Is l-tartaric(D-tartaric, levotartaric) used in the video or racemic one? Not clear but I guess that if D-meth is in the solid part should be l-tartaric(D-tartaric, levotartaric) right? With d-tartaric(L-tartaric, dextrotartaric) the D-meth should be in the liquid part. Thanks.
Is l-tartaric(D-tartaric, levotartaric) used in the video or racemic one? Not clear but I guess that if D-meth is in the solid part should be l-tartaric(D-tartaric, levotartaric) right? With d-tartaric(L-tartaric, dextrotartaric) the D-meth should be in the liquid part. Thanks
Wow, you must have already had that in the works. Wasn't it just yesterday I suggested this? Am I right in thinking an sn2 inversion could turn the L into D? Maybe only half and then another resolution, and again would give you 87.5% D and 12.5% L? Probably stop there. I know I have a paper that illudes to such, I'll look through my pdf's
The rotation of optically-active phenylalkamines of the formula:-- <FORM:0405520/IV/1> wherein R1 represents phenyl or a monohydroxyphenyl and R2 and R3 an alkyl group or hydrogen, is reversed by first acylating and then splitting off the acyl group. In the case of compounds containing two...
This CAS is d-tartaric(L-tartaric, dextrotartaric) in that case D-Meth should be in the liquid part so I guess video tartaric is levo isomer but maybe Im wrong, can someone confirm that please?
I always use d-tartaric(L-tartaric, dextrotartaric) and I assume that my D-meth is in the liquid parte.
Awesome!!! Its a joy to watch that video. High resolution, enjoyable music, no distorted voices and you see all the many different steps so you can reproduce yourself! And very nice that finally there is a video about tartaric acid.
I think many will share this video and reproduce the steps. Good job!
Is l-tartaric(D-tartaric, levotartaric) used in the video or racemic one? Not clear but I guess that if D-meth is in the solid part should be l-tartaric(D-tartaric, levotartaric) right? With d-tartaric(L-tartaric, dextrotartaric) the D-meth should be in the liquid part. Thanks.
This CAS is d-tartaric(L-tartaric, dextrotartaric) in that case D-Meth should be in the liquid part so I guess video tartaric is levo isomer but maybe Im wrong, can someone confirm that please?
I always use d-tartaric(L-tartaric, dextrotartaric) and I assume that my D-meth is in the liquid parte.
Ok racemic one then thanks a lot, and the other doubt that I have is why it's necessary to bring the freebase to a high acid PH? Wouldn't be better for that crystallization point to leave it quite acid but more close to PH6? Like 4-4.5?
Thanks.
I have been waiting for this one, specifically because all of the things you show have an interesting way of having a unfatbomable amout of pre-pre…all the way to which route you may have to use to get their this is the video i have been holding on to the edge of my seat waiting for, there’s something to be said for a series of reactions in which most of the folks watching understand the science or they don’t get the chemistry but have learned through repetition and don’t know what’s happening when they basify and even from their so many options, hcl gas and recrystalization is a preferred method when using say the birch reduction of HI/P reduction of ephedrine but this is not that, so I’m glad you showed this for all above mentioned that will benefit, I’m so close if I had a trustworthy source for very few things i am a grown man I can get glass and heating mantle etc, when I do I promise to pay homage in any way possible
Is l-tartaric(D-tartaric, levotartaric) used in the video or racemic one? Not clear but I guess that if D-meth is in the solid part should be l-tartaric(D-tartaric, levotartaric) right? With d-tartaric(L-tartaric, dextrotartaric) the D-meth should be in the liquid part. Thanks.
Ok racemic one then thanks a lot, and the other doubt that I have is why it's necessary to bring the freebase to a high acid PH? Wouldn't be better for that crystallization point to leave it quite acid but more close to PH6? Like 4-4.5?
Thanks.
When will you record the step by step of the chemical synthesis of LSD (Lysergic acid diethylamide) of MDMA (3,4-Methylenedioxymethamphetamine) or 2C-B (2-4-bromo-2,5-dimethoxyphenylethanamine)?
many users need such synthesis!
I have like 1L of this orange base oil, Its vacuüm b after put in reactor vessel And processed with methyl platina and bubbled with hydrogen.. need to destill it first?
I have like 1L of this orange base oil, Its vacuüm b after put in reactor vessel And processed with methyl platina and bubbled with hydrogen.. need to destill it first?
