Question

Identify the product A in the following reaction: 
\(\text{CH\(_3\)-CH\(_2\)-CH\(_2\)Cl + 2 KOH\(_{aq}\) $\xrightarrow{\Delta}$ A + 2 KCl + H\(_2\)O} \)
 

• A. CH\(_3\)-CH\(_2\)-CH\(_3\)OH
• B. CH\(_3\)-CH\(_2\)-C=O
• C. CH\(_3\)-CH\(_2\)-CH\(_3\)Cl
• D. CH\(_3\)-CH\(_2\)-C\(_2\)H\(_5\)

Hint:

In elimination reactions with alcohols, the halogen atom is replaced by a hydroxyl group. Heating with KOH leads to the formation of alcohols.

Answer 1

The Correct Option is D

Step 1: Analyzing the reaction.
This is a reaction between a primary alkyl halide (CH\(_3\)-CH\(_2\)-CH\(_2\)Cl) and aqueous potassium hydroxide (KOH) under heat. This results in an elimination reaction, where the chloride (Cl\(^-\)) is replaced by a hydroxyl group, forming an alcohol. The product is butane.

Step 2: Conclusion.
The correct product is (D) CH\(_3\)-CH\(_2\)-C\(_2\)H\(_5\).