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C(2)H(5)Cloverset("alc. KOH")rarrAoverse...

`C_(2)H_(5)Cloverset("alc. KOH")rarrAoverset(dil.H_(2)SO_(4))rarrB`.
Here A and B are

A

`C_(2)H_(4),C_(2)H_(5)OH`

B

`C_(2)H_(6),C_(2)H_(5)OH`

C

`C_(3)H_(8),C_(2)H_(5)OH`

D

`C_(2)H_(2),C_(2)H_(5)OH`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question, we need to analyze the reactions step by step. ### Step 1: Identify the Reactants The reactant given is ethyl chloride, which has the chemical formula C₂H₅Cl. ### Step 2: Reaction with Alcoholic KOH When ethyl chloride (C₂H₅Cl) reacts with alcoholic KOH, it undergoes an elimination reaction (dehydrohalogenation). In this reaction, the chlorine atom (Cl) is removed along with a hydrogen atom (H) from the adjacent carbon, leading to the formation of an alkene. **Reaction:** C₂H₅Cl + KOH (alc.) → C₂H₄ + KCl + H₂O Here, the product formed is ethene (C₂H₄), which is the compound A. ### Step 3: Reaction with Dilute H₂SO₄ Next, the product A (C₂H₄) reacts with dilute sulfuric acid (H₂SO₄). In this reaction, the alkene undergoes hydration, where water is added across the double bond. **Reaction:** C₂H₄ + H₂O (in presence of H₂SO₄) → C₂H₅OH The product formed is ethanol (C₂H₅OH), which is the compound B. ### Summary of Products - A (from the first reaction) is ethene (C₂H₄). - B (from the second reaction) is ethanol (C₂H₅OH). ### Final Answer A = C₂H₄ (Ethene) B = C₂H₅OH (Ethanol) ---
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