105.69DL freebase to 20g D meth hcl? Isn't that a low yeild I thought it should be 50%? Couldn't you keep extracting untill you get the 50% D isomer out?
105.69DL freebase to 20g D meth hcl? Isn't that a low yeild I thought it should be 50%? Couldn't you keep extracting untill you get the 50% D isomer out?
I'm thinking
D-meth-D-tartate
D-Meth-L-tartate
L-Meth-D-tartate
L-Meth-L-Tartate
Yes you neutralize it and do it again to pick up some of the former D,L as D,D. Then again.theres a link in my rating for this video.
V I was under the impression that you just use the D-tartaric acid and get D,D and L,D, but maybe you can't separate that.
105.69DL freebase to 20g D meth hcl? Isn't that a low yeild I thought it should be 50%? Couldn't you keep extracting untill you get the 50% D isomer out?
Everybody talking about the 20g yield, that was what was crystalized out of the ethanol solution. Left in the solution is the other 9 grams from the first extraction. Assumably, the solution will be reduced and recrystallized or evaporated to collect it
Yes I understand the difference he put in 105.69g of freebase racemic and got 20g of the d isomer methamphetime...my understanding was a full yeild would be around 52g of d isomer meth? Or is that incorrect
On the dextroamphetamine synth you posted you have 6g racemic and goes to 2.63g D isomer close to a 40-42% yeild(I'm not sure how much you gain after convert it from freebase to sulphate) I think maybe doing another wash of hot methanol would produce a higher yeild closer to that 50% mark? Or is this not possible
Yes I understand the difference he put in 105.69g of freebase racemic and got 20g of the d isomer methamphetime...my understanding was a full yeild would be around 52g of d isomer meth? Or is that incorrect
On the dextroamphetamine synth you posted you have 6g racemic and goes to 2.63g D isomer close to a 40-42% yeild(I'm not sure how much you gain after convert it from freebase to sulphate) I think maybe doing another wash of hot methanol would produce a higher yeild closer to that 50% mark? Or is this not possible
The in d-amph synthesis has different yield. You count "40-42%" from 6g of racemic. In theory, racemic mixture is assumed 50/50 of d- and l-isomers mixture. In practice, you can get absolutely different ratio like 60/40 or 30/70. It takes to carry out more precise research to explore dependences in synthesis and isomers ratio. We've decided that it is worth to show isomer separation procedure for BB Forum members cuz it is quite popular issue with methamphetamine/amphetamine synthesis.
I will simplify the task for you with a hydrogen chloride generator, just take hydrochloric acid and sour, and then we drive off water in a vacuum and that's it
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.
Spoiler: Pic.17
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.
Why is the video using DL tartaric and the above post is using D tartaric? I’m so confused. Thanks
I want to upset you, but recently I did a separation using l-tartaric acid and gave the resulting product for testing to people who understand it and they said that it was shit and not meth
I want to upset you, but recently I did a separation using l-tartaric acid and gave the resulting product for testing to people who understand it and they said that it was shit and not meth
If use D tartaric acid, the L would precipitate and d-amphetamine remains in the solution? If I use L tartaric then the opposite? And using DL tartaric acid would be same as L tartaric acid?
If use D tartaric acid, the L would precipitate and d-amphetamine remains in the solution? If I use L tartaric then the opposite? And using DL tartaric acid would be same as L tartaric acid?
Is this correct there's a lot of misinformation on what precipitates out with what isomer of tartaric acid...there's a few I've seen saying D tartaric precipitates the L meth is that incorrect?
I'm still not sure about the advantage of using ethanol 88% instead pure methanol, possibly is something obvious that I can't see but I think the small water part from ethanol don't helps to the process, I'm wrong?
Ok I'm just trying to adapt this method to the reagents that I have now on hands and that I always use it for that process.
I'm testing in small scale.
These are the amounts used:
- 35gr pure freebase
- 42gr L-tartaric
- 205ml Methanol
The amounts are like video ratio but methanol I decided to use a 12% less due ethanol in video is a 88% solution.
I added the acid to the flask later the methanol and finally the freebase.
Once I started stirring immediately happened what usually happened to me in the rest of the processes I have tried for this purpose the solid salts precipitated quickly forming a mass.
