E1cb reaction conditions
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 &amp; 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: $\ce {I^- &lt; Br^- &lt; Cl^- &lt; F^-}$. 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., &amp; Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.