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Author: Subject: Preparing "Neutral" Copper(II) Nitrate Solution for Aldehyde Synthesis
nimgoldman
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[*] posted on 5-11-2019 at 16:38
Preparing "Neutral" Copper(II) Nitrate Solution for Aldehyde Synthesis


I want to prepare a copper(II) nitrate solution for the synthesis of an aldehyde, e.g. benzaldehyde from benzyl chloride:



I was thinking about first simply dissolving copper in a small amount of conc. nitric acid, then diluting the solution with water.

The solution made in this way will however contain excess nitric acid, which might over-oxidize the aldehyde product to carboxylic acid (??).

Here are some walk-arounds I am thinking of:
  1. Isolate copper(II) nitrate first (lengthy procedure).
  2. Add copper metal to a solution of silver(II) nitrate (requires recycling the silver compound free of acid, same problem).
  3. Neutralize the excess acid with a base (might precipitate the copper as carbonate or hydroxide).
  4. Bubble air through the solution (neutralize the excess acid w. carbon dioxide).
  5. Boil off the excess nitric acid before dilution.


Maybe the little excess nitric acid might not be an issue in the first place, but I would like to be sure.
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DraconicAcid
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[*] posted on 5-11-2019 at 16:41


If you add excess copper(II) carbonate (basic or otherwise) to your nitric acid, and filter out the excess, you should have a bare minumum of excess nitric acid.



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AJKOER
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[*] posted on 5-11-2019 at 17:16


Try adding CuSO4 to aqueous KNO3 and freeze out the potassium sulfate hydrate!

Note, KNO3 is widely and cheaply available. If you cannot find any CuSO4, see http://www.sciencemadness.org/talk/viewthread.php?tid=151055... for a home prep path (requires NaHSO4).


[Edited on 6-11-2019 by AJKOER]
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elementcollector1
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[*] posted on 5-11-2019 at 18:01


Crystallize the nitrate out at low temperature and redissolve it? Or would some acid become trapped in the crystal structure?



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nimgoldman
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[*] posted on 5-11-2019 at 18:43


Quote: Originally posted by elementcollector1  
Crystallize the nitrate out at low temperature and redissolve it? Or would some acid become trapped in the crystal structure?


Yes this is valid. I pulled out the procedure from Armarego et. al. (Purification of Laboratory Chemicals) and they also describe recrystallization from conc. aqueous soln. (0.5 mL/g):

Quote:
Cupric nitrate trihydrate [10031-43-3 (3H2O), 3251-23-8 (anhydrous)] M 241.6, m 114o, b 170o(dec), d 4 20
2.0. Crystallise it from weak aqueous HNO3 (0.5ml/g) by cooling from room temperature. The anhydrous salt
can be prepared by dissolving copper metal in a 1:1 mixture of liquid NO2 and ethyl acetate and purified by
sublimation [Evans et al. J Chem Soc, Faraday Trans 1 75 1023 1979]. The hexahydrate dehydrates to the
trihydrate at 26o, and the anhydrous salt sublimes between 150 and 225o, but melts at 255-256o and is
deliquescent.


I would then let the crystals air-dry at mild temperature of 26-50 °C, crush and re-dry to get just the trihydrate (for stoichiometry), then quickly move the crystalline powder to a vacuum desiccator for final drying.

But I like the approach of DraconicAcid most - dissolution of basic copper carbonate in nitric acid - this way there is no hassle with elemental metals, no nitrogen dioxide fumes.

If copper carbonate is not available, then perhaps the approach of AJKOER. I have CuSO4 at hand, as well as NaNO3. I can get KNO3 as well.
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Sulaiman
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[*] posted on 5-11-2019 at 19:41


copper(II) nitrate decomposes at 256oC so you could just boil to dryness (anhydrous if required) ?



CAUTION : Hobby Chemist, not Professional or even Amateur
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nimgoldman
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[*] posted on 5-11-2019 at 22:01


Quote: Originally posted by Sulaiman  
copper(II) nitrate decomposes at 256oC so you could just boil to dryness (anhydrous if required) ?


Yes, but I am afraid some sublimation may happen and perhaps even some low level decomposition (unless very evened-out heating is provided - e.g. lab oven).

