See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the compound: All three structures represent 1,2-dimethylcyclopentane. There are two stereocenters (C1 and C2), giving possible stereoisomers: cis (meso or racemic pair) and trans. Step 2 - Analyze Structure (I): Both CH3 groups are on wedge bonds (both pointing up/toward viewer) and both H atoms are on dash bonds (pointing down/away). This means both methyl groups are on the same face — this is the cis-1,2-dimethylcyclopentane. Since C1 and C2 each have CH3 up and H down, C1 is (1R,2S) or equivalent — actually, for cis-1,2-dimethylcyclopentane the two stereocenters have opposite configurations making it a meso compound. Structure (I) is the meso (cis) form. Step 3 - Analyze Structure (II): At one carbon: H is on wedge (up) and CH3 is on dash (down); at the adjacent carbon: CH3 is on wedge (up) and H is on dash (down). This gives one methyl up and one methyl down — trans-1,2-dimethylcyclopentane. This is one enantiomer of the trans pair, e.g., (1R,2R). Step 4 - Analyze Structure (III): At one carbon: CH3 is on wedge (up) and H is on dash (down); at the adjacent carbon: H is on wedge (up) and CH3 is on dash (down). This also gives one methyl up and one methyl down — trans-1,2-dimethylcyclopentane. This is the other enantiomer of the trans pair, e.g., (1S,2S). Step 5 - Identify enantiomeric relationship: Structures II and III are both trans-1,2-dimethylcyclopentane but are non-superimposable mirror images of each other — they are enantiomers. Structure I is the meso compound (cis form), which is achiral and is not an enantiomer of II or III (it is a diastereomer of both). Step 6 - Why other options fail: - (a) I and II: I is meso/cis, II is trans — they are diastereomers, not enantiomers. - (b) I and III: Same reasoning — diastereomers. - (d) I, II and III: I is not an enantiomer of II or III. Therefore, the correct answer is C.