Why is an alkoxide released instead of hydroxide in the attack of trialkyloxyborane by hydroxide?

Question

In the mechanism of a hydroboration-oxidation, the oxidation part includes a nucleophilic attack of a hydroxide on a trialkyloxyborane ( $\ce{B(OR)_3}$ where $\ce{R}$ is an alkyl group).

It creates a tetrahedral intermediate consisting of a negative center of boron with four groups attached to it (the hydroxide nucleophile and the three alkoxides). This step is followed by the loss of an alkoxide as the leaving group. However, to my understanding, the alkoxide is a stronger base than the hydroxide. Thus, why doesn't the nucleophilic attack fail?

Edit: Is it because of steric comparisons between the size of the alkoxide versus the hydroxide (since boron is a relatively small atom)?

Answer 1

Both hydroxide and alcoholate can be displaced from the tetracoordinated intermediate. The difference is that displacing the hydroxide is just the back reaction of the inital attack while displacing the alcoholate will move the reaction forward.

You cannot argue that $\ce{X}$ is a stronger base than $\ce{Y}$ and therefore $\ce{X}$ will always get displaced. At best, you can argue that $\ce{X}$ is ten times stronger so the probability of $\ce{X}$ begin displaced is higher than that of $\ce{Y}$. For all intents and purposes of the argument, the basicities of (primary) alcoholates and hydroxide are too similar to make a real difference.