Which halide of formula `C_(4)H_(9)I` is capable of producing but -2-ene with alcoholic KOH solution ?

Options

A. 1-Iodobutane
B. 2-Iodobutane
C. 1-Iodo-2-methylpropane
D. 2-Iodo-2-methylpropane

Explanation

To determine which halide of formula C₄H₉I can produce but-2-ene when reacted with alcoholic KOH, we will analyze the given options step by step.

### Step 1: Identify the structure of the halides
The halides of formula C₄H₉I can be:
1. 1-iodobutane
2. 2-iodobutane
3. 3-iodobutane
4. 1-iodopropane (not applicable since it does not have 4 carbons)

### Step 2: Analyze the first option - 1-iodobutane
1-iodobutane has the structure:
- CH₃-CH₂-CH₂-CH₂I

When 1-iodobutane reacts with alcoholic KOH, it undergoes dehydrohalogenation, which involves the elimination of HI (hydrogen iodide). The elimination mechanism will yield the following:
- Removing the iodine (I) from the carbon attached to it and a hydrogen (H) from an adjacent carbon will lead to the formation of a double bond.

The product formed will be:
- CH₃-CH₂-CH=CH₂ (but-1-ene)

Since we want but-2-ene, this option does not work.

### Step 3: Analyze the second option - 2-iodobutane
2-iodobutane has the structure:
- CH₃-CH(I)-CH₂-CH₃

When 2-iodobutane reacts with alcoholic KOH, the elimination reaction will proceed as follows:
- The iodine (I) can be eliminated from the second carbon, and a hydrogen (H) can be eliminated from the adjacent carbon (the first or third carbon).

Following the Zaitsev's rule (more substituted alkene is favored), the hydrogen will be removed from the first carbon:
- This will lead to the formation of a double bond between the second and third carbons.

The product formed will be:
- CH₃-CH=CH-CH₃ (but-2-ene)

This is the desired product.

### Conclusion
After analyzing both options, we find that the halide capable of producing but-2-ene when reacted with alcoholic KOH is **2-iodobutane**.

### Final Answer
The correct halide is **2-iodobutane**.

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