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Author: Subject: Oxidation of salicylic acid to 1,2 benzoic acid
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[*] posted on 6-11-2013 at 09:54
Oxidation of salicylic acid to 1,2 benzoic acid


Because salicylic acid acts as a alcohol is it possible to use KMnO4 to further oxidize the hydroxyl group to a carboxylic acid, yielding MnO2 and Potassium 1,2 benzoic acid. Im just a beginner in organic chemistry. :D
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DraconicAcid
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[*] posted on 6-11-2013 at 09:59


Where would the extra carbon come from? Salicylic acid is a phenol- an oxidizing agent such as KMnO4 is not going to do the conversion ROH -> RCO2H.



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Lambda-Eyde
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[*] posted on 6-11-2013 at 09:59


No. If you study the structures of the two side by side (and understand what the structural formulae tell you) you should be able to figure out why. Also, such topics belong in the "Beginnings" section.



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6-11-2013 at 10:47
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[*] posted on 6-11-2013 at 14:57


I'm feeling quite patient this evening, so I've made a quick reaction scheme for what you are proposing, showing all of the carbon atoms in addition to the standard functional groups and terminal carbons. As you can see in red and as DraconicAcid pointed out, a carbon magically appears in your product, which is impossible.


salicyl oxidation.gif - 2kB

[Edited on 6-11-2013 by Hexavalent]




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deltaH
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[*] posted on 6-11-2013 at 23:32


In 2053 the nobel prize for chemistry shall have been awarded (sorry tenses get very complicated with time travel) for the development of a homogeneous catalyst that inserts CO resulting in the conversion of a phenolic group to an aromatic carboxylic acid.

In his acceptance speech, said winner will have gone on to state that inspiration for this came from a science forum some forty years prior ;)

EDIT:

Huh... I've just checked the rule book and apparently while I am not permitted to give the details of this, I am permitted of giving abstract clues about it... so here goes:

pic1.jpg - 6kB pic2.jpg - 13kBpic3.jpg - 6kB pic4.jpg - 4kB

[Edited on 7-11-2013 by deltaH]




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[*] posted on 7-11-2013 at 09:37


Alright, the OP's question has been answered, and this thread is now obsolete. [closed]

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