHO followed by H2/Pt
(d) 1° R—Br (2 mol) + potassium phthalimide followed by H3O+/heat

Answer: (c) 1° R—NH2 + RCHO followed by H2/Pt

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