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 of what aniline reacts with excess COCl2 and KOH to form, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are aniline (C6H5NH2), phosgene (COCl2), and potassium hydroxide (KOH).

2. **Understand Aniline's Structure**:
- Aniline consists of a benzene ring attached to an amino group (-NH2). The nitrogen atom has a lone pair of electrons that can act as a nucleophile.

3. **Nucleophilic Attack**:
- Aniline will act as a nucleophile and attack the electrophilic carbonyl carbon of phosgene (COCl2). The carbonyl carbon is positively charged due to the electronegativity of the oxygen atom.

4. **Formation of an Intermediate**:
- The attack results in the formation of an intermediate where the nitrogen of aniline forms a double bond with the carbonyl carbon, leading to the structure:
- C6H5-NH-C(=O)-Cl.

5. **Release of HCl**:
- The intermediate undergoes a rearrangement where a chloride ion (Cl-) is released, resulting in the formation of an acyl derivative of aniline.

6. **Role of KOH**:
- KOH acts as a base and abstracts a proton (H+) from the nitrogen atom of the intermediate, generating a negatively charged nitrogen.

7. **Final Product Formation**:
- The negatively charged nitrogen can then facilitate the release of another chloride ion, resulting in the formation of phenyl isocyanate (C6H5-N=C=O).

8. **Conclusion**:
- The final product of the reaction between aniline, excess COCl2, and KOH is phenyl isocyanate.

### Final Answer:
The reaction of aniline with excess COCl2 and KOH forms **phenyl isocyanate**.

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