See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Free radical benzylic halogenation under photochemical (hv) conditions. Step 1: Under hv (light), CBrCl3 undergoes homolytic cleavage of the C-Br bond (the weakest bond in CBrCl3, since C-Br bond dissociation energy is lower than C-Cl). This generates a Br• radical and •CCl3 radical. Step 2: The Br• radical abstracts a benzylic hydrogen from toluene (Ph-CH3) to form the resonance-stabilized benzylic radical (Ph-CH2•) and HBr. Benzylic hydrogens are preferentially abstracted because the resulting benzylic radical is highly stabilized by resonance with the aromatic ring. Step 3: The benzylic radical (Ph-CH2•) then reacts with another molecule of CBrCl3, abstracting the Br atom (again, the C-Br bond is weaker and more easily broken than C-Cl) to give Ph-CH2-Br and regenerate •CCl3, which propagates the chain. Step 4: The net result is that only the benzylic C-H is replaced by Br, giving benzyl bromide (Ph-CH2-Br). No ring substitution occurs as stated. Why other options fail: - (a) Ph-CH2-Cl: Chlorine abstraction from CBrCl3 would require breaking a stronger C-Cl bond; Br is selectively transferred because C-Br bond in CBrCl3 is weaker. - (c) Ph-CH2-CCl3: This would require radical coupling of Ph-CH2• with •CCl3, which is a termination step and not the major propagation product. - (d) Ph-CH2-CBrCl2: This would require addition of a CBrCl2 fragment, which is not the mechanism here. Therefore, the correct answer is B.