Normally in that case I would keep it at RT during 24 hours before filtering but I'm trying to follow the video method so now I'm refluxing it, with heating the mass turned practically all liquid again but not completely so I won't filter it in hot after reflux.
I will let it to RT and later filter it.
I will try to follow as good as possible the video process and update you once I finish it ;-)
Is this correct there's a lot of misinformation on what precipitates out with what isomer of tartaric acid...there's a few I've seen saying D tartaric precipitates the L meth is
*enantomers. The confusion is due to naming conventions. I-tartaric acid is D- tartaric acid also (-) or R,R, (2R,3R) here's where it Really gets wild. Racemic tartaric acid is 2R,3S and 2S,3R, NOT A MIXTURE OF D AND L. A mixture of the 2 meso configurations.
Wait ... so you used RACEMIC tartrate to resolve racemic meth? I guess you relied on the solubility differences between the tartrate salts to precipitate d-meth-l-tartrate but not l-meth-d-tartrate.
Did you do any quantification of the result? Polarimeter?
*enantomers. The confusion is due to naming conventions. I-tartaric acid is D- tartaric acid also (-) or R,R, (2R,3R) here's where it Really gets wild. Racemic tartaric acid is 2R,3S and 2S,3R, NOT A MIXTURE OF D AND L. A mixture of the 2 meso configurations.
I used L+ tartaric acid(Dextro tartaric, R isomer, d enantiomer...) because is what I had on hand and always used for that process and the D-meth would be in the liquid part. The reason or advantage to use racemic one in the video is totally unknown for me.
Wait ... so you used RACEMIC tartrate to resolve racemic meth? I guess you relied on the solubility differences between the tartrate salts to precipitate d-meth-l-tartrate but not l-meth-d-tartrate.
Did you do any quantification of the result? Polarimeter?
Would be extremely interesting that video's creator could give us light over the chemistry behind that reagents selection because in that case honestly I'm quite confused(not talking about tartaric enantiomers hahahah) do you think that could be possible?
Thanks.
I used L+ tartaric acid(Dextro tartaric, R isomer, d enantiomer...) because is what I had on hand and always used for that process and the D-meth would be in the liquid part. The reason or advantage to use racemic one in the video is totally unknown for me.
Ok I have the results of my tests trying to follow the video method but the similar reagents that I have on hand and that I usually use.
I divided 70gr of racemic freebase(not steam distilled and no acid-base washing made just extracted and dried from the NaBH4 P2P reduction) in 2 equal parts of 35gr each. With the first 35 gr I did video process but with some different reagents and with the other 35gr I did the method that I use to make that is similar to the video process:
- Method 1(Video method)
- 35gr pure freebase
- 42gr L-tartaric
- 205ml Methanol
The amounts are like video ratio but methanol I decided to use a 12% less due ethanol in video is a 88% solution.
I added the acid to the flask later the methanol and finally the freebase.
Once I started stirring immediately happened what usually happened to me in the rest of the processes I have tried for this purpose the solid salts precipitated quickly forming a mass.
I reflux it during one hour, with heating the mass come a thick liquid.
After the refluxing hour I keep it at RT around 4-5 hours until no new solids are visible created in the mixture.
I filtered it and keep in different pots the solid and the liquid part. -Liquid part
I get cold the liquid part in the freezer with a NaOH solution to avoid excess heat in the alkalinization step.
I added slowly the NaOH solution to the mixture until ph13 and add it to a separation funnel during 30 min.
I take out the aqueous bottom layer and keep it , the thin oil top layer is kept in a pot.
I extract with DCM the aqueous layer and add these DCM extractions to oil layer.
I dried it with anhydrous sodium sulfate and filtered it.
I evaporated the DCM and 10gr of D-meth freebase obtained(don't looks as clean as it should)
- Solid part
I added more or less the same ml of warm water than gr of the solids weight, they has been dissolved it with stirring and let it cool down.
I added slowly the cold NaOH solution until ph13 and add it to a separation funnel during 30 min.
The next steps are exactly the same as the liquid part process.
After DCM is evaporated I got 15gr of L-meth freebase quite "dirty" too
- Method 2(the one I usually do and similar to the video method)
Methanol and tartaric are mixed until complete dissolution, freebase is added while stirring and continue stirring vigorously 2-3min.
Let it at RT during 24 hours until the complete solid mass is formed(if in a couple of hours no solids are formed you should stir it again)
It's filtered and now are exactly the same steps as the other method.
