See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

This question asks in which pairs the second ion is more stable than the first ion. We analyze each pair using aromaticity (Hückel's rule: 4n+2 π electrons for aromatic stability) and carbocation/carbanion stability principles. Pair (i): Cyclopropenyl cation vs. cyclopropenyl anion. - Cyclopropenyl cation: 3-membered ring, 2 π electrons (n=0, 4(0)+2=2), aromatic → very stable. - Cyclopropenyl anion: 3-membered ring, 4 π electrons (4n with n=1), antiaromatic → very unstable. - Second ion (anion) is LESS stable than first (cation). Pair (i) does NOT qualify. Pair (ii): Cyclopentadienyl cation vs. cyclopentadienyl anion. - Cyclopentadienyl cation: 5-membered ring, 4 π electrons (4n, n=1), antiaromatic → unstable. - Cyclopentadienyl anion: 5-membered ring, 6 π electrons (4n+2, n=1), aromatic → very stable. - Second ion (anion) is MORE stable than first (cation). Pair (ii) QUALIFIES. Pair (iii): Cycloheptadienyl anion vs. cycloheptadienyl cation (tropylium-related). - Cycloheptadienyl anion (cycloheptatrienyl anion): 7-membered ring, 8 π electrons (4n, n=2), antiaromatic → unstable. - Cycloheptadienyl cation (tropylium cation): 7-membered ring, 6 π electrons (4n+2, n=1), aromatic → very stable. - Second ion (cation/tropylium) is MORE stable than first (anion). Pair (iii) QUALIFIES. Pair (iv): 1-hydroxycyclohexyl cation vs. 1-methylcyclohexyl cation. - 1-hydroxycyclohexyl cation: the positive charge on carbon bearing OH; lone pair on oxygen can donate into the empty p orbital, providing resonance stabilization, making it an oxocarbenium ion → relatively stable. - 1-methylcyclohexyl cation: tertiary carbocation stabilized by hyperconjugation/induction from methyl group. - The OH group's lone pair donation (resonance) generally provides more stabilization to an adjacent carbocation than a methyl group's hyperconjugation. Thus, first ion (OH-bearing cation) is more stable than second (methyl-bearing cation). - Second ion is LESS stable than first. Pair (iv) does NOT qualify. Therefore, only pairs (ii) and (iii) have the second ion more stable than the first. Therefore, the correct answer is B.