Why acidity of p-methoxybenzoic acid is more acidic than p-hydroxybenzoic acid?

Question

I know +M effect of $\ce{OCH3}$ is more than $\ce{OH}$, but in my book (ALLEN General organic chemistry), it is given that $\mathrm{p}K_\mathrm{a}$ of para-substituted benzoic acid containing $\ce{OCH3}$ is less than that of para-substituted benzoic acid containing $\ce{OH}$:

$\mathrm{p}K_\mathrm{a}$/Benzoic acid → $4.20$

$$ \begin{array}{lcccc} \hline \text{Substituent} & \mathrm{p}K_\mathrm{a} & \textit{o-} & \textit{m-} & \textit{p-} \\ \hline \cdots \\ \ce{OMe} & - & 4.09 & 4.09 & 4.47 \\ \ce{OH} & - & 2.98 & 4.08 & 4.58 \\ \cdots \\ \hline \end{array} $$

I am in doubt whether these data wrong or correct. If correct, what can be an explanation?

Answer 1

Both -$\ce{OCH3}$ and -$\ce{OH}$ groups have exhibited two effects on the aromatic ring: (1) Electron donating resonance or mesomeric effect (+M) and (2) Electron withdrawing inductive effect (-I). For both acids in hand, the electron withdrawing inductive effect (-I) is almost same since both -$\ce{OCH3}$ and -$\ce{OH}$ groups are 4 carbons away from the acid center. However, when -$\ce{OH}$ group is attached to either the para- or ortho-position, it has a more tendency to delocalize its lone pair electrons towards the aromatic ring than that of -$\ce{OCH3}$ group (Order of activating: -$\ce{O-} \gt$ -$\ce{OH} \gt$ -$\ce{OCH3}$; for additional learning, read articles on Hammett plots).

As a result, the electron density on the carbon para to -$\ce{OH}$ substitution group increases more than that on the carbon para to -$\ce{OCH3}$ substitution group. Thus, the electron density on carbonyl carbon of carboxylic group attached to those para-carbons also increases accordingly, and consequently, the polarity of $\ce{O-H}$ of carboxylic group decreases as a result. That polarity decrease causes the reduced acidity ($\mathrm{p}K_\mathrm{a}$s are $4.58$ and $4.47$, when para-substitutions are $\ce{OH}$ and $\ce{OCH3}$, respectively) compared to benzoic acid ($\mathrm{p}K_\mathrm{a}$ is $4.20$ when para-substitution is $\ce{H}$). The $\mathrm{p}K_\mathrm{a}$s of p-hydroxybenzoic acid is higher than that of p-methoxybenzoic acid here, because as explained above, p-$\ce{OH}$ group caused reduced polarity on $\ce{O-H}$ of carboxylic group than that caused by p-$\ce{OCH3}$ group.

Answer 2

I am directly going to address the source of your confusion: $\text{+M}$ of -$\ce{OCH3}$ is more than -$\ce{OH}$. You see, you are basically saying that the $\text{+I}$ effect of $\ce{CH3}$ in -$\ce{OCH3}$ is going to push more electron density towards the ring, hence, -$\ce{OCH3}$ is more activating than -$\ce{OH}$.

But look at the bigger picture, and let's evoke Bent's rule here.

You see, due to steric reasons,the $\ce{C_{(\text{benzene})}-O-C_{(\text{alkyl group})}}$ bond angle of $\ce{OCH3}$ is going to be more than the $\ce{C_{(\text{benzene})}-O-H}$ bond angle of the -$\ce{OH}$. Hence, the percent $\mathrm{s}$-character is going to go up for the central oxygen atom in $\ce{OCH3}$ relatively (to rationalize this, think of the bond angles associated with $\mathrm{sp^3, sp^2,}$ and $\mathrm{sp}$ hybridization and their respective $\mathrm{s}$-characters)

So now, due to increased $\mathrm{s}$-character on both sides of the $\ce{OCH3}$ oxygen, you get an overall more electronegative substituent for the ring as compared to $\ce{OH}$, as the electrons displaced towards the oxygen from the methyl in $\ce{OCH3}$ would still experience a lowering in the energy of the $\sigma_\ce{(O-CH3)}$ bond. Hence, as compared to $\ce{OH}$, the ring will be activated less by $\ce{OCH3}$. I feel that you can now complete the reasoning, by checking the stability of the conjugate bases.

Note: A better qualitative reasoning could have been provided via frontier MO theory, but I chose not to evoke it here. If the OP requests, I can supply the same as well.