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I have a selection of boronic acids that I want to reduce to borane forms. Some of the boronic acids in question are phenylboronic acid (C6H5B(OH)2), tolylboronic acid (C7H7B(OH)2), and p-phenylenediborinic acid (C6H4(B(OH)2)2). I want to convert them into phenylborane (C6H5BH2), tolylborane (C7H7BH2), and p-phenylenediborane (C6H4(BH2)2).

I think I can do this by dissolving the boronic acid in ether and adding Lithium Aluminum Hydride (LAH). Giving a reaction that looks something like:

Reduction of Phenylboronic acid

However, I am afraid that, given how aggressive a reducing agent LAH is I might get something like:

enter image description here

So I guess I have two questions:

  • Will LAH attack the boron-carbon bond shown in the second reaction diagram?
  • Can the reduction of the boronic acid be accomplished via a less aggressive reducing agent, perhaps Sodium BoroHydride?

Just a note: I am well aware that the boranes produced will be oxygen sensitive, either converting back to boric acid on exposure to oxygen, or catching fire.

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  • $\begingroup$ There are quite a lot of reducing agents out there. $\ce{DIBAL-H}$ is one of them. $\ce{Sn + HCl}$ is yet another. So is $\ce{NaBH4}$. So many out there. $\endgroup$
    – Pritt says Reinstate Monica
    May 15, 2017 at 15:11
  • $\begingroup$ I know there are a lot out there, but are they strong enough to reduce boronic acids? As I said in one of my questions: Can the reduction be accomplished with something less aggressive than LAH, like NaBH4. $\endgroup$
    – James Matta
    May 15, 2017 at 15:22
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    $\begingroup$ @PrittBalagopal Just because reducing agents exist doesn't mean they will be able to affect a particular transformation $\endgroup$
    – orthocresol
    May 15, 2017 at 19:05

1 Answer 1

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LAH will attack the boron-oxygen bonds without going after the boron-carbon bond, see here (Eq. Z, p. 142).

But it also deprotonates the hydroxyl groups as a strong base like a Grignard reagent. You need to prevent that. Convert your acid to an ester (OH -> OR), remove the water and then apply the LAH.

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    $\begingroup$ Upvote for the reference and the useful note, I hadn't thought about what would happen to the hydroxyl group once it was removed. Out of curiosity, if I do not convert to the ester first, will it just require more LAH and leave me with Aluminum oxide and Lithium oxide? Or will something more insidious happen? $\endgroup$
    – James Matta
    May 15, 2017 at 17:37
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    $\begingroup$ What's insidious is if you deprotonate the hydroxyl groups the boron loses its electrophilicity and you don't get the hydride transfer. Compare with a Grignard reagent which converts a carboxylate ester to an alcohol but the parent acid just gives a salt. $\endgroup$
    – Oscar Lanzi
    May 15, 2017 at 18:13
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    $\begingroup$ Ah, I see, thanks. As a physicist with a little more chemistry than most physicists under my belt I have gotten mixed up in liquid scintillator development but sometimes that little bit of extra knowledge isn't enough because of how hazy it is. $\endgroup$
    – James Matta
    May 15, 2017 at 18:55

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