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underground
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Birch reduction without ammonia
I want to know if there is any way to substitute the liquid ammonia in Birch reduction with something more easily available. After searching a bit i
found a reduction similar to Birch reduction called Benkeser reduction. Wikipedia says that is uses low molecular weight alkyl amines as the solvent
and as a source of protons.
Also will any way of the above reduction be more powerful than the other ? And in general what are their main differences
[Edited on 8-12-2017 by underground]
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clearly_not_atara
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A common modification uses diethyl ether with sodium hydroxide and ammonium nitrate. I believe the patent is titled "Novel dissolving-metal reduction
schema".
It's also possible to use ethylamine bp 21 C in an ice bath, only problem is that ethylamine is unpurchaseable (listed) and not easy to make.
Decarboxylation of beta-alanine is the OTC route but ketones do not catalyze the rxn and it's hard to collect the product due to its low bp.
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arkoma
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That is the basis for the "shake and bake" reduction of psuedoephedrine. I believe Coleman fuel can be used along with bone dry
NH4NO3 and NaOH
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underground
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Quote: Originally posted by clearly_not_atara  | A common modification uses diethyl ether with sodium hydroxide and ammonium nitrate. I believe the patent is titled "Novel dissolving-metal reduction
schema".
It's also possible to use ethylamine bp 21 C in an ice bath, only problem is that ethylamine is unpurchaseable (listed) and not easy to make.
Decarboxylation of beta-alanine is the OTC route but ketones do not catalyze the rxn and it's hard to collect the product due to its low bp.
|
| Quote: Originally posted by arkoma | | That is the basis for the "shake and bake" reduction of psuedoephedrine. I believe Coleman fuel can be used along with bone dry
NH4NO3 and NaOH |
With this way ammonia is produced (NH4NO3 + NaOH → NaNO3 + NH3 + H2O), so this is actually a Birch reduction. The pressure for the "shake and bake"
method is needed to liquify ammonia from ammonia gas that is produced from the above reaction. I dont think so diethyl ether is neccesary for that
reaction. It is just a solution of the disolving materials that need to be reduced .
[Edited on 8-12-2017 by underground]
[Edited on 8-12-2017 by underground]
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clearly_not_atara
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Diethyl ether allows the reaction to occur above the boiling point of ammonia by solvating it; this avoids using liquid ammonia, as you asked. I think
hydrocarbons might also work as arkoma mentioned.
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Melgar
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@not_atara: Does the patent say anything about a two-liter soda bottle? 
IIRC, ethylenediamine works reasonably well here, although it's prone to forming heterocycles and other side reactions.
Also, if you have a bunch of reactive shit at the bottom of the reaction vessel, you're better off using lithium than sodium. Because lithium will
float.
The first step in the process of learning something is admitting that you don't know it already.
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arkoma
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*cough* hydrocarbon solvents work. "Starting fluid" is mostly hexane.
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Melgar
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The fact that ether is a hydrogen bond acceptor certainly works in its favor, as far as dissolving ammonia.
The first step in the process of learning something is admitting that you don't know it already.
I'm givin' the spam shields max power at full warp, but they just dinna have the power! We're gonna have to evacuate to new forum software!
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underground
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| Quote: Originally posted by Melgar |
The fact that ether is a hydrogen bond acceptor certainly works in its favor, as far as dissolving ammonia. |
So if the trick is by disolving ammonia gas into a non-polar solvent, why then pressure is needed ? Also some other eco-friendly solvent (like
Limonene) will work ? Unfortunately i cant find any ammonia solubility info for non-polar solvents
[Edited on 9-12-2017 by underground]
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Nicodem
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Thread Moved
9-12-2017 at 08:34 |
Chemi Pharma
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here's a study about Birch reduction conditions with Lithium and ethylenediamine instead of liquid ammonia, as cited by @Melgar.
They also say that sodium can be used with ethylenediamine instead of lithium, but that's a powerless reduction system and can form polymeric
byproducts with naphthalene, for example.
