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guy
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Thanks that was a good explanation. So since this weird compound of manganese is stable in basic conditions, is there reason to believe that it has
oxygens around it?
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not_important
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OK
1) Remember that you can run the haloform reaction on alcohols and get a positive test for ketones, becase
RR`CHOH + OCl- => RR`CO + H2O + Cl-
Read the history of sugar chemistry, the oxidation of crabohydrates with halogen and base was often used.
So rather than the reduction of Mn4+, you most likely had the oxidation of the alcohol.
2) The oxidation by permanganat of alkenes to diols is thought to have a transition compound like
C-O O-
| \ /
| Mn
| / \
C-O O-
As does oxidations wih OsO4
So a complex of Mn and diols (or higher polyols) whould not be unexpected.
3) H2O2 is a strong oxidiser, about as strong as permanganate. At the same time, it can reduce some substances while oxidising itself to O2 and
water. So mixtures with H2O2 can give unexpected oxidation levels.
4) Mn2+ and Mn3+ appear to form stable aqueous complexes. While many of these include amino groups, some may only have H-O- and -O-C as the ligands
http://stratingh.eldoc.ub.rug.nl/FILES/root/FeringaBL/2004/M...
5) The colours seen are quite similar to those I made when attempting a good, strong cone 10 red. I was trying for Mn3+, but I know that sometimes I
was getting some Mn4+ About the only 'ligand' in these is oxygen and phosphorus, and as there's no water Mn3+ is a bit more stable than with aqueous
chemistry; plus going to Mn4+ doesn't mean its going to clump up as MnO2 (at least at low concentrations).
Try adding some borax to the red coloured stuff. Borates form fairly strong complexes with polyols, so it might affect the colour.
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guy
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Can anyone with measuring equiptment try finding how much glycerol per MnO2 or Mn2+ is needed? That would really help.
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chemoleo
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If there are truly ketones or aldehydes present, there are numerous great analytical tests that determine the rpesence of these. For instance dinitro
phenylhydrazine. I doubt this is what it is because these oxidations don't happen this rapidly, do they? Add some H2O2 to small amounts of EtOH in
NaOH, and see how long it takes till you smell Acetaldehyde. By the way, what happens if you heat the complex? Is it stable? What happens if you add a
surplus of oxidiser, to then heat it? Any change? This might tell whether oxidation of the ligand is detrimental to complex formation.
Do you have a dessicator? You could try crystallising something that way. But I think first of all these experiments have to be done
stoichiometrically, so far they are qualitative at best. Lots of work I know. Work with larger amounts, and it is easier.
Never Stop to Begin, and Never Begin to Stop...
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guy
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| Quote: | Originally posted by chemoleo
If there are truly ketones or aldehydes present, there are numerous great analytical tests that determine the rpesence of these. For instance dinitro
phenylhydrazine. I doubt this is what it is because these oxidations don't happen this rapidly, do they? Add some H2O2 to small amounts of EtOH in
NaOH, and see how long it takes till you smell Acetaldehyde. By the way, what happens if you heat the complex? Is it stable? What happens if you add a
surplus of oxidiser, to then heat it? Any change? This might tell whether oxidation of the ligand is detrimental to complex formation.
Do you have a dessicator? You could try crystallising something that way. But I think first of all these experiments have to be done
stoichiometrically, so far they are qualitative at best. Lots of work I know. Work with larger amounts, and it is easier. |
I tried boiling it, nothing happens, you can actually crystallize it. I tried boiling it in excess NaOCl and nothing happens. The problem is I have
no accurate measuring tools so I am asking anyone who has the time ad equiptment and actually cares about this .
[Edited on 7/28/2006 by guy]
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guy
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<b>Some New Information</b>
I just crystallized some of this product and washed it with alcohol. Its a bright orange solid.
Ok so I put some of this in a crucible and heated it. It decomposes quickly to MnO2 (black solid) and some smoke (water vapor + ??). The solid I
thought must have some Na2CO3, and I tested this by adding vinegar to it. It bubbled indicating CO3(2-).
What could I do with this? Measure the amount of Na+ present by precipitating some insoluble CO3(2-).
Measure the change in mass and calculate what constituted the missing mass through guess and check?
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not_important
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Can you dissolve a bit of it in plain water, get the original color, and get a bit of it to crystalize back from that solution? This might help
determine if it is stable on its own, or only in the presence of an excess of one of the reagents.
Going to MnO2 sure suggests it is Mn(IV) or higher, but see below.
(oops, see that you don't have measuring gear. But I'm leaving this anyway)
If you have a sensitive scale you could weigh a crucible, weight some of the orange crystals into that, heat, cool away from air/moisture, and weight
again to get an idea of how much organics and water are in it. The leach it out with distilled water, dry and weigh again for how much is Mn.
Evaporate the water solution to get the weight of the soluble stuff, then test that for Mn(II) - if the orange stuff is a complex of Mn(III) it could
got to II + IV when it decomposes.
Wish all my gear wasn't in storage.
