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

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Answer

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

### Step-by-Step Solution:

1. **Identify the Starting Material**:
The starting material is carboxyphenol, which is a phenolic compound with a carboxylic acid group (-COOH) attached to a benzene ring. The structure can be represented as:
\[ \text{C}_6\text{H}_4(\text{OH})(\text{COOH}) \]

2. **Understanding the Reaction**:
The reaction involves the deoxidation of carboxyphenol using zinc powder (Zn) as a reducing agent. Zinc is known to reduce functional groups, and in this case, it will primarily affect the hydroxyl (-OH) group of carboxyphenol.

3. **Mechanism of Deoxidation**:
When zinc powder is added to carboxyphenol, it donates electrons to the hydroxyl group. This leads to the formation of a phenoxyl radical. The reaction can be summarized as:
\[ \text{C}_6\text{H}_4(\text{OH})(\text{COOH}) + \text{Zn} \rightarrow \text{C}_6\text{H}_4(\text{O}^-) + \text{ZnO} + \text{H}^+ \]
Here, the hydroxyl group loses a hydrogen atom, resulting in the formation of a phenoxyl ion (C6H4O^-).

4. **Formation of Benzoic Acid**:
As the reaction proceeds, the phenoxyl ion can stabilize by accepting a proton (H^+) from the environment, leading to the formation of benzoic acid (C6H5COOH) as the main product:
\[ \text{C}_6\text{H}_4(\text{O}^-)(\text{COOH}) + \text{H}^+ \rightarrow \text{C}_6\text{H}_5(\text{COOH}) + \text{ZnO} \]

5. **Final Product**:
The final products of the reaction are benzoic acid and zinc oxide (ZnO). Therefore, the overall reaction can be summarized as:
\[ \text{Carboxyphenol} \xrightarrow{\text{Zn}} \text{Benzoic Acid} + \text{ZnO} \]

### Final Answer:
The product formed when carboxyphenol is deoxidized with zinc powder is **benzoic acid (C6H5COOH)** and zinc oxide (ZnO).

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