Propene is allowed to react with $\mathrm{m}$-chloroperoxybenzoic acid. The product $\left(\mathrm{A}\right)$ is then reduced with ${\mathrm{LiAlH}}_4$ in dry ether to give $\left(\mathrm{B}\right).$

${\mathrm{CH}}_3\mathrm{CH}={\mathrm{CH}}_2\stackrel{\mathrm{mCPBA}}{⟶}\mathrm{A}\underset{2·{\mathrm{H}}_3{\mathrm{O}}^{+}}{\overset{1.{\mathrm{LiAlH}}_4}{\to }}\mathrm{B}$

The structure of the product $\left(\mathrm{B}\right)$ is: