See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: The susceptibility of a carbon-halogen bond to nucleophilic substitution depends on the strength (and polarizability) of the C-X bond. A weaker, more polarizable C-X bond is more easily broken by an incoming nucleophile. Step 1 - Bond strength trend: As we go down the halogen group (F → Cl → Br → I), the C-X bond becomes progressively weaker because the halogen atom becomes larger and the overlap with carbon's orbitals decreases. Bond dissociation energies: C-F > C-Cl > C-Br > C-I. Step 2 - Polarizability trend: Larger halogens (I > Br > Cl > F) are more polarizable, meaning the electron cloud is more easily distorted, facilitating the transition state in nucleophilic substitution. Step 3 - Leaving group ability: The leaving group ability follows the order I⁻ > Br⁻ > Cl⁻ > F⁻, because iodide is the most stable leaving group (weakest base, most polarizable, most stable as an anion in solution). Step 4 - Evaluate each option: (a) 2-fluorobutane has the strongest C-F bond and fluoride is the worst leaving group — least susceptible. (b) 2-chlorobutane has a stronger C-Cl bond and chloride is a poorer leaving group than bromide or iodide. (c) 2-bromobutane has a weaker C-Br bond than C-Cl, but still stronger than C-I. (d) 2-iodobutane has the weakest C-I bond and iodide is the best leaving group among the four halogens listed. Step 5 - Conclusion: 2-iodobutane has the most susceptible carbon-halogen bond toward nucleophilic substitution because the C-I bond is the weakest and I⁻ is the best leaving group. Therefore, the correct answer is D.