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First off, it is a very small difference in $\mathrm{p}K_a$ that you are asking about. In fact, I would say that the two compunds have similar acidity. It is often difficult to explain such small differences, but what follows is a general approach that can be used to probe the question and to try and answer questions related to equilibria.

In this question we are comparing the effect of two different substituents on an equilibrium. To perform this comparison we must look at the effect of the substituent on both the starting material (the substituted phenol) and the product (the substituted phenoxide anion). If the substituent stabilizes the phenol it will tend to shift the equilibrium towards the phenol side making the phenol less acidic; if the substituent stabilizes the phenoxide it will tend to shift the equilibrium to the phenoxide side making the phenol more acidic.

From your other question we see that the p-methoxy group destabilizes the phenoxide anion to a greater extent than the p-methyl group. Based on this alone, we would say the substituent effect on the phenoxide side of the equilibrium should be such that the p-methoxy substituent pushes the equilibrium more strongly to the phenol side than the p-methyl group will. This would tend to make p-methoxyphenol less acidic than p-methylphenol.

Looking at the phenol side of the equilibrium, both the methoxy and the methyl groups are electron releasing through resonance, with the methoxy having a much stronger effect. The phenolic $\ce{OH}$ group is also strongly electron releasing through resonance. Perhaps having two strongly electron releasing groups (hydroxyl and methoxy) attached to the aromatic ring causes some destabilization due to electron repulsion associated with the high electron density in the ring. Certainly such an effect would be greater with methoxy than methyl. If true, then we could argue that the p-methoxy substituent also destabilizes the starting phenol (more then the methyl group) and this would tend to push the equilibrium to the product side making p-methoxyphenol more acidic than p-methylphenol.

If the substituent effects on the phenol and the phenoxide are comparable, then they would cancel and we might expect p-methoxyphenol and p-methylphenol to have comparable acidities and if the substituent effects cancel we would also expect both phenols to have $\mathrm{p}K_a$'s similar to phenol ($\mathrm{p}K_a$=10), which they do.

The approach used here is sound. I think the explanation of the substituent effects on the phenoxide are also sound. The explanation of the substituent effects on the phenol side is (admittedly) weak. Nonetheless, this explanation does correctly explain why p-methoxyphenol and p-methylphenol have comparable acidities and why these acidities are close to the acidity of phenol.

First off, it is a very small difference in $\mathrm{p}K_\mathrm{a}$ that you are asking about. In fact, I would say that the two compounds have similar acidity. It is often difficult to explain such small differences, but what follows is a general approach that can be used to probe the question and to try and answer questions related to equilibria.

In this question we are comparing the effect of two different substituents on an equilibrium. To perform this comparison we must look at the effect of the substituent on both the starting material (the substituted phenol) and the product (the substituted phenoxide anion). If the substituent stabilizes the phenol it will tend to shift the equilibrium towards the phenol side making the phenol less acidic; if the substituent stabilizes the phenoxide it will tend to shift the equilibrium to the phenoxide side making the phenol more acidic.

From your other question we see that the p-methoxy group destabilizes the phenoxide anion to a greater extent than the p-methyl group. Based on this alone, we would say the substituent effect on the phenoxide side of the equilibrium should be such that the p-methoxy substituent pushes the equilibrium more strongly to the phenol side than the p-methyl group will. This would tend to make p-methoxyphenol less acidic than p-methylphenol.

Looking at the phenol side of the equilibrium, both the methoxy and the methyl groups are electron releasing through resonance, with the methoxy having a much stronger effect. The phenolic $\ce{OH}$ group is also strongly electron releasing through resonance. Perhaps having two strongly electron releasing groups (hydroxyl and methoxy) attached to the aromatic ring causes some destabilization due to electron repulsion associated with the high electron density in the ring. Certainly such an effect would be greater with methoxy than methyl. If true, then we could argue that the p-methoxy substituent also destabilizes the starting phenol (more then the methyl group) and this would tend to push the equilibrium to the product side making p-methoxyphenol more acidic than p-methylphenol.

If the substituent effects on the phenol and the phenoxide are comparable, then they would cancel and we might expect p-methoxyphenol and p-methylphenol to have comparable acidities and if the substituent effects cancel we would also expect both phenols to have $\mathrm{p}K_\mathrm{a}$'s similar to phenol ($\mathrm{p}K_\mathrm{a}=10$), which they do.

