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Identification of primary, secondary and tertiary amines: NEET MCQs with answers

53 questions on this topic of Organic Compounds Containing Nitrogen (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. An organic compound 'A' having molecular formula C₃H₉N reacts with acetyl chloride to give compound 'B'. Compound 'A' also reacts with benzenesulphonyl chloride to form a product insoluble in aqueous KOH. How many replaceable hydrogen atoms are attached directly to the nitrogen atom in compound 'A'?

  1. 0
  2. 2
  3. 1
  4. 3
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Answer: (C) 1

Compound 'A' reacts with benzenesulphonyl chloride to form an alkali-insoluble sulphonamide, indicating that 'A' is a secondary amine. A secondary amine possesses exactly one hydrogen atom attached directly to the nitrogen atom.

medium

2. Given below are two statements: Statement I: Aniline upon reaction with nitrous acid at 0–5 °C followed by alkaline β-naphthol gives an orange-red dye, whereas N-methylaniline does not. Statement II: Secondary aromatic amines react with nitrous acid to form N-nitroso derivatives instead of diazonium salts. In the light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (A) Both Statement I and Statement II are correct

Aniline is a primary aromatic amine that undergoes diazotization to form benzenediazonium chloride, which couples with alkaline β-naphthol to yield 1-(phenyldiazenyl)naphthalen-2-ol, an orange dye. N-Methylaniline is a secondary amine; it reacts with nitrous acid via electrophilic nitrosation to form a yellow oily N-nitroso compound, not a diazonium salt, so it does not give the azo dye test.

hard

3. An organic compound 'X' (C₅H₁₃N) is treated with Hinsberg's reagent to produce a product soluble in aqueous KOH. When 'X' is treated with NaNO₂ and dilute HCl at 273–278 K, quantitative evolution of N₂ gas is observed. How many constitutionally isomeric primary amines of formula C₅H₁₃N are possible?

  1. 4
  2. 6
  3. 7
  4. 8
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Answer: (D) 8

Compound X is an aliphatic primary amine as it yields an alkali-soluble sulphonamide and evolves N₂ with nitrous acid. Pentyl amine has 8 constitutional isomers of 1° amines: pentan-1-amine, pentan-2-amine, pentan-3-amine, 3-methylbutan-1-amine, 3-methylbutan-2-amine, 2-methylbutan-1-amine, 2-methylbutan-2-amine, and 2,2-dimethylpropan-1-amine.

medium

4. An amine A gives a foul smell with CHCl3 and alcoholic KOH, and its Hinsberg product dissolves in alkali. Amine A is:

  1. A tertiary amine
  2. A quaternary ammonium salt
  3. A primary amine
  4. A secondary amine
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Answer: (C) A primary amine

A positive carbylamine test and an alkali-soluble Hinsberg product both indicate a primary amine.

NCERT Class 12

easy

5. Which of the following amines on treatment with nitrous acid (NaNO₂ + dilute HCl) at 0–5 °C produces a yellow oily nitrosamine?

  1. Aniline
  2. Ethanamine
  3. N-Methylaniline
  4. N,N-Dimethylaniline
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Answer: (C) N-Methylaniline

Secondary amines (both aliphatic and aromatic), such as N-methylaniline, react with nitrous acid to yield N-nitrosoamines, which separate out as yellow oily liquids. Primary aromatic amines form diazonium salts, while primary aliphatic amines liberate N₂ gas.

hard

6. A neutral organic liquid 'Y' has the molecular formula C₃H₉N. When treated with excess methyl iodide, one mole of 'Y' consumes exactly 2 moles of methyl iodide to yield a quaternary ammonium iodide salt. Compound 'Y' is:

  1. CH₃CH₂NHCH₃
  2. (CH₃)₂NH
  3. CH₃CH₂CH₂NH₂
  4. (CH₃)₃N
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Answer: (A) CH₃CH₂NHCH₃

Exhaustive methylation of amines requires (4 - n) moles of alkyl halide per mole of amine, where n is the number of alkyl/aryl groups attached to the nitrogen atom. A primary amine consumes 3 moles of CH₃I, a secondary amine consumes 2 moles of CH₃I, and a tertiary amine consumes 1 mole of CH₃I to form a quaternary salt. Therefore, 'Y' is a secondary amine of formula C₃H₉N, which is ethylmethylamine (CH₃CH₂NHCH₃).

medium

7. In the Hinsberg test, a primary amine gives a sulphonamide that is soluble in alkali because:

  1. It is ionic
  2. The hydrogen on nitrogen is strongly acidic due to the electron-withdrawing sulphonyl group
  3. It contains no hydrogen on nitrogen
  4. It is a tertiary amide
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Answer: (B) The hydrogen on nitrogen is strongly acidic due to the electron-withdrawing sulphonyl group

A primary amine forms an N-alkylbenzenesulphonamide whose N-H hydrogen is acidic because of the strongly electron-withdrawing sulphonyl group, so it dissolves in alkali.

