In the nitration of aniline, we observe the formation of a meta-substituted product (47%) almost as much as a para (52%).

However, can't the electrophile attack at the position where $\ce{-NH2}$ is already attached, so that after the attack, it will be stabilised by the mesomeric (+) effect of $\ce{-NH2}$? Maybe just after some time, the $\ce{-NH2}$ group will become $\ce{-NH3+}$ (acidic medium) and $\ce{NH3}$ would depart as a leaving group?

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    $\begingroup$ Can you try drawing arrows for it, because it doesn't really work that way. Ipso attack of an electrophile leads to a carbocation on C-2 which cannot be stabilised by the lone pair on N (and note that since the nitration is carried out in strong acid, that lone pair will be permanently protonated anyway). Next you cannot have "NH3" leaving and regenerating an uncharged benzene ring $\endgroup$ Commented May 7, 2017 at 10:31
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    $\begingroup$ @orthocresol I did try to. I meant to say if 'while' the electron rebound back to their position in the intermediate, -NH2 group protonates, and leaves? $\endgroup$ Commented May 7, 2017 at 10:33

2 Answers 2


Maybe just after some time, the $\ce{-NH2}$ group will become $\ce{−NH3+}$ (acidic medium) and $\ce{NH3}$ would depart as a leaving group?

No! You are confusing the leaving group oxonium ion with the ammonium ion. Simply put it, $\ce{R-NH2}$ does not leave as $\ce{R+}$ and $\ce{NH3}$, as you might expect from the hydroxyl group leaving mechanism.

But that does not mean there is no other way to make an $\ce{-NH2}$ group (which is a very bad leaving group) leave. You can instead make it leave as dinitrogen gas!

The way this works is by the reaction of $\ce{R-NH2}$ with $\ce{HNO2}$. If $\ce{R}$ is an:

  • aromatic group: a diazonium ion, $\ce{R-N2+}$, will be formed (stable at ice cold temperature). Further reactions such as Balz-Schiemann, Gattermann, Sandmeyer, etc. may be employed to replace the $\ce{-N2+}$ with another group, and make it leave as dinitrogen gas.
  • alkyl group: $\ce{R-N2+}$ will be formed but it will, being highly unstable (given that it does not enjoy the resonance stabilization of an aromatic ring), instantly decompose to release dinitrogen gas and form a classical carbocation $\ce{R+}$. This carbocation may rearrange, if necessary, and then receive an attack by a nucleophile.

This is how you make the bad leaving group, $\ce{-NH2}$, leave any aromatic ring or an alkyl group.

  • $\begingroup$ See the answer to this question chemistry.stackexchange.com/questions/31644/… $\endgroup$
    – Waylander
    Commented Apr 4, 2018 at 16:06
  • $\begingroup$ @Waylander Thanks for the link, though I don't know how that answers this question, or offers a correction to the answer I've given. Can you please explain? Thanks again. $\endgroup$ Commented Apr 4, 2018 at 16:09
  • $\begingroup$ Simply because it addresses the question of the role of the anilium ion in nitration, and specifically makes no mention of any loss of NH3 $\endgroup$
    – Waylander
    Commented Apr 4, 2018 at 17:50
  • $\begingroup$ @way I too have said that - without forming a diazonium ion - there's no way you could get the - NH2 group to leave the ring. So, I still don't see the problem... $\endgroup$ Commented Apr 5, 2018 at 0:12

With nitric acid as a nitrating agent, the difficulty is that the acid protonates the amino group and destroys its nucleophility.

However, other compounds may provide a nitryl ion moiety without a protic hydrogen, thus leaving the nitrogen open to nitration. A casual Google search does not turn up an example with aniline, but ammonia can be converted to nitramide at nitrogen by $\ce{N2O5}$:

$\ce{N2O5 + 2NH3 -> \underset{nitramide}{H2NNO2} + NH4NO3}$

The $\ce{NO2^+}$ portion of the $\ce{N2O5}$ nitrates a nitrogen atom on the ammonia molecule, and the displaced proton is scooped up by a second ammonia molecule to form a salt (ammonium nitrate) with the $\ce{NO3^-}$ part of the original nitrogen pentoxide.


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