- In the liquid part once I had all the DCM extractions I decided to make an acid-base washing, even the DCM color looks cleaner than the other one, to avoid to be "dirty" like the first method happened and I obtained 10gr of D-meth absolutely clean and pure(pic link at the bottom)
- In the solid part 13gr of enough clean L-meth freebase without acid-base washing obtained.
- Cleaning issue could be fixed during the crystallization stage too are many ways.
Obviously is very difficult to check how much D or L are in reality in the stuff only with Chiral analysis not easy accessible.
I also know that it is not the most correct to compare two methods using some different reagents but in reality they do practically the same function in the process.
In short, similar methods and yields the main differences are that in the second one you can avoid the reflux(so could be easier) and the final freebases looks cleaner than the others method.
When I will have free time again like this weekend(unfortunately not usual :-( only will be remaining to test in small scale the"mexican"method for meth isomer separation is quite different and I only did it in big scale although I finally decided to use the method explained for big scale too.
These would be the 3 methods that I did and I can talk about them with experience.
About when and how to make an acid-base washing instead steam distillation...etc etc needs a complete and long post the same for crystallization.
I'm thinking to post all the possible and more profitable options during the route from BMK(5449) to D-Meth HCL but this would be a long work and I don't know when I will have enough time again hahah at least is planned.
Ok I have the results of my tests trying to follow the video method but the similar reagents that I have on hand and that I usually use.
I divided 70gr of racemic freebase(not steam distilled and no acid-base washing made just extracted and dried from the NaBH4 P2P reduction) in 2 equal parts of 35gr each. With the first 35 gr I did video process but with some different reagents and with the other 35gr I did the method that I use to make that is similar to the video process:
- Method 1(Video method)
- 35gr pure freebase
- 42gr L-tartaric
- 205ml Methanol
The amounts are like video ratio but methanol I decided to use a 12% less due ethanol in video is a 88% solution.
I added the acid to the flask later the methanol and finally the freebase.
Once I started stirring immediately happened what usually happened to me in the rest of the processes I have tried for this purpose the solid salts precipitated quickly forming a mass.
I reflux it during one hour, with heating the mass come a thick liquid.
After the refluxing hour I keep it at RT around 4-5 hours until no new solids are visible created in the mixture.
I filtered it and keep in different pots the solid and the liquid part. -Liquid part
I get cold the liquid part in the freezer with a NaOH solution to avoid excess heat in the alkalinization step.
I added slowly the NaOH solution to the mixture until ph13 and add it to a separation funnel during 30 min.
I take out the aqueous bottom layer and keep it , the thin oil top layer is kept in a pot.
I extract with DCM the aqueous layer and add these DCM extractions to oil layer.
I dried it with anhydrous sodium sulfate and filtered it.
I evaporated the DCM and 10gr of D-meth freebase obtained(don't looks as clean as it should)
- Solid part
I added more or less the same ml of warm water than gr of the solids weight, they has been dissolved it with stirring and let it cool down.
I added slowly the cold NaOH solution until ph13 and add it to a separation funnel during 30 min.
The next steps are exactly the same as the liquid part process.
After DCM is evaporated I got 15gr of L-meth freebase quite "dirty" too
- Method 2(the one I usually do and similar to the video method)
Methanol and tartaric are mixed until complete dissolution, freebase is added while stirring and continue stirring vigorously 2-3min.
Let it at RT during 24 hours until the complete solid mass is formed(if in a couple of hours no solids are formed you should stir it again)
It's filtered and now are exactly the same steps as the other method.
- In the liquid part once I had all the DCM extractions I decided to make an acid-base washing, even the DCM color looks cleaner than the other one, to avoid to be "dirty" like the first method happened and I obtained 10gr of D-meth absolutely clean and pure(pic link at the bottom)
- In the solid part 13gr of enough clean L-meth freebase without acid-base washing obtained.
- Cleaning issue could be fixed during the crystallization stage too are many ways.
Obviously is very difficult to check how much D or L are in reality in the stuff only with Chiral analysis not easy accessible.
I also know that it is not the most correct to compare two methods using some different reagents but in reality they do practically the same function in the process.
In short, similar methods and yields the main differences are that in the second one you can avoid the reflux(so could be easier) and the final freebases looks cleaner than the others method.
