Question

Identify Z in the given sequence of reactions:

• A. Aromatic ketone
• B. Arene
• C. Aromatic aldehyde
• D. Aromatic carboxylic acid

Hint:

KMnO$_4$ cleaves alkenes to carboxylic acids, SOCl$_2$ forms acyl chlorides, Friedel–Crafts adds substituents to benzene, and Clemmensen reduction removes carbonyl oxygen.

Answer 1

The Correct Option is B

Concept: This is a multistep organic reaction sequence involving oxidation, acyl chloride formation, Friedel–Crafts acylation, and Clemmensen reduction.

Step 1: Oxidative cleavage of 2-butene.
Hot acidic $\text{KMnO}_4$ cleaves the double bond of 2-butene to form carboxylic acids: \[ \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3 \rightarrow 2\,\text{CH}_3\text{COOH} \] Thus, $X$ is ethanoic acid.

Step 2: Formation of acyl chloride.
Ethanoic acid reacts with $\text{SOCl}_2$ to form acetyl chloride: \[ \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COCl} \] Thus, $Y$ is acetyl chloride.

Step 3: Friedel–Crafts acylation.
Acetyl chloride reacts with benzene in presence of anhydrous $\text{AlCl}_3$ to form acetophenone: \[ \text{C}_6\text{H}_6 \rightarrow \text{C}_6\text{H}_5\text{COCH}_3 \]

Step 4: Clemmensen reduction.
Acetophenone is reduced using $\text{Zn-Hg}/\text{HCl}$ to ethylbenzene: \[ \text{C}_6\text{H}_5\text{COCH}_3 \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{CH}_3 \] Conclusion:
The final product $Z$ is ethylbenzene, which is an alkyl-substituted benzene ring. Hence, it is classified as an arene.