When ethanamide is treated with Br2 and NaOH in ethanolic solution we get

The product is methanamine.This is the Hofmann bromamide degradation: an amide treated with bromine and sodium hydroxide gives a primary amine with one carbon fewer.CH3CONH2 + Br2 + 4NaOH → CH3NH2 + Na2CO3 + 2NaBr + 2H2OThe carbonyl carbon is lost as carbonate, so ethanamide (two carbons) gives methanamine (one carbon), not ethanamine. No brominated compound is the final product.

Amines and Nitrogen Compounds Questions and Answers

Q: Aniline reacts with acetic anhydride in pyridine to give a product which reacts with Br2 in CH3COOH to get

The product is p-bromoacetanilide.Acetic anhydride in pyridine acetylates aniline to acetanilide, C6H5NHCOCH3. The acetyl group makes the nitrogen's lone pair less available to the ring, so the ring is only moderately activated, and bromination in acetic acid gives a single substitution, mainly at the para position (the bulky NHCOCH3 group hinders the ortho positions).The bromoaniline options leave out the acetyl group, which stays on until the amide is hydrolysed in a separate step. o-Bromoacetanilide is the minor product, and the meta isomer is not formed because NHCOCH3 directs ortho and para.

Q: Which of the following answers positively in carbylamine test?

The amine that gives a positive carbylamine test is benzenamine (aniline).The carbylamine (isocyanide) test, heating with chloroform and alcoholic KOH, is given only by primary amines, aliphatic or aromatic. They form isocyanides, recognised by their foul smell. Benzenamine, C6H5NH2, is a primary amine.N-ethylmethanamine and N-methylbenzenamine are secondary amines, and N,N-dimethylethanamine and N,N-dimethylbenzenamine are tertiary amines; none of them gives the test.

Q: Which of the following amine has the highest pKb value in aqueous phase?

The amine with the highest pKb is Benzenamine.A higher pKb means a weaker base. In aniline the nitrogen lone pair is delocalised into the benzene ring by resonance, so it is less available to accept a proton. Aniline's pKb is about 9.4.N-methylbenzenamine (about 9.3) is slightly stronger, because the methyl group pushes electrons onto nitrogen. The aliphatic amines, methanamine, ethanamine and N-methylmethanamine, are much stronger bases, with pKb values around 3.3.

Q: An aliphatic amine reacts with Hinsberg's reagent to give a product insoluble in NaOH. The amine is

The amine is N-methylethanamine, a secondary amine.Hinsberg's reagent is benzenesulphonyl chloride. A secondary amine gives an N,N-disubstituted sulphonamide with no hydrogen on nitrogen, so the product has no acidic hydrogen and is insoluble in alkali.Primary amines such as ethanamine give a sulphonamide with an acidic N-H, which dissolves in NaOH. Tertiary amines such as N,N-diethylethanamine do not react. Benzenamine and p-toluidine are primary and aromatic, not aliphatic.

Q: Benzenamine reacts with ethanoic anhydride to give

The product is N-phenylethanamide (acetanilide).Ethanoic anhydride acetylates the -NH2 group of benzenamine (aniline): one acetyl group replaces a hydrogen on nitrogen and ethanoic acid is released.C6H5NH2 + (CH3CO)2O → C6H5NHCOCH3 + CH3COOHBenzamide and N-phenylbenzamide would need a benzoyl group, from benzoyl chloride, not an acetyl group. Acylation takes place on nitrogen, not on the ring, so the ring-substituted products do not form.

Q: Which of the following amines does not form carbylamine?

The amine that does not form a carbylamine is N-Methylethanamine.The carbylamine (isocyanide) test, heating with chloroform and alcoholic KOH, works only for primary amines, aliphatic or aromatic, giving a foul-smelling isocyanide.N-methylethanamine, CH3CH2NHCH3, is a secondary amine and does not respond. Ethanamine, propan-1-amine, propan-2-amine and benzenamine (aniline) are all primary amines and give the test.

Q: Which one of the following compounds is strongly basic in aqueous medium?

The strongly basic compound is N-ethylethanamine (diethylamine).Aliphatic amines are stronger bases than ammonia, because alkyl groups push electron density onto the nitrogen. In water, secondary amines such as diethylamine are the strongest, owing to the combined effect of electron release and solvation of the ammonium ion.In benzenamine (aniline) and N,N-dimethylbenzenamine the nitrogen's lone pair is delocalised into the ring, making them much weaker bases than ammonia. Phenylmethanamine (benzylamine) is a primary amine, weaker than diethylamine, and ammonia is weaker than both aliphatic amines.

