The major product of the following reaction is ${\mathrm{CH}}_3\mathrm{CH}={\mathrm{CHCO}}_2{\mathrm{CH}}_3\stackrel{{\mathrm{LiAlH}}_4}{\to }$

$\begin{array}{l}{\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-COO{\mathrm{CH}}_3\\ \downarrow \mathrm{LAH}\\ {\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_2-\mathrm{OH}\end{array}$

LAH will not reduce double bond, if it is not conjugated with phenyl ring