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Answer any five questions… (Class 12 Chemistry 2024, Q32)

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ChemistryCBSE Class 12 · Board exam 2024, set 56/1/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2024: question 32

Set in the CBSE Class 12 board exam of 2024, paper 56/1/1, section E, question 32, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32 · 5 marks · Long answerAmines

Answer any five questions of the following :

  1. (a)

    N,N-diethyl-benzenesulphonamide is insoluble in alkali. Give reason.

    1 mark
  2. (b)

    Aniline does not undergo Friedel-Crafts reaction. Why ?

    1 mark
  3. (c)

    Write a simple chemical test to distinguish between methylamine and aniline.

    1 mark
  4. (d)

    Write the chemical reaction involved in Gabriel phthalimide synthesis.

    1 mark
  5. (e)

    How will you convert aniline to p-bromoaniline ?

    1 mark
  6. (f)

    Complete the following reaction : benzenediazonium chloride ()

    1 mark
  7. (g)

    Write the structures of A and B in the following reaction : benzoic acid A B

    1 mark

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 56/1/1.

Hints

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

Hint 1

Sulphonamides dissolve in alkali only if an H is left on N; amines are Lewis bases.

Hint 2

For aniline reactions, think of protecting –NH₂ by acetylation and of diazonium/azo chemistry; heating an ammonium salt of an acid gives an amide.

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) In N,N-diethylbenzenesulphonamide, , there is no hydrogen on the nitrogen atom, so it is not acidic and does not form a salt with alkali; hence it is insoluble.

    1 mark

  2. (b) Aniline, a Lewis base, reacts with the catalyst AlCl₃ (Lewis acid) to form a salt; the nitrogen acquires a positive charge, which strongly deactivates the ring, so the Friedel–Crafts reaction does not occur.

    1 mark

  3. (c) Azo dye test: treat each with NaNO₂ + dil. HCl at 273–278 K and then with alkaline β-naphthol (or phenol). Aniline forms benzenediazonium chloride which couples to give an orange/red dye; methylamine gives methanol with evolution of N₂ and no dye.

    1 mark

  4. (d) Gabriel phthalimide synthesis: phthalimide potassium phthalimide N-alkylphthalimide + sodium phthalate. i.e. ;

    1 mark

  5. (e) Protect –NH₂ by acetylation, brominate, then hydrolyse:

    1 mark

  6. (f) (nitrobenzene).

    1 mark

  7. (g) (A, benzamide) (B, aniline, by Hoffmann bromamide degradation).

Answer: (a) No H on N, not acidic; (b) forms salt with AlCl₃, ring deactivated; (c) azo dye test – aniline gives orange dye; (d) phthalimide → K-phthalimide → N-alkylphthalimide → RNH₂; (e) acetylation, bromination, hydrolysis; (f) nitrobenzene; (g) A = benzamide, B = aniline

Where the marks are

  • No acidic H on nitrogen (1 mark)
  • Salt formation with Lewis acid AlCl₃ (1 mark)
  • Correct distinguishing test with observation (1 mark)
  • Gabriel synthesis reaction (1 mark)
  • Three-step conversion of aniline to p-bromoaniline (1 mark)
  • Nitrobenzene as product (1 mark)
  • A = benzamide, B = aniline (1 mark) — any five parts counted

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 paper2024 board paper, set 56/1/1: 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.