Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: elupuu on March 27, 2013, 05:08:05 PM
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So I need to know the mechanism for the reduction of para-nitrophenol with NaBH4 with Pd/C catalyst.
I initally proposed a mechanism in which oxygen offers his lone pair to BH3 and hydride attacks the nitrogen. OBH3- is the leaving group and the reaction happens again for the other oxygen.
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Then why do you need the catalyst?
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That's what problematic, if the catalyst is needed and my route doesnt include it then my mechanism is flawed and I need to see the real mechanism.
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That's what problematic, if the catalyst is needed and my route doesnt include it then my mechanism is flawed and I need to see the real mechanism.
What product do you expect from this reduction?
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That's what problematic, if the catalyst is needed and my route doesnt include it then my mechanism is flawed and I need to see the real mechanism.
The product should be is 4-Hydroxyaminobenzene
What product do you expect from this reduction?
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What is your reaction solvent?
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The solvent is an aqueous solution of NaOH and the expected product is para-hydroxyaniline.
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OK thanks.
In this case is the aqueous NaOH and NaBH4 not generating hydrogen which is reducing the nitro group using the catalyst?
So effectively you are doing a hydrogenation with an internal hydrogen source instead of using a hydrogen cylinder.
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I dont believe generating hydrogen is possible with the conditions Im having. Also I found out NaBH4 is required in 4 equivalents.
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So I need to know the mechanism for the reduction of para-nitrophenol with NaBH4 with Pd/C catalyst.
I initally proposed a mechanism in which oxygen offers his lone pair to BH3 and hydride attacks the nitrogen. OBH3- is the leaving group and the reaction happens again for the other oxygen.
This mechanism does not seem to need Pd/C. I too thought BH4(-) was a hydrogen source. I didn't think borohydride reduced nitro compounds, but especially an acintro anion. Although I did not anticipate palladium accepting electrons from borohydride, that is the same net result as the addition of H2 to palladium. If need be, I'd write the addition of hydride to palladium followed by protonation of the anion. Use this reductant to convert a nitro group to an amine (over several steps).