The nitration of N,N-dimethylaniline with $\ce{H2SO4}$ and $\ce{HNO3}$ gives mainly the meta product, even though $\ce{-NMe2}$ is an ortho,para-directing group. Why is this so?


  • 1
    $\begingroup$ en.wikipedia.org/wiki/Tetryl - reaction can proceed in other way $\endgroup$
    – Mithoron
    Jul 30, 2015 at 22:55
  • $\begingroup$ The best way around this issue, from a reagent/yield point of view, is to use acetic anhydride to N-acetylate the primary amine. This gives acetanilide, the acetyl derivative of aniline. $\endgroup$
    – user19518
    Aug 30, 2015 at 0:13
  • $\begingroup$ Essentially a dupe: chemistry.stackexchange.com/questions/31644/… $\endgroup$
    – bon
    Feb 16, 2016 at 18:55

2 Answers 2


In the presence of these strong acids the $\ce{-NMe2}$ group is protonated, and the protonated form is electron-withdrawing via the inductive effect. This discourages attack at the electron-poor ortho position.

Under the conditions I know for that experiment, you get a mixture of para- and meta-product, but no ortho-product due to steric hindrance.


In these acidic conditions, the lone pair on the nitrogen will abstract a proton of the acid in an acid-base fashion. This protonated amine will then be considered a meta-directing group. It is a deactivator that will draw electron density away from the ring (because nitrogen, a relatively electronegative atom, will have a positive charge).

The best way to get around this is to N-acylate. This can be done by first reacting the mono-substituted amine with acetic chloride ($\ce{CH3COCl}$) with pyridine to get an amide. That way, the ring is still activated and the lone pair on the amide nitrogen will not interfere with acids. It will direct ortho-para (para favored due to sterics).


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