Seeing this mechanism from a waste paper in classroom. It seems like the final product is epoxide. Is this possible? I'm not sure what "E" means, but it is likely to be electric field.
No, the reaction mechanism as drawn is untenable. Bis trifluoroemethylamine is inherently quite acidic and a lousy nucleophile. It is more likely that there is a catalytic decomposition of bistrifluoromethylamine (which is quite unstable) with accompanying loss of HF.
Since the anion from bis-trifluoroemethylamine is a stable base and N is in high electron density, I first agree with that N can attack carbon bonding to oxygen.
Now opinions from you all make me wonder what is the second, which might be imagineed, step will going to?
My professor are curious now, too. He is curious about that if there is no proton there, like ionic liquid, is the reaction still the same as what we thought?
It seems to me that there is a reason that this was waste paper. It does not seem to contain anything of interest. The arrows used are not coherent.
At a best guess, it seems to be a mechanistic suggestion for graphite corrosion in the presence of bis-trifluoromethylamine. Given the indication of what appears to be an electric field, this would suggest an electrochemical cell.
If you were to react graphite with bis-trifluoromethylamine in the presence of an electric field, at best I would expect minimal formation of a charge transfer complex at the solid liquid interface.
An applied electric field of a high enough intensity effectively increases the degree of ionisation in solution. A conductor such as graphite, would result in charge induction at the surfaces and an induced electric potential. This may result in a change in the surface electrochemistry leading to differing charge transition complexes at the surface.