2-Chlorobutane on treatment with alcoholic KOH/`Delta` gives major product :

Options

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

Answer

To solve the problem of what major product is formed when 2-chlorobutane is treated with alcoholic KOH in the presence of heat, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant is 2-chlorobutane, which has the following structure:
```
CH3-CH(Cl)-CH2-CH3
```
- The reagent is alcoholic KOH, which is a strong base.

2. **Understand the Reaction Type**:
- The reaction will proceed via an elimination mechanism, specifically the E2 mechanism. This mechanism involves the simultaneous removal of a hydrogen atom and a leaving group (in this case, Cl) to form a double bond.

3. **Determine the Beta Carbons**:
- In the structure of 2-chlorobutane, the carbon atom bonded to the chlorine (the alpha carbon) is the second carbon. The beta carbons are the first and third carbons (C1 and C3).
- The structure can be represented as:
```
C1: CH3
C2: CH(Cl)
C3: CH2
C4: CH3
```

4. **Apply the Zaitsev's Rule**:
- According to Zaitsev's rule, the more substituted alkene will be the major product. This means that the hydrogen will be removed from the beta carbon that has fewer hydrogen atoms.
- Here, C1 (CH3) has 3 hydrogens, while C3 (CH2) has 2 hydrogens. Therefore, the hydrogen will be removed from C3.

5. **Elimination Process**:
- When the hydrogen is removed from C3, the chlorine leaves, leading to the formation of a double bond between C2 and C3.
- The resulting product will be:
```
CH3-CH=CH-CH3
```
- This compound is 2-butene.

6. **Conclusion**:
- The major product formed from the reaction of 2-chlorobutane with alcoholic KOH in the presence of heat is **2-butene**.

### Final Answer:
The major product is **2-butene**.

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