Chloroform when treated aniline and alcoholic KOH, the product formed is?

Options

A. Phenyl cyanide
B. Phenyl isocyanide
C. Chlorobenzene
D. Phenol

Answer

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

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We have aniline (C6H5NH2), which consists of a benzene ring with an amino group (-NH2).
- Chloroform (CHCl3) is the second reactant.
- Alcoholic potassium hydroxide (KOH) is also present, which acts as a base.

2. **Understand the Reaction Mechanism**:
- In the presence of KOH, the base will abstract a hydrogen atom from the amino group of aniline, resulting in the formation of a positively charged nitrogen species (C6H5NH3+).
- Chloroform (CHCl3) can undergo dehydrohalogenation in the presence of a strong base, leading to the formation of a reactive intermediate, dichlorocarbene (CCl2).

3. **Formation of the Intermediate**:
- The base (OH-) abstracts a hydrogen from the chloroform, leading to the release of HCl and the formation of CCl2, which has a lone pair of electrons on the carbon atom.

4. **Nucleophilic Attack**:
- The lone pair of electrons on the nitrogen atom of the aniline can now attack the electrophilic carbon of the dichlorocarbene (CCl2).
- This forms a new bond between the nitrogen and carbon, resulting in an intermediate compound.

5. **Elimination of Chlorine**:
- The intermediate undergoes rearrangement where one of the chlorine atoms is eliminated, leading to the formation of a double bond between the nitrogen and carbon.
- The nitrogen now has a positive charge, and the structure can rearrange to form a triple bond.

6. **Final Product**:
- The final product of this reaction is phenyl isocyanide (C6H5N≡C). This compound has a phenyl group attached to an isocyanide functional group.

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