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The cleavage of diborane was presented in a recent lecture, and it was said that the borane could be cleaved unsymmetrically if sterics allowed, or symmetrically if they did not, as shown in the image below:

enter image description here

However, no mechanism for this cleavage was provided, nor can it be found in Housecroft and Sharpe's "Inorganic Chemistry", our course textbook, nor can I find much about it online.

I have postulated such a mechanism:

enter image description here

Is it known how this cleavage occurs, and if it does indeed follow this mechanism? Is my attempt a reasonable proposition?

Jacob
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    I cannot discern the failure to account for charges in your mechanism. – Lighthart Apr 25 '16 at 16:57
  • @Lighthart, neither can I now actually; I guess I was having a bit of a mind-blank as I posted this; sorry about that, I'll edit the question appropriately. I'd still like to know if this is a reasonable mechanism though, or whether the correct one is known. – Jacob Apr 25 '16 at 17:40
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    Inorg. Chem., 1987, 26, 43 may or may not be relevant - I don't have the time to read in detail now. It does mention the word "intermediates" and is indexed under "inorganic reaction mechanisms" on SciFinder. Feel free to steal it to write an answer. – orthocresol Apr 25 '16 at 18:22
  • Related, unanswered: http://chemistry.stackexchange.com/questions/27826/why-is-diborane-cleaved-symmetrically-by-trimethylamine-but-unsymmetrically-by-a – orthocresol Apr 25 '16 at 18:24

1 Answers1

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Is it known how this cleavage occurs?

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.

enter image description here

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:

enter image description here

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.

Source: Boron Hydride Chemistry, Earl Muetterties

Berry Holmes
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