Aniline reacts with excess of phosgene and KOH to form :

Options

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B.
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D.

Answer

To solve the question of what compound is formed when aniline reacts with excess phosgene and KOH, we can break down the process step by step.

### Step-by-Step Solution:

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

2. **Structure of Aniline**:
- Aniline consists of a benzene ring (C6H5) with an amino group (NH2) attached to it.

3. **Reaction with Phosgene**:
- Phosgene (COCl2) is a carbonyl compound. The nitrogen atom in the amino group has a lone pair of electrons, which can attack the carbon atom of phosgene.
- The reaction can be represented as:
\[ \text{C6H5NH2} + \text{COCl2} \rightarrow \text{C6H5N}^+ \text{(C=O)} \text{Cl}^- \]
- Here, the nitrogen forms a bond with the carbon of phosgene, resulting in a positively charged intermediate.

4. **Formation of an Intermediate**:
- The intermediate formed is a benzamide derivative where the nitrogen is positively charged, and there is a chlorine atom and a carbonyl group (C=O) attached.

5. **Role of KOH**:
- KOH acts as a base in this reaction. The hydroxide ion (OH-) can deprotonate the positively charged nitrogen, leading to the formation of a neutral compound.
- The reaction can be represented as:
\[ \text{C6H5N}^+ \text{(C=O)} \text{Cl}^- + \text{OH}^- \rightarrow \text{C6H5N=C=O} + \text{H2O} + \text{Cl}^- \]

6. **Final Product**:
- The final product is an isocyanate derivative of aniline, specifically phenyl isocyanate (C6H5N=C=O).
- The overall reaction can be summarized as:
\[ \text{C6H5NH2} + \text{COCl2} + \text{KOH} \rightarrow \text{C6H5N=C=O} + \text{H2O} + \text{KCl} \]

### Conclusion:
The compound formed when aniline reacts with excess phosgene and KOH is **phenyl isocyanate (C6H5N=C=O)**.