The approach used here is sound. I think the explanation of the substituent effects on the phenoxide are also sound. The explanation of the substituent effects on the phenol side is (admittedly) weak. Nonetheless, this explanation does correctly explain why p-methoxyphenol and p-methylphenol have comparable acidities and why these acidities are close to the acidity of phenol.

First off, it is a very small difference in $\mathrm{p}K_a$ that you are asking about. In fact, I would say that the two compunds have similar acidity. It is often difficult to explain such small differences, but what follows is a general approach that can be used to probe the question and to try and answer questions related to equilibria.

In this question we are comparing the effect of two different substituents on an equilibrium. To perform this comparison we must look at the effect of the substituent on both the starting material (the substituted phenol) and the product (the substituted phenoxide anion). If the substituent stabilizes the phenol it will tend to shift the equilibrium towards the phenol side making the phenol less acidic; if the substituent stabilizes the phenoxide it will tend to shift the equilibrium to the phenoxide side making the phenol more acidic.

From your other question we see that the p-methoxy group destabilizes the phenoxide anion to a greater extent than the p-methyl group. Based on this alone, we would say the substituent effect on the phenoxide side of the equilibrium should be such that the p-methoxy substituent pushes the equilibrium more strongly to the phenol side than the p-methyl group will. This would tend to make p-methoxyphenol less acidic than p-methylphenol.

Looking at the phenol side of the equilibrium, both the methoxy and the methyl groups are electron releasing through resonance, with the methoxy having a much stronger effect. The phenolic $\ce{OH}$ group is also strongly electron releasing through resonance. Perhaps having two strongly electron releasing groups (hydroxyl and methoxy) attached to the aromatic ring causes some destabilization due to electron repulsion associated with the high electron density in the ring. Certainly such an effect would be greater with methoxy than methyl. If true, then we could argue that the p-methoxy substituent also destabilizes the starting phenol (more then the methyl group) and this would tend to push the equilibrium to the product side making p-methoxyphenol more acidic than p-methylphenol.

If the substituent effects on the phenol and the phenoxide are comparable, then they would cancel and we might expect p-methoxyphenol and p-methylphenol to have comparable acidities and if the substituent effects cancel we would also expect both phenols to have $\mathrm{p}K_a$'s similar to phenol ($\mathrm{p}K_a$=10), which they do.

The approach used here is sound. I think the explanation of the substituent effects on the phenoxide are also sound. The explanation of the substituent effects on the phenol side is (admittedly) weak. Nonetheless, this explanation does correctly explain why p-methoxyphenol and p-methylphenol have comparable acidities and why these acidities are close to the acidity of phenol.

First off, it is a very small difference in $\mathrm{p}K_a$ that you are asking about. In fact, I would say that the two compunds have similar acidity. It is often difficult to explain such small differences, but what follows is a general approach that can be used to probe the question and to try and answer questions related to equilibria.

In this question we are comparing the effect of two different substituents on an equilibrium. To perform this comparison we must look at the effect of the substituent on both the starting material (the substituted phenol) and the product (the substituted phenoxide anion). If the substituent stabilizes the phenol it will tend to shift the equilibrium towards the phenol side making the phenol less acidic; if the substituent stabilizes the phenoxide it will tend to shift the equilibrium to the phenoxide side making the phenol more acidic.

From your other question we see that the p-methoxy group destabilizes the phenoxide anion to a greater extent than the p-methyl group. Based on this alone, we would say the substituent effect on the phenoxide side of the equilibrium should be such that the p-methoxy substituent pushes the equilibrium more strongly to the phenol side than the p-methyl group will. This would tend to make p-methoxyphenol less acidic than p-methylphenol.

Looking at the phenol side of the equilibrium, both the methoxy and the methyl groups are electron releasing through resonance, with the methoxy having a much stronger effect. The phenolic $\ce{OH}$ group is also strongly electron releasing through resonance. Perhaps having two strongly electron releasing groups (hydroxyl and methoxy) attached to the aromatic ring causes some destabilization due to electron repulsion associated with the high electron density in the ring. Certainly such an effect would be greater with methoxy than methyl. If true, then we could argue that the p-methoxy substituent also destabilizes the starting phenol (more then the methyl group) and this would tend to push the equilibrium to the product side making p-methoxyphenol more acidic than p-methylphenol.

