3-methoxy 4,5 diethoxy amphetamine

[mr_grieves]

Ive used the search engine and pihkal, but I can't find any information on this compound.  I would prefer to avoid dimethylsulfate or methyliodide and thought thatthis compound could easily be made by
1. brominating eugenol
2. refluxing the resulting 5-bromo in an NaOH solution.
4. reacting the resulting 3-methoxy 4,5 dihydroxyeugenol with ethyl bromide (bp is like 108) to yield 3methoxy 4,5 diethoxy eugenol. from there it would be standard workup to the amphetamine. Any one think it would have appreciable potency.

  Or do you think i would be better off just ethoxylating the 4 position and brominating the 5 position to get 3-methoxy, 4-ethoxy, 5-bromoamphetamine.

[Chimimanie]

If you consider to synthetise your compound, by the same time try to synthetise EEM.

It use the same chemicals, but you must ethylate vanillin before bromination, to brominate at the 2- position instead of the 3.

Mhhh I was reading your post and I saw some flaw in your synthetic scheme:

1.Brominating eugenol

I dont think brominating eugenol will work well, because the bromine will most likely react with the  olefin on the side chain. You must start from vanillin instead.

2.Refluxing the resulting 5-bromo in an NaOH solution.

It does not work like that, you must use a copper salt and an inert atmosphere etc... This swap is not that easy! (Evidently you could think you can do it by an aryne intermediate, at outrageously high temperatures, but it will destroy your little molecule, for sure.....)

3.reacting the resulting 3-methoxy 4,5 dihydroxyeugenol with ethyl bromide (bp is like 108) to yield 3-methoxy 4,5 diethoxy eugenol. from there it would be standard workup to the amphetamine. Any one think it would have appreciable potency.

You must use a base too. Standard conditions are K2CO3 + bromide in acetone.

And BTW, you can make ethyl bromide easily from ethanol, KBr and H2SO4, like in

http://www.orgsyn.org/orgsyn/prep.asp?prep=cv1p0025

but using more H2SO4 to generate HBr in situ.

And not, the bp of ethyl bromide is 37-40°C.

Or do you think i would be better off just ethoxylating the 4 position and brominating the 5 position to get 3-methoxy, 4-ethoxy, 5-bromoamphetamine.

No, again, if you alkylate vanillin then brominate you will get the 2,4,5 isomer, if you brominate then alkylate you will get the 3,4,5.

What you should do, if you have not access to nitoethane (to make the amphetamine from the substituted benzaldehyde), nor an inert atmosphere (to make the bromo -> hydroxy swap), is simply to stay with easier alternative which look interesting:

....just ethylate eugenol then react it like usually to the amphetamine, to get the amphetamine analogue of MEPEA.

Please note than MEPEA is active at 300mg, a dose at which MDPEA is inactive.... By extrapolation you could conclude than MEA (4-ethoxy-3-methoxy-amphetamine) is more active than MDA.

That is only a guess, but what strike me is that such an easily obtainable compound as not yet been tested.

Maybee you will bee the first?