Skip to content

Amines are usually formed… (Class 12 Chemistry 2025, Q30)

questionbuddy.in/q/cbse-12-chemistry-2025-56-7-1-q30-amines-usually-formed-amides-imides

ChemistryCBSE Class 12 · Board exam 2025, set 56/7/1 · 4 marks · Case-based

CBSE Class 12 Chemistry 2025: question 30

Set in the CBSE Class 12 board exam of 2025, paper 56/7/1, section D, question 30, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q30 · 4 marks · Case-basedCompetency-basedAmines

Amines are usually formed from amides, imides, halides, nitro compounds, etc. They exhibit hydrogen bonding which influences their physical properties. In alkyl amines, a combination of electron releasing, steric and H-bonding factors influence the stability of the substituted ammonium cations in protic polar solvents and thus affect the basic nature of amines. Alkyl amines are found to be stronger bases than ammonia. Amines being basic in nature, react with acids to form salts. Aryldiazonium salts, undergo replacement of the diazonium group with a variety of nucleophiles to produce aryl halides, cyanides, phenols and arenes.
Answer the following questions :

  1. (a)

    How can you convert the following ? (i) Ethanoic acid to methanamine (ii) Propanenitrile to 1-aminopropane

    2 marks
  2. (b)

    Why is value of aniline more than that of methylamine ?

    1 mark
  3. (c)(i)

    Arrange the following in increasing order of their basic strength in aqueous solution : , ,

    1 mark

  4. OR

    (c)(ii)

    Give the structures of A and B in the following reaction :

    1 mark

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 56/7/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Ethanoic acid has two carbons but methanamine has one — which reaction removes a carbon?

Hint 2

Reducing –CN gives –CH₂NH₂.

Hint 3

Think of resonance in aniline and of solvation effects in water.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

  1. (a)(i) (Hoffmann bromamide degradation)

    1 mark

  2. (a)(ii)

    1 mark

  3. (b) In aniline, the lone pair on N is delocalised into the benzene ring by resonance, so it is less available to accept a proton. In methylamine, the +I effect of CH₃ increases electron density on N. Aniline is therefore a weaker base, and its is higher.

    1 mark

  4. (c)(i)

    This order results from the combined effect of +I, solvation (H-bonding) and steric factors in water.

    1 mark

  5. (c)(ii) A = (aniline); B = (benzenediazonium chloride).

Answer: (a)(i) CH₃COOH → CH₃CONH₂ → (Br₂/NaOH) CH₃NH₂; (ii) reduction with LiAlH₄ gives CH₃CH₂CH₂NH₂. (b) Resonance delocalises the N lone pair in aniline. (c)(i) (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH; (c)(ii) A = C₆H₅NH₂, B = C₆H₅N₂⁺Cl⁻

Where the marks are

  • Ethanoic acid → amide → Hoffmann degradation (1 mark)
  • Reduction of propanenitrile (1 mark)
  • Resonance reduces lone-pair availability in aniline (1 mark)
  • (c)(i) Correct order, OR (c)(ii) A (½ mark) and B (½ mark) (1 mark)

Revise the chapter

Practise this chapterBoard-style questions from Amines, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 56/7/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.