`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 question regarding the reaction of C₂H₅Cl (ethyl chloride) when heated with alcoholic KOH, we will follow these steps:

### Step 1: Identify the Reactants
The reactant in this case is C₂H₅Cl (ethyl chloride) and the reagent is alcoholic KOH (potassium hydroxide in alcohol).

**Hint:** Alcoholic KOH is a strong base that promotes elimination reactions rather than substitution.

### Step 2: Understand the Reaction Type
When C₂H₅Cl is heated with alcoholic KOH, it undergoes an elimination reaction (dehydrohalogenation) rather than a substitution reaction. This is because the alcoholic medium favors the formation of alkenes.

**Hint:** Look for the formation of a double bond in the product when a hydrogen halide is eliminated.

### Step 3: Determine the Elimination Mechanism
In this reaction, KOH will abstract a hydrogen atom from the β-carbon (the carbon adjacent to the one bonded to Cl), and the Cl atom will be eliminated as a chloride ion (Cl⁻). This results in the formation of an alkene.

**Hint:** Identify the β-hydrogen that will be removed to form the double bond.

### Step 4: Write the Reaction
The elimination of HCl from C₂H₅Cl leads to the formation of ethylene (C₂H₄).

The reaction can be summarized as follows:
\[ \text{C}_2\text{H}_5\text{Cl} + \text{KOH (alc)} \rightarrow \text{C}_2\text{H}_4 + \text{KCl} + \text{H}_2\text{O} \]

**Hint:** Remember to balance the equation and include all products formed.

### Step 5: Conclusion
Thus, when C₂H₅Cl is heated with alcoholic KOH, the product formed is C₂H₄ (ethylene).

**Final Answer:** C₂H₄ (ethylene) is produced.

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