Skip to content

Compound 'X' with molecular… (Class 12 Chemistry 2026, Q28)

questionbuddy.in/q/cbse-12-chemistry-2026-56-1-1-q28-compound-molecular-formula-reacts-aqueous

ChemistryCBSE Class 12 · Board exam 2026, set 56/1/1 · 3 marks · Short answer

CBSE Class 12 Chemistry 2026: question 28

Set in the CBSE Class 12 board exam of 2026, paper 56/1/1, section C, question 28, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q28 · 3 marks · Short answerHaloalkanes and Haloarenes

Compound 'X' with molecular formula reacts with aqueous KOH to give an alcohol. The rate of this reaction depends only on the concentration of the compound 'X'. When an optically active isomer 'Y' of the compound 'X' was treated with aqueous KOH solution, the rate of reaction was found to be dependent on concentration of compound 'Y' and aqueous KOH both.
(a) Write down the structural formula of both 'X' and 'Y'.
(b) Out of 'X' and 'Y', which one will undergo racemisation and why ?
(c) Out of 'X' and 'Y', which one will form product with inversion of configuration and why ?

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

Hints

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

Hint 1

A rate depending only on [substrate] suggests S_N1; which C₄H₉Br forms the most stable carbocation?

Hint 2

Which C₄H₉Br isomer has a chiral carbon?

Hint 3

Compare the stereochemical result of S_N1 with that of S_N2.

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) The rate depends only on [X], so X reacts by ; it is a tertiary bromide. **X = ** (2-bromo-2-methylpropane).

    The optically active isomer that reacts by (rate depends on both [Y] and [OH⁻]) must be chiral. **Y = ** (2-bromobutane).

    1 mark

  2. (b) X (the substrate) undergoes racemisation. In a planar carbocation forms, and can attack it equally from either face, giving a racemic mixture.

    1 mark

  3. (c) Y gives the product (butan-2-ol) with inversion of configuration. In the nucleophile attacks from the side opposite to the leaving Br⁻ (backside attack), which inverts the configuration (Walden inversion).

    1 mark

Answer: (a) X = (CH₃)₃CBr, Y = CH₃CH(Br)CH₂CH₃; (b) X (S_N1, planar carbocation); (c) Y (S_N2, backside attack)

Where the marks are

  • Structure of X (½ mark) and Y (½ mark)
  • X with reason: S_N1 / planar carbocation (½ + ½ marks); Y also accepted
  • Y with reason: S_N2 / backside attack (½ + ½ marks)

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 56/1/1: 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.