Author Topic: Processing 1 Mole Ketone Using Nitromethane  (Read 3735 times)

0 Members and 1 Guest are viewing this topic.

abacus

  • Guest
Processing 1 Mole Ketone Using Nitromethane
« on: November 23, 2002, 07:09:00 AM »
Here’s my virtually foolproof method to process hundreds of grams of ketone via al/hg using nitromethane in 1 day, and more importantly never fails.

The procedure is to reduce the nitromethane to methylamine in a methanol solution, recover the methylamine/methanol from the sludge, and use the recovered methylamine laden methanol to process a large amount of ketone with a small amount of aluminium.

The beauty of this method is that nitromethane can be used to make heaps of methylamine in a lot less time than making it from hexamine or formaldehyde, with a lot less hassle IMHO.

Reaction

There are two reaction equations that explain the al/hg using nitromethane as follows:

Hydroxide Mechanism
2 Al  +  6  H2O  >    2 Al(OH)3  +  6 H
6 H  +  CH3NO2  >   CH3NH2   + 2 H2O

2 Al + 4 H2O + CH3NO2 > CH3NH2 + 2 Al(OH)3

Which means 4 moles of water and 2 moles aluminium are needed with every mole nitromethane for that to be converted to methylamine.  This reaction occurs when sufficient water is present and is very exothermic.

Oxide Mechanism
2 Al  + 3  H2O  >     Al2O3  +  6 H
6 H  +  CH3NO2  >   CH3NH2   + 2 H2O

 2 Al + H2O + CH3N02 > CH3NH2 + Al2O3

Which means only 1 mole of water is required for every mole nitromethane to produce the methylamine.

My experimental observations seem to confirm both equations seem to work depending on how much water is in the reaction.

However I believe that the reduction of the ketone via imine to the amine is via the hydroxide mechanism as follows:

Methylamine Mechanism

Reaction Between Ketone And Methylamine
3 Ketone  +  3 Methylamine  > 3 Imine +  3 Water 

Now the imine is reduced to the amine only by the hydroxide mechanisms as follows;

Hydroxide Mechanism
6  H2O  +  2 Al    >   6 H  +  2 Al(OH)3
6 H  +  3 Imine   >   3 MDMA

3 R2CO + 3 CH3NH2 + 2 Al + 3 H2O > 3 R 2CHNHCH3 + 2 Al(OH)3

Which requires 1 mole water per mole ketone.

This is consistent with the typical MM type reaction because the “heet” that he used had enough water in it for the oxide  and hydroxide mechanism reductions of the nitromethane as well as the methylamine.

Scaling up the MM type reaction is very unpractical and prone to failures in my experience.  

The best way for a larger scale al/hg using nitromethane as the methylamine source is to make the methylamine separately and then carry out the al/hg with ketone as follows:

Stage 1:  Making Methylamine In Methanol Using Nitromethane And Al/Hg

Processing 1 mole of ketone will require 2.5-3 moles methylamine dissolved in methanol as follows:  

Procedure

Cut 162g aluminium flashing into small squares and place in a 2 neck 2L, or ideally 3L flask with 1500mls methanol and add 0.5g HgCl to amalgamate.  Let amalgamate for 5 –10 minutes.

Mix 54mls water with 183g (162mls) nitromethane in addition funnel and add 100mls methanol to form a solution.

Then pour in about 30mls of this nitromethane mixture into the reaction flask and let it heat up until refluxing. 

This will cause a vigorous reflux, then slowly drip the remaining nitromethane mixture in until it is all gone, this will take about 30 minutes or more.

Ideally place a hose from the reflux condenser to a 10-30% HCL bubbler solution to stop escaping methylamine gas from stinking the lab out.

Stirring isn’t really needed, just occasionally swirl the flask by hand to help break up the aluminium sludge.

Let reflux for about an hour until the reaction cools and reflux stops.  Then add 200ml cooled 50% NaOH solution.  This will start another vigorous reflux for an hour or so.

NB:  The NaOH helps finish the reaction, breaks up the aluminium sludge and then forces the methylamine gas out on heating.
 
