`PhOHunderset(Me_2SO_4)overset( 2 mol NaOH ) to P ` In given reaction product P is :

Options

A. `Ph-O-SO_2OMe`
B. `PhOMe`
C. `PhOSO_2OPh`
D. `PhMe`

Explanation

To solve the question regarding the reaction of phenol (PhOH) with dimethyl sulfate (Me2SO4) in the presence of sodium hydroxide (NaOH), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are phenol (C6H5OH) and dimethyl sulfate (Me2SO4).
- Sodium hydroxide (NaOH) is used as a base in this reaction.

2. **Understand the Reaction Type**:
- This reaction is an example of the Williamson ether synthesis, which involves the formation of ethers from alcohols and alkyl halides or alkyl sulfates in the presence of a base.

3. **Formation of Phenoxide Ion**:
- When phenol reacts with NaOH, it gets deprotonated to form the phenoxide ion (C6H5O⁻), which is a strong nucleophile.

4. **Nucleophilic Attack**:
- The phenoxide ion will attack the electrophilic carbon in dimethyl sulfate. This results in the substitution of one of the methyl groups in dimethyl sulfate with the phenoxide ion.

5. **Formation of Product**:
- The reaction leads to the formation of anisole (C6H5OCH3) and the by-products of the reaction, which include sodium sulfate (Na2SO4) and water (H2O).
- The balanced reaction can be written as:
\[ 2 \text{C}_6\text{H}_5\text{OH} + \text{Me}_2\text{SO}_4 \xrightarrow{2 \text{NaOH}} 2 \text{C}_6\text{H}_5\text{OCH}_3 + \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \]

6. **Identify Product P**:
- The product P formed in this reaction is anisole, which can also be represented as phenyl methyl ether (C6H5OCH3).

### Final Answer:
The product P is **anisole (C6H5OCH3)**.