$CH_3CH_2NH_2 + CHCl_3+ 3KOH \rightarrow (a) + (b) + 3H_2O$
$\begin{array}{l} Cl -\left( CH _2\right)_4- Cl \xrightarrow{\text { excess } NH _3} A \xrightarrow{ NaOH } \\ B + H _2 O + NaCl\end{array}$