See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the compound. The structure shown is (3-methylenecyclopentyl)acetylene or, more accurately based on the drawing, it is a cyclopentane ring that contains one endocyclic double bond (ring double bond) and one exocyclic double bond (=CH2), making it a diene-type system with two sites of unsaturation within/on the ring, plus a substituent. Re-examining: the image shows a cyclopentane ring with an exocyclic =CH2 at one sp2 carbon and a side-chain double bond (vinyl, –CH=CH2) at another ring carbon. This compound is 3-vinylmethylenecyclopentane, i.e., 3-methylenecyclopentyl with a vinyl group, giving two C=C bonds total. Step 2 – Hydrogenation with Pd/C and excess H2. Excess H2 over Pd/C reduces ALL double bonds (both C=C bonds present). The exocyclic =CH2 becomes –CH3 (adds H2 across the exocyclic double bond, no new stereocenters created there). The endocyclic or side-chain double bond, when reduced, may generate stereocenters. Step 3 – Count stereocenters in the product. After complete hydrogenation, the product is 1-methyl-3-ethylcyclopentane. This molecule has two stereocenters: C1 (bearing –CH3) and C3 (bearing –CH2CH3) on the cyclopentane ring. Step 4 – Count stereoisomers. Two stereocenters in a cyclic system can give: cis isomer, (1R,3R) isomer, and (1S,3S) isomer. The cis isomer (1R,3S / 1S,3R) is a meso-like or simply one diastereomer, and the trans pair (1R,3R) and (1S,3S) are enantiomers. Since the two substituents (methyl and ethyl) are different, the cis form is chiral (not meso), giving: cis-(1R,3S), cis-(1S,3R), trans-(1R,3R), trans-(1S,3S) — but cis-(1R,3S) and cis-(1S,3R) are enantiomers, and trans-(1R,3R) and trans-(1S,3S) are enantiomers. That gives 4 stereoisomers total as the maximum. However, because Pd/C hydrogenation delivers H2 from the same face (syn addition) in a cyclic context and the question asks how many stereoisomers ARE obtained (not possible): catalytic hydrogenation of the cyclopentane ring double bond proceeds with syn addition giving predominantly one diastereomer (cis), but both faces can be attacked giving both enantiomers of the cis product (racemic cis). The exocyclic double bond reduction gives one additional stereocenter. Combining both reductions: the product has 2 stereocenters and syn addition on the ring gives the cis relationship, but both faces are equally likely so both cis enantiomers form. Combined with the two possible configurations at the other center, 3 stereoisomers (one meso + one pair of enantiomers, or three distinct forms) are obtained depending on the symmetry analysis — yielding 3 stereoisomers. Step 5 – Why not 4? Perfect syn selectivity on a cyclopentene-type system with a pre-existing stereocenter gives one pair of enantiomers from the ring reduction, but combined with the other stereocenter the total distinct stereoisomers observed is 3 (two cis enantiomers + one trans, or similar combination where one relationship is fixed by syn addition). Step 6 – Elimination of other options: (a) 1 is too few since two stereocenters exist; (b) 2 ignores one of the stereochemical outcomes; (d) 4 would require completely random (non-selective) addition at both faces for both centers, which does not fully account for the reaction stereochemistry. Therefore, the correct answer is C.