See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The reactivity of alkenes toward electrophilic addition of HBr depends on the electron density of the double bond. Greater electron density (more electron-donating groups on the double bond carbons) stabilizes the developing carbocation in the transition state, increasing reactivity. Step 1 – Identify substituents on the double bond carbon (the substituted sp2 carbon) in each structure: - (a): Two methyl groups (CH3, CH3) — both are electron-donating by hyperconjugation and induction. - (b): One methyl group (CH3) and one methoxy group (OCH3) — OCH3 is a stronger electron donor than CH3 because oxygen donates electron density by resonance (lone pair donation into the double bond), making OCH3 a much better electron-donating group toward the pi system. - (c): One methyl group (CH3) and one methoxymethyl group (CH3OCH2) — the oxygen in CH3OCH2 is separated from the double bond by a CH2 group, so resonance donation to the pi bond is not possible; only weak inductive electron donation through the carbon chain occurs. Step 2 – Rank electron-donating ability toward the double bond: - OCH3 (direct resonance donation to C=C) > CH3 (hyperconjugation) > CH3OCH2 (inductive only, attenuated by intervening CH2) Step 3 – Compare reactivity: - Structure (b): has OCH3 directly on the vinyl carbon → strongest electron donation to the double bond → highest electron density on C=C → most reactive toward HBr (electrophile). - Structure (a): has two CH3 groups → moderate electron donation → second most reactive. - Structure (c): has CH3 and CH3OCH2 → the CH3OCH2 group is slightly more electron-withdrawing (or no better than CH3) toward the double bond compared to two CH3 groups in (a), because the oxygen's inductive withdrawal slightly offsets any weak donation, making (c) less reactive than (a). Step 4 – Order: b > a > c Why other options fail: - (a) a > b > c: Ignores the strong resonance donation of OCH3 directly on the double bond. - (c) b > c > a: Incorrectly ranks (c) above (a); CH3OCH2 does not donate as well as two CH3 groups. - (d) a > c > b: Ignores resonance donation of OCH3 entirely. Therefore, the correct answer is B.