How many isomers of molecular formula `C_(3)H_(9)N` will liberate `N_(2)` gas on treatment with `HNO_(2)`?

Options

A. One
B. Three
C. Two
D. Five

Answer

To determine how many isomers of the molecular formula C₃H₉N will liberate N₂ gas upon treatment with HNO₂, we can follow these steps:

### Step 1: Identify the Isomers of C₃H₉N
The molecular formula C₃H₉N can have several structural isomers. We can categorize them into primary, secondary, and tertiary amines:

1. **Propylamine (1-aminopropane)**: CH₃CH₂CH₂NH₂ (Primary amine)
2. **Isopropylamine (2-aminopropane)**: CH₃CH(NH₂)CH₃ (Secondary amine)
3. **N,N-Dimethylamine**: (CH₃)₂NCH₂CH₃ (Tertiary amine)
4. **Cyclopropylamine**: C₃H₆NH (considered in some contexts but not a direct isomer of C₃H₉N)

### Step 2: Determine Which Isomers Liberate N₂ Gas
When treated with HNO₂ (nitrous acid), only primary amines will liberate nitrogen gas (N₂). The reaction for primary amines with nitrous acid is as follows:

\[ RNH₂ + HNO₂ \rightarrow R-OH + N₂ + H₂O \]

Now, we can analyze our identified isomers:

- **Propylamine (1-aminopropane)**: This is a primary amine and will liberate N₂ gas.
- **Isopropylamine (2-aminopropane)**: This is a secondary amine and will **not** liberate N₂ gas.
- **N,N-Dimethylamine**: This is a tertiary amine and will **not** liberate N₂ gas.

### Step 3: Count the Isomers that Liberate N₂ Gas
From the analysis:
- The only isomer that liberates N₂ gas is **Propylamine**.

### Conclusion
Thus, there is **1 isomer** of C₃H₉N that will liberate N₂ gas upon treatment with HNO₂.

### Final Answer
The answer is **1**.

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