The compound diborane on alkylation loses the terminal $\ce{H}$-bonds rather than the bridge bonds indicating that the bridge bond is stronger but on the other hand it's length is found to be longer than terminal hydrogen bonds. But usually bond length is inversely proportional to bond strength.

What is the reason for this?

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    $\begingroup$ I don’t really see how you can conclude anything about bond strengths from one reaction. $\endgroup$
    – orthocresol
    Mar 26 '18 at 13:56
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    $\begingroup$ @ortho Why not? Does this method have known exceptions? Or is it just illogical? $\endgroup$ Mar 26 '18 at 14:12
  • $\begingroup$ Premise is false, also it's not first case with this weird idea about strength of those bridging bonds. No idea where it's stemming form. $\endgroup$
    – Mithoron
    Mar 26 '18 at 17:51
  • $\begingroup$ Is there an answer to this question with a proper reference? It still seems unsolved almost 2 years later $\endgroup$ Jan 27 '20 at 11:54
  • $\begingroup$ The answer by Oscar Lanzi Here is a good one, but the reference seems a bit shady (it has Wikipedia as its citations). I’m looking for a definitive answer on whether the bridge bond is stronger or weaker (In terms of bond enthalpy) than the terminal bonds. $\endgroup$ Jan 27 '20 at 12:26

As ocresol said, its not wise to conclude the bond strength from one reaction. Rather, considering the structure of the molecule is a more viable option.
Actually the bridged bond is weaker than the terminal $\ce{B-H}$ bonds. This is because of the presence of multielectron density region. Since it is a pair of shared electrons spread between two regions, so more volume and less the electron density. enter image description here

Image Source: Quora.
Also, the bond order of terminal $\ce{B-H}$ bonds is 1, whereas the bond order of the bridged bond is half of it ie. 0.5. And it is intuitive that bond strength is directly related to bond order.

  • $\begingroup$ Also, a source on quora claims that on alkylation, whichever H atom is released, the B-C bond will remain/transfer to a terminal position because that would be energetically more favorable. I’m not sure about it though. $\endgroup$
    – MollyCooL
    Mar 26 '18 at 14:44
  • $\begingroup$ So can anyone explain why the terminal bonds rather than the bridge bonds are lost during alkylation? $\endgroup$
    – M. S. L
    Mar 27 '18 at 10:02
  • $\begingroup$ @M.S.L Thats what I told you, its apparently not that the terminal bonds are broken. Even if the bridged hydrogen is released, the B-C bond formed because of alkylation, returns to the terminal position because it is energitically favourable. $\endgroup$
    – MollyCooL
    Mar 27 '18 at 11:02
  • $\begingroup$ Yes i read that..but i am sorry, i didn't quite get it why its more energetically favorable .Correct me if i am wrong but if the bridge is reformed again like you said that means the bridge renders some stability to the molecule and that would mean the bridge bond is more stronger than the terminal ones..but then again u said the bridge bonds are weaker ......I am sorry but what am i missing? $\endgroup$
    – M. S. L
    Mar 27 '18 at 12:20
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    $\begingroup$ Your last comment is not 100% rigorous either. The B–C bond does not move in isolation: moving a bridging methyl group to a terminal position means that you have to move a terminal hydrogen into the bridge. If the rearrangement is thermodynamically favoured, it only implies that the bridging B–C–B bond plus the terminal B–H bond, combined, are weaker than a terminal B–C bond plus a bridging B–H–B bond. $\endgroup$
    – orthocresol
    Mar 27 '18 at 20:57

Diborane is supposed to be a dimer of BH3, which is an electron-deficient species so it needs to look for a way to gain some respite from the drought of electron density. So it adjusts for this 3c-2e bonded system where it shares the B-H σ-bond electron density and shares it to the Boron atom of another BH3 unit, forming a dimer.

enter image description here

The 'bridging' bond has a bond order of 0.5 as @MollyCooL stated, which means it is certainly a weaker bond than the terminal ones, which have 'unity' bond order. However, in the case of alkylation reaction, the reacting unit being the actual species BH3, the dimer has to break its bridges, let BH3 react, and then when the product is formed, look for a way again to stabilize itself by forming that bridge bond again.

When we look at the final product, it seems like the bridge hasn't broken at all. But it has only reformed.

Here, the bond length does run in an inverse relation with the bond strength as per the image data (Wikipedia). The bridge is indeed weaker than the terminal bond. However it may not be true in all cases. (Thanks Martin for correcting)

P.S. I am a new contributor and not very experienced with the subject so please correct me if I am wrong.

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    $\begingroup$ Welcome to Chemistry.se! We use MathJax for mathematical and chemical formulae, if you want to know more, please have a look here and here. While it often appears that a longer bond is weaker, this is strictly speaking not true, and also only an observation which cannot be proved. $\endgroup$ Feb 1 '20 at 16:05
  • $\begingroup$ Thanks Martin. Noted, will edit the answer. BTW do you have any theoretical examples to back that up? $\endgroup$ Feb 1 '20 at 16:14

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