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Question

Compare the acidic stengths between benzylammonium ion and phenol.

I first tried to remove the H+ ion from both of them and tried to compare the relative stabilities of resulting conjugate bases, but I am not sure, which one would be more stable.

The first one would be a neutral compound and the second one would have resonating structures on which the negative charge gets mostly delocalised to carbon.

I have seen somewhere on the internet that benzylammonium ion has a $\mathrm{pK_a}$ value of 9.33 while phenol has a $\mathrm{pK_a}$ value of 10. If this data is correct, then how do we explain the acidity order?

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    $\begingroup$ You could as well take an orange and a banana and ask why one winds up heavier then another. $\endgroup$ – Mithoron May 5 at 13:25
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    $\begingroup$ You can't compare two "different" compounds when there are many factors and some of them contradict each other. In such cases we only have the data to look upon. It's true that many hypothesis can be made from the data, but those are nothing but humans being humans and acting like they know what is going on. In short: Data drives theory(many times wrong) and not the other way around. $\endgroup$ – Nisarg Bhavsar May 5 at 13:43
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    $\begingroup$ The problem is that, the explanation isn't going to serve anyone. Not even you. $\endgroup$ – Nisarg Bhavsar May 5 at 13:47
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    $\begingroup$ And even if someone gave an explanation it will boil down to the fact that the the ammonium ion's inductive effect is dominant over the the resonance in the phenoxide ion "as per the data". Sitting on a chair with armrest we can't predict which is more favourable a carbanion with -I of ammonium ion and -M of benzene or an oxygen anion with -M of benzene. $\endgroup$ – Nisarg Bhavsar May 5 at 13:54
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    $\begingroup$ @BuckThorn I just did. So benzylamine has a pKa of 8.82 and benzyl alcohol has a pKa of 15.4 . Even in this case, I don't know the reason behind this. I would appreciate if answerer explains this case too. But why should this be related to the original question? $\endgroup$ – Light Yagami May 8 at 9:25
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The acid strength of each compound can be explained, but the acidity order is much more difficult to compare, because the two compounds are only remotely connected. It is misleading to conclude that the mere presence of a phenyl group somehow connects these molecules. The similarity of the pK$_a$s is likely a coincidence.

The question needs a clear definition of pK$_a$. We can ask what the pK$_a$ of phenol is, or the pK$_b$ of phenoxide. When we ask for the pK$_a$ of benzyl amine, we want the pK$_a$ of benzylammonium ion, not the pK$_a$ of benzyl amine going to C$_6$H$_5$CH$_2$NH$^-$. There is a bit of flexibility (or sloppiness) in the terminology here, and also a conflict in the values presented: the question quotes 9.33 for benzylammonium ion (I found 9.34 - no big deal), but the comment by the OP gives 8.82 for benzyl amine. The values should be identical because we know what the OP means. It appears that 8.82 is incorrect.

The difference between these molecules is far more important than any similarities. For example, one is a neutral molecule that you can put into a bottle with 6 x 10$^{23}$ more; the other is a charged piece of a molecule that you can hardly bring next to another one.

The phenyl ring has an enormous effect on the acidity of phenol, due to the resonance effect. It stabilizes the oxygen anion by more than 5 orders of magnitude, compared to, e.g., methanol (pK$_a$ = 15.5) or benzyl alcohol (15.4) or ethanol (15.9).

(Interesting that phenylethanol (pK$_a$ = 14.81) is more acidic than benzyl alcohol. This suggests that the methyl group in ethanol is more electron-donating than the benzyl group in benzyl alcohol; this agrees with the discussion in #3 below, for phenethylammonium ion.)

The phenyl ring has a much smaller effect on the amine molecule-ion; no resonance, just inductive. The pK$_a$ of benzylammonium ion (9.34) can be compared with, e.g.:

  1. methylammonium ion (pK$_a$ = 10.66) Replacing the phenyl of benzylammonium ion with H reduces the K$_a$ by a factor of 21.

This is equivalent to saying that benzyl amine is less basic (pK$_b$ = 4.66) than methylamine (pK$_b$ = 3.34); therefore the protonated benzyl amine will give up an added proton more readily than methylamine. We can conclude that methyl is more electron-donating than benzyl.

  1. ethylammonium ion (pK$_a$ = 10.71) Replacing the phenyl of benzylammonium ion with a methyl group reduces the K$_a$ by a factor of 23. Thus, ethylamine, like methylamine, is more basic than benzyl amine because methyl is more electron-donating than phenyl. (But we knew that.)

  2. Phenethylammonium ion has a pK$_a$ = 9.73. Putting another CH$_2$ group between phenyl and nitrogen reduces K$_a$ by a factor of 2.5. This indicates that a benzyl group is less electron-donating than a methyl or ethyl group, in agreement with the comparison of the inductive effect between benzylammonium vs methyl- or ethylammonium ions and benzyl alcohol vs phenylethanol.

Benzyl amine and phenol are the neutral actors in this drama, and might react nicely in a one-to-one ratio. What would be the pH in solution? The graph shows the ratio of ion concentration to neutral molecule concentration for phenol and benzyl amine, separately, over the pH range of 5 to 9. They show contrasting slopes which cross at pH 7.33. At this pH, the ionic fraction of each compound is 0.21% of the whole - it doesn’t matter what concentration you started with if everything is soluble (benzyl amine is very soluble, but phenol, mp 43 C, is soluble in water slightly less than ~0.1 M in water at room temperature). Just add the same number of moles of each neutral compound to get 0.21% reaction and pH 7.33. What a coincidence! Practically the same pH as pure water.

enter image description here

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  • $\begingroup$ I sincerely thank you for your efforts into this answer. The values that I gave were just based upon a google search and so I knew that these are not reliable values :). $\endgroup$ – Light Yagami May 12 at 8:13
  • $\begingroup$ @Light Yagami: You're welcome; my pleasure. It was a good question - it stretched the imagination. $\endgroup$ – James Gaidis May 12 at 20:26

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