The simplest Methamphetamine synthesis from Amphetamine

BongMan

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i don't know what gone wrong but i lost almost 20 gm of pure amphetamine , i tried to neutralist it with NaOH, and extract with petroleum ether nothing was left on evaporation , tried again with DCM again nothing left on evaporation , guys anybody tried this method , if anyone going to try use small quantity ..
 

metux

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20gram is alot,i think you need more practise,but doint lose another 20gram mate,you can start 1gram scale
 

BongMan

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i tried with it 4-5 times so i lost total 20 gm.
 

Wael

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I do not understand step 6 in acid-base extraction. We are supposed to make the solution alkaline and extract, but add the acid first. Why do I ask an expert to explain this step to me?
 

G.Patton

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Hello, use Eng language in public post please.
In order to extract methamphetamine.
 

Wael

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Excuse me, sir, am I wrong in publishing the message? I did not mean to annoy, but I wanted to share my voice with the expert to understand the point of extraction from this synthesis
 

Raxd

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Can the same principle be used for MDMA from MDA free base ? If the solution of Formaldehyde is a water-based one,this would be a problem ? And if so , adding more anhydrous EtOH would solve the water content inbalance ?
 

Wael

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Hello, I think that water does not affect the reaction. It is important that the formaldehyde is of high purity and the concentration is 40% or 37%.
 

Mr Gonzo

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With the replacement of amalgam to LAH, on the Erowid page that is linked to this thread. Tech 3 I believe is Shulgin's, using formic acid and LAH in THF on MDA to MDMA.
Can D-amph be used in place of MDA to produce the D-meth product, using the formic acid procedure?
 

oneimone

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Raney nickel can be used insted of al/hg amalgam? always problem with fucking mercury
 

cokemuffin

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Would it also be possible to methylate the imine which forms from p2p and ammonia with ch3cl to get methamphetamine?
C8LRl0bMiZ
 

cokemuffin

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cokemuffin

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which imine does not exist, the 1st or 2nd one, cuz i read about reductive amination/alkylation on erowid and there showing those imines?
 

G.Patton

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I told about first one. This topic about another synthesis. Make new topic if you wanna discuss this issue.
 

cokemuffin

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if you can't methylate the imine i guess it's worthless
 

Germanium

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Can the same principle be used for 6MAPB/5MAPB from 6APB/5APB ?
 

redsm

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@G.Patton I believe one of the drawn intermediates on this synthesis (and from the original rhodium article) is incorrect.

(3), the representation of the amphetamine-formaldehyde imine intermediate possesses the same structure as the phenylacetone-methylamine imine that can be seen elsewhere

this does not seem correct to me for two reasons: first, drawing the standard mechanism for imine formation on the substrate produces an imine wherein the double bond is between N and the would-be methyl group on the methamphetamine, and NOT between N and the alpha-carbon as in the listed structure. and think about it: the double bond is generated when the lone pair of the amine kicks off the water formed on the formaldehyde oxygen as a leaving group, which must be done on the side of the formaldehyde (the would-be methyl group)

second, the synthesis is reported as stereospecific based on the starting substrate. that is, the reduction of d-amph will yield d-meth and the same for the l-isomer. structure as is shown (3) possesses no chiral centers, and its reduction is equivalent to the standard reductive amination of phenylacetone which yields a racemic product. the proposed structure I've given for the intermediate possesses a chiral center, which would be the same as the starting material's and therefore match the observed stereospecificity of the reaction.

if i am correct, this would clear up the issues regarding the stereospecificity of the reaction earlier in the thread. if i have made a mistake somewhere, i apologize, but i was curious as to why the reduction of the non-chiral (3) leads to a non-racemic mixture of products when a non-racemic amphetamine starting material is used.
 

G.Patton

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Hi. Firstly, a-carbon has more electron density than methyl carbon. Typical mechanism lead to such intermediate (hidden under Breaking Bad watermark, sry):
MEbZ1GwD3P

Secondly, do you have any reliable data about stereo-specific result of this synthesis?
 
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redsm

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As you said, I cannot see the final product of the mechanism because of the logo on the picture, but it looks like the double bond is formed between the N and Carbon C1 from the mechanism diagram. If my guess about the picture is correct, then the picture you've given supports my hypothesis that in the reaction between amphetamine and formaldehyde the imine intermediate has a double bond between N and the methyl carbon and NOT between N and the alpha carbon.

If the mechanism photo you have provided shows something else under the occluding logo, please send a link to where I can see the diagram without the logo, as I would like to see it, however it seems to support what I have said earlier.

However my mechanistic understanding is not top-tier and is certainly lacking in expertise. Do you have any idea on how the above mechanism can generate a double bond between N and the alpha carbon, given that the O that forms the H2O leaving group is attached to the formaldehyde C1 carbon from the mechanism diagram?

As for "reliable data" about the stereospecifity, I have none other than the original Rhodium article purports that the synthesis is stereospecific, as another member said early in the thread on page 1 of replies, here: https://www.erowid.org/archive/rhodium/chemistry/amphetamine.methylation.html

The above article states that forming imine with d-amphetamine and reducing gives d-methamphetamine. Given that the currently-shown structure (3) is identical to the phenylacetone-methylamine imine, the reduction of which yields a racemic product, it is clear that one of the two must be wrong, no?

Either the amphetamine-formaldehyde imine is different than the phenylacetone-methylamine imine, or the reaction is not stereospecific as is stated in the Rhodium source.

Someone with the adequate resources could perhaps test the reaction with different optical mixtures of amphetamine and report back, however I cannot gather this data.

Thank you for your replies.
 

redsm

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Unfortunately I still do not understand from the picture how the structure (3) is generated from amphetamine and formaldehyde reaction. Your picture shows methylamine attacking acetone, which is much more analogous to the formation of the phenylacetone-methylamine imine than what I believe to be the amphetamine-formaldehyde imine though they use the same mechanism. I cannot see how the double bond forms with C2 in our case when it must form with C1 in order to kick off the H2O leaving group, which is connected to C1 (the formaldehyde carbon that becomes the methyl group carbon).

I encourage you to draw out the full mechanisms of both the formation of imine from phenylacetone-methylamine and of amphetamine-formaldehyde to compare and see for yourself. Perhaps I am making an elementary mistake, but when I do so I get two different products on paper. For these products to be identical, there must be some step after the loss of H2O leaving group where the double bond on N shifts from C2 to the alpha carbon, which is not represented in the standard imine formation mechanism, but maybe this is the step I am missing.

I hope you can at least agree that either the given structure of the imine (3) is wrong or that the synthesis is not stereoselective.
 

NexusPrime

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rhodium is a mistake, the double bond is not there, see the example with benzaldehyde below
 
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