See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Reaction of cyclopentadiene with HCl: Cyclopentadiene (bicyclic diene, actually the monomer with one double bond pair) undergoes HCl addition. Cyclopentadiene has two double bonds in a five-membered ring. Addition of HCl follows Markovnikov's rule and gives the allylic chloride product (A): 5-chlorocyclopent-1-ene (3-chlorocyclopentene), where Cl adds to give an allylic/vinylic chloride. Specifically, HCl adds across one double bond of cyclopenta-1,3-diene to give 5-chlorocyclopent-1-ene (A) via 1,4-addition (allylic addition), placing Cl at C5 (the sp3 carbon) with a remaining double bond between C1 and C2. Step 2 - Reaction of (A) with NaHCO3/H2O to give (B) with molecular formula C5H8O: NaHCO3/H2O is a mild basic aqueous condition that can hydrolyze an allylic chloride to an allylic alcohol. 5-chlorocyclopent-1-ene hydrolyzes to give cyclopent-2-en-1-ol (B), which has the molecular formula C5H8O. This confirms B is an allylic alcohol: cyclopent-2-en-1-ol. Step 3 - Oxidation of (B) with Na2Cr2O7 to give (C): Na2Cr2O7 is an oxidizing agent. Oxidation of the secondary allylic alcohol cyclopent-2-en-1-ol gives the corresponding enone, cyclopent-2-en-1-one (2-cyclopentenone). This is a cyclopentenone ring with the double bond between C2 and C3 and a ketone at C1. Step 4 - Identifying the product among options: The end product (C) is cyclopent-2-en-1-one, which matches option (b): a cyclopentenone with one double bond conjugated with the carbonyl. Why other options fail: - (a) shows a cyclopentenone with double bond at a different position (non-conjugated enone), not the direct oxidation product. - (c) shows the alcohol (B) itself, not the oxidized product (C). - (d) shows a dienone or a structure with extended conjugation, which is not the product of simple alcohol oxidation. Therefore, the correct answer is B.