See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The starting material is a 1,2,4-trioxolane (ozonide), the product of ozonolysis of an alkene. In ozonolysis, an alkene R-CH=CR2 (where one carbon bears H and R', the other bears two R groups) is converted to an ozonide via the Criegee mechanism. The ozonide is a five-membered ring containing one ether oxygen and one peroxide (O-O) linkage. Reductive workup with dimethyl sulfide (Me2S) cleaves the ozonide, oxidizing Me2S to Me2S=O (DMSO) while reducing the ozonide to the corresponding carbonyl fragments. Reasoning: The ozonide structure shown has one carbon bearing H and R' (this carbon came from the R'CH= end of the original alkene) and one carbon bearing two R groups (this carbon came from the =CR2 end). Upon reductive cleavage with Me2S: - The carbon bearing H and R' gives an aldehyde: R'-CHO - The carbon bearing two R groups gives a ketone: R-C(=O)-R Both carbonyl compounds are produced simultaneously as Product A. Why other options fail: - Option (a) alone (R-C(=O)-R, ketone) is incomplete because the aldehyde fragment is also formed. - Option (b) alone (R'-CHO, aldehyde) is incomplete because the ketone fragment is also formed. - Option (c) (R-CO2H, carboxylic acid) would only form under oxidative workup conditions (e.g., H2O2, O3/H2O), not with the reductive Me2S workup. - Option (d) correctly states both (a) and (b) are formed, which matches the ozonolysis reductive workup mechanism. Therefore, the correct answer is D.