brew - 15-5-2008 at 19:48
I am learning chemistry and have begun by making various salts. I decided to place very thin copper wire in conc sulphuric acid and noticed that the
reaction was extremely slow. I read somewhere that adding H2O2 would kick things along and I slowly added some and the solution quickly turned blue
and small amount of gas was bubbling up. I am assuming that Hydrogen cation has disassociated from the H2O2 and has been reduced to H2 hence the
copper is itself being oxidised and able to form the copper sulphate. Is my understanding correct. Also whilst I am asking. I want to understand the
mechanism of the oxidation of toluene to benzaldehyde with the reagants; sulphuric acid, Mn02, copper sulphate pentahydrate and of course toluene. I
have searched and have found these reagants for this reaction, along with numerous alternative precedures etc, but have struggled to understand fully
the mechanism. Any suggestions where I may understand this reaction would be appreciated etc. So far I feel that the Mn02 is going to supply the
Oxygen but to do this I am not exactly sure how it is done. I also am confused regarding the copper sulphate. Is the sulphate anion deprotanating the
Hydrogen from the methyl componant of the toluene. I can find information regarding the oxidation of an alcohol but not from a methyl group on the
toluene. Any suggestions would be appreciated
bmc
chemrox - 15-5-2008 at 21:37
The solution you're working with is called "pirhana" among the inorganic crowd and is used for cleaning metals. It's dangerous stuff; be really
careful about disposal. DO NOT allow this stuff to come in contact with alcohols or ketones! The mechanism as I understand (and will probably be
corrected) is this: (unbalanced) H3O+ + H2O2 --> H2O + O+. It is one of the few ways to release O+ and it is a hellishly strong oxidizer. The
blue color is Cu++. The gas is probably O2 or O3 and maybe the two mixed. Witht the oxidation of toluene, it may be that a benzilic hydrogen is
abstracted ... an H* is pulled off and a methyl radical is activated which then pulls in an oxygen radical. Or it may be that a proton is pulled off
a carbocation is formed and that pulls in an oxygen ion. Is here a paper you read? Could you post it? Have you read through the thread on this
topic? It's quite extensive and guys more knowledgeable than I have contributed.
ShadowWarrior4444 - 15-5-2008 at 22:19
I would also like to note that in an earlier post of mine I mentioned that transition metals will catalyze the formation of Caro's Acid from H2O2 and
H2SO4. Or peroxydisulfuric acid--haven’t got around to checking which one it is yet. This may require fairly concentrated H2O2 though, 3% may not
work.
(It should look like a fluffy white precipitate if it does occur.) Though, now that I think about it... it may have been a bit of anhydrous Iron
Sulfate, though it formed with copper as well. This reminds me that I need to do a few more tests.
[Edited on 5-16-2008 by ShadowWarrior4444]
woelen - 15-5-2008 at 23:19
| Quote: |
| blue color is Cu++. |
Yes, but if there were blue color, then the solution must have contained a lot of water.
In concentrated sulphuric acid, copper(II) ions are colorless and a white solid is formed, anhydrous CuSO4. So, this most likely was not piranha's
solution, but a weak diluted version of it.
brew - 15-5-2008 at 23:26
...hence if the mechanism you have written is correct it is this O+ that is itself reduced by coppers valance electrons be it forming O3 in the
process, and if this is the case then the copper ion is able to free itself from its neighboring atoms and be in solution. This makes scence as
oxygens electronegativity in its free state is high let alone its attration for electrons when it is a O+ As far as disposal I am working on very
small scale and will look at ways of removing/reversing its reactivity perhaps. I will be careful. I have no interest in doing this again as I can
easily obtain copper sulphate at a good price.
And Just so I understand, and not relying on purely my understanding, why should I not let this stuff come in contact with alcohols and ketones? I
will accept your warning but want to know why for sure. This is in no way me disputing your warning etc.
As far as the oxidation of toluene, I dare say it is complicated with the reagants involved. It is probably something later on I will learn about -
yet when I dont understand something I get compelled to understand it, but I dare say I need to get the basics down pat before I delve to far into
organic- I will also search this forum again.
12AX7 - 15-5-2008 at 23:40
Organics, peroxides and acid get together badly. Think runaway oxidation. The one-two punch of peroxide's oxidizing capacity combined with
concentrated sulfuric acid's dehydrating power make for some quick, nonspecific combustion of organics, especially when warm (and combustion makes
heat, think about that).
Small amounts of H2O2 are fine and not likely to run away. The bubbles you are seeing are probably O2 from the spontaneous (possibly catalyzed)
decomposition of peroxide.
Tim
brew - 15-5-2008 at 23:48
The solution of sulfuric acid I added to the thin copper wire was initialy 98% - when little to no reaction was observed no gas evolved initially and
only a slight greying of color at 24 hours I then remembered that the addition of H2O2 would "kick things along abit"-cant recall where i got this
from in this forum. I added slowly and cautiously 6% hydrogen peroxide 5-8ml only hence the water and the blue color.
I might get some more copper and rinse it with solvent and soak it in NaOH for a little while and do so to perhaps make sure that the copper wasn't
coated with something that prevented the H3O from attacking and check for the formation of anhydrous copper sulphate - White precipitate when I add
the conc H2SO4 post rinsing copper of the NaOH
MagicJigPipe - 16-5-2008 at 02:13
Woelen, the addition of H2O2 will cause the solution to become "hydrous". Even 100% would decompose to some water, right?
woelen - 16-5-2008 at 04:30
Yes, but "some water" is not sufficient to allow blue aqueous copper(II) ion. I have done experiments with 98% sulphuric acid, to which already quite
some water was added (e.g. 20 ml of water to 100 ml of acid) and when hydrated copper(II) chloride is added, then this turns brown, due to
dehydration. The dehydrating power of sulphuric acid is remarkable.
But as I understood from brew, he added 6% H2O2, and then he might have added a lot of water. He mentions 5-8 ml, but this only tells something if the
amount of sulphuric acid is mentioned as well.
brew - 16-5-2008 at 14:55
Unfortunately-and hopefully will not be happening again - I didn't note the exact amount of 98%sulphuric acid used, at a reasonably close estimate I
would say 30ml.(my flask contains about 35ml) I see the importance of being more precise when I explore rather than doing bucket chemistry. Taking an
average value for the H2O2 = 6.5ml at 6% hence the approximate amount of water I added is 6.1 ml.
Regarding woelens' experiment I think I understand that sulphuric acid will protonate all H2O molecules inc-any associated with the copper(ll)
chloride. If there is no H20 left for the copper(ll) chloride to ion-dipole arrange with, then the brown color observed is a precipitated
copper(ll)chloride.Is this correct and if it is, I dare say very straightforward to most forum users - but very good material for me to understand
intermolecular forces.
brew - 16-5-2008 at 15:15
Re: my last post. If sulphuric acid has protonated the water molecules, then the copper cation would be in an ion-dipole arrangement with the sulphate
anion and the chloride anions would be ion-dipole to the H3O hence I dont think it would precipitate. I am abit confused with this. I dont understand
the color change. Has the sulphuric acid reduced the copper(ll) too copper(l)? I think this makes more scence and must be the reason for the color
change and not my initial assesment. Any help on this would be appreciated!