See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Under acid-catalyzed conditions (H+/Delta), the allylic alcohol undergoes an acid-catalyzed allylic rearrangement (or dehydration followed by reprotonation), generating a carbocation intermediate. The key feature is that the molecule is symmetric: both ends of the allylic system bear identical substituents — Ph and 1-naphthyl groups. When the OH is protonated and leaves, it generates a resonance-stabilized allylic carbocation. The two ends of the allylic carbocation are equivalent (both are Ph(1-naphthyl)CH-), so nucleophilic attack by water (or the counterion in the reprotonation/re-addition) at either end gives the same carbon skeleton. The new stereocenter(s) formed upon reprotonation are generated without any facial selectivity because the carbocation intermediate is planar and achiral (due to the symmetric allylic system). Since the two ends of the allylic cation are identical and the carbocation is planar, attack from either face is equally probable, producing a pair of enantiomers in equal amounts — a racemic mixture. The product has one new stereocenter (the carbon bearing OH after water re-addition) and the other carbon (bearing H via protonation) also becomes a stereocenter, but since both termini of the allylic cation are identical, the product obtained is the same compound regardless of which end is attacked. The planar carbocation intermediate has no chiral information, and both faces are equally accessible, giving a racemic (50:50) mixture of enantiomers. Why other options fail: (a) Meso compound requires an internal plane of symmetry within the product with two stereocenters of opposite configuration — not applicable here because the symmetric allylic cation gives enantiomers, not a meso form. (c) Diastereomers would require unequal amounts or different connectivity — not the case here. (d) Optically pure enantiomers would require a chiral intermediate or asymmetric induction — the planar carbocation provides no such selectivity. Therefore, the correct answer is B.