When I will have free time again like this weekend(unfortunately not usual :-( only will be remaining to test in small scale the"mexican"method for meth isomer separation is quite different and I only did it in big scale although I finally decided to use the method explained for big scale too.
These would be the 3 methods that I did and I can talk about them with experience.
About when and how to make an acid-base washing instead steam distillation...etc etc needs a complete and long post the same for crystallization.
I'm thinking to post all the possible and more profitable options during the route from BMK(5449) to D-Meth HCL but this would be a long work and I don't know when I will have enough time again hahah at least is planned.
I'm pretty impressed with your creativity. Looks method 2 might be perfect.
Is process for MDMA the same as for meth? If L-(+)-Tartaric acid CAS 87-69-4, then R isomer will be in salt, S isomer will be in freebase leftovers?
The proportion in molar mass shall be corrected accordingly - it is not a problem to calculate.
I'm pretty impressed with your creativity. Looks method 2 might be perfect.
Is process for MDMA the same as for meth? If L-(+)-Tartaric acid CAS 87-69-4, then R isomer will be in salt, S isomer will be in freebase leftovers?
The proportion in molar mass shall be corrected accordingly - it is not a problem to calculate.
It should be the same I never did it with MDMA because D-MDMA has worse or less funny effects than racemic. And yes with that tartatic cas S isomer will be in the liquid part.
It should be the same I never did it with MDMA because D-MDMA has worse or less funny effects than racemic. And yes with that tartatic cas S isomer will be in the liquid part.
Aren't you confused with L and D? If we talk about meth, then most powerful is D isomer (R enantiomer).
But in case of MDMA it is vice versa: people use L isomer (S enantiomer). Some sources even point this powerful S is pretty unusual for a meth family member.
Aren't you confused with L and D? If we talk about meth, then most powerful is D isomer (R enantiomer).
But in case of MDMA it is vice versa: people use L isomer (S enantiomer). Some sources even point this powerful S is pretty unusual for a meth family member.
Are you sure of that?Im not sure now you will make me investigate it hahaha you mean the "strong" isomer in MDMA is L?I think I read it the opposite thing somewhere I will check it but what Im completely sure that the supposed stronger isomer of MDMA is less funny than racemic.
100% sure.
What I read literally: L (S) isomer is more euphoric in case of MDMA, but D (R) has rather hallucinogen property.
Then, i found a study on mice, stating the difference between D and L:
Racemic MDMA (0.3-30 mg/kg), S(+)-MDMA (0.3-30 mg/kg), R(-)-MDMA (0.3-50 mg/kg) and saline vehicle (10 ml/kg) were comprehensively evaluated in fully automated and computer-integrated activity chambers, which were designed for mice, and provided a detailed analysis of the frequency, location...
Aren't you confused with L and D? If we talk about meth, then most powerful is D isomer (R enantiomer).
But in case of MDMA it is vice versa: people use L isomer (S enantiomer). Some sources even point this powerful S is pretty unusual for a meth family member.
100% sure.
What I read literally: L (S) isomer is more euphoric in case of MDMA, but D (R) has rather hallucinogen property.
Then, i found a study on mice, stating the difference between D and L:
Racemic MDMA (0.3-30 mg/kg), S(+)-MDMA (0.3-30 mg/kg), R(-)-MDMA (0.3-50 mg/kg) and saline vehicle (10 ml/kg) were comprehensively evaluated in fully automated and computer-integrated activity chambers, which were designed for mice, and provided a detailed analysis of the frequency, location...
Same goes for meth, but I think separating by precipitation is not as effective with meth as using the different solubilities in methanol of the two tartaric diastereomers.
So you acidify the racemic freebase with racemic tartrate (both D and L), precipitating D and L-meth tartrate.
Then you recrystallize those salts from methanol. L-meth tartrate is much more soluble in methanol and will stay in solution while D-meth crashes out and can be collected.
I have been following a professional chemist doing clandestine work on a smaller forum, and he has been working intensively with resolution of amphetamines for a while. let's just say itisn't really as easy as it seems. It won't be 100 %, you're lucky to reach 80, you'll loose product, and the only way to know for sure how well you did is by buying a polarimeter- either an automatic one for €10-20.000 or a cheaper manual one that takes forever to learn how to calibrate and master.
Same goes for meth, but I think separating by precipitation is not as effective with meth as using the different solubilities in methanol of the two tartaric diastereomers.
