# What is hydroboration oxidation in alkynes?

Jul 18, 2015

It's similar to for alkenes, but besides creating ${\left[B {\left(O H\right)}_{4}\right]}^{-}$, ${H}_{2} O$, and $O O {H}^{-}$ (hydrogen peroxide's conjugate base), instead of getting the three $m o l s$ of an alcohol, you have an enol. This enol can undergo keto-enol tautomerization.

In this case it is in basic (base-ic) conditions, with ${M}^{+} O {H}^{-}$ available within the reaction vessel.

An example of this is:

• First, the enol's $R - O H$ donates its proton to the base ($O {H}^{-}$ from ${M}^{+} O {H}^{-}$) to form an enolate. (${H}_{2} O$ forms, now)
• Then, the enolate's oxygen moves its electrons down to form a $\pi$ bond (double bond = 1 $\sigma$ + 1 $\pi$) and the $\pi$ bond down at the bottom donates its pi electrons to the resultant ${H}_{2} O$ that just formed, grabbing a proton off and reforming the base ($O {H}^{-}$ from ${M}^{+} O {H}^{-}$).

You then form a ketone or aldehyde, depending on the location of the triple bond. Also, remember that hydroboration is anti-Markovnikov.