See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: The starting material is a bis-tosylate (the wedge and dash bonds on the terminal carbons represent OTs leaving groups, as implied by the reaction context and the answer choices showing OTs groups). The reagent H2S/KOH generates HS⁻ (hydrosulfide anion) in situ, which is a good nucleophile. With KOH deprotonating H2S to give HS⁻, the nucleophile can perform SN2 reactions on the tosylate electrophilic centers. Step 1 – Identify the starting material: The starting material is (2R,6S)-2,6-dimethyl-4,4-dimethyl-1,7-heptanediol bis-tosylate (or the enantiomer/diastereomer shown), i.e., a diol ditosylate where both terminal carbons bearing methyl groups are activated as tosylates with defined stereochemistry. Step 2 – Reaction with HS⁻: The first equivalent of HS⁻ attacks one of the OTs carbons via SN2 (inversion), giving a thiolate intermediate after deprotonation by KOH. The resulting thiolate then undergoes intramolecular SN2 on the second OTs carbon (also with inversion), forming a cyclic thioether (thiane ring). Step 3 – Stereochemical outcome: Both steps proceed with inversion of configuration. The starting material has one OTs on a wedge and one on a dash (opposite faces). After double inversion (once at each carbon), the net result at each center is retention relative to the other. The geometry of the product thiane ring (chair or flat representation) gives trans relationship between the two methyl substituents at C2 and C6. In the drawn structure, both methyl groups appear on dashes (both pointing the same face downward in the ring), which represents the trans-2,6-dimethyl-4,4-dimethylthiane with both methyls in equivalent axial or equatorial positions consistent with the double-inversion SN2 pathway. Step 4 – Why not other options: (a) shows one methyl on a bold wedge and one on a dash — this would result from only one inversion or a different stereochemical relationship, inconsistent with double SN2. (c) and (d) show open-chain monosubstitution products where only one OTs was replaced by SH — these would form if the intramolecular cyclization did not occur, but given the favorable 6-membered ring formation and the conditions (KOH present to deprotonate), cyclization is strongly favored. (d) differs from (c) only in the configuration at the SH carbon, but both are incomplete reaction products. The thermodynamically and kinetically favored product under these conditions is the cyclic thioether via double SN2. Therefore, the correct answer is B.