In presence of NaOH, phenol react with `CHCl_(3)` to form o-hydroxy benzaldehyde.This reaction is called

Options

A. Riemer-Tiemann reaction
B. Sandmeyer reaction
C. Hoffmann degradation reaction
D. Gattermann aldehyde synthesis

Answer

To solve the question regarding the reaction of phenol with CHCl3 in the presence of NaOH to form o-hydroxy benzaldehyde, we can break down the process step by step.

### Step-by-Step Solution:

1. **Identify the Reactants and Conditions**:
- The reactants are phenol (C6H5OH) and chloroform (CHCl3).
- The reaction occurs in the presence of sodium hydroxide (NaOH), which acts as a base.

2. **Formation of Dichlorocarbene**:
- When phenol reacts with CHCl3 in the presence of NaOH, the NaOH deprotonates CHCl3, leading to the formation of a dichlorocarbene intermediate (CCl2).
- The reaction can be represented as:
\[ CHCl_3 + NaOH \rightarrow CHCl_2^- + H_2O \]
- The CHCl2^- ion undergoes internal repulsion, causing one chlorine atom to leave, forming the dichlorocarbene (CCl2).

3. **Nucleophilic Attack by Phenoxide Ion**:
- In the presence of NaOH, phenol is converted to its phenoxide ion (C6H5O^-).
- The phenoxide ion acts as a nucleophile and attacks the electrophilic carbon of the dichlorocarbene (CCl2).
- This step can be represented as:
\[ C_6H_5O^- + CCl_2 \rightarrow C_6H_5O-CCl_2 \]

4. **Formation of Ortho-Hydroxy Benzaldehyde**:
- The product formed from the nucleophilic attack is then hydrolyzed, leading to the formation of o-hydroxy benzaldehyde (C6H4(OH)CHO).
- The final step involves the elimination of a water molecule to stabilize the product, resulting in:
\[ C_6H_5O-CCl_2 \rightarrow C_6H_4(OH)CHO + HCl + H_2O \]

5. **Name of the Reaction**:
- This entire process is known as the **Reimann-Thiemann reaction**.

### Final Answer:
The reaction of phenol with CHCl3 in the presence of NaOH to form o-hydroxy benzaldehyde is called the **Reimann-Thiemann reaction**.

---