Past exam of the chemistry B course of the University of Cambridge / 2024 / Paper 1 / B2 / b / ii / Solution

The same base first makes $\mathrm{Ph}{\mathrm{C}}^{-}(\mathrm{OPh})\mathrm{CN}$. This time addition occurs at a carbon bearing hydrogen para to the nitro group, giving a sigma-H adduct. The attacked carbon temporarily bears both its original H and the new carbon substituent. Negative charge is again delocalized through the ring toward the nitro oxygens.

A further base removes the hydrogen on that aromatic addition site. Its C–H electrons form an exocyclic bond toward the incoming carbon, while that carbon expels phenoxide. This is a base-induced elimination across the new ring–side-chain bond. Redistribution of the nitronate/quinoid pi system restores an aromatic ring and gives the side-chain carbanion $\mathrm{p}-\mathrm{N}{\mathrm{O}}_2{\mathrm{C}}_6{\mathrm{H}}_4-{\mathrm{C}}^{-}(\mathrm{Ph})\mathrm{CN}$. protonation during acidic workup yields $\mathrm{p}-\mathrm{N}{\mathrm{O}}_2{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{CH}(\mathrm{Ph})\mathrm{CN}$.

This is vicarious nucleophilic substitution: the ring loses hydrogen while the nucleophile sacrifices its phenoxy leaving group. hydrogen is not expelled as hydride, and the product's side-chain hydrogen can be supplied on workup rather than moving directly from the ring in a compulsory hydride shift. Base and isotope experiments in the primary mechanistic work support reversible addition followed by base-induced elimination, while not uniquely choosing every E2/E1cb limit.^[ref] The displayed para product is consistent with access to the less crowded activated position.