Can I get some help in figuring out the mechanism for this reaction.
Adipic acid is a molecule with two carboxylic groups,but I can't see how that happens. My initial thoughts involved that the carbonyl group in cyclohexanone is protonated,but I couldn't get any further as I couldn't figure out what would happen after that.
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asked Jul 5, 2017 at 11:04
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2$\begingroup$ Are you sure you want to use cyclohexanone $\ce{ C6H10O}$, not cyclohexanol $\ce{C6H11OH}$ for this reaction? This huge picture must carry a subliminal message, but I'm asking just to be 100% sure we are on the same page. $\endgroup$– andselisk ♦Jul 5, 2017 at 11:10
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$\begingroup$ I am sure I want to use cyclohexanone, but does it really matter because cyclohexanone can be converted into cyclohexanol. $\endgroup$– Daenys TargaryenJul 5, 2017 at 12:05
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1$\begingroup$ From what I remember at industrial scale cyclohexanone and cyclohexanol mixture is rectified, and cyclohexanone is used in the production of caprolactam, whereas cyclohexanol is utilized for the synthesis of adipic acid by oxidation. BTW, you need a catalyst ($\ce{NH4VO3}$) with 50% $\ce{HNO3}$ for that. $\endgroup$– andselisk ♦Jul 5, 2017 at 12:12
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$\begingroup$ chemistry.stackexchange.com/questions/4416/oxidation-of-ketones $\endgroup$– MithoronJul 5, 2017 at 15:25
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1 Answer
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It think this reaction goes via following mechanism. 
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1$\begingroup$ How does the last step of the reaction occur? $\endgroup$ Jul 5, 2017 at 12:07
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1$\begingroup$ @Daenys Targarysn Further oxidation leads aldehyde to carboxylic acid. That's how the last step occurs. $\endgroup$ Jul 5, 2017 at 13:12
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2$\begingroup$ That is incorrect, at very least mechanism of initial attack is written terribly. $\endgroup$– MithoronJul 5, 2017 at 15:11
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$\begingroup$ @Mithoron Yes I am actually inclined to agree with you about the initial attack but its just what looked obvious to me .. $\endgroup$ Jul 6, 2017 at 4:01
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$\begingroup$ But the intermediate products formed while this transformation are correct so I just suggested this mechanism . $\endgroup$ Jul 6, 2017 at 4:07