Dextroamphetamine Synthesis (Nabenhauer, 1942)

G.Patton

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Introduction

At the moment, there are many ways to synthesize dextroamphetamine. They can be divided into 3 types: biosynthesis (using biomass), direct synthesis, and synthesis of racemic (the sum of l- and d-isomers) amphetamine followed by separation of optical isomers. In our case, a synthesis was chosen, in which simple reagents and rapid synthesis were used, which is maximally adapted to «home conditions». The process is as follows: we obtain the racemic amphetamine in the classical way, then we divide it into 2 optical isomers (l- and d-) by the Nabenhower method [US patent 2276508, Nabenhauer FP, "Method for the separation of optically active alpha-methylphenethylamine", published 17 March 1942, assigned to Smith Kline French].

Synthesis

Synthesis of amphetamine from P2NP via Al/Hg. First, we make an aluminum amalgam. It is needed in order to clean the aluminum from the strong oxide layer that forms when interacting with air. We take 14 g of aluminum foil and tear it with our hands into pieces of 2x2, 3x3 cm in size. Be sure to tear, not cut, to increase the surface area. Place into a 3-necked round-bottomed flask and fill the foil completely with water.
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Now we are preparing the mercury salt. We take a mercury thermometer from the pharmacy, wrap it in paper, break it at the bottom tip. Pour all the mercury into a glass, where then add 4 ml of nitric acid (70%). Do not forget that mercury vapors are hazardous to health! To initiate the reaction, the glass had to be heated to about 50 degrees, stirring occasionally. All the mercury dissolved about for 30 minutes, and an orange gas, nitrogen oxide (IV), was released from the glass. The reaction equation is as follows:
Hg + 4HNO3 ----> Hg(NO3)2 + 2NO2 + 2H2O
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Pipette 2 ml of the solution and place it in a round-bottom flask with foil. After about 5 minutes, the foil lost its shine, became dull, and a small layer of gray sludge (aluminum hydroxide) collected at the bottom of the flask.
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We drain the liquid and rinse the foil with water 3 times.
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Pour 30 ml of water into the flask, insert a thermometer into the left throat of the flask, insert a reflux condenser into the central throat, and insert a dropping funnel with 110 ml of 14% P2NP (it is phenylnitropropene; 1-phenyl-2-methyl-2-nitroethylene) solution into the right throat.
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Acetic essence is often used to produce hydrogen, but I "start" the reaction to produce hydrogen with water. Less acidic medium, which means less alkali must be added later. Many people ask the question: "How to remove this water?" There is no need to remove water anywhere, it reacts with aluminum and hydrogen is obtained:
2Al + 6H2O ---> 2Al (OH) 3 + 3H2
It is very important to remember that the reaction of P2NP reduction comes with a very! large exotherm. It is necessary to carefully monitor the temperature and prevent overheating above 60 degrees. Personally, I kept the temperature around 50-55 degrees. Failure to comply with this technology reduces the yield of the product and gives a colored (pink, orange) product. The infusion of the entire P2NP took about 50 minutes. Change the dropping funnel to a glass stopper. We got just such a gray solution.
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To increase the yield, you can heat the reaction mass at a temperature of 50-60 degrees for 30 minutes in a water bath.
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We cool the mixture to room temperature, put a plug instead of the thermometer, remove the reflux condenser. We make an alkali solution based on 1 part of sodium hydroxide - 2 parts of water. Dissolution proceeds with heating, so we wait until the solution cools down. It is not worth pouring solid alkali into the reaction mass or pouring a hot solution, as this reduces the yield, like any overheating. Pour alkali cooled to room temperature to the reaction mass until pH = 11-12, wait 30-40 minutes until all the floating aluminum dissolves, yellow oil floats up. At the same time, we also monitor the temperature. Reaction equations:
2Al + 2NaOH + 6H2O ==> 2Na[Al(OH4)] + 3H2
NaOH + Al (OH)3 ---> Na[Al(OH)4]

