Equilibrium with an E2 step

Question

E2 mechanisms are known to be irreversible in general, because the forward step takes 2 species and produces three species, and a trimolecular reverse step is unlikely, purely on entropic grounds. (As mentioned in this post.)

Now does that mean that any reaction that is reversible cannot have an E2 step, on the grounds of microscopic reversibility?

Particularly, I want to ask about this hypothetical mechanism, of an isomerisation:

I am not sure how to draw its mechanism. Clearly, some elimination reaction is taking place: it can be E1, E2 or E1cB. Because this is an alkene compound, E1 seems unlikely, only E2 and E1cB are possible. $\ce{MeS-}$ seems like a bad base for both E1cB and E2, so I can't say which one is better. Can I remove E2 from the options purely because the reaction is reversible?

[Note: This is an exercise question which asks me to draw the best possible mechanism if the intermediate detected was the one in the picture. It does not matter if the actual reaction happens in that way or not, the question is hypothetical]

Answer 1

I think your thought process is correct, but you can push it further.

The way I would solve this problem is to think about how you could go from the alkyne to either alkene.

Then, via microscopic reversibility, you would argue that the reverse mechanism--the elimination--is very similar.

The opposite of conjugate addition + protonation would be most similar to E1cb.