2-Mehtylbut-1-ene reacts wth mercuric acetate in presence of water to form a product, which on reduction with `NaBH_(4)` yield

Options

A. 2-Methylbutane-2-ol
B. 2-Methylbutane-1-ol
C. 3-Methylbutane-2-ol
D. none of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the Reactants
The reactant given is 2-methylbut-1-ene. Its structure can be represented as follows:

```
CH3
|
CH2=CH-CH2-CH3
```

### Step 2: Understand the Reaction Mechanism
The reaction involves mercuric acetate (Hg(OAc)₂) and water (H₂O). This is known as the oxymercuration-demercuration reaction. In this reaction, the alkene reacts with mercuric acetate in the presence of water to form an alcohol.

### Step 3: Apply Markovnikov's Rule
According to Markovnikov's rule, during the addition of H₂O to the alkene, the hydroxyl group (OH) will attach to the more substituted carbon atom. In this case, the double bond will break, and the OH group will attach to the second carbon (the more substituted carbon) of the 2-methylbut-1-ene.

### Step 4: Write the Product of Oxymercuration
After the reaction, the product will be:
```
CH3
|
CH2-OH
|
CH2-CH3
```
This product can be represented as 2-methylbutan-2-ol.

### Step 5: Reduction with Sodium Borohydride (NaBH₄)
The product formed from the oxymercuration step is then reduced with sodium borohydride (NaBH₄). However, since the product is already an alcohol (2-methylbutan-2-ol), the reduction step does not change the structure further.

### Final Product
The final product after the reduction will still be 2-methylbutan-2-ol.

### Conclusion
The correct answer is **2-methylbutan-2-ol**.