See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Carbocation stability is greatly enhanced by aromaticity. A carbocation that, upon formation, yields an aromatic system (satisfying Hückel's rule: 4n+2 π electrons in a planar, cyclic, fully conjugated system) will form most readily. Step 1 – Analyze option (b): Tropone (cyclohepta-2,4,6-trien-1-one) is a seven-membered ring with a carbonyl. Protonation of the carbonyl oxygen by HClO4 gives a hydroxyl-stabilized cation, but the resulting species is the hydroxytropylium cation (cycloheptatrienyl cation with OH), which has 6 π electrons and is aromatic. This is favorable, but the driving force competes with option (c). Step 2 – Analyze option (c): Tropolone is 2-hydroxytropon — a seven-membered ring bearing both a C=O and an adjacent OH group with a fully conjugated framework. When treated with HClO4, protonation occurs readily. The key feature is that loss of water (or protonation of the carbonyl) from tropolone generates the hydroxytropylium cation or directly the tropylium-type cation. More importantly, tropolone under acidic conditions can lose OH (as water) to directly form the fully aromatic tropylium cation (C7H7+), a 6 π-electron aromatic carbocation (7-membered ring, 6 π electrons: 4n+2 with n=1). The presence of both OH and C=O on adjacent carbons of the seven-membered conjugated ring means that protonation and loss of water yields the stable aromatic tropylium cation immediately. This is the most favorable pathway because the product carbocation is fully aromatic. Step 3 – Analyze option (a): Cyclopenta-2,4-dien-1-one with HClO4 would protonate the carbonyl. The resulting carbocation would be a cyclopentadienyl cation with 4 π electrons, which is antiaromatic — highly unstable. This does NOT form readily. Step 4 – Analyze option (d): 5-chlorocyclopenta-1,3-diene with AlCl3 would ionize to give the cyclopentadienyl cation (4 π electrons, antiaromatic) — also very unstable. Does not form readily. Step 5 – Compare (b) and (c): Both involve seven-membered rings that can give aromatic cations. However, tropolone (c) has an OH group directly on the ring adjacent to the carbonyl. Under acidic conditions (HClO4), the OH group facilitates direct formation of the tropylium-type aromatic carbocation by loss of water, making the carbocation formation even more facile than in plain tropone (b), where only protonation of carbonyl occurs without direct loss of a leaving group to give the fully delocalized aromatic cation as cleanly. Tropolone loses water to give the aromatic hydroxytropylium / tropylium cation more readily because it already has the OH leaving group positioned on the aromatic ring framework. Therefore, the correct answer is C.