When reflux stops, change the glassware setup for distillation.  Place hose from vacuum outlet on the bent receiver to HCL bubbler to again stop lost methylamine gas from stinking your lab out.  Care should be taken to stop any suck backs.

Turn on the hotplate and distil the methanol with methylamine from the reaction flask until the stillhead thermometer reads around 80-90DegC.   This ensures all of the methylamine and methanol has been recovered and there should be around 1400ml in the receiving flask with a small amount of elemental mercury at the bottom.

Don’t heat until completely dry as the sludge will heat up and smoke.

Wash the hardened sludge residue from the distillation flask with warm water.

The distilled methanol should now contain enough methylamine to animate well over 1 mole Ketone.

Ideally titrate a small sample of methanol/methylamine solution to neutral with 10% HCL to work out exactly how much methylamine is recovered.

I normally find I get a 6% methylamine solution.  This means I have almost 3 moles methylamine recovered in the methanol, which is perfect for the next step.

Time to this stage is around 5 hours.

Stage 2:  Animating 1 Mole Ketone

Use your favorite aluminium foil or flashing.  Foil will react faster and probably won’t need external heat to kick-start the reaction.

Cut 30-40g aluminium into small squares (for 0.3mm thick flashing) and place in a 2L flask. 

Mix 0.1-0.2g HgCl with 100 mls clean methanol and then add this to the flask containing the aluminium to amalgamate.

Then pour in the recovered 1400mls 6% methylamine methanol solution from the previous step, add 178g ketone, and 30mls water and stir.

If using thicker aluminium use external heat to bring to reaction to reflux.  

Magnetic stir bar stirring works very well as there is a very low amount of aluminium and sludge to deal with and quite a bit of solvent.

Let reflux for 3 hours (using sheet aluminium) with gentle external heat when needed and magnetic stir bar stirring.

NB:  If using thin foil, the reaction should proceed much faster.

Extraction And Acid Base Wash

Remove the leftover aluminium by pouring the methanol/sludge through a sieve.

Then filter the methanol/sludge/product solution through a Buchner funnel.

Scrape out the filter cake into a container, basify with 1L 30% NaOH and add 1L Xylene or Toluene, add to separating funnel and extract the left over product into the non polar layer then remove the bottom aqueous layer when the layers separate.  Set this first non-polar extraction aside.

Distill most of the filtered methanol solution away until 2 layers form then add 200mls 10% NaOH, then add the first non-polar extraction (xylene or toluene) extraction into the flask to dissolve the rest of the MDMA and pour back into the separating funnel.

Separate the NaOH layer from the xylene or toluene in the separating funnel.

Wash this toluene or xylene with 100mls 10% NaOH once or twice and remove bottom wash layers.

Add 1L 5-10% HCL to the separating funnel and shake.  This extracts the MDMA into the HCL solution.  Separate and extract non polar once more with 100mls water.

Combine HCL extractions back into the cleaned separating funnel and slowly basify with 10% NaOH solution until MDMA falls to the bottom of the separating funnel and pH >11

Yield of raw MDMA base usually around 160g.

Extract MDMA with 200ml DCM washes and combine. 

Time to this stage is around 10-12 hours.

Then, strip of DCM, water, and vacuum distill the freebase to be left with very pure MDMA base.  

Yield of distilled freebase is around 135g, which is molar yield of 70%.

Crystallize using your method of choice.

I prefer to dissolve the distilled MDMA freebase in 4 times the volume of acetone and gas without any need for drying with magnesium sulphate.  Then the gassed and filtered MDMA.HCL is re-crystallized from an equal weight of boiling ethanol.

Abacus


Rhodium

  • Guest
And the rest?
« Reply #1 on: November 23, 2002, 07:19:00 AM »
Don't take this as critizism, but have you investigated where the other 30% of your ketone went?
It seems very strange to me that you have 160g of raw freebase (amine, as it went through an acid-base extraction), but only isolate 135g pure MDMA. Is there 25g of polymerized goo left in your distillation flask? How strong is your vacuum?

abacus

  • Guest
quite right
« Reply #2 on: November 23, 2002, 07:31:00 AM »
Only get about 80-85% yeild MDMA from ketone and who knows where rest of ketone goes.

