Recently, we were learning about free radical halogenation in school. A set of notes given to us wrote that the rate-determining step is the propagation step. This seems to go against my understanding of the chemistry of this reaction. The step that likely has the highest activation energy should be the initiation step as it is the only step that involves bond-breaking solely. I have looked up various sources and they also seem to suggest that the rate-determining step ought to be the first step.

Advanced Organic Chemistry (5th ed.) by Carey and Sundberg states that:

Free radicals are usually highly reactive and the individual steps in a chain reaction typically have high absolute rate constants. However, the concentrations of the intermediates are low. The overall rates of reaction depend on the balance between the initiation and termination phases of the reaction, which start and end the chain sequence.

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    $\begingroup$ This will depend heavily on the concentrations involved. If the concentration is very low, the propagation step could be a lot slower because it's bimolecular. $\endgroup$
    – Zhe
    Commented Oct 2, 2018 at 13:30
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    $\begingroup$ It depends on halogen. $\endgroup$
    – Mithoron
    Commented Oct 2, 2018 at 14:36

1 Answer 1


It does seem odd as a blanket statement. However, the bond dissociation energies (BDEs) for Cl2 and Br2 are quite low, 243 and 193 kJ/mol, respectively, while BDEs for the C-H bond in alkanes are much higher (eg. 435 kJ/mol for methane). So, it's not necessarily the case that events involving only bond breaking will be the rate-determining step, particularly since, as Zhe points out, the propagation steps are bimolecular.

Bromine radicals are significantly less reactive than chlorine radicals, leading to interesting selectivity differences between chlorination and bromination reactions - https://youtu.be/c5enwHzAwmA

Edit: BDEs are taken from Klein, "Organic Chemistry", 3rd ed., Chapter 10.

  • $\begingroup$ But the difference here is that the propagation steps are overall still more energetically favourable because formation of bonds like the H-Cl bond or C-Cl bonds are quite exothermic, compensating for the energy used to break the C-H bonds. $\endgroup$ Commented Oct 5, 2018 at 23:48
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    $\begingroup$ The initiation step, breaking of the X2 bond, is not rate limiting nor does it participate in the thermochemical calculations. The initiation step need only supply a catalytic amount of halogen atom for the two propagation steps to proceed. Heed @Tan Yong Boon $\endgroup$
    – user55119
    Commented Jan 3, 2019 at 20:36

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