Felab
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The cyanide mystery
Today, I was trying to make some potassium cyanide. I did it by mixing potassium hydroxide, urea and wood filings as my reductant. I mixed all
thoroughly and I aplied heat with a blowtorch. Everithing was going well: ammonia was escaping from the reaction, the mixture was glowing a dull red
and so on. I washed the ashes with water and I tested the cyanide content by the addition of ferrous sulphate to some of the solution. It gave
positive so I added some alcohol to the rest of the solution, hoping that the cyanide would crush out. Then something realy weird happened. Nothing
solid precipitated but instead two layers of liquid separated out. I have no idea of what is the lower layer (which reacts with iron sulphate) made
of. If you have any theory of what could it be made of, please, tell me.
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fusso
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Why wood ash instead of other carbon source like sugar?
What colours are the 2 layers? Are they milky/cloudy?
What observations did you make when the lower layer react with FeSO4? Colour change? Anything precipitate out?
KCN is highly soluble in H2O so maybe not enough EtOH added or too much H2O is used
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AJKOER
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One of prior comments on SM (see http://www.sciencemadness.org/talk/viewthread.php?tid=23 ):
Quote: Originally posted by AJKOER  | OK, I an across a preparation path using available, mostly households, starting materials. To quote from Atomistry.com on HCN (link: http://carbon.atomistry.com/hydrocyanic_acid.html ):
"Bergmann has shown that when ammonia is passed over carbon heated to about 1300° C., 90 per cent, of it is converted into HCN. The reaction is
endothermic, its heat being -39,500 calories. A modification of this reaction is that of Roeder and Grunwald, who pass a mixture of ammonia, and
nitrous oxide over heated carbon, the reaction being:
2NH3 + N2O + 4C = 4HCN + H2O - 58,000 calories.
Owing to the heat of decomposition of nitrous oxide, which is endothermic, and the heat of formation of steam, it is not necessary to heat the carbon
to so high a temperature as in the former case; indeed the yield of hydrogen cyanide is nearly quantitative when the temperature of the carbon is but
450° C. "
........
[Edited on 4-10-2014 by AJKOER] |
--------------------------------------------------------------------
A bit harder, try this path (see https://chemiday.com/en/reaction/3-1-0-261):
CO + NH3 --500 C, Al2O3--> HCN + H2O
"Carbon monoxide react with ammonia to produce hydrogen cyanide and water. The technical method production hydrogen cyanide. This reaction takes place
at a temperature of 500-800°C, an overpressure. In this reaction, the catalyst is can be V2O5, CeO2 Al2O3, ThO2."
Text sounds like a translation.
At such high temperatures, I would expect:
NH3 + Heat ---> .H + .NH2
based on the action of hv on ammonia (see R1 at https://journals.ametsoc.org/doi/pdf/10.1175/1520-0469%28197... )
Then, subsequent reactions with CO forming HCN and H2O.
------------------------------------------------------------------
The above speculated radical mechanism is interesting, if correct, as the hydrogen atom radical can be formed at RT by the action of NaOH (or HCl) on
Aluminum metal where some .H radical could be imbued on the surface of the Aluminum. My prior related comment:
| Quote: Originally posted by AJKOER |
.......
Next, imbue the surface of Mg or Al with the hydrogen atom radical (from the traditional nascent hydrogen generation methods based on say Al/NaOH).
One may assume that the •H radical functional behaves (per its seemingly reversible formation reaction: e- + H+ = •H ) as apparently a (e-,H+)
pair acting on ions. For an example from 'Hydrometallurgy 2008: Proceedings of the Sixth International Symposium', p. 818, a commercial reductive
leaching equation, to quote:
" PbS + 2 •H = Pb + H2S (5) " (see https://books.google.com/books?id=1etfSdk55SYC&pg=PA818&... )
which I view functionally as follows:
Pb(+2)S(2-) + 2 (e-, H+) = Pb + H2S (g)
......
