12
$\begingroup$

If we take the example of salicylic acid, hydrogen bonding is present in the acid as follows:

enter image description here

Even after deprotonation, it has intramolecular hydrogen bonding as follows:

enter image description here

My question:

p-hydroxybenzoic acid ($\mathrm pK_\mathrm a = 4.54$) is less acidic than o-hydroxybenzoic acid ($\mathrm pK_\mathrm a = 2.97$), the reason being hydrogen bonding in the latter after deprotonation.

If the same reason accounts for the acidity of salicylic acid (o-hydroxybenzoic acid), then why is the H-bonding before deprotonation not considered (which should decrease the acidity of the compound)?

Improve this question
$\endgroup$
1

2 Answers 2

Reset to default
11
$\begingroup$

Some of the conformers of salicylic acid are given below, all of which are planar in configuration.$\mathrm{^{[1]}}$ enter image description here

enter image description here

Although the compound can adopt several conformations the first conformer(in the first image, the structure on the left) is of the lowest energy and has a strong intramolecular hydrogen bond. The second conformer(first image, right structure) is the conformer with the second lowest energy as described in the published paper as given below:

The molecule can adopt several conformations, although only one of them is highly populated and was assigned in IR experiments. The spectrum was recorded by Fiedler et al., $\mathrm{^{[2]}}$ in tetrachloride solution and indicated a strong intramolecular H-bond. The deviation of the OH −1 stretching frequency from that in phenol was 395 cm . The theoretically calculated deviation is 359 cm−1 at the B3LYP/6-311+G(d,p) level. The lowest-energy conformer is planar, the phenolic OH is a H-bond donor to the carbonyl oxygen of the syn carboxylic group. The =O...H distance and the =O...H–O bond angle were calculated at 176 pm and 145°, respectively. The second-most-stable conformer is higher in energy by 14.3 kJ/mol, where the phenolic OH is the H-bond donor to the syn carboxylic OH. Similar conclusions were drawn by Yahagi et al.,$\mathrm{^{[3]}}$ by interpreting the gas-phase IR frequencies of the phenolic OH.

Much more convincing is the fact that the lowest energy conformer has a fractional population distribution$\mathrm{^*}$ of nearly 99.7% which further supports the argument regarding the higher acidity of salicylic acid in comparison to its para isomer. The higher acidity of the compound thus is contributed by the first conformer. The lesser interaction of the carboxyl group’s acidic proton and the stability of the conjugate base due to hydrogen bonding thus increases the acidity of the compound.
Regarding detailed mechanism of acidity of salicylic acid you may refer this and this post in chemistry SE. You may further note that the other structures might have decreased the acidity of the compound if their population distributions would have been higher. The same has been mentioned in the published paper as given below:

In a former calculation by Nagy et al.$\mathrm{^{[4]}}$, the two conformers above were found also to be the most stable with MP2/6-31G*//HF/6-31G* energy separation of 13.7 kJ/mol and free energy difference of 12.1 kJ/mol at T = 298 K. All other conformers are much higher in free energy, supporting the estimate of Fiedler that the population of the lowest-energy form is 99.7%.

Notes and References

  1. Nagy, Peter. Competing Intramolecular vs. Intermolecular Hydrogen Bonds in Solution. Int. J. Mol. Sci. 2014, DOI: 10.3390/ijms151119562
  2. Fiedler, P.; Böhm, S.; Kulhánek, J.; Exner, O. Acidity of ortho-substituted benzoic acids: An infrared and theoretical study of the intramolecular hydrogen bonds. Org. Biomol. Chem. 2006, 4, 2003–2011. DOI: https://doi.org/10.1039/B601875K
  3. Yahagi, T.; Fujii, A.; Ebata, T.; Mikam, N. Infrared spectroscopy of the OH stretching vibrations of jet-cooled salicylic acid and its dimer in S0 and S1. J. Phys. Chem. A 2001, 105, 10673–10680. DOI: https://doi.org/10.1021/jp0126199
  4. Nagy, P.I.; Dunn, W.J, III.; Alagona, G.; Ghio, C. Theoretical studies of the 2- and 4-hydroxybenzoic acids with competing hydrogen bonds in the gas phase and aqueous solution. J. Phys. Chem. 1993, 97, 4628–4642. DOI: https://doi.org/10.1039/B601875K

