$A$ compound $(A)$ of boron reacts with $NMe_3$ to give an adduct $(B)$ which on hydrolysis gives a compound $(C)$ and hydrogen gas. Compound $(C)$ is an acid. Identify the compounds $A, B$ and $C$. Give the reactions involved.

(N/A) Compound $(A)$ of boron reacts with $NMe_3$ to form an adduct $(B)$,indicating that $(A)$ is a Lewis acid. Since $(B)$ on hydrolysis yields an acid $(C)$ and $H_2$ gas,$(A)$ is identified as $B_2H_6$,$(B)$ is the adduct $BH_3 \cdot NMe_3$,and $(C)$ is boric acid $(H_3BO_3)$.
The reactions are as follows:
$B_2H_6 + 2NMe_3 \rightarrow 2BH_3 \cdot NMe_3$
$(A) \text{ (Diborane)} + \text{Reactant} \rightarrow (B) \text{ (Adduct)}$
$BH_3 \cdot NMe_3 + 3H_2O \rightarrow H_3BO_3 + NMe_3 + 3H_2$
$(B) + \text{Hydrolysis} \rightarrow (C) \text{ (Boric acid)} + \text{Byproducts}$