The product of following reaction is
`CH_(3) underset(CH_3)underset(|)overset(CH_3)overset(|)C-CH = CH_(2) overset( (i) Hg(OAc)_2 , H_(2)O)underset((ii) NaBH_4) rarr`

Options

A. `CH_(3)- overset(CH_3)overset(|)underset(CH_3)underset(|)C-underset(OH)underset(|)CH-CH_(3)`
B. `CH_(3)-overset(CH_3)overset(|)underset(CH_3)underset(|)C-CH_(2)-CH_(2)OH`
C. `CH_(3)-overset(OH)overset(|)underset(CH_3)underset(|)C-overset(CH_3)overset(|)CH-CH_(3)`
D. `HOCH_(2) - overset(CH_3)overset(|)underset(CH_3)underset(|)C-CH_(2) - CH_(2)`

Explanation

To solve the reaction given in the question, we will follow these steps:

### Step 1: Identify the Reactants
The reactant in the given reaction is 2-methyl-2-pentene (CH₃)₂C=CH-CH₂-CH₃. We also have the reagents: Hg(OAc)₂ and H₂O followed by NaBH₄.

### Step 2: Understand the Reaction Mechanism
The reaction involves oxymercuration-demercuration. The first step (oxymercuration) involves the addition of mercuric acetate (Hg(OAc)₂) and water (H₂O) across the double bond. This results in the formation of a mercurinium ion intermediate.

### Step 3: Formation of the Mercurinium Ion
When the double bond of 2-methyl-2-pentene reacts with Hg(OAc)₂, a cyclic mercurinium ion is formed. The double bond attacks the mercury, leading to the formation of a three-membered ring structure with a positive charge on the mercury.

### Step 4: Nucleophilic Attack
Water (H₂O), which acts as a nucleophile, attacks the more substituted carbon of the mercurinium ion. This results in the formation of an alcohol intermediate. The product at this stage is 2-methyl-2-pentanol.

### Step 5: Demercuration
The second step involves the reduction of the mercurial intermediate using sodium borohydride (NaBH₄). This step replaces the mercury group with a hydrogen atom, leading to the final product.

### Step 6: Final Product
After the reduction, the final product of the reaction is 2-methyl-2-pentanol (CH₃)₂C(OH)CH₂CH₃.

### Final Answer
The product of the reaction is 2-methyl-2-pentanol.

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