= Solution
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 $\mathrm{OSO_2Me}$.
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. \b[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.
$$
\mathrm{{-}C(Me){=}CH{-}CH(OH){-}CH_2{-}CH{=}C(Cl)R}
\longrightarrow
\mathrm{{-}C(Me){=}CH{-}CH{=}CH{-}CH{=}C(Cl)R}.
$$
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.
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