Sciencemadness Discussion Board

Mescaline by aminoethylation of pyrogallol

Theoretic - 17-3-2016 at 07:46

A shorter route to mescaline might be possible, by selectively alkylating pyrogallol (1,2,3-hydroxybenzene) with X-CH2-CH2-NH2. With the middle 2-OH deprotonated,* the anion will react at the 5 position that is both activated and unhindered. The resulting species is neutral, thus less reactive (also due to steric crowding). Deprotonation is more difficult due to the new alkyl group. Under conditions of reduced reactivity, and non-excess X-CH2-CH2-NH2, the product will react less than the pyrogalloxide. It can then be methylated at the OH's to give mescaline.

*The reason the middle OH is deprotonated, is that the negative charge can be distributed around the ring without clashing with either of the other OH's. Also, both the other OH's can hydrogen bond to the middle O-; it's a five-membered ring that forms, but still an effect.

chemrox - 17-3-2016 at 10:43

I can almost type with gloves on. OK you have 3 activating o,p directors on the ring. Aren't you worried about a mixture of products? Would the addition proceed without AlCl3? (nicodem) Otherwise you might halogenate some of your OH. Also since 3,4,5-trimethoxy benzaldehyde is available... why go this route? For fun? In case there's a sudden shortage of better precursors? R-OCO would react similarly, why not start from 1,2,3-trimethoxy instead? Or from gallic acid?

unionised - 17-3-2016 at 11:53

Just how stable is chloro ethanamine in the first place?

If you dissolve pyrogallol in a base it will deprotonate and you end up with a negative charge distributed over the oxygens.
So ,any reaction with an alkyl halide will give you an ether


Delta - 17-3-2016 at 11:58

Is the oxygen anion a much stronger ring activator than hydroxyl groups? Because both the 6 and 4 positions are being activated by two hydroxyl groups each. Would the o,p-directing effects of the 1,3-hydroxyls out way the directing effect of the oxygen anion? And I agree with chemrox, wouldn't a Friedel-Crafts catalyst be necessary? Even if it weren't, wouldn't it be prudent to use to prevent o-alkylation, even if the R-O- is sterically hindered?

chemrox - 17-3-2016 at 16:47

"out weigh?" Sorry you probably aren't English speaking origin..good call on the catalyst/activator if it works that way. Nicodem please weigh in sir.

Delta - 17-3-2016 at 19:07

Quote: Originally posted by chemrox  
"out weigh?" Sorry you probably aren't English speaking origin..good call on the catalyst/activator if it works that way. Nicodem please weigh in sir.

No, I am an English speaker, however I am not sure why I typed that way and I have no excuse.

Delta - 17-3-2016 at 19:20

Apparently after reading up on the Williamson ether synthesis, o-alkylation would be a significant side reaction, as ring substitution tends to be a side reaction of the Williamson synthesis when involving phenoxides. So, I would anticipate a mixture of products, and possibly very low yield.
Perhaps 2,4-dinitrophenol would be a more suitable substrate than 1,2,3-trihydroxybenzene?

[Edited on 18-3-2016 by Delta]

Theoretic - 18-3-2016 at 07:03

The O- releasing charge into the ring matters greatly. There's even a 30-40-fold difference (see page 4) between Na and K phenoxides, in the right solvent.
A MClx Friedel-Crafts catalyst would not be necessary to make the reaction go. The substrate is already very reactive. It could be useful as a blocking group for the O- to stop it being alkylated (or halogenated, according to chemrox :P). Zn<sup>2+</sup> might do the same with its bulk.
XCH<sub>2</sub>CH<sub>2</sub>NH<sub>2</sub> are usually stored as salts. The free molecule slowly eliminates to the reactive aziridinium.
Trimethoxybenzaldehyde can be available, but it's also more watched because it can be used for this. The reducing agent can be unavailable, and sometimes even the nitromethane (I saw a thread on the Hive about that).
Delta: hard to see how 2,4 dinitrophenol would be a better substrate. It may possibly be a nucleophilic aromatic substitution substrate, but that's a whole different (and probably longer) series of steps.

[Edited on 18-3-2016 by Theoretic]

Delta - 19-3-2016 at 11:27

Ah, I did not know that about the O- activation! Good to know. My suggestion was made without the knowledge about the O- you just gave. When I suggested it, I was thinking that although deactivating, it would be a better director. In my mind I was thinking the amount of steps outweighed the risk of unwanted direction to the 4 or 6 position.

[Edited on 19-3-2016 by Delta]

unionised - 19-3-2016 at 12:49

Did anyone think about methylating first, then there's no hydroxy groups left to act as a competing nucleophile.

