JEE Chemistry Amines MCQs Set 24

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Multiple Choice Questions

Question 1: How can benzyl chloride be converted into 2-phenylethanamine in two steps?
(a) \( \ce{NH3} \), then \( \ce{CH3Cl} \)
(b) \( \ce{KCN} \), then \( \ce{LiAlH4} \) (or \( \ce{H2/Ni} \))
(c) \( \ce{NaNO2} \), then reduction
(d) \( \ce{KCN} \), then partial hydrolysis
Show Answer & Explanation

Answer: (b) \( \ce{KCN} \), then \( \ce{LiAlH4} \) (or \( \ce{H2/Ni} \))

Explanation:
1. Cyanide substitutes Cl, extending the chain by one carbon: \( \ce{C6H5CH2CN} \).
2. Reducing the nitrile gives \( \ce{C6H5CH2CH2NH2} \).

Teacher's Note:
1. Nitrile reduction gives a primary amine with the same carbon count.

Question 2: Propanoic acid reacts with \( \ce{SOCl2} \) to give A, A with \( \ce{NH3} \) gives B, and B with \( \ce{Br2/KOH} \) gives C. C is:
(a) Propylamine
(b) N-Methylethanamine
(c) Propanamide
(d) Ethylamine
Show Answer & Explanation

Answer: (d) Ethylamine

Explanation:
1. A = propanoyl chloride; B = propanamide.
2. Hofmann bromamide degradation removes one carbon: \( \ce{CH3CH2NH2} \).

Teacher's Note:
1. Hofmann degradation shortens the chain by one carbon.

Question 3: Benzamide is treated with \( \ce{Br2/NaOH} \) to give A, A is diazotised to B, and B with \( \ce{H3PO2} \) gives C. C is:
(a) Phenol
(b) Toluene
(c) Benzene
(d) Chlorobenzene
Show Answer & Explanation

Answer: (c) Benzene

Explanation:
1. A = aniline (Hofmann degradation).
2. B = benzenediazonium chloride; \( \ce{H3PO2} \) replaces \( \ce{N2+} \) by H, giving benzene.

Teacher's Note:
1. Hypophosphorous acid removes the diazonium group.

Question 4: An amine is obtained as the major product in which of the following reactions?
(a) 1-Chloro-1-methylcyclohexane treated with \( \ce{NH3} \)
(b) \( \ce{CH3CONHCH3} \) treated with \( \ce{LiAlH4} \)
(c) 2-Chlorobutane treated with \( \ce{NaNH2/NH3} \)
(d) None of these
Show Answer & Explanation

Answer: (b) \( \ce{CH3CONHCH3} \) treated with \( \ce{LiAlH4} \)

Explanation:
1. \( \ce{LiAlH4} \) reduces the amide to the amine \( \ce{CH3CH2NHCH3} \).
2. A tertiary halide and a strongly basic amide ion mainly give elimination (alkenes), not amines.

Teacher's Note:
1. Tertiary halides eliminate with basic nucleophiles.

Question 5: Ethyl iodide is treated with NaCN to give A, A is partially hydrolysed to B, and B is treated with \( \ce{NaOH + Br2} \) to give C. C is:
(a) Propylamine
(b) Methylamine
(c) Propanamide
(d) Ethylamine
Show Answer & Explanation

Answer: (d) Ethylamine

Explanation:
1. A = propanenitrile; B = propanamide (partial hydrolysis).
2. Hofmann degradation of propanamide gives ethylamine.

Teacher's Note:
1. The nitrile adds a carbon; Hofmann removes it.

Question 6: In the Hofmann bromamide degradation reaction, the number of moles of NaOH and \( \ce{Br2} \) used per mole of amine produced are:
(a) one mole of NaOH and one mole of \( \ce{Br2} \)
(b) four moles of NaOH and one mole of \( \ce{Br2} \)
(c) four moles of NaOH and two moles of \( \ce{Br2} \)
(d) two moles of NaOH and two moles of \( \ce{Br2} \)
Show Answer & Explanation

Answer: (b) four moles of NaOH and one mole of \( \ce{Br2} \)

Explanation:
1. \( \ce{RCONH2 + Br2 + 4NaOH -> RNH2 + Na2CO3 + 2NaBr + 2H2O} \).

Teacher's Note:
1. Balance the overall equation.

Question 7: Nitrobenzene is reduced with Sn/HCl to P, P is diazotised to Q, Q with CuCN/HCN gives R and R is hydrolysed to S. Which statement is not correct?
(a) P is \( \ce{C6H5NH2} \)
(b) R is \( \ce{C6H5CH2CN} \)
(c) Q is \( \ce{C6H5N2+Cl-} \)
(d) S is \( \ce{C6H5COOH} \)
Show Answer & Explanation

Answer: (b) R is \( \ce{C6H5CH2CN} \)

Explanation:
1. R is benzonitrile, \( \ce{C6H5CN} \), formed by the Sandmeyer reaction; \( \ce{C6H5CH2CN} \) would need a benzyl halide.
2. The other statements are correct, and S = benzoic acid.

