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Recently I came across this statement in my textbook:

Direct attachment of groups such as phenyl or vinyl to the carboxylic acid, increases the acidity of corresponding carboxylic acid contrary to the decrease expected due to resonance effect.

On researching online, I found this statement as an answer:

Although we expect a decrease in acidity due to the resonance effect of phenyl and vinyl groups, but because of the greater electronegativity of the the $\ce{sp^2}$ hybridized carbon which is directly attached to the carboxylic carbon.

My Thoughts

Please keep in mind that the carbonyl carbon(here carboxylic carbon) is also $\ce{sp^2}$ hybridized. So $\ce{sp^2}$ hybridized carbon is directly attached to the carboxylic carbon. Therefore there is no electronegativity difference between the two carbons and as a result of this only +R effect should work and direct attachment of groups such as phenyl or vinyl to the carboxylic acid should not increase the acidity of corresponding carboxylic acid rather they should decrease its acidity due destabilization (due to electron donating nature vinyl and phenyl group by means of resonance) of conjugate base so formed.

The question

Kindly explain it theoretically by giving plausible reasons as to why the statement in my textbook is true?

Note: avoid quoting pKa values

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    $\begingroup$ There is sp3-sp2 versus sp2-sp2 difference of electronegativity differences. $\endgroup$
    – Poutnik
    Apr 9, 2020 at 2:45
  • 1
    $\begingroup$ I didn't get you? $\endgroup$
    – Chemist
    Apr 9, 2020 at 4:08
  • $\begingroup$ sp2 carbon of e.g. acrylic acid has bigger electronegativity than sp3 carbon of e.g. acetic acid, what leads to the former causing lower carboxyl electron density, compared to the latter, and bigger carboxyl acidity. $\endgroup$
    – Poutnik
    Apr 9, 2020 at 10:10
  • $\begingroup$ Look at the carboxylate ($\ce{O-C-O}$) part of the molecule. It is an analogue of the allyl system where we have replaced the two terminal carbons with oxygens. There are 4 pi electrons in the carboxylate part of the ionized acid, so the HOMO is $\ce{\psi_{2}}$ of the allyl system. Note that in this orbital there is a node at the carboxylate carbon. This node effectively disconnects substituents attached to this carbon in a pi or resonance sense. Hence any resonance interaction between a substituent attached to the carboxylate carbon and the carboxylate fragment is minimized. $\endgroup$
    – ron
    Apr 13, 2020 at 17:21
  • $\begingroup$ The title of this question is incomplete, and too general. Please add citations for the statements you are referring to. $\endgroup$
    – Martin - マーチン ♦
    May 12, 2020 at 13:50

1 Answer 1

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When the cases of carboxilic acids are considered, we must note that the conjugation of $\ce{COO-}$ group is so strong in itself, that it's conjugation with the rest of the compound is rendered relatively weak. In such cases, the inductive effect is dominant.

Let us take the example of benzoic acid. The carbon to which $\ce{COOH}$ is attached shall be named as the key-carbon $\ce{C}$1. The conjugate base would be $\ce{PhCOO-}$, and the +I(Inductive effect) of $\ce{COO-}$ group will be dominant on $\ce{C}$1. Any group attached to this compound will further increase or decrease the acidic strength of the compound as a whole, depending upon the attached group's contribution towards electron density at $\ce{C}$1.

For example, in case of para-nitrobenzoic acid, electron density at $\ce{C}$1 decreases due to $\ce{NO2-}$'s strong -R (Mesomeric) Effect(partial positive charge is developed on $\ce{C}$1 which is stabilised by $\ce{COO-}$'s +I). Hence it turns out to be a strong acid.

The opposite case is that of para-methoxybenzoic acid. The electron density at $\ce{C}$1 increases due to +R of $\ce{OCH3}$ group and makes the conjugate base unstable, hence acid weaker. This can be cross checked by the $\mathrm{p}K_\mathrm{a}$ values of above mentioned two compounds as well.

Now coming to the compounds mentioned in the question, which are phenyl and vinyl, one must note that both of these would reduce the charge density on $\ce{C}$1 because of the $\mathrm{sp^2}$ hybridised Carbon atom which contributes to the overall electro-negativity of these two groups.

Long story short, the resonance effect between $\ce{COO-}$ and the rest of the compound is suppressed by a relatively stronger Inductive effect which works mutually between $\ce{COO-}$ and the rest of the compound.

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