= Solution
\b[D is $\mathrm{(MeO)_2CHCH_2CH_2CO_2Me}$, and E is $\mathrm{OHCCH_2CH_2CO_2Et}$.] D results from <transesterification>; E results from <acid-catalysed acetal hydrolysis>.
For D, <methoxide> attacks the <ester> carbonyl carbon and moves the carbonyl $\pi$ pair to <oxygen>. Collapse of the tetrahedral <reaction intermediate> reforms C=O and expels <ethoxide>. <Proton transfer> with the <methanol> <solvent> regenerates <methoxide>. Excess <methanol> favours replacement of the ethyl <ester> by a methyl <ester>; the dimethyl <acetal> survives these basic conditions.
For E, protonate one <acetal> <oxygen>, then break its C–O bond with the pair going to <oxygen>, releasing <methanol> and forming an <oxocarbenium ion>. <Water> attacks that carbon. Deprotonation gives a <hemiacetal>; protonation of its remaining methoxy <oxygen> permits loss of a second <methanol>. The hydroxyl <oxygen>'s <lone pair> forms C=O during this loss, and deprotonation gives the <aldehyde>, regenerating the acid <catalyst>.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025-ester-acetal-mechanisms.png]
{title=Original products and electron-flow steps for <transesterification> and <acid-catalysed acetal hydrolysis>}
The spectroscopy is consistent with these products. D has one <ester> carbonyl near $1745\,\mathrm{cm^{-1}}$ and six <carbon-13 NMR> environments: acetal CH, two distinct chain $\mathrm{CH_2}$ sites, carbonyl, ester $\mathrm{OCH_3}$, and the two equivalent acetal $\mathrm{OCH_3}$ groups. Thus seven carbon atoms give six resonances. E has separate <ester> and <aldehyde> carbonyl bands, approximately $1745$ and $1730\,\mathrm{cm^{-1}}$, and six distinct carbon atoms: two carbonyls, two chain methylenes and the two ethyl carbons.
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