Question

The product of following reaction is

A.
 

B.

C.

D.

Answer 1

Hint: The process in which alkene to alcohol is done in the presence of $Hg{{(OAc)}_{2}}$ sodium borohydride ($NaB{{H}_{4}}$) is called Oxymercuration-demercuration reaction. The major product is alcohol which is a result of the Markovnikov addition reaction.

Complete Step by Step Answer:
The oxymercuration-demercuration of alkenes provides us with an alternative pathway for the synthesis of Merkovnikov’s alcohol from any alkene. In this process, any alkene compound is treated with mercury(II) acetate $Hg{{(OAc)}_{2}}$and $NaB{{H}_{4}}$ the final product in which hydroxyl group ($-OH$) bonds to the more substituted carbon atom of the alkene compound.
The mechanism of this reaction contains more than one step. In the first step, an electrophile $HgOA{{c}^{+}}$ is generated and gives a cyclic mercurinium ion. A weakly nucleophilic water attacks on the most substituted carbon to open the three-membered cyclic mercurinium ion bridge followed by proton transfer to a solvent water molecule. The reason behind attacking more substituted carbon centres than the primary carbon centre is that the partial positive charge is better accommodated on tertiary carbon than on primary carbon.
Mechanism:

Finally, the organomercury intermediate is further reduced by sodium borohydride $(NaB{{H}_{4}})$ via hydride shift. Hence the major product is $3,3-dimethylbutan-2-ol$.

Therefore , option (A) is correct.

Note: Another process for the reduction of alkene to alcohol is known as hydroboration-oxidation. This is a reduction process of alkene followed by anti-Markovnikov regioselectivity. Here hydroxyl group is attached to the less substituted carbon centre or a carbon centre with less number of hydrogens.