Author Topic: resolution of racemic meth  (Read 5340 times)

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zero_nrg

  • Guest
resolution of racemic meth
« on: March 12, 2004, 06:59:00 PM »
Have been reading through the procedures for seperating a racemic mix of meth  into D and L form.  The easiest seems to be a procedure from Rusznak.  Question: Any reason why toluene, or some other non-polar, should not be substituted for the benzene?

[

https://www.thevespiary.org/rhodium/Rhodium/chemistry/amphetamine.resolution.html%5DA

mixture of 0.1 mole (13.52 g.) phenylisopropylamine (or 14.92 g. methamphetamine base) in 60 ml benzene, 0.05 mole d-tartaric acid (7.50 g.) in 30 ml water, and 2 g sodium hydroxide (reagent grade or titrated equivalent) in 3 ml water was kept 4 hours with intermittent shaking, and the organic phase evaporated to give 98% L-phenylisopropylamine. The aqueous phase was extracted with benzene at pH 13 and evaporated to give 96% D-enantiomer

Rhodium

  • Guest
Toluene can most often substitute for benzene
« Reply #1 on: March 12, 2004, 10:57:00 PM »
I see no reason why you couldn't use toluene instead, but I would not reccommend you to use just any non-polar you have lying around.


elfspice

  • Guest
benzene bad
« Reply #2 on: March 20, 2004, 02:15:00 AM »
a lot of old articles and reports talk about using benzene, it used to be common as hexane and toluene and dcm are now.

... base) in 60 ml benzene, 0.05 mole ... 98% L-phenylisopropylamine. The aqueous phase was extracted with benzene at pH 13 ...


I would think that hexane and the lesser naptha should work too, it looks to me like the principle of the method is you put the right ratio of sodium ions and d-tartaric acid in there, and an ionic equilibrium develops that gradually drives the l-meth into the freebase over a long period of time.

note that 2g of NaOH is 0.0501 moles, so that would read:

A mixture of 0.1 mole (13.52 g.) phenylisopropylamine (or 14.92 g. methamphetamine base) in 60 ml benzene, 0.05 mole d-tartaric acid (7.50 g.) in 30 ml water, and 0.05 (2g) sodium hydroxide (reagent grade or titrated equivalent) in 3 ml water

in short form:

0.1667 mol H2O + 0.5 mol NaOH
01.667 mol H2O + 0.5 mol D-Tartaric Acid (grape-sourced tartaric acid)
0.1 mol d,l meth HCl (ie 0.1 mol of HCl molecules in there too)

I think that if they say .1 mol of phenylisopropylamine and 60ml of benzene then they expect that .05 mol of the l isomer will happily dissolve into it. in this described procedure they are more or less saying ~6.5g of l-amphetamine in 60ml of benzene... that's a lot of stuff dissolved in there... i think if you use toluene probably 50% extra volume of solvent and hexane maybe twice as much. somebee with more experience than me can suggest appropriate differences, most expert cooks would know how much is enough toluene per gram of meth to dissolve it.

i understand there would be a way to then extract that l-meth into acid solution and then it can be isomerised into racemic again and the separation done again. I would think that the 3rd or 4th time would have the majority of it out... this would in effect approximately 150% increase the effect of the stuff, lower the doses needed etc etc.

barkingburro

  • Guest
i have a better question
« Reply #3 on: March 22, 2004, 05:57:00 PM »
why do you want to seperate the two? swibb has had very very good results from a 1/3 - 2/3 and  1/4 - 3/4 mix of meth. l and d respectively. swibb's 'personal' is generally of te mixed variety

zero_nrg

  • Guest
percentages obtained
« Reply #4 on: March 31, 2004, 10:22:00 PM »
Barkingburro:  how would one obtain a 75/25 mix of D/L (or any other ratio besides 50/50) if the racemic mix was not seperated first?  Also, based on recent experience, I am not sure if a racemic mix crystalizes like pure D (and probably L-but I have not tried it).  My limited experience with this is that of charliebrown in post

Post 460635 (missing)

