Maybe this will help you further I have not tested it personally yet. I could look in my folders maybe I can find the original PDF still
General Procedure for the Racemization Experiments. A
0.06 M solution of amine (100 mg) and thiol (1.2 or 0.2 equiv)
in benzene was refluxed for 2 h (stoichiometric condition) or 7
h (catalytic condition) in the presence of AIBN (an overall
quantity of 20 mol % of AIBN was divided into three equal
portions (when reaction time > 2 h) that were added successively each 2 h). After concentration, the residue was diluted
with HCl (1 M), and the solution was washed with Et2O. The
aqueous phase was basified with saturated sodium carbonate
and extracted with Et2O. The pure amine was isolated after
drying on MgSO4 and concentration
General Procedure for the Racemization Experiments. A
0.06 M solution of amine (100 mg) and thiol (1.2 or 0.2 equiv)
in benzene was refluxed for 2 h (stoichiometric condition) or 7
h (catalytic condition) in the presence of AIBN (an overall
quantity of 20 mol % of AIBN was divided into three equal
portions (when reaction time > 2 h) that were added successively each 2 h). After concentration, the residue was diluted
with HCl (1 M), and the solution was washed with Et2O. The
aqueous phase was basified with saturated sodium carbonate
and extracted with Et2O. The pure amine was isolated after
drying on MgSO4 and concentration
- By Sciencenutz
I have access to aibn and thiol but would appreciate some help with the proper ratios converted to say per kg of L-meth freebase? I'll post my results
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- By btcboss2022
I would reflux the mixture don't means that reflux will begins at toluene b.p and 0.06M is the concentration in grams will depend on the amount of solvent used.
- By Sciencenutz
here are the weights of all 3.
L methamphetime 149.23g/mol
Methyl thiogylcolate 106.14g/mol
AIBN 164.21g/mol
Therefore would it be 1 mol L-meth and it says 0.2 equal so 106.14x0.2=21.22g. And a 20% for aibn is 32.84g.
So say I did 149.23g L-meth freebase with 500ml toluene and 21.22g thiol and refluxed for 7 hours while every 2 hours add 10.94g (32.84/3=10.94g) of AIBN.
Would that be the correct numbers? I'm a bit confused still on this 0.06M and how to scale that up
L methamphetime 149.23g/mol
Methyl thiogylcolate 106.14g/mol
AIBN 164.21g/mol
Therefore would it be 1 mol L-meth and it says 0.2 equal so 106.14x0.2=21.22g. And a 20% for aibn is 32.84g.
So say I did 149.23g L-meth freebase with 500ml toluene and 21.22g thiol and refluxed for 7 hours while every 2 hours add 10.94g (32.84/3=10.94g) of AIBN.
Would that be the correct numbers? I'm a bit confused still on this 0.06M and how to scale that up
- By btcboss2022
500ml of toluene per mol? How you calculated?
The calculation of liters per mol of the paper process for 0.06M concentration should be:
149.23gr*0.06M=8.95gr of freebase per liter
149.23gr/8.95gr=16.67L of solvent(toluene) per mol of freebase
Is a very low concentration but in reactions between radicals many by-side products could be generated so I guess thats the reason of this low concentration.
I think that wouldn't be problems in use 10 times this concentration using 1.66L of toluene per mol of freebase.
Another point is why are you focused on the 7 hours process when you can do it in 2 hours?
The calculation of liters per mol of the paper process for 0.06M concentration should be:
149.23gr*0.06M=8.95gr of freebase per liter
149.23gr/8.95gr=16.67L of solvent(toluene) per mol of freebase
Is a very low concentration but in reactions between radicals many by-side products could be generated so I guess thats the reason of this low concentration.
I think that wouldn't be problems in use 10 times this concentration using 1.66L of toluene per mol of freebase.
Another point is why are you focused on the 7 hours process when you can do it in 2 hours?
Ok guys but has anyone tried heating with reflux in slightly acidic conditions?
Simple as.
I'm willing to bet it's gonna work the question is how well.
Simple as.
I'm willing to bet it's gonna work the question is how well.
- By HerrHaber
I bet it will but there is a reason why CHEM IS TRY so therefore someone needs to experiment this with a rack of test tubes and different concentrations, pH, temperatures and times of reaction... research work. "Nobody said it was easy, nobody said that it could be so hard, we're going back to the start" Coldplay-The Scientist.
