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I am trying to understand this mechanism for nitrobenzene reduction with $\ce{Sn/HCl}$. I do not understand why it implies that you pull off the $\ce{C-H}$ hydrogen instead of the $\ce{O-H}$ hydrogen in the methanol. Is it actually suggesting that the resulting $\ce{.CH2-OH}$ radical formed from the methanol eliminates to form formaldehyde or do you get a rearrangement where you form $\ce{(CH3-O)-}$? If the latter is the case, why not just pluck the proton from the alcohol group instead of breaking the $\ce{C-H}$ bond?

Also, it mentions that organic solvents are preferred for this reaction, why exactly is that over water?

nitrobenzene reduction to aniline with Sn/HCl

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  • $\begingroup$ I'm skeptical about the mechanism, but it's worth putting it out there that in radical reactions C–H bonds are easier to cleave than O–H bonds due to their lower bond dissociation energies. $\endgroup$
    – orthocresol
    Commented May 10, 2018 at 21:19

2 Answers 2

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I don't buy that mechanism. Like you noted, breaking a C-H bond in methanol seems unlikely, especially since the reaction is run in the presence of acid, so a ready source of protons is available.

Most of these reactions that involve a metal surface proceed through what is known as a single electron transfer (SET) mechanism. Reductions of nitro compounds, be they catalytic hydrogenations or metal surface - acid catalyzed reductions, usually proceed through nitroso and hydroxylamine intermediates. This link provides some discussion of the reaction and presents the following (more reasonable) mechanism where electrons are transferred from the metal surface to the various intermediates and the acid provides a ready source of protons. These reactions are usually run in an organic solvent in order to bring the organic nitro compound into solution.

enter image description here

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  • $\begingroup$ No problem, though I have to say, I really dislike SET mechanisms. There is too much unknown about them, they seem to involve some kind of magic... (And I also would suggest, that the nitroso intermediate will be skipped in this.) $\endgroup$
    – Martin - マーチン ♦
    Commented Aug 28, 2014 at 3:36
  • $\begingroup$ @Martin "the nitroso intermediate will be skipped" Why? See books.google.com/… $\endgroup$
    – ron
    Commented Aug 28, 2014 at 3:54
  • $\begingroup$ I would expect the oxygen to be protonated all the way, at least to the major extent. In your scheme you go from $$\ce{R-N(OH)(OH2)+ ->[-\ce{H2O;\ - H+}] R-N=O ->[+\ce{H+}] R-N=OH+ },$$ but I would assume that it proceeds like this: $$\ce{R-N(OH)(OH2)+ ->[-\ce{H2O}] R-N=OH+ <=>[\pm\ce{H+}] R-N=O}$$ So you can trap the nitroso compound, but it is not actually part of the mechanism. $\endgroup$
    – Martin - マーチン ♦
    Commented Aug 28, 2014 at 6:14
  • $\begingroup$ Is there any particular reason why you use zinc here? I assume it does not make a big difference, but the OP asked about tin. $\endgroup$
    – Martin - マーチン ♦
    Commented Aug 28, 2014 at 6:17
  • $\begingroup$ @Martin No particular reason, it was an available drawing of the mechanism and zinc, tin , mercury amalgam, etc. should all proceed by the same mechanism. $\endgroup$
    – ron
    Commented Aug 28, 2014 at 12:42
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In the mechanism you are considering, there are two options for reduction of R-$\ce{N^{.}H}$ (and similar for other neutral radical species)

  1. $\ce{CH3OH + R-N^{.}H -> RNH2 + C^{.}H_2OH }$

  2. $\ce{CH3OH + R-N^{.}H -> RNH2 + CH3O^{.}} $

In the first option (1) nitrogen oxidizes carbon - rare, but possible, as carbon has lower electronegativity. In option (2) nitrogen oxidizes oxygen - an element that has higher electronegativity, and this is unlikely.

In addition, in option (1) free radical is stabilized by forming 3-electron 2-center bond $\pi$-bond C...O, while in option (2) such stabilization cannot occur.

Taking proton (i.e. hydrogen cation) from $\ce{OH}$ is not an option for reduction, and $\ce{R-N^{.}H}$ is a radical, and expects hydrogen atom, not ion. Taking proton will produce $\ce{R-N^{+.}H2}$ and $\ce{CH3O-}$ and the first particle is not much better than the original $\ce{R-N^{.}H}$ while heterolitic bond breaking is generally less favorable, than radical one.

Proton may be accepted by negatively charged radical anions, like $\ce{R-N(O)O^{-.}}$ but after that they will seek a hydrogen atom for reduction.

Of course, it IS theoretically possible, that the radicals are actually reduced by tin and all the reaction happens on metal surface, and this is likely the case if the reaction happens in water, but in case it happens in organic solvent, especially reducing one like methanol, it is theoretically possible that it does participate in reduction. Until proper proofs, like trapping of the intermediate species, are obtained, any mechanism is no more then speculation anyway.

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