Symmetrical and Unsymmetrical Cleavage of Diborane by Amines

To determine the nature of the amine(s) $\text{X}$, we examine the cleavage behavior of diborane ($\text{B}_2\text{H}_6$) in the presence of Lewis bases.

Diborane undergoes two types of cleavage reactions depending on the steric hindrance and basicity of the attacking Lewis base:

1. Symmetrical Cleavage:
   Sterically bulky Lewis bases (such as tertiary amines, large phosphines, ethers, and $\text{CO}$) cause symmetrical cleavage of the diborane bridge, yielding monomeric Lewis acid-base adducts:
   $\text{B}_2\text{H}_6 + 2\text{L} \rightarrow 2\,\text{H}_3\text{B}\leftarrow\text{L}$
   For example, with trimethylamine ($(\text{CH}_3)_3\text{N}$):
   $\text{B}_2\text{H}_6 + 2(\text{CH}_3)_3\text{N} \rightarrow 2\,(\text{CH}_3)_3\text{N}\cdot\text{BH}_3$

2. Unsymmetrical Cleavage:
   Small, less sterically hindered Lewis bases cause unsymmetrical cleavage across the bridging system, generating an ionic salt consisting of a complex cation and a tetrahydroborate anion:
   $\text{B}_2\text{H}_6 + 2\text{X} \rightarrow [\text{BH}_2(\text{X})_2]^+[\text{BH}_4]^-$
   This mode of cleavage occurs with:

   • Ammonia ($\text{NH}_3$)
   • Primary amines (e.g., $\text{CH}_3\text{NH}_2$)
   • Secondary amines (e.g., $(\text{CH}_3)_2\text{NH}$)

Since the given reaction:
$2\text{X} + \text{B}_2\text{H}_6 \rightarrow [\text{BH}_2(\text{X})_2]^+[\text{BH}_4]^-$
represents the unsymmetrical cleavage of diborane, the amine $\text{X}$ must be a sterically unhindered base, which includes $\text{NH}_3$, $\text{CH}_3\text{NH}_2$, and $(\text{CH}_3)_2\text{NH}$. In contrast, $(\text{CH}_3)_3\text{N}$ is sterically hindered and undergoes symmetrical cleavage.

Therefore, the correct options are A, B, and C.