See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

## Solution: Carbocation Stability and Formation Enthalpy **Step 1: Identify the carbocations formed** - (I) Cyclohexyl chloride → secondary cyclohexyl carbocation - (II) Cyclohexene chloride → allylic carbocation (resonance stabilized) - (III) Benzyl chloride → benzyl carbocation (resonance stabilized) - (IV) Chlorobenzene → phenyl carbocation (very unstable) **Step 2: Rank carbocation stability** $$\text{Benzyl} > \text{Allylic} > \text{Secondary alkyl} > \text{Phenyl}$$ Benzyl and allylic carbocations have resonance stabilization. Phenyl cations are extremely destabilized due to sp² hybridization of the carbon bearing the positive charge. **Step 3: Relate stability to formation enthalpy** More stable carbocations require **less energy** to form, so $\Delta H$ is **more negative** (lower enthalpy). $$\text{Stability: } (III) > (II) > (I) > (IV)$$ $$\therefore \Delta H_3 < \Delta H_2 < \Delta H_1 < \Delta H_4$$ **Step 4: Verify with option B** $$\Delta H_4° > \Delta H_1° > \Delta H_2° > \Delta H_3°$$ This correctly shows phenyl (highest/most positive) through benzyl (lowest/most negative). **Answer: (B)** ✓