The reaction of 1 mol each of p-hydroxyacetophenone and methyl magnesium iodide will give:

Options

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B.
C.
D.

Answer

To solve the question regarding the reaction of 1 mole each of p-hydroxyacetophenone and methyl magnesium iodide, we will follow these steps:

### Step 1: Identify the Reactants
The reactants are p-hydroxyacetophenone and methyl magnesium iodide (CH3MgI).
- p-Hydroxyacetophenone has the structure:
\[ \text{C}_6\text{H}_4(\text{OH})\text{C(O)CH}_3 \]
- Methyl magnesium iodide is a Grignard reagent, which can be represented as CH3MgI.

### Step 2: Understand the Functional Groups
p-Hydroxyacetophenone contains a hydroxyl group (-OH) and a carbonyl group (C=O). The presence of the hydroxyl group makes it a suitable substrate for a reaction with a Grignard reagent.

### Step 3: Reaction Mechanism
When p-hydroxyacetophenone reacts with methyl magnesium iodide, the nucleophilic methyl group (CH3-) from the Grignard reagent attacks the carbonyl carbon of the ketone group in p-hydroxyacetophenone. This leads to the formation of a tetrahedral intermediate.

### Step 4: Proton Transfer
The tetrahedral intermediate will then undergo proton transfer. The hydroxyl group (-OH) in p-hydroxyacetophenone has an acidic hydrogen, which can be replaced by the methyl group from the Grignard reagent. This results in the formation of methane (CH4) as a byproduct.

### Step 5: Final Product Formation
The final product of the reaction will be an alcohol, specifically:
\[ \text{C}_6\text{H}_4(\text{OH})\text{C(CH}_3)_2\text{O} \]
This product is a tertiary alcohol where the original -OH group remains, and the carbonyl carbon is now bonded to two methyl groups.

### Conclusion
Thus, the reaction of 1 mole each of p-hydroxyacetophenone and methyl magnesium iodide will yield a tertiary alcohol and methane as a byproduct.