See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the transformation in step A: The starting material is limonene, which contains two double bonds – a trisubstituted endocyclic alkene (more electron-rich) and a terminal exocyclic alkene (less electron-rich). Reagent A converts only the trisubstituted endocyclic double bond to an epoxide while leaving the terminal vinyl (isopropenyl) group intact. A peracid (RCO3H, e.g., mCPBA) is the classic reagent for epoxidation of alkenes. Peracids preferentially epoxidize the more electron-rich, more substituted double bond, so the endocyclic trisubstituted alkene is selectively epoxidized, giving the monoepoxide with the isopropenyl group unchanged. This confirms A = RCO3H. Step 2 – Identify the transformation in step B: The monoepoxide intermediate still contains a terminal alkene (the isopropenyl =CH2 group). In the product, this terminal alkene has been converted to a methyl ketone (C=O on a methyl-bearing carbon), while the epoxide ring is preserved. Ozonolysis (O3) of a terminal alkene (CH2=C<) cleaves the double bond oxidatively. For the isopropenyl group CH2=C(CH3)–, ozonolysis followed by reductive or oxidative workup would cleave it; under standard ozonolysis with oxidative workup (or simply O3/Zn or O3/Me2S), the terminal =CH2 gives formaldehyde and the internal carbon becomes a ketone carbonyl. The product shows a methyl ketone (–C(=O)–CH3) at C4, consistent with ozonolysis of CH2=C(CH3)– which yields formaldehyde + a ketone at that carbon. The epoxide is not affected by O3 under these conditions. Therefore B = O3. Step 3 – Evaluate other options: Option (a) B = H2O2 would not cleave a double bond to give a ketone selectively. Option (b) B = HIO4 is used for diol cleavage, not direct alkene cleavage, and would require a diol intermediate. Option (d) reverses A and B: O3 first would attack both double bonds non-selectively and would not give a clean monoepoxide. Only option (c) A = RCO3H, B = O3 is consistent with both transformations. Therefore, the correct answer is C.