See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Carbocation stability increases with (i) resonance/conjugation, (ii) electron-donating groups (EDG) such as alkoxy (OMe) directly stabilizing via lone pair donation, (iii) hyperconjugation and inductive effects, and (iv) substitution (tertiary > secondary > primary). Step 2 - Analyze each option: Option (a): The positive charge is on a carbon that is part of an allylic system (CH2=C(CH3)-), and the OMe group is on the other end of the conjugated diene system (C(OMe)=CH-). The carbocation is allylic and vinylogously stabilized by OMe, but the positive charge is on a less substituted terminal carbon. Option (b): The positive charge is on a primary carbon (CH2+), adjacent to no alkene or OMe directly. The OMe and alkene are present but not directly conjugated with the cationic center. This is the least stable. Option (c): The positive charge is on a tertiary carbon that is also allylic (C+(CH3) adjacent to CH=CH2). However, the OMe group is on a saturated sp3 carbon (CH(OMe)) two carbons away and cannot donate electrons by resonance to the carbocation. Stabilization comes from tertiary character and allylic resonance only. Option (d): The positive charge is on a carbon that is simultaneously: (1) allylic with respect to the isopropylidene group (CH=C(CH3)2), AND (2) directly conjugated with the C(OMe)=CH- vinyl ether system. This means the OMe oxygen can donate its lone pair through the extended conjugated system directly to the carbocation center. The structure is: CH3-C(OMe)=CH-C+(CH3)-CH=C(CH3)2. The cationic carbon is tertiary, allylic on both sides, and in conjugation with the electron-rich OMe-bearing alkene, making it a highly stabilized oxocarbenium-type/extended allylic cation. Step 3 - Why (d) is best: In option (d), the carbocation benefits from: tertiary substitution, allylic resonance toward the isobutenyl side, AND direct conjugation with the methoxy-substituted alkene (vinyl ether resonance), allowing OMe oxygen lone pair donation all the way to the positive charge. This cumulative stabilization is the greatest among all options. Step 4 - Why others fail: (a) lacks tertiary character at the cation and OMe conjugation is less direct; (b) is primary and poorly stabilized; (c) has tertiary allylic stabilization but OMe is on sp3 carbon and cannot conjugate. Therefore, the correct answer is D.