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We pour everything into a separating funnel. We are waiting for the separation of layers. Take the oily fraction.
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We extract amphetamine from the sludge, washing it 3 times with 10 ml of petroleum ether. Divide the upper part with a separating funnel. Combine all extracts with "oil" and put them in ice water for cooling.
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A large drop of water remains in the bottom of the glass, which is separated on a separating funnel. Pour the top layer into a glass and dry over anhydrous magnesium sulfate. There we clean the amphetamine from the remnants of mercury and water.
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We filter the liquid from solid magnesium sulfate on a Buchner funnel.
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I take concentrated 98% sulfuric acid. Prepare a solution of sulfuric acid in acetone in a volume ratio of 1:10. I took the technical acetone, in the hardware store, and distilled it, taking away the "heads" and "tails". Then I dried it with anhydrous magnesium sulfate. Many people ask whether is it possible to make a solution in IPA. Yes, you can, but IPA (isopropyl alcohol) is evaporated longer than acetone.
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Drop by drop, carefully and with stirring, acidify with sulfuric acid in acetone to pH = 6. A white precipitate forms at the bottom.
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Cool the reaction mass in ice water, filter the precipitate on a Buchner funnel, rinse with 3 ml of cold acetone.
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Air dry the filtered product and weight.
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7.55 g (0.0205 mol) of amphetamine sulfate was obtained.
Calculations:
m (P2NP) = 110 * 0.14 = 15.4 g.
n (P2NP) = 15.4 / 163.17 = 0.094 mol.
n (amphetamine sulfate) = n (P2NP) = 0.094 mol.
n (amphetamine base)= 0.0205 * 2 = 0.0410 mol.
The reaction yield is 0.0410 /0.094 = 43.6 %.
You will carry out different reaction with sulfuric acid. Amphetamine free base 2 mole + 1 mole of sulfuric acid = 1 mole of amphetamine sulphate. When you count amph. sulphate, you have to multiply by two your mole result of amph. sulphate cuz you take 2 mole amph. base for one mole of amph sulfate.

Extraction of d-amphetamine

We've got racemic amphetamine. It contains 1 molecule of d-amphetamine per 1 molecule of l-amphetamine. Next, 6 g of racemate is taken and dissolved in 6 ml of water, an alkali solution is added to reach pH = 11.
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Extract with 5 ml of petroleum ether and warm the solution, add 2.45 g of d-tartaric acid in alcohol solution to the mixture. Then add alcohol until completely dissolved and cool with stirring. The l-amphetamine d-tartaric salt precipitates. The d-amphetamine remains in the solution. You can repeat procedure of cleaning precipitate of l-amphetamine d-tartaric salt by methanol to increase yield.
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We precipitate d-amphetamine with an additional amount of d-tartaric acid. We filter the precipitate, get the base of d-amphetamine, adding alkali to pH = 11.
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We acidify the d-amphetamine base to pH = 6 with a solution of sulfuric acid in acetone. This gives 2.63 g of d-amphetamine sulfate. Yield is 2.63 / 3 = 87.7%
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This extraction method valid to amphetamine, which was synthesized by any routes. There is another way to obtain dextroamphetamine.

Source

Nabenhauer, Fred P. "Method for the separation of optically active alpha-methylphenethylamine." U.S. Patent No. 2,276,508. 17 Mar. 1942.
 
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diogenes

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Hi, great explanation, I have been looking for this clarity for a long time. Just two questions. Pure H2SO4 is difficult to optain in my country, they are selling aquarium Ph adjuster containing H3SO4 (Orthosulphuric acid), would this also be good. What is the best commercially available sulphate recommended by you? What should be used for basifying the solution (commercially available). I think caustic soda comes in lower concentration, would this also be suffice? thank you for your help
 

G.Patton

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Hello. You can use diluted sulfuric acid (H2SO4), but in larger load. Sulfurous acid (H2SO3) isn't suitable because it's a weak acid and sulfuric acid (H2SO4) is strong. Caustic soda is the same as sodium hydroxide.
H3SO4 (Orthosulphuric acid) do not exist.
 
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diogenes

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Hi, sorry my bad I just stupidly confused Phosphoric acid (H3PO4) with Sulphuric acid (H2SO4). My question is basically where to get Sulphuric acid (H2SO4). Phosporic acid is easily available, but somehow I don`t find it as stimulating as the sulphate. (Your post finally clarifies and quantifies the difference an another topic, but even when adjusting for the dose it falls behind - I wonder whether absorption/excretion could be different - but this is off topic here.) So it would be great if I could try this separation method with D-sulphate in the end.

Another question, does the type or concentration of alcohol matter when adding the tartaric acid? thank you
 

G.Patton

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No matter.

You can find it as accumulator electrolyte in hardware store or auto shop, it contains ~36% sulfuric acid
 

diogenes

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Thank you for the information, it worked miraculously. I have evaporated the water and now have a pretty concentrated H2SO4. What is the proportion of sulphuric acid and aceton? Are they not going to make all kinds of new molecules when mixed together? Sorry if it is a dumb question but I`m a bit wary of mixing the two together. Also the NaOH solution for basifying: I have about 40% solution, does the concentration matter?
 

G.Patton

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Prepare a solution of sulfuric acid in acetone in a volume ratio of 1:10. (с)
Take 1 volume of acid and 10 volumes of acetone

It will be okay. You have to reach pH= 11-12, so, concentration not so important.
 

diogenes

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Thank you very much Patton. I will report back when I try this separation, I hope I will get relatively good yields.
 

G.Patton

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Good luck, man!
 
