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