Q: The correct formula of Hinsberg's reagent is

Hinsberg's reagent is C6H5SO2Cl, benzenesulphonyl chloride.It is used to distinguish primary, secondary and tertiary amines: primary amines give a sulphonamide soluble in alkali, secondary amines give one insoluble in alkali, and tertiary amines do not react.The other formulae do not represent benzenesulphonyl chloride: C6H5SOCl has one oxygen too few, C6H5SO3Cl one too many, and the dichloride formulae carry an extra chlorine.

31.

Aniline reacts with acetic anhydride in pyridine to give a product which reacts with Br₂ in CH₃COOH to get

A.o-bromoaniline
B.p-bromoaniline
C.p-bromoacetanilide
D.o-bromoacetanilide
E.m-bromoacetanilide

Exam Year(s)

KEAM - 2026,

Answer: C

The product is p-bromoacetanilide.

Acetic anhydride in pyridine acetylates aniline to acetanilide, C₆H₅NHCOCH₃. The acetyl group makes the nitrogen's lone pair less available to the ring, so the ring is only moderately activated, and bromination in acetic acid gives a single substitution, mainly at the para position (the bulky NHCOCH₃ group hinders the ortho positions).

The bromoaniline options leave out the acetyl group, which stays on until the amide is hydrolysed in a separate step. o-Bromoacetanilide is the minor product, and the meta isomer is not formed because NHCOCH₃ directs ortho and para.

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32.

Which of the following answers positively in carbylamine test?

A.N-ethylmethanamine
B.N,N-dimethylethanamine
C.benzenamine
D.N-methylbenzenamine
E.N,N-dimethylbenzeneamine

Exam Year(s)

KEAM - 2026,

Answer: C

The amine that gives a positive carbylamine test is benzenamine (aniline).

The carbylamine (isocyanide) test, heating with chloroform and alcoholic KOH, is given only by primary amines, aliphatic or aromatic. They form isocyanides, recognised by their foul smell. Benzenamine, C₆H₅NH₂, is a primary amine.

N-ethylmethanamine and N-methylbenzenamine are secondary amines, and N,N-dimethylethanamine and N,N-dimethylbenzenamine are tertiary amines; none of them gives the test.

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33.

Which of the following amine has the highest pK_b value in aqueous phase?

A.Methanamine
B.N-methylmethanamine
C.Ethanamine
D.N-Methylbenzenamine
E.Benzenamine

Exam Year(s)

KEAM - 2026,

Answer: E

The amine with the highest pK_b is Benzenamine.

A higher pK_b means a weaker base. In aniline the nitrogen lone pair is delocalised into the benzene ring by resonance, so it is less available to accept a proton. Aniline's pK_b is about 9.4.

N-methylbenzenamine (about 9.3) is slightly stronger, because the methyl group pushes electrons onto nitrogen. The aliphatic amines, methanamine, ethanamine and N-methylmethanamine, are much stronger bases, with pK_b values around 3.3.

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34.

An aliphatic amine reacts with Hinsberg's reagent to give a product insoluble in NaOH. The amine is

A.ethanamine
B.N-methylethanamine
C.benzenamine
D.N, N-diethylethanamine
E.p-toludine

Exam Year(s)

KEAM - 2026,

Answer: B

The amine is N-methylethanamine, a secondary amine.

Hinsberg's reagent is benzenesulphonyl chloride. A secondary amine gives an N,N-disubstituted sulphonamide with no hydrogen on nitrogen, so the product has no acidic hydrogen and is insoluble in alkali.

Primary amines such as ethanamine give a sulphonamide with an acidic N-H, which dissolves in NaOH. Tertiary amines such as N,N-diethylethanamine do not react. Benzenamine and p-toluidine are primary and aromatic, not aliphatic.

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35.

When ethanamide is treated with Br₂ and NaOH in ethanolic solution we get

A.ethanamine
B.bromoethanamide
C.bromoethane
D.methanamine
E.2-bromoethanamine

Exam Year(s)

KEAM - 2026,

Answer: D

The product is methanamine.

This is the Hofmann bromamide degradation: an amide treated with bromine and sodium hydroxide gives a primary amine with one carbon fewer.

CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O

The carbonyl carbon is lost as carbonate, so ethanamide (two carbons) gives methanamine (one carbon), not ethanamine. No brominated compound is the final product.

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36.

