organic chemistry - What is the mechanism of trialkylborane oxidation with hydroperoxide?

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?

Answer

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.