= Solution
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1-q3-acetal-lactonization.png]
{title=Benzylidene-acetal hydrolysis releases a diol that closes to an unsaturated gamma-lactone}
First hydrolyze the benzylidene <acetal>. Protonate one <acetal> <oxygen> and break its bond to the Ph-CH centre, giving one free <alcohol> and a benzylic <oxocarbenium ion>. Water attacks that centre; <proton> transfers give a <hemiacetal>. Protonate the remaining acetal-derived OR group and cleave it to release the second <alcohol> and \b[<benzaldehyde>].
The resulting unsaturated hydroxy <ester> is
$$
\mathrm{HO-CH(CH_2CH{=}CH_2)-C(CH_2OH){=}CH-C(O)OMe}.
$$
The OH on the allyl-substituted methine is gamma to the <ester> <carbonyl>. <Protonation> of that <carbonyl> enables intramolecular OH attack to give a five-membered <tetrahedral intermediate>. <Proton> transfer to methoxy and collapse expel <methanol>, followed by <deprotonation> to give the \b[alpha,beta-unsaturated gamma-lactone].
The other liberated OH remains CH$_2$OH. Both C=C units survive: one is the allyl substituent and the other becomes the <lactone>'s conjugated double bond. <Hydrolysis> to a hydroxy acid followed by <lactonization> is an alternative sequence under the aqueous acid; direct acid-catalysed intramolecular <ester> exchange gives the same atom mapping.
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