See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: SN2 reactions require backside attack by the nucleophile on the carbon bearing the leaving group. This requires that the carbon be accessible from the back side, and that the transition state (with nucleophile, central carbon, and leaving group collinear) be geometrically achievable. Step 1 - Identify each substrate: (a) Bromocyclohexane: secondary alkyl bromide on a cyclohexane ring. Backside attack is hindered but not impossible; SN2 is slow but not negligible. (b) 3-Bromocyclohex-1-ene: allylic secondary bromide. SN2 is facilitated by the allylic position; rate is reasonable. (c) 1-Bromonorbornene (bridgehead bromide on bicyclo[2.2.1]hept-2-ene): The bromine is at a bridgehead carbon of the norbornane/norbornene skeleton. The bridgehead carbon is locked in a rigid bicyclic framework. Backside attack is geometrically impossible because the bridge carbon cannot be approached from the back (Bredt's rule analog for SN2: the bridgehead carbon is surrounded by ring bridges on all sides, making the required linear Nu---C---LG transition state completely blocked). This is an extreme case of steric and geometric impossibility for SN2. (d) 3-Bromocyclohex-1-ene: similar to (b), an allylic secondary bromide; SN2 can occur. Step 2 - Why (c) is the answer: In bicyclo[2.2.1] systems, the bridgehead position is so sterically encumbered by the three bridges converging at that carbon that a nucleophile cannot approach from the back side under any practical conditions. The rate of SN2 at a bridgehead of a small bicyclic system is effectively zero (negligible). Step 3 - Why other options fail: (a) Cyclohexyl bromide is secondary and somewhat hindered, but SN2 can still occur (just slowly). (b) and (d) are allylic bromides where SN2 is actually facilitated relative to simple secondary halides. Therefore, the correct answer is C.