# Hydration via Hydroboration-Oxidation

## Key Questions

Hydroboration-oxidation is a method of making alcohols from alkenes.

#### Explanation:

It involves the addition of ${\text{BH}}_{3}$ to an alkene, followed by oxidation with alkaline hydrogen peroxide to form an alcohol.

The reaction is a Markovnikov addition of ${\text{BH}}_{3}$ to the alkene

On oxidation of the boron intermediate, the $\text{OH}$ group ends up on the less substituted carbon.

This is opposite to the position of the $\text{OH}$ group in the acid-catalyzed Markovnikov addition of water to an alkene, so the reaction is often called the anti-Markovnikov addition of water to an alkene.

The borane-THF complex (BTHF) is used for hydroboration for reasons of safety and convenience.

#### Explanation:

The active ingredient is borane, ${\text{BH}}_{3}$, but borane is a highly toxic gas.

Borane exists naturally as the dimer ${\text{B"_2"H}}_{6}$ (diborane), but diborane mixes easily with air and forms explosive mixtures.

Also, it ignites spontaneously in moist air at room temperature.

In a solution in THF, borane exists as a loose Lewis acid-base complex. This allows boron to have an octet and makes the reagent more stable.

The solution is commercially available in a 1 mol/L concentration in volumes from 25 to 800 mL.

It is much more convenient to work with the solution than with a gas. Even so, the solution must be stored at 2 to 8 °C, and it must have a stabilizer added.

Borane forms a more stable and more soluble Lewis acid-base complex with dimethyl sulfide:

"H"_3stackrelcolor(blue)("-")("B")"-"stackrelcolor(blue)(+)("S")("CH"_3)_2

It is available in concentrations of 2, 5, and 10 mol/L and in volumes from 25 mL to 18 L.

That should make it a more convenient reagent than the BTHF complex.

There is only one problem: It has the highly disagreeable smell of rotten cabbage!

• 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.