$B_2H_6 + NH_3 \xrightarrow[\text{low temperature}]{\text{excess} NH_3} B_2H_6. 2NH_3$
Diborane with ammonia gives $B _2 H _6 \cdot 2 NH _3$ that is formulated as $\left[ BH _2\left( NH _3\right)_2\right]^{+}\left[ BH _4\right]^{-}$which when heated to $473 K$, decomposes to give borazole.