Attachment: lithium-ethylenediamine reducing system.pdf (491kB)
This file has been downloaded 627 times
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underground
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After searching a bit around the forum i saw that it is not necessary to use liquid ammonia, instead gas ammonia will do the trick. I have seen many
people here in this forum successfully made the "lithium bronze" by just bubbling ammonia into a solvent (usually Et2O) containing lithium metal. The
"Lithium Bronze" looks to be lithium tetraamine (Li(NH3)4). Now some calculations.
Li + 4NH3 --> Li(NH3)4 +1e
Lets assume a molecule Y that weights 200g/mol needs +2e and 1 mol of NH3 to be reduced. So for each mole is needed 2 moles of Li and 9 moles of NH3
Li = 7 g/mol
Nh3 = 17g/mol
Y = 200g/mol
So for each 200g of Y molecule we need 14g of Li metal and I53g of NH3
Gas law says that 1 mole of any gas will occupy 22.4 liters
of volume at standard temp and pressure
Here are my thoughts.
A balloon can easily took 3Liters of air so 3liters of NH3 in gas stage is 2,28g
So a 3L ballon filled with NH3 (2,28g) can react with 0,20g of Li and recude 1,20g of Y
So what now if we put the 0,20g of Li, 1,2g of Y some Et2O and fill it with NH3 ? In theory it ill successfully reduce the the Y material right ?
These ay looks to be quite easy for amateaur chemists rather that dealing with liquit NH3 and Ethylenediamine/Diethylamine
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AJKOER
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| Quote: Originally posted by clearly_not_atara | A common modification uses diethyl ether with sodium hydroxide and ammonium nitrate. I believe the patent is titled "Novel dissolving-metal reduction
schema".
It's also possible to use ethylamine bp 21 C in an ice bath, only problem is that ethylamine is unpurchaseable (listed) and not easy to make.
Decarboxylation of beta-alanine is the OTC route but ketones do not catalyze the rxn and it's hard to collect the product due to its low bp.
|
Ethylamine may perhaps be made, among other products, following the path described in this paper ("ADDITION OF ETHYL RADICALS TO ETHYLENE", by J.A.
Pinder and D.J. Le Roy, available at http://www.nrcresearchpress.com/doi/pdf/10.1139/v57-085 ) to create the ethyl radical, .C2H5, or perhaps trying a one pot variation avoiding the
pumping of gases,...,where dry NH3 is also added to the system.
Suggested reaction path:
Hg(3P1) + H2 = Hg(1S0) + 2 .H
.H + C2H4 = .C2H5
.H + NH3 = .NH2 + H2 (Source: see http://aip.scitation.org/doi/abs/10.1063/1.3605242?journalCo... )
.C2H5 + .NH2 = C2H5NH2
Some other reactions:
.C2H5 + .C2H5 = C4H10
.C2H5 + .C2H5 = C2H4 + C2H6
.NH2 + .H = NH3
.....
Not easy to execute but reagents are available.
------------------------------------------------------
The above is a dated reference (1957) and a more recent work (1964) mentions a path to .C2H5 from the photolysis/heating of C2H6 in N2O claiming to
produce the ethyl radical (see http://www.nrcresearchpress.com/doi/pdf/10.1139/v64-401 ), to quote:
"or as the initiation process in the oxidation of molecules which readily decompose into free radicals, such as ethane (2).....
Castellion and Noyes (3) studied the photochemical reaction between nitrous oxide and ethane at temperatures up to 105 degrees. They concluded that
ethyl radicals formed in the system do not undergo reaction with nitrous oxide at these temperatures. "
The process may not be as efficient but does avoid the use of mercury vapors. Not sure if if the cited temperature is 105 F or C. My take looking at a
2004 work "Mechanism of the Photolysis of Ethane at 1470 Å" (see http://aip.scitation.org/doi/abs/10.1063/1.1725255?journalCo... ) and higher temperature decomposition research as well (see http://www.nrcresearchpress.com/doi/pdf/10.1139/v66-358):
Primary (85%):
C2H6 --Heat 105,light--> .H + .C2H5
.C2H5 --> .H + C2H4
.......
Secondary (15%):
C2H6 --Heat 105,light--> .CH3 + .CH3
2 .CH3 + 2 C2H6 --> 2 CH4 + 2 .C2H5
.C2H5 --> C2H4 + .H
.....
