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Which of these compounds - anilinium ion and para-fluoro anilium ion - is more acidic?

I thought the answer would the be first one since $\ce{-F}$ is an ortho-para activating group, so the electron density at the carbon at which $\ce{-NH3+}$ is attached would be more. I know that $\ce{-F}$ has a strong inductive effect, but since this is para with respect to $\ce{-NH3+}$, I thought I could say +M dominates. However the answer given was the second one. Why is this so and when can I neglect the inductive effect of halogen substituents in benzene to favour the mesomeric effect caused by them?


1 Answer 1


The best way to compare acidities of organic compounds is by drawing out the conjugate bases and figuring out which one is more stable. Molecules with more stable conjugate bases are more acidic, so doing the same here:

The conjugate bases are aniline and para-fluoro aniline respectively. The $\ce{-NH2}$ group shows +M (mesomeric effect), so on drawing out the resonance structures we find that the second resonance structure is more stable as the resulting negative charge is inductively withdrawn by fluorine group.

For halogens the inductive effect dominates over their positive mesomeric effect, especially in the case of fluorine (because it is small and the lone pairs experience a high effective nuclear charge,in case of large atoms like chlorine, bromine and iodine, the outer electrons are held weakly)

  • $\begingroup$ So the only place I can consider the +M effect is if I have electrophilic substitution happening? $\endgroup$
    – Reya
    Apr 26, 2017 at 10:54
  • 6
    $\begingroup$ The halogens are only weak activating groups,so +M effect can be considered in electrophilic substitutions. In general their inductive nature is more dominant than their ability to show +M effect $\endgroup$ Apr 26, 2017 at 11:01
  • $\begingroup$ But the inductive effect is virtually non-existent for the 3rd carbon atom and beyond, whereas the +M effect on o/p would still be present. $\endgroup$
    – xasthor
    May 3, 2017 at 6:56
  • 1
    $\begingroup$ @xasthor if you draw the second resonance structure,the negative charge will appear directly on the carbon bonded to fluorine. Here inductive effect is dominant. $\endgroup$ May 5, 2017 at 3:36

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