Chloroform when treated with aniline and alcoholic KOH forms -

Options

A. Phenyl cyanide
B. Phenyl isocyanide
C. Phenyl cyanate
D. Phenyl isocyanate

Answer

To solve the question of what product is formed when chloroform reacts with aniline and alcoholic KOH, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are chloroform (CHCl3) and aniline (C6H5NH2) in the presence of alcoholic KOH.

2. **Understand the Reaction Type**:
- This reaction is a type of carbylamine reaction, which occurs between chloroform and primary amines (like aniline) in the presence of a strong base.

3. **Mechanism of the Reaction**:
- Chloroform (CHCl3) reacts with the alcoholic KOH to generate a reactive intermediate, which is a carbanion (CCl3^-).
- The base (KOH) deprotonates the amine (aniline), allowing the nitrogen atom to have a lone pair that can attack the carbon of the carbanion.

4. **Formation of Isocyanide**:
- The nitrogen from aniline donates its lone pair to the carbon atom of the carbanion (CCl3^-), forming a bond. This results in the formation of phenyl isocyanide (C6H5N≡C).
- The reaction also involves the elimination of HCl (hydrochloric acid) as a byproduct.

5. **Final Product**:
- The final product of the reaction is phenyl isocyanide (C6H5N≡C).

### Conclusion:
The product formed when chloroform is treated with aniline and alcoholic KOH is **phenyl isocyanide**.