Then when distilling freebase get around 83% of starting raw MDMA.  I recently found I was letting my oil bath temperature getting too high with a crap vacuum source seemingly "burning" the base when distilling and I hope to improve distilling yield in the future. 

I admit I need a better vacuum source as I use a water aspirator @ 20mmHG which is marginal for freebase distillation, comes over at 165DegC.

BTW over all yield from 1 kilo sassafrass is 475g MDMA.HCL after crap vacuum distillation AND recrystallising so I don't complain really.

EDit:  But I think the "kilos" i purchase may be a bit big coz I end up with 1050mls Isosafrole from them and after checking notes I got 550g MDMA.HCL from the last one after vacuum distillation of freebase, gassing in acetone AND recrystalling (twice) in ethanol.

Abacus

BlingBling

  • Guest
Very descriptive, thank you.
« Reply #3 on: November 23, 2002, 07:42:00 AM »
Very descriptive, thank you.  :)


>>Scrape out the filter cake into a container, basify with 1L 30% NaOH and add 1L Xylene
>>or Toluene, add to separating funnel and extract the left over product into the non polar
>>layer then remove the bottom aqueous layer when the layers separate.  Set this first
>>non-polar extraction aside.

I may be reading this wrong, but are you saying to basify the filter cake?? I may be reading this wrong..


Rhodium

  • Guest
Of course not, you're doing a great job!
« Reply #4 on: November 23, 2002, 07:47:00 AM »
Of course not, you're doing a great job! But I just become so curious when reading detailed writeups like this, I always feel a drive to optimize every reaction I come across, even if it takes me half a day to optimize something from 85 to 95% and the savings just amounts to 5-10 bucks, I still do it.  ;)

abacus

  • Guest
do what I said
« Reply #5 on: November 23, 2002, 07:58:00 AM »
Bling

Do what you read.  There is quite a bit of product trapped in the filter cake.  I found its up to 20-30g.

You can either boil the sludge up with more methanol to extract it OR do as I do which is basify the sludge and extract that 20-30g into non polar as I said.

Abacus

BlingBling

  • Guest
gotcha, basifty the cake..
« Reply #6 on: November 23, 2002, 08:07:00 AM »
This is different, and sounds like fun. Do you actually witness the mdma freebase oil drop out of the aq. layer in the step below?!

>>Combine HCL extractions back into the cleaned separating funnel and slowly basify with
>>10% NaOH solution until MDMA falls to the bottom of the separating funnel and pH >11

If it wasn't saturday night, I'd be half way to the bat cave right now. :p
This method of isolating the freebase oil is very interesting and sounds like a big time saver!  :)

Moriarty

  • Guest
Stoichiometry? I believe that's the correct term
« Reply #7 on: November 23, 2002, 09:13:00 AM »
I don't mean to "knit pick" but the reaction mechanism looks a little different, I believe.

The production of aluminum hydroxide (a biproduct):

                   Hg++
4Al + 12H2O -----> 4Al(OH)3 + 2H2 + 8H+ + 8e-*

*The reality is that a great deal of that "H+" will become H2(g).  I placed some of that in there.  It's just a guess.  I'm leaving out the production of H2(g) in future equations and won't try to account for electrons.  I don't know the status of the electronegativity in the reaction environment or how much H+ ion (simply free electrons and protons) is actually available.  I don't know.  I probably wouldn't understand ther math if there was some so I'm not going to mention it.

It is in this environment that reductions occur.  In organic chemistry a reduction is the addition of hydrogen or the removal of oxygen.  That doesn't really apply above but I want to focus on the reduction of three (3) reactants (all organic).  The ketone (3,4-Methylenedioxyphenyl-2-prpanone), nitromethane and the intermediary imine.

