See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

The compound is 1,3,5-trimethylcyclohexane. Step 1: Identify stereocenters. C1, C3, and C5 each bear a methyl group and are stereocenters (each carbon is attached to two different ring portions and a methyl group vs H). Step 2: Maximum possible stereoisomers = 2^n = 2^3 = 8, but symmetry reduces this number. Step 3: Analyze the symmetry. The molecule has a plane of symmetry through C1 (or through the ring depending on configuration). The three stereocenters at C1, C3, C5 are related by the molecular symmetry (C3v symmetry of the ring positions). Step 4: Enumerate distinct stereoisomers. The possible combinations are: (i) all three methyls on the same face (all cis to each other): (1α,3α,5α) — this is the all-cis isomer, which has a plane of symmetry, making it achiral (meso-like). (ii) Two methyls on one face and one on the other: e.g., (1α,3α,5β). Due to the symmetry of the 1,3,5 positions, all arrangements with two on one face and one on the other are equivalent — but we need to check chirality. The (1α,3α,5β) isomer and (1α,3β,5α) and (1β,3α,5α) are all related by the C3 symmetry and are the same compound. Similarly (1α,3β,5β) and its permutations are all the same compound. So the distinct stereoisomers are: (a) all-up: (1α,3α,5α) — achiral, (b) all-down: (1β,3β,5β) — achiral, and these two are actually mirror images of each other; since the molecule with all three up has a plane of symmetry (C3v), it is achiral and identical to its mirror image. Wait — re-examining: (1α,3α,5α) has C3v symmetry and is achiral (same as its mirror image all-β which is also achiral). These two are the same compound (superimposable). So that gives 1 achiral compound. (c) (1α,3α,5β): two up, one down — this isomer has a plane of symmetry through C5 and the midpoint of C2-C4 bond, making it achiral. (d) There is no other distinct arrangement. So total distinct stereoisomers: the all-same (cis) isomer (achiral) and the two-up-one-down isomer (achiral) = 2 stereoisomers total. The all-cis (all on same face) is one compound, and the 1,3-cis,5-trans (two on one face, one on other) is another compound. Both are achiral due to internal planes of symmetry. Thus total stereoisomers = 2. Therefore, the correct answer is A.