Question

The reaction of following with ${\mathrm{H}}_3{\mathrm{N}}_3{\mathrm{B}}_3{\mathrm{Cl}}_3$(A) within${\mathrm{LiBH}}_4$ in tetrahydrofuran gives inorganic benzene (B), Then (A) reacts with (C) forming${\mathrm{H}}_3{\mathrm{N}}_3{\mathrm{B}}_3{\left(\mathrm{Me}\right)}_3$. Compounds (B) and (C) are

• A. Borazine and $\mathrm{MeBr}$
• B. Boron nitride and $\mathrm{MeBr}$
• C. Borazine and $\mathrm{MeMgBr}$
• D. Diborane and $\mathrm{MeMgBr}$

Answer 1

The correct option is C

Borazine and $\mathrm{MeMgBr}$

The explanation for the correct option

C. Borazine and $\mathrm{MeMgBr}$

• $$\begin{gathered} {\mathrm{H}}_3{\mathrm{N}}_3{\mathrm{B}}_3{\mathrm{Cl}}_3\left(\mathrm{s}\right)+{\mathrm{LiBH}}_4\left(\mathrm{s}\right)\underset{\mathrm{T}.\mathrm{H}.\mathrm{F}}{\overset{\mathrm{In}\mathrm{tetrahydrofurane}}{\to }}{\mathrm{B}}_3{\mathrm{N}}_3{\mathrm{H}}_6\left(\mathrm{s}\right)+\mathrm{LiCl}\left(\mathrm{s}\right)+{\mathrm{BCl}}_3\left(\mathrm{s}\right) \\ \left(\mathrm{A}\right)(\mathrm{Inorganic}\mathrm{Benzene})\left(\mathrm{B}\right) \end{gathered}$$
• $$\begin{gathered} {\mathrm{B}}_3{\mathrm{N}}_3{\mathrm{H}}_3{\mathrm{Cl}}_3\left(\mathrm{s}\right)+3\mathrm{MeMgBr}\left(\mathrm{s}\right)\to {\mathrm{B}}_3{\mathrm{N}}_3{\mathrm{H}}_3{\left({\mathrm{CH}}_3\right)}_3\left(\mathrm{s}\right)+3\mathrm{MgBrCl}\left(\mathrm{s}\right) \\ \left(\mathrm{A}\right)\left(\mathrm{C}\right) \end{gathered}$$

Hence option (C) is correct.,Compounds (B) and (C) are Borazine and $\mathrm{MeMgBr}$.