= Solution
\b[The first product is $\mathrm{BrCH=C(CH_3)CH_2CH_2OCH_3}$, with the original <alkene> geometry retained.] <methoxide> attacks the primary saturated <carbon> bearing <bromine> from the backside. Its <oxygen> <lone pair> forms the new C–O bond as the C–Br bonding pair leaves with <bromide>. The other <bromine> is vinylic, directly bonded to an sp$^2$ <carbon>: ordinary <SN2 reaction> is unfavourable there, and ionization would require an unstable vinyl cation. Thus the primary alkyl <bromide> is replaced selectively, rather than replacing the <vinylic halide> or moving the double bond. There is no observable inversion label at the attacked <carbon> because it is a methylene.
\b[The second product is <sodium 4-propoxybutane-1-sulfonate>, $\mathrm{Na^+\,{}^-O_3S(CH_2)_4OCH_2CH_2CH_3}$.] <sodium hydride> first removes the <alcohol> <proton>, giving <propoxide> and <hydrogen>. The <propoxide> <oxygen> then attacks the primary ring <carbon> bonded to the ring <oxygen>; simultaneous cleavage of that carbon–oxygen bond opens the <sultone>. The departing ring <oxygen> remains attached to <sulfur> as part of the resonance-stabilized <sulfonate> anion. The <carbon> attached directly to <sulfur> is not the substitution site, because cleavage there would not expel the same stable <sulfonate> <leaving group>. This is carbon-centred ring opening of a cyclic <sulfonate ester>, not addition across an S=O bond.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1-substitution-mechanisms.png]
{title=Original electron-pair arrows for selective primary-carbon substitution and propoxide opening of the four-carbon <sultone>}
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