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Draw the structure of the… (Class 12 Chemistry 2025, Q26A)

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ChemistryCBSE Class 12 · Board exam 2025, set 56/1/1 · 3 marks · Short answer

CBSE Class 12 Chemistry 2025: question 26(A)

Set in the CBSE Class 12 board exam of 2025, paper 56/1/1, section C, question 26(A), for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q26(A) · 3 marks · Short answerHaloalkanes and Haloarenes

Draw the structure of the major monohalo product for each of the following reaction : (1 + 1 + 1 = 3)
(a) 1-chloro-4-ethylbenzene () with , Heat ?
(b) 1-methylcyclohexene + HBr ?
(c) 4-(hydroxymethyl)phenol () with HCl, Heat ?

Figure in the paper: (a) benzene ring with Cl at one para position and CH2-CH3 at the other; (b) cyclohexene ring with a CH3 on a doubly bonded carbon; (c) benzene ring with OH at one para position and CH2-OH (HO-H2C) at the other. See it in the official paper.

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

Hints

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

Hint 1

Halogen with heat on an alkylbenzene works by free radicals. Which C–H gives the most stable radical?

Hint 2

For HBr addition, think about which carbocation is more stable.

Hint 3

Which –OH can be replaced by HCl, the alcoholic one or the phenolic one?

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) with heat brings about free-radical substitution at the benzylic carbon, which forms the most stable radical. Product: [1-bromo-1-(4-chlorophenyl)ethane].

    1 mark

  2. (b) HBr adds according to Markovnikov's rule: H goes to the =CH– carbon and Br to the methyl-bearing carbon, via a tertiary carbocation. Product: 1-bromo-1-methylcyclohexane, a cyclohexane ring with Br and on the same carbon.

    1 mark

  3. (c) Only the benzylic –OH is replaced by Cl. The phenolic –OH has partial C=O double bond character because of resonance and is not substituted. Product: [4-(chloromethyl)phenol].

    1 mark

Answer: (a) Cl–C₆H₄–CHBr–CH₃; (b) 1-bromo-1-methylcyclohexane; (c) HO–C₆H₄–CH₂Cl

Where the marks are

  • Benzylic bromination product (1 mark)
  • Markovnikov product, 1-bromo-1-methylcyclohexane (1 mark)
  • 4-(chloromethyl)phenol, with the phenolic OH unchanged (1 mark)

Revise the chapter

Practise this chapterBoard-style questions from Haloalkanes and Haloarenes, one at a time, marked against the answer key and marking points.Back to the full paper2025 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.