Why does this show E1CB mechanism?

[![2006 IITJEE Previous Year Question](https://i.stack.imgur.com/IGyY6.png)](https://i.stack.imgur.com/IGyY6.png)


The following is a solution to a "match the following" question from IITJEE 2006.


I'm confused why does it show E1CB mechanism? Isn't the mechanism exclusively for highly acidic Hydrogen atom, which eventually creates a carbanion intermediate? I don't know why the following setup creates such a carbanion intermediate.




I agree the info is confusing. If the major product is deuterated it implies a rapid exchange at the benzyl position forming a carbanion that did NOT eliminate. The kinetics would be of interest. Extensive studies of these compounds generally support an E2 mechanism. In any case were the mechanism E1CB formation of the carbanion should give more immediate elimination since the deuterated compound should react much more slowly. A reference should have been cited. Unfortunately I have no access to chem abstracts to research in depth but one reference that pops up states that deuterium exchange does not take place.


If this is a fabricated question it is more than unfortunate it is professional malfeasance. One far out possibility is a paper was misread and a 1-phenylethyl bromide was reacted. This compound might have an active benzyl H that exchanges faster than the E2 reaction. It might be an interesting reaction to study.


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Conditions for E1CB mechanism are as follows: -


1. The leaving group must be bad (Fluorine, etc.)
2. The carbanion should be resonance stabilized.


Since the leaving group in the question is Bromine (which is a good leaving group), this will not show E1CB. Apparently, the question is wrong.


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