See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

To determine chirality, a molecule must be non-superimposable on its mirror image (no internal plane of symmetry, no improper rotation axis). Step 1 - Analyze Structure I (bicyclic anhydride): Structure I is a bicyclic compound (norbornene-based) with an anhydride bridge. It has two CH3 groups and two H atoms on specific stereocenters. Examining the molecule: it possesses a plane of symmetry (or C2 axis) that makes the two halves mirror images of each other internally, rendering it a meso-like or achiral compound. The molecule has an internal symmetry element, so it is ACHIRAL. Step 2 - Analyze Structure II (dichloro diol): Structure II has two stereocenters: C1 bears (CH2OH, Cl, H) and C2 bears (Cl, H, CH2OH). Both stereocenters are present. Checking for internal symmetry: the two ends of the molecule are CH2OH groups, but the stereochemistry at the two centers as drawn does not create an internal mirror plane (they are not in a meso relationship). Therefore, Structure II is CHIRAL. Step 3 - Analyze Structure III (Fischer projection, aldopentose-like): Structure III: CH2OH - [HO|H] - [HO|H] - [HO|H] - CHO. This is an aldotetrose or aldopentose with all three OH groups on the left. With CHO at the bottom and CH2OH at the top, there are three stereocenters all with the same configuration (all OH on left). Checking for internal symmetry: the top and bottom ends are different (CH2OH vs CHO), so no internal mirror plane exists. All three stereocenters have the same relative configuration but the molecule is asymmetric due to the different end groups. Structure III is CHIRAL. Step 4 - Analyze Structure IV (Fischer projection, alditol-like): Structure IV: CH2OH - [HO|H] - [HO|H] - [HO|H] - CH2OH. Both ends are CH2OH (identical), giving this molecule a potential internal mirror plane. With all OH on the left and identical end groups, there is an internal plane of symmetry making it a meso compound. Structure IV is ACHIRAL (meso). Step 5 - Analyze Structure V (2,6-dimethylcycloheptanone): Structure V is a cycloheptanone with CH3 groups at the two alpha carbons. One CH3 is on a wedge (coming toward viewer) and the other is on a dash (going away from viewer) - these are on opposite faces. This trans relationship means there is NO internal plane of symmetry bisecting the ring through the carbonyl, because the two methyl groups are on opposite faces. The molecule cannot be superimposed on its mirror image. Structure V is CHIRAL. Step 6 - Compile results: Chiral structures: II, III, V. Achiral structures: I (internal symmetry in bicyclic framework), IV (meso alditol). Step 7 - Match to answer choices: (a) I, II, III - incorrect (I is achiral) (b) II, III, V - correct (c) II, III - incorrect (misses V) (d) I, II - incorrect (I is achiral, misses III and V) Therefore, the correct answer is B.