JEE Chemistry Amines MCQs Set 27

Chemistry Objective Questions and Answers: Amines

Explore reliable objective questions for Amines tailored for JEE learners. Utilizing these Chemistry multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

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Multiple Choice Questions

Question 1: Which of the following is a Sandmeyer reaction?
(a) C6H5N2Cl + HBF4, Δ → C6H5F
(b) C6H5N2Cl + Cu/HCl → C6H5Cl
(c) C6H5N2Cl + Cu2Cl2/HCl → C6H5Cl
(d) C6H5N2Cl + C6H6/NaOH → C6H5–C6H5
Show Answer & Explanation

Answer: (c) C6H5N2Cl + Cu2Cl2/HCl → C6H5Cl

Explanation:
1. The Sandmeyer reaction uses cuprous halide (Cu2Cl2/HCl, Cu2Br2/HBr or CuCN) to replace N2⁺.
2. With copper powder and HCl the reaction is the Gattermann reaction.
3. HBF4/Δ is the Balz-Schiemann reaction, and the last one is the Gomberg-Bachmann reaction.
4. Answer: (c)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 2: Which of the following amines cannot be formed by Gabriel phthalimide synthesis?
(a) p-Toluidine
(b) Benzylamine
(c) Isopropylamine
(d) sec-Butylamine
Show Answer & Explanation

Answer: (a) p-Toluidine

Explanation:
1. Gabriel synthesis uses the SN2 reaction of potassium phthalimide with an alkyl halide.
2. An aryl halide cannot undergo SN2, so aromatic amines such as p-toluidine cannot be made this way.
3. Benzylamine is formed from benzyl halide; secondary alkyl halides react poorly but can be used, unlike aryl halides.
4. Answer: (a)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 3: Which of the following conversions is incorrect?
(a) CH3CN → CH3CH2NH2 (Na/C2H5OH or LAH)
(b) CH3CONH2 → CH3CH2NH2 (Br2/KOH)
(c) CH3CONH2 → CH3COOH (NaNO2 + HCl)
(d) CH3CONH2 → CH3CN (P2O5)
Show Answer & Explanation

Answer: (b) CH3CONH2 → CH3CH2NH2 (Br2/KOH)

Explanation:
1. The Hofmann bromamide reaction removes one carbon atom.
2. Acetamide (2 C) gives methylamine, CH3NH2, not ethylamine.
3. The other three conversions are correct.
4. Answer: (b)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 4: Aniline cannot be obtained by:
(a) C6H5NO2 with conc. H2SO4, electrolysis
(b) C6H5Cl with NaNH2, Δ
(c) C6H5COCl with NaN3, then KOH
(d) C6H5OH with NH3, dry ZnCl2
Show Answer & Explanation

Answer: (a) C6H5NO2 with conc. H2SO4, electrolysis

Explanation:
1. Electrolytic reduction of nitrobenzene in strong acid gives p-aminophenol (through Bamberger rearrangement), not aniline.
2. Chlorobenzene with NaNH2 (benzyne), the Curtius rearrangement of benzoyl azide, and the ammonolysis of phenol over ZnCl2 all give aniline.
3. Answer: (a)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 5: Nitrobenzene is reduced with Zn + NH4Cl to give compound A. A is:
(a)
(b)
(c)
(d)
Show Answer & Explanation

Answer: (d)

Explanation:
1. In neutral medium (Zn + NH4Cl) nitrobenzene is reduced only to the hydroxylamine stage.
2. A is N-phenylhydroxylamine, C6H5NHOH.
3. Answer: (d)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 6: Aniline → (H2SO4) A → (453–473 K) B. Compound B is:
(a)
(b)
(c)
(d)
Show Answer & Explanation

Answer: (d)

Explanation:
1. Aniline and H2SO4 give anilinium hydrogen sulphate (A).
2. On heating at 453–473 K it rearranges to the para sulphonic acid.
3. B is sulphanilic acid (4-aminobenzenesulphonic acid), which exists as a zwitterion.
4. Answer: (d)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 7: Aniline reacts with Br2/H2O to give A. A is:
(a)
(b)
(c)
(d) Both (a) and (b)
Show Answer & Explanation

Answer: (c)

Explanation:
1. The NH2 group strongly activates the ring at all ortho and para positions.
2. In water the reaction cannot be stopped at a mono stage and gives a white precipitate of 2,4,6-tribromoaniline.
3. Answer: (c)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 8: The reagents needed to convert benzamide into acetanilide (C6H5CONH2 → C6H5NHCOCH3) are:
(a) KOH, Br2; LiAlH4
(b) KOH, Br2; CH3COCl
(c) KOH, Br2; Ni/H2; CH3COCl
(d) HONO; (CH3CO)2O
Show Answer & Explanation

Answer: (b) KOH, Br2; CH3COCl

Explanation:
1. Hofmann bromamide reaction (KOH/Br2) turns benzamide into aniline.
2. Acetylation with CH3COCl then gives acetanilide.
3. Answer: (b)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 9: Benzaldehyde + aniline → product. The product is called:
(a) Azo dye
(b) Schiff's reagent
(c) Schiff base
(d) Glyptal (a polymer)
Show Answer & Explanation

