Which of the following product `(s)` will be obtained when isopropylamine is treated with sodium nitrite and hydrochloric acid ?

Options

A. `CH_(3)-underset(underset(OH)|)(C)H-CH_(3)`
B. `CH_(2)-CH=CH_(2)`
C. `CH_(3)-overset(overset(Cl)|)(C)H-CH_(2)`
D. All of these

Explanation

To determine the products obtained when isopropylamine is treated with sodium nitrite (NaNO2) and hydrochloric acid (HCl), we can follow these steps:

### Step 1: Identify the Structure of Isopropylamine
Isopropylamine has the following structure:
- Chemical formula: \( \text{C}_3\text{H}_9\text{N} \)
- Structural formula:
```
H H
| |
H - C - C - NH2
| |
H H
```

### Step 2: Reaction with Sodium Nitrite and Hydrochloric Acid
When isopropylamine is treated with sodium nitrite and hydrochloric acid, it forms a diazonium salt. The reaction can be summarized as follows:
- The amine group (\(-NH_2\)) is converted to a diazonium ion (\(R-N_2^+\)):

\[ \text{C}_3\text{H}_9\text{N} + \text{NaNO}_2 + \text{HCl} \rightarrow \text{C}_3\text{H}_9\text{N}_2^+ \text{Cl}^- + \text{NaCl} + \text{H}_2\text{O} \]

### Step 3: Formation of Products
The diazonium salt is unstable and can decompose to release nitrogen gas (\(N_2\)). The decomposition can lead to different products depending on the conditions:
1. **Elimination Reaction**: The loss of nitrogen gas can lead to the formation of propene:
\[ \text{C}_3\text{H}_9\text{N}_2^+ \rightarrow \text{C}_3\text{H}_6 + N_2 + HCl \]
(Propene: \( \text{CH}_3\text{CH}=\text{CH}_2 \))

2. **Nucleophilic Attack**: The carbocation formed can react with chloride ions (\(Cl^-\)) to form isopropyl chloride:
\[ \text{C}_3\text{H}_9^+ + Cl^- \rightarrow \text{C}_3\text{H}_7\text{Cl} \]
(Isopropyl chloride: \( \text{C}_3\text{H}_7\text{Cl} \))

3. **Further Reactions**: The carbocation can also react with hydroxide ions (\(OH^-\)) to form isopropanol:
\[ \text{C}_3\text{H}_9^+ + OH^- \rightarrow \text{C}_3\text{H}_8\text{O} \]
(Isopropanol: \( \text{C}_3\text{H}_8\text{O} \))

### Step 4: Summary of Products
From the above reactions, we can conclude that the following products can be formed:
- Propene (\( \text{CH}_3\text{CH}=\text{CH}_2 \))
- Isopropyl chloride (\( \text{C}_3\text{H}_7\text{Cl} \))
- Isopropanol (\( \text{C}_3\text{H}_8\text{O} \))

### Final Answer
The products obtained when isopropylamine is treated with sodium nitrite and hydrochloric acid include:
1. Propene (\( \text{CH}_3\text{CH}=\text{CH}_2 \))
2. Isopropyl chloride (\( \text{C}_3\text{H}_7\text{Cl} \))
3. Isopropanol (\( \text{C}_3\text{H}_8\text{O} \))