See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Carbocation stability follows the order: tertiary > secondary > primary > methyl. Additional stabilization comes from hyperconjugation and inductive effects of alkyl substituents. Step 1: Analyze each option. - Option (d): (+)CH3 is a methyl carbocation — the least stable, no alkyl groups to donate electron density. - Option (b): A primary carbocation on a carbon that has only one alkyl substituent (primary). Slightly more stable than methyl but still very unstable. - Option (c): An exocyclic primary carbocation (-CH2+) attached to a bicyclic (hydrindane) framework. Although the bicyclic system provides some hyperconjugation, the carbocation center itself is primary (only one carbon attached to the cationic carbon from the ring, and the ring carbon). This is still a primary carbocation. - Option (a): A tertiary carbocation at the ring junction or ring carbon of cyclohexane bearing a methyl group — specifically, a 1-methylcyclohexyl cation. The positive charge is on a carbon bonded to three carbon groups (two ring carbons and one methyl), making it a tertiary carbocation. Step 2: Compare stabilities. - Option (a) is tertiary: three alkyl groups attached to the carbocation center provide maximum hyperconjugation and inductive stabilization. - Options (b) and (c) are primary carbocations. - Option (d) is a methyl carbocation. Step 3: Why other options fail. - (b) and (c) are primary carbocations — significantly less stable than a tertiary carbocation. - (d) is a methyl (essentially no stabilization) — least stable. - (a) tertiary carbocation on a cyclohexane ring with an extra methyl group is the most stable among the four. Therefore, the correct answer is A.