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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 •CH3 + Al(lll) + 3 CN- + 3 H2O
......
[Edited on 10-11-2018 by AJKOER][/rquote]
where the corresponding net reaction in the current case could be:
Net: Al + 3 CH2CONH2 --?--> 3 •CH2 + Al(lll) + 3 CN- + 3 H2O
indicating a possible, but small, cyanide presence.
In the current experiment, the action of the hydroxyl radical on elemental carbon is a path to CO:
•OH + C = •H + CO (EDIT see comments and references at http://www.sciencemadness.org/talk/viewthread.php?tid=97845#... )
Also, given the possible presence of CO, from the same referenced SM thread above:
[Edited on 4-10-2014 by AJKOER][/rquote]
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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.
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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........
.......
Some possible paths to a cyanide presence. Assuming the presence of CO, I would suggest also the simple path:
CO + •NH2 = •CONH2 (or •CN + H2O)
•H2N + CO = •H2NCO (or •NC + H2O)
•H2NCO + •CONH2 = H2NC(O)C(O)NH2 (oxamide)
(Or: •NC + •CN = NCCN )
Per Wiki on Cyanogen (https://en.wikipedia.org/wiki/Cyanogen ) to quote:
“Cyanogen is the anhydride of oxamide:
H2NC(O)C(O)NH2 → NCCN + 2 H2O
…….
Like other cyanides, cyanogen is very toxic, as it readily undergoes reduction to cyanide, …”
Given the theory presented above, the observed pale blue coloration developed in the synthesis following the introduction of pure aqueous NH3 in
distilled water, may be due to HCN or a harmless bluish tinge created from the addition of aqueous ammonia to an aluminum salt (see https://mysite.du.edu/~jcalvert/phys/alumin.htm and search on 'bluish').
In any event, perform the experiment in a well vented area.
I believe the reduction in color upon boiling with added vinegar, suggests a successful preparation with a minor threat of HCN.
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[Edited on 4-10-2014 by AJKOER] |