Step 1: Identify the first step on toluene. Toluene reacts with acetyl chloride $CH_3COCl$ in the presence of anhydrous $AlCl_3$, which is a Friedel-Crafts acylation. Step 2: Decide the position of attack. The $-CH_3$ group is ortho and para directing, so the acetyl group adds mainly at the para position, giving para-methyl acetophenone, $p\text{-}CH_3C_6H_4COCH_3$. Step 3: Oxidise with alkaline $KMnO_4$. Hot alkaline $KMnO_4$ is a strong oxidant that converts benzylic side chains into $-COOH$ groups. Both the methyl side chain and the acetyl side chain are oxidised to carboxylic acid groups. Step 4: Write the oxidised intermediate. \[ p\text{-}CH_3C_6H_4COCH_3 \rightarrow p\text{-}HOOC\text{-}C_6H_4\text{-}COOH \] giving two $-COOH$ groups in the para arrangement. Step 5: Acidify. Dilute $H_2SO_4$ protonates the carboxylate salt to release the free dicarboxylic acid. Step 6: Name the product. The product is benzene-1,4-dicarboxylic acid, known as terephthalic acid, the answer marked option 4. \[ \boxed{\text{Terephthalic acid}} \]