(charliebrown: "kinetics of racemic meth crystallization", Stimulants)
where the mix just crystalizes all at once into very small crystals (note exception below).  Would it make sense that a racemic mixture could not form large crystals as the isomeric configurations in space are different for D and L ? 
Another question, after gassing a batch of racemic freebase in toluene the crystals were vac filtered and the mother liquor poured back into the beaker where the base was originally gassed.  There was perhaps a gram of tiny white crystals floating around in the toluene and the toluene was again gassed with HCl to preciptate more crystals if any.  Not only did no more crystals precipitate, but the ones that were floating around in the toluene disappeared into the toluene.  No matter, just left the beaker sitting in the sink as the remaining crystals were not worth the effort to recover.  After approx. 24 hrs looked at this beaker again and to my surprise the most gorgeous absolutely clear crystals had formed on the bottom of the flask.  Not huge crystals but certainly larger than than the tiny white crytals originally swimming around in the toluene.  The gas was generated with H2SO4 on NaCl and equipped with calcium chloride drying tube which was removed not too long after the gassing procedure started (clogged up).  Could water be responsible for the disappearance of the tiny white crystals in the toluene which then precipitated as the liquid cooled after any non-polar crap was solvated by the toluene?  The original reduction on P2P with NaBH4 via the Labtop dry method.  -zero

barkingburro

  • Guest
not an expert but
« Reply #5 on: April 04, 2004, 11:45:00 PM »
it has appeared to swibb that the d and l isomers x-talize in different layers in ice cold anyhydrous ipa. which is which swibb is not really sure of, but one seems to make a longer flat rectangle of a xtal and the other is more like a cubish xtal.  the cubish xtals come out looking kinda like a q-bert board and the others kinda like 2x4's.

swibb was also assuming that u had a single batch of racemic with a known % mixture and could then adjust ratio by adding additional 100% d-. not trying to come across as an expert in seperation just wonderin y go through the trouble

Rhodium

  • Guest
no spontaneous isomer separation
« Reply #6 on: April 05, 2004, 01:35:00 AM »
No, spontaneous isomer separation rarely happens. Either it is two different crystal shapes (less likely) or an impurity (more likely).


zero_nrg

  • Guest
Louis Pasteur used a micrscope and tweezers to
« Reply #7 on: April 28, 2004, 08:26:00 PM »
Louis Pasteur used a micrscope and tweezers to seperate racemic tartaric acid into the D and L form.  Those crystals must be very different in their geometry which, as Rhod indicates, is rare. -zero

Rhodium

  • Guest
To be precise, it was sodium ammonium tartrate
« Reply #8 on: April 28, 2004, 09:15:00 PM »
[Louis Pasteur] discovered that sodium ammonium tartrate crystallizes in two enantiomorphic forms. After mechanical separation of the different crystals, using tweezers under a microscope, their aqueous solutions rotated polarized light into different directions! The grand old man in the field of optical rotation, Jean Baptiste Biot (1774-1862), insisted that Pasteur had to repeat his experiments in public. Pasteur was successful because two fortunate circumstances worked together. First, he had selected sodium ammonium tartrate which is one of the very few salts of tartaric acid that forms enantiomorphic crystals which can be separated manually; second, he did his crystallization at temperatures below 26°C (79°F); at higher temperatures, only the racemate crystallizes.

Taken from

Journal of Receptor and Signal Transduction Research 19, 15-39 (1999)

(http://home.t-online.de/home/kubinyi/chanc-tx.pdf)


zero_nrg

  • Guest
seperation anxiety
« Reply #9 on: June 06, 2004, 09:44:00 PM »
Excuse my general chem. and A/B ignorance, but regarding the Rusznak procedure above, can anyone say definitively whether the freebase or the HCl is used in the procedure?  If the HCl is used, as elfspice indicates, why is the weight given for the freebase?  Also, I´m assuming that  " (or 14.92 g. methamphetamine base) in 60 ml benzene" means to dissolve the freebase in the benzene (doubt the HCl dissolves in benzene).  Molar equivalents of D-tartaric acid and NaOH will react to form a salt (tried this last night to satisfy my curiosity).  So even if the  freebase were more reactive wouldn´t the D-tartaric acid first react with the NaOH to form a salt simply because they are in the same phase? If the tartaric acid did react with the FB first, wouldn´t it just be liberated by the basic conditions in the aqueous layer and float up with the benzene?  When I first read this procedure I thought that the write-up was just sloppy and what was meant was to take the FB dissolve it into the benzene, add the tartaric acid in water, wait and shake 4 hrs, seperate in funnel, now use the 2 grams NaOH in 3 ml H2O to liberate the D-freebase which dissolved into the aqueous layer as it reacted with the acid.  Very confused.  -zero

Rhodium

  • Guest
the freebase should be used
« Reply #10 on: June 07, 2004, 01:50:00 AM »
Elfspice incorrectly assumed that the NaOH was added to the solution to freebase the added amine hydrochloride. It is not. The freebase amine is to be used, and the purpose of the NaOH is to make the monosodium salt of the tartaric acid (which is a dicarboxylic acid).

