Amines NEET PYQs (2005–2026) with Solutions

10 solved Amines PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumPreparation Methods

   Select the reagents that reduce nitriles to primary amines.

   1. AOption A: B, D and E only
   2. BOption B: A, C and D only
   3. COption C: A, D and E only
   4. DOption D: A, B and C only

   Show answer & explanationHide answer & explanation

   Correct answer: (B) A, C and D only

   Explanation

   LiAlH₄/H₂O, H₂/Ni and Na(Hg)/C₂H₅OH reduce nitriles (R–CN) to primary amines (R–CH₂NH₂). Sn/HCl and Br₂/NaOH do not. Hence option (B) is correct.

2. Question 2 (NEET 2026)

   NEET 2026MediumPreparation of aniline

   Identify the reactions which give aniline as the major product. A. $\mathrm{C_6H_5CN} \xrightarrow{\mathrm{LiAlH_4}}$ B. $\mathrm{C_6H_5CONH_2} \xrightarrow{\mathrm{KOH,\ Br_2}}$ C. $\mathrm{C_6H_5NO_2} \xrightarrow{\mathrm{NaBH_4}}$ D. $\mathrm{C_6H_5NHCOCH_3} \xrightarrow[\Delta]{\mathrm{HCl,\ H_2O}}$ Choose the correct answer from the options given below.

   1. AOption A: A and B only
   2. BOption B: B and D only
   3. COption C: A and C only
   4. DOption D: C and D only

   Show answer & explanationHide answer & explanation

   Correct answer: (B) B and D only

   Explanation

   A: $\mathrm{LiAlH_4}$ reduces benzonitrile to benzylamine, $\mathrm{C_6H_5CH_2NH_2}$. The nitrile carbon is kept, so aniline is not formed. B: Hofmann bromamide degradation of benzamide with $\mathrm{Br_2}$/KOH gives an amine with one carbon fewer: $\mathrm{C_6H_5CONH_2 \rightarrow C_6H_5NH_2}$ (aniline). This gives aniline. C: $\mathrm{NaBH_4}$ is a mild reducing agent and does not reduce the nitro group to $\mathrm{-NH_2}$. Nitrobenzene is reduced to aniline by Sn/HCl, Fe/HCl or $\mathrm{H_2}$/Pd. So this does not give aniline. D: acid hydrolysis of acetanilide (an amide) on heating gives aniline (as its salt) and acetic acid. This gives aniline. So B and D only: option (B).

3. Question 3 (NEET 2026)

   NEET 2026MediumSubstituent effects on acidity and basicity

   Given below are two statements: Statement-I : Oxidation of p-nitrotoluene with acidic $\mathrm{KMnO_4}$ 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.

   1. AOption A: Both Statement-I and Statement-II are correct
   2. BOption B: Both Statement-I and Statement-II are incorrect
   3. COption C: Statement-I is correct but Statement-II is incorrect
   4. DOption D: Statement-I is incorrect but Statement-II is correct

   Show answer & explanationHide answer & explanation

   Correct answer: (A) Both Statement-I and Statement-II are correct

   Explanation

   Statement-I: acidic $\mathrm{KMnO_4}$ oxidises the side-chain $\mathrm{-CH_3}$ to $\mathrm{-COOH}$, giving p-nitrobenzoic acid. The electron-withdrawing $\mathrm{-NO_2}$ group (−I, −R) stabilises the carboxylate anion, so p-nitrobenzoic acid ($\mathrm{p}K_a \approx 3.4$) is a stronger acid than benzoic acid ($\mathrm{p}K_a \approx 4.2$). Statement-I is correct. Statement-II: Sn/HCl reduces $\mathrm{-NO_2}$ to $\mathrm{-NH_2}$; neutralisation then frees the amine, p-toluidine ($\mathrm{p\text{-}CH_3C_6H_4NH_2}$). The electron-donating $\mathrm{-CH_3}$ group (+I, hyperconjugation) increases electron density on N, so p-toluidine ($\mathrm{p}K_b = 8.92$) is more basic than aniline ($\mathrm{p}K_b = 9.38$). Statement-II is correct. So the answer is option (A).

4. Question 4 (NEET 2025)

   NEET 2025MediumPhysical & Chemical Properties

   The correct order of decreasing basic strength of the given amines is :

   1. AOption A: N-methylaniline > benzenamine > ethanamine > N-ethylethanamine
   2. BOption B: N-ethylethanamine > ethanamine > benzenamine > N-methylaniline
   3. COption C: N-ethylethanamine > ethanamine > N-methylaniline > benzenamine
   4. DOption D: benzenamine > ethanamine > N-methylaniline > N-ethylethanamine

   Show answer & explanationHide answer & explanation

   Correct answer: (C) N-ethylethanamine > ethanamine > N-methylaniline > benzenamine

   Explanation

   Aliphatic amines are more basic than aromatic amines because the lone pair on nitrogen in aromatic amines is delocalized into the benzene ring. Among aliphatic amines, secondary amines are generally more basic than primary amines due to the +I effect of alkyl groups. Therefore, the decreasing order is: N-ethylethanamine > ethanamine > N-methylaniline > benzenamine.

