Aniline reacts with excess of ` COCl_2 and KOH ` to form :

Options

A. `C_6 H_5 OH`
B. `C_6 H_5 NCO`
C. `C_6 H_5 NHCOCl`
D. `C_6 H_5 COCl`

Answer

To solve the question regarding the reaction of aniline with excess `COCl2` (thionyl chloride) and `KOH`, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are aniline (C6H5NH2) and thionyl chloride (COCl2).
- Aniline is a primary amine with the structure C6H5NH2.

2. **Reaction with Thionyl Chloride**:
- When aniline reacts with thionyl chloride, the amine group (-NH2) will react with the carbonyl group (C=O) of thionyl chloride.
- This reaction leads to the formation of an intermediate compound where the nitrogen atom of aniline forms a bond with the carbon atom of the carbonyl group, resulting in the formation of an acylated amine.

3. **Formation of the Intermediate**:
- The reaction can be represented as:
\[ C6H5NH2 + COCl2 \rightarrow C6H5NHC(O)Cl \]
- Here, C6H5NHC(O)Cl is the intermediate formed.

4. **Role of KOH**:
- The presence of KOH (potassium hydroxide) in the reaction acts as a base. It will deprotonate the acylated amine, leading to the formation of a more stable compound.
- The hydroxide ion (OH-) will abstract a proton from the nitrogen, resulting in a positively charged nitrogen atom.

5. **Final Product Formation**:
- After deprotonation, the intermediate will rearrange and eliminate a chloride ion (Cl-) due to the presence of KOH.
- The final product can be represented as:
\[ C6H5N=C=O \]
- This compound is known as an isocyanate.

6. **Conclusion**:
- The final product formed from the reaction of aniline with excess `COCl2` and `KOH` is **benzene isocyanate** (C6H5N=C=O).

### Final Answer:
The product formed is **benzene isocyanate (C6H5N=C=O)**.

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