Methods for Synthesis of Methamphetamine by Yogi

Discussions of any and all sorts of mind-warping chemicals, pills, booze, mind machines, trippy stuff, current street prices, importing pharmaceuticals, smart drugs, nutrients, herbs, and altered states. Please try to keep your conversations "theoretical" in nature, since this is an open conference.

Moderator: murderman143

Post Reply
User avatar
Site Admin
Posts: 402
Joined: Thu Jun 01, 2017 7:35 am
Location: The Interweb

Methods for Synthesis of Methamphetamine by Yogi

Post by Helladamnleet » Fri Aug 25, 2017 12:50 pm


There've been a lot of synthesis methods proposed on UseNet for synthesizing methamphetamine. Thus far, I haven't seen one that I imagine would work.

One, from Phrack magazine, is the "tried and true method" for prepping meth from Vick's nasal inhalers. Vick's nasal inhalers contain "l-desoxyephedrine," another name for "l-methamphetamine." The l- isomer of methamphetamine is the relatively inactive one, usable as a (mild) nasal decongestant. The d- isomer is the one that everyone wants and that Uncle Sam has declared is just too cool for anyone except doctors.

The procedure described would extract the l-meth froom the inhalers and collect it and that's it. I'm sorry, but the Isomer Fairy can't wave her magick wand and reverse the chirality of the molecule. The only way to change between the two isomers is to oxidize the l-meth into phenylacetone, condense it with methylamine, then reduce it. Sorry, but soaking inhalers in HCl then separating the "juice" with Et2OH just won't do it. You'll get l-meth and that's that.

A more credible souding one mentions that "methamphetamine is prepared by the calalytic reduction of pseudoephedrine in acetic acid" blahblahblah and then goes on to describe, not catalytic reduction via acetic acid, but reduction with sodium borohydride. I'm sorry to say that no method attempting to directly reduce (pseudo)ephedrine's hydroxyl group is going to work. You can't expose it to a strong acid, or a weak acid, or sodium borohydride, or even lithium aluminum hydride and expect it to reduce at all. As with the Vick's Inhalers "recipe," you get a lot of SOMETHING, but it ain't d-meth. All you'll be left with is your (pseudo)ephedrine and a bunch of acid, lithium, and/or sodium and lotsa hydrogen gas. This is because the hydroxyl group (the OH in ephedrine) is on a very acidic carbon (the first carbon away from the ring) and a hydroxyl group is very basic. If the hydroxyl were on the second carbon from the ring (the carbon with the amine group, the NH2 or NHCH3), there might be some chance, but it's not and there's not. You're not getting a basic group off an acidic carbon without a fight, and acids, borohydride, and LiAlhydride aren't gonna fight that hard.


One of the easiest ways to make methamphetamine is from amphetamine. Of course, this assumes you have amphetamine in the first place, but let's just pretend you have some and you want to spice it up a bit. The difference between amphetamine and methamphetamine is the addition of a single methyl group (CH3) to the amino group sticking off the middle carbon atom in the chain. Fortunately, substituting amines is really simple. Vaporize your amine (your amphetamine) with a bunch of vaporized chloromethane (CH3Cl, a solvent) and some gaseous pyridine... voila, the amino group takes the methyl from the chloromethane and lets a hydrogen go. The hydrogen joins the liberated chlorine, and the resulting HCl is soaked up by the pyridine. The pyridine is optional. Adding it drives the reaction a bit by pulling the excess HCl out of the equation, but it's not neessary.

Assuming you don't have amphetamine lying around, an easy synthesis with a very high yield is to reduce the condensation product of phenylacetone and methylamine. The benefit of this method is that different amines can be used to produce novel N-alkyl amphetamines (ethamphetamine, tert-butylamphetamine, etc)

Making it from ephedrine or pseudoephedrine is possible. The only difference between methamphetamine and (pseudo)ephedrine is that damn alpha-hydroxy group. Reacting your ephedrine with thionyl chloride replaes the OH with Cl to produce N-methyl-alpha-chloroamphetamine as an intermediate. Hydrogenating this product is easy: use lithium aluminum hydride, sodium borohydride, or even hydrogen gas with nickel or platinum metal as a catalyst. The product of this step is N-methylamphetamine and HCl. Evaporate off the water and you have methamphetamine hydrochloride.

A surprisingly simple synthesis is possible from the amino acid phenylalanine, which is available at health food stores for about $14 for 100 tablets. Phenylalanine is 2-amino-3-phenylpropanoic acid, which is more or less amphetamine with a COOH where the CH3 should be at the end of the chain. Thionyl chloride will replace the OH with a Cl, which falls off and is replaced by H when you give it lithium aluminum hydride, sodium borohydride, or hydrogen gas and nickel/platinum. If you use hydrogen and metal for that step, you'll have to reduce the carbonyl group with one of the hydrides, so best save time + effort and use them and do both reductions at once. When that carbonyl is reduced, you now have amphetamine. Go back up to that first one I mentioned for upgrading amphetamine into methamphetamine.

Note that azll of these (and probably anything anyone ever comes up with) will give you a mix of d- and l- isomers. The d- is cool, the l- is shit, remember. If you have time, energy, and equipment, you can separate the two and reprocess the l- into d- by oxidizing it and re-aminating it as described in the "critique" of the Phrack synthesis.


H /\\ / \ NH / \\ / \ / || | | || | | || | CH3 amphetamine \ // \// H /\\ / \ NCH3 / \\ / \ / || | | || | | || | CH3 methamphetamine \ // has a CH3 at N that amphetamine doesn't \//

OH | H /\\ / \ NCH3 / \\ / \ / || | | || | | ephedrine and pseudoephedrine || | CH3 the difference is whether the OH \ // points up or down \//

H /\\ / \ NH / \\ / \ / || | | || | | || | C=O phenylalanine \ // | compare to amphetamine \// OH

Direct hydrogenation over Pd or Pd on a carrier is well known and facile. You add a little perchloric, phosphoric or sulphuric acid, which esterifies the-OH group that you're complaining about.

Thus making the intermediate halide via SOCl2, like you mentioned, is unecessary.

Hydrogenation starting with (-) ephedrine, whether direct or via the halide, will give d-meth. If you start with dl-ephedrine, you get dl-meth.
[IP: logged]

Post Reply

Who is online

In total there is 1 user online :: 0 registered, 0 hidden and 1 guest

Most users ever online was 79 on Thu Sep 20, 2018 11:09 am

Users browsing this forum: No registered users and 1 guest


Total posts 557
Total topics 301
Total members 152
Our newest member TanYaoW