give mechanism of hydroboration of 3-methyl-2-butene followed by oxidative work up with H2O2/NaOH?

Answer 1

I believe the name should be 2-methyl-2-butene.

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

Answer 2

the easiest way to see it is for it to be drawn for you. pick up Ege's orgo text and it explains it step by step. if you don't like your text get another one. they all cover the same stuff - they just do it differently so find one that works for you