sec`-` Butyl chloride on boiling with alc. `KOH` gives `…………` as the main product.

Options

A. `1-`Butene
B. `2-`Butene
C. `1-`Butanol
D. `2-`Butanol

Explanation

To determine the main product formed when sec-butyl chloride is boiled with alcoholic KOH, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of Sec-Butyl Chloride**:
Sec-butyl chloride has the molecular formula C4H9Cl. Its structure can be represented as:
\[ \text{CH}_3 - \text{CH}(\text{Cl}) - \text{CH}_2 - \text{CH}_3 \]
This indicates that the chlorine atom is attached to the second carbon of the butyl group.

2. **Reaction with Alcoholic KOH**:
When sec-butyl chloride is treated with alcoholic KOH, it undergoes a dehydrohalogenation reaction. In this reaction, the KOH acts as a base, removing a hydrogen atom and the chlorine atom (as HCl) from the sec-butyl chloride.

3. **Formation of Alkenes**:
The removal of H and Cl leads to the formation of alkenes. In this case, the possible alkenes that can be formed are:
- But-1-ene (from the first and second carbon)
- But-2-ene (from the second and third carbon)

4. **Applying Saytzeff's Rule**:
According to Saytzeff's rule (or Zaitsev's rule), the more substituted alkene will be the major product. In this case, but-2-ene is more substituted than but-1-ene because it has two alkyl groups attached to the double bond (the second carbon has two methyl groups).

5. **Conclusion**:
Therefore, the main product formed when sec-butyl chloride is boiled with alcoholic KOH is **but-2-ene**.

### Final Answer:
The main product is **but-2-ene**.

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