Teacher's Note:
1. Sandmeyer cyanation keeps the same number of carbons on the ring side.

Question 8: Aniline is diazotised to A, A is treated with \( \ce{H3PO2} \) to give B, B reacts with \( \ce{CH3Cl/AlCl3} \) to give C, and C with \( \ce{Cl2} \) (1 mole, light, 383 K) gives D, which on aqueous KOH gives benzyl alcohol. C and D are:
(a) Toluene and chlorotoluene
(b) Ethylbenzene and benzyl chloride
(c) Toluene and benzyl chloride
(d) Benzene and benzyl chloride
Show Answer & Explanation

Answer: (c) Toluene and benzyl chloride

Explanation:
1. B = benzene (diazonium group replaced by H); C = toluene by Friedel-Crafts methylation.
2. Light and heat chlorinate the side chain giving \( \ce{C6H5CH2Cl} \); aqueous KOH hydrolyses it to benzyl alcohol.

Teacher's Note:
1. Light means side-chain halogenation.

Question 9: Propanoic acid is treated with \( \ce{Br2} \) and red P to give X, which reacts with \( \ce{NH3} \) to give Y. Y is:
(a) Alanine
(b) Lactic acid
(c) Ethylamine
(d) Propylamine
Show Answer & Explanation

Answer: (a) Alanine

Explanation:
1. Hell-Volhard-Zelinsky reaction brominates the alpha carbon: X = 2-bromopropanoic acid.
2. Ammonia replaces Br by \( \ce{NH2} \): Y = 2-aminopropanoic acid (alanine).

Teacher's Note:
1. HVZ gives alpha-halo acids.

Question 10: Benzene is nitrated to A, A is brominated with \( \ce{Br2/Fe} \) to B, B is reduced by Sn/HCl to C, C is diazotised to D, and D is boiled with dilute \( \ce{H2SO4} \) to give E. E is:
(a) 3-Bromophenol
(b) 4-Bromophenol
(c) 2-Bromophenol
(d) Phenol
Show Answer & Explanation

Answer: (a) 3-Bromophenol

Explanation:
1. A = nitrobenzene; B = 3-bromonitrobenzene (meta director).
2. C = 3-bromoaniline; D = its diazonium salt, which hydrolyses to 3-bromophenol.

Teacher's Note:
1. The nitro group directs bromine to the meta position.

Question 11: Propanoic acid is heated with \( \ce{NH3} \) to give A, A is treated with \( \ce{Br2/KOH} \) to give B, B is treated with \( \ce{HNO2} \) to give C, and C is oxidised with \( \ce{K2Cr2O7/H2SO4} \) to give D. D is:
(a) Propanoic acid
(b) Ethanoic acid
(c) Ethanal
(d) Methanoic acid
Show Answer & Explanation

Answer: (b) Ethanoic acid

Explanation:
1. A = propanamide, B = ethylamine, C = ethanol (HNO2 on a primary aliphatic amine).
2. Oxidation of ethanol by acidified dichromate gives ethanoic acid.

Teacher's Note:
1. Primary aliphatic amine plus HNO2 gives the alcohol.

Question 12: In the sequence X + \( \ce{NaNO2} \) + 2HCl (273-278 K) to 3-methylbenzenediazonium chloride, then \( \ce{HBF4} \) to Y, then \( \ce{NaNO2/Cu} \), heat to 3-nitrotoluene, X is:
(a) 3-Methylaniline (m-toluidine)
(b) 4-Methylaniline
(c) 2-Methylaniline
(d) N-Methylaniline
Show Answer & Explanation

Answer: (a) 3-Methylaniline (m-toluidine)

Explanation:
1. Diazotisation requires a primary aromatic amine.
2. The diazonium salt retains the substituent pattern, so X must be m-toluidine.

Teacher's Note:
1. The diazonium group replaces the NH2 at the same position.

Question 13: Nitrobenzene is reduced with Fe/HCl to A, A is treated with \( \ce{HNO2} \) at 273 K to give B, and B couples with phenol to give C. C is:
(a) Diphenylamine
(b) 4-Hydroxyazobenzene (p-hydroxyazobenzene)
(c) 4-Aminoazobenzene
(d) 2,4,6-Tribromophenol
Show Answer & Explanation

Answer: (b) 4-Hydroxyazobenzene (p-hydroxyazobenzene)

Explanation:
1. A = aniline; B = benzenediazonium chloride.
2. Coupling with phenol in mildly alkaline medium occurs at the para position, giving the azo dye \( \ce{C6H5N=NC6H4OH} \).

Teacher's Note:
1. Azo coupling occurs para to the activating group.

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