Testing new d-Meth routes(for me) without P2P or P2NP

btcboss2022

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Now that I have some time between productions I will test these routes without P2P or P2NP that I never done before because these new precursors wasn't available or I didn't search them:

2-Bromo-4'-methylpropiophenone->2-Bromopropiophenone->Ephedrone->d-ephedrone->d-Secondary amine->d-Meth

2-Bromo-4'-methylpropiophenone->2-Bromopropiophenone->Ephedrone->l-ephedrone->l-ephedrine->d-Meth

2-Bromo-4'-methylpropiophenone->2-Bromopropiophenone->Ephedrone->dl-ephedrone->dl-ephedrine->dl-Meth->d-Meth


As you can see, there may be some variations depending on which product is chosen for enantiomer resolution.
However, it is clear that the sooner this is done, the fewer reagents will be used due to the reduced amount of product left to process.
It is important to consider the difficulty of this process for each product and the potential impacts on the chirality of subsequent processes.
I'm waiting some ordered reagents to finish the routes but they should work properly.
If anyone already has experience with any of them, any suggestions or tip will be welcome.
Thanks.
 

Acidosis

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This scheme is correct, some time ago I got that same route from a Polish source of friends, I use these reagents with success.
 

Paranoia

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I paid 220 euro and submit a @btcboss2022 ice sample lets see soon the results, can take up to 2 months (sample is on the way)
 

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Propiophenone->2-Bromopropiophenone->Ephedrone->Ephedrine->dl-Meth->d-Meth
 

azides

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Yeah I thought this looked more correct.

You could also do a slight variation and make PPA and make D amph if I remember or 4-mar for thr advanced
 

btcboss2022

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After small tests the easiest way seems:

2-Bromo-4'-methylpropiophenone->4'-methylephedrone->Ephedrone->dl-Ephedrine->dl-Meth->d-Meth
 

OrgUnikum

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How are you going to clip the 4-methyl off the aromatic ring?

Also the resolution of the racemate would be better done on the ephedrine, This works much better then then on Meth, the second chiral center giving more hold to cling on. I think here tartaric acid will work without issues.
 
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hoboking

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so 4-Methylmethamphetamine would be the end product?that is pretty weak drug compared to meth
 

OrgUnikum

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Naaa, he clearly says he wants d-Meth
 

btcboss2022

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Oh thanks I thought that was more difficult to do it in ephedrine I remember to read that somewhere I don't know where now hahah
It seems like you've read my mind hahaha over the 4-methyl issue I have a dilemma on that point.
The idea was an oxidative demethylation to selectively oxidize the methyl group to a methylene group but I would be in the same situation I think wont be easier with methylene so from 3-4 days ago I was thinking to take it off from 2-Bromo-4'-methylpropiophenone just heating it with NaOH solution, so the new scheme would be like this:

2-Bromo-4'-methylpropiophenone->2-Bromopropiophenone->Ephedrone->dl-Ephedrine->l-Ephedrine->d-Meth

Thanks.
 

OrgUnikum

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Get some propiophenone (no problem in Poland), brominate with NBS at the 2- position (with tosic acid it is said to be almost quantitative), do the substitution with methylamine (add a tiny amount of NaI if it is bitchy) and then the optical resolution, followed by the NaBH4 reduction of the keto group to an alcohol. Or first reduce to the alc and then resolve? Whatever works best,
Everything tried and easy and mostly very high yielding reactions. All chemicals needed are easy to source. Make 4-MMC from the 4-methyl this makes more sense.
For the resolution a chiral acid is needed, tartaric comes to mind first, but ephedrine was in the olde days the most preferred chiral resolution agent for acids. A literature search WHAT they resolved big style back then will for sure bring up some nowadays easy available chiral acids. As what worked one way must work the other way round too.

good luck
 

btcboss2022

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Hello,

Thanks a lot for the info I found a supply for propiophenone and bromine so I will do it as you say.
About resolution I did hundreds of them with d-tartaric with racemic meth, mdma and some days ago I did it with 4MMC too.
But now I ordered O,O-Dibenzoyl-2R,3R-tartaric acid seems the most effective one and I will try it with racemic ephedrine. I will update you.
Thanks.
 

