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Which of the following processes is responsible for the fact that free radical bromination of methane is slower than free radical chlorination?

A. initiation
B. hydrogen abstraction
C. halogen abstraction
D. termination
E. entropy

1 Answer

4 votes

The correct answer is B. hydrogen abstraction.

In free radical halogenation of hydrocarbons, such as methane, the reaction proceeds through three steps: initiation, propagation, and termination. In the propagation step, a hydrogen atom is abstracted from the hydrocarbon molecule by the halogen radical, forming a new carbon-centered radical and a hydrogen halide molecule.

The bond dissociation energy (BDE) of the C-H bond in methane is 105 kcal/mol, while the BDE of the C-Cl bond in chloromethane is 83 kcal/mol and the BDE of the C-Br bond in bromomethane is 68 kcal/mol.

Because the C-H bond in methane is stronger than the C-Cl and C-Br bonds, it requires more energy to break, making hydrogen abstraction in bromination slower than in chlorination. This is because the C-Br bond is weaker than the C-Cl bond, so the energy released from the reaction of Br with H is lower than that released from the reaction of Cl with H.

answered
User Andres Gardiol
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