One Step Dextroamphetamine Synthesis
- Thread starter G.Patton
- Start date
Kinda sucks when the reagents are more expensive than the end product.
Product: (4S,5R)-(-)-4-METHYL-5-PHENYL-2-OXAZOLIDINONE
Quantity: 5kg
Price: USD $8597.00
Product: (4S,5R)-(-)-4-METHYL-5-PHENYL-2-OXAZOLIDINONE
Quantity: 5kg
Price: USD $8597.00
- By Pororo
Chinese offer it for 500$ per 1 kg. But purity could be questionable.
- By handle
Dear, here is our best quotation for you:
1.Product:(4S,5R)-4-Methyl-5-Phenyloxazolidin-2-One, CAS:16251-45-9.
2.Price:USD1739.6/kg including shipping fee, door to door based on 5kg.(Total price is USD8695)
3.Package: 5kg/bag*1 bag, 1 carton
(2layers *0.24mm/layer food grade/pharma grade PET inner bags+1layer*0.24mm/layer
food grade/pharma grade PET Aluminium foil bag+5ply corrugated carton)
4.Delivery: Right away from our warehouse after receiving the payment.
5.Payment: T/T in advance.
6.Validity: 10 working days.
1.Product:(4S,5R)-4-Methyl-5-Phenyloxazolidin-2-One, CAS:16251-45-9.
2.Price:USD1739.6/kg including shipping fee, door to door based on 5kg.(Total price is USD8695)
3.Package: 5kg/bag*1 bag, 1 carton
(2layers *0.24mm/layer food grade/pharma grade PET inner bags+1layer*0.24mm/layer
food grade/pharma grade PET Aluminium foil bag+5ply corrugated carton)
4.Delivery: Right away from our warehouse after receiving the payment.
5.Payment: T/T in advance.
6.Validity: 10 working days.
Thank you for an interesting route dear. It looks it can work with Kipp gas generator to run the synth.
I.e., flask is charged with 100 g of (4S,5R)-(-)-4-methyl-5-phenyl-2-oxazolidinone and 5.66 g of Pd/C.
Kipp is charged with 36.37 g of Zn. Acid added and synth goes naturally under standard pressure and RT until all Zn reacted.
Questionable topic is how quickly will the catalyst turns inert and how it is affected by main prec purity?
I.e., flask is charged with 100 g of (4S,5R)-(-)-4-methyl-5-phenyl-2-oxazolidinone and 5.66 g of Pd/C.
Kipp is charged with 36.37 g of Zn. Acid added and synth goes naturally under standard pressure and RT until all Zn reacted.
Questionable topic is how quickly will the catalyst turns inert and how it is affected by main prec purity?
I did another attempt to understand this transformation, but all I got as the result of hydration in Pd/C presence is a transformation of oxazolidinone to aminoketone, i.e.:
(4S,5R)-(-)-4-methyl-5-phenyl-2-oxazolidinone+H2 = (1R,2S)-2-amino-1-phenylpropan-1-one
But I totally don't understand HOW it is transformed to amphetamine? Can some one explain this topic, please?
(4S,5R)-(-)-4-methyl-5-phenyl-2-oxazolidinone+H2 = (1R,2S)-2-amino-1-phenylpropan-1-one
But I totally don't understand HOW it is transformed to amphetamine? Can some one explain this topic, please?
- By G.Patton
Hi. Have you carried out this synthesis?
I can't answer you for sure. I assume, that H2 hydrated oxygen atoms and CO2 gas is released as one of the reaction product.
I can't answer you for sure. I assume, that H2 hydrated oxygen atoms and CO2 gas is released as one of the reaction product.
- By Pororo
I gonna try it on occasion. But I want to understand the principle first.
If it works precisely as it is said in the document, then I assume, reaction must be stopped periodically to remove CO2. It is not a problem.
The problem is: hydration of oxazolidinone will probably bring us to aminoketone, when O molecule released, then H2O formed and our catalyst is done((((
Or there is something totally different?
If it works precisely as it is said in the document, then I assume, reaction must be stopped periodically to remove CO2. It is not a problem.
The problem is: hydration of oxazolidinone will probably bring us to aminoketone, when O molecule released, then H2O formed and our catalyst is done((((
Or there is something totally different?
Has anybody got a success with this synth? I don't sleep, don't eat, thinking and thinking again and again.
First question is about Pd/C 10% (50 % water wet). This as I know refers to the water suspension. It looks simple, but it is not: what I can not confirm is the weight. As it is said, for the experiment Pd/C was used as below: "0.34 g 10% palladium on-carbon (50% water wet)"
Does it mean 0.34g of 10% Pd/C + 0.34g of water turned to the suspension? Or it is 0.34g of ready suspension i.e. 0.17g of Pd/C + 0.17g of water?
Second difficult question about degassing. CO2 appears as the side product of this reaction. Some relatively small part of it will be dissolved in ethanol, as I estimate around 15%. Another 10% will fill the reactor. For the rest of CO2 degassing is required. Even if H2 applied under 2 atm, it will not solve the problem (or we need to apply H2 under 10 atm that's crazy story ahahah).
The question is HOW the fuck to determine the point to lock H2 and degas the reactor from CO2?
First question is about Pd/C 10% (50 % water wet). This as I know refers to the water suspension. It looks simple, but it is not: what I can not confirm is the weight. As it is said, for the experiment Pd/C was used as below: "0.34 g 10% palladium on-carbon (50% water wet)"
Does it mean 0.34g of 10% Pd/C + 0.34g of water turned to the suspension? Or it is 0.34g of ready suspension i.e. 0.17g of Pd/C + 0.17g of water?
Second difficult question about degassing. CO2 appears as the side product of this reaction. Some relatively small part of it will be dissolved in ethanol, as I estimate around 15%. Another 10% will fill the reactor. For the rest of CO2 degassing is required. Even if H2 applied under 2 atm, it will not solve the problem (or we need to apply H2 under 10 atm that's crazy story ahahah).
The question is HOW the fuck to determine the point to lock H2 and degas the reactor from CO2?
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