Amines

36 NEET-UG practice questions on Amines, part of Chemistry. The first 25 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

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   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

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   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

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   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

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   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

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   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

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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

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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

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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

    

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

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11. Question 11 · Hard

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

    

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    1. A2,4,6-Tribromoaniline
    2. B4-Bromoaniline
    3. CBromobenzene
    4. DBenzyl bromide

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12. Question 12 · Easy

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

    

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    1. APhenyl isocyanide
    2. BChlorobenzene
    3. CBenzonitrile
    4. DN-Phenylmethanamide

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13. Question 13 · Hard

    Arrange the amines in increasing order of basic strength. A. Aniline B. Benzylamine C. 4-Nitroaniline D. 4-Methylaniline

    1. AB, D, A, C
    2. BC, A, D, B
    3. CA, C, D, B
    4. DC, D, A, B

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14. Question 14 · Medium

    In the balanced equation for the Hofmann bromamide degradation of an amide, RCONH₂ + Br₂ + x NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O, what is the value of x?

    1. A1
    2. B3
    3. C2
    4. D4

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15. Question 15 · Medium

    Given below are two statements: Statement I: Aniline is a weaker base than ammonia, because the lone pair on its nitrogen is delocalised over the benzene ring. Statement II: In aqueous solution, the order of basic strength of the ethylamines is (C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂. Choose the correct answer.

    1. ABoth Statement I and Statement II are incorrect
    2. BStatement I is incorrect but Statement II is correct
    3. CStatement I is correct but Statement II is incorrect
    4. DBoth Statement I and Statement II are correct

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16. Question 16 · Easy

    With benzenesulphonyl chloride, which amine gives a sulphonamide that is insoluble in aqueous alkali?

    1. APropan-1-amine
    2. BN-Ethylethanamine
    3. CN,N-Diethylethanamine
    4. DEthanamine

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17. Question 17 · Medium

    Benzenediazonium chloride is added to phenol in a mildly alkaline solution. An orange dye forms. What is it?

    1. A4-Hydroxyazobenzene
    2. B4-Aminoazobenzene
    3. CChlorobenzene
    4. DDiphenyl ether

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18. Question 18 · Medium

    Consider these statements about tests for amines. A. Primary amines, aliphatic or aromatic, give the carbylamine test. B. Tertiary amines react with benzenesulphonyl chloride to give sulphonamides. C. The sulphonamide from a primary amine is soluble in alkali. D. Secondary amines give the carbylamine test. Which of them are correct?

    1. AC and D only
    2. BA and C only
    3. CA and D only
    4. DA, B and C only

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19. Question 19 · Medium

    Direct nitration of aniline with a mixture of concentrated HNO₃ and H₂SO₄ gives a large amount of the meta product. Why?

    1. AIn the strongly acidic mixture aniline is protonated to the anilinium ion, which is meta-directing
    2. BThe –NH₂ group is itself a meta-directing group
    3. CThe nitronium ion attacks nitrogen first and then moves to the meta position
    4. DThe –NH₂ group blocks the ortho and para positions by steric hindrance

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20. Question 20 · Medium

    Assertion A: Amines have lower boiling points than alcohols of comparable molecular mass. Reason R: Amines cannot form intermolecular hydrogen bonds. Choose the correct answer.

    1. AA is correct but R is not correct
    2. BBoth A and R are correct and R is the correct explanation of A
    3. CBoth A and R are correct but R is not the correct explanation of A
    4. DA is not correct but R is correct

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21. Question 21 · Easy

    Benzenediazonium chloride is treated with hypophosphorous acid (H₃PO₂) and water. What is the organic product?

    1. AAniline
    2. BChlorobenzene
    3. CPhenol
    4. DBenzene

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22. Question 22 · Hard

    Benzyl chloride is treated with ethanolic KCN to give P. P is reduced with LiAlH₄ to give Q. Q is treated with NaNO₂ and dilute HCl at 273-278 K. What is the main organic product?

    1. APhenylmethanol
    2. BPhenylmethanamine
    3. C2-Phenylethanediazonium chloride
    4. D2-Phenylethanol

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23. Question 23 · Easy

    Propanamide is warmed with bromine and aqueous NaOH. Which amine is formed?

    1. APropan-1-amine
    2. BEthanamine
    3. CMethanamine
    4. DN-Methylethanamine

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24. Question 24 · Medium

    Given below are two statements: Statement I: In aqueous solution, (CH₃)₂NH is a stronger base than (CH₃)₃N. Statement II: In the gas phase, (CH₃)₃N is a weaker base than CH₃NH₂. Choose the correct answer.

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

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25. Question 25 · Hard

    Arrange these steps in the correct order to convert benzene into 1,3,5-tribromobenzene. A. Treat with bromine water B. Nitrate with concentrated HNO₃ and H₂SO₄ C. Diazotise with NaNO₂ and HCl at 273-278 K, then treat with H₃PO₂ D. Reduce with Sn and HCl, then neutralise

    1. AB, D, C, A
    2. BB, D, A, C
    3. CD, B, A, C
    4. DB, A, D, C

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