The reaction of `p-HOC_(6)H_(4)COOH` with excess `Br_(2)` forms

Options

A.
B.
C.
D.

Answer

To solve the question regarding the reaction of para-hydroxybenzoic acid (p-HOC₆H₄COOH) with excess bromine (Br₂), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of Para-Hydroxybenzoic Acid**:
- Para-hydroxybenzoic acid has a hydroxyl group (-OH) and a carboxylic acid group (-COOH) attached to a benzene ring at the para position (1,4 positions).
- The chemical structure can be represented as:
```
OH
|
C6H4 - COOH
```

2. **Reaction with Excess Bromine**:
- When para-hydroxybenzoic acid reacts with excess bromine, the carboxylic acid group undergoes decarboxylation. This means that the -COOH group is removed, releasing carbon dioxide (CO₂).
- The reaction can be represented as:
```
p-HOC₆H₄COOH + Br₂ → p-HOC₆H₄Br + CO₂ + HBr
```

3. **Formation of Brominated Product**:
- After the decarboxylation, the resulting compound (which is para-bromophenol) can further react with excess bromine.
- The hydroxyl group on the benzene ring is ortho/para directing, which means that bromine will add to the 2 and 4 positions of the ring.
- This leads to the formation of 2,4,6-tribromophenol.

4. **Final Product**:
- The final product of the reaction is therefore 2,4,6-tribromophenol.
- The chemical structure can be represented as:
```
Br
|
Br-C6H2(OH)-Br
```

5. **Conclusion**:
- Thus, the correct answer to the question is that the reaction of para-hydroxybenzoic acid with excess bromine forms 2,4,6-tribromophenol.

### Final Answer:
The reaction of para-hydroxybenzoic acid with excess bromine forms **2,4,6-tribromophenol**.

---