Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / Section A / 19 / b / Solution

Styrene, benzoyl peroxide and TEMPO provide the required radical addition–trapping sequence. Heating the peroxide breaks its O–O bond into benzoyloxy radicals. Before decarboxylation, $\mathrm{PhC}{\mathrm{O}}_2^{\bullet }$ can add to styrene's terminal carbon, generating the resonance-stabilized radical $\mathrm{PhC}{\mathrm{O}}_2\mathrm{C}{\mathrm{H}}_2\dot{\mathrm{C}}\mathrm{HPh}$. Coupling with the oxygen-centred TEMPO radical then gives $\mathrm{PhC}{\mathrm{O}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{CH}(\mathrm{Ph})-\mathrm{O}-\mathrm{TEMPO}$, the benzoate-protected alkoxyamine A.

The connectivity distinguishes this route from initiators that leave a carbon-based radical fragment at the chain end. Competition with benzoyloxy-radical decarboxylation affects yield, so this is a plausible preparation rather than a claim of exclusive product formation. Subsequent benzoate cleavage reveals B's hydroxyl group without requiring rupture of its alkoxyamine linkage.