Past exam of the chemistry B course of the University of Cambridge / 2024 / Paper 1 / B4 / e

The alcohol oxygen attacks sulfur of methanesulfonic anhydride. Cleavage of the anhydride S–O bond releases methanesulfonate, and the amine base removes the O–H proton, giving the corresponding mesylate. This activation preserves the substrate C–O bond while replacing the poor hydroxide leaving group by $\mathrm{OS}{\mathrm{O}}_2\mathrm{Me}$.

The base then removes a hydrogen from the adjacent methylene on the side toward the vinyl chloride. In a favorable antiperiplanar arrangement, C–H electrons form the new C=C bond while the C–OMs bond breaks. An E2 elimination creates the alkene between the former alcohol carbon and that methylene. It links the two neighboring unsaturated regions into the extended conjugated triene segment, while the pre-existing carbon skeleton and vinyl chloride are retained.The left-hand polyenoate portion is unchanged. The base is diisopropylethylamine; it accepts acid during mesylation and promotes elimination. The product's E/Z mixture concerns the newly formed alkene, and no single geometry needs to be assigned.