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2-chloro-2-phenylpropionic acid and phthalic acid

Between 2-chloro-2-phenylpropionic acid and hydrogen phthalate, which is the more acidic compound?

I'd say the $\ce{H}$ on the second one is more acidic as the $\ce{-COOH}$ group is directly attached to the aromatic group. Is that correct?

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    $\begingroup$ Two negative charges in the same molecule is destabilizing. $\endgroup$ – ron Sep 25 '16 at 15:25
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In the second molecule, the charge on the carboxylate ion is not stabilised through resonance, so we can rule out any resonance stabilisation. Moreover, the fact that the benzene ring is in fact electron donating within the $\pi$ system doesn't help either.

More importantly, however, (as it was pointed out to me by @orthocresol and @hippalelectryon in chat) one can expect intramolecular hydrogen bonding.

Moreover, after the second deprotonation, the second molecule would have two negative charges in close proximity to each other, and this, again, is obviously destabilising.

A combination of these factors would make the second molecule a weaker acid.

In the first case, however, the inductive electron withdrawing effect would increase the acidity, (and the benzene ring is far enough to not have an effect).

Thus, I would say it is the first compound that is more acidic.


@Hippalectryon used a software to compute a pKa of 3.87 for the first one, and Wikipedia records 5.51 for the second one, which confirms my claim.

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  • $\begingroup$ You could include the computed value for the first one and the measured value for the second one... $\endgroup$ – DHMO Sep 25 '16 at 15:23
  • $\begingroup$ @user34388 good idea! feel free to edit the answer and add that in :) $\endgroup$ – getafix Sep 25 '16 at 15:29
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As a very, very general rule, the two $\mathrm{p}K_\mathrm{a}$ values of a diacid will differ by a value of at least $1$ (likely more). Take for example phosphoric acid and aspartic acid whose $\mathrm{p}K_\mathrm{a}$ values are the following:

$$\begin{array}{ccccc}\hline \text{acid} & \mathrm{p}K_\mathrm{a,1} & \mathrm{p}K_\mathrm{a,2} & \mathrm{p}K_\mathrm{a,3} & \text{Source}\\ \hline \ce{H3PO4} & 2.148 & 7.198 & 12.319 & [1]\\ \text{aspartic acid} & 1.99 & 3.90 & 9.90 & [2] \\ \hline \end{array} $$

From that alone we would expect the monoacid to have a higher $\mathrm{p}K_\mathrm{a}$ than the diacid. And there are more contributing factors, as partly noted in getafix’s answer:

  • Chloroacetic acid already has a lower $\mathrm{p}K_\mathrm{a}$ value than benzoic acid ($2.9$ versus $4.2$) implying that the molecule on the left (which is pretty similar to chloroacetic acid) has a lower one than the right-hand one (closer to benzoic acid).

  • A chlorine atom is electron-withdrawing ($-I$). A non-conjugating phenyl ring would also be slightly electron-withdrawing but chances are that the two carboxylic acid groups don’t deviate too strongly from planarity making the general effect more of a $+M$ one.

  • A (very strong) cyclic hydrogen bond can form between the two acid groups on the right-hand molecule leading to a strong stabilisation of the monoanion.

  • The two anions would even be in very close proximity, which isn’t good (they are better separated in aspartic acid).


Sources:

[1]: https://en.wikipedia.org/wiki/Phosphoric_acid

[2]: http://academics.keene.edu/rblatchly/Chem220/hand/npaa/aawpka.htm

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@Lydia Hernandez. 2-chloro-2-phenylpropionic acid is certainly more acidic than hydrogen phthalate. The reason is very simple, A -COOH group has a well defined conjugated system of its own. It is because of this conjugated system the acidic behavior of this group is defined. On the other hand whenever a -COOH group is attached to the benzene ring directly it (benzene ring) disturbs the conjugation of -COOH group, thus the leaving of H becomes difficult. This disturbance is known as "cross Conjugation". You can observe the same with the help of following figures - enter image description here enter image description here You can verify the answer via pH or pKa values.

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