Hydration via Hydroboration-Oxidation

Key Questions

What is hydroboration oxidation in alkenes?

Answer:

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

Explanation:

It involves the addition of $"BH"_3$ to an alkene, followed by oxidation with alkaline hydrogen peroxide to form an alcohol.

The reaction is a Markovnikov addition of $"BH"_3$ to the alkene

On oxidation of the boron intermediate, the $"OH"$ group ends up on the less substituted carbon.

This is opposite to the position of the $"OH"$ 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.

Ernest Z. · 2 · Jul 20 2015
Why is the borane–THF complex used for hydroboration?

Answer:

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

Explanation:

The active ingredient is borane, $"BH"_3$ , but borane is a highly toxic gas.

Borane exists naturally as the dimer $"B"_2"H"_6$ (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!

Ernest Z. · 6 · Dec 1 2014
What is hydroboration oxidation in alkynes?
It's similar to for alkenes, but besides creating $[B(OH)_4]^(-)$ , $H_2O$ , and $OOH^(-)$ (hydrogen peroxide's conjugate base), instead of getting the three $mols$ 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 $M^(+)OH^(-)$ available within the reaction vessel.

An example of this is:

First, the enol's $R-OH$ donates its proton to the base ( $OH^(-)$ from $M^(+)OH^(-)$ ) to form an enolate. ( $H_2O$ 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 $H_2O$ that just formed, grabbing a proton off and reforming the base ( $OH^(-)$ from $M^(+)OH^(-)$ ).

You then form a ketone or aldehyde, depending on the location of the triple bond. Also, remember that hydroboration is anti-Markovnikov.
Truong-Son N. · 1 · Jul 18 2015