Salicylic acid on treatment with bromine water will give

Options

A. 2-bromo-6-hydroxybenzoic acid
B. 2,4,6-tribromophenol
C. 2,6-dibromobenzoic acid
D. 1,3-dibromo-6-hydroxybenzoic acid

Answer

To solve the question regarding the reaction of salicylic acid with bromine water, we will follow these steps:

### Step 1: Understand the Structure of Salicylic Acid
Salicylic acid is an aromatic compound with the following structure:
- It has a carboxylic acid group (-COOH) and a hydroxyl group (-OH) attached to a benzene ring.
- The structure can be represented as follows:

```
OH
|
C6H4 - C(=O)OH
```

### Step 2: Identify the Reactivity of the Functional Groups
- The -OH group is an activating group, which means it increases the reactivity of the benzene ring towards electrophilic substitution reactions.
- The -COOH group is a deactivating group, which means it decreases the reactivity of the benzene ring.

### Step 3: Reaction with Bromine Water
- When salicylic acid is treated with bromine water (Br2 in H2O), the bromine acts as an electrophile.
- The reaction will primarily occur at the ortho (2-position) and para (4-position) positions relative to the -OH group due to its activating nature.

### Step 4: Determine the Product
- The bromination will lead to the substitution of hydrogen atoms on the benzene ring with bromine atoms.
- The product formed will have bromine atoms at the 2, 4, and 6 positions of the benzene ring, resulting in 2,4,6-tribromophenol.

### Step 5: Write the Final Product
- The final product of the reaction is 2,4,6-tribromophenol, which can be represented as:

```
Br
|
Br - C6H2 - OH
|
Br
```

### Conclusion
The correct answer to the question is that salicylic acid on treatment with bromine water will give 2,4,6-tribromophenol.