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Carboxylic acid is much more acidic than phenol ($\mathrm{pK}_\mathrm{a}$ difference of ~6). I wonder why? In terms of resonance, there are 5 resonance structures for phenol, and only 2 in carboxyl group.

My thought is 4 out of 5 structures in phenol are minor (lone pair electrons at carbon, less stable), compared to the equivalent resonance in carboxyl group. Is that really the sole answer to the large $\mathrm{pK}_\mathrm{a}$ difference? I am not so convinced.

Although these resonance (4/5) structures are minor in phenol, I think the combined delocalization effect should make it more acidic than a carboxylic acid.

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    $\begingroup$ The facts are what they are. Carboxylic acids are more acidic than phenol. If I'm a negative charge, perhaps I'm happier on two electronegative oxygens rather than one oxygen and three carbons. $\endgroup$
    – user55119
    Jul 22, 2020 at 14:48
  • $\begingroup$ This question is not exactly a duplicate. OP is asking why there is more effective delocalisation and some practical proof for this. $\endgroup$ Jul 24, 2020 at 9:00

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Since you haven't mentioned the carboxylic acid to be measured, I assume the carboxylic acid to be $\ce{HCOOH}$. It has a $\mathrm pK_\mathrm a$ of $3.75$

Now, as you mentioned, phenol (or carbolic acid) has a $\mathrm pK_\mathrm a$ of $10.0$. There is a difference of about $6$ in the two values. This is because of the effective delocalization that takes place in formic acid compared to that of phenol. The best way to check for this is by computing the charges on the atoms in their respective conjugate bases and then check the stabilities of the two.

So first, we check the delocalization on the phenoxide ion.

The Mulliken partial charges on the atoms ($\ce{C-1}$ as carbon connecting to oxygen and we move counter-clockwise. $\ce{H-2}$ signifies the hydrogen bonded to $\ce{C-2}$)

$$ \begin{array} {c} \begin{array}{lr} \text{Atom} & \text{Charge} \\ \hline \ce{C-1} & 0.307 \\ \ce{C-2} & -0.246 \\ \ce{C-3} & -0.204 \\ \ce{C-4} & -0.243 \\ \ce{C-5} & -0.204\\ \ce{C-6} & -0.246 \\ \end{array} \begin{array}{lr} \text{Atom} & \text{Charge} \\ \hline \ce{H-1} & 0.103 \\ \ce{H-2} & 0.100 \\ \ce{H-3} & 0.089 \\ \ce{H-4} & 0.100 \\ \ce{H-5} & 0.103\\ \ce{O} & -0.659\\ \end{array} \end{array} $$

Now, we check the delocalization on the formate ion,

$$ \begin{array}{lr} \text{Atom} & \text{Charge} \\ \hline \ce{C} & 0.363 \\ \ce{H} & -0.083 \\ \ce{O-1} & -0.639 \\ \ce{O-2} & -0.639 \\ \end{array} $$

Comparing the two, we can see that the more effective delocalization has taken place in formic acid since the net charge on an individual oxygen atom is less ($-0.659$ in phenoxide ion and $-0.639$ in the formate ion)

Due to this, we can say that the net effective charge is much better delocalized in formic acid which means that the conjugate base is better stabilized. Even though there were five possible resonance structures for the phenoxide ion, there do not contribute to the final resonance hybrid as well as the two equivalent resonance structures for the formate ion.

Since order of conjugate base stability is directly proportional to acidity. Hence formic acid would be more acidic than phenol.

Note: These values were calculated on WebMO with geometry optimization using Gamess DFT B3LYP/3-21G

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