See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 – Bromohydrin formation from (Z)-2-butene with Br2/H2O. (Z)-2-Butene has the two methyl groups on the same side (cis). The bromonium ion forms by anti addition of Br2 to the double bond. Because of the Z-geometry and the symmetry of the alkene, the bromonium ion intermediate is achiral (it has a plane of symmetry). Water then attacks anti to bromine at either carbon of the bromonium ion. Because the molecule is symmetric, attack at C2 and attack at C3 give the same set of products. The anti-addition of water to the bromonium ion from (Z)-2-butene gives the erythro bromohydrin, which is the (2R,3S) + (2S,3R) pair — i.e., a racemic mixture of the two enantiomeric anti-addition products. More precisely: the bromonium ion from (Z)-2-butene (both methyls same side) opens with water anti, producing the threo bromohydrin as a racemic mixture: one enantiomer is (2R,3S)-3-bromo-2-butanol and the other is (2S,3R)-3-bromo-2-butanol (these are enantiomers of the erythro/threo pair depending on convention, but the key point is a racemic mixture of anti-addition products). Step 2 – Intramolecular SN2 with methoxide. Methoxide (a base) deprotonates the OH of the bromohydrin to give an alkoxide. The alkoxide then performs an intramolecular SN2 on the adjacent carbon bearing Br. SN2 proceeds with inversion at the carbon bearing Br. Step 3 – Stereochemical outcome. For (2R,3S)-3-bromo-2-butanol: the alkoxide at C2 (R) attacks C3 (S) with inversion → C3 becomes R → product is (2R,3R)-2,3-dimethyloxirane. For (2S,3R)-3-bromo-2-butanol: the alkoxide at C2 (S) attacks C3 (R) with inversion → C3 becomes S → product is (2S,3S)-2,3-dimethyloxirane. (2R,3R)- and (2S,3S)-2,3-dimethyloxirane are enantiomers of each other (trans-epoxide). They are produced in equal amounts from the racemic bromohydrin mixture. Step 4 – Identifying structures I, II, and III. Structure (I) shows trans-2,3-dimethyloxirane with one specific arrangement of wedge/dash bonds: H and CH3 on wedges (one enantiomer of the trans epoxide). Structure (II) shows the same trans-2,3-dimethyloxirane but mirrored: the other enantiomer. Structure (III) shows cis-2,3-dimethyloxirane (meso compound), which would require syn addition — not produced by the anti-addition mechanism here. Since the bromohydrin is formed as a racemic mixture, and each enantiomer undergoes intramolecular SN2 with inversion to give one enantiomer of the trans-epoxide, the final product is an equal mixture of (I) and (II) — the racemic trans-2,3-dimethyloxirane. Why other options fail: (a) Only (I): wrong, because both enantiomers of the bromohydrin are formed equally. (b) Only (II): wrong, same reason. (c) Only (III): wrong, the meso (cis) epoxide would require syn addition, which does not occur via the bromonium mechanism. Therefore, the correct answer is D.