Theoretic - 21-3-2016 at 11:18

unionised: indeed methylation would do that (and it was probably on a few people's minds), however as Delta said, the 4 and 6 positions have two activating groups each. But the desired 5 position has only one.
So, deprotonation of the 2-OH would activate the 5 position preferentially. Methylating all three OH groups would make this impossible.
However, if there is a methylating agent that would not affect the middle OH (for steric reasons perhaps), that would be ideal. -Eliminating deprotonation of the other OHs that would activate 4 and 6 positions, and stopping O-alkylation on two of the OH's, leaving only the more hindered middle one.

[Edited on 21-3-2016 by Theoretic]

unionised - 21-3-2016 at 12:09

Once you deprotonate any of the hydroxyls, you create a phenoxide ion; that's going to attack the alkyl chloride and produce (at least mainly) an ether you don't want.
(It might have some effect as an ethanolamine type antihistamine or something.)

So you have to rely on the electron donating ability of the three methoxy groups.
For better or worse, that gives you the wrong isomer, rather than mescaline.
Aparently this stuff
is inactive.
And, even without a catalyst the halo amine is unstable WRT polymerisation etc.
If you add a typical FC type catalyst like AlCl3 it will just trash the amine.

If I thought this scheme had a cat in hell's chance of getting anywhere I'n not have contributed to it, since I don't want to be seen helping a cook's handbook.

Theoretic - 21-3-2016 at 18:32

Did you even read the part about the middle OH being hindered, or selective methylation of the OH groups?
Or complexation of the O-by the metal halide catalyst?
And what are you talking about, 'trash the amine'? Since when was complexation called trashing?
(AlCl<sub>3</sub> will probably take off the chloride instead anyway)
Yes, the free halo amine is unstable; but as I pointed out it is stored as the HCl salt. It can be added to the mix as it is consumed. It's not necessary to have so much of it that it'll react with itself, instead of the more reactive phenoxide.

Using the word 'cook' is just derogatory. That word, ah, vaguely implies, a 'meth cook', in other words someone making quite a destructive drug. There's a huge difference between a psychedelic like mescaline, and meth (something that's mainly abused). What's the point of being that concerned with how one ...appears... to puritans?

Lastly, what kind of a creepy metaphor is 'a cat's in hell'? What's wrong with 'snowball's'? This is worse than schrodinger's 'cat in a cyanide box' metaphor.

[Edited on 22-3-2016 by Theoretic]

unionised - 22-3-2016 at 13:43

Once the amine has stuck to the metal it's not going to be very effective, and nor is the metal. I call that combination trash.
Re. "(AlCl3 will probably take off the chloride instead anyway)" .
You think that chloride is a better base than the amine here?
Are you sure?

You would learn more if you looked into English idiom, rather than making insulting assumptions

Mescaline and methamphetamine are roughly equally illegal (i.e. neither of them is permitted in most jurisdictions ) and thus your distinction between them is a moot point. It's not "puritanism" that matters; it's the law.

Good luck getting the right phenol group to deprotonate.
But it doesn't matter.
The outcome will be that you get one of two different ethers- that are not mescaline.

The only way to stop ether formation is (ironically) to methylate (or some such) the hydroxy groups.

Theoretic - 22-3-2016 at 15:56

Yes, in an ionizing solvent the chloride can compete, because it has a full negative charge and this reaction course forms a separated ion pair. The amine meanwhile reacts with the positive charge on the carbon left behind, so now both the amine and the chloride have reacted with an electrophile. So there's an enthalpy and an entropy advantage.
And the chloro amine doesn't need AlCl<sub>3</sub> for activation. It eliminates Cl- spontaneously. If not much AlCl<sub>3</sub> is present, it'll all react with the phenolic O- instead.

Sticking it out at me that you 'don't wan to be seen helping a cook', means assuming 'it's the law that matters' (above all else). That is morally dangerous. There is a distinction between law and morality, and it's unsettling how you let the former set itself above the latter.
Anyway, you actually believe that if other forum members see you 'helping a cook's book' (or whatever) by abstractly talking about a possible synthesis, this will land you in jail?

So you didn't come up with that cat saying. But you still went ahead and picked it over the snowball version. It's still creepy, and you still used it to insult me, so there you go; don't be surprised by how I react.

Lastly, hadn't I already said why mainly the middle OH will deprotonate? And that AlCl<sub>3</sub> will cover the O-? And the hydrogen bonds from neighbouring OH's (that'd create some deactivation of O- even without AlCl<sub>3</sub>;)? And about selective methylation? You didn't bother to read it though. Venting your spleen, rather than actually reading what's been posted.

"you're a 'cook' I don't even want to be seen helping you, I'm only bothering to talk because your cooking wouldn't even cook" - That is what you sound like. Why me even bother to reply, when you're trolling for a reaction and dispensing your own chemical opinions, rather than reading the replies.