Acetone on treatment with `CH_(3)-Mg-I` and on further hydrolysis gives

Options

A. Isopropyl alcohol
B. Primary alcohol
C. Acetic acid
D. 2 - methyl 2 - propanol

Answer

To solve the question regarding the reaction of acetone with methyl magnesium iodide (a Grignard reagent) followed by hydrolysis, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are acetone (CH₃COCH₃) and methyl magnesium iodide (CH₃MgI).

### Step 2: Understand the Reaction Mechanism
Grignard reagents act as nucleophiles. In this case, the methyl group (CH₃⁻) from methyl magnesium iodide will attack the electrophilic carbon of the carbonyl group (C=O) in acetone.

### Step 3: Nucleophilic Attack
The nucleophilic methyl group attacks the carbonyl carbon of acetone, leading to the formation of a tetrahedral intermediate. The structure of the intermediate can be represented as follows:
- The carbonyl oxygen now has a negative charge (O⁻), while the carbon that was part of the carbonyl becomes bonded to the methyl group.

### Step 4: Formation of the Addition Product
The addition product formed is:
- (CH₃)₂C(OH)CH₃, which is 2-methyl-2-propanol (also known as tert-butyl alcohol).

### Step 5: Hydrolysis of the Addition Product
Upon hydrolysis, the negatively charged oxygen (O⁻) in the tetrahedral intermediate will abstract a proton (H⁺) from water, resulting in the formation of the alcohol:
- (CH₃)₂C(OH)CH₃.

### Step 6: Final Product
The final product after hydrolysis is 2-methyl-2-propanol.

### Conclusion
Thus, the product of the reaction of acetone with methyl magnesium iodide followed by hydrolysis is 2-methyl-2-propanol.

### Answer
The correct answer is 2-methyl-2-propanol.

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