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Tsjerk
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As you need the nitrite in catalytic amounts you could reduce a bit of nitrate with lead which is very doable in small amounts.
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AJKOER
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Actually, rather than focusing on concentrating Nitric acid, instead try adding N2O (or, a nitrite, see http://cires1.colorado.edu/jimenez/AtmChem/CHEM-5151_S05_L7.... ) to even dilute HNO3 and treat with UV/strong sunlight in the presence of the
substance to be nitrated. The targeted photolysis product is the nitrate radical, which while short lived, I suspect may be a stronger nitrating agent
that your super strength HNO3! More precisely, NO3• can add to double bonds or engage in hydrogen abstraction (see page 4 at https://www3.nd.edu/~ndrlrcdc/Compilations/Ino/Ino.pdf ).
The underlying mechanics:
N2O + H2O + hv ---> N2 + OH- + •OH
•OH + HNO3 ⇆ H2O + NO3•
Rate constant for the forward reaction estimated as (8.6 ± 1.3) × E07 L mol-1 s-1
Reference: see "Equilibrium constant of the reaction .OH+HNO3←→H2O+NO3. in aqueous solution" by Poskrebyshev, G. A.; Neta, P.; Huie, R. E.,
published in Journal of Geophysical Research: Atmospheres, Volume 106, Issue D5, pp. 4995-5004, 03/2001. Link: http://adsabs.harvard.edu/abs/2001JGR...106.4995P
With respect to my possible reactivity claims, some specifics, see, for example, "Reactivity trends in reactions of the nitrate radical (NO3) with
inorganic and organic cloud water constituents" by Hartmut Herrmann et al. To quote from the abstract:
"The rate coefficients of the reactions of NO3 with acetone (reaction 2), tetrahydrofuran (reaction 3), and hydrated formaldehyde (reaction 4) were
determined at T = 298 K to be k2 = (4.4 ± 0.6)· 103 L mol−1 s−1, k3 = (2.1 ± 0.1 )· 107 L mol−1 s−1, and k4 = (7.8 ± 1.2)· 105 L
mol−1s−1, respectively. For the reactions of NO3 with the alcohols methanol (reaction 5), ethanol (reaction 6 ), 1-propanol (reaction 7),
2-propanol (reaction 8), and 2-methyl-2-propanol (reaction 9 ) the following rate constants at T = 298 K were obtained: k5 = (5.1 ± 0.3)·105 L
mol−1 s−1, k6 = (2.4 ±0.5)·106 L mol−1 s−1, k7 = (3.2 ± 0.1)· 106 L mol−1 s−1, k8 = (3.7 ± 0.9)·106 L mol−1 s−1, and k9 = (6.6
± 0.7)·104 L mol−1 s−1, respectively."
Link: http://www.sciencedirect.com/science/article/pii/00167037949...
[Edited on 2-3-2017 by AJKOER]
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LD5050
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Quote: Originally posted by Tsjerk  | | As you need the nitrite in catalytic amounts you could reduce a bit of nitrate with lead which is very doable in small amounts.
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Tsjerk is there a write up or some more info on this that I can read up on? I would be interested in this if I have the materials.
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Tsjerk
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AJOEKER, as lovely theoretic your post always are, I wouldn't put your advice in practice. I wouldn't add N2O to any reaction that doesn't need N2O,
especially in this case as it is a oxidation reaction and not a nitration. It would just give more benzoic acid.
LD5050, I'm on my phone and don't want to search, but have a look on this board, it is discussed in detail.
[Edited on 2-3-2017 by Tsjerk]
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AJKOER
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My suggestion was either N2O or nitrite (more light sensitive per my cited reference than dinitrogen oxide, and nitrates also, but much less photo
active) under photolysis employing dilute HNO3. Interestingly, note the prior comment below by Amos:
| Quote: Originally posted by Amos | | Tsjerk is correct; if you even read my original benzaldehyde thread you'd have seen all the papers that achieved even better results than mine using
dilute nitric acid and a small amount of nitrite. |
Could it be that fluorescent lab lights were photo-catalysts for benzaldeyde in the presence of nitrite, or a trace transition metal impurity
producing hydroxyl radicals in a Fenton-like fashion, or even the action of (or in combination with) air/O2 and light on any present transition metal
ions?
Related research with organics, see http://www.publish.csiro.au/en/pdf/EN10004 .
[Edit] The general effect of radiation, light or micro wave irradiation or electric current on N2O I would described as:
N2O + e- ----> N2 + O-
which can be oxidizing, but in the presence of water:
O- + H2O = OH- + •OH (See reaction [14] at https://pdfs.semanticscholar.org/d696/b35956e38351dd2eae6706... )
where the equilibrium is far to the right except under highly alkaline conditions (or no water).
Also, my suggested trial use of N2O is due primarily to its low price and wide availability of 8 gram cartridges sold in major stores for the purpose
of beverage infusion, which is also cool, but perhaps not healthy for sun bathers/drinkers as I once speculated on SM. Note, no credit card generally
required, disclosure of mailing address, possible watch list, etc., which may be associated with various internet (or sometimes store) purchases of
various chemicals (including NaNO2) by home chemists.
[Edited on 2-3-2017 by AJKOER]
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