Past exam of the chemistry B course of the University of Cambridge / 2023 / Paper 1 / B2 / c / iii

base removes an alpha-amide proton from G, giving a carbon nucleophile. Its intramolecular attack on the nitroaryl ipso carbon, which currently bears the ether oxygen, forms a five-membered cyclic Meisenheimer complex. Count its atoms along the existing tether: alpha carbon, the original phenyl carbon bearing the side chain, its adjacent phenoxy carbon, ether oxygen, and attacked nitroaryl ipso carbon. The ortho nitro group again stabilizes the temporarily dearomatized anion.

Rearomatization now breaks the nitroaryl–O bond, leaving phenoxide on the original ring. proton transfer gives $(2-\mathrm{HO}{\mathrm{C}}_6{\mathrm{H}}_4)-\mathrm{CH}(2-\mathrm{N}{\mathrm{O}}_2{\mathrm{C}}_6{\mathrm{H}}_4)-\mathrm{C}(=\mathrm{O}){\mathrm{N}}_{\mathrm{morph}}$. This is a Truce-Smiles rearrangement: the nitroaryl group migrates from oxygen to carbon, while the original phenoxy oxygen stays on its original ring and becomes OH. No nitro-group migration or amide cleavage is needed. G and H have identical formulas; only the bond connections and a proton position change. carbonate need not produce a high equilibrium enolate concentration: a small population can be consumed by favourable intramolecular cyclization and rearomatization. The specified transformation alone does not establish its detailed proton-transfer sequence.