[Edited on 4-10-2018 by AJKOER] |
So the surface .H could be further enlisted in other (unspecified) reactions with CO per above via the radical reaction:
.H + NH3 = H2 + .NH2
which are the same radicals (.H, .NH2) postulated above to be formed at high temperatures that eventually result in a HCN product.
[Edited on 9-11-2018 by AJKOER]
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unionised
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| Quote: Originally posted by AJKOER | One of prior comments on SM (see http://www.sciencemadness.org/talk/viewthread.php?tid=23 ):
| Quote: Originally posted by AJKOER | OK, I an across a preparation path using available, mostly households, starting materials. To quote from Atomistry.com on HCN (link: http://carbon.atomistry.com/hydrocyanic_acid.html ):
"Bergmann has shown that when ammonia is passed over carbon heated to about 1300° C., 90 per cent, of it is converted into HCN. The reaction is
endothermic, its heat being -39,500 calories. A modification of this reaction is that of Roeder and Grunwald, who pass a mixture of ammonia, and
nitrous oxide over heated carbon, the reaction being:
2NH3 + N2O + 4C = 4HCN + H2O - 58,000 calories.
Owing to the heat of decomposition of nitrous oxide, which is endothermic, and the heat of formation of steam, it is not necessary to heat the carbon
to so high a temperature as in the former case; indeed the yield of hydrogen cyanide is nearly quantitative when the temperature of the carbon is but
450° C. "
........
[Edited on 4-10-2014 by AJKOER] |
--------------------------------------------------------------------
A bit harder, try this path (see https://chemiday.com/en/reaction/3-1-0-261):
CO + NH3 --500 C, Al2O3--> HCN + H2O
"Carbon monoxide react with ammonia to produce hydrogen cyanide and water. The technical method production hydrogen cyanide. This reaction takes place
at a temperature of 500-800°C, an overpressure. In this reaction, the catalyst is can be V2O5, CeO2 Al2O3, ThO2."
Text sounds like a translation.
At such high temperatures, I would expect:
NH3 + Heat ---> .H + .NH2
based on the action of hv on ammonia (see R1 at https://journals.ametsoc.org/doi/pdf/10.1175/1520-0469%28197... )
Then, subsequent reactions with CO forming HCN and H2O.
------------------------------------------------------------------
The above speculated radical mechanism is interesting, if correct, as the hydrogen atom radical can be formed at RT by the action of NaOH (or HCl) on
Aluminum metal where some .H radical could be imbued on the surface of the Aluminum. My prior related comment:
| Quote: Originally posted by AJKOER |
.......
Next, imbue the surface of Mg or Al with the hydrogen atom radical (from the traditional nascent hydrogen generation methods based on say Al/NaOH).
One may assume that the •H radical functional behaves (per its seemingly reversible formation reaction: e- + H+ = •H ) as apparently a (e-,H+)
pair acting on ions. For an example from 'Hydrometallurgy 2008: Proceedings of the Sixth International Symposium', p. 818, a commercial reductive
leaching equation, to quote:
" PbS + 2 •H = Pb + H2S (5) " (see https://books.google.com/books?id=1etfSdk55SYC&pg=PA818&... )
which I view functionally as follows:
Pb(+2)S(2-) + 2 (e-, H+) = Pb + H2S (g)
......
[Edited on 4-10-2018 by AJKOER] |
So the surface .H could be further enlisted in other (unspecified) reactions with CO per above via the radical reaction:
.H + NH3 = H2 + .NH2
which are the same radicals (.H, .NH2) postulated above to be formed at high temperatures that eventually result in a HCN product.
[Edited on 9-11-2018 by AJKOER] |
An alternative observation, based in actual chemistry and which actually answers the question.
Strong solutions of potassium carbonate will not mix with ethanol.
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AJKOER
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See this prior SM thread http://www.sciencemadness.org/talk/viewthread.php?tid=9128 where there is reference to a patent. To quote:
"I have made several attempts to make KOCN from urea and KOH based on the guidance of US patent 3935300 (Feb 20, 1974), and encouragement from Hilski.
My urea is PTS brand fertilizer. KOH is technical."
