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Author: Subject: Help with oxidation of aldehyde
postart
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[*] posted on 29-6-2010 at 18:39
Help with oxidation of aldehyde



Nice forum you folks have here, well here goes. I have a mixture of 2-phenylpropanal and phenyl2propanone the end goal is to isolate the ketone. I am planning to oxidize the aldehyde with %5 NaOH/ %3 H2O2 but I'm not sure of what measurements of NaOH and H2O2 to use. If any chemist can help it would be greatly appreciated. Once the aldehyde has been oxidized the solution will be made acidic until red and the resulting carboxylic acid removed by filtration. The aldehyde/ketone mixture has not been analyzed but should be in the ballpark of 38%aldehyde/62%ketone theoredically.

Thanks!
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mnick12
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[*] posted on 29-6-2010 at 19:24


Erm this should be in the beginings section.
And for isolating your phenylacetone you could use the bisulfite adduct. Then acidify to get your phenylacetone.

Just out of curiousity what do you plan on doing with your phenylacetone?
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postart
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[*] posted on 29-6-2010 at 19:58


I thought both the aldehyde and ketone form a bisulfite addition compound.

[Edited on 30-6-2010 by postart]
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Melgar
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[*] posted on 29-6-2010 at 20:07


Hmm... what might he be planning to do with phenylacetone? Methinks he has some METHod in mind already...
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Ozone
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[*] posted on 29-6-2010 at 20:11


"...theoredically."

I hope this is really a clever pun regarding the questions on red-uction which are sure to follow.

O3




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postart
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[*] posted on 29-6-2010 at 20:39


Actually I'm interested in anticoagulants.
[Edited on 30-6-2010 by postart]

[Edited on 30-6-2010 by postart]
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turd
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[*] posted on 29-6-2010 at 22:40


Quote: Originally posted by postart  

Once the aldehyde has been oxidized the solution will be made acidic until red and the resulting carboxylic acid removed by filtration.

What? This makes no sense. You should make it basic and partition between water and an organic solvent. The salt of the acid will go into the aqueous phase, the ketone into the organic phase. Dry the organic phase over Na2SO4, strip solvent under vacuum, distill, done.

For oxidation of aldehyde to acid check the literature, this must have been studied a million times. And as others noted: this is material for beginnings.

Last point: you don't have justify yourself. Doing real chemistry is good enough.

[Edited on 30-6-2010 by turd]
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postart
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[*] posted on 30-6-2010 at 18:46


Mol weights:
NaOH g/mol= 39.99711
H2O2 g/mol= 34.0147
2-phenylpropanal g/mol= 134.1751

grams/per mol of 2-PhenylPropanal @ 1.1 molar excess:
NaOH- 43.996821 grams/per mol 2-PhenylPropanal
H2O2- 37.41617 grams/per mol 2-PhenylPropanal

340.0/g 2-PhenylPropanal needs oxidized
340g/2-PhenylPropanal = 2.534 mol
111.487944414/g NaOH @ 1.1 mol ratio
94.81257478 /g H2O2 @ 1.1 mol ratio

average from above:
340g 2-PhenylPropanal
111g NaOH
95g H2O2

note:
1000ml 3% H2O2 = 30g H2O2
1000ml 10% NaOH = 100g NaOH

So 1110ml 10% NaOH + 3200ml 3%H2O2 will oxidize the 340g 2-Phenylpropanal. Once this has been done 560grams of ketone can be distilled out. right? Advice an critique welcomed.
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Sandmeyer
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[*] posted on 30-6-2010 at 19:42


one could distill, or make bisulfite adduct, aldehydes react much faster than ketones.. Then the aldehyde can be treated with H2SO4 to get more honey.

[Edited on 1-7-2010 by Sandmeyer]




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postart
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[*] posted on 30-6-2010 at 19:47


Thats how the mixtur was formed. Now I'm trying to isolate the rearanged ketone by oxidizing the left over aldehyde.
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Nicodem
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[*] posted on 1-7-2010 at 00:36


Please open referenceless threads only in the Beginnings section! Also, what is the point in opening a new thread on an already discussed subject? Next, time use one of the existing threads on the topic.

What is the reference for the oxidation of aldehydes with H2O2/NaOH?

...moving to Beginnings.
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DJF90
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[*] posted on 1-7-2010 at 05:52


I would expect Baeyer Villiger oxidation with these conditions. I'm not even sure they're suitable reagents for the oxidation of aldehyde to acid.
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postart
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[*] posted on 1-7-2010 at 09:31


The Baeyer Villiger oxidation uses H2O2 to oxidize a ketone to an ester does this mean my thinking is wrong and the ketone will just be reduced to an ester?:(
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DJF90
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[*] posted on 1-7-2010 at 13:33


For a start, the ketone will be oxidised to an ester, not reduced. And yes, that is correct, unless you can find a magic paper saying your reagents will oxidise aldehydes to acids, which, mechanistically at least, seems unlikely.
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postart
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[*] posted on 8-7-2010 at 12:52


Well oxidation is out. So, I am thinking an addition of 1.1mol sodium bisulfite for every mole of hydrotropic aldehyde. I'm thinking the 1.1 molar excess of bisulfite will ensure all the aldehydic material is gone. The addition product filtered out and solvent distilled to yeild ketone. Sound good?
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DJF90
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[*] posted on 8-7-2010 at 13:43


Why is oxidation out? Not all oxidising agents react the same wayto those which you suggest, and by using oxidation, separation of the ketone and the resulting carboxylic acid could be effected using base extraction (Basify, extract ketone into organic layer, acid remains in aqueous as carboxylate salt).
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postart
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[*] posted on 8-7-2010 at 18:01


I just thougt the bisulfite addition would be easier.
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postart
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[*] posted on 13-7-2010 at 09:34


So I'm going to add a 1.1 molar excess of bisulfite to the mixture. As I understand this will cause the aldehyde(hydrotropic) to form an addition product that can be filtered out. I have heard the aldehyde will form the addition product more readily, but could somone explaine how this works out rather than forming an addition product of some of the ketone and aldehyde?


Thanks!
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DJF90
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[*] posted on 13-7-2010 at 16:16


Its all about reactivity, and reversibility.
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[*] posted on 12-9-2010 at 00:28


Actually, I don't believe you do have a mixture that includes the aldehyde.

Although the popular writeup of this method includes Rh.'s comments to the effect that it would produce a mixture of the aldehyde & ketone (which, to be fair, is a very reasonable assumption) in this case it turns out that the end product contains no aldehyde.

Look for an article by a Russian guy named Danilov wherein it is speculated that in this case the reaction proceeds to completion because the ketone forms more stable complexes with the H2SO4 than does the aldehyde, thereby removing the ketone from the reaction soon after formation, but leaving the aldehyde to undergo further reaction. For the method under discussion, it is stated quite clearly that 'only ketone can be found in these complexes, especially in the case of concentrated sulphuric acid.'

On the other hand, in my experience attempting to isolate the ketone from the acid complexes and subsequently formed polymerization products is a real bi0tch.
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[*] posted on 28-12-2015 at 17:44


If the aldehyde is currently in solution with the said ketone, my question is, is the bisulfite adduct of phenylacetone soluble in water? I do know that some ketone bisulfite adducts based on molecular size are insoluble on water, and as we know the bisulfite adduct of 2-phenylpropanal is already soluble in water.



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