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

G is the diaryl ether $(2-\mathrm{N}{\mathrm{O}}_2{\mathrm{C}}_6{\mathrm{H}}_4)-\mathrm{O}-(2-\mathrm{C}{\mathrm{H}}_2\mathrm{C}(=\mathrm{O}){\mathrm{N}}_{\mathrm{morph}}{\mathrm{C}}_6{\mathrm{H}}_4)$. Here $N_{morph}$ denotes the nitrogen of the intact morpholine ring; the phenoxy oxygen and phenylacetamide chain are ortho on their original ring. potassium carbonate first generates phenoxide from F. Its oxygen attacks the fluorine-bearing carbon of 1-fluoro-2-nitrobenzene, forming a Meisenheimer complex. Delocalization of its negative charge into the ortho nitro group stabilizes this addition intermediate. Fluoride elimination restores aromaticity and completes nucleophilic aromatic substitution.

This first step forms an O–aryl bond, rather than immediately arylating the less acidic alpha carbon. The neutral product has formula ${\mathrm{C}}_{18}{\mathrm{H}}_{18}{\mathrm{N}}_2{\mathrm{O}}_5$, obtained from the two substrates by net loss of HF. The connectivity is shown beside B in the independently drawn diagram above. The amide-substrate O-to-C migration strategy has direct primary precedent.^[ref]