what product is formed hydroboration reduction?

Hydroboration of alkenes forms alcohols.

The general form of the hydroboration of alkenes mechanism is as follows:

First step is the attack of the alkene on BH₃, which then forms a four membered ring intermediate of partial bonds. It is because of this intermediate that hydroboration forms the anti-Markovnikov product. The boron atom is highly electrophilic because of its empty p orbital and forms a slight bonding interaction with the pi bond. Since some electron density from the double bond is going towards bonding with the boron, the carbon opposite the boron is slightly electron deficient, left with a slightly positive charge. Positive charges are best stabilized by more highly substituted carbons, so the carbon opposite the boron tends to be the most highly substituted. Once the transition state breaks down, BH2 is attached to the least substituted carbon.

Peroxide then removes the borane and replaces it with the alcohol to form the anti-markovnikov product.

Thanks & Regard

Aarti Gupta

askiitians Faculty