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diogenes

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Hi again :) I was looking at the numbers in the procedure as still waiting for some solvents to arrive. When starting the D-amphetamine purification, the 6ml of water given seems way too little. The solubility of the sulphate salt is given as 8.8 part water to one part amph. sulphate. Is it not a typo, did you not mean 60ml? Here is a link, but haven`t double-checked in the Merck:

Reading the original patent, Nabenhauer seems to say that if deficient amount of D-Tartaric acid is added, then the L-isoform crystalises first (as described by you above), however, if D-tartaric acid is added in excess, then it is the D-isoform which crystalizes first. Can you, by any chance, explain the chemistry behind this? Why is one isomer quicker to crystalize than the other? Why does it change when excess tartaric acid is added?

I`d really appreciate if you could shed some light on this, I could not find a source, but perhaps this is because it is obvious for someone with more chemistry knowledge/experience. thank you
 

G.Patton

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You can add water until it dissolves completely with minimum water. After, you'll add alkaline water solution with water, which will help to dissolve it as well.
Simply say: firstly, stereoselective reaction is carried out on more "easier" and low-energy way with l-isomer and after goes by second way on more high-energy way.
 
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SergMarsian

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Hello, since there is no dextroamphetamine on the market and I need it, there is a need to make it for myself on my own, I am a beginner, so I have some questions, as I understand it first with the use of tartaric acid with methanol - L amphetamine falls into the precipitate and then you need to take a solution without this precipitate and add without alcohol just d tartaric acid and add alkali until the PH reaches 11 and only then it can be filtered and get a dry base of D amphetamine, after which acidification with sulfuric acid in acetone and filtration to get dry pure d amphetamine, right?

Do need tartaric acid D(-) or D(+) ?
 

G.Patton

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Hi, no difference.
better to take ethanol
"an additional amount of d-tartaric acid." Means d-tartaric acid alcohol solution, sorry for the inaccuracy.
affirmative
 

SergMarsian

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Thank you very much for the answers, another question about the alcoholic solution of d-tartaric acid, d tartaric acid with ethanol (96%) in what proportion to interfere? until the tartaric acid dissolves?
 

diogenes

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Hi Serg, I think it precipitates when you add the extra D-tartaric acid solution, and this is when you filter, so you get a more or less dry precipitate which is D-tartaric-D-amphetamine. Then you add some alkaline solution to free the base from the tartaric acid, so you get a thin film of D-amphetamine base on top of a water solution of Sodium and tartarate. Then you have to extract the base with as little solvent as possible to make it very dense, and then add Sulphuric acid/Acetone to this to precipitate D-amphetamine sulphate.
 

diogenes

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Patton, would it not be better to use 100% ethanol, drying over some MgSO4 and distilling it?

Serg, we are pretty much in the same boat, I got into chemistry facing the same problem that only racemic amphetamine is available. That said, Patton makes it sound so easy, because it is easy once you are on his level in chemistry. There are many fine details you have to get right, e.g. having pure reagents anhydrous acetone, sulphuric acid etc.ILet me know how you get on.
 

G.Patton

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Yes, it's better to increase rxn yield
 

ASheSChem

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i don't understand how you calculate yielding :x
 

G.Patton

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We took 6g of racemate = 3 g of D-amph+3 g of L-amph
We got
2.63g yield = x%
3.0g theory = 100%

2.63*100/3=87.7%
 

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Thanks for an interesting study. But I'm surprised with your tartaric acid proportions.
6g of amph sulphate is 3.49 g of racemic amph base. Then why do you use 2.45 g of d-tartaric? It shall come in molar proportions 1:4, i.e. 0.97g of d-tartaric is enough???
 
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Pororo

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I got your idea. I don't know why you believe you are going to get bitartrate. May be because of ethanol.

If I understand terms correctly, tartrate means 1:2, but bitartrate means 1:1 molar proportions!!!

But I don't use ethanol. I use other solvents.

What I know, crystallization stops as soon as 0.25 moles of acid consumed with yield 95%. Then, if 0.25 moles of opposite acid is added, precipitation will start again with yield 95%.
And now, when 0.25 of L and 0.25 of D were applied, if I try to add more D or L or DL acid to mother liquor - nothing happens. It looks there is no base, ph is also acidic.
For the prove, may be, may be, on occasion. Not sure when I will do amph separation again, end of September may be.
 
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Pororo

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I think, the point is here mr. Patton, and it is very simple: sequence fucking matters!!!

1. If you have a flask with with 1 mole of tartaric acid, d or l or dl, whatever, and start adding 1 mole of corresponding base in small portions, in the result you will get mostly bitartrate and some tartrate (+ some unreacted acid probably) with salt's resulting PH = 2.
2. This is what I do! If you have a flask with with 1 mole of base and start adding 0.5 mole of tartaric acid in small portions, in the result you will get mostly tartrate and some bitartrate (+ some unreacted acid.) with salt's resulting PH = 5.0-5.5.
Or 1 mole of racemic base and 0.25 mole of d or l acid if you are targeting to separate them
 
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