The reaction of ${\mathrm{H}}_3{\mathrm{N}}_3{\mathrm{B}}_3\mathrm{C}{\mathrm{l}}_3\left(\mathrm{A}\right)$ with $\mathrm{L}\mathrm{i}\mathrm{B}{\mathrm{H}}_4$ in tetrahydrofuran gives inorganic benzene $\left(\mathrm{B}\right)$. The reaction is given below.  Further, the reaction of compound  $\left(\mathrm{A}\right)$ reacts with $\left(\mathrm{C}\right)$ leads to ${\mathrm{H}}_3{\mathrm{N}}_3{\mathrm{B}}_3{\left(\mathrm{M}\mathrm{e}\right)}_3$ .

The reaction is given beow:

 $\underset{\left(\mathrm{A}\right)}{{\mathrm{B}}_3{\mathrm{N}}_3{\mathrm{H}}_3{\mathrm{CI}}_3}+{\mathrm{LiBH}}_4⟶\underset{\left(\mathrm{B}\right)}{{\mathrm{B}}_3{\mathrm{N}}_3{\mathrm{H}}_6}+\mathrm{LiCI}+{\mathrm{BCI}}_3$
$\underset{\left(\mathrm{A}\right)}{{\mathrm{B}}_3{\mathrm{N}}_3{\mathrm{H}}_3{\mathrm{CI}}_3}+\underset{\left(\mathrm{C}\right)}{3\mathrm{MeMgBr}}⟶{\mathrm{B}}_3{\mathrm{N}}_3{\mathrm{H}}_3{\left({\mathrm{CH}}_3\right)}_3+3\mathrm{MgBrCI}$