What will be the major product when m-cresol is reacted with propargyl bromide `(HC-=C-CH_(2)Br)` in Presence of `K_(2)CO_(3)` in acetone?

Options

A.
B.
C.
D.

Explanation

To determine the major product when m-cresol reacts with propargyl bromide in the presence of K2CO3 in acetone, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- m-Cresol (also known as 3-methylphenol) has the structure:
\[ \text{C}_6\text{H}_4(\text{OH})(\text{CH}_3) \]
- Propargyl bromide has the structure:
\[ \text{HC} \equiv \text{C} - \text{CH}_2\text{Br} \]

2. **Understand the Reaction Conditions**:
- K2CO3 acts as a base in this reaction.
- Acetone is used as a solvent.

3. **Mechanism of the Reaction**:
- The reaction follows the Williamson ether synthesis mechanism, which involves the formation of an alkoxide ion from m-cresol.
- K2CO3 will deprotonate the hydroxyl group (-OH) of m-cresol to form the alkoxide ion:
\[ \text{C}_6\text{H}_4(\text{O}^-) (\text{CH}_3) + \text{K}^+ \]

4. **Nucleophilic Attack**:
- The alkoxide ion acts as a nucleophile and attacks the electrophilic carbon in propargyl bromide. The carbon attached to the bromine is a good electrophile due to the presence of the leaving group (Br).
- The reaction occurs via an SN2 mechanism, where the nucleophile attacks from the backside, leading to the displacement of the bromine atom.

5. **Formation of the Ether**:
- The product formed from this reaction will be an ether. The structure of the ether can be represented as:
\[ \text{C}_6\text{H}_4(\text{O})(\text{CH}_3) - \text{CH}_2 - \text{C} \equiv \text{C} - \text{H} \]
- Therefore, the major product is:
\[ \text{C}_6\text{H}_4(\text{O})(\text{CH}_3) - \text{CH}_2 - \text{C} \equiv \text{C} \]

### Final Product:
The major product of the reaction is:
\[ \text{3-Methylphenyl propargyl ether} \]