4-MMC (mephedrone) synthesis. Complete video tutorial.
- Thread starter Marvin "Popcorn" Sutton
- Start date
Greetings.
Despite reading a lot, I still have doubts.
Can i dissolve methylamine hcl in a NaOH solution?
Would I get aqueous methylamine + NaCl?
Is the use of aqueous methylamine with NaCl possible in this mephedrone synthesis? If not possible, how to wash off the NaCl?
Could someone good in stoichiometry calculate the amount of grams of methylamine hcl and aqueous NaOH solution to obtain 40% aqueous methylamine??
Thank you very much!!!
Despite reading a lot, I still have doubts.
Can i dissolve methylamine hcl in a NaOH solution?
Would I get aqueous methylamine + NaCl?
Is the use of aqueous methylamine with NaCl possible in this mephedrone synthesis? If not possible, how to wash off the NaCl?
Could someone good in stoichiometry calculate the amount of grams of methylamine hcl and aqueous NaOH solution to obtain 40% aqueous methylamine??
Thank you very much!!!
CH3NH2·HCl (67.52g) + NaOH (40g) → CH3NH2 (31.06g) + NaCl (58.44g) + H2O (18.01g)
200ml of 40% aqueous methylamine => 80g CH3NH2 dissolved in 200g H2O
If I use (174g of methylamine hcl + 67% aqueous NAOH solution)
CH3NH2.HCL 174g
NAOH 103g + H20 154g
I obtain (200ml of 40% aqueous methylamine):
CH3NH2 80g
NACL 150g
H20 200g
Would it be possible?
200ml of 40% aqueous methylamine => 80g CH3NH2 dissolved in 200g H2O
If I use (174g of methylamine hcl + 67% aqueous NAOH solution)
CH3NH2.HCL 174g
NAOH 103g + H20 154g
I obtain (200ml of 40% aqueous methylamine):
CH3NH2 80g
NACL 150g
H20 200g
Would it be possible?
- By malignoalfa
REAGENTS
CH3NH2·HCl 67,518 g/mol
NaOH 40 g/mol
PRODUCTS
CH3NH2 31.0571 g/mol
NaCl 58.4428 g/mol
H20 18 g/mol
I decided to do some calculations myself to clarify the matter for myself; maybe it will clarify for others.
Methylamine aq 40% → 1 CH3NH2 (31g) + 2.5 H2O (45g)
31g divided by 31g + 45g = 40%
The density of Methylamine 40% (w/w) in water is 0.897 g/mL, that is, 200ml has 179.4 grams.
40% methylamine = 71.76g (2.31 mol)
60% water = 107.64g
I have seen recommendations to use a 1:5 ratio of NaOH to H2O and add dropwise to an ice bath aqueous solution of methylamine.hcl. Perhaps the use of a lower concentration of NaOH for basification has the objective of making the reaction more peaceful, not heating, not releasing methylamine in gas; but I can't say for sure.
To clarify, the 1:5 ratio of NaOH to H2O refers to a 31% concentration.
1 mol NaOH => 40g
5 mol H20 => 90g
40g divided by 40g + 90g = 31%
Methylamine.hcl has a water solubility of 1g/mL (at 20 °C)
1 mol CH3NH2.HCL => 67.52g would need 4 mols of H20 (72g/72ml) to completely dissolve, a concentration of 48%:
67.52g divided by 67.52g + 72g = 48%
The balanced reaction equation is:
CH3NH2·HCl + NaOH → CH3NH2 + NaCl + H2O
The equation involving solutions of methylamine.hcl with NaOH solution above would look like this:
(1 CH3NH2·HCl + 4 H2O) (48% aq) + (1 NaOH + 5 H2O) (31% aq) → 1 CH3NH2 + 1 NaCl + 10 H2O
To obtain 2.31 mol of methylamine equivalent to 200ml of Methylamine 40% (w/w) in water, we would have:
(2.31 CH3NH2·HCl + 9.24 H2O) (48% aq) + (2.31 NaOH + 11.55 H2O) (31% aq) → 2.31 CH3NH2 + 2.31 NaCl + 23.1 H2O
In grams, we would have (rounded):
(156g CH3NH2·HCl + 166g H2O) + (93g NaOH + 208g H2O) → 72g CH3NH2 + 135g NaCl + 416g H2O
We would have 72g of methylamine with 135g of NaCl dissolved in 416g (ml) of H20.
An aqueous solution of 33% methylamine + NaCl, being 11% methylamine and 22% NaCl.
72g + 135g divided by 72g + 135g + 416g = 33%
72g/623g (11%) + 135g/623g (22%) = 33%
I don't know if excess water would disturb the synthesis of this topic, but just removing excess water with a drying agent without removing NaCl would not solve it, as we would have solubility problems (maximum saturation). I couldn't tell if the NaCl would precipitate or if the methylamine would be released as a gas, or perhaps both.
