Alkene Reactions: Hydroboration using BH3, H2O2, and NaOH

Borane (BH3) in the presence of hydrogen peroxide (H2O2) and sodium hydroxide (NaOH) react with alkenes to form alcohols:

This hydroboration reaction adds an alcohol functional group in an anti-Markovnikov addition, adding the OH to the carbon atom that has the most hydrogen atoms (or the least substituted carbon). This reaction adds stereochemistry to the molecule as the alcohol group and paired hydrogen group are added in syn addition:

Syn Addition

No stereocenters

The reaction mechanism is depicted below:

In the first step, electrons from the alkene bond react with borane in a syn addition of the B and H atom. Borane is added to the least substituted carbon in this step.

In the second step, sodium hydroxide (NaOH) activates hydrogen peroxide (H2O2) by deprotonating H2O2, resulting in negatively charged OOH- ionically paired with Na+.

In the third step, the Na+ OOH- moiety attacks the boron, binding to it causing it to have 4 bonds and a negative charge.

In the fourth step, the bond between the negatively charged Boron atom and the least substituted carbon is migrated to the oxygen atom, as it is better able to carry the negative charge. The bond between the two oxygens from the peroxide moiety is broken and the second oxygen breaks off, regenerating NaOH.

In the fifth step, the regenerated NaOH attacks the boron, causing the boron-oxygen bond to break and give the electrons to the O atom bound to the alkane. This deprotonated alcohol is known as an alkoxide.

In the sixth step, a water molecule protonates the alkoxide, resulting in a final product of an alcohol formed from an alkene.

The reagents used for this reaction are BH3, THF (or H2O), NaOH, and H2O2.