I would argue that any formed cyanate, OCN-, can be converted to (CN)2 or CN- by the hydrogen atom radical:
OCN- + .H = OH- + .CN or .OH + CN-
.CN + .CN = (CN)2
(CN)2 + H2O --> H2C2O4 + NH3 (which could produce an insoluble oxalate precipitate)
.H + .CN = HCN
.CN + H2O = .OH + HCN
Prior thread comment (see http://www.sciencemadness.org/talk/viewthread.php?tid=87030 ) especially details on Clancy method:
| Quote: Originally posted by AJKOER | Here is a good historical account of methods, mostly at elevated temperatures, see https://www.911metallurgist.com/blog/how-to-make-cyanide .
-------------------------------
Did come up with a 'radical' idea based on the possible action of the hydrogen atom radical on cyanate at low pH absence oxygen. Likely paths include:
.H + OCN- = OH- + .CN
Or: .H + OCN- = .OH + CN- (most probable being reverse of Eq (12) in the Haynes reference below)
Or: .H + OCN- = .OHCN- (which could interestingly be akin to the chlorine .OHCl- radical anion)
If .CN is created, possible further reactions:
.CN + .CN --> (CN)2
(CN)2 + H2O --> H2C2O4 + NH3 (which could produce an insoluble oxalate precipitate)
.H + .CN = HCN
.CN + H2O = .OH + HCN
......
Some background on the cyano radical see https://en.wikipedia.org/wiki/Cyano_radical and https://www.sciencedirect.com/science/article/pii/0010218077... Also, see https://aip.scitation.org/doi/10.1063/1.463945 for other suggested (unexpected?) reactions.
-------------------------------------
Sources of .H include:
Fe --> Fe(ll) + 2 e-
Fe(ll) --> Fe(lll) + e-
H+ + e- = .H
.OH + H2 = H2O + .H
and also, heating H2 to high temperatures, and photolysis of pure hydrogen,....or ELECTROLYSIS of water in the presence of an oxygen scavenger.
----------------------------------------
Performing a search on electrolysis involving cyanates finds an old mining method to recover ore claiming 'Cyanide Regeneration' employing an oxygen
scavenger (KI is recommended). See Engineering and Mining Journal, Volume 90, Ocober 8, 1910, discussing the 'The New Clancy Cyanide Patents', at
https://books.google.com/books?id=wIogAQAAMAAJ&pg=PA701&... .
The author of the process notes a major well known obstacle in working with cyanates, to quote:
"when cyanates are allowed to stand for any considerable time, or are retained in solution, they are converted into ammonia and potassium carbonate,
and are transposed by the lime used in the cyanide process into insoluble carbonates or by the action of sulphuric acid in the ore into sulphates—so
that the original cyanide finds itself eventually in the residue dumps in the form of sulphates or insoluble carbonates."
The Clancy method further involves also the addition of an amide compound to quote:
"To show the action of these amide compounds, take, for example, urea (carbamide) and add this to a solution of a cyanate (such, for example, as
potassium cyanate) and electrolyze this mixture between inert electrodes. After a few minutes the said mixture will be transformed into one which will
dissolve the precious metal in alkaline solution. "
The role of amide in recent work ("Bifurcate localization modes of excess electron in aqueous Ca(2+)amide solution revealed by ab initio molecular
dynamics simulation: towards hydrated electron versus hydrated amide anion." by Zhang R, Bu Y., see https://www.ncbi.nlm.nih.gov/pubmed/27351489 ) suggests a promotion of solvated electron activity. See also https://www.sciencedirect.com/science/article/pii/0146572484... .
Bottom line, one may find this old Clancy method as a path to convert CNO- to CN-, which I discovered based on my 'radical' idea.
Related chemistry, see https://www.911metallurgist.com/chemistry-cyanogen-compounds... .
[Edited on 11-8-2018 by AJKOER] |
Bottom line, replace wood filings as the reductant with any convenient path to the hydrogen atom radical. Note, removing dissolved oxygen is needed
as:
.H + O2 = .HO2 = H+ + .O2- (pKa = 4.88)
resulting in the consumption of the required hydrogen atom radical.
