JEE Main 2023 · 8 April · morning · Single correct · +4 / −1

Choose the halogen which is most reactive towards S_N 1 reaction in the given compounds (A, B, C D)

JEE Main PYQ figure
  1. (a) A-Br_((b)) ; B-I₍ₐ₎ ; C-Br₍ₐ₎ ; D-Br₍ₐ₎
  2. (b) A-Br₍ₐ₎ ; B-I₍ₐ₎ ; C-Br₍ₐ₎ ; D-Br₍ₐ₎
  3. (c) A-Br_((b)) ; B-I_((b)) ; C-Br_((b)) ; D-Br_((b))
  4. (d) A-Br₍ₐ₎ ; B-I₍ₐ₎ ; C-Br_((b)) ; D-Br₍ₐ₎
Answer and solution

Answer (d) A-Br₍ₐ₎ ; B-I₍ₐ₎ ; C-Br_((b)) ; D-Br₍ₐ₎

Solution

Chemistry ma’am · 3 steps

For each compound, ask which halogen leaves behind the more stable, flat carbocation.

  1. What SN1 needs. In S_N1 the halide leaves first and a carbocation forms. The halogen whose carbocation is better stabilised (by resonance or by alkyl groups) and can become planar reacts fastest.
    CompoundCarbocation from (a)Carbocation from (b)More reactive
    Asecondary benzylic, resonance with the ringplain secondaryBr₍ₐ₎
    Bprimary allylic, resonance with C=Cprimary, double bond too far awayI₍ₐ₎
    Cbridgehead, cannot become planartertiary, ring carbon free to flattenBr_((b))
    DtertiaryprimaryBr₍ₐ₎
  2. Bridgehead exception. In C, Br₍ₐ₎ sits on a tertiary carbon, but that carbon is a bridgehead of the bicyclic ring. The rigid cage stops it from flattening to the sp² shape a carbocation needs, so it hardly ionises at all.
  3. So. The answer is A–Br₍ₐ₎, B–I₍ₐ₎, C–Br_((b)), D–Br₍ₐ₎, which is option (d).

The idea to keep. SN1 reactivity follows carbocation stability: benzylic and allylic beat plain primary or secondary, and tertiary beats primary. A bridgehead carbocation is the exception: it cannot go planar, so it barely forms.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Halogen Derivatives — Haloalkanes & Haloarenes

Asked about 120 times in JEE Main since 2020. This question is from 2023.