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I came across this question in an exercise book. Can anyone give detailed answer (mechanism) of this reaction?
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I am unable to proceed. I thought there maybe some redox reaction involved where $$\ce{Fe^3+ -> Fe^2+}$$ maybe taking place but then nitrobenzene would have to be oxidised. Can someone please help ?

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I'd go more for an oxidation, but on the ring (hydroxylation). My guess comes from this paper here : "The Hydroxylation of Aromatic Nitro Compounds by Alkalies" by O. C. Dermer.

This is very harsh (and somehow old) chemistry.

First, for m-dinitrobenzene and sym-trinitrobenzene, hydroxylation is reported in your conditions, that is to say alcaline conditions with the help of potassium ferricyanide as oxidizing agent (reactions A and B, see reference in the paper).

Second, the authors report the reaction C (with less than 4% yield though, according to my calculations), with a small amount of p-hydroxyazobenzene as a side-product. They did this reaction by mixing only nitrobenzene with potassium hydroxide, so I would expect the same product (o-nitrophenol) with maybe a better yield in your conditions.

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First, since there are no electrophiles that are formed, the only reaction that comes to mind is nucleophilic aromatic substitution.

$\ce{NaOH}$ does a displacement reaction with $\ce{K3Fe(CN)6}$, yielding the $\ce{CN-}$ ion.

The cyanide ion attacks the benzene ring, giving three possible resonance structures:

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The hydrogen attached to the carbon with the cyanide ion is accepted by the strong base $\ce{NaOH}$.

Next, the nitrile undergoes acid-catalysed hydrolysis to form a carboxylic acid:

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  • $\begingroup$ I thought it was tough to remove ligands inside a complex. Can you explain how/why CN- comes out? $\endgroup$
    – Black Jack 21
    Commented Feb 2, 2017 at 11:49
  • $\begingroup$ Btw, you left your mechanism at amide, no need to edit it to acid though. I know the mechanism $\endgroup$
    – Black Jack 21
    Commented Feb 2, 2017 at 11:51
  • $\begingroup$ While $\ce{CN-}$ is possible to be displaced by $\ce{OH-}$, it is a much stronger ligand. Therefore I suspect that this reaction would not happen to a large extent. What's more, nucleophilic aromatic substitution generally occurs with aryl halides since $\ce{X-}$ is a much better leaving group. $\endgroup$
    – Linus Choy
    Commented Feb 2, 2017 at 11:53
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    $\begingroup$ I'm afraid there are some real problems with this loss of a hydride. If CN- attacks nitrobenzene, the subsequent step must expel H- not H+. Do you have any reference for this reaction? Nucleophilic substitution of hydrides is not unknown (Chichibabin reaction), but it doesn't make much sense here. cc @AlphaRomeo $\endgroup$
    – orthocresol
    Commented Feb 2, 2017 at 12:15
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    $\begingroup$ Aside from that there are some errors in the mechanisms. In the first mechanism, the first arrow should be a forward arrow $\to$ not a resonance arrow $\leftrightarrow$. The third structure is missing a double bond between C and N. In the second mechanism, the amide has to be protonated first before H2O can attack, and NH2 has to be protonated before it can leave. $\endgroup$
    – orthocresol
    Commented Feb 2, 2017 at 12:20

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