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(i) Answer the following… (Class 12 Chemistry 2026, Q33b)

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ChemistryCBSE Class 12 · Board exam 2026, set 56/5/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2026: question 33(b)

Set in the CBSE Class 12 board exam of 2026, paper 56/5/1, section E, question 33(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(b) · 5 marks · Long answerHaloalkanes and Haloarenes

(i) Answer the following : (3)
(I) Which isomer of is most reactive towards reaction ?
(II) Predict the alkene that would be formed by dehydrohalogenation of 1-Bromo-1-methylcyclohexane.
(III) Although chlorine shows strong – I effect, yet it is ortho/para-directing in electrophilic aromatic substitution reactions. Why ?
(ii) Write the major product in the following reactions : (2)
(I) 2 chlorobenzene
(II) 4-nitrocumene (p-isopropylnitrobenzene)

Figure in the paper: (I) chlorobenzene (benzene ring with Cl). (II) benzene ring with CH(CH3)2 at one position and O2N (nitro) at the para position. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 56/5/1.

Hints

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

Hint 1

S_N1 reactivity depends on carbocation stability. Elimination usually gives the more substituted alkene.

Hint 2

For chlorobenzene, separate the effect that controls reactivity (–I) from the one that controls orientation (resonance).

Hint 3

Na in dry ether joins two aryl groups. Br₂ with heat attacks the side chain at the benzylic carbon, not the ring.

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. (i)(I) tert-Butyl bromide [, 2-bromo-2-methylpropane] is the most reactive isomer of towards . It forms the most stable, tertiary carbocation .

    1 mark

  2. (i)(II) Elimination of HBr with alcoholic KOH follows Saytzeff's rule: the more substituted alkene is the major product. The major product is 1-methylcyclohexene (C=C inside the ring at C-1, carrying the ). Methylenecyclohexane (exocyclic ) forms as the minor product.

    1 mark

  3. (i)(III) The –I effect of Cl withdraws electrons, so it deactivates the ring. However, the lone pairs on Cl are donated to the ring by resonance (+R effect), which raises the electron density mainly at the ortho and para positions. As a result, the arenium ion (carbocation intermediate) formed by attack at o/p positions is extra stabilised, since in one structure the positive charge lies on the carbon bearing Cl and is delocalised onto Cl. So substitution is directed to the ortho and para positions.

    1 mark

  4. (ii)(I) Fittig reaction gives biphenyl:

    1 mark

  5. (ii)(II) On heating, brominates by a free-radical route at the benzylic C–H, which forms a stable tertiary benzylic radical. The major product is 2-bromo-2-(4-nitrophenyl)propane:

    1 mark

Answer: (i)(I) tert-Butyl bromide, (CH₃)₃CBr; (II) 1-methylcyclohexene (major); (III) +R effect of Cl stabilises o/p arenium ions; (ii)(I) biphenyl, C₆H₅–C₆H₅; (II) p-O₂N–C₆H₄–C(Br)(CH₃)₂.

Where the marks are

  • Identifies tert-butyl bromide / 2-bromo-2-methylpropane (1 mark)
  • Gives 1-methylcyclohexene (Saytzeff product) as the alkene; methylenecyclohexane also accepted (1 mark)
  • Explains o/p direction by the +R effect stabilising the carbocation intermediate (1 mark)
  • Writes biphenyl as the Fittig product (1 mark)
  • Writes the benzylic bromide p-O₂N–C₆H₄–C(Br)(CH₃)₂ (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 paper2026 board paper, set 56/5/1: 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.