OrgUnikum

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. 2022 May 13;27(10):3134.
doi: 10.3390/molecules27103134.

Optical Resolution of Two Pharmaceutical Bases with Various Uses of Tartaric Acid Derivatives and Their Sodium Salts: Racemic Ephedrine and Chloramphenicol Base​


here you go! Article is free on pub-med here:
 

Paranoia

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I paid 220 euro and submit a @btcboss2022 ice sample lets see soon the results, can take up to 2 months (sample is on the way)
 

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OrgUnikum

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And if you have the correct stereoisomer of methcathione should be able to do the reduction to d-methamphetamine directly, without preparing the (pseudo)ephedrine first. This should also benefit overall yields.

Nevertheless the best way would be in my eyes to go with BMK -> P2P - reductive amination with (S)-(−)-α-Methylbenzylamine and NaBH4 -> d-N-benzylamphetamine -> debenzylation and monomethylation one pot with Pd/C and paraformaldehyde, done as CTH style reaction with formate salts as hydrogen donors -> d-methamphetamine.
The drawbck being the high costs of Pd/C, but still worth it regarding the value of the product.
 

btcboss2022

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Directly?
Until I know should be like this:
Propiophenone->2-Bromopropiophenone->Ephedrone->d-ephedrone->d-Secondary amine->d-Meth.
Right?
 
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OrgUnikum

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Propiophenone->2-Bromopropiophenone->Ephedrone->d-ephedrone->d-Meth
ephedrone is methcathione and I believe this can be reduced directly to methamphetamine. I mean you reduce once to ephedrine and then reduce again? I wlll look it up, I am sure though that the Birch (Lithium or sodium metal in liquid ammonia) would do it. What reduction system had you in mind for the ephedrine reduction?

Why did you write secondary amine instead of ephedrine? Methcathione as ephedrine as methamphetamine are all secondary amines as the have the methyl group on the amine.
 

OrgUnikum

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Ok looked it up and no problem, H3PO2/H3PO3/red phosphorus + iodine reduce carbonyls to the hydrocarbon say the =O of the methcathione will be removed. As its a double reduction the amounts of phosphorus compound and iodine will have to be adjusted, but thats all about it.

And btw. even H3PO4 together with NaI or KI will work although with diminished yields.

H3PO3/I or H3PO2/I is the best way, RP/I being quite violent and H3PO4/NaI or KI has lower yields
 

btcboss2022

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I thought in the past that a reductive amination with nabh4 and methylamine of the methcathinone could give N-methylamphetamine.
Theoriccally reacting with methylamine, the ketone group of methcathinone is aminated, introducing the N-methyl group necessary for N-methylmethamphetamine.
The reducing agent nabh4 facilitate the reduction of the ketone to a secondary amine.
But the secondary amine from that process is not exactly N-methylmethamphetamine is N-methyl-2-amino-propan-1-one.
 

OrgUnikum

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I just realized that Ephedrone is prone to the keto–enol tautomerism and will probably racemize in solution. This sounds bad but actually you might have stumbled here on something really neat. As you can still get the d-isomere to crystallize out of solution as when not dissolved the racemization won't happen. And with d-isomere crystallized out you can use Viedma ripening or a similar technique and you end up with 100% d-ephedrone salt.

And you can throw this salt in portions directly into the H3PO3/I reaction where it gets instantly reduced to d-Meth with no time to racemize, the strongly acidic conditions will help to prevent it too. Means you will not loose any of your product by separating the d-, all will be turned into d-.

And no this does not work for Meth or Amph, this only works for compounds which racemize by themselves or can be made to racemize by adding something (a ultra-strong base or so) to the solution.
 

btcboss2022

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So in that case to have the d-ephedrone in the solid part I must use l-tartaric and I guess it was a typo mistake but would be H3PO2/I reduction.
Thanks a lot for your help.
 
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