See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: The reaction involves base-catalyzed isomerization of an alpha-hydroxy aldehyde. Compound (A) is Ph–CH(OH)–CHO (2-phenyl-2-hydroxyacetaldehyde), which is an alpha-hydroxy aldehyde. Under basic conditions (HO⁻), this undergoes an intramolecular 1,2-hydride shift (analogous to a Meerwein–Ponndorf–Verley or Cannizzaro-type 1,2-shift) known as the Lobry de Bruyn–Alberda van Ekenstein rearrangement or, more precisely here, a simple base-catalyzed tautomeric/retro-aldol-aldol type isomerization. Step-by-step reasoning: 1. Compound (A) is Ph–CH(OH)–CHO: it has a hydroxyl group alpha to the aldehyde carbonyl. 2. Under basic conditions, the base abstracts the alpha-hydrogen (the H on the carbon bearing OH), generating an enolate or facilitating a 1,2-hydride shift. 3. A 1,2-hydride shift from the carbon bearing OH to the adjacent carbonyl carbon converts the aldehyde function to an alcohol and oxidizes the alcohol carbon to a ketone carbonyl: Ph–CH(OH)–CHO → Ph–C(=O)–CH₂OH. 4. This gives Ph–C(=O)–CH₂–OH, which is phenacyl alcohol (2-hydroxyacetophenone), option (c). 5. (A) (Ph–CH(OH)–CHO) and (B) (Ph–C(=O)–CH₂OH) are constitutional isomers with the same molecular formula C₈H₈O₂, consistent with the problem statement. 6. Why other options fail: - (a) Ph–CH₂–COOH is phenylacetic acid; it is an isomer but requires oxidation state change not achievable by simple base-catalyzed isomerization. - (b) Ph–C(=O)–OCH₃ is methyl benzoate; different connectivity and requires transesterification conditions. - (d) H–C(=O)–CH₂–O–Ph is phenoxyacetaldehyde; requires C–O bond formation, not a simple isomerization. 7. Therefore (B) = Ph–C(=O)–CH₂–OH, which is option (c). Therefore, the correct answer is C.