See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: The reaction of benzyl alcohols with HBr proceeds via an SN1 mechanism, forming a benzylic carbocation intermediate. The stability of the benzylic carbocation determines the reactivity. Electron-donating groups (EDGs) on the aromatic ring stabilize the positive charge through resonance and induction, increasing reactivity. Electron-withdrawing groups (EWGs) destabilize the carbocation, decreasing reactivity. Step 1: Identify the substituents and their electronic effects on the benzylic carbocation. - (I) No substituent: baseline reactivity. - (II) para-NO2: strong EWG by resonance and induction. Strongly destabilizes the benzylic carbocation. Least reactive. - (III) para-OCH3 (methoxy): strong EDG by resonance (+M effect). Strongly stabilizes the benzylic carbocation via resonance donation into the ring. Most reactive. - (IV) para-Br: Br is weakly electron-withdrawing by induction (-I) but weakly electron-donating by resonance (+M). Net effect is slightly deactivating, making it slightly less reactive than the unsubstituted (I) but more reactive than the strongly deactivating NO2 group. Step 2: Rank the reactivities. - Most reactive: III (para-OCH3, strong EDG stabilizes carbocation most) - Second: I (unsubstituted, baseline) - Third: IV (para-Br, slight net deactivation but much less than NO2) - Least reactive: II (para-NO2, strong EWG destabilizes carbocation most) Order: III > I > IV > II Step 3: Why other options fail. - (b) III > I > II > IV: Places II above IV, but NO2 is a far stronger deactivator than Br, so II should be last, not IV. - (c) I > III > IV > II: Places I above III, but methoxy (EDG) stabilizes the carbocation better than no substituent, so III should be more reactive than I. - (d) I > III > II > IV: Same error as (c) with I above III, and also incorrectly places II above IV. Therefore, the correct answer is A.