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guy
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You can only dissolve it in basic solution or it will decompose into MnO2. Its like manganate where it is only stable in alkaline solution. I
crystallized before by evaporating most of the water but not all (made sure to not get any NaOH precipitating out).
Another thing I wanted to do if I had equiptment was to titrate the decomposition product to see how much Na2CO3 was in there.
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guy
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Actually I did some more experiments and I was wrong. THe solid is SPARINGLY soluble in water and almost insoluble in basic. In acid it will
disproportionate to dissolve to form a brown-cinnamon solution and MnO2. The solid is DIFFERENT than the solution it came out of.
The brown solution will react with H2O2 to form Mn2+. It will react with NaOH to very slowly form MnO2. It does not react with NaOCl or H2O2 in
basic solution.
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not_important
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Good job, guy.
Here's some related information, from Sidwick's Chemical Elements and Their Compounds.
Mn(III) oxide has two forms. The hydrated one, MnO(OH) is formed when wet MnO is exposed to air, espeically if bases are present, when MnCl2 is
oxidised by air in the presence of excess NH4Cl, and several other ways. It can be grey, brown, or black depending on how it was made. It will
generate chlorine if treated with HCl.
Mn(III) forms complexes with H3PO4, oxalic acid, malonic acid, and other oxyo compounds. These generally are a deep red, red-violet, or red-brown in
solution, the pure solid compounds are usdually red or green, some are yellow or yellow-brown ,and others red-brown.
These all seem to contain the structure
-O
\ /
Mn
/ \
-O
A test for Mn(III) might be as follows. If you have very concentrated H3PO4, 98%, or fairly concentrated H3PO4 and 98% H2SO4 (mix a cc of each), try
dissolving a bit of the mystery compound in that. In strong acid the colour will be violet, if the solution is diluted then grey-gren MnPO$ will
precipitate.
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guy
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How light is the color of Mn(III) phosphate complex? I am trying to test for Mn(III) after I acidified the red solution. When the red solution is
acidified with acetic acid, it forms MnO2 and a brown solution (manganese(III) acetate?). I added some phosphoric acid (from shower cleaner) and it
formed a very light-green precipitate.
I suspect this compound might oxidize the glycerol ligands in acid to form an aldehyde or ketone (glycerol can do either or both), but I don't have
the neccesary chemicals to do it!
[Edited on 8/3/2006 by guy]
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not_important
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I'm not sure that the precipitate is MnO2, it could be MnO(OH); it's not simple to tell.
The several descriptions of MnPO4 just say grey or grey-green, as 'dark grey' is used for other things I would assume it is fairly light in shade.
The Mn(II) phospates are all pink to red.
Mn(III) acetate isn't stable in water, the brown may be yet another complex or colloidal manganese hydroxide/oxides. I suspect that MnO(OH) would
give the phosphate, MnPO4 is very insoluable which is mainly why it can be made in water.
I think you could get the oxidation even in somewhat basic solution, and getting carboxylic acids isn't out of the question.
Off the top of my head, the only way that comes to mind to investigate the organics would be to make 10 or 20 grams of the complex, hit it with
trisodium phosphate and sodium carbonate to drop out all the manganese while still in an alkaline solution. Then filter and distill, simple ketones
and aldehydes would come over with the steam; I don't think any of the hydroxy-(acids,ketone,aldehydes) would. Evaporate the remainder of the
carbonate/phosphate solution, check for smell and then try extracting with acetone to see if there is relatively non-volatile organics.
If you have strong sulphuric acid, try dropping a small crystal of the orange stuff into 1cc H2SO4 + icc tile cleaner. Mn(III) forms sulfate
complexes as well as phosphate, if the acid is strong enough.
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guy
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Here's my guess on the compound based on my observations so far.
Edit:
Deleted pic.
[Edited on 8/4/2006 by guy]
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Nicodem
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Just a few comments on mechanism drawing:
The net electron transfer is of two electrons so the Mn(V) should go to Mn(III). So your mechanism is wrong in showing MnO2 as the product. However it
is true that fast disproportionations lead to MnO2 as the end Mn species in neutral of weakly acidic media, but you omited that. There also lack an
electron pair transfer arrow from H-C bond to C-O bond. Also, you should not depict the "glycerolmanganic acid" in the dissociated form or else the
protonation step makes no sense.
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
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guy
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Thanks for pointing that out. New drawing
[Edited on 8/5/2006 by guy]
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not_important
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Nice drawings, what are you using to do them?
I've done more researching, and it still looks as if you have a complex of Mn(III), Mn(III) and Mn(II), or just maybe Mn(III) and Mn(IV).
Interestingly complexes with Mn(III) often have several Mn atoms in each complex, similar with the mixed oxidation state cases. I've seen refernces,
full article tucked away behind 'for fee' walls, of complexes based around 4 Mn(III) ions
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guy
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| Quote: | Originally posted by not_important
Nice drawings, what are you using to do them?
I've done more researching, and it still looks as if you have a complex of Mn(III), Mn(III) and Mn(II), or just maybe Mn(III) and Mn(IV).
