See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 – Analyze molecule (a): The cyclohexane ring has two adjacent (1,2) stereocenters, both with H shown on the same face (both bold wedges upward). The substituents are CO2CH2CH2OH at C1 and CO2H at C2. Since both H atoms are on the same face, the two substituents are on the same face as well — this is the cis isomer. Wait, re-examining: the H atoms are drawn as bold wedges (coming toward viewer) at both C1 and C2, meaning the H's are cis to each other, so the larger groups are also on the same side — cis-1,2-disubstituted cyclohexane with two different substituents (CO2CH2CH2OH ≠ CO2H). A cis-1,2-disubstituted cyclohexane with two DIFFERENT groups has two non-superimposable stereocenters and no internal mirror plane, making it chiral. It has 2 chiral centers (even number). Therefore (a) matches (p) Chiral and (s) Even number of chiral centers → a-p,s. Step 2 – Analyze molecule (b): This is a symmetric dimer: two identical trans-1,2-disubstituted cyclohexane units connected by an ethylene glycol diester bridge. Both rings show the same stereochemical pattern (both H's on same face at each ring). The molecule has an internal mirror plane (C2 symmetry or mirror plane bisecting the linker), making the two chiral centers cancel each other — this is a meso compound. A meso compound is achiral overall despite having chiral centers. It has 4 chiral centers total (even number). Therefore (b) matches (q) Achiral, (r) Meso, and (s) Even number of chiral centers → b-q,r,s. Step 3 – Analyze molecule (c): This is a sawhorse drawing of tartaric acid. The configuration shown has OH on the left-top carbon pointing left and OH on the right carbon pointing right (anti arrangement of OH groups). In meso-tartaric acid, the two OH groups are on opposite sides in the eclipsed conformation giving an internal mirror plane. Examining the drawing: left carbon has OH (up-left) and CO2H, right carbon has OH (down-right) and CO2H, with H's going the other directions. This anti/opposite arrangement of OH groups corresponds to meso-tartaric acid. The molecule has an internal mirror plane, making it achiral (meso). It has 2 chiral centers (even number). Therefore (c) matches (q) Achiral, (r) Meso, and (s) Even number of chiral centers → c-q,r,s. Step 4 – Analyze molecule (d): This sawhorse drawing shows: top-left H, top-right CO2H, bottom-left HO, bottom-right CO2H, with H on the front carbon and H on the back carbon. This arrangement places both OH and CO2H groups such that the two stereocenters are non-equivalent in their spatial arrangement — the OH groups are on the same side (syn), which corresponds to either (+)- or (-)-tartaric acid (the chiral, optically active form). This molecule lacks an internal mirror plane and is therefore chiral. It has 2 chiral centers (even number). Therefore (d) matches (p) Chiral and (s) Even number of chiral centers → d-p,s. Summary of matches: - (a): Chiral (p), Even number of chiral centers (s) → a-p,s - (b): Achiral (q), Meso (r), Even number of chiral centers (s) → b-q,r,s - (c): Achiral (q), Meso (r), Even number of chiral centers (s) → c-q,r,s - (d): Chiral (p), Even number of chiral centers (s) → d-p,s Therefore, the correct answer is a-p,s; b-q,r,s; c-q,r,s; d-p,s.