CH3NH2·HCl 67,518 g/mol
NaOH 40 g/mol
PRODUCTS
CH3NH2 31.0571 g/mol
NaCl 58.4428 g/mol
H20 18 g/mol
I decided to do some calculations myself to clarify the matter for myself; maybe it will clarify for others.
Methylamine aq 40% → 1 CH3NH2 (31g) + 2.5 H2O (45g)
31g divided by 31g + 45g = 40%
The density of Methylamine 40% (w/w) in water is 0.897 g/mL, that is, 200ml has 179.4 grams.
40% methylamine = 71.76g (2.31 mol)
60% water = 107.64g
I have seen recommendations to use a 1:5 ratio of NaOH to H2O and add dropwise to an ice bath aqueous solution of methylamine.hcl. Perhaps the use of a lower concentration of NaOH for basification has the objective of making the reaction more peaceful, not heating, not releasing methylamine in gas; but I can't say for sure.
To clarify, the 1:5 ratio of NaOH to H2O refers to a 31% concentration.
1 mol NaOH => 40g
5 mol H20 => 90g
40g divided by 40g + 90g = 31%
Methylamine.hcl has a water solubility of 1g/mL (at 20 °C)
1 mol CH3NH2.HCL => 67.52g would need 4 mols of H20 (72g/72ml) to completely dissolve, a concentration of 48%:
67.52g divided by 67.52g + 72g = 48%
The balanced reaction equation is:
CH3NH2·HCl + NaOH → CH3NH2 + NaCl + H2O
The equation involving solutions of methylamine.hcl with NaOH solution above would look like this:
(1 CH3NH2·HCl + 4 H2O) (48% aq) + (1 NaOH + 5 H2O) (31% aq) → 1 CH3NH2 + 1 NaCl + 10 H2O
To obtain 2.31 mol of methylamine equivalent to 200ml of Methylamine 40% (w/w) in water, we would have:
(2.31 CH3NH2·HCl + 9.24 H2O) (48% aq) + (2.31 NaOH + 11.55 H2O) (31% aq) → 2.31 CH3NH2 + 2.31 NaCl + 23.1 H2O
In grams, we would have (rounded):
(156g CH3NH2·HCl + 166g H2O) + (93g NaOH + 208g H2O) → 72g CH3NH2 + 135g NaCl + 416g H2O
We would have 72g of methylamine with 135g of NaCl dissolved in 416g (ml) of H20.
An aqueous solution of 33% methylamine + NaCl, being 11% methylamine and 22% NaCl.
72g + 135g divided by 72g + 135g + 416g = 33%
72g/623g (11%) + 135g/623g (22%) = 33%
I don't know if excess water would disturb the synthesis of this topic, but just removing excess water with a drying agent without removing NaCl would not solve it, as we would have solubility problems (maximum saturation). I couldn't tell if the NaCl would precipitate or if the methylamine would be released as a gas, or perhaps both.
I couldn't find a theoretical chemical reaction for this process. Could anyone indicate?
- By G.Patton
Have you looked around Methcatinones section? https://bbgate.com/threads/mephedro...haloketone-in-ethyl-acetate-1-10-kg-scale.39/
There are reaction schemes
There are reaction schemes
I tried react 100g 2-bromo4methylpropiophenone with proportion 200ml methyloamine 40% and 300ml nmp. When last portion OF nmp was added, solution suddenly became yellow and clear, after 20min OF stirring it stays same, not browny and foggy like in the video. What I did wrong?
- By StarWars
Your freebaasee is good what equipment u use
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- By Czeczeczecze
Same equipment like in the video, but my flask is 6000ml. I’ll try again in 2 days, maybe i did some stupid error. When i was washing my solution with water; water layer had smell OF methyloamine, I don’t know if this is important
- By Czeczeczecze
Ok, thanks for help. I’ll try again and post results. So if my solution stays yellow after nmp addition it’s still OK? I read that Brown color is an indicator of succesfull amination
- By Czeczeczecze
My experiment resulted in nice white Powder but yield was 30% any ideas why so low? And another problem, when solution was in frezeer, after 4 hours it was completely frozen
We brominate with the absorption of hydrobromic acid. Then we replace with methylamine and carry out the selection with further purification.
My synthesis takes approximately 3 hours from start to salt isolation. + crystallization.
Tried to crystallize in different ways. Nitrogen, thermoses, drying cabinets.
My synthesis takes approximately 3 hours from start to salt isolation. + crystallization.
Tried to crystallize in different ways. Nitrogen, thermoses, drying cabinets.
- By Czeczeczecze
Can You specify Your synthesis? I want to obtain similar results. What equipment is needed etc
