How do I theoretically predict the acidity order of the following para-halophenols?
(experimental $\mathrm{p}K\mathrm{a}$ values beside each)

  • parafluorophenol (9.92)
  • parachlorophenol (9.41)
  • parabromophenol (9.17)
  • paraiodophenol (9.30)

Experimental acidic strength order: parabromophenol > paraiodophenol > parachlorophenol > parafluorophenol

At first, I thought of simply using the -I effect to determine the stability of these phenol derivatives. Since -I effect is shown best by fluoro group, therefore charge delocalisation would be best in the fluoro derivative's anion. But then at the same time, there's +M effect of fluoro group too. This answer emphasises on the dominance of inductive effect of chloro group over its mesomeric effect. Is this applicable to fluorine too?

I also thought of intermolecular hydrogen bonding that would be strongest in parafluorophenol decreasing its acidity.

Quora answers here associate the higher acidity of 4-chlorophenol to the presence of vacant d- orbitals in chlorine atom, but "d-orbitals'" usage by a third period element is debatable.


1 Answer 1


I think that, this acidity order can be justified by considering a combined effect of +R effect, -I effect and intermolecular hydrogen bonding (where it is dominant).

For 4-fluorophenol, the +R effect of $\ce F$ is significant but weaker than -I effect. The electron pair on the $\ce{2p}$ orbitals of $\ce F$ can easily delocalise with the benzene ring because of the stronger $\ce{2p\pi -2p\pi}$ overlap, which makes $\ce {+R} $ effect significant . That's why it increases the electron density on the carbon near the $\ce{-OH}$ group to such an extent which decreases its ability to donate protons, greatly decreasing its acidity.
Considering intermolecular hydrogen bonding also, the acidity of the compound decreases further due to strongest intermolecular hydrogen bonding between the halogens.

For 4-chlorophenol, the +R effect is negligible due to much weaker $\ce{3p\pi -2p\pi}$ overlap and thus the $\ce{-I}$ effect becomes more prominent. But $\ce {Cl}$ is electronegative enough to form strong intermolecular hydrogen bonding($\ce{HO-C6H4-Cl...H-O-C6H4-Cl...)}$. That's why, it becomes difficult to remove hydrogen atoms from $\ce{-OH}$ group, decreasing its acidity.

For 4-bromophenol and 4-iodophenol, there is no strong hydrogen bonding due to much less electronegative $\ce{Br}$ and $\ce{I}$ atoms, and also +R effect is ineffective due to very weak $\ce{4p\pi -2p\pi}$ and $\ce{5p\pi -2p\pi}$ overlap. Only -I effect is acing for these two compounds in favour of acidity. As -I effect of $\ce{Br}$ is larger than $\ce{I}$, the acidity of p-bromophenol is greater than the p-iodophenol.

  • $\begingroup$ What about HOMO/LUMO iodine as we go down the group the HOMO of the halogen increases in energy, thus decreasing the energy gap. Of course there is something wrong with my reasoning as the pKa values will not match the order through HOMO/LUMO,, but what's wrong ? $\endgroup$ Commented Feb 26, 2018 at 9:44
  • $\begingroup$ @Avatar Shiny,Can you please write your logic properly in the comment box. I am not getting it properly what you are thinking about HOMO-LUMO . $\endgroup$
    – Soumik Das
    Commented Feb 26, 2018 at 10:32
  • $\begingroup$ @Soumik what I mean is that, as we go down the the group the halogens become softer their HOMO's are less strongly attracted down the group this leads to an increase in the HOMO's energy . For a good molecular orbital to form the HOMO/LUMO gap must be small. LUMO's are high in energy as they are unoccupied. So a high energy HOMO helps in reducing the gap . So by HOMO/LUMO we are getting an opposite trend to the observed acidity order of $\pu{+R}$ effect . So how is it that while seeing the nucleophilicity order we see iodene's HOMO energy and ignore the size of orbitals and over here $\endgroup$ Commented Feb 26, 2018 at 11:35
  • $\begingroup$ We are supposed to see the size of orbitals $\endgroup$ Commented Feb 26, 2018 at 11:35
  • 4
    $\begingroup$ I effect is stronger than R effect for all halogens. Therefore all halogens are deactivating. $\endgroup$
    – Archer
    Commented Mar 3, 2018 at 15:50

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