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 / 35 / c / i / Solution

K is ethyl 4-(4-methylbenzoyl)-1H-pyrrole-2-carboxylate. AlCl${}_3$ activates the aroyl chloride to an acylium-like electrophile. Pyrrole attacks through C4, forming a sigma complex; deprotonation restores aromaticity and work-up releases the ketone from its Lewis-acid complex.

Unsubstituted pyrrole normally favours an $\alpha$ position, but the ester at C2 changes the regioselectivity. Attack at C5, or at C3, permits important sigma-complex contributors with positive charge at the ester-bearing C2, which is strongly destabilized by the electron-withdrawing ester. C4 substitution avoids that particularly unfavourable contributor. The C2 ester therefore directs the acylation towards C4, overriding the simple unsubstituted-pyrrole rule. Combining the starting ester with a p-toluoyl group while replacing a ring hydrogen gives ${\mathrm{C}}_{15}{\mathrm{H}}_{15}\mathrm{N}{\mathrm{O}}_3$, as required.