5. Question 5 (NEET 2025)

   NEET 2025EasyDiazonium Salts

   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 :

   1. AOption A: Both Statement I and Statement II are correct
   2. BOption B: Both Statement I and Statement II are incorrect
   3. COption C: Statement I is correct but Statement II is incorrect
   4. DOption D: Statement I is incorrect but Statement II is correct

   Show answer & explanationHide answer & explanation

   Correct answer: (A) Both Statement I and Statement II are correct

   Explanation

   Benzenediazonium salts are prepared by diazotisation of aniline using nitrous acid at low temperature (273–278 K). These salts are unstable and decompose readily in the dry state. Direct iodination of benzene is difficult. Therefore, iodobenzene is commonly prepared by treating benzenediazonium salt with KI. Hence, both Statement I and Statement II are correct, so option (A) is correct.

6. Question 6 (NEET 2024)

   NEET 2024MediumPreparation Methods

   Given below are two statements: Statement I : Aniline does not undergo Friedel-Crafts alkylation reaction. Statement II : Aniline cannot be prepared through Gabriel synthesis. In the light of the above statements, choose the correct answer from the options given below:

   1. AOption A: Statement I is correct but Statement II is false.
   2. BOption B: Statement I is incorrect but Statement II is true.
   3. COption C: Both Statement I and Statement II are true.
   4. DOption D: Both Statement I and Statement II are false.

   Show answer & explanationHide answer & explanation

   Correct answer: (C) Both Statement I and Statement II are true.

   Explanation

   Aniline does not undergo Friedel-Crafts alkylation because the amino group forms a complex with Lewis acid catalysts such as AlCl₃, reducing ring reactivity. Gabriel synthesis is suitable for preparing aliphatic primary amines but not aromatic amines like aniline. Therefore, both statements are correct. Hence, option (C) is correct.

7. Question 7 (NEET 2024)

   NEET 2024HardPreparation Methods

   Identify the major product C formed in the following reaction sequence:

   1. AOption A: butanamide
   2. BOption B: α-bromobutanoic acid
   3. COption C: propylamine
   4. DOption D: butylamine

   Show answer & explanationHide answer & explanation

   Correct answer: (C) propylamine

   Explanation

   Propyl iodide reacts with NaCN to form butanenitrile. Partial hydrolysis of nitrile gives butanamide. Treatment with Br₂/NaOH causes Hofmann bromamide degradation: $$RCONH_2 \rightarrow RNH_2$$ with loss of one carbon atom. Thus butanamide gives propylamine. Hence, option (C) is correct.

8. Question 8 (NEET 2023)

   NEET 2023MediumClassification & Nomenclature

   Which of the following reactions will NOT give primary amine product?

   1. AOption A: Reaction (1)
   2. BOption B: Reaction (2)
   3. COption C: Reaction (3)
   4. DOption D: Reaction (4)

   Show answer & explanationHide answer & explanation

   Correct answer: (C) Reaction (3)

   Explanation

   Hofmann bromamide reaction of acetamide gives methylamine, a primary amine. Reduction of nitrile (CH3CN) with LiAlH4 also gives a primary amine. Reduction of acetamide with LiAlH4 gives ethylamine, which is also a primary amine. However, reduction of isocyanide (CH3NC) gives a secondary amine, therefore reaction (3) does NOT give a primary amine.

9. Question 9 (NEET 2023)

   NEET 2023MediumDiazonium Salts

   Identify the product in the following reaction:

   1. AOption A: Option 1
   2. BOption B: Option 2
   3. COption C: Option 3
   4. DOption D: Option 4

   Show answer & explanationHide answer & explanation

   Correct answer: (B) Option 2

   Explanation

   Benzene diazonium chloride reacts with Cu₂Br₂/HBr to form bromobenzene (Sandmeyer reaction). Bromobenzene reacts with Mg/dry ether to form phenyl magnesium bromide, which on hydrolysis with water gives benzene.

10. Question 10 (NEET 2021)

    NEET 2021MediumDiazonium Salts: Preparation, Reactions and Importance

    The reagent 'R' in the given sequence of chemical reactions is:

    1. AOption A: CuCN/KCN
    2. BOption B: $H_2O$
    3. COption C: $CH_3CH_2OH$
    4. DOption D: HI

    Show answer & explanationHide answer & explanation

    Correct answer: (C) $CH_3CH_2OH$

    Explanation

    Aryl diazonium salts are reduced to benzene derivatives when treated with ethanol. Here, tribromobenzenediazonium chloride reacts with ethanol to replace the diazonium group by hydrogen, forming 1,3,5-tribromobenzene. Hence, R = C2H5OH.