Choose the correct option for the following reactions.

The starting alkene is:

$$\mathrm{3,3\text{-}dimethylbut\text{-}1\text{-}ene}$$

For formation of product $$\mathrm{A}$$, the reagents used are:

$$\mathrm{Hg(OAc)_2 \cdot H_2O \ followed\ by\ NaBH_4}$$

This reaction is:

$$\mathrm{Oxymercuration\text{-}Demercuration}$$

The reaction proceeds through a mercurinium ion intermediate and does not involve carbocation rearrangement.

Water attacks the more substituted carbon of the double bond.

Hence, the hydroxyl group attaches according to:

$$\mathrm{Markovnikov\ addition}$$

Therefore, product $$\mathrm{A}$$ is a Markovnikov product.

For formation of product $$\mathrm{B}$$, the reagents used are:

$$\mathrm{(BH_3)_2 \ followed\ by\ H_2O_2/OH^-}$$

This reaction is:

$$\mathrm{Hydroboration\text{-}Oxidation}$$

Boron attaches to the less substituted carbon due to steric and electronic factors.

During oxidation, boron is replaced by:

$$\mathrm{-OH}$$

Thus, the hydroxyl group finally appears on the less substituted carbon.

Hence, hydration occurs through:

$$\mathrm{Anti\text{-}Markovnikov\ addition}$$

Therefore, product $$\mathrm{B}$$ is an anti-Markovnikov product.

Thus:

$$\mathrm{A = Markovnikov\ Product}$$

$$\mathrm{B = Anti\text{-}Markovnikov\ Product}$$