See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Free radical stability follows the order based on resonance and hyperconjugation. Step 1: Evaluate option (a). Bromine is indeed more selective and less reactive in free radical halogenation due to the higher activation energy required and a more endothermic hydrogen abstraction step. This is a correct statement. Step 2: Evaluate option (b). Chlorine is less selective and more reactive in free radical halogenation because it reacts exothermically with lower activation energy, attacking primary, secondary, and tertiary C-H bonds with less discrimination. This is a correct statement. Step 3: Evaluate option (c). Benzyl free radical (PhCH2•) is stabilized by resonance delocalization into the aromatic ring, making it more stable than a simple secondary (2°) free radical which is only stabilized by hyperconjugation. This is a correct statement. Step 4: Evaluate option (d). The vinyl free radical (CH2=CH•) has the radical on an sp2 carbon, which holds electrons more tightly and provides no resonance stabilization to the radical. The allyl free radical (CH2=CH-CH2•) is stabilized by resonance delocalization over three carbons (two resonance structures). Therefore, allyl free radical is MORE stable than vinyl free radical, not less. Option (d) states the opposite — that vinyl free radical is more stable than allyl free radical — which is incorrect. Why other options fail: Options (a), (b), and (c) are all factually correct statements about halogen selectivity/reactivity and free radical stability, so they cannot be the incorrect statement. Therefore, the correct answer is D.