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Amines

12 NEET-UG practice questions on Amines, part of Chemistry. The first 12 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

  1. Question 1 · Medium

    Given below are two statements: Statement I: The sulphonamide formed from a primary amine and benzenesulphonyl chloride dissolves in aqueous alkali. Statement II: Tertiary amines do not react with benzenesulphonyl chloride. Choose the correct answer.

    1. ABoth Statement I and Statement II are correct
    2. BStatement I is correct but Statement II is incorrect
    3. CBoth Statement I and Statement II are incorrect
    4. DStatement I is incorrect but Statement II is correct
    Show answer

    Answer: A · Both Statement I and Statement II are correct

    The N-H of a primary sulphonamide is acidic because of the strongly electron-withdrawing sulphonyl group, so it dissolves in alkali. Tertiary amines have no N-H to replace, so they do not react.

    • B: Statement II is right, since a tertiary amine has no replaceable hydrogen.
    • C: Both statements are right, so neither can be rejected.
    • D: Statement I is right, since the primary sulphonamide has an acidic hydrogen.

    Remember this

    Hinsberg test: primary product dissolves in alkali, secondary product does not, tertiary amine does not react.

  2. Question 2 · Easy

    Gabriel phthalimide synthesis fails to give which of these primary amines?

    1. AAniline
    2. BBenzylamine
    3. CPropan-1-amine
    4. DEthanamine
    Show answer

    Answer: A · Aniline

    The phthalimide anion must displace halide from an alkyl halide by nucleophilic substitution. Aryl halides do not undergo this reaction, so aromatic primary amines cannot be prepared this way.

    • B: Benzyl halides are very reactive, so this amine forms readily.
    • C: This forms from a primary alkyl halide, the ideal substrate.
    • D: This forms from bromoethane, an easy SN2 substrate.

    Remember this

    Gabriel synthesis gives aliphatic primary amines only, since aryl halides resist nucleophilic substitution.

  3. Question 3 · Medium

    Assertion A: Aniline does not undergo Friedel-Crafts alkylation. Reason R: Aniline is a weaker base than ethanamine. Choose the correct answer.

    1. ABoth A and R are correct and R is the correct explanation of A
    2. BA is not correct but R is correct
    3. CBoth A and R are correct but R is not the correct explanation of A
    4. DA is correct but R is not correct
    Show answer

    Answer: C · Both A and R are correct but R is not the correct explanation of A

    Both statements are true. Aniline fails in Friedel-Crafts reaction because it forms a salt with the Lewis acid AlCl₃, which puts a positive charge on nitrogen and deactivates the ring. Its weaker basicity compared with ethanamine, from resonance, is not the reason.

    • A: R is true but is not why the reaction fails; salt formation with AlCl₃ is.
    • B: A is correct, as aniline does not undergo this reaction.
    • D: R is correct, since resonance lowers the basicity of aniline.

    Remember this

    Aniline fails in Friedel-Crafts because its nitrogen binds AlCl₃, deactivating the ring and tying up the catalyst.

  4. Question 4 · Medium

    Match List-I with List-II for reactions of benzenediazonium chloride. List-I (Reagent) A. Cu₂Cl₂ with HCl B. H₃PO₂ with H₂O C. Warm water D. HBF₄, then heat List-II (Main organic product) I. Benzene II. Chlorobenzene III. Phenol IV. Fluorobenzene

    1. AA-II, B-I, C-III, D-IV
    2. BA-II, B-III, C-I, D-IV
    3. CA-IV, B-I, C-III, D-II
    4. DA-III, B-I, C-II, D-IV
    Show answer

    Answer: A · A-II, B-I, C-III, D-IV

    Cu₂Cl₂/HCl is the Sandmeyer reaction, giving chlorobenzene. Hypophosphorous acid replaces the diazo group by hydrogen, and warm water gives phenol. HBF₄ forms the diazonium fluoroborate, which on heating gives fluorobenzene.

