In the chemical reaction, `CH_3CH_2NH_2 + CHCl_3 + 3KOH to A + B + 3H_2O`
The compounds A and B respectively are

Options

A. `CH_3CH_2CONH_2` and 3KCl
B. `C_2H_5NC` and `K_2CO_3`
C. `C_2H_5NC` and `3KCl`
D. `C_2H_5CN` and 3KCl

Explanation

To solve the reaction \( CH_3CH_2NH_2 + CHCl_3 + 3KOH \to A + B + 3H_2O \), we need to identify the products A and B formed in this reaction.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- Ethylamine (\( CH_3CH_2NH_2 \))
- Chloroform (\( CHCl_3 \))
- Potassium hydroxide (\( KOH \))

2. **Recognize the Type of Reaction**:
- This reaction is known as the **Carbylamine Reaction**. It is a characteristic reaction of primary amines with chloroform in the presence of a strong base (like KOH).

3. **Mechanism of the Reaction**:
- The strong base \( KOH \) deprotonates the ethylamine, forming an ethylamine anion.
- The ethylamine anion then reacts with chloroform to form a reactive intermediate called **dichlorocarbene**.

4. **Formation of Products**:
- The dichlorocarbene reacts with the ethylamine to form an **alkyl isocyanide**. This is the compound A.
- The other product formed in the reaction is **potassium chloride (KCl)**, which is compound B.

5. **Final Products**:
- Therefore, the products of the reaction are:
- A: Alkyl isocyanide (specifically, ethyl isocyanide, \( CH_3CH_2N \equiv C \))
- B: Potassium chloride (\( KCl \))

### Conclusion:
The compounds A and B respectively are:
- A: Ethyl isocyanide (\( CH_3CH_2N \equiv C \))
- B: Potassium chloride (\( KCl \))