= Solution
Number the cyclopentanone ring $p_1$ at its <carbonyl>, then $p_2$ at the methyl-bearing <carbon>, $p_3$ at the <mesylate>-bearing <carbon>, $p_4$ at the <carbon> joined to <cyclohexane> $h_2$, and $p_5$ at the remaining methylene. \b[B has a $p_2{=}p_3$ <enone> and the $h_1{=}h_6$ <enamine>.] <triethylamine> removes the <hydrogen> at $p_2$ while the $p_3$–O bond breaks to release <mesylate>. An <E2 elimination>, with approximately <antiperiplanar geometry> in the puckered ring, gives the conjugated cyclopentenone. Elimination towards $p_4$ instead would give a nonconjugated <alkene>.
The <enamine> <carbon> $h_6$ now attacks <enone> beta <carbon> $p_3$ intramolecularly. <nitrogen> donates into $h_1$ to give the <iminium ion>, while the <enone> <electrons> move onto the <ketone> <oxygen>. <protonation> of the <ketone> <enolate> gives C. \b[The second new bond is $h_6{-}p_3$; the first was $h_2{-}p_4$.] These two connections account for the bridged skeleton without a concerted <cycloaddition> or an aldol step. The formula of B is $\mathrm{C_{16}H_{23}NO}$: relative to A, elimination removes $\mathrm{CH_4O_3S}$ overall, namely methanesulfonic <acid (chemistry)> after <proton transfers>.
\Image[/past-exam-of-the-chemistry-b-course-of-the-university-of-cambridge/2023/paper-1-missing-intermediates.png]
{title=Original atom-numbered reconstruction of intermediate B and the diaryl ether G; coloured bonds identify the subsequent cyclization and aryl migration}
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