Question

Find the product Z in the series of reactions. \[ \text{Phenol} \xrightarrow{\text{Zn dust}} X \xrightarrow{\text{CH}_3\text{Cl / anhyd. AlCl}_3} Y \xrightarrow{\text{alkaline KMnO}_4} Z \]

• A. benzaldehyde
• B. benzene
• C. benzoic acid
• D. benzophenone

Hint:

Any alkyl benzene on oxidation with KMnO$_4$ gives benzoic acid.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a multi-step organic synthesis involving reduction, Friedel-Crafts alkylation, and side-chain oxidation of an aromatic ring.
Step 3: Detailed Explanation:
1. Step 1 (Reduction): Phenol reacts with Zinc dust. The oxygen is removed as $ZnO$, leaving benzene.
\[ C_{6}H_{5}OH + Zn \rightarrow C_{6}H_{6} (X) + ZnO \]
Product $X$ is Benzene.
2. Step 2 (Friedel-Crafts Alkylation): Benzene reacts with methyl chloride in the presence of anhydrous $AlCl_{3}$.
\[ C_{6}H_{6} + CH_{3}Cl \xrightarrow{AlCl_{3}} C_{6}H_{5}CH_{3} (Y) + HCl \]
Product $Y$ is Toluene.
3. Step 3 (Oxidation): Toluene is treated with alkaline $KMnO_{4}$. Strong oxidizing agents oxidize any alkyl side chain on a benzene ring (provided there is at least one benzylic hydrogen) completely to a carboxylic acid group.
\[ C_{6}H_{5}CH_{3} \xrightarrow{KMnO_{4}/OH^{-}} C_{6}H_{5}COOH (Z) \]
Product $Z$ is Benzoic acid.
Step 4: Final Answer:
The final product $Z$ is benzoic acid.