As far as I know, many electrophilic aromatic substitution reactions fail to undergo the usual way in aniline. This is due to $\ce{-H2N}$ group being basic (pKa of $\ce{NH3}$ = 38), even it is stabilized by the aromatic ring (resonance). Therefore, it often acts as a Lewis / Bronsted base (becoming a deactivating group), rather than being an activating group. Therefore, reactions like Friedel-Crafts reactions, nitration, sulphonation etc. will not undergo, unless the $\ce{-H2N}$ undergoes acylation.

e.g. Failure of Friedel-Crafts alkylation:

I am pondering why phenol does not behave the same way as aniline does. In my mind, fast proton transfer can happen during nitration and sulphonation (pKa of phenol = 10). It can also behave as Lewis Base and attack AlCl3 (presence of lone pair electrons). (or any other activating groups that the aromatic carbon is directly bonded to an atom that has lone pair electron)

But it seems the truth is that phenol just behaves as a strong activating group. What is wrong with my deductions?


This post says phenol actually behaves the same as aniline (according to the first sentence from OP), echoed by this website.

This website and Organic Chemistry 9th edition by Wade Jr. only says aniline fails, but not mentioning phenol.

  • $\begingroup$ Phenol can protonate on oxygen but the protonation is reversible. $\endgroup$ – user55119 Aug 7 '20 at 19:56
  • $\begingroup$ @user55119 I know the -H2N in aniline is much more basic. But e.g. nitration, even with leveling effect, pKa of H3O+ is 1.7 (highest among all acids in solution), compared to 10 in phenol, so I think it is still 10^8 times more favorable for protonation of phenol, which is huge. Please correct me if I am incorrect $\endgroup$ – 234ff Aug 8 '20 at 1:56
  • $\begingroup$ Are you saying that the oxygen of phenol is 10^8 times more basic than the nitrogen in aniline? if so, I think not. Is ammonia more basic than water? $\endgroup$ – user55119 Aug 8 '20 at 3:33
  • $\begingroup$ I know aniline is much more basic, so the protonation is irreversible. I mean in the solution of nitration (conc. HNO3 + conc. H2SO4), H3O+ is present, which has a pKa of 1.7. When pKa = 1.7 compared to pKa of phenol = 10.0, the equilibrium favors the phenol being protonated by around 10^8 times, which I think can be said irreversible too. This causes me to wonder why phenol does not behave as aniline in this case. Sorry for causing misunderstanding $\endgroup$ – 234ff Aug 8 '20 at 3:47
  • 1
    $\begingroup$ Compare pKa of H2O (15.7) with phenol (10). H3O+ (-1.7 !!!!) with PhOH2+, whatever value it has. You are comparing apples and oranges. $\endgroup$ – user55119 Aug 8 '20 at 14:07

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