Reimer Tiemann reaction is useful for the preparation:

Options

A. Benzaldehyde
B. Salicylaldehyde
C. Toluene
D. Acetophenone

Explanation

### Step-by-Step Solution:

1. **Understanding the Reimer-Tiemann Reaction**:
The Reimer-Tiemann reaction involves the reaction of phenol with chloroform (CHCl3) in the presence of a strong base, typically potassium hydroxide (KOH).

2. **Starting Materials**:
The starting material for the Reimer-Tiemann reaction is phenol (C6H5OH).

3. **Reaction Conditions**:
The reaction is conducted in the presence of KOH and chloroform. KOH acts as a base to deprotonate phenol.

4. **Formation of Carbene**:
When KOH deprotonates phenol, it generates a phenoxide ion. The chloroform reacts with KOH to produce a dichlorocarbene (CCl2), which is a reactive intermediate.

5. **Nucleophilic Attack**:
The phenoxide ion, which is electron-rich, attacks the electron-deficient carbene. This results in the substitution of the carbene at the ortho position relative to the hydroxyl group on the phenol.

6. **Formation of Salicylaldehyde**:
After the nucleophilic attack, further rearrangements and eliminations occur, leading to the formation of salicylaldehyde (C7H6O), which has an aldehyde (CHO) group at the ortho position relative to the hydroxyl (OH) group.

7. **Conclusion**:
Therefore, the Reimer-Tiemann reaction is useful for the preparation of salicylaldehyde.

### Final Answer:
The correct answer is **salicylaldehyde** (option B).

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