What will be the product , when carboxy phenol, obtained by Reimer Tiemann's process, is deoxidised with Zn powder ?

Options

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Answer

To solve the question regarding the product formed when carboxyphenol (obtained via the Reimer-Tiemann process) is deoxidized with zinc powder, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Carboxyphenol Structure**:
- Carboxyphenol is a compound that contains both a hydroxyl group (-OH) and a carboxylic acid group (-COOH) attached to a benzene ring. Its structure can be represented as:
\[ \text{C}_6\text{H}_4(\text{OH})(\text{COOH}) \]

2. **Understand the Deoxidation Process**:
- Deoxidation (or reduction) with zinc powder generally involves the removal of oxygen or the addition of hydrogen. In this case, we are reducing the carboxyphenol.

3. **Reaction with Zinc Powder**:
- When carboxyphenol is heated with zinc powder, the -COOH group can be reduced. The reaction involves the removal of the oxygen from the carboxylic acid group, leading to the formation of a benzene ring with a carboxyl radical.

4. **Formation of Benzoic Acid**:
- The carboxyl radical can then stabilize and form benzoic acid (C6H5COOH) as the final product. The overall reaction can be summarized as:
\[ \text{C}_6\text{H}_4(\text{OH})(\text{COOH}) \xrightarrow{\text{Zn, heat}} \text{C}_6\text{H}_5(\text{COOH}) + \text{ZnO} \]

5. **Conclusion**:
- The product formed when carboxyphenol is deoxidized with zinc powder is **benzoic acid**.

### Final Answer:
The product when carboxyphenol obtained by the Reimer-Tiemann process is deoxidized with zinc powder is **benzoic acid**.

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