Treatment of n-propylamine with nitrous acid gives

Options

A. `CH_(3)CH=CH_(2)`
B. `CH_(3)-CH_(2)-CH_(2)OH`
C. `H_(3)C-underset(OH)underset(|)(C)H-CH_(3)`
D. none of these

Answer

To solve the question regarding the treatment of n-propylamine with nitrous acid, we will follow these steps:

### Step 1: Identify the Reactants
The reactant in this case is n-propylamine, which has the chemical structure CH3-CH2-CH2-NH2.

### Step 2: Understand the Reaction
n-Propylamine is a primary amine. When primary amines react with nitrous acid (HNO2), they typically form diazonium salts. The nitrous acid can be generated in situ from sodium nitrite (NaNO2) and hydrochloric acid (HCl).

### Step 3: Write the Reaction
The reaction can be summarized as follows:
1. n-Propylamine reacts with nitrous acid to form n-propyldiazonium chloride:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 + \text{HNO}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{N}_2^+\text{Cl}^- \]

### Step 4: Hydrolysis of Diazonium Salt
The diazonium salt formed is unstable and will undergo hydrolysis when treated with water, leading to the formation of a primary alcohol. The hydrolysis reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{N}_2^+\text{Cl}^- + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{N}_2 + \text{HCl} \]

### Step 5: Identify the Product
The product of this reaction is propanol (CH3-CH2-CH2-OH), which is a primary alcohol.

### Conclusion
Thus, the treatment of n-propylamine with nitrous acid gives propanol.

### Final Answer
The product is CH3-CH2-CH2-OH (propanol), which matches with option B.

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