Find out the major products from the following reactions.

The starting compound is 2-methylbut-1-ene.

$$CH_2=C(CH_3)CH_2CH_3$$

Both reactions convert the alkene into an alcohol by adding $$H$$ and $$OH$$ across the double bond. The difference lies in the position at which the $$OH$$ group is introduced.

For reaction A, the reagents are $$BH_3/THF$$ followed by $$H_2O_2/OH^-$$.

This is the hydroboration-oxidation reaction.

Hydroboration-oxidation follows anti-Markovnikov addition. Therefore, the $$OH$$ group is introduced at the less substituted carbon of the double bond.

The product formed is

$$HOCH_2CH(CH_3)CH_2CH_3$$

which is 2-methylbutan-1-ol.

Thus, product A is a primary alcohol.

For reaction B, the reagents are $$Hg(OAc)_2,\ H_2O$$ followed by $$NaBH_4$$.

This is the oxymercuration-demercuration reaction.

Oxymercuration-demercuration follows Markovnikov addition without carbocation rearrangement. Therefore, the $$OH$$ group is introduced at the more substituted carbon of the double bond.

The product formed is

$$CH_3C(OH)(CH_3)CH_2CH_3$$

which is 2-methylbutan-2-ol.

Thus, product B is a tertiary alcohol.

Therefore,

$$A=\text{2-methylbutan-1-ol}$$

and

$$B=\text{2-methylbutan-2-ol}$$

Hence, the correct answer is

$$\text{Option A}$$