`C_(2)H_(5)Cl` on heating with alcoholic KOH will produce

Options

A. `C_(2)H_(5)OH`
B. `C_(2)H_(4)`
C. `C_(2)H_(2)`
D. `C_(2)H_(6)`

Answer

To solve the problem, we need to understand the reaction of ethyl chloride (C2H5Cl) when it is heated with alcoholic KOH. The process involves elimination reactions, specifically dehydrohalogenation.

### Step-by-Step Solution:

1. **Identify the Reactants:**
- The reactant is ethyl chloride (C2H5Cl).
- The reagent is alcoholic KOH.

2. **Understand the Role of Alcoholic KOH:**
- Alcoholic KOH acts as a strong base. In this context, it will facilitate the elimination of hydrogen chloride (HCl) from ethyl chloride.

3. **Mechanism of Reaction:**
- The alcoholic KOH will abstract a hydrogen atom from the β-carbon (the carbon adjacent to the carbon bonded to the chlorine).
- This leads to the formation of a double bond between the two carbon atoms, resulting in the elimination of HCl.

4. **Write the Reaction:**
- The elimination reaction can be represented as follows:
\[ \text{C2H5Cl} \xrightarrow{\text{alcoholic KOH}} \text{C2H4} + \text{HCl} \]
- Here, C2H4 is ethylene (or ethene), which is an alkene.

5. **Final Product:**
- The product of the reaction is ethylene (C2H4).

### Conclusion:
When C2H5Cl is heated with alcoholic KOH, it produces C2H4 (ethylene) and HCl.

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