# E1cb reaction conditions

I'm wondering why an E1cb reaction needs a poor leaving group like -OH attached to one of the carbons.

I thought the conjugate base of the leaving group must be stable enough not to de-protonate one of the newly formed alkene's H's. But if this is the case, why can't a good leaving group like Br be used in E1cb which would neither de-protonate the alkene's H's.

I'm wondering why an E1cb reaction needs a poor leaving group like -OH attached to one of the carbons.

In an E1cb reaction, the rate-determining step (RDS) is the $$\ce {C-H}$$ bond cleavage. According to Carey & Sundberg (2007), p. 552, for the irreversible E1cb reaction (when the hydrogen abstracted is very acidic), the rate of the reaction may increase in the following order of leaving group: $$\ce {I^- < Br^- < Cl^- < F^-}$$. Why does the reaction rate increase as the leaving group becomes "lousier"?

From kinetics, we known that the rate of a reaction is determined by the slowest step, the RDS. In the E1cb, the RDS involves removal of a proton to form a carbanion. This step would be accelerated if the $$\ce {C-H}$$ bond is weakened and when the carbanion experiences greater stabilisation of the negative charge. Poor leaving groups, such as $$\ce {OH^-}$$ and $$\ce {F^-}$$, are often able to provide inductive stabilisation of the negative charge, e.g. $$\ce {-OH}$$ and $$\ce {-F}$$ substituents are electron-withdrawing in nature. This would thus help to speed up the RDS and thus, the overall reaction rate would be faster with these poor leaving groups.

Note: The explanation for this is similar to that for this question: Why are fluorides more reactive in nucleophilic aromatic substitutions than bromides?

Paraphrased: If this is the case, can E1cb reactions still involve good leaving groups, such as $$\ce {Br^-}$$?

On the same page, the authors also conclude that:

E1cb reactions generally require both carbanion stabilization and a relatively poor leaving group

If the leaving group becomes a better one, e.g. if the $$\ce {F^-}$$ becomes $$\ce {I^-}$$, then the $$\ce {C-X}$$ cleavage may be more advanced than the $$\ce {C-H}$$ bond cleavage, resulting in the reaction becoming E2, E1-like or even E1. Thus, if good leaving groups are present, likely the elimination will not proceed by E1cb but by the other pathways.

Update

I just realised someone had a similar question answered already: Why must the leaving group in E1cb be poor? You can refer to that answer for a different perspective. But ultimately, the essence of the correct explanation is well-contained in my answer above.

Reference

Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.