See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Dehydration of Cyclohexylmethanol **Step 1: Identify the substrate** Cyclohexyl-$CH_2OH$ is a primary alcohol attached to a cyclohexane ring. **Step 2: Determine the dehydration mechanism** Under conc. $H_2SO_4$, this undergoes E1 elimination: - Protonation: $-CH_2OH + H^+ \rightarrow -CH_2^+OH_2$ - Water loss: Forms carbocation intermediate **Step 3: Carbocation rearrangement (key step)** The primary carbocation ($-CH_2^+$) is highly unstable. The adjacent cyclohexane ring undergoes a **hydride shift**: - A hydride from the ring migrates to the primary carbocation - This generates a more stable secondary benzylic-type carbocation on the ring **Step 4: Deprotonation** The carbocation loses a proton from the adjacent ring carbon, forming a **double bond between the ring and the side chain**: $$\text{Cyclohexyl-CH}_2\text{OH} \rightarrow \text{Cyclohexene=CH}_2 \text{ (exocyclic alkene)}$$ **Answer: (B)** The exocyclic methylene product is favored because the hydride shift stabilizes the intermediate carbocation, and the resulting exocyclic double bond is more stable than an endocyclic one.