detailed solution

Correct option is A

Two alcohols are produced. ${\mathrm{RCOOR}}^{\mathrm{\prime }}\stackrel{{\mathrm{LiAlH}}_4}{⟶}{\mathrm{RCH}}_2\mathrm{OH}+{\mathrm{R}}^{\mathrm{\prime }}\mathrm{OH}$.

The treatment of an open chain ester with ${\mathrm{LiAlH}}_4$​ followed by acid hydrolysis produces two alcohols.
For example, reduction of methyl acetate with lithium aluminum hydride produces methanol and ethanol.

${\mathrm{CH}}_3{\mathrm{COOCH}}_3+4\left[{\mathrm{H}}^{+}\right]\to {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}+{\mathrm{CH}}_3\mathrm{OH}$

This reduction can also be carried out by using sodium in the presence of alcohol, or hydrogen gas in the presence of copper chromite catalyst under pressure at $200-{300}^{\circ }\mathrm{C}$.