Nullification of Caffeine in Codeine CWE??

[adder]

In 6-membered pyridine derivatives the pKa drops with more nitrogen atoms as they all act as if they were electron-withdrawing groups built into the ring, so the basicity drops similarly as it does when you put an EWG onto pyridine or anilines. In caffeine I suppose the 6-membered ring as a whole withdraws electron density from the imidazole ring (the amide moiety at C5 is electron-withdrawing, the imide at C4 is very mildly donating at best if at all).

Imidazole being more basic than pyridine is due to electron-rich -NH- moiety I suppose (pyrrole much more activated than benzene for instance), however, no wonder they both have lower pKa (their conjugate acids have, I mean) than amines as the basic nitrogen atoms are sp2-hybridized and part of the aromatic ring, in amines alkyl groups are donating to the nitrogen atom (although sterics also come into play).

 

[belligerent drunk]

Imines have slightly lower pkas (~9). Why is this so?

Imines are sp2 hybridized, which in simple terms means higher electronegativity, thus lower basicity. Nitriles are less basic (lower c. acid pKa) still.

 

[belligerent drunk]

Here's where I'm confused:
-imidazole pka is 7. Why is this slightly lower than normal imines?
-pyridine pka is 5.5. Why is this still even lower?

Imidazole's nitrogens could be imagined as a part of an amidine. So maybe by the same mechanism (the second nitrogen stabilizing) amidine is more basic than would be expected of an sp2 hybridized nitrogen in an aromatic ring, like pyridine. After imidazole protonation, the two NH become equivalent, and the positive charge is pretty much smeared all over the place.

And yeah, like adder said, the 6-membered ring with amide and imide moieites is probably quite strongly electron-withdrawing, lowering the electron density at that basic center. The sp2 nitrogen is also in direct conjugation with the ketone in 6th position, which (in my mind) makes it more of an amide, which obviously would be non-basic.

E: aryl groups are typically electronwithdrawing, so maybe a nitrogen that is a part of an aromatic ring has lower basicity, because its surroundings are rather electronwithdrawing?

It's a sp2-hybridized i.e. planar nitrogen atom, and part of an aromatic heterocycle. Protonating it would destroy the aromaticity.

No. Not in the case of sp2 nitrogens. In the case of pyridine (pKa = 5.5), the lone pair is not a part of the aromatic system, the pi double bond is, so protonating the lone pair does not affect the aromatic system. In the case of pyrrole (pKa = 0.5), the lone pair is a part of the aromatic system.

 

[Nagelfar]

evaporate it down to a powder

I have a basic question about the evaporation procedure for a CWE, do we know how much of the active ingredient in the solution is lost in the water vapor that runs off, is it negligible or dependent on the compound, etc? Is avoiding bringing it to a boil proper to avoid pyrolysis, etc?

 

[aced126]

It is dependant on when the compound itself starts to evaporate or boil. If the compound boils at a lower temperature than water, then this technique cannot be used.