Hi. :)
It's time for PrimoPyro's questionable synthesis of the week, assembled from materials found at Rhodium's website. This time it's a rather simple discussion, and I hope this to bee a viable option. Thank you for reading this, and thank you for any data you share. :)
The basic synthesis involves benzyl chloride, or phenyl methyl chloride, forming a grignard intermediate, and reaction with acetonitrile to form a phenyl-2-imino-propane intermediate with magnesium halide. This is either decomposed by hydrolysis to P2P, or reduced with NaBH4 to amphetamine in one pot.
This synthesis deals with the reduction to amphetamine. The mother document is:
https://www.thevespiary.org/rhodium/Rhodium/chemistry/amphetamine.html (https://www.thevespiary.org/rhodium/Rhodium/chemistry/amphetamine.html)
By switching the functional groups, as suggested by rhodium in that document, you can then use benzyl chloride and acetonitrile, instead of phenyl-acetonitrile and methyl iodide.
So this one pot synthesis is established. No questions here. Wonderful. Let's move on.
I'd like you to look at:
https://www.thevespiary.org/rhodium/Rhodium/chemistry/amphetamine.methylation.html (https://www.thevespiary.org/rhodium/Rhodium/chemistry/amphetamine.methylation.html)
and at the first synthesis, using formaldehyde + reducing agent. Example 3, with the LiAlH4 is the subject here.
I want to know if it is feasable, to combine this step with the one pot amphetamine synthesis. I wonder a few things specifically:
1.Could the addition of formaldehyde step, be done in the same pot as the previous amphetamine formation?
2.In the reduction of the N-formyl-amphetamine, is NaBH4 a suitable reducing agent? If so, one could use additional NaBH4 from the amphetamine forming reaction to reduce the N-formyl-amphetamine. I dont mean to add excess in the first reaction, I mean merely to be able to use the same reducing agent for both reactions in a single pot.
3.If NaBH4 cannot bee used for the N-formyl-amphetamine reduction, then my next question is: Can LiAlH4 bee used as the reducing agent for the reduction of the magnesium halide derivative to amphetamine, instead of NaBH4, in the first reaction?
I dont want to use two different reducing agents, Id much rather use the same one for each step, and most desireably, in a cleaned up one pot synthesis.
In any answer, if it doesnt work for the one pot, please act as if the question were re-asked, just applying to using it in a seperate reaction.
Thank you again for reading this. :)
PrimoPyro
Vivent Longtemps la Ruche!
I see that no one can really answer my question yet? 46 views, two are my own, so 44 people have viewed the thread; no replies. Ok. I will continue to wait.
Meanwhile, I'd like to add that I found this little gem:
https://www.thevespiary.org/rhodium/Rhodium/chemistry/grignardimine.html (https://www.thevespiary.org/rhodium/Rhodium/chemistry/grignardimine.html)
I remember this concept was brought up once before, by tomjuan, but this is somewhat different, as he wanted to take it back too many steps too far, and thus lost efficiency, and the idea was canned.
Ah, this paragraph was written over a half hour later than the last, after I utfse for the reactions I wanted to talk about. Wow, I know a lot more about this now, it's kinda cool. Most of my questions have been answered. :)
There seems to be some debate on the order to add the reagents together: benzyl magnesium chloride to acetald(methyl)imine, or the reverse? Flat out, which is the correct way?
Basically the desired synthesis for SWIPP would bee thus:
To take BzCl and form the grignard in THF with pure Mg and perhaps a small iodine crystal if needed. Nothing worth talking about here.
Form the imine between methylamine and acetaldehyde, in THF solvent, with in situ drying via anhydrous zeolite. This is chilled quickly and immediately the addition is begun between the BzMgCl and MeAm=Et, in whichever order is correct.
Also, as a seperate idea, an addition if you will:
I was wondering if this would solve all the issues of getting methylamine freebase in THF, and of the water product of imine formation: THF is dried first of course, then magnesium alkoxide is solvated in the THF, followed by addition of methylamine sulfate.
