See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The Cannizzaro reaction and intramolecular cyclization. When an aldehyde lacking alpha-hydrogens is treated with concentrated NaOH, it undergoes the Cannizzaro reaction (disproportionation): one aldehyde molecule is oxidized to a carboxylate and another is reduced to an alcohol. For o-phthalaldehyde (option c), both CHO groups lack alpha-hydrogens relative to the aromatic ring. Under concentrated NaOH, an intramolecular Cannizzaro reaction occurs: one CHO is oxidized to COO- (carboxylate) and the other CHO is reduced to CH2OH (primary alcohol), all within the same molecule. This gives the sodium salt of 2-(hydroxymethyl)benzoic acid. Upon acidification, the free acid 2-(hydroxymethyl)benzoic acid is liberated, which spontaneously undergoes intramolecular esterification (lactonization) between the carboxylic acid and the adjacent hydroxymethyl group (-CH2OH) to form the five-membered lactone ring: phthalide (isobenzofuran-1(3H)-one). Why other options fail: - Option (a): Methyl ester with adjacent carboxylic acid - NaOH would saponify the ester to give phthalic acid dianion; acidification gives phthalic acid, not phthalide directly under these mild conditions. - Option (b): 2-carboxybenzaldehyde - NaOH could cause Cannizzaro on the aldehyde, but the carboxylic acid is already present; the aldehyde would be reduced to alcohol giving 2-(hydroxymethyl)benzoic acid, which upon acidification would give phthalide. However, option (b) could also work, but the intramolecular Cannizzaro from o-phthalaldehyde (c) is the classic and more straightforward route. Actually, re-examining: with option (b), the aldehyde could undergo intermolecular Cannizzaro (no intramolecular partner aldehyde), giving a mixture. With option (c), the intramolecular Cannizzaro is highly favored and clean, making it the correct answer. - Option (d): Phthalic acid - NaOH gives phthalate salt; acidification gives phthalic acid, which forms phthalic anhydride (six-membered transition state is less favored for five-membered lactone) - actually phthalic acid forms phthalic anhydride (a six-membered anhydride) not phthalide. The intramolecular Cannizzaro reaction of o-phthalaldehyde (option c) with concentrated NaOH, followed by acidification, cleanly produces 2-(hydroxymethyl)benzoic acid which spontaneously lactonizes to phthalide. Therefore, the correct answer is C.