`underset((A))(C_4H_8Cl_2)underset("Heat")overset(alc.KOH)rarrBoverset(HgSO_4,H_2SO_4)rarrC` .
C given idoform test. Hence A can be

Options

A. 1,1 - dichlorobutane
B. 1,2 - dichlorobutane
C. 2,2 - dichlorobutane
D. Any of these

Explanation

To solve the problem, we need to determine the possible structure of compound A (C4H8Cl2) based on the reactions described and the final product C that gives the iodoform test. Here’s the step-by-step solution:

### Step 1: Understanding the Reaction
The compound C4H8Cl2 is reacting with alcoholic KOH under heat. This suggests a dehydrohalogenation reaction where KOH acts as a base to remove hydrogen halides (HCl) from the compound.

**Hint:** Look for how the removal of halogens and protons leads to the formation of a double or triple bond.

### Step 2: Possible Structures for A
The molecular formula C4H8Cl2 indicates that A could be a dichloroalkane. The two common structures for dichlorobutane are:
1. 1,1-Dichlorobutane
2. 1,2-Dichlorobutane
3. 2,2-Dichlorobutane

**Hint:** Consider the positions of the chlorine atoms and how they can affect the stability of the resulting alkenes or alkynes.

### Step 3: Elimination Reaction
When A undergoes elimination with alcoholic KOH, it can lose two halogen atoms and two hydrogen atoms. This can lead to the formation of either a double bond (alkene) or a triple bond (alkyne).

- For 1,1-Dichlorobutane:
- The reaction would yield 1-butyne (C4H6).

- For 1,2-Dichlorobutane:
- The reaction would yield 1-butyne (C4H6) or 2-butyne (C4H6).

- For 2,2-Dichlorobutane:
- The reaction would yield 2-butyne (C4H6).

**Hint:** Identify which of these products can further react with HgSO4 and H2SO4 to form a ketone.

### Step 4: Reaction with HgSO4 and H2SO4
The product formed after the elimination reaction (either 1-butyne or 2-butyne) will react with HgSO4 in the presence of H2SO4 to form a ketone.

- The reaction of 1-butyne would yield butan-2-one (a methyl ketone).
- The reaction of 2-butyne would yield butan-2-one (also a methyl ketone).

**Hint:** Remember that only methyl ketones give the iodoform test.

### Step 5: Iodoform Test
Since the final product C gives the iodoform test, it must be a methyl ketone. The only methyl ketone that can be formed from the structures of A is butan-2-one.

### Conclusion
Based on the analysis, the possible structures for A that lead to a product C giving the iodoform test are:
- 1,1-Dichlorobutane
- 1,2-Dichlorobutane
- 2,2-Dichlorobutane

Thus, the answer is that A can be either 1,1-Dichlorobutane, 1,2-Dichlorobutane, or 2,2-Dichlorobutane.

**Final Answer:** A can be 1,1-Dichlorobutane, 1,2-Dichlorobutane, or 2,2-Dichlorobutane.