Past exam of the chemistry B course of the University of Cambridge / 2022 / Paper 1 / B3 / d

Distinguish the original thiirane sulfur $S_o$ from the sulfur $S_n$ supplied by thiocyanate. sulfur attack by ${}^{-}{\mathrm{S}}_{\mathrm{n}}-\mathrm{C}\equiv \mathrm{N}$ at one thiirane carbon opens its C–$S_o$ bond with inversion. This produces a thiolate at the neighbouring carbon and a C–$S_n$–CN group at the attacked carbon. The OH-bearing remote carbon is unaffected.

The original thiolate $S_o^{-}$ now attacks the nitrile carbon intramolecularly. Shift one C≡N pi pair onto nitrogen. The resulting cyclic imidodithiocarbonate has the five-membered sequence ring carbon–$S_o$–C(=N${}^{-}$)–$S_n$–adjacent ring carbon; its two carbon atoms are also part of the original six-membered carbocycle, exactly giving the hinted bicyclic intermediate.

Collapse can reform C≡N while breaking the nitrile-carbon–$S_n$ bond. This transfers the CN group to $S_o$ and generates $S_n^{-}$ as the thiolate on the other ring carbon. That thiolate attacks the remaining C–$S_o$ bond from the back, expelling ${}^{-}{\mathrm{S}}_{\mathrm{o}}-\mathrm{C}\equiv \mathrm{N}$ and reclosing the thiirane. Each thiirane carbon has undergone one inversion, so the new sulfur bridge lies on the opposite face; the OH configuration is retained. The thiocyanate sulfur has entered the ring and the original ring sulfur has left as thiocyanate. This atom-labelled exchange mechanism explains how the same elemental formula can conceal sulfur replacement; isotope labelling would be needed to test that exchange experimentally. A single epimerization at the OH carbon would give the wrong stereochemical outcome.