Since I need just the aqueous solution (free of nitric acid), I think the suggestions provided so far will work okay.
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G-Coupled
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[*] posted on 5-11-2019 at 22:04


What might Copper Nitrate decompose into at such temps, I wonder?

[Edited on 6-11-2019 by G-Coupled]
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[*] posted on 5-11-2019 at 22:49


Quote: Originally posted by DraconicAcid  
If you add excess copper(II) carbonate (basic or otherwise) to your nitric acid, and filter out the excess, you should have a bare minumum of excess nitric acid.

I second this, this is what I have done in the past.




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[*] posted on 6-11-2019 at 04:39


I would not boil in air/oxygen contact an acidic transition metal salt, in general, due to possible basic salt formation!

In the case of Cu(NO3)2, there exists a basic nitrate salt per Atomistry (see http://copper.atomistry.com/cupric_nitrate.html ). To quote:

"The heat of formation of the anhydrous salt from its elements is 71.49 Cal.; that in solution is 81.96 Cal. Thomsen's value for the heat of formation of the hexahydrate from the anhydrous salt and liquid water is 21-18 Cal., and from its elements and water 92.94 Cal. It yields a green, basic salt, Cu(NO3)2,3Cu(OH)2. "

Per another source (see https://www.cambridge.org/core/journals/proceedings-of-the-r... ) to quote:

"The only basic nitrate of copper appears to be Cu(NO3)2. 3Cu(OH)2. The product obtained by heating the trihydrate to 100° has this composition, and is not Cu(NO3)2. "

[Edited on 6-11-2019 by AJKOER]
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[*] posted on 6-11-2019 at 08:45


Nimgoldman, do you have a published procedure with copper-nitrate what you try to follow? Could you post this procedure? I'm curious about this reaction.
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AJKOER
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[*] posted on 6-11-2019 at 08:57


Quote: Originally posted by AJKOER  

I would not boil in air/oxygen contact an acidic transition metal salt, in general, due to possible basic salt formation!
...


where the net reaction can be express as:

4 Cu(l)/Fe(ll)/Co(ll)... (aq) + O2 + 2 H+ --> 4 Cu(ll)/Fe(lll)/Co(lll) + 2 OH- (see http://www.sciencemadness.org/talk/viewthread.php?tid=81800#... )

Prior comment outlying the chemistry:

Quote: Originally posted by AJKOER  

............
First, the action of water on an aqua cupric complex:

[Cu(H2O)6]2+ (aq) + H2O (l) = [Cu(H2O)5(OH)]+ (aq) + H3O+ (aq)

Second, a redox couple equilibrium reaction leading to a presence of cuprous:

Cu(ll) + Fe(ll) = Cu(l) + Fe(lll)

which is acted on by oxygen bubbles in near boiling water consuming H+ based on the net reaction derived from a 2013 radical reaction supplement, "Impacts of aerosols on the chemistry of atmospheric trace gases: a case study of peroxides radicals"', by H. Liang1, Z. M. Chen1, D. Huang1, Y. Zhao1 and Z. Y. Li, link: https://www.google.com/url?sa=t&source=web&rct=j&... :

R24 O2(aq) + Cu+ → Cu2+ + O2− ( k = 4.6xE05 )
R27 O2− + Cu+ + 2 H+ → Cu2+ + H2O2 ( k = 9.4xE09 )
R25 H2O2 + Cu+ → Cu2+ + .OH + OH− ( k= 7.0 xE03 )
R23 .OH + Cu+ → Cu2+ + OH− ( k = 3.0×E09 )

Net reaction: O2 + 4 Cu+ + 2 H+ → 4 Cu2+ + 2 OH-

Electrolysis reference: See p. 7 at https://www.utc.edu/faculty/tom-rybolt/pdfs/electrochem2014.... for the reverse reaction with 2 H+ adding to each side. Alternate source of the above reaction, per my records, but access to the full article is no longer free, see: https://www.researchgate.net/publication/262451840_Review_of... .