The disappearance of the optical activity of a substance is not always related
to the disappearance of the chirality of the molecule. It can also be caused by reversible transformations of one of the enantiomers into the other, until a mixture is reached
equimolecular of both enantiomers (a racemic). This transformation
called racemization can take place spontaneously (that is, under the action of ter¬
molecules, see further examples for optically active derivatives of bifenil), or under the action of acid or bacid catalysts (v. vol. II, "Stereo-chemistry II")
And from vol. 2 stereo chemistry, reffering totartric acid, but we can draw some conclusions that the phenyletylamine we are working on also has simillar parameters which it does the same things, who knows maybe its the same.Its not an choh group but you never know.
Racemization. By heating (+)-tartrio acid
with water, at 160°, it is largely converted into meso-tartaric acid (by racemization of a single CHOH group). Under the same conditions, but at 175°, a higher proportion of (it)-tartaric acid is formed (through the racemization of both CHOH groups). This reaction is strongly catalyzed by bases; upon heating with a sodium hydroxide solution, racemization takes place at 100°. HCl acid dil. it also has a catalytic action, but weaker (the temperature required to observe the reaction is 120-140°).
This racemization reaction is used to obtain meso-tartaric and (±)-tartaric acids. Racemizations of this kind are also observed with other α-substituted acids.
to the disappearance of the chirality of the molecule. It can also be caused by reversible transformations of one of the enantiomers into the other, until a mixture is reached
equimolecular of both enantiomers (a racemic). This transformation
called racemization can take place spontaneously (that is, under the action of ter¬
molecules, see further examples for optically active derivatives of bifenil), or under the action of acid or bacid catalysts (v. vol. II, "Stereo-chemistry II")
And from vol. 2 stereo chemistry, reffering totartric acid, but we can draw some conclusions that the phenyletylamine we are working on also has simillar parameters which it does the same things, who knows maybe its the same.Its not an choh group but you never know.
Racemization. By heating (+)-tartrio acid
with water, at 160°, it is largely converted into meso-tartaric acid (by racemization of a single CHOH group). Under the same conditions, but at 175°, a higher proportion of (it)-tartaric acid is formed (through the racemization of both CHOH groups). This reaction is strongly catalyzed by bases; upon heating with a sodium hydroxide solution, racemization takes place at 100°. HCl acid dil. it also has a catalytic action, but weaker (the temperature required to observe the reaction is 120-140°).
This racemization reaction is used to obtain meso-tartaric and (±)-tartaric acids. Racemizations of this kind are also observed with other α-substituted acids.
- By btcboss2022
I don't think that works for racemization, descarboxilation or hydrolisis won't racemize a L enantiomer freebase.
no wonder they call you the boss
if it needs under 90 i would use benzen , but if 90 dosent do the trick go toluene, xylene and so on
if it needs under 90 i would use benzen , but if 90 dosent do the trick go toluene, xylene and so on
- By HerrHaber
For short it's all about equilibrium and time of reaction I think is the main driving force (catalyst is also a must). I remember not being let to leave a reaction overnight because my prof promissed it will racemize and I was working in the stereochemistry lab so equilibrium was the enemy there.
- By btcboss2022
I will do it with water, NaOH and HCL too the 3 routes ok?After results we talk again.
- By btcboss2022
Who calls me the boss?ahahah first time I see this I don't think so hahah but thanks anyway ;-)
- By btcboss2022
Thats my plan:
I have L-Meth HCL I will take 45gr(15+15+15) and I Will turn It on freebase.
Once I have the L-meth freebase I Will do the tartaric separation process again to try to have the Max amount of L enantiomer.
I Will divide the reseparated L-meth freebase in 3 equal parts and I Will do the 3 routes as detailed there.
Once finished I Will do again the tartaric sepation process and we Will see the results.
Thanks.
I have L-Meth HCL I will take 45gr(15+15+15) and I Will turn It on freebase.
Once I have the L-meth freebase I Will do the tartaric separation process again to try to have the Max amount of L enantiomer.
I Will divide the reseparated L-meth freebase in 3 equal parts and I Will do the 3 routes as detailed there.
Once finished I Will do again the tartaric sepation process and we Will see the results.
Thanks.
