NEET Amines PYQs & Practice Questions

10 free NEET Chemistry questions on Amines, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Classification and Nomenclature · Basicity of Amines · Preparation of Amines · Chemical Reactions of Amines · Electrophilic Aromatic Substitution · Preparation of Amines – Hofmann Bromamide Reaction · Identification and Tests for Amines · Basic Strength of Amines · Diazonium Salt Preparation and Reactions · Identifying a Tertiary Amine · Strongest Base Among Substituted Amines in Water · Reagent That Adds a Carbon While Forming a Primary Amine

Q1NEET 2026Acidity and Basicity - Effect of Substituents (Assertion-Reason)
Given below are two statements: Statement-I: Oxidation of p-nitrotoluene with acidic KMnO4 gives an acid that is stronger than benzoic acid. Statement-II: Reduction of p-nitrotoluene with Sn/HCl followed by neutralization gives an amine that is more basic than aniline. In light of the above statements, choose the most appropriate answer from the options given below.
- A. Both Statement-I and Statement-II are incorrect.
- B. Statement-I is correct but Statement-II is incorrect.
- C. Statement-I is incorrect but Statement-II is correct.
- D. Both Statement-I and Statement-II are correct.
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Answer: C — Statement-I is incorrect but Statement-II is correct.
Oxidation of p-nitrotoluene with acidic KMnO4 oxidizes the methyl group to -COOH, giving p-nitrobenzoic acid. The electron-withdrawing -NO2 group stabilizes the carboxylate conjugate base, making it a stronger acid than benzoic acid -- Statement-I is correct. Reduction of p-nitrotoluene's -NO2 group with Sn/HCl (followed by neutralization) gives p-toluidine (p-methylaniline). The electron-donating methyl group increases electron density on nitrogen, making p-toluidine more basic than aniline -- Statement-II is correct. Both statements are correct.
Q2NEET 2026Preparation of Aniline - Comparing Methods
Identify the reactions which give aniline as the major product. A. PhCN --[LiAlH4]--> ? B. PhCONH2 --[KOH, Br2]--> ? C. PhNO2 --[NaBH4]--> ? D. PhNHCOCH3 (acetanilide) --[HCl, H2O, Δ]--> ? Choose the correct answer from the options given below.
- A. B and D only
- B. A and C only
- C. C and D only
- D. A and B only
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Answer: —
A: PhCN + LiAlH4 reduces the nitrile to a primary amine with an extra carbon, giving benzylamine (PhCH2NH2), not aniline. B: PhCONH2 (benzamide) + KOH/Br2 undergoes the Hofmann bromamide degradation, directly giving aniline (PhNH2) with loss of the carbonyl carbon as CO2 -- gives aniline. C: NaBH4 is a mild reducing agent and does not reduce aromatic nitro groups under normal conditions, so nitrobenzene is NOT reduced to aniline by NaBH4. D: Acetanilide (PhNHCOCH3) undergoes acidic hydrolysis with HCl/H2O/heat to give aniline and acetic acid -- gives aniline. So only B and D give aniline.
Q3NEET 2025Basic Strength of Amines
The correct order of decreasing basic strength of the given amines is : N-methylaniline, benzenamine, ethanamine, N-ethylethanamine
- A. N-methylaniline > benzenamine > ethanamine > N-ethylethanamine
- B. N-ethylethanamine > ethanamine > benzenamine > N-methylaniline
- C. N-ethylethanamine > ethanamine > N-methylaniline > benzenamine
- D. benzenamine > ethanamine > N-methylaniline > N-ethylethanamine
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Answer: B — N-ethylethanamine > ethanamine > benzenamine > N-methylaniline
Aliphatic amines are more basic than aromatic amines due to +I effect without resonance delocalization of lone pair. Among aliphatic amines, secondary (diethylamine) > primary (ethanamine). Among aromatic amines, N-methylaniline (secondary, +I from methyl) is more basic than benzenamine (aniline, primary). Order: N-ethylethanamine > ethanamine > N-methylaniline > benzenamine.
Q4NEET 2025Diazonium Salt Preparation and Reactions
Given below are two statements: Statement I: Benzenediazonium salt is prepared by the reaction of aniline with nitrous acid at 273-278 K. It decomposes easily in the dry state. Statement II: Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI. In the light of the above statements, choose the most appropriate answer from the options given below :
- A. Both Statement I and Statement II are correct
- B. Both Statement I and Statement II are incorrect
- C. Statement I is correct but Statement II is incorrect
- D. Statement I is incorrect but Statement II is correct
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Answer: —
Both statements are correct: benzenediazonium salt is indeed prepared from aniline + NaNO2/HCl at 273-278K and is unstable/decomposes in dry form; and direct iodination of benzene is difficult, so iodobenzene is made via the Sandmeyer-type reaction of the diazonium salt with KI.

