malignoalfa

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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?
 

malignoalfa

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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?
malignoalfaREAGENTS

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.
 

aa1178251182

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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.
malignoalfaDid you synthesize 4-MMC with methylamine hydrochloride?
 

Chery

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Why is no one using methylamine in methanol, i thought an anhydrous medium would increase the yield?
 

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This video represent a simple synthesis way of 4-MMC (mephedrone) in a small scale from 4'-methylpropiophenone. There is total synthesis video manual which explain everything and gives opportunity to make 4-MMC to every beginner of organic synthesis.

Equipment and glassware: View attachment 6900

Reagents:
  • 4'-Methylpropiophenone (cas 5337-93-9) - 100g;
  • Benzene – 600ml;
  • Bromine (Br2) - 100g;
  • N-methylpyrrolidone - 300ml;
  • Methylamine free base 40% (aqueous solution) - 200 ml;
  • Hydrochloric Acid 35% - 55ml;
  • Acetone - 1000ml;

Description of the 4-MMC (mephedrone) synthesis video:
From 0:00 to 0:48
It is said at the start of this video that laboratory and reagents temperature (except for those that need to be cooled) should be moderate, about 20-25ºC.

Next, the preparation of the reactor to work is described:
"2000ml Round bottom flask is placed on a magnetic stirrer and secured to a retort stand and clamp for securing apparatus. A magnetic stirrer work is checked. Reflux condenser is installed on the central flask neck and a gas outlet hose is fit to it. The hose is installed carefully, without kinks. The end of the hose is immersed into a canister with 15% aqueous sodium bicarbonate solution (to absorb odors)."

Then, the reagents loading is come :
""4-Methylpropiophenone 100g and benzene 300ml are poured into the flask and stired. Bromine 100g and benzene 100ml are poured into the drip funnel. A thermometer is installed and bromine is dropped into a reaction mixture. A gentle heating is turned on and the temperature in the flask is increased to 35 ºC, then the heater is turned off."


From 0:48 to 3:23
Bromination reaction. The bromination is started by a chemist after the all reagents loading. Bromine solution is added in small portions to benzene in the flask, so that the reaction mixture has time to discolour. After the bromine addition, the mixture is stirred on a magnetic stirrer for addition 30 minutes.
"Bromine is added drop by drop to the mixture in four equal portions after the switching off the heater . Each further bromine portion is added after discoloring of previous one."
"The reaction mixture is stirred for additional 30 minutes after complete bromine addition"
The 4-Methylpropiophenone bromination gives 2-bromo-4-Methylpropiophenone in benzene.


From 3:23 to 6:16
Reaction mixture is
washed in separatory funnel and excess pressure is revealed through a tap. The funnel is twisted in a circular motion, for collecting the remaining water and draining it, the solution of 2-Bromo-4-Methylpropiophenone in benzene is left for the following synthesis.
" Water 200 ml is added after 30 minutes."
"The mixture is poured into the separatory funnel. Layers are separated clearly. The bottom layer is disposed, the top layer is needed."
"The washing is repeated. Water 200 ml is poured and shaken well. Layers are separated clearly. The bottom layer is drained.
Do not forget to release the pressure by opening the tap."
"The funnel is rotated in a circular motion to collect all water in the bottom layer."


From 6:17 to 9:02
Methylamination reaction. The solution of 2-Bromo-4-Methylpropiophenone in benzene is poured back into the flask, methylamine is poured. The reaction mixture is heated to 40 ºC. NMP is added in small portions, make sure theat a temperature is not exceeded 60 ºC. In case the temperature is exceeded 60ºC, the following addition have to be carried out when temperature is droped. After all NMP has been added, the reaction mixture is stirred for 15 minutes. A rich amber color of the reaction mixture can be observed during the methylamination process. This is a sign that the reaction is over.

2-Bromo-4-methylpropiophenone methylamination gives mephedrone free base (4-MMC).
"The obtained 2-Bromo-4Methylpropiophenone in benzene is poured back into a clean flask and 40% methylamine 200ml (aqueous solution) is added."
"A reflux condenser, drip funnel and thermometer are installed onto the flask. The stirrer is turned on and flask is heated to 40 ºC."
"When the temperature is reached 40ºC, the heater is turned off and N-methylpyrrolidone 300 ml is added from the drip funnel, make sure that the temperature is not exceed 60 ºC. A short exothermic reaction takes place and the temperature can rise after addition of N-methylpyrrolidone."
"Reaction mixture is stirred for 15 minutes after all N-methylpyrrolidone is added."


