How to properly generate Methylamine gas to Methanol

[G.Patton]

Best Method
Through experimentation, it was determined that the best method for extracting the MeNH2 from the aqueous solution is to raise the temperature of the solution while stirring. Gas is produced immediately upon stirring at standard temperature, and pressure and the solution begins to boil at 60 ºC. Copious amounts of MeNH2(g) can be obtained by gradually increasing the temperature of the solution between 60 ºC and 80 ºC at normal pressure. A reflux condenser and a gas washing tube filled with anhydrous MgSO4 to pre-dry the gas and 3A molecular sieve to provide a final drying are sufficient to remove any water vapor. (Note, use of NaOH to dry the gas is not recommended for the following reason: NaOH will form a hard moist cake at the vapor/NaOH interface. This thin cake will eventually impede gas flow, raising line pressure, and causing a joint to pop or explode--escaping toxic noxious MeNH2 fumes will quickly render the workplace uninhabitable. This is a fact based on experience, not idle speculation, so avoid the posted suggestion to use NaOH to dry MeNH2 gas: that suggestion is unsound and although a person can “get by” in the short term, in the long term it will eventually lead to catastrophe.)

As the temperature of the solution increases to 80 ºC, water vapor is observed condensing in the lower half of the reflux condenser. After a time at the same temperature, the production of MeNH2(g) begins to decrease. At this point, the partially spent solution is allowed to cool and is pumped out of the reactor flask (a peristaltic pump is ideal) and into a plastic (HDPE) carboy for further treatment to recover the remaining MeNH2. The reactor is reloaded with fresh 40 % solution and stirring and heating are applied, as above, until the temperature reaches 80 ºC and gas production diminishes, whereupon this partially spent solution is added to the contents of the carboy, and the reactor reloaded. In this fashion, a substantial amount of 40 % solution can be processed, without having to take the gas apparatus apart, nor expose the work area to MeNH2 fumes, nor with any solid residue remaining in the boiling flask.

The partially spent solution can then be further treated to obtain practically 100% of the remaining MeNH2. This is accomplished by addition of muriatic acid according to the reaction MeNH2(aq) + HCl(aq) = MeNH2•HCl(aq). The reactor should be maintained in an ice bath during acid addition because there is substantial heat generated by acid addition. Upon neutralization, the MeNH2•HCl solution is brought to a boil, the water and any MeNH2 vapor recovered by condensation, and the dry MeNH2•HCl can then be reacted with saturated NaOH solution to generate MeNH2(g) according to the reaction:
​

MeNH2•HCl(s) + NaOH(aq) = MeNH2 (g) + NaCl(aq)+ H2O​

There is an advantage obtained by first boiling off the major part of the MeNH2(g), prior to acid addition, since substantially less acid is consumed and substantially less hydrochloride salt is produced, therefore less NaOH is needed to convert the hydrochloride to a gas, etc., and less mess and hassle overall. In fact, a person could produce all the MeNH2 he or she might need by simple stirring and boiling of the initial 40 % solution, easily unloading and reloading the reactor, and save the task of reacting the spent solution with muriatic acid for some later date.

Absorption of MeNH2 in MeOH
It is assumed that members reading this post understand that one of the reasons for generating MeNH2 gas is to absorb that gas in cold, stirred MeOH. By weighing the MeOH before and after gas dissolution, the amount of MeNH2 recovered can be calculated, which is required for additional syntheses. Please note that some people suggested using a dispersion tube when absorbing MeNH2(g) into MeOH. That advice is not warranted and is unsound because it will increase line pressure on the system, which could lead to catastrophe. Do not follow those suggestions-do not use a dispersion tube. MeNH2(g) is readily absorbed in cold MeOH. The b.p. of MeNH2 is -6 ºC, so a salt water/ice bath is sufficient to condense vapors. Besides, MeNH2(g) is absorbed in MeOH at just about any temperature short of hot. 1/2” OD polyethylene tubing from the hardware store without any sort of dispersion device on the end of the tubing is sufficient for this purpose.

