See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The Baeyer-Villiger (BV) oxidation converts a ketone into an ester (or lactone) by insertion of an oxygen atom between the carbonyl carbon and one of its adjacent carbons. The peracid (here MMPP) attacks the carbonyl to form a Criegee intermediate, then the better migrating group migrates to oxygen. Step 1 – Identify the substrate. The ketone is 2-R-cyclohexan-1-one, an asymmetric cyclic ketone. The carbonyl is flanked by two different carbon groups: (i) the ring CH2 on one side and (ii) the ring CH(R) on the other side. Step 2 – Determine migratory aptitude. In the Criegee intermediate, the group that migrates is the one that can better stabilize partial positive charge in the transition state. Migratory aptitude order: tertiary C > secondary C > primary C > methyl. The carbon bearing R is secondary (more substituted) compared to the unsubstituted ring CH2 (also secondary but without the extra R substituent making it less substituted). More precisely, the C bearing R is a secondary carbon with an alkyl substituent R, making it migrate preferentially over the simple –CH2– carbon. Step 3 – Determine ring expansion outcome. In cyclic ketones undergoing BV, oxygen is inserted between the carbonyl and the migrating carbon, expanding the ring by one atom. Starting from a 6-membered ring (cyclohexanone), migration of the C(R) carbon inserts O between C1(carbonyl) and C2(R), producing a 7-membered lactone (oxepan-2-one skeleton) with R remaining on the carbon now alpha to the ring oxygen (i.e., R is on the carbon adjacent to the –O– in the 7-membered ring). Step 4 – Match to options. Option (b) shows a 7-membered lactone ring (oxepan-2-one) with R on the carbon adjacent to the ring oxygen, which is exactly the product of migration of the R-bearing carbon. This is the major product. Why other options fail: - Option (a) shows a 7-membered lactone with an exocyclic O-R, which is not the BV product structure. - Option (c) shows a 6-membered lactone, which would result from no ring expansion or migration of the less-substituted carbon, and would be the minor product. - Option (d) shows no oxidation/rearrangement product, just addition of OH, which is not the BV outcome. Therefore, the correct answer is B.