= Solution
The first step is an endo-selective <Diels-Alder reaction> of the pyrone diene with the electron-poor acrylamide. Number the pyrone $O1-C2(=O)-C3(OH)=C4-C5=C6$, and the acrylamide carbons $C_\alpha$ and $C_\beta$, with $C_\alpha$ next to its <carbonyl>. The favored polar connection is \b[C3–C$_\alpha$ and C6–C$_\beta$], leaving C4=C5 and the original C2–O1 lactone bridge intact.
Lithium allyloxide, generated from <n-Butyllithium> and allyl <alcohol>, attacks the bridging lactone <carbonyl>. Collapse of the <tetrahedral intermediate> cleaves its acyl–oxygen bond. <proton transfer> converts the former O1 bridge into an OH substituent at C6 and leaves the former pyrone <carbonyl> as an allyl <ester> attached to C3. The morpholine <amide> is retained.
Thus \b[C is allyl 1,4-dihydroxy-2-(morpholine-4-carbonyl)cyclohex-5-ene-1-carboxylate], with the <stereochemistry> described below. In this product numbering C1 is the former pyrone C3, C2 is the acrylamide $C_\alpha$, C3 is $C_\beta$, and C4 is the former pyrone C6.
\Image[/past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1-pyrone-product.png]
{title=Predicted ring-opened pyrone cycloadduct with explicit relative stereochemistry}
The bridge fixes the former C3 carboxyl group and C6 oxygen on the same face. An endo acrylamide places its <amide> substituent on that face too. Accordingly \b[CO$_2$allyl at C1, the C2 <amide> and the C4 OH are cis; the C1 OH is on the opposite face]. The alternative face of attack gives the <enantiomer>, so achiral reagents give a racemate. The $1731\ \mathrm{cm^{-1}}$ band is consistent with the newly formed <ester>, while the strong lower-frequency band is consistent with the retained, strongly polarized <amide> <carbonyl>. These bands support carbonyl-class retention; they do not independently prove the predicted regioisomer or <stereochemistry>.
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