Question

What is $Z$ in the following reactions?

$B{Cl}_3+H_2\underset{450}{\overset{Cu.Al}{\to }}X+HCl$

$X\stackrel{methylation}{\to }Z$

• A. $\left(C{H}_3\right)B{H}_2$
• B. ${\left(C{H}_3\right)}_4{B}_2{H}_2$
• C. ${\left(C{H}_3\right)}_3{B}_2{H}_3$
• D. ${\left(C{H}_3\right)}_6{B}_2$

Answer 1

The correct option is B ${\left(C{H}_3\right)}_4{B}_2{H}_2$

A mixture of $BC{l}_3$ and $H_2$ at low pressure when subjected to silent electric discharge gives $B_2{H}_6$.

$BC{l}_3+H_2\stackrel{Cu.Al}{\to }{B}_2{H}_6+HCl$

$B_2{H}_6\stackrel{methylation}{\to }(C{H}_3{)}_4{B}_2{H}_2$

It has two coplanar $B{H}_2$ groups. There are 4 terminal and 2 bridge H atoms methylation of $B_2{H}_6$ gives $(C{H}_3{)}_4{B}_2{H}_2$.

The hybridisation of B is $s{p}^3$.

Hence, option $B$ is correct.