Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 7 / Solution

Open the cyclic ester at its acyl–oxygen bond. The resulting delta-hydroxy acid isOxidation-state reversal at the benzylic alcohol gives $\mathrm{H}{\mathrm{O}}_2\mathrm{C}-\mathrm{CH}(\mathrm{nPr})-\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}(\mathrm{iPr})-\mathrm{C}(=\mathrm{O})\mathrm{Ph}$. This identifies the key Michael disconnection between the central $\mathrm{C}{\mathrm{H}}_2$ and $\mathrm{CH}(\mathrm{iPr})$ groups:The nucleophile is the enolate or enamine of the phenyl ketone. The electrophile's propyl substituent must be alpha to its ester carbonyl; using a beta-propyl acrylate would move the substituent to the wrong ring position.

Disconnect $\mathrm{PhCOC}{\mathrm{H}}_2\mathrm{iPr}$ further to the indicated acetophenone. Crossed aldol addition to acetone followed by dehydration gives $\mathrm{PhCOCH}=\mathrm{CM}{\mathrm{e}}_2$; selective conjugate reduction or controlled alkene hydrogenation then gives $\mathrm{PhCOC}{\mathrm{H}}_2\mathrm{CHM}{\mathrm{e}}_2$. This avoids requiring an efficient SN2 alkylation with a secondary isopropyl halide. The alpha-propyl acrylate can be obtained by alpha-hydroxymethylation/dehydration of ethyl pentanoate, $\mathrm{Et}{\mathrm{O}}_2\mathrm{CC}{\mathrm{H}}_2\mathrm{nPr}$, using formaldehyde.

In the forward direction, perform the Michael coupling, reduce the phenyl ketone selectively with sodium borohydride while retaining the ordinary ester, hydrolyse the ester, and cyclize the delta-hydroxy acid by controlled acid-catalysed lactonization or an activated-acid coupling. The six-membered ring then has the order carbonyl–$\mathrm{CH}(\mathrm{nPr})$–$\mathrm{C}{\mathrm{H}}_2$–$\mathrm{CH}(\mathrm{iPr})$–$\mathrm{CH}(\mathrm{Ph})$–O. This atom order checks all three substituent positions. No relative configuration is specified, so the scheme does not imply a single diastereomer from the Michael addition or reduction.