Benzenamine reacts with ethanoic anhydride to give

A.benzamide
B.N-phenylbenzamide
C.N-phenylethanamide
D.o-aminoacetophenone
E.p-ethylbenzamide

Exam Year(s)

KEAM - 2026,

Answer: C

The product is N-phenylethanamide (acetanilide).

Ethanoic anhydride acetylates the -NH₂ group of benzenamine (aniline): one acetyl group replaces a hydrogen on nitrogen and ethanoic acid is released.

C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH

Benzamide and N-phenylbenzamide would need a benzoyl group, from benzoyl chloride, not an acetyl group. Acylation takes place on nitrogen, not on the ring, so the ring-substituted products do not form.

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37.

The increasing order of boiling point of the following amines is

A.\(\mathrm{C_2H_5N(CH_3)_2 \lt (C_2H_5)_2NH \lt C_4H_9NH_2}\)
B.\(\mathrm{C_4H_9NH_2 \lt C_2H_5N(CH_3)_2 \lt (C_2H_5)_2NH}\)
C.\(\mathrm{C_2H_5NH(CH_3)_2 \lt C_4H_9NH_2 \lt (C_2H_5)_2NH}\)
D.\(\mathrm{(C_2H_5)_2NH \lt C_4H_9NH_2 \lt C_2H_5N(CH_3)_2}\)
E.\(\mathrm{(C_2H_5)_2NH \lt C_2H_5N(CH_3)_2 \lt C_4H_9NH_2}\)

Exam Year(s)

KEAM - 2026,

Answer: A

The increasing order is \(\mathrm{C_2H_5N(CH_3)_2 \lt (C_2H_5)_2NH \lt C_4H_9NH_2}\).

The three amines are isomers (C₄H₁₁N), so their boiling points depend on intermolecular hydrogen bonding. A primary amine has two N-H hydrogens and bonds most strongly; a secondary amine has one; a tertiary amine has none.

So the tertiary amine boils lowest, then the secondary, then the primary. The other orders place the tertiary amine above an amine that can hydrogen-bond, or put the secondary above the primary.

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38.

Which of the following amines does not form carbylamine?

A.Ethanamine
B.Benzenamine
C.Propan-2-amine
D.Propan-1-amine
E.N-Methylethanamine

Exam Year(s)

KEAM - 2026,

Answer: E

The amine that does not form a carbylamine is N-Methylethanamine.

The carbylamine (isocyanide) test, heating with chloroform and alcoholic KOH, works only for primary amines, aliphatic or aromatic, giving a foul-smelling isocyanide.

N-methylethanamine, CH₃CH₂NHCH₃, is a secondary amine and does not respond. Ethanamine, propan-1-amine, propan-2-amine and benzenamine (aniline) are all primary amines and give the test.

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39.

Which one of the following compounds is strongly basic in aqueous medium?

A.Benzenamine
B.N-ethylethanamine
C.Phenylmethanamine
D.N,N-Dimethylbenzenamine
E.Ammonia

Exam Year(s)

KEAM - 2026,

Answer: B

The strongly basic compound is N-ethylethanamine (diethylamine).

Aliphatic amines are stronger bases than ammonia, because alkyl groups push electron density onto the nitrogen. In water, secondary amines such as diethylamine are the strongest, owing to the combined effect of electron release and solvation of the ammonium ion.

In benzenamine (aniline) and N,N-dimethylbenzenamine the nitrogen's lone pair is delocalised into the ring, making them much weaker bases than ammonia. Phenylmethanamine (benzylamine) is a primary amine, weaker than diethylamine, and ammonia is weaker than both aliphatic amines.

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40.

The correct formula of Hinsberg's reagent is

A.\(\mathrm{C_{6}H_{5}SO_{2}Cl}\)
B.\(\mathrm{C_{6}H_{5}SOCl}\)
C.\(\mathrm{C_{6}H_{5}SOCl_{2}}\)
D.\(\mathrm{C_{6}H_{5}SO_{3}Cl}\)
E.\(\mathrm{C_{6}H_{5}SO_{2}Cl_{2}}\)

Exam Year(s)

KEAM - 2026,

Answer: A

Hinsberg's reagent is C₆H₅SO₂Cl, benzenesulphonyl chloride.

It is used to distinguish primary, secondary and tertiary amines: primary amines give a sulphonamide soluble in alkali, secondary amines give one insoluble in alkali, and tertiary amines do not react.

The other formulae do not represent benzenesulphonyl chloride: C₆H₅SOCl has one oxygen too few, C₆H₅SO₃Cl one too many, and the dichloride formulae carry an extra chlorine.

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