In 3-nitroaniline, if it were subjected to an electrophilic substitution reaction, the electrophile will most likely attack at (b) or (d) as mentioned in my textbook. 3-nitroaniline To me, it seems that the only role of the nitro group is that the electrophile will not attack at (a) due to steric hindrance. So my question is, why is the -M effect of the nitro group ($\ce{-NO2}$) discarded and only the +M effect of $\ce{-NH2}$ plays a role? Nitro group is meta directing, so why wouldn't the product form at (c) as well? It is as if the nitro group doesn't even exist. In better words, why is the effect of the ring activator, $\ce{-NH2}$, more dominant than the effect of the ring deactivator, $\ce{-NO2}$?

  • 2
    $\begingroup$ You answered the question yourself. The process is electrophilic aromatic substitution. Clearly, the resonance contribution of the amino group carries the day over the electron-withdrawing nature of the nitro group. $\endgroup$
    – user55119
    Mar 2, 2018 at 23:36
  • $\begingroup$ Yes but why? @user55119 $\endgroup$
    – dr.drizzy
    Mar 3, 2018 at 7:19
  • 1
    $\begingroup$ Whatever is going to react with the aromatic ring is an electrophile (electron-loving: NO2+, Br+, R(CO)+, etc.). Electron donation via resonance from the amino group increases electron density at the o,p-positions. Resonance of the nitro group deactivates the electron density of the ring. Therefore, the amino group will dominate. In a competitive electrophilic aromatic substitution reaction, the electrophile reacts faster with aniline than with nitrobenzene. The relative rates of reaction would be aniline>m-nitroaniline>nitrobenzene. $\endgroup$
    – user55119
    Mar 3, 2018 at 20:18

1 Answer 1


$\ce{-NH2}$ activates the ring due to its +M effect, increasing electron density in the ring at (b) and (d). On the other hand, $\ce{-NO2^{-}}$ causes de-activation because of its -M and -I effect and the electron density would be higher at (c) relatively.

Now since in activated rings, the rate of electrophilic aromatic substitution is higher than those in deactivated rings, the effect of $\ce{-NH2}$ as ortho-para director is seen more during product formation and not of $\ce{NO2^{-}}$ majorly. The electrophile will attack at (c), but since its rate will be quite slow, it would not form the major product.

  • $\begingroup$ What about the -M effect of $\ce{-NO2}$? $\endgroup$
    – dr.drizzy
    Mar 3, 2018 at 7:37
  • $\begingroup$ It would just not take place because $\ce{-NO2^{2-}}$ is highly unstable. $\endgroup$
    – Tanuj
    Mar 3, 2018 at 7:40
  • $\begingroup$ $\ce{-NO2}$ is a very strong ring deactivator, stronger deactivator than $\ce{-C=O}$ even, right? @Tanuj $\endgroup$
    – dr.drizzy
    Mar 3, 2018 at 7:44
  • $\begingroup$ Yup ! The strongest I would say. $\endgroup$
    – Tanuj
    Mar 3, 2018 at 7:45
  • $\begingroup$ @raajsuriya You see , $\ce{-NO2^{-}}$ is strongly deactivating , so it decreases rate of electrophilic aromatic substitution. $\endgroup$
    – Tanuj
    Mar 3, 2018 at 7:50

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