Answer: SN2 attack by the alpha-carbanion on the alpha carbon bearing the leaving group, forming a three-membered ring (episulfone) that extrudes SO 2.
- A SN2 attack by the alpha-carbanion on the alpha carbon bearing the leaving group, forming a three-membered ring (episulfone) that extrudes SO<sub>2</sub>
- B A perfectly ordinary E2 elimination occurring directly across the adjacent alpha and beta carbon atoms according to most researchers in the majority of cases studied
- C A straightforward free radical chain process that is propagated continuously by sulfonyl radicals as widely reported in standard practice
- D An SN1-type ionisation of the sulfone substrate followed by a subsequent carbocation rearrangement step under most conditions encountered
Correct answer: A. SN2 attack by the alpha-carbanion on the alpha carbon bearing the leaving group, forming a three-membered ring (episulfone) that extrudes SO<sub>2</sub>
Explanation: In the Ramberg-Backlund reaction, a base generates a carbanion alpha to the sulfone; intramolecular SN2 displaces the halide to form an episulfone (thiirane-1,1-dioxide); the episulfone then loses SO₂ to give the alkene.
SN2 proceeds in a single step with backside attack and inversion of configuration, while SN1 forms a planar carbocation intermediate first, leading to racemisation.
Concept context
Study carbon-halogen compounds: how they are made, how they react via SN1/SN2 and elimination, their stereochemistry, and their environmental impact.