[Edited on 10-11-2018 by AJKOER]
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clearly_not_atara
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Fwiw cyanides are easily produced via alkali decomposition of formaldoxime. The method has the particular advantage that there is almost no chance of
releasing any HCN.
This decomposition occurs also for other aldoximes but the nitriles tend to be further hydrolysed to acids. CN- resists hydrolysis due to its
negative charge.
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Felab
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| Quote: |
Why wood ash instead of other carbon source like sugar? What colours are the 2 layers? Are they milky/cloudy? What observations did you make when the
lower layer react with FeSO4? Colour change? Anything precipitate out? KCN is highly soluble in H2O so maybe not enough EtOH added or too much H2O is
used
|
I used wood filings because it was what I had on hand.
The layer I think contains cyanide is a bit milky.
It turned blue when FeSO4 was added. And yes, it precipitated prussian blue.
Yes, but if I had used to little EtOH then the cyanide would have dissolved into it instead of generating two layers.
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AJKOER
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| Quote: Originally posted by clearly_not_atara | Fwiw cyanides are easily produced via alkali decomposition of formaldoxime. The method has the particular advantage that there is almost no chance of
releasing any HCN.
This decomposition occurs also for other aldoximes but the nitriles tend to be further hydrolysed to acids. CN- resists hydrolysis due to its
negative charge. |
I did notice this article (see 'Excision of CN− and OCN− from acetamide and some amide derivatives triggered by low energy electrons' by Constanze
Koenig-Lehmann, et al, abstract at https://pubs.rsc.org/en/Content/ArticleLanding/2008/CP/b8121... ). To quote in part:
"Low energy electron attachment to acetamide and some of its derivatives shows unique features in that the unimolecular reactions of the transient
anions are remarkably complex, involving multiple bond cleavages and the formation of new molecules. Each of the three compounds acetamide
(CH3C(O)NH2), glycolamide (CH2OHC(O)NH2) and cyanoacetamide (CH2CNC(O)NH2) shows a pronounced resonance located near 2 eV and decomposing into CN−
along a concerted reaction forming a neutral H2O molecule and the corresponding radical (methyl and methoxy). "
Per the above, I would express the basis of a possible reaction as:
Al --> Al(lll) + 3 e-
3 x [ CH3C(O)NH2 + e- --> .CH3 + CN- + H2O ]
Net: Al + 3 CH3C(O)NH2 + 3 e- --> 3 .CH3 + Al(lll) + 3 CN- + 3 H2O
Some gas formation could be observed:
.CH3 + .CH3 = C2H6
,,,,,
[Edited on 10-11-2018 by AJKOER]
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Dan Vizine
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| Quote: Originally posted by Felab | | Today, I was trying to make some potassium cyanide. I did it by mixing potassium hydroxide, urea and wood filings as my reductant. I mixed all
thoroughly and I aplied heat with a blowtorch. Everithing was going well: ammonia was escaping from the reaction, the mixture was glowing a dull red
and so on. I washed the ashes with water and I tested the cyanide content by the addition of ferrous sulphate to some of the solution. It gave
positive so I added some alcohol to the rest of the solution, hoping that the cyanide would crush out. Then something realy weird happened. Nothing
solid precipitated but instead two layers of liquid separated out. I have no idea of what is the lower layer (which reacts with iron sulphate) made
of. If you have any theory of what could it be made of, please, tell me. |
Good luck. I've tried this general scheme with poor results in every case. I can smell the cyanide but...
Every case where this is discussed it's always with the same general result, ideas, theories, guesses. Can anyone recall one of these that ended with
" And here's my product, a fair yield of what is believed to be reasonably pure KCN"? I can't.
There are much easier ways.
"All Your Children Are Poor Unfortunate Victims of Lies You Believe, a Plague Upon Your Ignorance that Keeps the Youth from the Truth They
Deserve"...F. Zappa
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