See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Phthalamide is the cyclic imide derived from phthalic anhydride and ammonia, i.e., isoindole-1,3(2H)-dione. When treated with NaOH (aqueous base), the following occurs: Step 1: Phthalamide (cyclic imide) undergoes base hydrolysis. NaOH acts as a nucleophile/base to hydrolyze the imide C-N bonds. Step 2: The cyclic imide has two amide (C-N) bonds. Under basic conditions, NaOH hydrolyzes both C-N bonds of the imide ring. This ring-opening hydrolysis converts the cyclic imide into the dicarboxylate (phthalate) salt plus ammonia. Step 3: The product is the disodium salt of phthalic acid (sodium phthalate), where both carboxylic acid groups are present as sodium carboxylate (-COONa) groups on the ortho positions of the benzene ring. However, looking at option (c), it shows a benzene ring fused to a six-membered ring with two ONa groups and two C=O groups - this represents the disodium phthalate where both carboxylate groups have been converted to -COONa (i.e., the ring has opened and both amide nitrogens replaced by ONa oxygens, giving the ortho-dicarboxylate disodium salt). Step 4: Why other options fail: - Option (a): Shows an ester-linked dimer, not the product of simple NaOH hydrolysis of phthalamide. - Option (b): Shows a diol (reduction product), not relevant to base hydrolysis of an imide. - Option (d): Shows a dilactone, which would require acidic conditions or specific esterification, not NaOH treatment. Step 5: The major product of treating phthalamide with NaOH is the disodium phthalate salt - the ring-opened hydrolysis product with two -COONa groups, corresponding to option (c). Therefore, the correct answer is C.