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)
- (a) A-Br_((b)) ; B-I₍ₐ₎ ; C-Br₍ₐ₎ ; D-Br₍ₐ₎
- (b) A-Br₍ₐ₎ ; B-I₍ₐ₎ ; C-Br₍ₐ₎ ; D-Br₍ₐ₎
- (c) A-Br_((b)) ; B-I_((b)) ; C-Br_((b)) ; D-Br_((b))
- (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.
- 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.
Compound Carbocation from (a) Carbocation from (b) More reactive A secondary benzylic, resonance with the ring plain secondary Br₍ₐ₎ B primary allylic, resonance with C=C primary, double bond too far away I₍ₐ₎ C bridgehead, cannot become planar tertiary, ring carbon free to flatten Br_((b)) D tertiary primary Br₍ₐ₎ - 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.
- 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.