$CH _{3}- CH _{2}- C \equiv N$$\xrightarrow{{LiAl{H_4}}}$$CH _{3}- CH _{2}- CH _{2}- NH _{2}$
$\left( A \right)\xrightarrow{{\operatorname{Re}duction}}\left( B \right)\xrightarrow{{HN{O_2}}}{C_2}{H_5}OH$
Number of moles of $NaOH $ used in above Hoffmann bromamide reaction is
In the given reaction what will be the product