When we do nitrosation of phenols, we get para-nitrosophenol as the major product, which tautomerises to quinone mono oxime. This article states that it largely exists in the quinonoid form.

My question is, why? I know that nitroso form is less stable than the oxime form, and enol is less stable than keto, but on the other hand we are losing the aromaticity also. Why is being non aromatic getting more favourable than being aromatic in this case?

I read the post shared by @Safdar in the comments, but this answer there says that both forms are aromatic. Is this true?

  • 2
    $\begingroup$ Possible dupe: chemistry.stackexchange.com/questions/125617/… ? I guess the same principle applies here.. $\endgroup$ Commented Feb 17, 2021 at 8:23
  • $\begingroup$ @Safdar I agree that it is similar, but in that post both rings are aromatic, but I doubt that it quinonoid form is aromatic in this case, as you can see in the link I have given. $\endgroup$
    – V.G
    Commented Feb 17, 2021 at 9:40

1 Answer 1


Two factors come into play: (1) the inherent difference in resonance energy is not as great as you might think, and (2) the greater tendency for a dissociable proton to bind more tightly to an oxime function rather than a phenol function appears to counterbalance the reduced $\pi$-electron bonding difference.

Less than meets the eye

Suppose you were to use the Hückel model to estimate the pi-electron binding energy for several eight-carbon conjugated systems. By that method I find the following results:

  • For the conjugated straight chain $\ce{CH2=(CH=)_6CH2}$ the total $\pi$-electron binding energy works out to $9.52\beta$, where in this theory $2\beta$ represents one isolated $\pi$ bond.

  • For styrene, with a $\pi$-electron pattern matching the nitroso-phenol, the result is $10.53\beta$. A difference of about one $\beta$ unit is typical when an aromatic ring is compared with the opened ring.

  • For para-xylylene whose quinonoid $\pi$-electron structure matches the oxime tautomer described here, we get $9.92\beta$.

Of course the aromatic structure is most stable, but not all non-aromatic structures are created equal. The Hückel-model calculations indicate that even though the quinonoid structure is typically called non-aromatic, in fact it captures a significant part (about $40$% based on the numbers) of the extra $\pi$-electron stabilization we usually associate with an aromatic ring. The energetic difference between the two competing $\pi$-electron configurations is thereby less than the "aromatic" and "non-aromatic" labels would seem to imply.

Swimming upstream

Now let's see where the dissociable proton is more stable. To do that we look up the $pK_a$ values typically associated with each prospective acidic functiin:

The oxime function is clearly holding the proton more tightly; placing the acidic proton there instead of on the phenol function saves $2.2$ powers of ten in the equilibrium constant.

We cannot directly compare this with the Hückel-model calculations from the previous section, because that model is only semi quantitative and a more complex calculation would be needed to capture the effects of the heteroatoms in the conjugated structure. But the results do suggest that the combination of the quinonoid structure capturing part of what is usually aromatic $\pi$-electron stabilization with the preference of a dissociable proton to bind to the oxime function leads the oxime tautomer to win out over the nitroso-phenol.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.