
$(a)$ $CH_3CH_2NH_2+ HI$
$(b)$ $(CH_3)_3N + HBr$


Product of the above reaction is
$[P]\xrightarrow{\begin{subarray}{l}
(i)\,NAN{O_2}/HCl,\,0 - {5^o}C \\
(ii)\,\beta - napthol/NaOH
\end{subarray} }Colored\,\,Solid$
$[P]\xrightarrow{{B{r_2}/{H_2}O}}{C_7}{H_6}NB{r_3}$
The compound $[P]$ is
