44
$\begingroup$

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?

Reaction

$\endgroup$
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Tetryl - reaction can proceed in other way $\endgroup$ – Mithoron Jul 30 '15 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 '15 at 0:13
  • $\begingroup$ Essentially a dupe: chemistry.stackexchange.com/questions/31644/… $\endgroup$ – bon Feb 16 '16 at 18:55
36
$\begingroup$

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.

$\endgroup$
20
$\begingroup$

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).

$\endgroup$
-1
$\begingroup$

There is form of $\ce{NH3+}$ ion when sulfuric acid is in prior meta ($47\%$) but when sulphuric acid is in excess meta ($90\%$) . due to formation of $\ce{NH3+}$ ion ring become meta directing

$\endgroup$
  • 4
    $\begingroup$ This is a start for a potentially good answer, but you should elaborate. We'll polish the grammar. $\endgroup$ – M.A.R. Feb 16 '16 at 16:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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