The magnesium alkoxide reacts with the methylamine sulfate to form methylamine freebase, magnesium sulfate (a drying agent 8) ) and an alcohol, instead of water. All are formed in completely anhydrous conditions, as the THF is dry, and no water is formed. Then upon imine formation with acetaldehyde, the water is sucked up by the anhydrous MgSO4.
The discussion is not about the difficulties of preparing Mg alkoxides.....yet. This thread is for their USE. Thank you.
So people, whatcha think?
PrimoPyro
Vivent Longtemps la Ruche!
Your ideas are often so big and general, that it is hard to give a short, easy answer, like those I use to give. It would also help to add specific questions 1, 2, 3 etc to be answered.
In Post 18602 (missing)
(Rhodium: "Re: a revolution in amphetamine synthesis ??", Chemistry Discourse), Post 20327 (missing)
(Scooby_Doo: "Re: P2Pol from acetaldehyde", Chemistry Discourse), Post 103438 (not existing) and Post 202649 (https://www.thevespiary.org/talk/index.php?topic=11489.msg20264900#msg20264900)
(Rhodium: "Meth via grignard rxn (Gazz Chim Italiana)", Novel Discourse) you will find discussions on the BnMgCl + acetaldehyde methylimine reaction.
For your idea about using Mg(OR)2 as drying agent for a Grignard is not good - the formed alcohol would react with the grignard reagent, forming a hydrocarbon (PhCH2MgCl gives toluene for example).
To go back to your questions in the first post:
Yes, the reaction can probably be tweaked somehow to be a one-pot reaction mixture, but you would lose a lot of yield/efficiency, compared to the 1-2 acid/base extractions you skip inbetween. The NaBH4 reduction of the magnesium imine proceeds in methanol, and thus a LiAlH4 reduction cannot be performed afterwards without thorougly evaporating the reaction mixture first, and then you might as well do an acid/base extraction at that point.
In an environment, where you have isolated the N-formylamine, LiAlH4 is usually used to reduce the amide. However, I have never seen LiAlH4 being used for a reductive amination like the preceeding step. If you do not isolate the formylamine (from amine + HCOOH) but instead make the imine (from amine + HCHO) you could in an anhydrous environment reduce it to the secondary amine, but you might get problems getting just the right number of alkylations (you want an N-methyl, with 1 eqv HCHO you might get 50/50 of dimethylamphetamine and amphetamine, even if you would expect 100% N-methylamphetamine). NaBH4 does not reduce amides (like N-formylamine) alone without additives like transition metal salts (Ni/Co/Cu) or lewis acids.
It is hard to say before it has been tried, but in your position, I would rather do it all stepwise first in a triel, and then later try to skip on the isolation steps, and see if the yield goes through the floor, or if the desired product is obtained.
There was an error on my https://www.thevespiary.org/rhodium/Rhodium/chemistry/amphetamine.methylation.html (https://www.thevespiary.org/rhodium/Rhodium/chemistry/amphetamine.methylation.html)
page which is now corrected. I wrote that the formylamine was formed upon treatment with formaldehyde, while the correct answer is the amphetamine formylimine. I have corrected the pictures and the text now.
I would advise against trying to generate methylamine from its salts in situ in this already sensitive reaction. Make chilled etheral equimolar solutions of acetaldehyde and methylamine, and mix them cooled to 0°C over silica gel or molecular sieves with swirling. This should form an unstable solution of acetaldehyde methylimine, which must be kept cold and used immediately.
i did read, after a google search, that ethanol can be acidified with potassium permaganate to yield acetaldehyde,
Hmm.. That doesn't sound quite right. Where did you read that? KMnO4 will oxidize ethanol mostly to acetic acid... You need a weaker oxidizing agent like PCC, or otherwise special conditions to stop at the aldehyde point. KMnO4 will indeed oxidize ethanol to acetaldehyde, but unless you can remove the aldehyde as it is formed or otherwise stop the oxidizing agent from working, it will easily oxidize the aldehyde again to the carboxylic acid.