The reduction of the nitromethane and ketone occurs in the reaction environment created by the oxidation shown above, so we pick up there:

4Al(OH)3 + 6H+ + CH3NO2 -----> CH3NH 2 + 2H2O*

*I'm cutting out several imtermediary steps.  I was also not real careful to balence anything.  Sorry :P !

In much the same way:

2Al(OH)3 + 6H+ + CH2O2C6H5CH2COCH3 -----> CH2O2C6H5CH2CH(NCH3)CH3+ + H2O

Reduction of imine to amine:

2Al(OH)3 + CH2O2C6H5CH2CH(NCH3)CH3+ +3H+ -----> MDMA

Correct me if I'm wrong but the reaction mechanism, though ol' Moriarty took the shortcuts, looks more like that than it did in the first post.  I am only posting so that we may all better understand the process.  I probably made errors and hope some smarter bee will correct me but a deaper understanding is what I desire and I believe we all need.  So, was that shortcut description accurate?  Let a bee know if you are in the know and have the time.  Thanks 8) !

By the way, I don't believe Al2O3 is ever created in the amalgamation.  That offends my understanding of the reaction.

Who wants to play cops and dope fiends?

Protium

  • Guest
Balancing equations
« Reply #8 on: November 23, 2002, 10:09:00 AM »
You should really go back and balance those equations Moriarty.  I share your opinion that the Al2O3 is not formed, but I also see no evidence of the hydroxide sludge playing any significant role in the reaction, as you said, it is a byproduct.  The way your equations are laid out are rather confusing on this point.  Even if the co-efficients of the equations are not balanced out, you should at least reflect the same molecules on both sides of the equation, or it won't be too good for illustrating the reaction.

I believe we all had a rather tiresome debate on this issue a couple of weeks back, I won't bother to link the thread but you can pull it up.

Pr(+)tium

abacus

  • Guest
That debate was pointless IMHO
« Reply #9 on: November 23, 2002, 10:36:00 AM »
I studied that debate and many others.  I have spent almost 2 kilos ketone, and 3 months of experiments, and many hours trying to work out the water, oxide, hydroxide etc issues to finally end up with this procedure that works for me every single time.

Maybe the oxide isn't formed and all the water does is help the aluminium break down.  

I know for a fact that when no water is present the aluminium swells up to fill the flask and when the right molar amount is present for the hydroxide pathway, the hydroxide sludge fills the solvent stopping the reaction half way anyway giving crap yields of methylamine.

I believe the nitro "burns" the alumninium giving the oxide as the by product, then if heaps water is present this converts to the hydroxide (as some oxides do in the presence of water)

In any case, I don't care about the process and feel free to correct my chemistry, but I point out that this works for me every single time.

All I care about is the end result and this works for me.

Abacus

Moriarty

  • Guest
Okay, I might do that tomorrow
« Reply #10 on: November 23, 2002, 10:43:00 AM »
Tonight, it's late brother.  Moriarty's tired.  You very well might see me do just that (balence those equations) tomorrow in this post.  I know I laid it out in a fairly confused manner.  I kept picking up each reduction at Al(OH)3 becuase I wanted to write between the margins, so to speak.  What I was trying to accomplish was to bring the reader back to the original inorganic oxidation that makes the reductions occur (there are more than one).  This is redox.  An oxidation occurs and that creates a reduction.  So, if I had done a better job, I'd of included every reactant in the reaction and simply showed the progression, line-by-line.  This I did not do for two (2) reasons.  One, my chem is rusty and I didn't want to risk making embarrassing mistakes.  Two, that would have taken several lines.  If I showed everything that happens, I would have lost the ability to clearly show (detracted from the reactions I wanted to talk about) the reductions of each reactant as they occur because they might have been lost in the "mess" somewhere.  In truth, I was just lazy but legitimately didn't want to focus on anything but the reductions and their cause.  I know I confused it but just go back to the right side of the reaction between Al and H2O and that's where I was hoping you'd pick up.  In reality, most use methanol and that's a whole different deal.  Sorry if that was confusing.  Pick it up at the right of the amalgamation and that will better illustrate my ill-conceived point.  I'll fix it tomorrow.