Suck-back control
Suck-back occurs when the amount of gas being generated is insufficient to compensate for the amount of gas being absorbed. When the production of gas diminishes, suck-back will occur. The suck-back of MeOH with MeNH2 gas can be rapid and violent. Continual suck-back indicates that it is time to replace the spent solution in the reactor with fresh 40 % solution. Suck-back is controlled by stop-cocks to relieve line pressure. However, a trap must be installed between the gas drying tube and the receiver for those inevitable times when the operator looks the other way and valuable MeNH2/MeOH solution is sucked into the system. The trap must be larger than the volume of MeOH in the receiver, so that nothing is lost and nothing can reach the reactor. If MeOH were ever to be sucked back into the hot reactor where the temperature is above the b.p. of MeOH, the resulting explosion of glassware will get everybody’s attention. But be assured that this is not possible with the described set-up.​

trap and stop-cocks​

​

 

[JustBrowsing223]

Best Method
Through experimentation, it was determined that the best method for extracting the MeNH2 from the aqueous solution is to raise the temperature of the solution while stirring. Gas is produced immediately upon stirring at standard temperature, and pressure and the solution begins to boil at 60 ºC. Copious amounts of MeNH2(g) can be obtained by gradually increasing the temperature of the solution between 60 ºC and 80 ºC at normal pressure. A reflux condenser and a gas washing tube filled with anhydrous MgSO4 to pre-dry the gas and 3A molecular sieve to provide a final drying are sufficient to remove any water vapor. (Note, use of NaOH to dry the gas is not recommended for the following reason: NaOH will form a hard moist cake at the vapor/NaOH interface. This thin cake will eventually impede gas flow, raising line pressure, and causing a joint to pop or explode--escaping toxic noxious MeNH2 fumes will quickly render the workplace uninhabitable. This is a fact based on experience, not idle speculation, so avoid the posted suggestion to use NaOH to dry MeNH2 gas: that suggestion is unsound and although a person can “get by” in the short term, in the long term it will eventually lead to catastrophe.)

As the temperature of the solution increases to 80 ºC, water vapor is observed condensing in the lower half of the reflux condenser. After a time at the same temperature, the production of MeNH2(g) begins to decrease. At this point, the partially spent solution is allowed to cool and is pumped out of the reactor flask (a peristaltic pump is ideal) and into a plastic (HDPE) carboy for further treatment to recover the remaining MeNH2. The reactor is reloaded with fresh 40 % solution and stirring and heating are applied, as above, until the temperature reaches 80 ºC and gas production diminishes, whereupon this partially spent solution is added to the contents of the carboy, and the reactor reloaded. In this fashion, a substantial amount of 40 % solution can be processed, without having to take the gas apparatus apart, nor expose the work area to MeNH2 fumes, nor with any solid residue remaining in the boiling flask.

The partially spent solution can then be further treated to obtain practically 100% of the remaining MeNH2. This is accomplished by addition of muriatic acid according to the reaction MeNH2(aq) + HCl(aq) = MeNH2•HCl(aq). The reactor should be maintained in an ice bath during acid addition because there is substantial heat generated by acid addition. Upon neutralization, the MeNH2•HCl solution is brought to a boil, the water and any MeNH2 vapor recovered by condensation, and the dry MeNH2•HCl can then be reacted with saturated NaOH solution to generate MeNH2(g) according to the reaction:
​

MeNH2•HCl(s) + NaOH(aq) = MeNH2 (g) + NaCl(aq)+ H2O​

There is an advantage obtained by first boiling off the major part of the MeNH2(g), prior to acid addition, since substantially less acid is consumed and substantially less hydrochloride salt is produced, therefore less NaOH is needed to convert the hydrochloride to a gas, etc., and less mess and hassle overall. In fact, a person could produce all the MeNH2 he or she might need by simple stirring and boiling of the initial 40 % solution, easily unloading and reloading the reactor, and save the task of reacting the spent solution with muriatic acid for some later date.

Absorption of MeNH2 in MeOH
It is assumed that members reading this post understand that one of the reasons for generating MeNH2 gas is to absorb that gas in cold, stirred MeOH. By weighing the MeOH before and after gas dissolution, the amount of MeNH2 recovered can be calculated, which is required for additional syntheses. Please note that some people suggested using a dispersion tube when absorbing MeNH2(g) into MeOH. That advice is not warranted and is unsound because it will increase line pressure on the system, which could lead to catastrophe. Do not follow those suggestions-do not use a dispersion tube. MeNH2(g) is readily absorbed in cold MeOH. The b.p. of MeNH2 is -6 ºC, so a salt water/ice bath is sufficient to condense vapors. Besides, MeNH2(g) is absorbed in MeOH at just about any temperature short of hot. 1/2” OD polyethylene tubing from the hardware store without any sort of dispersion device on the end of the tubing is sufficient for this purpose.