In solution, the monosodium d-tartrate then preferentially forms a water-soluble salt with d-methamphetamine while the l-methamphetamine (still remaining in its freebase form) can be extracted with a non-polar solvent.

The reason all this happens is that because of the molecular shape of d-tartaric acid is such that in solution it fits better together with d-methamphetamine than with l-methamphetamine, given a choice (but will freely form a salt with both forms if enough acid is present, therefore one carboxylic acid group on each tartaric acid molecule is 'disabled' by the addition of one equivalent tof NaOH).



latest addition (06-13-04): A stereoisomeric confusion by me corrected above (thanks for spotting it, bio!).
I incorrectly assumed that the preferentially formed salt was l-meth d-tartrate (as this is the salt isolated in many resolutions through fractional crystallizations). The correct interpretation of Rusznak's procedure has been entered above.

Also, note the following important discrepancy between the isomer designations:

The dextro isomer of tartaric acid is the d, (+), L or S isomer; the levo isomer is the l, (-), D or R isomer.
The dextro isomer of (meth)amphetamine is the d, (+), D or S isomer; the levo isomer is the l, (-), L or R isomer.


zero_nrg

  • Guest
Thanks Rhod, have two more questions: would it
« Reply #11 on: June 07, 2004, 03:53:00 AM »
Thanks Rhod, have two more questions: would it make sense  to react the NaOH with the D-tartaric acid before adding it to the freebase in benzene?  Also, why doesn´t the NaOH react with both carboxylic acid groups on some tartaric acid molecules?  Appreciate your insight.  -zero

Rhodium

  • Guest
monosodium tartrate equilibrium
« Reply #12 on: June 07, 2004, 05:26:00 AM »
I'd think it would be best to stick to the procedure as is, i.e. prepare the monosodium tartrate in situ.

The reason no disodium tartrate is formed is that the remaining carboxylic acid group becomes less acidic when monosodium tartrate has formed, so at equilibrium the result will be a solution containing only monosodium tartrate, because if a disodium tartrate molecule meets a free tartaric acid molecule, they'll immediately react to form two molecules of monosodium tartrate.


Rhodium

  • Guest
O,O'-Dibenzoyl-R,R-Tartaric Acid Meth Resolution
« Reply #13 on: June 09, 2004, 03:47:00 PM »
Optical Resolution in Two Immiscible Solvents in the Presence of an Intermediate Solvent
Optical Resolution of Methamphetamine by O,O'-Dibenzoyl-R,R-Tartaric Acid in Dichloroethane-Water-Methanol Solvent System

David Kozma and Elemer Fogassy

Synthetic Communications 29(24), 4315-4319 (1999)

(https://www.thevespiary.org/rhodium/Rhodium/chemistry/meth.resolution.di-bz-tartrate.html)

Abstract
A new optical resolution method was developed by the use of a two phase three component solvent system and demonstrated on the example of the resolution of the N-methyl-amphetamine with O,O'-Dibenzoyl-R,R-tartaric acid. By adding methanol as an intermediate solvent to the two phase water-dichloroethane solvent system the solubility and the optical purity of the precipitated salt can be adjusted in wide ranges. By this way a highly efficient one step resolution process can be accomplished.


bio

  • Guest
isomerization method
« Reply #14 on: June 12, 2004, 07:02:00 AM »
Now that I have all this L meth piling up an isomerization procedure is needed that is preferably simple and  has a high conversion ratio.

Once tried long ago a method from a Helvita Chimica Acta article from the 1920's, maybe Emde's stuff. It was very brief and supposed to convert a pure isomer of ephedrine into an equal percentage  racemic mixture......... Heat at 100 C degrees in a sealed tube with 25% HCl for 60 hours............ Simple as that if memory serves me correctly. I did it, but realized too late that my starting ingredient was really not the pure enantiomer and abandoned the project after that.

So the question is.......would this work for methamphetamine? A german bee should be able dig up the HCA article or is there another way quicker, could hardly be easier. Also I remember at the time why 25% HCl and not the 37% ?

Rhodium

  • Guest
Some racemization confusions cleared up
« Reply #15 on: June 12, 2004, 07:23:00 PM »
Now that I have all this L meth piling up an isomerization procedure is needed that is preferably simple and  has a high conversion ratio.

UTFSE. It's also in the Stimulants Forum FAQ.

Once tried long ago a method from a Helvita Chimica Acta article from the 1920's, maybe Emde's stuff. It was very brief and supposed to convert a pure isomer of ephedrine into an equal percentage  racemic mixture......... Heat at 100 C degrees in a sealed tube with 25% HCl for 60 hours...

