See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

We need to count the total number of stereoisomers for each compound. **Concept:** For compounds with multiple double bonds (E/Z isomerism) or stereocenters (chiral centers, cis/trans in rings), count all possible stereoisomers = 2^n where n = number of independent stereoelements, then subtract for any internal symmetry or constraints. --- **(a) CH3-CH=CH-CH=N-OH (but-2-en-1-al oxime or crotonaldehyde oxime)** This molecule has TWO double bonds that can show geometric isomerism: 1. CH3-CH=CH- : E or Z (cis/trans) 2. -CH=N-OH : E or Z (syn/anti) Total = 2 × 2 = 4 stereoisomers → matches **(q)** --- **(b) 1,3,5-trimethylcyclohexane** The molecule has three stereocenters (C1, C3, C5), each bearing a methyl group on the ring. However, due to the molecular symmetry (C3v-like), many combinations are identical. Possible configurations: - All three methyls cis (all up or all down): these two are mirror images of each other but the molecule has a plane of symmetry when all are on same side — actually the all-cis isomer (1,3,5-cis,cis,cis) is achiral (has a plane of symmetry). That counts as 1 isomer. - Two up, one down (and two down, one up): Due to the threefold symmetry of the ring, all 'two up one down' arrangements are equivalent. The resulting molecule is chiral → gives 1 pair of enantiomers = 2 isomers. Total distinct stereoisomers = 1 (all-cis, meso-like achiral) + 2 (enantiomeric pair) = but wait, reconsidering: all-cis gives 1 (achiral), the mixed arrangement gives an enantiomeric pair = 2. So total = 1 + 1 = 2 distinct stereoisomers? Actually the standard result for 1,3,5-trimethylcyclohexane: there are 2 stereoisomers total — the all-cis (achiral) and the 1,3-cis,5-trans (which exists as one achiral form due to the plane of symmetry through C5). This gives total = 2 stereoisomers → matches **(p)** --- **(c) CH3-CH=CH-CH=CH-CH=CH-CH3 (octa-2,4,6-triene)** This compound has THREE C=C double bonds, each capable of E/Z isomerism: 1. C2=C3: E or Z 2. C4=C5: E or Z 3. C6=C7: E or Z Maximum = 2^3 = 8, but we check for any that become identical due to symmetry. The molecule has a symmetric structure (CH3 on both ends), so some isomers may be identical. The symmetric nature means that swapping configurations of bond 1 and bond 3 simultaneously may give the same compound. After accounting for symmetry: 8/2 + (symmetric ones) = 6 distinct stereoisomers → matches **(r)** Specifically: EEE, ZZZ (these are symmetric, each unique) = 2; EEZ = ZEE (same due to symmetry) = 1 pair counted as 2 structures but they are identical → so EEZ gives 1; ZZE = EZZ → 1; EZE → 1; ZEZ → 1. Total = 2 + 1 + 1 + 1 + 1 = 6 stereoisomers. --- **(d) CH3-CH=CH-CH=CH-CH=CH-Ph (1-phenylhepta-1,3,5-triene or similar)** This compound also has THREE C=C double bonds capable of E/Z isomerism: 1. First C=C: E or Z 2. Second C=C: E or Z 3. Third C=C: E or Z Now the molecule is ASYMMETRIC (one end is CH3, other end is Ph), so no two isomers coincide due to molecular symmetry. Total = 2^3 = 8 stereoisomers → matches **(s)** --- **Summary of matches:** - (a) → (q): 4 stereoisomers - (b) → (p): 2 stereoisomers - (c) → (r): 6 stereoisomers - (d) → (s): 8 stereoisomers Therefore, the correct answer is a-q; b-p; c-r; d-s.