 Solution

The hydroxylactam is a cyclic hemiaminal. Its alkoxide can open to the corresponding aldehyde/amidate, $\mathrm{MeNHCO}-\mathrm{CH}(\mathrm{OBn})-\mathrm{C}{\mathrm{H}}_2-\mathrm{CHO}$ in its protonated representation. This exposes an aldehyde without reducing or replacing the lactam carbonyl.

The ketophosphonate anion attacks that aldehyde. Formation and collapse of the phosphorus-containing intermediate give the predominantly E enone $\mathrm{MeNHCO}-\mathrm{CH}(\mathrm{OBn})-\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}=\mathrm{CH}-\mathrm{COC}{\mathrm{H}}_3$ and phosphate by Horner-Wadsworth-Emmons reaction. The amide nitrogen then closes by intramolecular aza-Michael addition at the enone beta carbon. Protonation of the ketone enolate gives the observed five-membered lactam with a $-\mathrm{C}{\mathrm{H}}_2\mathrm{COC}{\mathrm{H}}_3$ substituent.

A folded, relatively unencumbered closure places OBn and the developing side chain in outward-facing positions; the expected major product has OBn and the C5 acetonyl substituent cis. With the OBn wedge retained, show the new C5–CH${}_2$COCH${}_3$ bond as a solid wedge. The closure is diastereoselective, while the original wavy hemiaminal centre is erased by aldehyde formation. This facial prediction rests on the preferred folded conformer rather than retention of either starting hemiaminal epimer.