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Topic: oxidation  (Read 4428 times)

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Offline Ankesh

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oxidation
« on: July 21, 2010, 04:45:00 AM »
what would be the product for:
            (O)
C6H5CN  :rarrow:
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Offline SVXX

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Re: oxidation
« Reply #1 on: July 21, 2010, 12:01:07 PM »
Benzamide perhaps? That is, C6H5CONH2. Specify the reagent..
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Offline sjb

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Re: oxidation
« Reply #2 on: July 21, 2010, 01:04:51 PM »
Wouldn't benzamide be simply hydration though, so no net redox?

Benzonitrile oxide? PhC#N(+)O(-)? What definition of oxidation are you using?
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Offline 408

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Re: oxidation
« Reply #3 on: July 21, 2010, 04:50:19 PM »
Depending on the oxidation reagent it could be any of the above or simply CO2, H2O and N2...
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Offline discodermolide

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Re: oxidation
« Reply #4 on: July 22, 2010, 12:46:31 AM »
Quote from: Ankesh on July 21, 2010, 04:45:00 AM
what would be the product for:
            (O)
C6H5CN  :rarrow:

You will get a nitrile oxide.
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Offline Ankesh

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Re: oxidation
« Reply #5 on: July 23, 2010, 12:40:29 PM »
oxidation is in the presence of hot KMnO4....sorry for not specifying the reagents before.
Also tell the products in presence of neutral Kmno4
« Last Edit: July 23, 2010, 01:13:35 PM by Ankesh »
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Offline Ankesh

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Re: oxidation
« Reply #6 on: July 23, 2010, 02:13:20 PM »
Please reply...
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Offline orgopete

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Re: oxidation
« Reply #7 on: July 24, 2010, 08:12:07 AM »
I am uncertain of the results of this reaction. The conditions are quite vigorous. KMnO4 reacts via radical mechanisms when hot. It also decomposes to O2 and MnO2 as I recall. Since base catalyzed hydrolyses of nitriles and amides are difficult due to the basicity of the anion of ammonia, I am going to guess that KMnO4 either oxidizes the nitrile (to the nitrile oxide) or the amide. That should make the hydrolysis easier and the product to be benzoic acid.

If I understand the mechanism correctly for KMnO4 oxidations, the products of the oxidation are MnO2, H2O, and KOH. I could also imagine that another more labile functional group becoming oxidized in the presence of a nitrile.
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