See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the target compound and work backward. The final product is isobenzofuran-1(3H)-one (also called phthalide or 1,3-dihydroisobenzofuran-1-one), which is a bicyclic compound consisting of a benzene ring fused to a five-membered lactone (cyclic ester) ring. Step 2 - Identify compound Y. Y is formed from X by treatment with H+ and heat (acid-catalyzed lactonization/cyclization). For Y to cyclize to phthalide under acidic conditions, Y must be 2-(hydroxymethyl)benzoic acid (or its equivalent) — a compound with a carboxylic acid and a hydroxymethyl group ortho to each other on a benzene ring. Under H+/heat, the OH of the -CH2OH attacks the carboxylic acid to form the lactone ring. Step 3 - Identify compound X. X is converted to Y by treatment with concentrated KOH. If X is ortho-bis(hydroxymethyl)benzene [i.e., 1,2-bis(hydroxymethyl)benzene], KOH could perform a Cannizzaro-type or oxidation-reduction reaction. However, looking more carefully: X must be a compound that upon treatment with conc. KOH gives a compound with both a -COOH (or carboxylate) and a -CH2OH group ortho on a benzene ring. Actually, working from option (a): ortho-bis(hydroxymethyl)cyclohexadiene (which is essentially the cyclohexadiene (non-aromatic) form with two CH2OH groups ortho to each other). Treatment with 2 moles of PCC in CH2Cl2 oxidizes both primary alcohols: one -CH2OH -> -CHO and the other -CH2OH -> -CHO, giving ortho-phthalaldehyde (benzene-1,2-dicarbaldehyde). Wait — but PCC oxidizes primary alcohols to aldehydes. Re-examining: option (a) shows ortho-bis(hydroxymethyl)benzene reacted with 2 moles PCC. 2 moles PCC would oxidize both -CH2OH groups to -CHO groups, giving o-phthalaldehyde (2-formylbenzaldehyde / benzene-1,2-dicarbaldehyde). This is compound X. Step 4 - Confirm X -> Y -> product. X = o-phthalaldehyde. Treatment with conc. KOH: a Cannizzaro reaction occurs between the two aldehyde groups intramolecularly (or intermolecularly). The intramolecular Cannizzaro reaction of o-phthalaldehyde with KOH gives the potassium salt of 2-(hydroxymethyl)benzoate, i.e., Y = 2-(hydroxymethyl)benzoic acid after acidification. Then Y under H+/heat undergoes lactonization to give phthalide. This is consistent. Step 5 - Why other options fail. (b) 2 moles DIBAL-H on a diester/ketone substrate would give different reduction products, not leading cleanly to X = o-phthalaldehyde. (c) 2H2/Pd-BaSO4 (Rosenmund conditions) on diacid chloride would give dialdehyde, but the substrate shown has an acid chloride and a -CH2COCl (not a simple acyl chloride), giving a different product. (d) CH3MgBr on a ketone gives a tertiary alcohol, not an aldehyde needed for X. Therefore, the correct answer is A.