See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: The enthalpy of a free-radical halogenation reaction depends on (1) the strength of the C–H bond being broken and (2) the strength of the H–X bond being formed. The overall reaction enthalpy is approximately: ΔH = BDE(C–H broken) – BDE(H–X formed) – BDE(X–X broken) + BDE(C–X formed), but for comparing these reactions the key factors are the type of halogen (Cl vs Br) and the type of C–H bond broken (allylic primary vs allylic tertiary). Step 1 – Identify the C–H bond broken in each reaction. All four reactions proceed via free-radical allylic halogenation under hv. (a) Propene: the allylic C–H broken is a primary allylic C–H (~BDE ~88 kcal/mol). (b) Propene + Br2: same primary allylic C–H broken. (c) 2-Methylpropene (isobutylene): the allylic C–H broken is on the =CH2 group, but the radical formed is a tertiary allylic radical. The BDE for a tertiary allylic C–H is lower (~83–85 kcal/mol) than primary allylic, making bond breaking easier (less endothermic step). (d) 2-Methylpropene + Br2: same tertiary allylic C–H as (c), but with Br2. Step 2 – Compare halogen: Cl2 vs Br2. BDE(H–Cl) ≈ 103 kcal/mol; BDE(H–Br) ≈ 87 kcal/mol. BDE(Cl–Cl) ≈ 58 kcal/mol; BDE(Br–Br) ≈ 46 kcal/mol. Net contribution from halogen: H–Cl formation releases more energy than H–Br, and Cl–Cl requires more energy to break, but the net effect is that chlorination is significantly more exothermic than bromination for comparable C–H bonds. The difference in H–X BDE (103 vs 87 = 16 kcal/mol) dominates. Step 3 – Compare C–H bond: tertiary allylic vs primary allylic. A tertiary allylic C–H is weaker than a primary allylic C–H (BDE ~83 vs ~88 kcal/mol), so less energy is needed to break it, making the reaction more exothermic by ~5 kcal/mol. Step 4 – Rank all four reactions. (a) Primary allylic + Cl2: moderately exothermic (strong H–Cl formed, but stronger C–H broken). (b) Primary allylic + Br2: least exothermic (weak H–Br formed, strong C–H broken). (c) Tertiary allylic + Cl2: MOST exothermic (strong H–Cl formed AND weaker C–H broken — both factors favor exothermicity). (d) Tertiary allylic + Br2: moderately exothermic (weaker C–H broken but weak H–Br formed). Step 5 – Why other options fail. (a) Uses Cl2 (good) but primary allylic C–H (stronger, less favorable) → less exothermic than (c). (b) Uses Br2 (weaker H–Br) and primary allylic C–H → least exothermic. (d) Uses Br2 (weaker H–Br) despite tertiary allylic C–H → less exothermic than (c). Option (c) combines both favorable factors: Cl2 (forming strong H–Cl) and the weakest C–H bond (tertiary allylic in isobutylene), making it the most exothermic reaction. Therefore, the correct answer is C.