NCERT Class 12

medium

8. Given below are two statements: Statement I: Aliphatic primary amines react with cold nitrous acid (HNO₂) to form aliphatic diazonium salts, which are stable at room temperature. Statement II: Primary amines can be distinguished from secondary and tertiary amines by warming them with carbon disulphide followed by mercuric chloride. In light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (C) Statement I is incorrect but Statement II is correct

Statement I is incorrect because aliphatic diazonium salts are highly unstable even at 0–5 °C and decompose quantitatively into carbocations, evolving N₂ gas and forming alcohols. Statement II is correct because Hoffmann's mustard oil reaction (CS₂ + HgCl₂) is specific for primary amines to yield alkyl isothiocyanates with a mustard oil-like smell.

medium

9. Given below are two statements: Statement I: Tertiary aliphatic amines dissolve in cold, dilute nitrous acid to form soluble trialkylammonium nitrite salts. Statement II: Secondary aliphatic amines do not react with nitrous acid because they lack acidic alpha-hydrogens. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct because tertiary aliphatic amines form soluble nitrite salts [R₃NH]⁺NO₂⁻ with cold nitrous acid. Statement II is incorrect because secondary amines readily react with nitrous acid via the lone pair on nitrogen to form insoluble N-nitrosoamines.

easy

10. A mixture of three amines X, Y, and Z was treated with benzenesulphonyl chloride. - Amine X reacted to form a precipitate that dissolved in aqueous NaOH. - Amine Y reacted to form a precipitate that remained insoluble in aqueous NaOH. - Amine Z did not react and remained as an unreacted organic layer. Which classification represents X, Y, and Z correctly?

  1. X = 1° amine, Y = 3° amine, Z = 2° amine
  2. X = 2° amine, Y = 1° amine, Z = 3° amine
  3. X = 3° amine, Y = 2° amine, Z = 1° amine
  4. X = 1° amine, Y = 2° amine, Z = 3° amine
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Answer: (D) X = 1° amine, Y = 2° amine, Z = 3° amine

Primary amines form N-alkylbenzenesulphonamides containing an acidic N-H proton, making them soluble in alkali. Secondary amines form N,N-dialkylbenzenesulphonamides lacking an acidic proton, remaining insoluble in alkali. Tertiary amines have no hydrogen on nitrogen and do not react.

easy

11. An aliphatic amine 'A' with the formula C₃H₉N reacts with benzenesulphonyl chloride to form a precipitate that is completely insoluble in aqueous sodium hydroxide. The structural formula of 'A' is:

  1. CH₃CH₂CH₂NH₂
  2. CH₃-CH(NH₂)-CH₃
  3. (CH₃)₂CH-NH-CH₃
  4. CH₃-NH-CH₂CH₃
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Answer: (D) CH₃-NH-CH₂CH₃

Reaction of a secondary amine with benzenesulphonyl chloride gives an N,N-dialkylbenzenesulphonamide which lacks an acidic hydrogen on the nitrogen atom. Consequently, this derivative is insoluble in aqueous NaOH. CH₃-NH-CH₂CH₃ (ethylmethylamine) is a secondary amine of formula C₃H₉N.

medium

12. Consider the distinction between aniline and N,N-dimethylaniline. Which of the following reagents can be used to distinguish them chemically? (I) Carbylamine test (CHCl₃ + alc. KOH) (II) Azo dye test (NaNO₂/HCl at 0–5 °C followed by alkaline β-naphthol) (III) Hinsberg's reagent (C₆H₅SO₂Cl + aqueous KOH) Choose the correct option:

  1. (I) only
  2. (I) and (II) only
  3. (II) and (III) only
  4. (I), (II) and (III)
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Answer: (D) (I), (II) and (III)

Aniline is a primary aromatic amine while N,N-dimethylaniline is a tertiary aromatic amine. Aniline gives a positive carbylamine test, forms a scarlet azo dye with alkaline β-naphthol, and reacts with Hinsberg's reagent to form a base-soluble sulphonamide. N,N-Dimethylaniline does not give the carbylamine test, does not yield a diazonium salt to form the typical azo dye with β-naphthol under these conditions, and does not react with Hinsberg's reagent. Thus, all three can distinguish them.

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