Suck-back control
Suck-back occurs when the amount of gas being generated is insufficient to compensate for the amount of gas being absorbed. When the production of gas diminishes, suck-back will occur. The suck-back of MeOH with MeNH2 gas can be rapid and violent. Continual suck-back indicates that it is time to replace the spent solution in the reactor with fresh 40 % solution. Suck-back is controlled by stop-cocks to relieve line pressure. However, a trap must be installed between the gas drying tube and the receiver for those inevitable times when the operator looks the other way and valuable MeNH2/MeOH solution is sucked into the system. The trap must be larger than the volume of MeOH in the receiver, so that nothing is lost and nothing can reach the reactor. If MeOH were ever to be sucked back into the hot reactor where the temperature is above the b.p. of MeOH, the resulting explosion of glassware will get everybody’s attention. But be assured that this is not possible with the described set-up.​

View attachment 4274 View attachment 4275
trap and stop-cocks​

​

G.Patton@G.Patton Sorry for the long reply, this will be a 2 part question.
1. Does this means just boiling 40% methylamine hcl aq solution (1000ml H2O + 400 g methylamine hcl? If thats the case than maybe you got an idea how much methylamine hcl would stay in the spent solution? Im just trying to understand how efficient the process would be.

So the apparatus would be something like - 2L 2neck roundbottom with thermometer and 50cm liebig condeser attached to gas drying tube and finished with molecular sieve drying flask , suckback controll and hoes/tube that goes into watter? (I want to get 40% aq solution for 4mmc synthesis)

Also, is the molecular sieve drying needed for making aq. solution?

2. If i would go the aq. NaOH route, i imagine the apparatus as follows, 2L 2neck rbf with pressure equalised dropping funnel in one and condenser in the other neck, continued by drying and suckback part, is that correct?
Maybe you could share what should be the methylamine hcl and watter ratio in reaction flask, as well as NaOH watter ratio in the dropping funnel?

I would be super happy for any info that could help me with part of the project. I simply dont have an option to safely purchase 40% solution.

 

[G.Patton]

@G.Patton Sorry for the long reply, this will be a 2 part question.
1. Does this means just boiling 40% methylamine hcl aq solution (1000ml H2O + 400 g methylamine hcl? If thats the case than maybe you got an idea how much methylamine hcl would stay in the spent solution? Im just trying to understand how efficient the process would be.

So the apparatus would be something like - 2L 2neck roundbottom with thermometer and 50cm liebig condeser attached to gas drying tube and finished with molecular sieve drying flask , suckback controll and hoes/tube that goes into watter? (I want to get 40% aq solution for 4mmc synthesis)

Also, is the molecular sieve drying needed for making aq. solution?

2. If i would go the aq. NaOH route, i imagine the apparatus as follows, 2L 2neck rbf with pressure equalised dropping funnel in one and condenser in the other neck, continued by drying and suckback part, is that correct?
Maybe you could share what should be the methylamine hcl and watter ratio in reaction flask, as well as NaOH watter ratio in the dropping funnel?

I would be super happy for any info that could help me with part of the project. I simply dont have an option to safely purchase 40% solution.

JustBrowsing223

Does this means just boiling 40% methylamine hcl aq solution (1000ml H2O + 400 g methylamine hcl? If thats the case than maybe you got an idea how much methylamine hcl would stay in the spent solution? Im just trying to understand how efficient the process would be.

Hello, no. There is written about methylamine water solution, not about methylamine hydrochloride. There is big difference. If you wanna get methylamine free base, you need add there an alkali. There is the reaction equation for you:

MeNH2•HCl(s) + NaOH(aq) = MeNH2 (gas) + NaCl(aq)+ H2O​

2L 2neck rbf with pressure equalised dropping funnel in one and condenser in the other neck, continued by drying and suckback part, is that correct?​

You need equimolar amount (1:1 in moles with methylamine*hcl) of NaOH in about 50% aq sln.

