Which of the following products are formed when 1-propanamine is treated with`NaNO_(2) + HCI`?

Options

A.
B.
C.
D.

Explanation

To determine the products formed when 1-propanamine is treated with NaNO2 and HCl, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant is 1-propanamine (C3H9N), which has the structure CH3-CH2-CH2-NH2.
- The other reactants are sodium nitrite (NaNO2) and hydrochloric acid (HCl).

2. **Formation of Nitrous Acid**:
- When NaNO2 is mixed with HCl, it produces nitrous acid (HNO2):
\[ \text{NaNO}_2 + \text{HCl} \rightarrow \text{HNO}_2 + \text{NaCl} \]

3. **Reaction of 1-Propanamine with Nitrous Acid**:
- The nitrous acid (HNO2) will react with 1-propanamine to form a diazonium salt. The reaction can be represented as:
\[ \text{R-NH}_2 + \text{HNO}_2 \rightarrow \text{R-N}_2^+ + \text{H}_2\text{O} \]
- For 1-propanamine, this results in the formation of propyl diazonium ion (C3H7N2^+).

4. **Decomposition of Diazonium Salt**:
- The diazonium ion is unstable and can decompose to form different products. The decomposition can lead to the formation of nitrogen gas (N2) and a carbocation:
\[ \text{R-N}_2^+ \rightarrow \text{R}^+ + \text{N}_2 \]
- The carbocation can then react with water (hydrolysis) to form alcohols or can rearrange to form alkenes.

5. **Possible Products**:
- **Product 1**: Hydrolysis of the carbocation leads to the formation of propanol (C3H7OH).
- **Product 2**: If the carbocation rearranges and loses a hydrogen, it can form propene (C3H6).
- **Product 3**: A more stable carbocation can form, leading to another alcohol or alkene.

6. **Final Products**:
- The final products from the reaction of 1-propanamine with NaNO2 and HCl are:
- Propanol (C3H7OH)
- Propene (C3H6)
- Other possible isomers or rearrangements.

### Conclusion:
The products formed when 1-propanamine is treated with NaNO2 and HCl include propanol and propene, among other possible products.