Equilibrium with an E2 step

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?

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]

• Note also that the entropy argument might not apply if you were, say, in methanethiol as a solvent. Well, maybe not methanethiol exactly since it has a very low boiling point.
– Zhe
Jan 7 '21 at 21:20