See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the starting material: Phthalic anhydride is the cyclic anhydride formed from phthalic acid (benzene-1,2-dicarboxylic acid). Step 2 – Reaction with PCl5 to give (A): PCl5 converts the cyclic anhydride to the corresponding diacid dichloride, i.e., phthaloyl dichloride (benzene-1,2-dicarbonyl dichloride, –COCl groups at both ortho positions). Step 3 – Reaction with LiAlH4 to give (B): LiAlH4 reduces both acid chloride groups fully to primary alcohols. Product (B) is o-xylylene diol, i.e., benzene-1,2-dimethanol (each –COCl becomes –CH2OH). Step 4 – Reaction with PCC to give (C): PCC (pyridinium chlorochromate) is a mild oxidant that oxidises primary alcohols to aldehydes without over-oxidising. Both –CH2OH groups are oxidised to –CHO. Product (C) is phthalaldehyde (benzene-1,2-dicarbaldehyde). Step 5 – Reaction with HO-/Delta to give (D): Treatment of an aromatic dialdehyde (phthalaldehyde) with aqueous base (HO-) and heat promotes the Cannizzaro reaction (since there are no alpha-H atoms on the aldehyde carbons relative to the ring context; actually benzaldehyde-type aldehydes undergo Cannizzaro). In the intramolecular Cannizzaro reaction of phthalaldehyde, one aldehyde is oxidised to a carboxylate (–CO2-) and the other is reduced to a primary alcohol (–CH2OH). Product (D) is therefore the salt of 2-(hydroxymethyl)benzoic acid, bearing one –CH2OH and one –CO2- group on the benzene ring. This matches option (a): benzene ring with –CH2OH and –CO2- substituents. Why other options fail: (b) Both groups as –CO2- would require oxidation of both aldehydes, which does not occur in Cannizzaro (one oxidised, one reduced). (c) Both groups as –CH2OH would mean both are reduced, also not Cannizzaro. (d) Both groups as –CH2O- (alkoxide) is chemically unreasonable under these conditions. Therefore, the correct answer is A.