Assuming that no rearrangement is taking place, then how many hydrocarbons are obtained from the reaction of 2-chloropentane with isopropyl chloride in the presence of sodium. Do not include stereoisomers.

Answer

To determine how many hydrocarbons are obtained from the reaction of 2-chloropentane with isopropyl chloride in the presence of sodium, we can follow these steps:

### Step 1: Identify the Reactants
- The reactants are 2-chloropentane and isopropyl chloride.
- **2-chloropentane** has the structure:
```
CH3-CH(Cl)-CH2-CH2-CH3
```
- **Isopropyl chloride** has the structure:
```
CH3-CH(Cl)-CH3
```

### Step 2: Understand the Reaction Type
- The reaction between alkyl halides in the presence of sodium is known as **Wurtz reaction**.
- In this reaction, two alkyl halides react to form a new alkane by coupling the alkyl groups and eliminating sodium halide.

### Step 3: Possible Coupling Reactions
- The possible combinations of the alkyl groups from the two reactants are:
1. The alkyl group from 2-chloropentane (C5H11) with itself.
2. The alkyl group from isopropyl chloride (C3H7) with itself.
3. The alkyl groups from both reactants combining.

### Step 4: Write the Possible Products
1. **From 2-chloropentane with itself**:
- Product: **n-hexane** (C6H14)
- Structure:
```
CH3-CH2-CH2-CH2-CH2-CH3
```

2. **From isopropyl chloride with itself**:
- Product: **diisopropyl ether** (C6H14)
- Structure:
```
CH3-CH(CH3)-CH2-CH(CH3)-CH3
```

3. **From 2-chloropentane and isopropyl chloride**:
- Product: **3-methylhexane** (C7H16)
- Structure:
```
CH3-CH(CH3)-CH2-CH2-CH2-CH3
```

### Step 5: Count the Unique Hydrocarbons
- The unique hydrocarbons formed from the reactions are:
1. n-hexane
2. diisopropyl ether
3. 3-methylhexane

### Conclusion
- Therefore, the total number of unique hydrocarbons obtained from the reaction is **3**.

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