From 9:03 to 11:33
Mephedrone free base washing. The free base is washed with sodium bicarbonate solution 200ml 5% and water 3 x 200ml . The mixture is shaken well, an excess gas pressure is released through a tap. Layers are separated clearly in the separatory funnel. The lower aqueous layer is drained into a waste, the upper organic layer with a free base is left for further work. A pinch of sodium chloride is added to solution for residual water removing from a turbidity. It is left for some time after well stirring to layer separating and a water residue is drained. The pure free mephedrone base is obtained.
"Then sodium bicarbonate solution 100ml 5% and benzene 200ml are added, stirred well, poured into a separatory funnel. Layers are separated clearly."
"The lower water layer is disposed off"
"Water 200 ml is poured to mephedrone free base in benzene (top layer), stirred well. Layers are separated clearly. The lower aqueous layer is disposed off."
"Mephedrone free base in benzene is washed with water three times."
"If mephedrone free base solution in the benzene looks cloudy after washing, some sodium chloride is added and stirred in a circular motions."
"The mixture is left to stand for a couple of minutes, the rest of the water with the sodium chloride is collected at the bottom layer. This water residue is disposed off. Pure mephedrone free base in benzene is remained in the funnel."


From 11:34 to 12:38
Acidification. Mephedrone free base is converted to hydrochloride salt in this part of the synthesis. Hydrochloric acid is added drop by drop to 4MMC free base with a constant stirring. The solution is become hot after acidification and precipitate is appeared. This precipitate is mephedrone hydrochloride. It is necessary to measure pH constantly during the acidification procedure. The acidification is stopped when pH 5-5.5 is reached. The beaker with mixture is covered with a film and put into a freezer for a few hours.
"Mephedrone base in benzene is poured into a wide beaker, put it on a magnetic stirrer. A stirring bar is placed inside and stirring is started. A drip funnel is placed above the beaker and hydrochloric acid 55 ml is added at 1-2 drops per second rate."

"pH is measured every 10 seconds to reach 5-5.5 pH. Mephedrone hydrochloride is formed in the beaker and the temperature is rise during the acidification procedure. Mephedrone hydrochloride can be dissolved, this situation is normal. Crystalls are appeared during a cooling procedure."

"The stirrer is turned off when pH is 5-5.5, the beaker is covered with a cling film and put it in a freezer for 4 hours. The freezer temperature shouldn't be lower than -15 ºC."


From 12:39
Filtration and washing.
The
filtration apparatus is prepared: Buechner funnel with a paper filter is placed on a Buechner flask in the final part of the video. The filter is moistened with acetone for a better contact with a funnel surface. The mixture is stirred well so that the precipitate does not stick to a beaker and poured into the funnel. The mephedrone remainder is washed off with an additional portions of acetone and transferred to a Buechner funnel.

Then a vacuum pump is connected to Buechner flask via a hose. The powder is thickened slightly and cracks are appeared during the filtration process. A small amount of acetone is added to the Buechner funnel, switching off the vacuum at the same time, for the better washing and purity level. The product is washed with acetone until transparent flushes is started to drip from the funnel.

The powder is dried in Buechner funnel with a switched on vacuum until mephedrone hydrochloride powder is become crumbly after washing with acetone. The mephedrone is air dried in a dry room and weighed.
"The resulting precipitate is filtered under vacuum and washed on the filter with acetone.
The long edge of the funnel spout in the opposite direction from the side outlet of the flask is unfolded."
"The filter have to be moistened with acetone."
"The mephedrone powder residue in the beaker is washed off with portion of acetone."
"Hose from a vacuum pump."
"Magnet from a magnetic stirrer."
"Mephedrone is dried on the air. The theoretical yield of mephedrone hydrochloride is 144g from 100g of 4-methylpropiophenone, but considering losses, you get ~100g (maybe more)."
Marvin Popcorn SuttonThis is cooll man...youre awesome teach us with completed video tutorial...cherrsss
 

OttoBulletproof

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Can Methylamine HCL be used or does it have to be Freebase?
 

