See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the compounds. All three structures are Fischer projections of 2,4-dimethylhexanedial (a hexanedial with methyl groups at C3 and C4, using the internal carbons as stereocenters). The molecule has CHO at both ends and two stereocenters bearing CH3 groups. Step 2 - Analyze Compound I. In compound I, the CH3 group at C3 is on the LEFT (H3C on left) and the CH3 group at C4 is on the RIGHT. In Fischer projection convention, this means C3 is (S) and C4 is (R) — or vice versa — giving an anti relationship. However, because the molecule has a C2 symmetry axis (the top and bottom halves are mirror images of each other with CHO at both ends), this is actually a meso compound. The internal mirror plane makes C3 and C4 have opposite configurations, and the molecule is achiral. Step 3 - Analyze Compound II. In compound II, C3 has H on left and CH3 on right, and C4 has H3C on left and H on right. This is identical in connectivity to compound I — CH3 is on the right at C3 and on the left at C4 — which is the same relative arrangement as compound I (just read from the opposite end). This is also a meso compound with an internal mirror plane. Step 4 - Confirm I and II are identical (or both meso). Compounds I and II both have the methyl groups on opposite sides in the Fischer projection (one left, one right), making both meso compounds. They are in fact the same compound (identical, not just meso), or at minimum both are meso and achiral. Step 5 - Analyze Compound III. In compound III, both CH3 groups are on the RIGHT side (C3: H on left, CH3 on right; C4: H on left, CH3 on right). This gives both stereocenters the same configuration. However, because the molecule is a dialdehyde (symmetric end groups CHO on both ends), the molecule with both methyls on the same side also possesses an internal mirror plane bisecting the C3-C4 bond. This makes compound III also a meso compound. Step 6 - Conclusion. All three compounds (I, II, and III) are meso compounds — they contain stereocenters but have internal planes of symmetry, making them achiral. Therefore none of the compounds is chiral. Why other options fail: (a) All compounds are chiral — incorrect, all are meso/achiral. (c) I and II are meso compounds — partially correct but ignores III, which is also meso/achiral. (d) I and II are diastereomers, and III is a meso compound — incorrect; I and II are both meso, not a chiral diastereomeric pair. (e) I and II are chiral — incorrect, both are meso. Therefore, the correct answer is B.