`CH_(3)-CH_(2)Cloverset("alcoholic KOH")rarrA` Compound `Aoverset(Br_(2))rarrB`. Compound `Bunderset(Delta)overset(Zn)rarrC`. Compound C is

Options

A. Acetylene
B. Ethane
C. Ethene
D. Methane

Answer

To solve the problem step by step, let's analyze the reactions involved:

1. **Starting Compound**: The starting compound is Ethyl Chloride (CH₃-CH₂Cl).

2. **Reaction with Alcoholic KOH**:
- Ethyl chloride (CH₃-CH₂Cl) reacts with alcoholic KOH. This reaction is a dehydrohalogenation reaction, where the KOH acts as a base and removes a hydrogen atom and the chlorine atom from the ethyl chloride.
- The product formed is Ethylene (CH₂=CH₂) and potassium chloride (KCl) along with water (H₂O).
- **Compound A**: CH₂=CH₂ (Ethylene)

3. **Reaction of Compound A with Br₂**:
- Ethylene (Compound A) reacts with bromine (Br₂). This is an addition reaction where bromine adds across the double bond.
- The product formed is 1,2-Dibromoethane (CH₂Br-CH₂Br).
- **Compound B**: CH₂Br-CH₂Br (1,2-Dibromoethane)

4. **Reaction of Compound B with Zinc (Δ)**:
- 1,2-Dibromoethane (Compound B) reacts with zinc in the presence of heat. This reaction is a dehalogenation reaction where zinc removes the bromine atoms.
- The product formed is Ethylene (CH₂=CH₂) again, and zinc bromide (ZnBr₂) is generated as a byproduct.
- **Compound C**: CH₂=CH₂ (Ethylene)

Thus, the final answer is that Compound C is Ethylene (CH₂=CH₂).

### Summary of Compounds:
- **Compound A**: Ethylene (CH₂=CH₂)
- **Compound B**: 1,2-Dibromoethane (CH₂Br-CH₂Br)
- **Compound C**: Ethylene (CH₂=CH₂)