The action of bromine water (excess) on salicylic acid results in the formation of :

Options

A.
B.
C.
D.

Explanation

To solve the question regarding the action of bromine water (excess) on salicylic acid, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of Salicylic Acid**:
- Salicylic acid has the following structure:
- A hydroxyl group (-OH) at the ortho position relative to a carboxylic acid group (-COOH).
- The molecular formula is C7H6O3.

2. **Understand the Reaction with Bromine Water**:
- Bromine water (Br2 in H2O) acts as an electrophilic brominating agent. In the presence of excess bromine, the aromatic ring of salicylic acid undergoes bromination.

3. **Mechanism of Bromination**:
- The hydroxyl group (-OH) is a strong activating group and directs bromination to the ortho and para positions of the benzene ring.
- The reaction will lead to the substitution of hydrogen atoms on the benzene ring with bromine atoms.

4. **Formation of the Brominated Product**:
- In the presence of excess bromine, the -OH group will direct bromination to the ortho and para positions.
- The final product will have three bromine atoms substituted on the benzene ring: one at the para position to the -OH group and two at the ortho positions.

5. **Final Structure of the Product**:
- The final structure will be:
- A benzene ring with three bromine atoms (Br) at the ortho and para positions relative to the -OH group.
- The carboxylic acid group (-COOH) remains unchanged.

6. **Select the Correct Option**:
- Among the given options, the correct product will have:
- One -OH group,
- Three -Br groups (two at ortho positions and one at para position),
- The -COOH group.