Typically the leaving group for E1cb is poor (like -OH or -OR) but why must this be the case? The substrate appears in the rate equation so surely a good leaving group would be beneficial?


1 Answer 1


There is a range of elimination reactions with E1cb at one end, E1 at the other end and E2 in between. It is not uncommon for these different reaction pathways to compete with one another. For example, in some elimination reactions the E1 and E2 pathways can operate in competition with one another. An activation energy is associated with each of these 3 reaction pathways. Whichever pathway has the lowest activation energy will be the major pathway followed. By changing solvent, reaction temperature, relative strength of the nucleophile, relative strength of the base, leaving group stability, etc., we can raise or lower the activation energy for each of these 3 pathways and shift a reaction towards one side of this mechanistic range or the other.

Typically the leaving group for E1cb is poor (like -OH or -OR) but why must this be the case?

The E1 mechanism involves ejecting a leaving group in its first step, while the E1cb mechanism involves removing a proton in its first step. Let's consider how changing the leaving group can shift an elimination reaction towards one pathway or the other. To a first approximation, changing a leaving group will not affect how hard or easy it is to remove the proton. So it is reasonable to assume that the activation energy for the E1cb process doesn't change as we vary our leaving group. If we use a better leaving group (make our leaving group more stable) that means that we have made ejecting the leaving group a lower energy pathway and the E1 process will become more favorable relative to the other elimination mechanisms. If we change our leaving group to one that is an extremely poor leaving group (make our leaving group less stable), then ejecting it becomes a higher energy process and the E1 reaction becomes less competitive with the other reaction pathways. Saying this last sentence differently, when we use a poor leaving group we raise the activation energy for the E1 mechanism. Since the rate of the E1cb process is not affected by the leaving group, its rate remains unchanged. Consequently, a poor leaving group will disfavor the E1 process making the E1cb process more competitive. If the leaving group is bad enough, we can disfavor the E1 process so much that we wind up pushing our reaction all the way over to the E1cb side.

  • $\begingroup$ Your answer cleared my confusion. Thanks a lot!! $\endgroup$
    – Natasha J
    Commented Jun 24, 2023 at 16:54

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