Phenol is heated with `CHCl_(3)` and alcoholic KOH when salicylaldehyde is produced. This reaction is known as :

Options

A. Rosenmund's reaction
B. Reimer-Tiemann reaction
C. Friedel-Craft's reaction
D. Sommelet reaction

Answer

To solve the question regarding the reaction of phenol with CHCl3 and alcoholic KOH to produce salicylaldehyde, we can break it down step by step:

### Step-by-Step Solution:

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

2. **Understand the Reaction Conditions**:
- The reaction takes place under heating conditions, which is essential for facilitating the reaction between phenol and chloroform in the presence of alcoholic KOH.

3. **Recognize the Type of Reaction**:
- This reaction is a named reaction known as the "Reimer-Tiemann reaction." It involves the ortho-formylation of phenols.

4. **Mechanism Overview**:
- In the presence of KOH, chloroform reacts to form a dichlorocarbene intermediate, which then attacks the ortho position of the phenol ring. This leads to the formation of salicylaldehyde (C6H4(OH)(CHO)).

5. **Identify the Product**:
- The product formed from this reaction is salicylaldehyde, which has both a hydroxyl (OH) group and an aldehyde (CHO) group on the benzene ring.

6. **Conclusion**:
- Therefore, the reaction of phenol with CHCl3 and alcoholic KOH to produce salicylaldehyde is known as the "Reimer-Tiemann reaction."

### Final Answer:
The reaction is known as the **Reimer-Tiemann reaction**.