[Edited on 14-4-2018 by AJKOER]


For those interested in the possible underlying mechanics of the reaction in acid conditions (pH <4.88, as superoxide, •O2-, exists at pH only above 4.88), I suggest:

4 x [ Cu(l) —> Cu(ll) + e- ]
2 x [ H+ + e- = •H ]
•H + O2 —> •HO2
•HO2 + e- —> HO2-
H2O = H+ + OH-
H+ + HO2- = H2O2
e- + H2O2 —> OH- + •OH
•H + •OH —> H2O

Net reaction in acidic conditions:

4 Cu(l) (aq) + O2 + 2 H+ —> 4 Cu(ll) (aq) + 2 OH-

-----------------------------------------------------

Some sources suggest the process continues with Cu(ll) going to Cu(llll):

4 Cu(ll) (aq) + O2 + 2 H+ —> 4 Cu(lll) (aq) + 2 OH- (see https://books.google.com/books?id=6TImAgAAQBAJ&pg=PA105&... )


[Edited on 7-11-2019 by AJKOER]
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nimgoldman
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[*] posted on 6-11-2019 at 11:54


Quote: Originally posted by Pumukli  
Nimgoldman, do you have a published procedure with copper-nitrate what you try to follow? Could you post this procedure? I'm curious about this reaction.


Yes the procedure is taken from here:

https://www.prepchem.com/synthesis-of-benzaldehyde/

Source: "Cohen, Julius Berend. A class-book of organic chemistry. Vol. 1. Macmillan and Co., limited, 1920."

Alternatively, you can also use Sommelet reaction to convert the benzyl halide to aldehyde.
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[*] posted on 6-11-2019 at 17:27


Quote: Originally posted by AJKOER  
Quote: Originally posted by AJKOER  

I would not boil in air/oxygen contact an acidic transition metal salt, in general, due to possible basic salt formation!
...


where the net reaction can be express as:

4 Cu(l)/Fe(ll)/Co(ll)... (aq) + O2 + 2 H+ --> 4 Cu(ll)/Fe(lll)/Co(lll) + 2 OH- (see http://www.sciencemadness.org/talk/viewthread.php?tid=81800#... )


No, it's Cu(2+) + 2 H2O --> 2 H(+) + Cu(OH)2, with the copper(II) hydroxide combining with copper(II) nitrate/chloride/sulphate/whatever to give a basic salt. I know you think that everything is a free radical redox reaction, but this isn't one.




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[*] posted on 7-11-2019 at 02:53


Quote: Originally posted by Pumukli  
Nimgoldman, do you have a published procedure with copper-nitrate what you try to follow? Could you post this procedure? I'm curious about this reaction.


I was curious about the reaction also. Apparently the Cu(II) is reduced to Cu(I) chloride and the nitrate to nitrite. I guess the reaction is driven by the insolubility of Cu(i) chloride.

Apparently copper(ii) nitrate can be used for a variety of oxidation reactions see:

Attachment: copper-nitrate-oxidiser-gao2018.pdf (1.1MB)
This file has been downloaded 411 times





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[*] posted on 12-11-2019 at 14:39


Quote: Originally posted by DraconicAcid  
......

No, it's Cu(2+) + 2 H2O --> 2 H(+) + Cu(OH)2, with the copper(II) hydroxide combining with copper(II) nitrate/chloride/sulphate/whatever to give a basic salt. I know you think that everything is a free radical redox reaction, but this isn't one.


Actually, Cu3+ is a suggested non-radical based active agent occurring in lieu of the hydroxyl radical. To cite a source "Degradation of contaminants by Cu+-activated molecular oxygen in aqueous solutions: Evidence for cupryl species (Cu3+)" at https://www.ncbi.nlm.nih.gov/pubmed/28246040, to quote:

"Copper ions (Cu2+ and Cu+) have shown potential as Fenton-like activators for the circumneutral removal of organic contaminants from aqueous solutions. However, the major active species (cupryl species (Cu3+) versus hydroxyl radical (OH)) produced during the activation of hydrogen peroxide by Cu+ remain unclear. In this study, Cu+-O2 oxidation, in which hydrogen peroxide is produced via the activated decomposition of dissolved molecular oxygen, was used to degrade sulfadiazine, methylene blue, and benzoic acid. The results showed that both sulfadiazine and methylene blue could be efficiently degraded by Cu+-O2 oxidation in a wide effective pH range from 2.0 to 10.0. Quenching experiments with different alcohols and the effect of Br- suggested that Cu3+ rather than OH was the major active species"

I speculatively agree with your suggested products involving Cu2+ and OH- that may be sourced via Cu3+ formation.

[Edited on 12-11-2019 by AJKOER]
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