Compound $(A)$ of boron reacts with NMe $_{3}$ and gives an adduct $(B)$ thus compound $(A)$ is Lewis

acid. Since $(B)$ on hydrolysis gives an acid $(C)$ and $\mathrm{H}_{2}$ gas, therefore (A) is $\mathrm{B}_{2} \mathrm{H}_{6}$, [B] isan

adduct $2 \mathrm{BH}_{3} \mathrm{NM} e_{3}$ and $(C)$ is boric acid.

Reactions are as follows:

$\quad \mathrm{B}_{2} \mathrm{H}_{6}+2 \mathrm{NM} e_{3} \rightarrow 2 \mathrm{BH}_{3} \mathrm{NMe}_{3}$

Diborane $(A)$ Adduct  $(B)$ $\quad$ Adduct $(B)$

$\mathrm{BH}_{3} \mathrm{NM} e_{3}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{3} \mathrm{BO}_{3}+\mathrm{NM} e_{3}+6 \mathrm{H}_{2}$