See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

## Solution: Carbocation Formation in Dehydration **Step 1: Identify the substrate structure** The starting material has a cyclopentane ring with a secondary alcohol ($-OH$) on a side chain bearing a methyl group. **Step 2: Understand dehydration mechanism** Under acidic conditions ($H^+$), the $OH$ group protonates and leaves as $H_2O$, forming a carbocation. The reaction follows **carbocation rearrangement** (Wagner-Meerwein shift) to form the most stable carbocation. **Step 3: Evaluate possible carbocations** - **Primary carbocation** (initial): Forms on the carbon bearing $OH$ — highly unstable - **Secondary carbocation** (rearranged): A hydride or methyl shift can move the positive charge to an adjacent carbon **Step 4: Apply hydride/methyl shift** A **hydride shift** occurs from the adjacent methyl-bearing carbon to the ring, generating a secondary carbocation on the **6-membered ring** (after rearrangement). Alternatively, expansion to a 6-membered ring carbocation increases ring stability. **Step 5: Stability consideration** Option (B) shows a secondary carbocation on a 6-membered ring with the methyl substituent. This is: - More stable than primary carbocations (A, C, D) - Achievable through a favorable 1,2-hydride shift - Consistent with carbocation rearrangement preference for **secondary** intermediates **Answer: (B)** — The secondary carbocation on the expanded 6-membered ring is the most thermodynamically stable intermediate formed via Wagner-Meerwein rearrangement.