Answer: (c) Schiff base

Explanation:
1. A primary amine condenses with an aldehyde, losing water, to form an imine (azomethine) C6H5CH=NC6H5.
2. Such an imine is called a Schiff base.
3. Answer: (c)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 10: The benzoylation of the –NH2 group is known as:
(a) Balz-Schiemann reaction
(b) Schotten-Baumann reaction
(c) Gabriel phthalimide reaction
(d) Friedel-Crafts reaction
Show Answer & Explanation

Answer: (b) Schotten-Baumann reaction

Explanation:
1. Acylation of amines or phenols with benzoyl chloride in the presence of aqueous NaOH is the Schotten-Baumann reaction.
2. Answer: (b)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 11: CH3CONH2 → (PCl5) A → (LiAlH4, then H2O) B. The final product B is:
(a) A primary alcohol
(b) A primary amine
(c) A secondary amine
(d) A carboxylic acid
Show Answer & Explanation

Answer: (b) A primary amine

Explanation:
1. PCl5 dehydrates acetamide to acetonitrile, CH3CN (A).
2. LiAlH4 reduces the nitrile to the primary amine, CH3CH2NH2 (B).
3. Answer: (b)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 12: Which of the following compounds gives a product with a foul smell when heated with chloroform and an alkali?
(a) (CH3)2NH
(b) CH3NH2
(c) CH3CONH2
(d) (CH3)3N
Show Answer & Explanation

Answer: (b) CH3NH2

Explanation:
1. Only primary amines give the carbylamine (isocyanide) reaction.
2. CH3NH2 is the only primary amine in the list.
3. Answer: (b)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 13: Nitrobenzene on reduction with Fe + HCl gives:
(a) Phenylhydroxylamine
(b) p-Aminophenol
(c) Azoxybenzene
(d) Aniline
Show Answer & Explanation

Answer: (d) Aniline

Explanation:
1. Reduction in strongly acidic medium with a metal and acid (Fe/HCl, Sn/HCl) goes all the way to the amine.
2. The product is aniline.
3. Answer: (d)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 14: Aniline cannot be obtained by the Gabriel phthalimide synthesis due to:
(a) Bulky nature of the benzene ring
(b) Aryl halides do not undergo nucleophilic substitution under these conditions
(c) Aryl halides do not undergo electrophilic substitution
(d) Both (a) and (c)
Show Answer & Explanation

Answer: (b) Aryl halides do not undergo nucleophilic substitution under these conditions

Explanation:
1. The Gabriel synthesis needs an SN2 reaction of potassium phthalimide with the halide.
2. In an aryl halide the C–Cl bond has partial double-bond character, and the ring blocks back-side attack, so nucleophilic substitution is very difficult.
3. Answer: (b)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 15: Which one can give propanamine on reaction with LiAlH4?
(a) Propyl cyanide
(b) N-Methylpropanamide
(c) Propanenitrile
(d) Both (a) and (c)
Show Answer & Explanation

Answer: (c) Propanenitrile

Explanation:
1. LiAlH4 reduces R–C≡N to R–CH2NH2.
2. Propanenitrile, CH3CH2CN, gives CH3CH2CH2NH2.
3. Propyl cyanide is butanenitrile and gives butylamine, and N-methylpropanamide gives the secondary amine N-methylpropanamine.
4. Answer: (c)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 16: Compound A with Br2 + NaOH gives (CH3)2CH–CH2–NH2. The structure of A is:
(a)
(b)
(c)
(d)
Show Answer & Explanation

Answer: (b)

Explanation:
1. The Hofmann bromamide reaction gives an amine with one carbon fewer than the amide.
2. The amine has four carbons, so the amide must have five: (CH3)2CH–CH2–CONH2.
3. Answer: (b)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 17: The isocyanide (carbylamine) test is used for the detection of:
(a) Primary amines
(b) Secondary amines
(c) Only aniline
(d) Tertiary amines
Show Answer & Explanation

Answer: (a) Primary amines

Explanation:
1. Only primary amines (aliphatic or aromatic) convert to isocyanides with CHCl3 and KOH, giving a foul smell.
2. Answer: (a)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 18: Which one of the following is soluble in NaOH?
(a) N-Ethylbenzenesulphonamide
(b) N,N-Diethylbenzenesulphonamide
(c) N-Ethylaniline
(d) N-Methylethanamine
Show Answer & Explanation

Answer: (a) N-Ethylbenzenesulphonamide

Explanation:
1. N-Ethylbenzenesulphonamide, C6H5SO2NHC2H5, has an acidic N–H (the SO2 group withdraws electrons), so it dissolves in alkali as the sodium salt.
2. The N,N-diethyl compound has no N–H, and the amines are not acidic.
3. Answer: (a)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 19: Benzenediazonium chloride + X (H⁺) gives a yellow dye. X is:
(a) Phenol
(b) Aniline
(c) Benzene
(d) Benzenesulphonic acid
Show Answer & Explanation

