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

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?

## 2 Answers

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.

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.