# Inductive effects of halogens

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.

• 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? – orthocresol Oct 12 '15 at 15:43
• @orthocresol So does the +R effect for all halides at the para position become more significant than the -I effect? – Ayan Gangopadhyay Oct 13 '15 at 2:07
• 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. – orthocresol Oct 13 '15 at 17:00
• The less electronegative halides would more easily donate their lone pairs. – Ayan Gangopadhyay Oct 13 '15 at 18:52
• 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. – orthocresol Oct 13 '15 at 19:06