UWe9o12jkied91d

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chuckmcgill

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has to be freebase, just mix hcl with naoh
UWe9o12jkied91ddo you have a good formula for making 40% aqueous methlyamine from its HCL and NaOH? I tried based on a technique in here that uses 40g methylamine HCL in 100ml water mixed with 17g Naoh in 50ml water. That seemed to not produce satisfactory results. The molar ratios seem off too but the poster said it is the only ratio he found to work
 

cartelloszetas

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My mmc thanks bb forum for help. U are the best :)
 

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tajira

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Can someone confirm these proportions for 100g of 2Bromo4Methylpropiophenone
2Bromo4Methylpropiophenone - 100g
DCM - 300ML (3ML PER G)
NMP - 300ML
METHYLAMINE - 200ml
 

HerrHaber

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Can someone confirm these proportions for 100g of 2Bromo4Methylpropiophenone
2Bromo4Methylpropiophenone - 100g
DCM - 300ML (3ML PER G)
NMP - 300ML
METHYLAMINE - 200ml
tajiraIt may well work but the whole thing comes with practice thus don’t jump into big quantities (100g is ok if you have the other 900g left).
 

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Hello, I have a problem. I tried the synthesis in various ways, on NMP extracting with teluene and ethyl acetate, on acetate only, almost always with an excess of methylamine. Usually it is possible to collect freebase well from the primary mixture using extraction, sometimes the problem is with extraction with ethyl acetate with water alone, sometimes it can not separate after animation, especially when it is used for extraction from RM on NMP, toluene as a typical nonpolar solvent for example picks up the freebase from NMP RMs well, which in turn is well miscible with water, so I think it can be considered the most efficient solvent for this reaction, NMP i mean. In syntheses, they usually describe that you need to heat RM on NMP for 40 minutes, but due to the miscibility with water, it can be assumed that the reaction proceeds here even faster and more efficiently than on ethyl acetate. It is also known that the reaction is exothermic, so when the temperature starts to drop, the reaction is theoretically over. Okay, but still my problem is that how would I not do it when acidifying, salt does not want to show up, I drop acid to pH 5.5-6, mix and mix and nothing, possibly darkens, but nothing it does not precipitate, and it should have swelled and become thick a long time ago. If such a mixture is then given for cooling or evaporation, nothing appears, only possibly black tar is made. And so practically every time. Recently, I just figured out that there is some sediment as after acidification I add some alkali like baking soda or caustic. Why is that, can someone help me solve this puzzle and tell me what I'm doing wrong?
 

UWe9o12jkied91d

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Hello, I have a problem. I tried the synthesis in various ways, on NMP extracting with teluene and ethyl acetate, on acetate only, almost always with an excess of methylamine. Usually it is possible to collect freebase well from the primary mixture using extraction, sometimes the problem is with extraction with ethyl acetate with water alone, sometimes it can not separate after animation, especially when it is used for extraction from RM on NMP, toluene as a typical nonpolar solvent for example picks up the freebase from NMP RMs well, which in turn is well miscible with water, so I think it can be considered the most efficient solvent for this reaction, NMP i mean. In syntheses, they usually describe that you need to heat RM on NMP for 40 minutes, but due to the miscibility with water, it can be assumed that the reaction proceeds here even faster and more efficiently than on ethyl acetate. It is also known that the reaction is exothermic, so when the temperature starts to drop, the reaction is theoretically over. Okay, but still my problem is that how would I not do it when acidifying, salt does not want to show up, I drop acid to pH 5.5-6, mix and mix and nothing, possibly darkens, but nothing it does not precipitate, and it should have swelled and become thick a long time ago. If such a mixture is then given for cooling or evaporation, nothing appears, only possibly black tar is made. And so practically every time. Recently, I just figured out that there is some sediment as after acidification I add some alkali like baking soda or caustic. Why is that, can someone help me solve this puzzle and tell me what I'm doing wrong?
MephistoList the steps taken succinctly, in order if you want a good answer.
Check your methylamine, it might steel reek to your human nose but it might be 15%.Percentage HCl used? Dried freebase before acid? Used acetone(cold) 1:1 in relation to rm+extraction solvent?Forgot about an acidified portion for 2 weeks in the back of the freezer? Kissed your cousin?
 

Albert A

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Hello everyone, Can someone please explain to me why the methylamination reaction is made simultaneously in two solvents benzene and NMP ?
(Not in one solvent).
 

PolandoEmingarnte

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Hello everyone, Can someone please explain to me why the methylamination reaction is made simultaneously in two solvents benzene and NMP ?
(Not in one solvent).
Albert Abecause u made 2b4m and then 4mmc
 

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I’m producing a yield today and I thought it’ll be helpful to give give an old method I used in my beginning stages as a chemist. It’s a simple method I’ve found for beginners and you don’t need much equipment.

Synthesizing commercial simple 4-bromoketone.

I will use methylamine in 40% aqueous solution w/ dcm for a simple Nucleuphile with Sn2 substitiotion.

2bromo4methylpropiophenone, dcm, and 2.5x moles the equivalent of methylamine needs a really good stir while keeping a constant cool temperature with an ice bath.