So you acidify the racemic freebase with racemic tartrate (both D and L), precipitating D and L-meth tartrate.
Then you recrystallize those salts from methanol. L-meth tartrate is much more soluble in methanol and will stay in solution while D-meth crashes out and can be collected.
I have been following a professional chemist doing clandestine work on a smaller forum, and he has been working intensively with resolution of amphetamines for a while. let's just say itisn't really as easy as it seems. It won't be 100 %, you're lucky to reach 80, you'll loose product, and the only way to know for sure how well you did is by buying a polarimeter- either an automatic one for €10-20.000 or a cheaper manual one that takes forever to learn how to calibrate and master.
All we know is a low yield of D-meth that is one of the reasons of its price.
I never saw racemic meth in the market or in the street honestly.
I will make an adittional test that I never did before.
I never did it simply because when I cristallized the racemic freebase I got the standard racemic crystals(not so big and no transparents) and after the tartaric process I got the big and transparent shards that people wants so enough working for me but now I want to check how much is working.
I have cristallized Dmeth and Lmeth from the other day test, I will take 5gr of each turn them freebase and make again the tartaric process separately.
In theory should be a big difference in the amounts of freebase obtained from solid and liquid part.
I will update you, also I will check if there is freebase remaining in the methanol after turn the mixture alkali and extract.
All we know is a low yield of D-meth that is one of the reasons of its price.
I never saw racemic meth in the market or in the street honestly.
I will make an adittional test that I never did before.
I never did it simply because when I cristallized the racemic freebase I got the standard racemic crystals(not so big and no transparents) and after the tartaric process I got the big and transparent shards that people wants so enough working for me but now I want to check how much is working.
I have cristallized Dmeth and Lmeth from the other day test, I will take 5gr of each turn them freebase and make again the tartaric process separately.
In theory should be a big difference in the amounts of freebase obtained from solid and liquid part.
I will update you, also I will check if there is freebase remaining in the methanol after turn the mixture alkali and extract.
What you guys are missing is this;
D-meth-D-tartrate
D-meth-L-tartrate
L-meth-L-tartrate
L-meth-D-tartrate
4 isomers. Once you crystallize the D,D- meth, you neutralize the remainder and hit it again and collect the former D,L, which you've re-acidified to D,D. THEN YOU DO IT AGAIN!
1/4(FIRST PASS) ,3/16 second pass ( 1/4 of 3/4 that is remaining), 3/64 third pass ( 1/4 of 3/16) you have now THEORETICALLY collected 31/64 of the entire rm.
IE 48.44% (theoretical idealism)
What you guys are missing is this;
D-meth-D-tartrate
D-meth-L-tartrate
L-meth-L-tartrate
L-meth-D-tartrate
4 isomers. Once you crystallize the D,D- meth, you neutralize the remainder and hit it again and collect the former D,L, which you've re-acidified to D,D. THEN YOU DO IT AGAIN!
1/4(FIRST PASS) ,3/16 second pass ( 1/4 of 3/4 that is remaining), 3/64 third pass ( 1/4 of 3/16) you have now THEORETICALLY collected 31/64 of the entire rm.
IE 48.44% (theoretical idealism)
I understand what you mean about the process but I dont get exactly what you mean with neutralize the remainder once crystallized. What is the remainder for you?Neutralize until what pH?
Can I ask you an example please?
Thanks and sorry.
I understand what you mean about the process but I dont get exactly what you mean with neutralize the remainder once crystallized. What is the remainder for you?Neutralize until what pH?
Can I ask you an example please?
Thanks and sorry.
I mean neutralize the acid salt . Conjugate the Lewis acid. Remove the tartate Moity, then add it back. The remainder being the reaction mixture minus the crystallized fraction of D,D. The solution will be about pH 7 or slightly lower. That is what a neutralization is.
I mean neutralize the acid salt . Conjugate the Lewis acid. Remove the tartate Moity, then add it back. The remainder being the reaction mixture minus the crystallized fraction of D,D. The solution will be about pH 7 or slightly lower. That is what a neutralization is.
I never did it simply because when I cristallized the racemic freebase I got the standard racemic crystals(not so big and no transparents) and after the tartaric process I got the big and transparent shards that people wants so enough working for me but now I want to check how much is working.
so this synthesis is for D-meth
it is the same D-meth as at RP/I/Pseudoephedrine synthese?
so actually the path through P2P method leads to a recamicmeth?
