Considering the above chemical reaction, identify the product X:

Step 1: Identify the Starting Material

4-methyl anisole = para-methyl anisole = 1-methoxy-4-methylbenzene ($\ce{CH3-C6H4-OCH3}$, methyl and methoxy groups para to each other).

Step 2: Alkaline KMnO4 Oxidation

Hot alkaline $\ce{KMnO4}$ oxidizes alkyl side chains on aromatic rings to $\ce{-COOH}$: The $\ce{-CH3}$ group is oxidized to $\ce{-COOH}$. The $\ce{-OCH3}$ group is not oxidized under these conditions.

Product X = 4-methoxybenzoic acid (anisic acid): $\ce{HOOC-C6H4-OCH3}$ (para).

Key Points:

• Alkaline $\ce{KMnO4}$: oxidizes $\ce{-CH3}$ (or any alkyl) on benzene ring to $\ce{-COOH}$
• $\ce{-OCH3}$ (methoxy ether) is NOT oxidized under these conditions
• Both $\ce{-CH3}$ and $\ce{-C2H5}$ etc. on ring all give $\ce{-COOH}$