Question

The reaction of $H_3{N}_3{B}_{3}C{l}_3\left(A\right)$ with $LiB{H}_4$ in tetrahydrofurah gives inorganic benzene (B). Further, the reaction of (A) with (C) leads to $H_3{N}_3{B}_3(Me{)}_3$. Compounds (B) and (C) respectively, are:

• A. Boron nitride and $MeBr$
• B. Borazine and $MeMgBr$
• C. Borazine and $MeBr$
• D. Diborane and $MeMgBr$

Answer 1

The correct option is B Borazine and $MeMgBr$

$\underset{\left(A\right)}{B_3{N}_3{H}_{3}Cl3}+LiB{H}_4\to \underset{B}{B_3{N}_3{H}_6}+LCI+BC{I}_3$


$\underset{A}{B_3{N}_3{H}_{3}C{l}_3}+\underset{C}{3MeMgBr}\to B_3{N}_3{H}_3(C{H}_3{)}_3+3MgBrCl$
So, B is borazine and C is $C{H}_{3}MgBr$