Kinetics for E1cb reactions

Asked Apr 10 '19 at 15:51

Active Apr 10 '19 at 15:51

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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)?

asked Apr 10 '19 at 15:51

Som V. Tambe

I would think that this depends on the substrate. – Zhe Apr 10 '19 at 19:14
1

See my answer here: https://chemistry.stackexchange.com/q/105485/44877 – Tan Yong Boon Apr 10 '19 at 22:55
@TanYongBoon oh ok so the rate is proportional to the [R-X] only right? – Som V. Tambe Apr 11 '19 at 01:27
@SomV.Tambe Since the rate-determining step involves both C-H bond breaking, wouldn't both base concentration and substrate concentration be factored in? – Tan Yong Boon Apr 11 '19 at 01:30
@TanYongBoon but then why is the reaction known as Unimolecular? – Som V. Tambe Apr 11 '19 at 01:30
@SomV.Tambe May I know where you heard that from? – Tan Yong Boon Apr 11 '19 at 01:32
@TanYongBoon https://en.m.wikipedia.org/wiki/E1cB-elimination_reaction – Som V. Tambe Apr 11 '19 at 01:35
I just checked. According to Carey & Sundberg (2007) Advanced Organic Chemistry, the E1cB mechanism involves second order kinetics. And I think it makes perfect sense as E1cB involves both base and the substrate reacting in the formation of the carbanion intermediate. Does that make sense to you? – Tan Yong Boon Apr 11 '19 at 15:54
Did you even read the section of the wikipedia page on chemical kinetics of E1cB? I think there is clearly a line stating that the reaction follows second order kinetics. – Tan Yong Boon Apr 11 '19 at 15:56

Kinetics for E1cb reactions

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