This is about methylamine free base in aq solution:
You need rbf with a reflux condenser and a gas washing tube after his filled with anhydrous MgSO4 to pre-dry the gas and 3A molecular sieve to provide a final drying are sufficient to remove any water vapor. You can add there instantly via first neck or use two neck rbf (it's better option).​

 

[G.Patton]

Then 5 seconds strong bubbling occured inside the Methanol jar and then it stopped

ResearcherIt's quite strange cuz you have excesses of pressure.

And when I wanted to open the stop cork with tubing in erlenmeyer flask (to put more Methylamine solution inside)
the methanol solution sucked back inside this erlenmayer and I lost 0.5 liters of methanol

You have to use Pressure Equalizing Dropping Funnel, trap between condenser and receiver flask with MeOH and stop-cock btw reaction flask with NaOH and condenser.

When you want to stop addition of MeNH2 to receiver flask, close stop-cock btw reaction flask with NaOH and condenser and then open drip funnel tap. Also, make connection with atmosphere from receiver flask via additional stop-cock.​

 

[hunter12]

It's quite strange cuz you have excesses of pressure.

You have to use Pressure Equalizing Dropping Funnel, trap between condenser and receiver flask with MeOH and stop-cock btw reaction flask with NaOH and condenser.

When you want to stop addition of MeNH2 to receiver flask, close stop-cock btw reaction flask with NaOH and condenser and then open drip funnel tap. Also, make connection with atmosphere from receiver flask via additional stop-cock.​

G.PattonDon't you have atleast a diagram of the setup? I guess it would be of great help

 

[diogenes]

Hi, what if I have MeNH2-HCl as starting material, can I just simply add equimolar alkali and have a pretty concentrated water or methanol solution depending on which solvent I disolve HCl salt. With methanol it is not ideal because a lot of water will be generated. In aquous solution there will be a lot of NaCl disolved. Could this salty water solution be used in synthesis or it would interfere with the synthesis. e.g.when adding it to 2Br4Methylpropiophenone in 4MMC synthesis, this would the easiest method without the need for bubbling. Or I could just distill the whole thing with enoguth H2O to disolve the bubbling gas until the water starts coming over. Anyone with experience in this? Experts? I would really appreciate some help, as Im completely new to this but would like to make some MDMD. Perhaps when using methanol as a starting solvent it will form a separate the saturated water from the methanol with Mehylamine.

 

[41Dxflatline]

Hi, what if I have MeNH2-HCl as starting material, can I just simply add equimolar alkali and have a pretty concentrated water or methanol solution depending on which solvent I disolve HCl salt. With methanol it is not ideal because a lot of water will be generated. In aquous solution there will be a lot of NaCl disolved. Could this salty water solution be used in synthesis or it would interfere with the synthesis. e.g.when adding it to 2Br4Methylpropiophenone in 4MMC synthesis, this would the easiest method without the need for bubbling. Or I could just distill the whole thing with enoguth H2O to disolve the bubbling gas until the water starts coming over. Anyone with experience in this? Experts? I would really appreciate some help, as Im completely new to this but would like to make some MDMD. Perhaps when using methanol as a starting solvent it will form a separate the saturated water from the methanol with Mehylamine.

diogenesUse a drying agent?

 

[diogenes]

Use a drying agent?

41Dxflatline41Dxflatline: I have thought of this, but e.g. starting from 250g of MeNH2 HCl, which is roughly 3.7mol, there would be 3.7 mol water created which is roughly 67ml quite a lot to dry... my thinking was that since NaCl solubility is much lower in methanol than water (about 20X less per volume), I would get essentially brine, which would soak up 24g from the 3.7mol NaCl (175g) and become saturated (brine). The remaining 151g NaCl would be in the methanol (solubiliy 14g/l) I couldn`t find the exact solubility of methylamine, but is significantly more than the NaCl, let`s say 20-30% w/w, so it would precipitate most of the NaCl, at least 151-14=137 g. The question is whether (despite the higher solubility would it precipitate even more or the waper pressure would increase and just fume out of solution. I could do the whole reaction in the freezer, or just use salt water ice (-10) to keep methylamine liquid (-6C). If it would push most of it out then (Ideally I would end up with methylamine saturated methanol. Just be mixing the ingreadientst together and putting everything in the freezer, a far easiers way than bubbling if it works. But this is where my knowledge ends and should probably look into laws of solubility. I could quickly decant, suck the methyl alcohol from the now frozen brine (freezer - 20), and keeping the methanol in the freezer, even the increase in volume would tell me how much methylamine I tave, but I could also measure the weight cold. To final drying I could drop in some mol sieves in case there is still some water left.

