$\mathrm{R}-\mathrm{NH}_2 \stackrel{\mathrm{NaNO}_2+\mathrm{HCl}}{\longrightarrow} \mathrm{R}-\mathrm{OH}+\mathrm{N}_2+\mathrm{H}_2 \mathrm{O}$
Thus, $\left(\mathrm{CH}_2\right)-\mathrm{NH} \rightarrow 2^{\circ}$ amine, do not give alcohol with $\mathrm{NaNO}_2$ \& $\mathrm{HCl}$.