Selective reduction of $\mathrm{PhCH}=\mathrm{CHCO}_2 \mathrm{Et}$ to $\mathrm{PhCH}=\mathrm{CHCH}_2

Question

Selective reduction of $\mathrm{PhCH}=\mathrm{CHCO}_2 \mathrm{Et}$ to $\mathrm{PhCH}=\mathrm{CHCH}_2 \mathrm{OH}$ is performed using

Professor PC Thomas Classes · Answer

$\mathrm{LiAlH}_4$ reduces esters to alcohols but leaves isolated or conjugated (in some conditions) double bonds alone. [FLAG: check exact reagents]