JEE Main · 2020 · 09 Jan 2020 · Shift 2 · p-Block Elements

Trichloroborazine, ${\mathrm{H}}_3{\mathrm{N}}_3{\mathrm{B}}_3\mathrm{C}{\mathrm{l}}_3$, reacts with $\mathrm{LiB}{\mathrm{H}}_4$ to have its chlorines replaced by hydrogen, giving borazine ${\mathrm{H}}_3{\mathrm{N}}_3{\mathrm{B}}_3{\mathrm{H}}_3$, the so-called inorganic benzene whose ring structure mirrors that of benzene. To set methyl groups on boron in place of the chlorines a carbanion-like methyl source is required, and the Grignard reagent methylmagnesium bromide, $\mathrm{MeMgBr}$, does exactly this to form ${\mathrm{H}}_3{\mathrm{N}}_3{\mathrm{B}}_3(\mathrm{Me}{)}_3$. The two compounds are thus borazine and $\mathrm{MeMgBr}$.