1
$\begingroup$

Why is 4-hydroxyaniline more basic than 2-hydroxyaniline? Hydroxyl is an electron donating group, and their resonance structures suggest that they both isolate a negative formal charge on the carbon that is bonded to the amine group (which is also an electron donating group) so it would be more unstable. However, there is a clear difference in basicity between the two, the former having a pKb of approximately 8.53 while the latter has a pKb of 9.28. What accounts for this huge difference?

I also have another question. 3-hydroxyaniline avoids putting the negative formal charge on the carbon that is bonded to the amine group, and it has a pKb of 9.83. This makes sense because that means 3-hydroxyaniline is more stable in terms of resonance and thus its conjugate acid would want to liberate more protons than the other two, rendering 3-hydroxyaniline less basic than the other two. However, aniline has a pKb of 9.37, which means that it is more basic than 3-hydroxyaniline. However, hydroxyaniline has a higher electron density than aniline so shouldn't it be more basic than aniline?

|improve this question
$\endgroup$
  • $\begingroup$ This is just my own conjecture: Perhaps, the fact that the two groups are so close together in 2-hydroxaniline allows hydrogen bonds to form between the lone pair on the nitrogen and the hydrogen on the oxygen. Thus, this decreases the basicity of the nitrogen and it is less able to donate the electron pair to acids. $\endgroup$ – Tan Yong Boon Dec 31 '17 at 2:42
  • $\begingroup$ There could be intramolecular h-bonding effects, a simpler (easier to accept) answer would be based on inductive effects. OH is excellent at pi donation but is sigma withdrawing. A 2-hydroxyl group is more electron-withdrawing than a 4-hydroxyl group. (Or less electron-donating, if you want to be precise.) This also explains why the 3-hydroxy compound is less basic: resonance donation to the carbon bearing the amine is lower, so the inductive withdrawal wins. Btw from a nomenclature perspective I think it’s aminophenol, not hydroxyaniline, but that’s not really important. $\endgroup$ – orthocresol♦ Dec 31 '17 at 4:21
  • $\begingroup$ @orthocresol But if we were to subscribe to the "closer to the OH- to the NH2-, the less basic it is" argument, then how would we explain that m-aminophenol is less basic than p-aminophenol? $\endgroup$ – Tan Yong Boon Dec 31 '17 at 4:34
  • $\begingroup$ @TanYongBoon how do you explain that phenol reacts with electrophiles at o,p but not meta? Same concept, just a slightly different context. It’s already in OP’s post, actually: in the resonance forms for 3-hydroxyaniline, the formal negative charge never resides on the carbon with the amino group. You can quantify these things, e.g. Hammett parameters. $\endgroup$ – orthocresol♦ Dec 31 '17 at 4:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.