By the way, I'm glad people (referring to the chemist who started this thread) help to refine the process.  Not a great deal of refined chemical equations are necessary to really explain that and I respect that for what it is.  You get results, and help others to get results by sharing your experiences, and that's cool.  I just want us to all understand it better.  I include myself in that, of course, as my post was more of a question than an answer.  Thanks for the reaction details.  Let's break this reaction down with methanol tomorrow.  Cool?

Oh, yeah, as for Al2O3 being created, absolutely not!  Maybe very small amounts but I think not!  Not that the old Brightstar write up is gospel but the chemistry behind it is solid and he notes that the Al2O3 that coats the Al foil is destroyed in the amalgamation and it is inferred that it must occur first for the mercuric chloride to coat the foil and foster proper amalgamation rates.  I don't want to get into that old thread discussion but there are a series of reasons why aluminum oxide is not created.  It's not.  Period.

P.S. Mods: Sorry if I lead this thread off point but I believe my post has intrinsic value.  I hope you agree.

Who wants to play cops and dope fiends?

terbium

  • Guest
Muddied water.
« Reply #11 on: November 24, 2002, 07:25:00 PM »
Correct me if I'm wrong but the reaction mechanism, though ol' Moriarty took the shortcuts, looks more like that than it did in the first post.
Yes, your description is confused to say the least.
 
I am only posting so that we may all better understand the process.
You certainly didn't achieve this.

I probably made errors and hope some smarter bee will correct me but a deaper understanding is what I desire and I believe we all need.
Then read Abacus' description in the first post in this thread or:

Post 348175

(terbium: "Aluminum reductant stoichiometry.", Chemistry Discourse)

Post 348185

(terbium: "Imine/Aluminum reduction stoichiometry.", Chemistry Discourse)


 

Baseline Does Not Exist.

terbium

  • Guest
More muddy water.
« Reply #12 on: November 24, 2002, 07:39:00 PM »
Tonight, it's late brother.  Moriarty's tired.
OK.

You very well might see me do just that (balence those equations) tomorrow in this post.
It has already been done, by Abacus in the first post in this thread and by me in an earlier thread.


I know I laid it out in a fairly confused manner.  I kept picking up each reduction at Al(OH)3 becuase I wanted to write between the margins, so to speak.  What I was trying to accomplish was to bring the reader back to the original inorganic oxidation that makes the reductions occur (there are more than one).  This is redox.  An oxidation occurs and that creates a reduction.
Yes, we know what "redox" means. Still, all you seem to be doing is confusing things with lots of words.

So, if I had done a better job, I'd of included every reactant in the reaction and simply showed the progression, line-by-line.
Abacus and I have already done this.
 
This I did not do for two (2) reasons.  One, my chem is rusty and I didn't want to risk making embarrassing mistakes.  Two, that would have taken several lines.  If I showed everything that happens, I would have lost the ability to clearly show (detracted from the reactions I wanted to talk about) the reductions of each reactant as they occur because they might have been lost in the "mess" somewhere.
Both Abacus and I have clearly shown all the details without any loss of clarity.

In truth, I was just lazy but legitimately didn't want to focus on anything but the reductions and their cause.  I know I confused it but just go back to the right side of the reaction between Al and H2O and that's where I was hoping you'd pick up.
Yes.

In reality, most use methanol and that's a whole different deal.  Sorry if that was confusing.  Pick it up at the right of the amalgamation and that will better illustrate my ill-conceived point.
Yes, you are confused again.

Let's break this reaction down with methanol tomorrow.  Cool?
Abacus and I have already done that.

Oh, yeah, as for Al2O3 being created, absolutely not!  Maybe very small amounts but I think not!  Not that the old Brightstar write up is gospel but the chemistry behind it is solid and he notes that the Al2O3 that coats the Al foil is destroyed in the amalgamation and it is inferred that it must occur first for the mercuric chloride to coat the foil and foster proper amalgamation rates.  I don't want to get into that old thread discussion but there are a series of reasons why aluminum oxide is not created.  It's not.  Period.
Yes, you shouldn't get into this discussion because you are confused again.