Also, for both paths:
.R + NH3 --> RH + .NH2
.H + NH3 = .NH2 + H2 (Source: see http://aip.scitation.org/doi/abs/10.1063/1.3605242?journalCo... )
.H + .H --> H2
.C2H5 + .NH2 = C2H5NH2
and certainly other products.
[Edited on 11-12-2017 by AJKOER]
[Edited on 12-12-2017 by AJKOER]
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unionised
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| Quote: Originally posted by AJKOER | | Quote: Originally posted by clearly_not_atara | A common modification uses diethyl ether with sodium hydroxide and ammonium nitrate. I believe the patent is titled "Novel dissolving-metal reduction
schema".
It's also possible to use ethylamine bp 21 C in an ice bath, only problem is that ethylamine is unpurchaseable (listed) and not easy to make.
Decarboxylation of beta-alanine is the OTC route but ketones do not catalyze the rxn and it's hard to collect the product due to its low bp.
|
Ethylamine may perhaps be made, among other products, following the path described in this paper ("ADDITION OF ETHYL RADICALS TO ETHYLENE", by J.A.
Pinder and D.J. Le Roy, available at http://www.nrcresearchpress.com/doi/pdf/10.1139/v57-085 ) to create the ethyl radical, .C2H5, or perhaps trying a one pot variation avoiding the
pumping of gases,...,where dry NH3 is also added to the system.
Suggested reaction path:
Hg(3P1) + H2 = Hg(1S0) + 2 .H
.H + C2H4 = .C2H5
.H + NH3 = .NH2 + H2 (Source: see http://aip.scitation.org/doi/abs/10.1063/1.3605242?journalCo... )
.C2H5 + .NH2 = C2H5NH2
Some other reactions:
.C2H5 + .C2H5 = C4H10
.C2H5 + .C2H5 = C2H4 + C2H6
.NH2 + .H = NH3
.....
Not easy to execute but reagents are available.
[Edited on 11-12-2017 by AJKOER] |
So, you think that an impractical method for making ethylamine from ammonia helps solve the problem of doing a Birch reduction without ammonia.
While you ponder that, perhaps you could look at the practical rate of production of ammonia from your scheme- Lets say you have a kilowatt of
photons at the right frequency.
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clearly_not_atara
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You've been posting on sciencemadness for fourteen years, and yet you still think it's worthwhile to argue with AJKOER about radical reactions? :p
Thermal decarboxylation of beta alanine is OTC and should be quite easy, although you'll need to pyrolyse a lot of it probably. But turning a stream
of hot ethylamine gas into a flask of liquid ethylamine is harder than it sounds.
http://pubs.rsc.org/-/content/articlelanding/1951/qr/qr95105...
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unionised
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| Quote: Originally posted by clearly_not_atara | You've been posting on sciencemadness for fourteen years, and yet you still think it's worthwhile to argue with AJKOER about radical reactions? :p
|
I wouldn't want anyone to think I was a quitter.
(And it seems he only started in 2011)
[Edited on 11-12-17 by unionised]
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underground
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Hey guys what do you think about my above post ? If these will work it will be very simple and easy. To make things even more simple i believe we can
use sodium magnesium oxide aggregate instead of lithium.  I dont think so magnesium oxide will do any damage to the reaction or hurt
somewhere.
[Edited on 11-12-2017 by underground]
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AJKOER
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Unionised:
Look again, I now have two cited paths to the ethyl radical  .
Note, the first paper focus is not on the extent of the ethyl radical formation, per se, but the results of successive radical collisions creating
certain products. For example, the odds of two .C2H5 species meeting up is low, and is different from a .C2H5 meeting, say .NH2, which could be
sourced from an excess of ammonia reacting with .R or .H radical.
