CH₃CH₂Cl alcoholic KOH → A Br₂/CCl₄ → B Zn alcohol/Δ → C. Here 'C' is:

Solution for the Given Reaction Sequence

Step 1: Dehydrohalogenation (Alcoholic KOH)

• Reactant: CH₃CH₂Cl (ethyl chloride)
• Reagent: Alcoholic KOH

Reaction: The reaction with alcoholic KOH leads to the elimination of HCl, forming ethene (CH₂=CH₂) as product A.

CH₃CH₂Cl → CH₂=CH₂ (A)  

Step 2: Addition of Bromine in CCl₄

• Reactant: CH₂=CH₂ (ethene) (A)
• Reagent: Br₂ in CCl₄

Reaction: Ethene reacts with bromine to undergo an electrophilic addition reaction, forming 1,2-dibromoethane (CH₂Br-CH₂Br) as product B.

CH₂=CH₂ + Br₂ → CH₂Br-CH₂Br (B)

Step 3: Reaction with Zn/Alcohol

• Reactant: CH₂Br-CH₂Br (1,2-dibromoethane) (B)
• Reagent: Zn/Alcohol, followed by heating (Δ)

Reaction: Zinc in alcohol reduces 1,2-dibromoethane to ethyne (C₂H₂) by removing two bromine atoms as ZnBr₂. The final product C is ethyne.

CH₂Br-CH₂Br → C₂H₂ (C)

Final Products

• A: Ethene (CH₂=CH₂)
• B: 1,2-Dibromoethane (CH₂Br-CH₂Br)
• C: Ethyne (C₂H₂)

Answer: The final product C is ethyne (C₂H₂).