We were taught that the kinetics of a reaction is determined by the reactions ‘rate determining step’ which is also the slowest step of the reaction. For E1 as well as E2 reactions the slowest steps are easily determined by the mechanism and I derived the rate laws for those.

In the case of E1cb reactions, which step can be taken as the slowest step?

Is it the one in which the acidic hydrogen is extracted by the base or is it the one where the elimination takes place (carbanion intermediate)?

  • $\begingroup$ I would think that this depends on the substrate. $\endgroup$
    – Zhe
    Apr 10, 2019 at 19:14
  • 1
    $\begingroup$ See my answer here: chemistry.stackexchange.com/q/105485/44877 $\endgroup$ Apr 10, 2019 at 22:55
  • $\begingroup$ @TanYongBoon oh ok so the rate is proportional to the [R-X] only right? $\endgroup$ Apr 11, 2019 at 1:27
  • $\begingroup$ @SomV.Tambe Since the rate-determining step involves both C-H bond breaking, wouldn't both base concentration and substrate concentration be factored in? $\endgroup$ Apr 11, 2019 at 1:30
  • $\begingroup$ @TanYongBoon but then why is the reaction known as Unimolecular? $\endgroup$ Apr 11, 2019 at 1:30


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