Peter Sykes mentions in the book Guidebook to Mechanism in Organic Chemistry (6th ed.) on p. 137 that:

Highly reactive aromatic compounds, such as phenol, are found to undergo ready nitration even in dilute nitric acid, and at a far more rapid rate than can be explained on the basis of the concentration of $\ce{^⊕NO2}$ that is present in the mixture. This has been shown to be due to the presence of nitrous acid in the system which nitrosates the reactive nucleus via the nitrosonium ion, $\ce{^⊕NO}$ […]

However, there was no mention as to why the nitration of more reactive aromatic compounds would result in there being nitrous acid in the system. I understand that the conventional explanation for why the nitration of activated benzene derivatives is more rapid is that there is a stronger attraction between the electrophile and the nucleophilic benzene ring due to the greater amount of electron density in the ring. This explanation using the presence of the nitrosonium ion seems interesting but I don't see how nitrous acid could be generated due to there being a more reactive aromatic compound being nitrated. Could someone enlighten me on this?


Possibly the electrophile is molecular $\ce{HNO3}$. Nitric acid is strong but not super-strong ($\mathrm{p}K_\mathrm{a} = -1.4$), so solutions on the order of a mole per liter have a significant amount of molecular $\ce{HNO3}$. As $\ce{NO2(OH)}$ this can transfer its nitryl ($\ce{NO2+}$) ion moiety to the aromatic ring in exchange for a proton to make the nitrated ring and water. Obviously molecular nitric acid is less electrophilic than the straight-out nitryl ion, so an activated aromatic ring is needed for this to work.

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    $\begingroup$ But the book mentions specifically that NO+ is generated $\endgroup$ Apr 25 '18 at 12:43
  • $\begingroup$ Not sure how that is, unless some of the organic compound is oxidized first. And nitrogen(III) species would disproportionate or be oxidized by air, no? Molecular nitric acid is in the solution with no side reactions, ergo a more parsimonious explanation. $\endgroup$ Apr 25 '18 at 13:04

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