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Q5NEET 2023Preparation of Amines
Which of the following reactions will NOT give primary amine as the product? (1) CH3CN --(i) LiAlH4 (ii) H3O+--> Product (2) CH3NC --(i) LiAlH4 (ii) H3O+--> Product (3) CH3CONH2 --(i) LiAlH4 (ii) H3O+--> Product (4) CH3CONH2 --Br2/KOH--> Product
- A. CH3CN with LiAlH4/H3O+
- B. CH3NC with LiAlH4/H3O+
- C. CH3CONH2 with LiAlH4/H3O+
- D. CH3CONH2 with Br2/KOH
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Answer: A — CH3CN with LiAlH4/H3O+
Reduction of an isocyanide (CH3NC) with LiAlH4 followed by hydrolysis gives a secondary amine (CH3-NH-CH3), not a primary amine. Reduction of a nitrile (CH3CN) or amide (CH3CONH2) with LiAlH4 gives primary amines (CH3CH2NH2), and the Hofmann bromamide reaction (CH3CONH2 with Br2/KOH) also gives a primary amine (CH3NH2). So option 2 is the one that does NOT give a primary amine.
Q6NEET 2022Kjeldahl's Method
The Kjeldahl's method for the estimation of nitrogen can be used to estimate the amount of nitrogen in which one of the following compounds? (1) Pyridine (nitrogen in aromatic ring) (2) Aniline (benzene with NH2 group) (3) Azobenzene (benzene-N=N-benzene) (4) Nitrobenzene (benzene with NO2 group)
- A. Pyridine
- B. Aniline (benzene with NH2 group)
- C. Azobenzene
- D. Nitrobenzene
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Answer: A — Pyridine
Kjeldahl's method is not applicable to nitrogen present in nitro and azo groups, or nitrogen present in a ring (like pyridine). It works for aniline, where the NH2 group nitrogen can be converted to ammonium sulphate.
Q7NEET 2022Reactions of Nitriles and Diazonium Salts
The product formed from the following reaction sequence is: Benzonitrile (Ph-CN) treated with (i) LiAlH4, H2O (ii) NaNO2 + HCl (iii) H2O
- A. Benzyl diazonium chloride (Ph-CH2-N2+Cl-)
- B. Benzyl chloride (Ph-CH2-Cl)
- C. Benzyl alcohol (Ph-CH2-OH)
- D. Benzamide (Ph-CO-NH2)
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Answer: B — Benzyl chloride (Ph-CH2-Cl)
Benzonitrile is reduced by LiAlH4/H2O to benzylamine (Ph-CH2-NH2). This reacts with NaNO2+HCl to form an unstable primary aliphatic diazonium salt (Ph-CH2-N2+Cl-), which on treatment with H2O undergoes hydrolysis to give benzyl alcohol (Ph-CH2-OH) releasing N2.
Q8NEET 2022Diazonium Salt Formation - Aliphatic vs Aromatic Amines
Given below are two statements: Statement I: Primary aliphatic amines react with HNO2 to give unstable diazonium salts. Statement II: Primary aromatic amines react with HNO2 to form diazonium salts which are stable even above 300 K. In the light of the above statements, choose the most appropriate answer from the options given below:
- A. Both Statement I and Statement II are correct.
- B. Both Statement I and Statement II are incorrect.
- C. Statement I is correct but Statement II is incorrect.
- D. Statement I is incorrect but Statement II is correct.
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Answer: B — Both Statement I and Statement II are incorrect.
Statement I is correct: aliphatic diazonium salts are highly unstable and decompose immediately even at low temperatures. Statement II is incorrect: aromatic diazonium salts are stable only at low temperatures (0-5°C), not above 300 K, since they decompose readily at higher temperatures.

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Q9NEET 2022Reaction Sequence - Nitrile to Diazonium Hydrolysis
The product formed from the following reaction sequence is: Benzonitrile (Ph-CN) -- (i) LiAlH4, H2O (ii) NaNO2 + HCl (iii) H2O -->
- A. Benzamide (Ph-CO-NH2)
- B. Phenethyl diazonium chloride (Ph-CH2-CH2-N2+Cl-)
- C. Benzyl chloride (Ph-CH2-Cl)
- D. Benzyl alcohol (Ph-CH2-OH)
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Answer: C — Benzyl chloride (Ph-CH2-Cl)
LiAlH4 reduces the nitrile (PhCN) to a primary amine, PhCH2NH2 (benzylamine). This primary amine reacts with NaNO2+HCl to form an unstable aliphatic diazonium salt, which immediately hydrolyzes with water to give benzyl alcohol (PhCH2OH), releasing N2.
Q10NEET 2021Hinsberg's Reagent Test
Identify the compound that will react with Hinsberg's reagent to give a solid which dissolves in alkali.
- A. CH3CH2-N(CH3)-CH2CH3 (tertiary amine, N,N-diethyl-N-methylamine type)
- B. CH3CH2-NO2 (nitroethane)
- C. CH3CH2-NH-CH3 (secondary amine, N-methylethanamine)
- D. CH3CH2-NH2 (primary amine, ethanamine)
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Answer: C — CH3CH2-NH-CH3 (secondary amine, N-methylethanamine)
Primary (1°) amine reacts with Hinsberg reagent (benzenesulfonyl chloride) to give a precipitate that is soluble in alkali, since the N-H proton remaining is acidic (sulfonamide N-H).

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