= Solution
A undergoes a <Robinson annulation>. For an unambiguous connectivity description, call the original ring <carbonyl> <carbon> C1 and the side-chain-bearing alpha <carbon> C6. Label the appended chain $\mathrm{C6{-}a{-}b{-}c(=O){-}d}$, where d is initially methyl. Base makes the enolate at d, which attacks C1 intramolecularly to close the six-membered ring $\mathrm{C1{-}C6{-}a{-}b{-}c{-}d}$. The <carbonyl> $\pi$ electrons move to <oxygen>, and protonation gives the beta-hydroxy <ketone>.
An <E1cb elimination> then removes the <proton> on d and the hydroxy group on C1, giving the conjugated enone \b[with C1=d and a <carbonyl> at c]. The original ring methyl remains on the <carbon> next to C1 on the other side. The two rings share the C1–C6 bond. The molecular formula is consistent with loss of one <water> molecule from the diketone.
\Image[/past-exam-of-the-chemistry-b-course-of-the-university-of-cambridge/2025/paper-1-annulation.png]
{title=Numbered fused enone product and the ring-closing aldol step in the Robinson annulation}
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