In the reaction, $$ 2 \mathrm{X}+\mathrm{B}_2 \mathrm{H}_6…
In the reaction,
$$
2 \mathrm{X}+\mathrm{B}_2 \mathrm{H}_6 \longrightarrow\left[\mathrm{BH}_2(\mathrm{X})_2\right]^{+}\left[\mathrm{BH}_4\right]^{-}
$$
The amine (s) $X$, is (are)
$\mathrm{NH}_3$
$\mathrm{CH}_3 \mathrm{NH}_2$
$\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$
$\left(\mathrm{CH}_3\right)_3 \mathrm{~N}$
Solution
Small amines such as $\mathrm{NH}_3 \cdot \mathrm{CH}_3 \mathrm{NH}_2$ and $\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$ give unsymmetrical cleavage of diborane according to following reaction,
$\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NH}_3 \longrightarrow {\left[\mathrm{H}_2 \mathrm{~B}\left(\mathrm{NH}_3\right)_2\right]^{+}\left[\mathrm{BH}_4\right]^{-}}$
Large amines, such as $\left(\mathrm{CH}_3\right)_3 \mathrm{~N}$ give symmetrical cleavage of diborane according to following reaction,
\(\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{~N}\left(\mathrm{CH}_3\right)_3 \longrightarrow 2 \mathrm{H}_3 \mathrm{~B} \longleftarrow \mathrm{N}\left(\mathrm{CH}_3\right)_3\)