See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: The reaction of benzyl alcohols with HBr proceeds via an SN1 mechanism (or SN1-like mechanism with benzylic carbocation stabilization). The rate-determining step is the formation of the benzylic carbocation intermediate. Therefore, substituents on the aromatic ring that stabilize the carbocation (electron-donating groups, EDGs) increase reactivity, while substituents that destabilize it (electron-withdrawing groups, EWGs) decrease reactivity. Step 1: Identify the substituents and their electronic effects. - (I) Unsubstituted benzyl alcohol: reference compound, benzylic carbocation is stabilized by resonance with the ring. - (II) 4-nitro (NO2): strong electron-withdrawing group (EWG) via both induction and resonance. Strongly destabilizes the benzylic carbocation. Lowest reactivity. - (III) 4-methoxy (OCH3): strong electron-donating group (EDG) via resonance (+M effect). Strongly stabilizes the benzylic carbocation. Highest reactivity. - (IV) 4-bromo (Br): Br is weakly electron-withdrawing by induction (-I) but weakly electron-donating by resonance (+M). Net effect is slightly electron-withdrawing overall, making the carbocation slightly less stable than unsubstituted benzyl. Reactivity less than (I) but greater than (II). Step 2: Rank the reactivities. - Highest: III (strong EDG, OCH3 stabilizes carbocation most) - Next: I (unsubstituted, reference) - Next: IV (Br is weakly deactivating, slightly less stable carbocation than I) - Lowest: II (strong EWG, NO2 destabilizes carbocation most) Order: III > I > IV > II Step 3: Match with options. Option (a): III > I > IV > II — matches our ranking. Why other options fail: - (b) III > I > II > IV: incorrectly places II above IV; NO2 is a stronger EWG than Br, so II should be less reactive than IV. - (c) I > III > IV > II: incorrectly places I above III; OCH3 (EDG) should make III more reactive than I. - (d) I > III > II > IV: same error as (c) plus incorrect ordering of II and IV. Therefore, the correct answer is A.