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 '19 at 19:14
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    $\begingroup$ See my answer here: chemistry.stackexchange.com/q/105485/44877 $\endgroup$ – Tan Yong Boon Apr 10 '19 at 22:55
  • $\begingroup$ @TanYongBoon oh ok so the rate is proportional to the [R-X] only right? $\endgroup$ – Som V. Tambe Apr 11 '19 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? $\endgroup$ – Tan Yong Boon Apr 11 '19 at 1:30
  • $\begingroup$ @TanYongBoon but then why is the reaction known as Unimolecular? $\endgroup$ – Som V. Tambe Apr 11 '19 at 1:30

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