Why rate of aromatic nucleophilic substitution of Ortho-fluoronitrobenzene is greater than para-fluoronitrobenzene?

Kindly note question is about why rate of ortho is greater than para position. Not meta position.

This is previous year KVPY question but they had not asked reason and just asked comparison of rate.

  • $\begingroup$ I think it's probably due to the inductive effect of the nitro group. The effect is strongest at the ortho position $\endgroup$
    – S R Maiti
    Apr 27, 2021 at 14:27
  • $\begingroup$ Please do not use formatting on the title. $\endgroup$ Apr 27, 2021 at 14:28

1 Answer 1


Reaction mechanism of aromatic nucleophilic substitution is as given -

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Source : https://www.masterorganicchemistry.com/2018/08/20/nucleophilic-aromatic-substitution-nas/

The rate determining step is the formation of the C-Nu bond, which causes the development of a negative charge on the carbon atom at ortho position to leaving group (fluorine). Presence of electron-withdrawing groups helps to stabilize the negative charge, and hence increase the rate of reaction.

The -NO2 substituent exerts not only a -M effect (resonance) but also a strong -I effect (inductive effect, which pulls the negative charge away from the ring). Inductive effect decreases rapidly as the distance increases, so its effect is much more pronounced for ortho than for para (Note that the initial negative charge is developed at the ortho position). As a result orthofluoronitrobenzene has a higher rate of reaction than parafluoronitrobenzene.


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