The major product of the following reaction is:

Step 1: Identify Starting Material

From the image: an ethylbenzene with $\ce{-CH2CH3}$ group. Treated with alkaline $\ce{KMnO4}$ then $\ce{H3O^+}$.

Step 2: Alkaline KMnO4 Oxidation of Alkyl Side Chain

Hot alkaline $\ce{KMnO4}$ oxidizes ANY alkyl side chain on benzene ring to $\ce{-COOH}$, regardless of chain length.

Product = benzoic acid ($\ce{C6H5COOH}$).

Key Points:

• Alkaline $\ce{KMnO4}$: all alkyl chains on benzene $\rightarrow$ $\ce{-COOH}$ (the carbon attached to ring becomes the carboxyl carbon)
• Whether the side chain is $\ce{-CH3}$, $\ce{-CH2CH3}$, or $\ce{-C(CH3)_3}$, the product is always the corresponding benzoic acid
• Exception: if the benzylic carbon has NO H (e.g., $\ce{-C(CH3)_3}$), no oxidation occurs