If the substituent effects on the phenol and the phenoxide are comparable, then they would cancel and we might expect p-methoxyphenol and p-methylphenol to have comparable acidities and if the substituent effects cancel we would also expect both phenols to have $\mathrm{p}K_a$'s similar to phenol ($\mathrm{p}K_a$=10), which they do.

The approach used here is sound. I think the explanation of the substituent effects on the phenoxide are also sound. The explanation of the substituent effects on the phenol side is (admittedly) weak. Nonetheless, this explanation does correctly explain why p-methoxyphenol and p-methylphenol have comparable acidities and why these acidities are close to the acidity of phenol.

Key Point: Resonance will only occur when it stabilizes a molecule.

Just because resonance can occur does not mean that resonance will occur. Consider the case of cyclobutadiene, resonance interaction would produce a square molecule that is destabilized (antiaromatic). Therefore the molecule adopts a rectangular geometry to avoid the destabilizing resonance interaction.

I've drawn two of the resonance structures for the ionized form of p-methoxyphenol below. Resonance would destabilize the ionized form because of resonance structures like A where the electron donating resonance effect of the methoxy group places two negative charges close to one another. Rather than destabilize the molecule, the methoxy group simply favors a conformation such that the oxygen lone pairs are twisted out of the plane of the aromatic p-orbitals and cannot overlap with them most of the time.

On the other hand, resonance structures like B are stabilized by the electron withdrawing inductive effect of the methoxy group. If we examine resonance structure C, the counterpart of resonance structure B for the p-methylphenol analogue, we see that since the methyl group is inductively electron releasing, resonance structure C is destabilized.

Conclusion: Since both the methoxy and methyl group release electrons through resonance and since this would destabilize the ionized, anionic phenols, resonance is minimized through conformational effects. Therefore, the smaller inductive effects of the two substituents control the situation with the methoxy stabilizing and the methyl destabilizing the ionized form. Hence, p-methoxyphenol is slightly more acidic than p-methylphenol.

First off, it is a very small difference in $\mathrm{p}K_a$ that you are asking about. In fact, I would say that the two compunds have similar acidity. It is often difficult to explain such small differences, but what follows is a general approach that can be used to probe the question and to try and answer questions related to equilibria.

In this question we are comparing the effect of two different substituents on an equilibrium. To perform this comparison we must look at the effect of the substituent on both the starting material (the substituted phenol) and the product (the substituted phenoxide anion). If the substituent stabilizes the phenol it will tend to shift the equilibrium towards the phenol side making the phenol less acidic; if the substituent stabilizes the phenoxide it will tend to shift the equilibrium to the phenoxide side making the phenol more acidic.

From your other question we see that the p-methoxy group destabilizes the phenoxide anion to a greater extent than the p-methyl group. Based on this alone, we would say the substituent effect on the phenoxide side of the equilibrium should be such that the p-methoxy substituent pushes the equilibrium more strongly to the phenol side than the p-methyl group will. This would tend to make p-methoxyphenol less acidic than p-methylphenol.

Looking at the phenol side of the equilibrium, both the methoxy and the methyl groups are electron releasing through resonance, with the methoxy having a much stronger effect. The phenolic $\ce{OH}$ group is also strongly electron releasing through resonance. Perhaps having two strongly electron releasing groups (hydroxyl and methoxy) attached to the aromatic ring causes some destabilization due to electron repulsion associated with the high electron density in the ring. Certainly such an effect would be greater with methoxy than methyl. If true, then we could argue that the p-methoxy substituent also destabilizes the starting phenol (more then the methyl group) and this would tend to push the equilibrium to the product side making p-methoxyphenol more acidic than p-methylphenol.

If the substituent effects on the phenol and the phenoxide are comparable, then they would cancel and we might expect p-methoxyphenol and p-methylphenol to have comparable acidities and if the substituent effects cancel we would also expect both phenols to have $\mathrm{p}K_a$'s similar to phenol ($\mathrm{p}K_a$=10), which they do.

The approach used here is sound. I think the explanation of the substituent effects on the phenoxide are also sound. The explanation of the substituent effects on the phenol side is (admittedly) weak. Nonetheless, this explanation does correctly explain why p-methoxyphenol and p-methylphenol have comparable acidities and why these acidities are close to the acidity of phenol.