[gloinddark]

Although I think I am familiar with the E1cB mechanism I cannot understand certain things, namely:
-why should the leaving group be a poor one for this mechanism to proceed?
-why is it called E1 when it looks more like an E2 elimination?

I would really appreciate your help.

[gloinddark]

where cB = conjugate base, of course...

i would really love to hear some sort of response even if to tell me off. pre exam paranoia.

[azmanam]

The LG needs to be poor, because if it were good, E2 would be more favorable.  When the base takes the proton, the LG is not good enough to leave concurrently.  Thus the mechanism stops at the CB intermediate.  Then in a second step, the anion collapses to the alkene and kicks off the LG.

It is E1 because the rds is unimolecular.  That is, the rate determining step involves only one molecule.  That means the deprotonation (which is bimolecular) is fast, and the elimination is slow.

[gloinddark]

thanks