- Joined
- Mar 15, 2022
- Messages
- 650
- Solutions
- 1
- Reaction score
- 662
- Points
- 93
- Deals
- 8
I think something wrong with graph.
If you use L+-tartaric acid(cheaper and easier to find)the D enantiomer(amph,Meth) will be in the solution not in the solid as there appears.
If you use D-tartaric acid(more expensive and difficult to find) the D enantiomer(amph,Meth) will be in the solid.
Thanks.
If you use L+-tartaric acid(cheaper and easier to find)the D enantiomer(amph,Meth) will be in the solution not in the solid as there appears.
If you use D-tartaric acid(more expensive and difficult to find) the D enantiomer(amph,Meth) will be in the solid.
Thanks.
G.Patton
Expert
- Joined
- Jul 5, 2021
- Messages
- 2,745
- Solutions
- 3
- Reaction score
- 2,920
- Points
- 113
- Deals
- 1
His scheme is correct. Salt is precipitated into the solid state. No difference, d-amphetamine l-tartaric acid salt or l-amphetamine d-tartaric salt. You can change acid and get difference isomer in the solid state with following procedures.
P.S. same for methamphetamine.
P.S. same for methamphetamine.
↑View previous replies…
- Language
- 🇬🇧
- Joined
- Mar 3, 2024
- Messages
- 25
- Reaction score
- 18
- Points
- 3
- By Ihml
I’m not Patton, but: it’d produce only D-meth of the ephedrine that was used was derived from a plant, since plants naturally only produce L-ephedrine (which turns into the opposite enantiomer after reaction - D-meth in this case). If chemically synthesized ephedrine was used, than you’d get racemic meth, because ephedrine synthesis isn’t stereospecific.