Anyway, there could be a fault in this reasoning, and I`m thinking of simply distilling the methylamine from water or , precooling the receiving flast with salt water ice and then attaching a dry ice aceton cold trap to make sure no methylamine gas would escape; That said, it might be easier to just generate MeNH2 with Patton`s method, let it cool to near zero in a graham going into a receiving flahofsk in salt water ice, and dry acetone trap for safety.

I just thought there is a well-known simple method is so obvious that no one is posting it and the experts will definitely know about think. It would be even easier if I could drill a hole into an ord freezer. If no one knows an even simpler method then I`ll try these ones and feed back here.

 

[Tweaker]

41Dxflatline: I have thought of this, but e.g. starting from 250g of MeNH2 HCl, which is roughly 3.7mol, there would be 3.7 mol water created which is roughly 67ml quite a lot to dry... my thinking was that since NaCl solubility is much lower in methanol than water (about 20X less per volume), I would get essentially brine, which would soak up 24g from the 3.7mol NaCl (175g) and become saturated (brine). The remaining 151g NaCl would be in the methanol (solubiliy 14g/l) I couldn`t find the exact solubility of methylamine, but is significantly more than the NaCl, let`s say 20-30% w/w, so it would precipitate most of the NaCl, at least 151-14=137 g. The question is whether (despite the higher solubility would it precipitate even more or the waper pressure would increase and just fume out of solution. I could do the whole reaction in the freezer, or just use salt water ice (-10) to keep methylamine liquid (-6C). If it would push most of it out then (Ideally I would end up with methylamine saturated methanol. Just be mixing the ingreadientst together and putting everything in the freezer, a far easiers way than bubbling if it works. But this is where my knowledge ends and should probably look into laws of solubility. I could quickly decant, suck the methyl alcohol from the now frozen brine (freezer - 20), and keeping the methanol in the freezer, even the increase in volume would tell me how much methylamine I tave, but I could also measure the weight cold. To final drying I could drop in some mol sieves in case there is still some water left.

Anyway, there could be a fault in this reasoning, and I`m thinking of simply distilling the methylamine from water or , precooling the receiving flast with salt water ice and then attaching a dry ice aceton cold trap to make sure no methylamine gas would escape; That said, it might be easier to just generate MeNH2 with Patton`s method, let it cool to near zero in a graham going into a receiving flahofsk in salt water ice, and dry acetone trap for safety.

I just thought there is a well-known simple method is so obvious that no one is posting it and the experts will definitely know about think. It would be even easier if I could drill a hole into an ord freezer. If no one knows an even simpler method then I`ll try these ones and feed back here.

diogenesMDMA can be done wet. But yes set temp to -10 when you add the methylamine HCl.

 

[41Dxflatline]

Methamphetamine synthesis from P2P by NaBH4 reduction. Medium-Scale.

Introduction I represent to BB audience Methamphetamine synthesis method of 1-phenyl-2-propanon (P2P) reductive amination by NaBH4. Following method allows obtaining and large batches of product. The hardest problem of this method is the heat generating during exothermic reaction of imine...

bbgate.com

Here's an example of methylamine being used as a salt in a synthesis

 

[workworkwork]

"A reflux condenser and a gas washing tube filled with anhydrous MgSO4 to pre-dry the gas and 3A molecular sieve to provide a final drying are sufficient to remove any water vapor. "

Is really necessary dry the methylamine gas with this 3 methods?
1st - condenser with very cold water to remove most of the water from the methylamine gas
2nd - the tube with the gas coming from the condenser is inserted in a flask filled with MgSO4
3rd - dry the methylamine in methanol solution with 3A molecular sieves

can I remove the MgSO4 step or it comes with so much water from the condenser that is better use the 3 methods to remove the water?

 

[Idontcarethatmuch]

Can you store the in Methanol solved Methylamine as long as you keep the temperature low enough

Is there a problem in concentrating as much M-Gas in Methanol as possible (how much would that be?) and later deluting it with more (cold) methanol if needed?

Seems like a chemical I shouldn't "just try it"^^

 

[WillD]

It is convenient to use TEA (triethylamine) to obtain methylamine gas in alcohol