See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand cyclopropane stereoisomerism basics: In 1,2-disubstituted cyclopropanes, the ring is rigid, so substituents can be cis or trans to each other, giving rise to geometrical isomerism. Additionally, chiral centers on the ring can give optical isomers. Step 2 - Analyze compound (a): 1-methyl-2-ethylcyclopropane with substituents H and CH3 on C1, H and Et on C2. This is 1-methyl-2-ethylcyclopropane. The two ring carbons bearing different substituents (C1: CH3, H and C2: Et, H) are stereocenters. The compound can exist as cis and trans isomers (geometrical isomerism, p). Both cis and trans forms are chiral (no plane of symmetry in either), so optical isomers exist (q). Hence a matches p, q. Step 3 - Analyze compound (b): Same connectivity as (a) - 1-methyl-2-ethylcyclopropane but drawn with different arrangement (CH3 and H on one carbon, Et and H on the other, both H's on the same face). This is essentially the same compound type: 1-methyl-2-ethylcyclopropane. It shows geometrical isomerism (cis/trans, p) and optical isomerism (q). Hence b matches p, q. Step 4 - Analyze compound (c): 1-deuterio-1H-2-methyl-3-ethylcyclopropane (apex carbon has D and H, left carbon has CH3 and H, right carbon has Et and H). The presence of D (deuterium) at the apex carbon makes it a stereocenter distinct from H. The compound has three different substituents around the ring: D/H at apex, CH3 at one ring carbon, Et at another. This gives geometrical isomers (cis/trans relationships, p) and since chiral centers are present with no compensating symmetry, optical isomers exist (q). Hence c matches p, q. Step 5 - Analyze compound (d): 1,1,2,3-trimethylcyclopropane... actually the apex carbon has CH3 and H, left carbon has CH3 and H, right carbon has CH3 and H - this is 1,2,3-trimethylcyclopropane with an additional H at apex... Re-reading: apex has CH3 (up) and H, left has CH3 and H, right has CH3 and H. This is 1,2,3-trimethylcyclopropane. For 1,2,3-trimethylcyclopropane: geometrical isomers exist (all-cis, or one trans, p). The all-cis isomer (1,2,3-trimethylcyclopropane with all methyls on same face) has a plane of symmetry (the plane bisecting through C1 and midpoint of C2-C3), making it a meso-like compound with a plane of symmetry (r). It does NOT have optical isomerism because of the plane of symmetry. Hence d matches p, r. Step 6 - Check against center of symmetry (s): None of these cyclopropane compounds have a center of inversion (centrosymmetry), so s is not matched by any. Step 7 - Compile matches: a: p (geometrical isomerism), q (optical isomerism) b: p (geometrical isomerism), q (optical isomerism) c: p (geometrical isomerism), q (optical isomerism) d: p (geometrical isomerism), r (contains plane of symmetry - the all-cis isomer is achiral) Therefore, the correct answer is a-p,q; b-p,q; c-p,q; d-p,r.