Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / Section I / 34 / c / iii

KOH hydrolyses the thiazole ethyl ester to a carboxylate; acidification gives the acid. Silver(I) then activates the neighbouring alkyne towards intramolecular six-membered O-cyclization. The carboxylate oxygen attacks the alkyne carbon farther from the thiazole, so the new ring follows carbonyl C–O–distal alkyne C–proximal alkyne C–thiazole C4–thiazole C5. A vinyl-silver intermediate is protonated to give the fused pyrone G.

The remote terminal carboxylic acid survives. Ester hydrolysis changes E's formula by $-{\mathrm{C}}_2{\mathrm{H}}_4$ to ${\mathrm{C}}_{16}{\mathrm{H}}_{12}\mathrm{FN}{\mathrm{O}}_4\mathrm{S}$; the subsequent addition across the alkyne changes connectivity without altering that formula. This distinguishes the lactonizing carboxyl group at thiazole C5 from the side-chain acid, which is not the group closing the observed fused ring.