See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Heterolytic Cleavage Analysis **Heterolytic cleavage** occurs via heterolysis: $R-X \rightarrow R^+ + X^-$ The C-Cl bond breaks to form a carbocation intermediate. Ease depends on **carbocation stability**. ## Carbocation Stability Order: $$\text{3° > 2° > 1° > CH_3^+}$$ ## Analysis of Each Compound: | Compound | Structure | Carbocation Type | Stability | |----------|-----------|------------------|-----------| | **I** | $CH_3-Cl$ | $CH_3^+$ (methyl) | Lowest | | **II** | $CH_3-CH_2-Cl$ | $CH_3CH_2^+$ (primary) | Low | | **III** | $(CH_3)_2CH-Cl$ | $(CH_3)_2CH^+$ (secondary) | Moderate | | **IV** | $(CH_3)_3C-Cl$ | $(CH_3)_3C^+$ (tertiary) | **Highest** ✓ | ## Why IV is Correct: Compound IV forms a **tertiary carbocation** stabilized by: - Three alkyl groups donating electron density through the inductive effect - Maximum hyperconjugation from three C-H bonds This makes the C-Cl bond most easily cleaved. **Answer: (D) IV**