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In this question by Jay, one of the sub parts of the total question has the following reaction identified as $\ce{E1CB}$ pathway: $$\ce{Ph-CH2-CH2Br} \text{ on treatment with } \ce{C2H5OD/C2H5O-} \\ \text{gives } \ce{Ph-CD=CH2} \text{ as the major product.} $$

Now, I know $\ce{E1CB}$ mechanism to be an elimination mechanism which involves strong base and weak LG which I found well illustrated in the following example:

enter image description here

From Wikipedia

In extrapolating the above scheme into the question's case, I can't understand the reaction at all. Firstly $\ce{Br}$ is a good leaving group, if $\ce{E_1CB}$ is to happen, I think it must be taken that $\ce{Ph}$ is the leaving group. Secondly, let's leave the leaving group part and focus on the hydrogens, how did the hydrogen turn into a deuterium in the final product?

I have seen the user TRC's answer where they simply claim it is because of the strong base, I feel the logic is flawed, since it could have been $\ce{E_2}$ as well in that case due to the bromine existing. So, I don't find the argument fully convincing.

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    $\begingroup$ The question is ambiguous for lab purpose. But in theory you should read first 2 comments of TRC below his answer. E1cb mechanism can only produce carbanion intermediate. $\endgroup$
    – Jay
    Aug 12 at 2:18
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    $\begingroup$ I agree with @Jay - questions like these are only good in theory. In practice you will of course have E2 as well - which will give the non-deuterium elimination product. My answer was only in accordance to the scenario given in the question itself, the only possible reason to explain both the reactant, reagent and already given product. $\endgroup$
    – TRC
    Aug 12 at 3:52
  • $\begingroup$ This would be a good read to understand the difference. $\endgroup$ Aug 12 at 17:21

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