# Why is diborane cleaved symmetrically by trimethylamine but unsymmetrically by ammonia?

I know that soft bases cleave diborane symmetrically but hard ones don't. However, what makes ammonia harder than trimethylamine? And why does hardness/softness affect this?

• Ooh, this question is a little old but I once read somewhere that this particular case was because of sterics. I can't remember the source - either Massey's Main Group Chemistry or Greenwood & Earnshaw's Chemistry of the Elements. – orthocresol Nov 20 '15 at 21:31
• Greenwood & Earnshaw's Chemistry of the Elements does indeed mention sterics as the reason but doesn't go into much detail. It's on page 165 – bon Nov 20 '15 at 22:53

The mechanism proposed for the cleavage reaction involves an initial attack by the donor on one boron atom in diborane, leading to cleavage of one $\ce{B-H-B}$ bridge.
This is followed by the attack of a second donor molecule, cleaving the remaining $\ce{B-H-B}$ bridge. If the boron atom in the pendant $\ce{BH3}$ group is attacked, two moles of adduct result, whereas attack on the atom already carrying a donor produces cation and borohydride:
Support for this view comes from the fact that the intermediate ($\text{I}$) is detectable by tensimetric titration of amine-borane with diborane and by $\ce{^{11}B}$ nmr.