Interestingly complexes with Mn(III) often have several Mn atoms in each complex, similar with the mixed oxidation state cases. I've seen refernces,
full article tucked away behind 'for fee' walls, of complexes based around 4 Mn(III) ions |
I used ChemDraw, unfortunetely its a trial version.
Did you find the information about the Mn(III) complex from Google scholar? I doubt this is a +3 because Woelen tried MnO2,
<b>Na2S2O8</b>, NaOH, glycerol and it had the same results. Therefore it has to be at least +5. It will work with OCl- but OCl- oxidizes
the glycerol faster than the MnO2 so it wont work as well.
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not_important
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Ah, but Mn(IV) and higher get reduced by polyols. Given the very low solubility of MnO2 and hydrated kin, I wonder if the oxidiser isn't converting a
bit of it to a higher state that gets the Mn into solution; a bit like a trace of Cr(II) solublizing CrCl3. And the complex gives enough stability to
Mn(III) that it can stay around. We need ESR gear.
Both online and books on hand research. None of the mentioned classes of Mn(IV) and higher complexes are the proper colur, most are greenish. And
most of them require nitrogen containing ligands as well, while there is mention of purely oxo complxes of Mn II and III. When making Mn(OAc)3,
Mn(OAc)2 and acetic acid stablize the III state.
Wish I had easy access to my gear. Be interesting to try to make some Mn(OAc)3 and add gylcerol.
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guy
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Are you thinking that a higher oxidation state of Mn oxidized some of the glycerol and itself being reduced to 3+? Is that possible in basic
conditions?
I guess I could try making MnO(OH) and adding glycerol (in basic conditions). MnOOH is made by Mn(OH)2 and shaking with oxygen? I guess I could try
that tommorow.
[Edited on 8/5/2006 by guy]
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guy
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<b>EVIDENCE for +3 state</b>
Mn(OH)2 was precipitated from a solution of Mn(OAc)2. This was slightly brown due to oxidation from air (MnOOH). More NaOH was added along with
glycerol. No apparent reaction until the air was bubbled into the mixture forming the same red solution!
My guess for why this reaction worked with MnO2 as the starting reagent was becaused MnO2 was oxidized to a higher state, which then oxidized the
glycerol, reducing itself to MnOOH and reacting with glycerol.
My hypothetical structure for this is:
When pH is lowered the Glycerolate get re-protonated leaving MnOOH again (which can be confused for MnO2)
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Nicodem
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And why not like in the attachment?
The Mn(III) salts are not oxy salts. For example, Mn(OAc)3 and not MnO(OAc) so I don't see why such a complex would have to have any Mn=O bonds.
[Edited on 5-8-2006 by Nicodem]
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guy
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| Quote: | Originally posted by Nicodem
And why not like in the attachment?
The Mn(III) salts are not oxy salts. For example, Mn(OAc)3 and not MnO(OAc) so I don't see why such a complex would have to have any Mn=O bonds.
[Edited on 5-8-2006 by Nicodem] |
Because how can it get rid of the oxy groups in a really high pH? The usual Mn(III) salts are formed in low pH's.
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not_important
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Ok, thanks guy! I suspected the possibility of Mn(III) simply because none of the complexes of higher oxidation states had colours in the right range.
This isn't going to be easy, Mn forms polynuclear complexes. As an example, even though it's done at 100 K, here is one with two Mn(III) and four
Mn(II)
http://scripts.iucr.org/cgi-bin/paper?gk2005&buy=1
Note that the ligands ketones and carboxylic acids, previous to finding this and your experiment with Mn(OH)2, I wondered if oxyidised glycerol is at
least part of the complex, even though Mn does complex with polyols. This doesn't discourage me from thinking that.
edit -
I suspect Mn(III) or a mixed state complex. But it is possible that there are higher states, complexing may lower the cost of getting there enough
that even air oxidation can reach them; think of the Co(III) complexes.
[Edited on 6-8-2006 by not_important]
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Nicodem
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| Quote: | Originally posted by guy
Because how can it get rid of the oxy groups in a really high pH? The usual Mn(III) salts are formed in low pH's. |
But you can't say acetic acid makes for a low pH.
Also, the complex in the reference Not_important provided, the Mn(III) is not in the form of an oxo salt. Furthermore, I can't imagine how can an oxo
salt (having covalent bonds besides the ionic charge) be suitable to interact with complexing ligands at all. Can, for example BiO(+) cation, form a
complex with some ligands? I don't know much about inorganic chemistry, so I could be wrong.
Edit: A Mn(III) complex would not behave like a Mn(III) salt at high pH. For example, CuCl2 will precipitate Cu(OH)2 at high pH in water solution, but
its complex [Cu(NH3)4]Cl2 will not.
[Edited on 6-8-2006 by Nicodem]
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
Read the The ScienceMadness Guidelines!
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guy
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Ah ok I see they you are right Nicodem. I was thinking they were covalent like in higher oxidation states like MnO4-. So the complex makes more
sense without the oxo-ligands.
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