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 / i / Solution

The thioamide first undergoes sulfur alkylation at the bromomalonate carbon, displacing bromide. Intramolecular attack of nitrogen on one ester carbonyl and ethanol loss produce a five-membered thiazolinone-type intermediate. Tautomerization gives ethyl 2-(3-fluorophenyl)-4-hydroxythiazole-5-carboxylate. For unambiguous atom assignment, write the ring in order as S1–C2(aryl)–N3–C4(OH)–C5(CO${}_2$Et): the thioamide carbon becomes C2, its nitrogen becomes N3, the attacked ester carbonyl becomes C4 and the original bromomalonate methine becomes C5.

Thus the correctly numbered intermediate is ethyl 2-(3-fluorophenyl)-4-hydroxythiazole-5-carboxylate. Triflic anhydride activates the hydroxyl oxygen; triethylamine removes its proton to give D = ethyl 2-(3-fluorophenyl)-4-(trifluoromethanesulfonyloxy)thiazole-5-carboxylate. This aryl triflate undergoes the subsequent palladium-catalysed alkynyl coupling at C4. The explicit ring atom mapping prevents interchanging the 4-hydroxy and 5-ester positions.