form of the hydroboration of alkenes mechanism is as follows:
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 (ie. it wants electrons), 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.
then removes the borane and replaces it with the alcohol to
form the anti-markovnikov product.
of the hydroboration mechanism: