When 2-hydroxybenzoic acid (salicylic acid) is treated with bromine water, the product formed is:

Options

A.
B.
C.
D.

Answer

To solve the question about the product formed when 2-hydroxybenzoic acid (salicylic acid) is treated with bromine water, we can follow these steps:

### Step 1: Understand the Structure of 2-Hydroxybenzoic Acid
2-hydroxybenzoic acid, commonly known as salicylic acid, has the following structure:
- A benzene ring (C6H5)
- A hydroxyl group (-OH) at the 2-position
- A carboxylic acid group (-COOH) at the 1-position

The structure can be represented as follows:
```
OH
|
C6H4-COOH
```

### Step 2: Identify the Reaction with Bromine Water
When salicylic acid is treated with bromine water, bromination occurs. The presence of the hydroxyl group (-OH) on the benzene ring activates the ring towards electrophilic substitution reactions.

### Step 3: Determine the Positions of Bromination
The hydroxyl group is an ortho/para director, meaning that bromine will preferentially add to the ortho (2-position) and para (4-position) positions relative to the hydroxyl group. In this case, the reaction will lead to bromination at the following positions:
- 2-position (ortho to -OH)
- 4-position (para to -OH)
- 6-position (ortho to -COOH)

### Step 4: Write the Structure of the Product
The product formed from the bromination of salicylic acid will be 2,4,6-tribromophenol. The structure can be represented as follows:
```
Br
|
Br-C6H2(OH)-COOH
|
Br
```
Where the bromine atoms are located at the 2, 4, and 6 positions of the benzene ring.

### Step 5: Conclusion
The final product of the reaction between 2-hydroxybenzoic acid and bromine water is 2,4,6-tribromophenol.

### Final Answer
The product formed is **2,4,6-tribromophenol**.

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