Both work and both are good.
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4-MMC (mephedrone) synthesis. Complete video tutorial.
C10H11BrO + CH5N → C11H15NO + HBr
As can be seen from the equation, one molecule of 2-bromo-4-methylpropiophenone produces one molecule of mephedrone.
C10H11BrO : 227.0977 g/mol
C11H15NO : 177.2429 g/mol
227.0977/177.2429=1.28127953
The coefficient is 1.28127953
This means that from 100g of 2-bromo-4-propiophenone, you will get 78.0469818g of mephedrone base.
The mass of mephedrone hydrochloride will increase by 1.19 g/cm3 and the coefficient will be 1.27273446
C11H15NO+HCl=177.2429+1.19=178.4329
227.0977/178.4329=1.27273446
100g(C10H11BrO)/1.27273446=78.5709849g(C11H15NO)
The theoretical yield of the reaction of 2-bromo-4-methylpropiophenone was 78.5709849g mephedrone hydrochloride. But it should be considered that it is difficult to achieve the theoretical yield, usually the loss is 20%.
As can be seen from the equation, one molecule of 2-bromo-4-methylpropiophenone produces one molecule of mephedrone.
C10H11BrO : 227.0977 g/mol
C11H15NO : 177.2429 g/mol
227.0977/177.2429=1.28127953
The coefficient is 1.28127953
This means that from 100g of 2-bromo-4-propiophenone, you will get 78.0469818g of mephedrone base.
The mass of mephedrone hydrochloride will increase by 1.19 g/cm3 and the coefficient will be 1.27273446
C11H15NO+HCl=177.2429+1.19=178.4329
227.0977/178.4329=1.27273446
100g(C10H11BrO)/1.27273446=78.5709849g(C11H15NO)
The theoretical yield of the reaction of 2-bromo-4-methylpropiophenone was 78.5709849g mephedrone hydrochloride. But it should be considered that it is difficult to achieve the theoretical yield, usually the loss is 20%.
Last edited by a moderator:
Not qualitatively purified, or little time for the reaction. More details are needed.
- By whitelabel
Greetings
4mp to 4mmc successfully this way for me but starting 4mp was yellow so bromination done as per process.
Resultant mephedrone was a pasty powder and not crystal like salt as in video. Power was also low even after crystallization. Any thoughts on the pasty powder and low power. Yield was about 120 gms recrystallized from 400 ml 4mp.
Is there any way to purify the 4mp we get to start with or is that fine?
Would a superior product be possible if 4mp is brominated using NBs and PTSA, purified with a solvent nd then make the nmp synthesis from that 2b4m?
Thanks in advance. Tc
4mp to 4mmc successfully this way for me but starting 4mp was yellow so bromination done as per process.
Resultant mephedrone was a pasty powder and not crystal like salt as in video. Power was also low even after crystallization. Any thoughts on the pasty powder and low power. Yield was about 120 gms recrystallized from 400 ml 4mp.
Is there any way to purify the 4mp we get to start with or is that fine?
Would a superior product be possible if 4mp is brominated using NBs and PTSA, purified with a solvent nd then make the nmp synthesis from that 2b4m?
Thanks in advance. Tc
I didn’t add acetone, just MMC freebase in benzene, and I dripped in 30% HCl at pH 5.5. I added the acid slowly. Could acetone be the issue? I forgot, because I usually acidify with DCM+water, so acetone didn’t come to mind. How did you acidify the beauty in the picture? Thank you very much.
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- By William D.
If you use aq. hydrochloric acid, you need a solvent that binds water. This is acetone or alcohol (ethanol, propanol)
Hello friends, where I can find the NMP?
This method can use 1451-82-7?
Of course you can, in which case you skip the first bromination step and proceed to the methylamination step.
@Marvin "Popcorn" Sutton can we scale this up using a 50litre batch reactor and drop the bromine Benzene mix from the reactors dropping funnel without any issues?
Bromine mix in an enclosed dropping funnel is a litle scary
Bromine mix in an enclosed dropping funnel is a litle scary