Proceed backward ; tribromobenzene is produced by boiling compound $$Z$$  with \[{{C}_{2}}{{H}_{5}}OH;Z\]    in turn is obtained by diazotisation of $$Y,$$  so $$Y$$  and $$Z$$  should have \[-N{{H}_{2}}\]  and \[-{{N}_{2}}Cl\]  groups respectively, in addition to three $$Br$$  atoms. Hence $$X$$  should be \[{{C}_{6}}{{H}_{5}}N{{H}_{2}}\]

NOTE : $$POC{l_3}$$  is a dehydrating agent. Hence \[{{C}_{6}}{{H}_{5}}CON{{H}_{2}}\xrightarrow{POC{{l}_{3}}}{{C}_{6}}{{H}_{5}}CN+{{H}_{2}}O\]

On reduction with $$Zn$$  and $$HCl,$$  \[{{C}_{6}}{{H}_{5}}{{N}_{2}}Cl\]    forms aniline as the main product.

\[C{{H}_{2}}=C{{H}_{2}}\xrightarrow[CC{{l}_{4}}]{B{{r}_{2}}}\]    \[\underset{\left( X \right)}{\mathop{BrC{{H}_{2}}-C{{H}_{2}}Br}}\,\xrightarrow{2NaCN}\]
\[\underset{\left( Y \right)}{\mathop{NCC{{H}_{2}}C{{H}_{2}}CN}}\,\xrightarrow{LiAl{{H}_{4}}}\]      \[\underset{\begin{smallmatrix} \left( \text{1, 4- Diaminobutane} \right) \\ \left( Z \right) \end{smallmatrix}}{\mathop{{{H}_{2}}NC{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}N{{H}_{2}}}}\,\]

Structural isomers of $${C_3}{H_9}N$$  are

\[\begin{align} & C{{H}_{3}}C{{H}_{2}}CN\xrightarrow{{{H}^{+}}/{{H}_{2}}O}\underset{\left( A \right)}{\mathop{C{{H}_{3}}C{{H}_{2}}COOH}}\,\xrightarrow[\Delta ]{N{{H}_{3}}} \\ & \underset{\left( B \right)}{\mathop{C{{H}_{3}}C{{H}_{2}}CON{{H}_{2}}}}\,\xrightarrow[NaOBr]{\text{Hoffmann bromamide reaction}}\underset{\left( C \right)}{\mathop{C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}}}\, \\ \end{align}\]

$$\mathop {{C_2}{H_5}N{H_2} + CHC{l_3}}\limits_{{\text{(Ethyl isocyanide)}}} + 3KOH \to $$       $${C_2}{H_5}N \equiv C + 3KCl + 3HCl$$