Phenol reacts with chloroform in the presence of aq. KOH at 340 K followed by hydrolysis of the resulting product giving salicyladehdye
The above reaction is called:

Options

A. Reimer-Tiemann reaction
B. Wurtz Reaction
C. Fries Rearrangement
D. Kolbe's Synthesis

Answer

To solve the question regarding the reaction of phenol with chloroform in the presence of aqueous KOH, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants in this reaction are phenol (C6H5OH), chloroform (CHCl3), and aqueous potassium hydroxide (KOH).

2. **Understand the Reaction Conditions**:
- The reaction occurs at a temperature of 340 K. The presence of aqueous KOH indicates that it acts as a base in this reaction.

3. **Deprotonation of Phenol**:
- Phenol can be deprotonated by KOH to form phenoxide ion (C6H5O^-). This is the first step in the reaction mechanism.

4. **Formation of Intermediate**:
- The phenoxide ion then reacts with chloroform (CHCl3) in the presence of KOH. This leads to the formation of an intermediate compound, which is a substituted benzyl chloride.

5. **Formation of Salicylaldehyde**:
- The intermediate undergoes further reaction with KOH, leading to the removal of a chlorine atom and the introduction of an aldehyde group (CHO) at the ortho position of the benzene ring. This results in the formation of salicylaldehyde (C6H4(OH)(CHO)).

6. **Hydrolysis**:
- Finally, the product undergoes hydrolysis, which involves the addition of water (H2O) and results in the formation of salicylaldehyde.

7. **Name the Reaction**:
- The overall reaction is known as the **Reimer-Tiemann reaction**, which is a well-known reaction in organic chemistry that introduces an aldehyde group into the ortho position of phenols.

### Final Answer:
The reaction described is called the **Reimer-Tiemann reaction**.

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