Author Topic: bromosafrole to MDMA via metallic methylamides  (Read 862 times)

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ning

  • Guest
bromosafrole to MDMA via metallic methylamides
« on: June 16, 2004, 09:23:00 PM »
I was searching on the internet the other day, and came across 2 interesting facts:

1. Zinc and aluminum alkyls are not very difficult to make. (pyrophoric, yes, but not too hard to manage)

2. If methylamine is bubbled through a solution of metal alkyl, an exchange reaction yields the metal methylamide.

As luck would have it, ethyl bromide, isopropyl bromide aren't too hard to prepare either. So I realized that this was a possible path to MDMA from bromosafrole.

1. Make zinc alkyl by refluxing zinc powder with ethyl bromide.
2. Bubble gaseous methylamine through so-formed alkyl to yield Zn(NHMe)2 and ethane.
3. add bromosafrole slowly to methylamide solution, keeping temperature below zero.
4. Quench mixture and extract MDMA. Party time!

I would expect a decent yield by this process.

One really promising point for the industrially-minded bee is that aluminum alkyls are very common and cheap industrial products, used for plastic production if I remember correctly. This would make the whole thing trivially easy for large-scale production.

The same operation could bee performed with CaH2 as well, I think, and many other strong organometallics.

Maybe even CaC2 could be used. Even though it's a weaker base than amide, it's product is a gas. So it could be possible to take advantage of the equilibrium by keeping the temperature of the solution below the boiling point of methylamine, and simply allowing the acetylene formed to evaporate away, thereby pushing equilibrium towards formation of the amide.

Thoughts?


Rhodium

  • Guest
That's a very roundabout synthesis of isosafrole
« Reply #1 on: June 16, 2004, 10:19:00 PM »
Once upon a time, I also thought it was that simple:



Rhodium (posted 10-15-98)
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The yields are probably pretty low when using the olefin, but shouldn't they be much improved if using a halosafrole instead?
 
drone 342 (posted 10-16-98)
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That's exactly what I originally thought, though LR believes that due to some acidity of the benzylic protons, a strong base like lithium amide or lithium methylamide is going to catalyze the formation of isosafrole.

Originally, what got me thinking in this direction was the Sn2 reactions (which is what you're proposing), can be greatly improved kinetically by two ways: getting a better leaving group, and by using a stronger base. Iodine is about as good as you can do clandestinely for a leaving group (if you can think of a better one, say so), but the base (ammonia or methylamine) could use a little tweeking. By using a metal amide (pKa is around 45), you're gonna have a very fast, very nice reaction. However, a side reaction that may happen is the abstraction of the benzylic proton, followed by the leaving of the halide, resulting in isosafrole.


There is a lot of reason to believe this is the case; there are examples in the literature where a strong base did this to safrole-based compound. However, I still remain somewhat skeptical that this cannot be avoided. I still believe reasonable conditions exist where this side-reaction scenario would hardly occur, and the result would be a very high-quality final product in good yields.

-drone #342

Quoted from

Post 122758

(dormouse: "Grignard reagents and methylenedioxy rings", Serious Chemistry)







ning

  • Guest
Ah, yes, the competitive elimination bit
« Reply #2 on: June 16, 2004, 10:24:00 PM »
I wish someone had done/could do experimentation on this.