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- In the reaction
$2 {X}+\mathrm{B}_2 \mathrm{H}_6 \rightarrow\left\lfloor\mathrm{BH}_2(\mathbf{X})_2\right\rfloor^{+}\left\lfloor\mathrm{BH}_4\right]^{-}$ the amine(s) ${X}$ is$(are)$
$(A)$ $\mathrm{NH}_3$ $(B)$ $\mathrm{CH}_3 \mathrm{NH}_2$ $(C)$ $\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$ $(D)$ $\left(\mathrm{CH}_3\right)_3 \mathrm{~N}$
$(A,D,B)$
$(A,C,D)$
$(A,B,C)$
$(B, C,D)$
Solution
Small amines such as $NH _3, CH _3 NH _2$ and $\left( CH _3\right)_2 NH$ give unsymmetrical cleavage of diborane according to following reaction.
$B _2 H _6+2 NH _3 \rightarrow\left[\left[ H _2 B \left( NH _3\right)_2\right]^{+}\left[ BH _4\right]^{-}\right.$
Large amines, such as $\left( CH _3\right)_3 N$ gives Symmetrical cleavage of diborane according to following reaction.
$B _2 H _6+2 N \left( CH _3\right)_3 \rightarrow 2 H _3 B \leftarrow N \left( CH _3\right)_3$
