See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: The rotation barrier about a C=C (double bond) in alkenes is determined by the strength of the pi bond. The pi bond must be broken (or partially broken) for rotation to occur. Substituents that delocalize electron density away from the pi bond reduce the effective pi bond order and lower the rotation barrier, while electron-donating groups that push electron density into the pi bond (or groups that don't conjugate) maintain or increase it. Step 1 - Analyze compound (I): CH2=CH2 (ethylene). This is a simple alkene with no substituents. The pi bond is a pure, unperturbed C=C pi bond. Rotation barrier for ethylene is approximately 65 kcal/mol. Step 2 - Analyze compound (II): CH3O-CH=CH2 (methyl vinyl ether). The methoxy group (-OCH3) is an electron-donating group via resonance (lone pair donation into the pi system). This donation increases electron density in the pi bond and raises the double bond character (partial C=C bond order increases due to the OCH3 lone pair donation: CH3O-CH=CH2 <-> CH3O+=CH-CH2-). This resonance donation strengthens the pi bond, thereby INCREASING the rotation barrier compared to simple ethylene. Step 3 - Analyze compound (III): CH3O-CH=CH-C(=O)-OEt. This molecule has both the electron-donating methoxy group (pushing electrons in) AND an electron-withdrawing ester group -C(=O)-OEt (pulling electrons out via conjugation). The ester group conjugates with the C=C, delocalizing the pi electrons across an extended conjugated system (CH3O-CH=CH-C(=O)-OEt). This extended conjugation spreads the pi electron density over more atoms, reducing the localized pi bond character of the central C=C. When a pi bond is part of a larger conjugated system with an electron-withdrawing group pulling density away, the effective bond order of any single C=C in that system is reduced compared to an isolated alkene. Therefore, the rotation barrier about the indicated C=C bond in (III) is LOWER than in (I). Step 4 - Order comparison: - Compound (II): OCH3 donates into C=C → increased pi bond character → highest barrier - Compound (I): unperturbed ethylene → intermediate barrier - Compound (III): OCH3 donates but ester withdraws and extends conjugation → pi bond delocalized → lowest barrier Thus the order is: II > I > III Why other options fail: - (b) III > II > I: Incorrect; III has the most delocalized pi bond, not the strongest. - (c) III > I > II: Incorrect; II has the highest barrier due to lone pair donation increasing pi bond strength. - (d) II > I > III: Wait - this is actually option (d), but the given answer is (a) I > II > III. Re-evaluation: The given answer is (a) I > II > III. Let me reconsider. For compound (II): CH3O-CH=CH2. The oxygen lone pair donation into the double bond creates resonance: partial positive on oxygen, partial negative on terminal CH2. This resonance actually means the C=C has some single bond character in the resonance contributor, which could be interpreted as LOWERING the rotation barrier because the pi electrons are delocalized onto oxygen. The C=C bond order decreases due to delocalization onto O. For compound (III): The ester group withdraws electrons through conjugation, further delocalizing the pi system. But the question is about which C=C bond specifically. In III, the C=C is flanked on one side by OCH3 (donor) and on the other by the ester (acceptor) - this push-pull (captodative-like) conjugation actually STRENGTHENS the pi bond or at least maintains delocalization in both directions, but the net effect on the specific C=C depends on interpretation. With the given answer being (a) I > II > III: Ethylene has the highest barrier (no delocalization reducing pi bond), OCH3 in II delocalizes pi electrons onto O (reduces C=C pi bond order slightly), and in III both OCH3 and ester conjugation further spread the pi electrons, giving the lowest barrier. This interpretation: any conjugation/delocalization reduces the localized C=C pi bond strength → lower barrier. More substituents capable of conjugation → more delocalization → lower barrier. I (no conjugating groups) > II (one conjugating OCH3) > III (two conjugating groups: OCH3 and ester). Therefore, the correct answer is A.