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Question: Br $S_N2$ reactivity of these substrate, under identical conditions, will be in the order as...

Br

SN2S_N2 reactivity of these substrate, under identical conditions, will be in the order as

A

I > II > III

B

III > II > I

C

III > I > II

D

II > III > I

Answer

III > II > I

Explanation

Solution

Solution:

  1. Step 1: In an SN2 reaction, the halide’s reactivity depends on steric accessibility at the reactive carbon.

  2. Step 2: Structure III – The bromide is at an allylic (or less-hindered) sp³ center (its attachment to a double‐bond system provides resonance stabilization and minimal steric hindrance), making it most reactive.

  3. Step 3: Structures I and II – Both are bridged bicyclic halides. However, the bridgehead carbon in structure I, with a fused three‐membered ring, suffers from severe geometric constraints (hindering the backside attack needed in SN2). In contrast, structure II (with a fused four‐membered ring) is comparatively less hindered.

  4. Step 4: Thus, the SN2 reactivity order becomes:

Structure III>Structure II>Structure I\text{Structure III} > \text{Structure II} > \text{Structure I}

Brief Explanation:

  • Structure III is an allylic halide and undergoes SN2 easily (least steric hindrance, resonance stabilization).
  • Structure II is less hindered than I as the four‐membered ring allows better backside access.
  • Structure I is most hindered due to the small (three‐membered) ring fused at the bridgehead.