What is the difference between recamicmeth and D-meth?
Is there any theoretical basis how you calculate all the percentage of methanol and l-tartaric? I don't have any doubt about your percentage, just want to understand this topic better.
Thanks.
Is there any theoretical basis how you calculate all the percentage of methanol and l-tartaric? I don't have any doubt about your percentage, just want to understand this topic better.
Thanks.
It's not my method, the method says 85 parts of freebase, 100 parts of tartaric and 1000 parts of methanol. If I'm not wrong is based in 1:1 moles of freebase and tartaric.
It's not my method, the method says 85 parts of freebase, 100 parts of tartaric and 1000 parts of methanol. If I'm not wrong is based in 1:1 moles of freebase and tartaric.
35gr of FB and 41,2gr of L-tartaric is not exactly 1:1, but rather 1:1.17 in case of meth.
It is not a problem for me to calculate the weight of MDMA FB and tartaric acid in the same proportions.
But methanol in this case is a real quest. I have a good education, but I'm hellish zero in chemistry - hahaha, and I simply don't understand methanol designation...
35gr of FB and 41,2gr of L-tartaric is not exactly 1:1, but rather 1:1.17 in case of meth.
It is not a problem for me to calculate the weight of MDMA FB and tartaric acid in the same proportions.
But methanol in this case is a real quest. I have a good education, but I'm hellish zero in chemistry - hahaha, and I simply don't understand methanol designation...
Is about solubility but I'm sure that methanol is only useful when you are using D-tartaric or DL tartaric, using L-tartaric is more easy with water I did it many times.
700gr+1.4L water
972gr FB
Works fine, the new improvement is that now I know is important to do it several times until no oil appears when turns liquid layer alkali.
Is about solubility but I'm sure that methanol is only useful when you are using D-tartaric or DL tartaric, using L-tartaric is more easy with water I did it many times.
700gr+1.4L water
972gr FB
Works fine, the new improvement is that now I know is important to do it several times until no oil appears when turns liquid layer alkali.
If I understand correctly, there shall be following scheme for MDMA:
we use FB and L-(+)-Tartaric acid CAS 87-69-4 in molecular proportions A (for FB):B(for acid) where B <= A/2
the reaction is done in water, approximately 1.4:1 of the FB volume
as the reaction finished, we have S-isomer as FB leftovers + some small quantity of unreacted FB R-isomer on top layer ...
... and the salt of most part of R-isomer dissolved in water
separate two layers with funnel,
water + salt of R-isomer is thrown away
top layer is used for the further purification and crystallization.
Probably, petroleum ether can be used to help separation, as neither salt, nor unreacted L-tartaric (if presents) aren't soluble in it?
Hopefully, I will get some inspiration and free time to do it in the nearest future.
Ouch! As I remember, Shulgin suggests proportions of S:R in 2:1 as the best!!!
Probably we can simply acidify our FB with L-tartaric in proportions close to 4:1 in water to remove excess of R?
Looks very easy, very simple.
If I understand correctly, there shall be following scheme for MDMA:
we use FB and L-(+)-Tartaric acid CAS 87-69-4 in molecular proportions A (for FB):B(for acid) where B <= A/2
the reaction is done in water, approximately 1.4:1 of the FB volume
as the reaction finished, we have S-isomer as FB leftovers + some small quantity of unreacted FB R-isomer on top layer ...
... and the salt of most part of R-isomer dissolved in water
separate two layers with funnel,
water + salt of R-isomer is thrown away
top layer is used for the further purification and crystallization.
Probably, petroleum ether can be used to help separation, as neither salt, nor unreacted L-tartaric (if presents) aren't soluble in it?
Hopefully, I will get some inspiration and free time to do it in the nearest future.
For MDMA scheme is moles of freebase 1:1 moles of tartaric solution, if you use L-tartaric the must keep the solid precipitate, the liquid part has the enantiomer and salt not desired.
You must solve and set alkali 12-13 ph the solid precipitate to make fb appears.
This fb is with the enantiomer that you want isolated.
The liquid part could be alkalinized but adding enough water too to get the fb of the other enantiomer if you want to mix them in ratios.
I can test it next week but is in this way.
How to make Japanese white rabbit sputum ingot can be reduced to methamphetamine at home, simple operation or no catalysis. Please give us some guidance.
