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What is the mechanism of the attached reaction? How could nosylate do the dehydrogenate the species?

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The reaction is from this synthesis (46th IChO preparatory problem 26): enter image description here

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    $\begingroup$ I think this is quite a complex mechanism, but I would assume that the nitro group is being reduced. Could you add a reference as a starting point? $\endgroup$ – Martin - マーチン Jan 6 '15 at 8:09
  • $\begingroup$ Edited my original post. $\endgroup$ – Marko Jan 6 '15 at 9:53
  • $\begingroup$ Thanks, that will help put it into context, but I am still thinking that a suitable mechanism is hard to propose. I tried tracking down the original paper, which they used as source, but was unsuccessful due to language barrier. $\endgroup$ – Martin - マーチン Jan 6 '15 at 11:20
  • $\begingroup$ So I guess that it would be rather written as a standard red-ox reaction. $\endgroup$ – Marko Jan 6 '15 at 12:28
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    $\begingroup$ @Martin Without knowing the fate of the sulfonate reagent, I don't feel confident in any mechanism. If there were some transition metal to mediate the dehydrogenation, the nitro group could be the terminal reductant. The reaction is under basic conditions (hydroxide) which makes me think it's going through an enolate. I wonder if somehow the O-sulfonate is formed, and then an elimination occurs to form a carbonyl with formation of sulfinate. That's my best guess, but there are no similar reactions in SciFinder. $\endgroup$ – jerepierre Jan 7 '15 at 16:16
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This is the reaction mechanism that I propose for this reaction. Though it does not show the complete reduction of the 3-nitrobenzene sulfonate, it could be reduced further through the nitroso-hydroxylamine pathway. This means that in total, a single nitrosyl anion could oxidize three molecules of the starting material.

Extraction of a hydrogen by the nitro group. The hydrogen is extremely acidic compared to an ordinary aliphatic $\ce{C-H}$ bond due to the massive conjugation in the resulting ion:

step 1

Amide-imidic acid tautomerization:

step 2

Resonance structures of the molecule, showing its extensive conjugation, and hence stability:

step 3

Radical reaction pathway and formation of the product. Every atom in product is $\mathrm{sp^2}$ hybridized and is conjugated over the entire system. It is much more stable than the starting material, and is the driving force of the entire reaction:

step 4

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