Phenol `+ CH_(2) O overset("Alkali") to A ` , Here, [A] is :

Options

A. O & p- Hydroxy benzyl alcohol
B. m - Hydroxy Benzaldehyde
C. Aniline
D. None of these

Answer

To solve the question, we need to determine the product (A) formed when phenol reacts with formaldehyde (CH₂O) in the presence of an alkali.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants are phenol (C₆H₅OH) and formaldehyde (CH₂O).

2. **Understand the Role of Alkali**:
In the presence of an alkali, phenol will deprotonate to form the phenoxide ion (C₆H₅O⁻). This is because phenol is slightly acidic and will react with the base to form the corresponding phenoxide ion.

\[ \text{C₆H₅OH} + \text{OH}⁻ \rightarrow \text{C₆H₅O}⁻ + \text{H₂O} \]

3. **Resonance Stabilization of Phenoxide Ion**:
The phenoxide ion is resonance-stabilized, which means that the negative charge can be delocalized over the aromatic ring. This delocalization increases the nucleophilicity of the ortho and para positions.

4. **Nucleophilic Attack on Formaldehyde**:
The nucleophilic oxygen in the phenoxide ion can attack the electrophilic carbon in formaldehyde. The attack will primarily occur at the ortho and para positions of the aromatic ring.

\[ \text{C₆H₅O}⁻ + \text{CH₂O} \rightarrow \text{C₆H₅(OH)CH₂O}⁻ \]

5. **Proton Transfer and Final Product Formation**:
After the nucleophilic attack, a proton (H⁺) will be extracted from water, leading to the formation of a hydroxymethyl group (-CH₂OH) at the ortho and para positions of the phenol. This results in the formation of ortho and para hydroxybenzyl alcohols.

The final product A can be represented as:
\[ \text{C₆H₄(OH)(CH₂OH)} \quad \text{(Ortho and Para Hydroxybenzyl Alcohol)} \]

### Conclusion:
The product A formed from the reaction of phenol and formaldehyde in the presence of alkali is ortho and para hydroxybenzyl alcohol.