= Solution
Let $R=\mathrm{CH_2CH(Ph)-O-TEMPO}$, so B is $\mathrm{HO-R}$. Its hydroxyl group initiates <ring-opening polymerization> of lactide, giving
$$
\boxed{\mathrm{C=H-[OCH(Me)CO]_{2n}-O-CH_2CH(Ph)-O-TEMPO}.}
$$
Each lactide contributes two lactyl repeat units. Dry <monomer>, exclusion of water and an alcohol-compatible catalyst such as tin(II) 2-ethylhexanoate under heating provide a practical route. <Alcohol>/metal-alkoxide exchange followed by lactide coordination and acyl–oxygen cleavage extends the <ester> chain while retaining a terminal hydroxyl group.
Heating C under oxygen-free conditions homolytically cleaves its benzylic C–O bond to the nitroxide, releasing a benzylic radical and TEMPO. The radical adds styrene repeatedly; reversible trapping by TEMPO controls the active radical concentration. D is therefore a \b[poly(lactide)–<polystyrene> <block copolymer>] with junction $\mathrm{PLA-OCH_2CH(Ph)-[CH_2CH(Ph)]_m-O-TEMPO}$ and the hydroxyl end of PLA retained. This is <nitroxide-mediated polymerization>; a temperature around 120–130 °C is a representative condition for a TEMPO/styrene system, rather than an exact universal optimum.
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