I have previously cited amine formation via photolysis with extra ordinary results on SM, on a thread that you, apparently, did not participated. Here
is an extract:
"Related source (absent the benefical effect of N2O to create more hydroxyl radicals) see K. Ogura, "Photolysis of CH4, NH3, H2O mixture: formation of
methylamine and ethylenediamine", in Journal of Photochemistry and Photobiology A: Chemistry, Volume 49, Issues 1–2, September 1989, Pages 53-61,
https://doi.org/10.1016/1010-6030(89)87105-9, link: http://www.sciencedirect.com/science/article/pii/10106030898... . Here is the abstract:
"The photolysis of the mixture CH4, NH3, H2O was performed with a low pressure mercury lamp at 100 °C and atmospheric pressure. Of the products, the
major nitrogen-containing compounds were methylamine and ethylenediamine; the maximum selectivity of the two amines exceeded 99% in the mol per cent
of N-containing products. The other products were methanol, ethane and hydrogen, but the formation of methanol and ethane decreased rapidly with an
increase in the amount of added ammonia. This was attributed to the preferential reactivity of methyl radicals with NH2 rather than with OH and/or
CH3. The formation rates of NH2, CH3NH2 and NH2C2H4NH2 are discussed."
[Edited on 11-12-2017 by AJKOER]
[Edited on 12-12-2017 by AJKOER]
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arkoma
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| Quote: Originally posted by unionised | | Quote: Originally posted by clearly_not_atara | You've been posting on sciencemadness for fourteen years, and yet you still think it's worthwhile to argue with AJKOER about radical reactions? :p
|
I wouldn't want anyone to think I was a quitter.
(And it seems he only started in 2011)
[Edited on 11-12-17 by unionised] |
Oh my that was a good laugh after a hard day!
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Corrosive Joeseph
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@ underground - You might like this attachment........
"Reductions with Lithium in Low Molecular Weight Amines and Ethylenediamine"
J. Org. Chem. 2000, 65, 7098-7104
There is also information on the forum about making isopropylamine from glyphosate weedkiller
/CJ
Attachment: Reductions with Lithium in Low Molecular Weight Amines and Ethylenediamine.pdf (122kB)
This file has been downloaded 620 times
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underground
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Ok thank you Corrosive Joeseph for your reply, i will study it
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unionised
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| Quote: Originally posted by AJKOER | Unionised:
Look again, I now have two cited paths to the ethyl radical .
[Edited on 12-12-2017 by AJKOER] |
Yes; but nobody asked for even one path.
If someone asks you for a recipe for stew with dumplings would you give a reply involving photolysis and some sort of reactive oxygen free radical?
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clearly_not_atara
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| Quote: Originally posted by unionised | | Quote: Originally posted by AJKOER | Unionised:
Look again, I now have two cited paths to the ethyl radical .
[Edited on 12-12-2017 by AJKOER] |
Yes; but nobody asked for even one path.
If someone asks you for a recipe for stew with dumplings would you give a reply involving photolysis and some sort of reactive oxygen free radical?
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All cooking does start with photosynthesis, after all.
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Corrosive Joeseph
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You are most welcome.
If you really want to study, you will probably also like this attachment -
"Reduction of C=X to CHXH by Dissolving Metals and Related Methods" - J.W. Huffman
Repost, but it is just tooo good.
Mandatory reading on your reaction includes -
"Reductions in Organic Chemistry" - Milos Hudlicky
http://cnqzu.com/library/Anarchy%20Folder/Chemistry/Crystali...
and
"Preparative Organic Chemistry" - Wilgand and Hilgetag
http://library.sciencemadness.org/library/books/Weygand_and_...
See also Melgar's attachment and my post in this thread.
"Reduction with alkali metal and alcohol" - https://www.sciencemadness.org/whisper/viewthread.php?tid=75...
Over and out......
/CJ
Attachment: Reduction of C=X to CHXH by Dissolving Metals and Related Methods - J.W. Huffman.pdf (2.9MB)
This file has been downloaded 520 times
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underground
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After searching a bit more it looks like Birch reduction also needs some source of hydrogen usually taken from alcohols. Will just Et2O provide the
hydrogen needed or must be something like ethanol/methanol. Ammonia itself cannot provide that free hydrogen ? Also what if i bubble ammonia into
anhydrous isopropyl alcohol ? Will this isopropyl alcohol/ammonia solution be effective for a birch reduction ?
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macckone
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underground, the hydrogen can be donated by the alcohol being reduced since birch is generally used to reduce an alcohol. Residual water can also be
a donor, but too much will kill the reaction by converting the lithium to hydroxide without reduction of the target compound.
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