Question

What happens when Diborane reacts with Lewis base?

Answer 1

1. Diborane (${\mathrm{B}}_2{\mathrm{H}}_6$), when treated with Lewis base, undergoes cleavage or bond-breaking to form Borane (${\mathrm{BH}}_3$) addition products.
   For example-
   $$\begin{gathered} {\mathrm{B}}_2{\mathrm{H}}_6+2\mathrm{CO}\to 2{\mathrm{BH}}_3.\mathrm{CO} \\ \mathrm{Diborane}\mathrm{Borane}\mathrm{addition}\mathrm{product} \\ (\mathrm{Borane}\mathrm{carbonyl}) \end{gathered}$$
2. The reaction of diborane with Lewis base Ammonia occurs in two steps.
   i) Ammonia reacts with diborane forming ${\mathrm{B}}_2{\mathrm{H}}_6.{\mathrm{NH}}_3$ whose ionic form is $[{\mathrm{BH}}_2{\left({\mathrm{NH}}_3\right)}_2{]}^{+}{\left[{\mathrm{BH}}_4\right]}^{-}$. ${\mathrm{B}}_2{\mathrm{H}}_6+{\mathrm{NH}}_3\to {\mathrm{B}}_2{\mathrm{H}}_6.{\mathrm{NH}}_3$
   ii) When $[{\mathrm{BH}}_2{\left({\mathrm{NH}}_3\right)}_2{]}^{+}{\left[{\mathrm{BH}}_4\right]}^{-}$ is further heated, it gives Borazine, ($2{\mathrm{B}}_3{\mathrm{H}}_6{\mathrm{N}}_3$) with the elimination of Hydrogen from the Lewis base. $3{\mathrm{B}}_2{\mathrm{H}}_6+6{\mathrm{NH}}_3\stackrel{473\mathrm{K}}{\to }2{\mathrm{B}}_3{\mathrm{N}}_3{\mathrm{H}}_6+12{\mathrm{H}}_2$
3. Therefore, Diborane reacts with Lewis base, forming Borane addition products by undergoing cleavage.