See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: The rotation barrier about a C=C double bond is related to the pi bond order (pi bond strength) at that bond. A higher pi bond order means a higher barrier to rotation. The pi bond order at the indicated C=C bond is reduced when the pi electrons are delocalized away from that bond into conjugated systems on both sides. More extensive conjugation (especially with electron-donating groups) spreads electron density across more atoms, lowering the pi bond order at the central bond and thus lowering the rotation barrier. Conversely, less conjugation means more localized pi electron density at the C=C bond and a higher rotation barrier. Step 1 - Analyze compound (I): Ph-CH=CH2. One end of the C=C bond is conjugated with the phenyl ring; the other end (CH2) has no conjugation partner. The pi electrons are delocalized into only one aromatic ring. The C=C bond retains relatively high pi bond character because delocalization is only on one side. Step 2 - Analyze compound (II): Ph-CH=CH-Ph. Both ends of the C=C bond are conjugated with phenyl rings. The pi electrons are delocalized into two phenyl rings symmetrically. This bilateral conjugation lowers the pi bond order at the central C=C bond more than in (I), giving a lower rotation barrier than (I). Step 3 - Analyze compound (III): Ph-CH=CH-(4-OCH3-C6H4). Both ends are conjugated with aryl groups, similar to (II), but one aryl group bears a para-methoxy (OCH3) substituent, which is a strong electron-donating group via resonance. The OCH3 group donates electron density into the ring and further into the conjugated pi system, increasing the degree of charge-transfer/conjugation across the C=C bond. This greater delocalization further reduces the pi bond order at the indicated C=C bond, giving an even lower rotation barrier than (II). Step 4 - Order the rotation barriers: Greatest pi bond localization (highest barrier) is in (I), where only one-sided conjugation occurs. (II) has two phenyl groups pulling pi density, lowering the barrier further. (III) has the same two-sided conjugation as (II) plus enhanced electron donation from OCH3, further lowering the barrier. Therefore: Rotation barrier: I > II > III. Why other options fail: - (b) III > II > I: Incorrect; more conjugation lowers the barrier, not raises it. - (c) III > I > II: Incorrect; (I) cannot have a lower barrier than (II) since (II) has two-sided conjugation. - (d) II > I > III: Incorrect; (I) with one-sided conjugation has a higher barrier than (II) with two-sided conjugation. Therefore, the correct answer is A.