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According to this question, the stoichiometry of the reduction of a ketone is such that sodium borohydride can reduce up to 4 ketones by forming a boron-alkoxide complex.

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The next step would involve the hydrolysis of the complex, and I've been searching through textbooks to no avail. What would the mechanism for this hydrolysis look like?

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I don’t know whether the mechanism of hydrolysis of the boronic acid esters has been studied. However, the dissociative mechanism immediately suggests itself if only by comparison to similar structures.

Hydrolysis mechanism

Since the workup is typically written $\ce{H+/H2O}$, you may consider some buffered source of external protons. These can protonate any of the alcoholates which can then dissociate. (The $\ce{B=O}$ double bond I drew in should be considered no more than additional stabilisation from oxygen lone pairs; boron is electron-deficient by definition.) Another water molecule can associate: the first bond is cleaved. Repeat three times and end up with four molecules of alcohol and tetrahydroxidoborate $\ce{[B(OH)4]-}$.

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  • $\begingroup$ 3rd species, boron should be negatively charged. Are those arrows dative bonds? $\endgroup$ – orthocresol Nov 16 '16 at 21:57
  • $\begingroup$ Yes and yes. Which I chose mainly to avoid adding even more charges … $\endgroup$ – Jan Nov 16 '16 at 21:59
  • $\begingroup$ @orthocresol The wrong charge also explains why the chemical warning wouldn’t go away ;) $\endgroup$ – Jan Nov 16 '16 at 22:01
  • $\begingroup$ You could just drop the double bond in the second intermediate... $\endgroup$ – Zhe Nov 16 '16 at 22:09
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For an acidic work up:

  1. Protonate an oxygen.
  2. Fragment the tetraalkyl borate into the trialkyl borate and your product alcohol.
  3. Add water to the trialkyl borate and remove the proton.
  4. Repeat.
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