For all the info you ever wanted about permanganate oxidations, see for example this excellent review:
Fatiadi, Alexander J. "The Classical Permanganate Ion: Still a Novel Oxidant in Organic Chemistry" Synthesis 1987; 85-127 DOI:
10.1055/s-1987-27859 (http://dx.doi.org/10.1055/s%2D1987%2D27859)
... And since I'm bored right now, how about some more info about the industrial preparation of acetaldehyde...
From Ullman's Encyclopedia of Industrial Chemistry, 6th ed:
Dehydrogenation of Ethanol. In the first work on ethanol dehydrogenation, published in 1886, ethanol was passed through glass tubes at 260 °C.
CH3CH2OH (l) ® CH3CHO (l) + H2 (g) DH = + 82.5 kJ/mol
Improved yields are obtained in the presence of catalysts such as platinum, copper, or oxides of zinc, nickel, or cobalt. In later patents, zinc and chromium catalysts [34], oxides of rare earth metals [35], and mixtures of copper and chromium oxides [36] have been reported. The lowest amounts of decomposition products are obtained using copper catalysts. Frequent regeneration of the catalysts is required, however.
Process Description. Ethanol vapor is passed at 260 – 290 °C over a catalyst consisting of copper sponge or copper activated with chromium oxide in a tubular reactor [37]. A conversion of 25 – 50 % per run is obtained. By washing with alcohol and water, acetaldehyde and ethanol are separated from the exhaust gas, which is mainly hydrogen. Pure acetaldehyde is obtained by distillation; the ethanol is separated from water and higher-boiling products by distillation and flows back to the reactor. The final acetaldehyde yield is ca. 90 %. Byproducts include butyric acid, crotonaldehyde, and ethyl acetate.
Oxidation of Ethanol. Oxidation of ethanol is the oldest and the best laboratory method for preparing acetaldehyde. In the commercial process, ethanol is oxidized catalytically with oxygen (or air) in the vapor phase.
CH3CH2OH (g) + 1/2 O2 (g) ® CH3CHO(l) + H2O (l) DH = - 242.0 kJ/mol
Copper, silver, and their oxides or alloys are the most frequently used catalysts [38].
For an example of a simultaneous oxidation – dehydrogenation process, see [39].
Veba-Chemie Process (Fig. (1)). Ethanol is mixed with air and passed over a silver catalyst at 500 – 650 °C (c). The temperature depends on the ratio of alcohol to air and the flow rate of the gas through the catalyst. Alcohol conversion varies between 50 and 70 % and the yield is between 97 and 99 % depending on the reaction conditions. Acetaldehyde and unreacted alcohol are removed from the waste gas by washing with cold alcohol (e) and separated by fractional distillation (h); after concentration the alcohol returns to the reactor. Heat formed in the reaction is utilized for steam production using a waste-heat recovery system immediately after the reaction zone.
The waste gas consists mainly of nitrogen, hydrogen, methane, carbon monoxide and carbon dioxide; it is burned as lean gas with low calorific value in steam generators. Small amounts of acetic acid are obtained as a byproduct.
[34] SU 287 919, 1970.
[35] Heavy Minerals Co., US 2 884 460, 1955 (V.
I.Komarevsky).
[36] Knapsack-Griesheim, DE 1 097 969, 1954 (W. Opitz, W.
Urbanski); DE 1 108 200, 1955 (W. Opitz, W. Urbanski).
[37] W. L. Faith, D. B. Keyes, R. L. Clarks: Industrial
Chemicals, 3rd ed., J. Wiley & Sons, New York 1965, p. 2.
[38] Shell Development Co., US 2 883 426, 1957 (W.
Brackman).Eastman Kodak Co., US 3 106 581, 1963 (S.
D.Neely).Veba-Chemie, DE 1 913 311, 1969 (W. Ester, W.
Hoitmann).
[39] Petroleum Refiner 36 (1957) 249.
-SpicyBrown