See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The reaction converts a terminal alkene (C=CH2) into a ketone (C=O), specifically transforming the exocyclic methylenyl group -C(CH3)=CH2 into an acetyl-type carbonyl -C(CH3)=O. This means the =CH2 group is oxidatively cleaved and replaced by =O, while the substituted carbon retains its methyl group and becomes a carbonyl carbon. Step 1: Identify the transformation. The starting material has an exocyclic alkene: -C(CH3)=CH2. The product has a ketone: -C(CH3)=O. This is an oxidative cleavage of a C=C double bond where the terminal =CH2 is converted to =O (formaldehyde would be the other fragment), and the internal carbon bearing the methyl group becomes a carbonyl (ketone/aldehyde carbon). Step 2: Ozonolysis with reductive workup (O3/Zn, H2O). Ozonolysis cleaves C=C double bonds. For a trisubstituted or 1,1-disubstituted alkene R2C=CH2, ozonolysis with reductive workup (Zn/H2O or Zn/AcOH) gives R2C=O (ketone) + H2C=O (formaldehyde). This exactly matches the observed transformation: -C(CH3)=CH2 → -C(CH3)=O + CH2O. The lactone ring is stable to ozonolysis conditions. Step 3: Evaluate other options. - (b) HIO4: Periodate cleaves 1,2-diols (vicinal diols), not simple alkenes. The substrate has no diol, so HIO4 would not react. - (c) CrO3: This is an oxidizing agent for alcohols (to ketones/aldehydes) or for allylic/benzylic positions. It does not cleave isolated alkene double bonds to give ketones directly. - (d) Cold dil. KMnO4: This reagent performs syn-dihydroxylation of alkenes (Baeyer's reagent), giving a 1,2-diol, not a ketone. It does not cleave the double bond under cold dilute conditions. Step 4: Conclusion. Only O3/Zn(H2O) (ozonolysis with reductive workup) cleanly converts a terminal alkene R2C=CH2 into the corresponding ketone R2C=O plus formaldehyde, matching the product shown. Therefore, the correct answer is A.