Answer: (b) Aniline

Explanation:
1. In acidic medium, the diazonium ion couples with aniline at the para position to give p-aminoazobenzene, a yellow azo dye.
2. Phenol couples in alkaline medium to give an orange dye.
3. Answer: (b)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 20: (CH3)2CH–CONH2 → (Br2/NaOH) X → (HNO2) Y. X and Y are:
(a) X = (CH3)2CHCH2NH2; Y = (CH3)2CHCH2OH
(b) X = CH3CH=CHNH2; Y = CH3CH=CHOH
(c) X = (CH3)2CHNH2; Y = (CH3)2CHOH
(d) X = CH3CH2CH2NH2; Y = CH3CH2CH2OH
Show Answer & Explanation

Answer: (c) X = (CH3)2CHNH2; Y = (CH3)2CHOH

Explanation:
1. Hofmann bromamide reaction removes the carbonyl carbon: isobutyramide gives isopropylamine (X).
2. A primary aliphatic amine with HNO2 gives the alcohol (via the diazonium ion), so Y is propan-2-ol.
3. Answer: (c)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 21: The correct route for converting p-chloroaniline to p-chlorobenzylamine is:
(a) Alkylation, KCN, H2/Pt
(b) Diazotisation, CuCN, H2/Pt
(c) Oxidation, H2/Pt, hydrolysis
(d) Diazotisation, H2O/H⁺, Sn/HCl
Show Answer & Explanation

Answer: (b) Diazotisation, CuCN, H2/Pt

Explanation:
1. Diazotisation of the NH2 group and the Sandmeyer reaction with CuCN give p-chlorobenzonitrile.
2. H2/Pt then reduces the C≡N to CH2NH2, giving p-chlorobenzylamine (the aryl chlorine is not touched).
3. Answer: (b)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 22: Pyrrolidine reacts with excess CH3I. The compound formed is:
(a)
(b)
(c)
(d)
Show Answer & Explanation

Answer: (b)

Explanation:
1. The secondary amine is methylated step by step: first the N-methyl tertiary amine and then, with excess CH3I, the quaternary ammonium salt.
2. Nitrogen carries two CH3 groups and two ring bonds: N,N-dimethylpyrrolidinium iodide.
3. Answer: (b)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 23: Propionic acid → (SOCl2) B → (NH3, Δ) C → (KOH/Br2) D. The structure of D is:
(a)
(b)
(c)
(d)
Show Answer & Explanation

Answer: (b)

Explanation:
1. Propanoyl chloride (B) and then propanamide (C) are formed.
2. The Hofmann bromamide reaction removes one carbon, giving ethylamine, CH3CH2NH2 (D).
3. Answer: (b)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 24: Benzoic acid → (PCl5) P → (H2, Pd/BaSO4) Q → (50% KOH) R. R is:
(a) Acid + alcohol
(b) Ester
(c) Aldol
(d) Salt + alcohol
Show Answer & Explanation

Answer: (d) Salt + alcohol

Explanation:
1. PCl5 gives benzoyl chloride (P) and Rosenmund reduction gives benzaldehyde (Q).
2. Benzaldehyde has no alpha-hydrogen, so with 50% KOH it undergoes the Cannizzaro reaction.
3. Products: sodium benzoate (a salt) and benzyl alcohol.
4. Answer: (d)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 25: Which of the following compounds will give the Hoffmann bromamide reaction?
(a) Ph–CO–NH–CH3
(b) (CH3)3N
(c) Ph–NH2
(d) CH3CH2CONH2
Show Answer & Explanation

Answer: (d) CH3CH2CONH2

Explanation:
1. The Hofmann bromamide reaction needs a primary amide, RCONH2.
2. Propanamide fits and gives ethylamine.
3. N-Methylbenzamide is a secondary amide, and the others are amines.
4. Answer: (d)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 26: Phthalimide → (KOH) A → (R–Cl) B → (KOH, H2O) C. C cannot be:
(a) CH3CH2NH2
(b) Ph–CH2NH2
(c) (CH3)2CH–NH2
(d) Ph–NH2
Show Answer & Explanation

Answer: (d) Ph–NH2

Explanation:
1. This is the Gabriel phthalimide synthesis of primary amines.
2. The halide R–Cl must undergo SN2, so aryl halides cannot be used.
3. Aniline, therefore, cannot be formed.
4. Answer: (d)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

Question 27: Phthalic anhydride on strong heating with NH3 gives:
(a)
(b)
(c)
(d)
Show Answer & Explanation

Answer: (b)

Explanation:
1. Ammonia opens the anhydride to the amide-acid, which cyclises with loss of water on strong heating.
2. The product is phthalimide.
3. Answer: (b)

Teacher's Note:
1. Learn the named reactions of amines and diazonium salts and the reagent used in each; trace the sequence one step at a time.

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