Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 6 / Solution

Open the lactone formally at its acyl–oxygen bond, then replace the resulting secondary alcohol by its ketone precursor. With $R=\mathrm{CH}(\mathrm{Me})\mathrm{CH}(\mathrm{Me}{)}_2$, the useful sequence of retrosynthetic intermediates isDisconnect the last compound as a Michael addition product of the methyl-side enolate of $\mathrm{RCOC}{\mathrm{H}}_3$ and ethyl methacrylate:The substituted carbon of the methacrylate becomes the methyl-bearing carbon alpha to the lactone carbonyl. The ketone carbon becomes the oxygen-bearing ring carbon. This checks the six-membered ring connectivity.

The required branched methyl ketone itself comes from butan-2-one by ethyl-side aldol addition to acetone, followed by dehydration and selective hydrogenation:Choose and isolate the aldol regioisomer shown. For the subsequent Michael step, use a bulky base at low temperature to favour its less hindered methyl-side kinetic enolate. Reduce the resulting ketone selectively with $\mathrm{NaB}{\mathrm{H}}_4$ while retaining the ester, hydrolyse the ester, then acidify and promote intramolecular esterification with water removal. The hydroxy acid closes to the required six-membered lactone. There is no specified stereochemical relationship to enforce, but a stereochemically pure target would require an additional selectivity or resolution step.