See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Concept: The Beckmann rearrangement of an oxime under acidic conditions (H2SO4) proceeds via migration of the anti-group relative to the leaving hydroxyl. The key mechanistic feature for isotope-labeling questions is tracking the origin of the carbonyl oxygen in the amide product. Step 1 - Identify the starting material: The starting material is diphenyl ketone oxime (benzophenone oxime) where the oxygen of the N-OH group bears an O-18 label (written as N=C with O-18-H on nitrogen). Step 2 - Beckmann rearrangement mechanism: Under H2SO4, the oxime undergoes protonation of the hydroxyl oxygen (O-18), generating a good leaving group (H2O-18 departs). This is the critical step: the O-18 labeled water leaves as H2O-18 (consistent with the H2O-18 shown as reagent/byproduct in the reaction arrow, confirming O-18 leaves with the departing water). Step 3 - Nitrilium ion and water attack: After loss of H2O-18 (the original oxime oxygen), a phenyl group migrates to give a nitrilium-type intermediate (Ph-C≡N+-Ph). Water from the solvent (H2O-18 is listed as reagent, but the key point is what water attacks the nitrilium) then attacks. Since the reaction medium contains H2O-18 as solvent, the nucleophilic water attacking the electrophilic carbon of the nitrilium ion is H2O-18, incorporating O-18 into the carbonyl of the amide product. Step 4 - Product formation: The amide formed is Ph-NH-C(=O-18)-Ph, i.e., the carbonyl oxygen in the product carries the O-18 label because it came from H2O-18 (the labeled water in the reaction medium that attacked the nitrilium intermediate). Step 5 - Why other options fail: - (a) Ph-NH-C(=O)-Ph: This would be correct if the carbonyl oxygen were unlabeled, but since H2O-18 is the nucleophile attacking the nitrilium, the product oxygen is O-18 labeled. Option (a) is wrong. - (c) (Ph)2-C(OH)-NH2: This is not the Beckmann rearrangement product; it would require no migration, which is incorrect. - (d) Ph-CH2-NH-Ph: This is a reduction product (reductive amination type), not from Beckmann rearrangement under these conditions. Therefore, the correct answer is B.