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In this oxidation step, where alkyl shift happens for 3 times to kick out the OH- to obtain the boron-ester. But why does this happen? Won't the boron atom become less stable as the empty p orbital is regenerated? Won't this cause the OH- (strong nucleophile) to attack the boron (electrophile) again?

What is the mechanism of trialkylborane oxidation with hydroperoxide? (I have read this but sorry I don't get it. before this reaction, isn't the boron atom also bonded to an oxygen atom? Why would this reaction make it more stable?)


But why does this happen?

Well, if it was a plain old carbon in the middle, this migration wouldn't happen. It would sit there forever. Boron isn't carbon, though: it has fewer protons, is less electronegative, and doesn't like holding onto electron density all that much. That electron density gets pushed onto the adjacent carbons, which make them rather more nucleophilic, and happy to jump next door into the $\sigma_\ce{O-O}^*$ orbital.

You can see, after all, that there's something about that boron that's not quite perfect: it has a formal negative charge. While this shouldn't be taken as a sign of great instability, it does suggest that boron won't mind losing that formal negative charge too much.

Won't the boron atom become less stable as the empty p orbital is regenerated?

The energetics of this migration aren't just about the boron atom, it's about the system as a whole. You are also breaking a weak O–O bond, forming a stronger C–O bond in replacement, and so on. And indeed, even if this one step is energetically unfavourable, it might not even matter as long as the overall reaction is spontaneous.

Is the overall reaction energetically favourable? I'll leave it to you to think about it, based on the types of bonds that are broken and formed over the course of the entire reaction. You already saw some of the considerations in the linked question. (I don't get what you mean by the boron atom being bonded to oxygen before the reaction; it isn't, as it has three bonds to carbon.)

Won't this cause the $\ce{OH-}$ (strong nucleophile) to attack the boron (electrophile) again?

Well, yes, but not much happens if you go down that route. It's far more interesting if $\ce{OOH-}$ attacks the boron again, because that means the second alkyl group can migrate. That is precisely the mechanism that you have learnt.

  • $\begingroup$ Phenol synthesis using cummene peroxide? $\endgroup$ – user96208 Aug 9 '20 at 13:59
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    $\begingroup$ @AnindyaPrithvi that's under acidic conditions. Not the same thing. By protonating the peroxide you get a much better leaving group, and the driving force for migration doesn't need to be so strong. This migration is done under basic conditions, and you are probably aware that $\ce{OH-}$ is not (usually) a good leaving group. $\endgroup$ – orthocresol Aug 9 '20 at 16:37
  • $\begingroup$ Got it, so if it were given that we have acidic condition in this question as well..then? $\endgroup$ – user96208 Aug 9 '20 at 16:49
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    $\begingroup$ @AnindyaPrithvi If you had acidic conditions you would never get a four-coordinate boron to begin with, so that's pretty much it. You'd have water floating around which is quite happy on its own, so isn't inclined to add to the trivalent boron. Maybe eventually you might hydrolyse the borane; I know this happens with carboxylic acids, but I don't know how stable they are to acids in general. $\endgroup$ – orthocresol Aug 9 '20 at 16:56

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