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This is the mechanism of trialkylborane. In this mechanism I can't understand why should the -OH group leave and the -R group enter in its place to form borate ester. Can someone please explain me that step?

  • $\begingroup$ The carbon migrates because peroxide bonds are generally weak and dissociate quite easily. $\endgroup$
    – Gokul
    Apr 28, 2015 at 16:57
  • $\begingroup$ Also, the mechanism is drawn poorly. In the species labeled "Unstable intermediate," the left arrow should not point towards the boron or the bond; the arrow head should terminate at the left oxygen. $\endgroup$
    – Zhe
    Oct 19, 2016 at 17:47
  • $\begingroup$ @Zhe Yes. In the upper mechanism, curved arrows are drawn in a wrong way. The arrowhead should be placed pointing towards boron in two of the curved arrows. $\endgroup$ Mar 11, 2018 at 12:38

1 Answer 1


One aspect is the release of the stable anion $\ce{OH-}$.

Another aspect is that $\ce{B-O}$ bonds are much stronger than $\ce{B-C}$ bonds because the empty p-orbital of boron overlaps with one p-orbital with a lone electron pair of oxygen.

$\ce{B-C}$ $D_{298}^ \circ = 448 \pm 29\ \mathrm{kJ\ mol^{-1}}$

$\ce{B-O}$ $D_{298}^ \circ = 809\ \mathrm{kJ\ mol^{-1}}$

values taken from “Bond Dissociation Energies”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.


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