Reaction of ethylamine with chloroform in alcoholic KOH gives

Options

A. `C_(2)H_(5)NC`
B. `CH_(3)NC`
C. `CH_(3)CN`
D. `CH_(3)NC`.

Answer

To solve the question regarding the reaction of ethylamine with chloroform in alcoholic KOH, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- Ethylamine (C2H5NH2) can be represented as CH3-CH2-NH2.
- Chloroform (CHCl3) is represented as CHCl3.

2. **Understand the Reaction Conditions**:
- The reaction occurs in the presence of alcoholic KOH, which acts as a base.

3. **Deprotonation of Ethylamine**:
- The alcoholic KOH will deprotonate the ethylamine. The nitrogen atom in ethylamine has a lone pair and can act as a base.
- The reaction can be represented as:
\[ CH3-CH2-NH2 + KOH \rightarrow CH3-CH2-NH + H2O \]

4. **Formation of Carbene**:
- The deprotonated ethylamine will react with chloroform. The chloroform will lose a hydrogen atom and form a carbene (C:).
- The reaction can be represented as:
\[ CHCl3 + CH3-CH2-NH \rightarrow CHCl2 + CH3-CH2-NH-C: \]

5. **Nucleophilic Attack**:
- The carbene (C:) will act as an electrophile and will be attacked by the nucleophilic nitrogen of the ethylamine.
- This results in the formation of an intermediate compound:
\[ CH3-CH2-NH-CCl2 \]

6. **Formation of Isocyanide**:
- The intermediate undergoes further reactions, leading to the formation of isocyanide (also known as carbylamine).
- The final product can be represented as:
\[ CH3-CH2-NC \]
- The isocyanide structure is CH3-CH2-N≡C.

### Final Answer:
The reaction of ethylamine with chloroform in alcoholic KOH gives isocyanide (ethyl isocyanide).