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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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 ..
BongMan20gram is alot,i think you need more practise,but doint lose another 20gram mate,you can start 1gram scale
 

Wael

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Introduction

There are many procedures out there for the production of N-methyl-amphetamines (methamphetamines) from various starting materials, such as phenyl-2-propanone (P2P), phenylacetaldehyde or ephedrine, but what if you already have an amphetamine and wanted to add a methyl group to amino group? If you used the first reaction that comes to mind for the conversion, to alkylate the amphetamine with methyl iodide or dimethylsulfate, you would be disappointed, as you would get a mixture of products, most important the N, N-dimethyl-amphetamine (of very low activity), as once the amphetamine has been methylated to methamphetamine, the molecule is much more susceptible to another alkylation, and thus the dimethyl- amphetamine is formed much faster than the remaining amphetamine is alkylated to methamphetamine. Actually, in the reaction mix you would find unreacted amphetamine, N-methylamphetamine, N, N-dimethyl- amphetamine and even some of a quaternary N, N, N-trimethylamphetammonium salt.

To avoid this happening, we must usually resort to indirect methods of introducing the methyl group. One way is to react the amphetamine with formaldehyde (either as an aqueous solution, or as paraformaldehyde) to get the amphetamine formaldehyde imine, which can then be reduced to N-methylamphetamine using a several different reducing agents, for example Al/Hg or Pt/H₂.

In this topic, I described the simplest synthesis of methamphetamine (4) from amphetamine salt via Imine (3) with help of Al/Hg reduction.

Equipment and glassware:
  • 5 L Round bottom flask;
  • Reflux condenser;
  • 1 L Separatory funnel;
  • Laboratory grade thermometer (0 °C to 100°C);
  • HCl gas apparatus;
  • Boiling chips;
  • Measuring cylinders 100 and 500 mL;
  • Vacuum source;
  • 2000 mL x1; 250 mL x3; 500 mL x3; 100 x2 Beakers;
  • Glass rod and spatula;
  • Laboratory scale (0.01-500 g is suitable);
  • Buchner flask (2 L) and funnel (or small Schott filter);
  • Retort stand and clamp for securing apparatus;
  • Ice water bath;
  • Glass rod;
  • Rotary evaporator (optional);
  • pH indicator paper;
  • 1 L Erlenmeyer flask.

Reagents:

  • 1 mole Amphetamine salt;
  • ~1 L 20% Sodium hydroxide (NaOH) water solution;
  • ~700 mL DCM or petrolium ether;
  • 1 mole (81 mL 37% or 75 mL 40%) Aqueous formaldehyde (CH2O);
  • 350 mL EtOH (ethanole 96-98%);
  • ~70 g of Al foil;
  • 1.62 g Mercury(II)nitrate (Hg(NO3)2);
  • ~500 mL 20 % Hydrochloric acid aq solution (HCl);
  • ~200 g Sodium or Magnesium sulphate (Na2SO4 or MgSO4) anhydrous.

Procedure


  1. You have to get 1 mole amphetamine free (1) base by addition salt of your amphetamine salt (sulphate or phosphate) to 20% NaOH water solution to pH 12. Stirr it for 15 min and extract amphetamine free base by DCM or petrolium ether 3 x 75 mL.
  2. Prepare Al amalgam. You have to use approximately ~70 g of Al foil and 1.62 g mercury(II)nitrate (Hg(NO3)2) to make amalgam as described in following topic.
  3. A mixture of 1 mole amphetamine freebase (136 g) and 1 mole aqueous formaldehyde (2) (81 mL 37% or 75 mL 40%) in 350 mL EtOH (ethanole) is poured into a 5 L round bottom flask with reflux condenser and an excess of aluminum amalgam, which was prepared in advance.
  4. Imine (3) is reduced for approximately two hours, to keep reaction temperature below ~50-60 °C. Cooling have to be applied if the reaction becomes too violent.
  5. 1 L Of cooled distilled water is added to the reaction mass, aluminum hydroxide solids are filtered off.
  6. The whole reaction mass is treated with 20 % HCl to pH 3 and extracted by DCM or petrolium ether 3 x 75 mL. DCM extracts are combined. Resulting water layer from filtration is alkalinized by 20% NaOH aq to pH 12 and extracted with DCM or petrolium ether 3 x 75 mL. Organic layers with methamphetamine (4) are combined (from first and second extraction of alkilized water layer), dried over Na2SO4 or MgSO4 anhydrous and concentrated in vacuum.
  7. Preparation of methamphetamine hydrochloride is carried out by bubbling of dry HCl gas via organic (DCM with methamphetamine) layer. The methamphetamine hydrochloride precipitate is filtered off by suction filtration, dried. Also, you can grow Ice methamphetamine by following methods.
G.Patton
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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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?
WaelHello, use Eng language in public post please.
In order to extract methamphetamine.
 

Wael

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Hello, use Eng language in public post please.
In order to extract methamphetamine.
G.PattonExcuse 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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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 ?
RaxdHello, 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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G.Patton

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

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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@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.
redsm
and NOT between N and the alpha-carbon as in the listed structure.
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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Hi. Firstly, a-carbon has more electron density than methyl carbon. Typical mechanism lead to such intermediate (hidden under Breaking Bad watermark, sry):
View attachment 27578
Secondly, do you have any reliable data about stereo-specific result of this synthesis?
G.PattonAs 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.
 

G.Patton

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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
, but it looks like the double bond is formed between the N and Carbon C1 from the mechanism diagram.
It is C2 in our case, carbon atom, bounded with methyl and benzyl. I'll send you another picture if you don't see on this:
GyZWpblFXn

Link: https://www.chemtube3d.com/nucleophilic-substitution-at-the-carbonyl-group-imine-formation/


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.
I have great doubts about stereoselectivity of this synthesis but I can't state that it's false without experiments.
 

redsm

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It is C2 in our case, carbon atom, bounded with methyl and benzyl. I'll send you another picture if you don't see on this:
View attachment 27794
Link: https://www.chemtube3d.com/nucleophilic-substitution-at-the-carbonyl-group-imine-formation/



I have great doubts about stereoselectivity of this synthesis but I can't state that it's false without experiments.
G.PattonUnfortunately 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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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.
redsmrhodium is a mistake, the double bond is not there, see the example with benzaldehyde below
 
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