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Recently the other week, I was asked the following question:

Would you expect 4-aminophenol to have a higher or lower pKa than 4- bromophenol when measured in water?

And I thought about this on the way home, and decided to try and justify this using resonance.

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My feeling was that Nitrogen couldn't handle 4 bonds and an extra lone pair on its atom, so the negative charge won't leave the benzene ring, whereas the Bromine could handle the double bond (I might have been mistaken with the negative charge on the Bromine - if so, do let me know please.)

Using this, I thought that this is the reason why 4-aminophenol would have a higher pKa than 4-bromophenol as its resonance form is not as extended as Bromine's, which is at a lower energy level to be able to handle the extra negative charge, plus all the carbons in the ring have 4 bonds as normal.

Is this a correct answer and justification?

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    $\begingroup$ Forget about octet expansion, it's a myth. The answer is simpler than that, anyway. NH2 donates electrons to the ring, whereas Br withdraws electrons. The phenoxide conjugate base is stabilised by donating electrons to the ring, so it's much easier if the other substituent helps it out by taking electrons away (Br), and it's much harder if the other substituent fights back instead (NH2). $\endgroup$ – orthocresol Jun 9 at 22:18
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Answer to this question is, as orthocresol pointed out, simpler than what you think. As you know, $\ce{-Br}$ is deactivate the ring (even though it is o,p-directing) while $\ce{-NH2}$ is great ring activator. What that means is regardless of the mesomeric effect, compared to $\ce{-H}$, $\ce{-Br}$ is overall electron withdrawing and $\ce{-NH2}$ is electron donating (remember, $\ce{-NH2}$ is a Lewis base).

Said that, I'd say basically, $\mathrm{pH}$ of a phenol ($\mathrm{p}K_\mathrm{a} = 9.95$) is depended on how stable phenolate ion is after it donated $\ce{H+}$. Electron withdrawing groups stabilize phenolate ion (e.g., $\ce{-NO2}$ group, p-nitrophenol: $\mathrm{p}K_\mathrm{a} = 7.1$) while electron donating groups destabilize it (e.g., $\ce{-OCH3}$ group, p-methoxyphenol: $\mathrm{p}K_\mathrm{a} = 10.2$). Thus, one can expect that p-bromophenol should have a lower $\mathrm{p}K_\mathrm{a}$ than that of phenol (meaning more acidic than phenol) while p-aminophenol should have a higher $\mathrm{p}K_\mathrm{a}$ than that of phenol (meaning more basic than phenol). This is true in reality. Actual $\mathrm{p}K_\mathrm{a}$ values of p-bromophenol and p-aminophenol are $9.34$ and $10.3$, respectively ($9.34 \lt 9.95 \lt 10.3$).

Sources for $\mathrm{p}K_\mathrm{a}$ values: Evans pKa Table and Wikipedia.

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In this situation, considering the distance between the $\ce{-OH}$ group and the substituents we can ignore inductive effects and focus on the mesomeric ones. $\ce{-Br}$ has a $+M_s$ effect, same as the $\ce{-NH2}$ group. To make an $\ce{-OH}$ group more acidic we need electron withdrawing groups as the slight positive charge built on the oxygen atom will facilitate the proton leaving. In order to find the more acidic molecule we look for the one with the most electron withdrawing substituent(or the least electron donating in this case). Consider the strength of the substituents' conjugation with the ring. Bromine is a huge atom, this difference in atomic radius compared to carbon gives their orbitals a hard time overlapping(I'm simplifying here for convenience's sake) and consequently a weaker $+M_s$ effect than $\ce{-NH2}$. Considering that, we can argue that 4-bromophenol has a lower pKa(is more acidic) than 4-aminophenol. And indeed, a quick google search validates my answer, the pKa of 4-bromophenol is 9.17 and the pKa of 4-aminophenol is 10.30 for the $\ce{-OH}$ group.

Sources: https://pubchem.ncbi.nlm.nih.gov/compound/4-bromophenol#section=Dissociation-Constants https://en.wikipedia.org/wiki/4-Aminophenol

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[sorry for posting this as an answer, but it isn't possible to include an image when posted as a comment]

Do we also need to consider hydrogen bonding effects in the conjugate base, since the measurement is in water? Presumably forming the phenolate would increase the basicity of the amine through the inductive effect, and this could be stabilized by hydrogen bonding to solvent. I have no idea the strength of this effect, but thought it worth considering.

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  • $\begingroup$ "it isn't possible to include an image when posted as a comment": this isn't true. One can upload an image via the uploader within an answer field, copy obtained URL and attach inline link to the image using [<image name>](<URL>) syntax (discard the answer afterwards). $\endgroup$ – andselisk Jun 10 at 7:26
  • $\begingroup$ Or you can ask a new question by clicking Ask Question. Include a link to this question if it helps to provide context. $\endgroup$ – Loong Jun 10 at 10:22
  • $\begingroup$ My bad, didn’t realise $\endgroup$ – PCK Jun 10 at 11:58

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