    • B: This swaps the products of H₃PO₂ and warm water.
    • C: This swaps chlorobenzene and fluorobenzene.
    • D: This swaps the products of Cu₂Cl₂ and warm water.

    Remember this

    The diazo group can be replaced by Cl, Br, CN (Sandmeyer), F (HBF₄), H (H₃PO₂) or OH (warm water).

  5. Question 5 · Medium

    Follow this sequence. Aniline + (CH₃CO)₂O → P P + Br₂ in CH₃COOH → Q (major) Q + H₃O⁺, heat → R What is R?

    1. A2-Bromoaniline
    2. B4-Bromoaniline
    3. C2,4,6-Tribromoaniline
    4. D3-Bromoaniline
    Show answer

    Answer: B · 4-Bromoaniline

    Acetylation gives acetanilide, in which the -NHCOCH₃ group is a weaker activator than -NH₂. So bromination stops at one bromine, mainly at the para position because of the group's bulk. Hydrolysis then restores -NH₂.

    • A: The ortho product is minor because the bulky acetamido group hinders that position.
    • C: This forms directly from aniline and bromine water, without protection.
    • D: The acetamido group directs ortho and para, not meta.

    Remember this

    Acetylation tames the -NH₂ group, allowing monosubstitution at the para position.

  6. Question 6 · Medium

    Arrange these in increasing order of basic strength in aqueous solution. A. NH₃ B. (C₂H₅)₂NH C. C₂H₅NH₂ D. (C₂H₅)₃N

    1. AA, C, D, B
    2. BC, A, D, B
    3. CA, C, B, D
    4. DA, D, C, B

    Answer and explanation in LearnFree account

  7. Question 7 · Easy

    A foul-smelling isocyanide forms when chloroform and alcoholic KOH are warmed with which amine?

    1. ADiethylamine
    2. BN-Methylethanamine
    3. CEthanamine
    4. DN,N-Dimethylethanamine

    Answer and explanation in LearnFree account

  8. Question 8 · Medium

    Propanamide, 7.3 g, is completely converted by Hofmann bromamide degradation. What mass of amine forms? (C = 12, H = 1, N = 14, O = 16)

    1. A3.1 g
    2. B5.9 g
    3. C4.5 g
    4. D7.3 g

    Answer and explanation in LearnFree account

  9. Question 9 · Hard

    Follow this sequence. Nitrobenzene + Sn, HCl, then base → P P + NaNO₂, HCl at 273-278 K → Q Q + CuCN, KCN → R R + H₃O⁺, prolonged heating → S What is S?

    1. ABenzamide
    2. BBenzoic acid
    3. CPhenol
    4. DBenzaldehyde

    Answer and explanation in LearnFree account

  10. Question 10 · Medium

    Arrange the amines drawn in the order of increasing basic strength in water. A. Structure A B. Structure B C. Structure C D. Structure D

    Figure: MedExams
    Four amines labelled A to D: one alkylamine and three arylamines with different ring substituents.
    1. AA, B, C, D
    2. BB, A, D, C
    3. CA, D, B, C
    4. DA, B, D, C

    Answer and explanation in LearnFree account

  11. Question 11 · Hard

    The sequence shown converts the amide into X and then into Y. What is Y?

    Figure: MedExams
    Reaction scheme: benzamide with bromine and sodium hydroxide gives X, and X is diazotised and treated with copper(I) bromide to give Y.
    1. A2,4,6-Tribromoaniline
    2. B4-Bromoaniline
    3. CBromobenzene
    4. DBenzyl bromide

    Answer and explanation in LearnFree account

  12. Question 12 · Easy

    Identify P, the foul-smelling product of the test shown.

    Figure: MedExams
    Reaction scheme: an aromatic amine heated with chloroform and potassium hydroxide gives an unknown product P.
    1. APhenyl isocyanide
    2. BChlorobenzene
    3. CBenzonitrile
    4. DN-Phenylmethanamide

    Answer and explanation in LearnFree account

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