I did research, about this one but the data and other things prove chlorination is faster than bromination, because chlorine is more reactive. But, some data says reaction of chlorine is more exothermic then bromine, that is the reason. But, if this is the case, then according to me bromination should be faster. Because, less energy is needed to start bromination and less energy will be given out in formation hydrogen bromide. And how reaction being more exothermic or less exothermic determines the rate of reaction?

  • $\begingroup$ The energy to start the reaction shouldn't be a key factor here; one photon is enough to initiate the reaction for thousands of molecules. Also, you don't create hydrogen bromide in bromination. $\endgroup$ – Hippalectryon Mar 21 '16 at 16:41
  • $\begingroup$ If you are asking about free radical halogenation, see this earlier answer. $\endgroup$ – ron Mar 21 '16 at 16:51
  • $\begingroup$ I don't understand. $\endgroup$ – ahmed Mar 21 '16 at 17:51
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    $\begingroup$ Halogenation of what? An alkane? Alkene? Another functional group? $\endgroup$ – jerepierre Mar 21 '16 at 17:57
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    $\begingroup$ What do you mean by link? $\endgroup$ – ahmed Mar 21 '16 at 18:53

According to the Hammond postulate, a highly exothermic reaction is more likely to have a low transition state barrier than a less exothermic reaction.

Lower barriers lead to faster reactions.

This is a multistep mechanism, so for this to be relevant the rate-determining step. According to this lecture the rate limiting step is the hydrogen abstraction from the alkane, forming a hydrogen-halide bond.

The bond dissociation energy for HCl is 431.8 kJ/mol and 365.7 kJ/mol for H-Br. The formation of the H-Cl is this much more exothermic, and the reaction will be faster.

  • $\begingroup$ For the question why chlorination should be faster than bromination? Answer should be Because chlorination is more exothermic. Not chlorine is more reactive. Right? $\endgroup$ – ahmed Mar 21 '16 at 18:57
  • $\begingroup$ Hammond's postulate says it is very difficult to separate those two circumstances. $\endgroup$ – Lighthart Mar 21 '16 at 21:26

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