Baseline Does Not Exist.

Osmium

  • Guest
Finally someone is doing this MeNO2 prereduction.
« Reply #13 on: November 25, 2002, 11:42:00 AM »
Finally someone is doing this MeNO2 prereduction.
You can reuse the post-reduction solvent too, after distillation. I always kept and reused the rotovapped EtOH solvent which I used, it contained a lot of MenH2 according to its smell, even though it was distilled under vacuum.

Thank for the writeup abacus. I'm glad people are finally realising how well this reaction works. Since the MeOH/MeNH2 mixture is distilled you can use RC fuel bought OTC without cleaning it up! A great simplification for many of the bees!

Only thing I don't quite like yet is the relatively small amount of water you add to the amination reaction, and the relatively big solvent volume of ~1400ml for 1 mole of ketone. 30-40g Al is roughly 1-1.5 moles, so you should have 3 times (1-1.5) moles of water present. I used 94% azeotropical EtOH as the solvent, and added the MeNH2 as a 40% solution in water, so there was quite a bit more water present in my case and the results were still excellent yield-wise.
Also, losses of freebase from distillation of an A/B cleaned raw product will usually be in the lower percentage range, 3% or so being typical when you use proper vacuum and glassware.

I can also recommend the following crystallization procedure: dissolve freebase in IPA, gas with HCl, and then crash out the product with excess acetone. Store at -20°C for some time and filter. Save those mother liquors, still some goodies in them!

This does not mean I think your writeup sucks! Not at all! I like it alot!

I'm not fat just horizontally disproportionate.

Osmium

  • Guest
Almost forgot, try this: Add the ketone dropwise ...
« Reply #14 on: November 25, 2002, 11:48:00 AM »
Almost forgot, try this:
Add the ketone dropwise over several hours to the Al/Hg/MeNH2 amination. This way every ketone molecule will have a huge excess of MeNH2 available to react with, my results using this 'starved feed' ketone addition method were pretty good.

I'm not fat just horizontally disproportionate.

UKBEE

  • Guest
you guys
« Reply #15 on: November 25, 2002, 07:59:00 PM »
LOOK into getting NaBH4 instead of HG/Al + methylamine.

Swims smurf friend got excellent yields with distilled ketone...


100g of ketone == over 99g of mdma.hcl.. 

I love the smell of Ketone in the morning.

Chromic

  • Guest
Yes, but...
« Reply #16 on: November 25, 2002, 08:45:00 PM »
An optimized Al/Hg will give the same yields. I've gotten back 1g mdma.hcl for each g of ketone. Although I agree, if you can get NaBH4 easily... go for it. It's much less messy.

terbium

  • Guest
Methylamine from nitromethane.
« Reply #17 on: November 25, 2002, 10:40:00 PM »
Although I agree, if you can get NaBH4 easily... go for it. It's much less messy.
Could borohydride also be used to reduce nitromethane to methylamine? If not then the Al/Hg reduction still would be a good way to obtain the methanolic methylamine.

Baseline Does Not Exist.

BlingBling

  • Guest
More FUn!
« Reply #18 on: November 25, 2002, 11:34:00 PM »
So say you make methylamine via al/hg and end up with 1000ml's of a 6% MeAm/meoh solution. You would just use that basic methylamine laden methanol in the NaBH4 reduction??

How much ketone and boro would be needed to make use of the 1000ml's of MeAm solvent?

Osmium

  • Guest
> Could borohydride also be used to reduce ...
« Reply #19 on: November 26, 2002, 12:09:00 AM »
> Could borohydride also be used to reduce nitromethane to
> methylamine?

No. For that reaction to take place you need different reducing agents, like BH3 under anhydrous conditions, presence of hydrogenation catalysts etc. And even if it worked, it wouldn't be economical to reduce nitromethane with NaBH4, since such a big excess of methylamine is required in that reaction.

I'm not fat just horizontally disproportionate.