JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

Ozonolysis (followed by reductive workup with Zn/H_2O) cleaves C=C bonds and converts each alkene carbon into a carbonyl. We can reverse-engineer the parent alkene from the fragments obtained. Analysing the products: HCHO (2 moles ): Two equivalents of formaldehyde indicate the presence of two terminal or exocyclic =CH_2 groups in the parent molecule. CH_3-CO-CO-CH_3 (biacetyl): To get this discrete fragment, the parent alkene must contain the structural unit -C(CH_3)=C(CH_3)-, i.e., an endocyclic double bond in which both sp^2 carbons bear methyl substituents. H-CO-CO-CHO (mesoxaldehyde): This arises from a fragment like =CH-C(=CH_2)-CH=, where cleavage of the adjacent C=C bonds generates three consecutive carbonyl groups (and also contributes to the formaldehyde count from the exocyclic =CH_2). Evaluating the options: The most diagnostic product is biacetyl, so the parent structure must contain the -C(CH_3)=C(CH_3)- unit. Option (1): One methyl group is on the endocyclic double bond and the other methyl is on a separate sp^3 carbon. Cleavage cannot give biacetyl. Option (2): Only a single methyl group is present, so biacetyl cannot be formed. Option (4): The methyl groups lie on opposite ends of the ring (across a 1,4-diene-type arrangement). Ozonolysis would place these methyls in different fragments, so biacetyl cannot be formed. Option (3): The left side of the ring contains an endocyclic double bond in which both carbons carry methyl groups, giving exactly the required -C(CH_3)=C(CH_3)- arrangement. Cleavage of this bond together with the adjacent ring double bonds releases a discrete CH_3-CO-CO-CH_3 fragment. The right side of the molecule carries exocyclic =CH_2 groups, which on cleavage produce the two moles of HCHO, while the intervening carbons give rise to H-CO-CO-CHO. Hence, the correct structure that yields all three ozonolysis products is option (3).