See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Carbocation stability is governed by resonance delocalization and electron donation. Step 1: Identify the nature of each cation. - Option (a): The positive charge is on the aromatic ring at a position ortho to the carbon bearing the -OH group. This is an arenium-type (cyclohexadienyl) cation or a ring carbocation, but the ring is shown as aromatic (benzene), meaning it is a sigma-complex type or vinyl cation on the ring - generally very unstable. - Option (b): The positive charge is on a non-aromatic cyclohexadiene ring carbon (the ring has only two double bonds shown, making it a cyclohexadienyl/non-aromatic system). The -OH group is on the carbon attached to the ring. - Option (c): The positive charge is directly on the ring carbon that bears the -CHOH substituent on a benzene ring. This is a destabilized arrangement because placing charge on an sp2 aromatic carbon that is also bearing a substituent is unfavorable. - Option (d): The positive charge is at the bottom of the benzene ring (para to the -CHOH group), again a ring carbocation on an aromatic system. Step 2: Evaluate option (b) more carefully. In option (b), the structure is a hydroxymethyl-substituted cyclohexadienyl cation (non-aromatic ring with the positive charge at the allylic/allylic position). The positive charge on the cyclohexadienyl system is stabilized by the adjacent pi system (allylic delocalization across the diene), AND crucially, the oxygen of the -OH group on the adjacent carbon can donate electron density through resonance (oxocarbenium stabilization). The oxygen lone pairs can overlap with the empty p orbital, forming a highly stabilized oxocarbenium ion: >C=OH(+), where the positive charge is delocalized onto oxygen. Step 3: Why (b) is most stable. In structure (b), the carbocation is at C1 of the cyclohexadiene ring, and the -OH-bearing carbon is directly attached. The oxygen lone pair can donate into the empty p orbital at C1, giving an oxocarbenium resonance structure (C=O(+)H), which is a very strong stabilization. Additionally, the allylic pi system of the cyclohexadiene further delocalizes the charge. This combination of oxygen resonance donation plus allylic delocalization makes (b) the most stable cation. Step 4: Why other options fail. - (a), (c), (d): Positive charges placed on aromatic ring carbons disrupt aromaticity or are in positions where oxygen lone-pair donation is geometrically or electronically less effective. Disrupting the aromatic sextet costs significant energy. - (b) does not disrupt an aromatic ring (it is already non-aromatic/cyclohexadienyl), so the stabilization from O lone pair donation is not offset by loss of aromaticity. Therefore, the correct answer is B.