After the third addition remove the ice bath and stir for 6 hours more while keeping a constant room temperature. Here will be noticed a layer turn from yellow-dark yellow-orangish.

Next add a precooled solution of NaOH until it reaches 11 ph.

Adding cooled sodium hydroxide into the solution, this will cool the natural thermal effects and preserve the proper reaction process.

Maneuver a water hose like device with hydrochloric acid solution to recycle some methylamine.

Give a good wash 2-3 time with brine solutions, and extract with aqueous diluted hydrochloric acid.

Wash off acidic layer with toulene and evaporate to obtain white crystalish powder consistant uniformed product. Next carefuly extract with vacum about 3-4 times with chlorofom, then evaporate the cholorofm to obtain the pure mephedrone. You can disolve it in hot IPA/water and let it evaporate slowly, it will create small crystals.

It can be also be obtain better with dibenzoyl-I-tartaric acid to resolute the reacemic mixture but think it’ll be easier for beginners with the first mechanism using IPA.

With dibenzoyl-I-tartaric acid you will have to add portions 1:1 mole of the acid per mole of amine in ethyl acetate and salt will precipitate. Then just follow up carefully with some A/B to obtain between 70-90% enantiomery pure mephedrone.

“You absolutely have to acidify the finished product with isopropanolic/hcl in order to react it because heating cathinones with acids racemizes them”.
 

PolandoEmingarnte

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I’m producing a yield today and I thought it’ll be helpful to give give an old method I used in my beginning stages as a chemist. It’s a simple method I’ve found for beginners and you don’t need much equipment.

Synthesizing commercial simple 4-bromoketone.

I will use methylamine in 40% aqueous solution w/ dcm for a simple Nucleuphile with Sn2 substitiotion.

2bromo4methylpropiophenone, dcm, and 2.5x moles the equivalent of methylamine needs a really good stir while keeping a constant cool temperature with an ice bath.

After the third addition remove the ice bath and stir for 6 hours more while keeping a constant room temperature. Here will be noticed a layer turn from yellow-dark yellow-orangish.

Next add a precooled solution of NaOH until it reaches 11 ph.

Adding cooled sodium hydroxide into the solution, this will cool the natural thermal effects and preserve the proper reaction process.

Maneuver a water hose like device with hydrochloric acid solution to recycle some methylamine.

Give a good wash 2-3 time with brine solutions, and extract with aqueous diluted hydrochloric acid.

Wash off acidic layer with toulene and evaporate to obtain white crystalish powder consistant uniformed product. Next carefuly extract with vacum about 3-4 times with chlorofom, then evaporate the cholorofm to obtain the pure mephedrone. You can disolve it in hot IPA/water and let it evaporate slowly, it will create small crystals.

It can be also be obtain better with dibenzoyl-I-tartaric acid to resolute the reacemic mixture but think it’ll be easier for beginners with the first mechanism using IPA.

With dibenzoyl-I-tartaric acid you will have to add portions 1:1 mole of the acid per mole of amine in ethyl acetate and salt will precipitate. Then just follow up carefully with some A/B to obtain between 70-90% enantiomery pure mephedrone.

“You absolutely have to acidify the finished product with isopropanolic/hcl in order to react it because heating cathinones with acids racemizes them”.
marty€mcflyafter 1 you can do the synthesis much faster, easier and you don't need labo equipment at all. synthesis can be done in 1.5 hours tartaric acid haha probably hydrochloric acid. no need to use ipa acetone or ethyl acetate in the synthesis. You write for a beginner where you have how much what do you have to add there is no freebase cleanup at all and naoh only burns freebase is another thing.
 

marty€mcfly

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I have access to a lab & lab equipment now so, h don’t use it anymore but, it’s nice to share!
 

Albert A

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I have access to a lab & lab equipment now so, h don’t use it anymore but, it’s nice to share!
marty€mcflyHow did the prescription change under more favorable conditions in the lab?

does the type of solvent ( DCM or NMP) if keeping the heat limit not affect the cleanliness of the final product?
 
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marty€mcfly

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How did the prescription change under more favorable conditions in the lab?

does the type of solvent ( DCM or NMP) if keeping the heat limit not affect the cleanliness of the final product?
Albert AI can’t say what changes the color chemically because I could be wrong but each solvent needs a different care. I just know the methods I use, and keeping a constant temperature with an ice bath keeps the color light.

“I can’t say that the lab makes it better because I’ve struck gold in less professional settings”. I -can-recant-that!
 
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