A mixture of `1-`chloropropane and `2-`chloropropane when treated with alcoholic `KOH`, it gives

Options

A. prop-1-ene
B. prop-2-ene
C. a mixture of prop-1-ene and prop-2 ene
D. propanol.

Explanation

To solve the problem of what a mixture of 1-chloropropane and 2-chloropropane gives when treated with alcoholic KOH, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Compounds**:
- The first compound is **1-chloropropane (C3H7Cl)**, which has the structure:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} \]
- The second compound is **2-chloropropane (C3H7Cl)**, which has the structure:
\[ \text{CH}_3\text{CHCl}\text{CH}_3 \]

2. **Understand the Reaction with Alcoholic KOH**:
- Alcoholic KOH promotes **elimination reactions** (specifically, beta-elimination). This means that we will be removing a hydrogen atom from the beta carbon (the carbon adjacent to the carbon with the leaving group, Cl) and the leaving group (Cl) itself.

3. **Reaction of 1-Chloropropane**:
- For **1-chloropropane**, the beta hydrogens are on the second carbon (CH2). When we remove one of these beta hydrogens, the reaction can be represented as follows:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{KOH (alc)} \rightarrow \text{CH}_3\text{CH}=\text{CH}_2 + \text{KCl} + \text{H}_2\text{O} \]
- The product formed is **prop-1-ene** (or simply propene).

4. **Reaction of 2-Chloropropane**:
- For **2-chloropropane**, the beta hydrogens are on both adjacent carbons. The reaction can be represented as:
\[ \text{CH}_3\text{CHCl}\text{CH}_3 + \text{KOH (alc)} \rightarrow \text{CH}_3\text{CH}=\text{CH}_2 + \text{KCl} + \text{H}_2\text{O} \]
- Again, the product formed is **prop-1-ene**.

5. **Conclusion**:
- Both reactions yield the same product, which is **prop-1-ene**. Therefore, when a mixture of 1-chloropropane and 2-chloropropane is treated with alcoholic KOH, the product is **prop-1-ene**.

### Final Answer:
The mixture of 1-chloropropane and 2-chloropropane when treated with alcoholic KOH gives **prop-1-ene**.

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