The Wikipedia page on inductive effects states that the order of inductive effects of halogens is as follows:

$$\ce{-F} > \ce{-Cl} > \ce{-Br} > \ce{-I}$$

From this, and from the fact that electronegativity of fluorine is higher than that of chlorine, I guessed that the acidity of p-fluorophenol is more than p-chlorophenol.

But the $\mathrm{p}K_\mathrm{a}$ values are:

  • p-fluorophenol: 9.8
  • p-chlorophenol: 9.4

Where did I go wrong? Why does this discrepancy occur? Detailed explanations are very much welcome.

  • 3
    $\begingroup$ The inductive effect is not very strong when the halide is so far away from the phenol group, so you have to look at the resonance effect. The halides all donate electron density. Which do you think is the best at donating electron density? $\endgroup$ Oct 12, 2015 at 15:43
  • $\begingroup$ @orthocresol So does the +R effect for all halides at the para position become more significant than the -I effect? $\endgroup$ Oct 13, 2015 at 2:07
  • 2
    $\begingroup$ Well, it depends on the context. The fact that the halides are deactivating overall should tell you that the -I outweighs the +R in general. My point is that, the decreasing trend in -I is not important in this question because 1) it's too many carbons away, so the trend is very weak; 2) as you said, it obviously leads to the wrong prediction regarding the $\text{p}K_\text{a}$'s. And you're not answering my question. $\endgroup$ Oct 13, 2015 at 17:00
  • $\begingroup$ The less electronegative halides would more easily donate their lone pairs. $\endgroup$ Oct 13, 2015 at 18:52
  • 1
    $\begingroup$ Not just that, but also the overlap with the benzene ring. Think about the Lewis acidity of the boron trihalides. If you're still confused, here's the answer: chemistry.stackexchange.com/questions/10288 Replace "the central boron atom" with "the benzene ring" and there's the answer to your question. $\endgroup$ Oct 13, 2015 at 19:06

3 Answers 3


This is due to the fact that we also consider +M effect . In fluorine it’s 2p orbital that participated in overlapping but in chlorine it’s 3 p orbital. 3p of chlorine does not overlap efficiently with 2p of benzene carbon so +m decreases in chlorine thus acidic nature increases


All your observations are correct.But one thing you seem to be missing is the presence of vacant d orbitals in Chlorine.I am sorry but I couldn't find pictures as such but I am sure you will understand after drawing out the resonance structures.

So when the lone pair on oxygen is delocalized and is at para position,then the d orbital of chlorine can "take up" and stabilize the compund.Flourine can't do the same.

  • 1
    $\begingroup$ This is the picture you want. i.imgur.com/p3CWF47.jpg But d orbitals aren't a good explanation. $\endgroup$ Oct 12, 2015 at 16:19
  • $\begingroup$ @Karan Singh If it was at the ortho position then I would have agreed but at such a long distance it doesn't seem very accurate. $\endgroup$ Oct 13, 2015 at 2:04
  • $\begingroup$ @orthocresol I am a student myself and I recently learned this.But can you tell why d orbitals aren't a good explanation? $\endgroup$ Oct 13, 2015 at 13:44
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    $\begingroup$ chemistry.stackexchange.com/a/5242/16683 $\endgroup$ Oct 13, 2015 at 16:59

Order of -I in Halogens is −F > −Cl > −Br > −I Order of +M in Halogens is −F > −Cl > −Br > −I

So whenever halogens are in resonance (as in p-fluorophenol and p-chlorophenol), we take combined effect as: −F < −Cl < −Br < −I and consider halogens as an overall EWG.

Hence, p-chlorophenol is more acidic than p-fluorophenol.

  • 2
    $\begingroup$ This is a wrong answer, your second line contradicts itself. Halogens are considered overall as electron-withdrawing groups. Right. Then the withdrawing effect is inductive effect, so shouldn't the order of -I in halogens be considered, giving F > Cl? Think it over. Also, when you answer, including a source, from a good textbook or online resource, is a great idea. $\endgroup$
    – TRC
    Jul 13, 2021 at 14:29

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