    $\mathrm{*}$ Note that here fractional population distribution refers to the population of each conformer in the compound. You may find more information about it on the Wikipedia article.
Improve this answer
$\endgroup$
6
  • 2
    $\begingroup$ So, if I got this right, the answer here is experimental fact? Or did I miss the theoretical explanation? $\endgroup$
    – Cathartic Encephalopathy
    Commented Jun 2, 2021 at 13:23
  • 1
    $\begingroup$ @Buraian Yes here the answer which I have posted only contains experimental facts about the population distribution of the conformers while the mechanism of acidity that has been explained briefly is theoretical. $\endgroup$
    – Rishi
    Commented Jun 2, 2021 at 14:23
  • 1
    $\begingroup$ I don't understand your argument about acidity of the ortho-isomer (which is the main question of OP). What does lesser interaction of the acidic proton have to do with higher acidity of ortho vs. para? Because there is no interaction in para-isomer either. Then you have mentioned the stability of the conjugate base due to H-bonding: does this mean that the conjugate base is more stable than the acid? There is H-bond in the acid as well, as you show in the first conformer. Also, is this part from a paper, or are you concluding it from the paper? $\endgroup$
    – S R Maiti
    Commented Jun 2, 2021 at 17:51
  • 2
    $\begingroup$ @ShoubhikRMaiti (1)The question of OP is that at the end Why the hydrogen bonding is to be considered(or why not) even before the deprotonation? (not about ortho vs para)(2) The block quoted texts are from a openly accesssable review paper(ref 1).(3) By lesser interaction I mean it is not involved in Hydrogen Bonding as in conformer 2(Refer statement 1 of this comment) $\endgroup$
    – Rishi
    Commented Jun 2, 2021 at 18:11
  • $\begingroup$ @ShoubhikRMaiti I apologise for not reading your comment properly, I missed the last 2 questions in your comment. (1) ” does this mean that the conjugate base is more stable than the acid?” The first conformer’s high population results from the fact that it is the most stable conformer but I do not know whether the conjugate base is less stable than the acid but I think the data for that should be there in one of the other references which are not openly accessible. (Continued... $\endgroup$
    – Rishi
    Commented Jun 4, 2021 at 16:03
6
$\begingroup$

The other answer has given the major conformer of the protonated form of the o-hydroxybenzoic acid, but I believe there is a part of your question that still remains to be answered. You asked why the hydrogen bond makes the ortho-isomer more acidic than the para-isomer, when the hydrogen bond is already present in the protonated form of the ortho-isomer.

For an explanation of this, let's consider the ionization equilibrium of the ortho-acid:

deprotonation/protonation equilibrium of o-hydroxybenzoic acid

(I am considering only the major conformer as mentioned in the other answer)

Now, there is hydrogen bond in both the protonated and deprotonated form. But, consider the strength of the hydrogen bond in both cases. In the protonated acid, the H-bond donor is a neutral $\ce{>C=O:}$. In the deprotonated state, the H-bond donor is an oxygen of a carboxyl anion. The negative charge is delocalised over both of the oxygen, so the H-bond donor $\ce{O}$ in this case has some extra electron density compared to the neutral $\ce{>C=O}$. So, it can form a stronger H-bond.

This means the H-bond is stronger in the deprotonated state than in the protonated state of the o-hydroxybenzoic acid. So, the presence of H-bond still makes the ortho-isomer a stronger acid (than the para-isomer where there is no intramolecular hydrogen bond), even if the H-bond can exist in both protonated and deprotonated forms.

Improve this answer
$\endgroup$
2
  • 3
    $\begingroup$ This answer imo seems more easily understandable and simpler, answers should always strike a balance in accuracy and accessibility. (Not saying this is inaccurate) $\endgroup$
    – napstablook
    Commented Jun 27, 2021 at 9:10
  • 1
    $\begingroup$ It kind of would be fun to look at this with calculations, see if it actually get stronger. Well, "see", more like guess the electronic structure, do some QTAIM, and guess some more... anyway, I've mentioned the number I would look at here: How to identify hydrogen bonds and other non-covalent interactions from structure considerations? $\endgroup$
    – Martin - マーチン ♦
    Commented Feb 2, 2022 at 22:24

Not the answer you're looking for? Browse other questions tagged or ask your own question.