It won't produce racemic ephedrine, it rather converts L-ephedrine to an equal mixture of L-ephedrine and D-pseudoephedrine (or D-ephedrine to an equal mixture of D-ephedrine and L-pseudoephedrine). This corresponds to racemizing the alcoholic chiral center while leaving the amine chiral center untouched.

Acids will not produce DL-Ephedrine regardless of how you do it, for that you need to use a strong base:

Post 480374 (missing)

(Rhodium: "Racemization of Optically Active Ephedrines", Stimulants)


So the question is.......would this work for methamphetamine?

No, the procedure is specific for amino alcohols, not simple amines.

A german bee should be able dig up the HCA article or is there another way quicker, could hardly be easier.

With TFSE you will find that I have already posted Emde's articles in this forum:

Post 401416

(Rhodium: "Hermann Emde's Ephedrine Chemistry Treatise", Stimulants)


Also I remember at the time why 25% HCl and not the 37% ?

I assume that using a higher concentration will result in some of the ephedrine being converted to chloroephedrine.


Rhodium

  • Guest
Please note that
« Reply #16 on: June 13, 2004, 02:27:00 AM »
I have made an important correction and addition to

Post 511856

(Rhodium: "the freebase should be used", Stimulants)



Rhodium

  • Guest
Pope-Peachey Optical Resolution of Meth
« Reply #17 on: October 05, 2004, 03:49:00 AM »
Study of the Mechanism of the Optical Resolution of N-Methylamphetamine via Diastereoisomeric Salt Formation by the Pope-Peachey Method
David Kozma, Zoltán Madarász, Maria Acs and Elemér Fogassy

Tetrahedron Asymmetry, 5(2), 193-194 (1994)

(https://www.thevespiary.org/rhodium/Rhodium/pdf/meth.chiral-sep.pope-peachy.pdf)

Abstract
During the optical resolution of racemic N-methylamphetamine by half an equivalent of R,R-tartaric acid in the presence of half an equivalent of hydrochloric acid the tartrate salt of the R-base precipitates with a low optical purity. The optical purity of the precipitated salt is increased by enantiomer exchange between the solid and solvent phase.


dwarfer

  • Guest
Unsubstantiated separation technique
« Reply #18 on: October 05, 2004, 02:47:00 PM »
[?534585]

?It is my opinion that at least two isomers of
MethAmphetamine can be separated in great degree
by a technique discovered serendipitously while
doing as described in the referenced thread.

In my mind it even makes sense that the re-precipitation
temperature would be separate among the isomers, since
the melt temperatures are.

The technique, if valid, works best  if the isomers are
visually distinguishable from one another in conformation.
(otherwise you would have to keep a close eye on the thermometer.)


<<Marve is looking for the MP for the (-)- MA.HCl if anybody knows.. ::)


methyl_ethyl

  • Guest
crossed polaroids
« Reply #19 on: October 05, 2004, 06:02:00 PM »
Louis Pasteur used a micrscope and tweezers to seperate racemic tartaric acid into the D and L form.  Those crystals must be very different in their geometry which, as Rhod indicates, is rare. -zero

More than likely the separation was done not based on visible differences in crystal geometry, however done through separation of visible crystals through crossed polaroids.

I have never actually tried separating crystals using crossed polarized light microscopy, however I think that it could be done.  When looking through a properly set up microscope utilizing crossed polarized optics it is possible (depending on which way your polarizer is set up) to only view dextrorotary, or the levorotary crystals i.e. turning the polarizer over the light source at 90° angles from each other, unless the crystals are birefringent, they will only be visible at one of the two poles, thus making a fairly easy separation.  For example all of the  visible crystals viewed under the initial pole could be moved toward one side of the microscope field.  Then rotate your polarizer 90° and move all of the visible crystals to the opposite field.  After a few repetitions of this all crystals will be seperated according to their visibility under crossed porlarization.

Now we have two groups of crystals separated by their visiblility under crossed polaroids.  I believe it is not possible to know at this point which "pile" of crystals is d- or l-, however you do know that these piles are opposite to one another as far as rotation is concerned, thus a solution of the crystals should be analysed via spectropolarimetry in order to determine which is d- and which is l-.

If I have time, (ha ha) it would be a nice to re-create this experiment and post some pictures.  I used to be very intrigued by crossed polarization microscopy and spectropolarimetry, although very simple concepts, it is quite amazing what the data generated can tell you about the molecule that you are analyzing.  Then again I am easily amused,  ;)

regards,

methyl_ethyl