An organic compound 'A' having molecular formula `C_(2)H_(3)N` on reduction gave another compound 'B' Upon treatment with nitrous acid, 'B' gave ethyl alcohol on warming with chloroform and alcoholic KOH, it formed an offensive smelling compound 'C'. The compound 'C' is :

Options

A. `CH_(3)C-=N`
B. `CH_(3)CH_(2)N-=C`
C. `CH_(3)CH_(2)NH_(2)`
D. `CH_(3)CH_(2)OH`

Answer

To solve the problem step by step, we will analyze the information given and deduce the structure of the compounds involved.

### Step 1: Identify Compound A
The molecular formula of compound A is \( C_2H_3N \). To determine the structure, we can calculate the degree of unsaturation (double bond equivalent) using the formula:

\[ \text{DBE} = \frac{(2C + 2 + N - H)}{2} \]

Substituting the values:

\[ \text{DBE} = \frac{(2 \times 2 + 2 + 1 - 3)}{2} = \frac{(4 + 2 + 1 - 3)}{2} = \frac{4}{2} = 2 \]

This indicates that compound A can have either 2 double bonds or 1 triple bond. Given that it contains nitrogen, it is likely to be a cyanide compound, specifically **ethyl cyanide** (or **ethyl nitrile**), which has the structure \( CH_3CH_2C \equiv N \).

### Step 2: Reduction of Compound A to Compound B
When ethyl cyanide is reduced, it typically converts to an amine. Therefore, upon reduction, compound B is **ethylamine** (\( CH_3CH_2NH_2 \)).

### Step 3: Reaction of Compound B with Nitrous Acid
When ethylamine reacts with nitrous acid (\( HNO_2 \)), it forms a diazonium salt. However, since ethylamine is a primary amine, the diazonium salt is unstable and decomposes to form **ethyl alcohol** (\( CH_3CH_2OH \)) and nitrogen gas.

### Step 4: Reaction of Ethyl Alcohol with Chloroform and Alcoholic KOH
The final step involves warming ethyl alcohol with chloroform and alcoholic KOH. This reaction is known as the **carbylamine reaction** (or isocyanate formation). In this reaction, ethyl alcohol reacts to form an isocyanate compound.

The structure of the resulting compound C is **ethyl isocyanate** (\( CH_3CH_2NCO \)), which is known to have an offensive smell.

### Conclusion
The compound C formed is **ethyl isocyanate**.

### Summary of Steps
1. Identify compound A as ethyl cyanide (\( CH_3CH_2C \equiv N \)).
2. Reduce compound A to form compound B, ethylamine (\( CH_3CH_2NH_2 \)).
3. Treat compound B with nitrous acid to produce ethyl alcohol (\( CH_3CH_2OH \)).
4. React ethyl alcohol with chloroform and alcoholic KOH to form compound C, ethyl isocyanate (\( CH_3CH_2NCO \)).