See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: KMnO4 in hot aqueous (acidic or basic) conditions is a strong oxidizing agent that can oxidize aldehydes to carboxylic acids and also hydrolyze acetals under acidic workup (H3O+). Step 1 - Identify the starting material: The starting material is piperonal (3,4-methylenedioxybenzaldehyde). It contains a methylenedioxy acetal group (-O-CH2-O-) bridging the 3 and 4 positions of the benzene ring, and an aldehyde (-CHO) at the 1 position. Step 2 - Reaction with KMnO4/H2O/heat: Hot acidic KMnO4 is a powerful oxidant. It oxidizes the aldehyde group (-CHO) to a carboxylic acid (-COOH). Additionally, under hot acidic aqueous conditions (H3O+), the methylenedioxy acetal group (-OCH2O-) is hydrolyzed. The acetal -OCH2O- hydrolyzes to give two free phenolic -OH groups (a catechol unit), releasing formaldehyde (HCHO), which is then further oxidized by KMnO4 to CO2 and water. Step 3 - Product formed: The benzene ring now bears two free -OH groups at positions 3 and 4 (catechol) and a -COOH group at position 1, giving 3,4-dihydroxybenzoic acid (protocatechuic acid), which is option (c). Step 4 - Why other options fail: - Option (a): This would be the product if only the aldehyde were oxidized and the methylenedioxy ring remained intact, but under hot acidic KMnO4 conditions, the acetal is also hydrolyzed. - Option (b): This shows incomplete oxidation products (both -OH and -CH2OH present) which is not consistent with strong oxidizing conditions. - Option (d): This shows an epoxide and retention of the aldehyde, which is inconsistent with the strong oxidizing and acidic conditions. Therefore, the correct answer is C.