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4-Bromophenol vs 4-aminophenol pKa in water comparison confusion
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Mark Hoheisel
4-Bromophenol vs 4-aminophenol pKa in water comparison confusion
[sorry for posting this as an answer, but it isn't possible to include an image when posted as a comment]
Do we also need to consider hydrogen bonding effects in the conjugate base, since the measurement is in water? Presumably forming the phenolate would increase the basicity of the amine through the inductive effect, and this could be stabilized by hydrogen bonding to solvent. I have no idea the strength of this effect, but thought it worth considering.
[sorry for posting this as an answer, but it isn't possible to include an image when posted as a comment]
Do we also need to consider hydrogen bonding effects in the conjugate base, since the measurement is in water? Presumably forming the phenolate would increase the basicity of the amine through the inductive effect, and this could be stabilized by hydrogen bonding to solvent. I have no idea the strength of this effect, but thought it worth considering.
In this situation, considering the distance between the $\ce{-OH}$ group and the substituents we can ignore inductive effects and focus on the mesomeric ones. $\ce{-Br}$ has a $+M_s$ effect, same as the $\ce{-NH2}$ group. To make an $\ce{-OH}$ group more acidic we need electron withdrawing groups as the slight positive charge built on the oxygen atom will facilitate the proton leaving. In order to find the more acidic molecule we look for the one with the most electron withdrawing substituent(or the least electron donating in this case). Consider the strength of the substituents' conjugation with the ring. Bromine is a huge atom, this difference in atomic radius compared to carbon gives their orbitals a hard time overlapping(I'm simplifying here for convenience's sake) and consequently a weaker $+M_s$ effect than $\ce{-NH2}$. Considering that, we can argue that 4-bromophenol has a lower pKa(is more acidic) than 4-aminophenol. And indeed, a quick google search validates my answer, the pKa of 4-bromophenol is 9.17 and the pKa of 4-aminophenol is 10.30 for the $\ce{-OH}$ group.
In this situation, considering the distance between the $\ce{-OH}$ group and the substituents we can ignore inductive effects and focus on the mesomeric ones. $\ce{-Br}$ has a $+M_s$ effect, same as the $\ce{-NH2}$ group. To make an $\ce{-OH}$ group more acidic we need electron withdrawing groups as the slight positive charge built on the oxygen atom will facilitate the proton leaving. In order to find the more acidic molecule we look for the one with the most electron withdrawing substituent(or the least electron donating in this case). Consider the strength of the substituents' conjugation with the ring. Bromine is a huge atom, this difference in atomic radius compared to carbon gives their orbitals a hard time overlapping(I'm simplifying here for convenience's sake) and consequently a weaker $+M_s$ effect than $\ce{-NH2}$. Considering that, we can argue that 4-bromophenol has a lower pKa(is more acidic) than 4-aminophenol. And indeed, a quick google search validates my answer, the pKa of 4-bromophenol is 9.17 and the pKa of 4-aminophenol is 10.30 for the $\ce{-OH}$ group.
Answer to this question is, as orthocresol pointed out, simpler than what you think. As you know, $\ce{-Br}$ is deactivate the ring (even though it is o,p-directing) while $\ce{-NH2}$ is great ring activator. What that means is regardless of the mesomeric effect, compared to $\ce{-H}$, $\ce{-Br}$ is overall electron withdrawing and $\ce{-NH2}$ is electron donating (remember, $\ce{-NH2}$ is a Lewis base).
Said that, I'd say basically, $\mathrm{pH}$ of a phenol ($\mathrm{p}K_\mathrm{a} = 9.95$) is depended on how stable phenolate ion is after it donated $\ce{H+}$. Electron withdrawing groups stabilize phenolate ion (e.g., $\ce{-NO2}$ group, p-nitrophenol: $\mathrm{p}K_\mathrm{a} = 7.1$) while electron donating groups destabilize it (e.g., $\ce{-OCH3}$ group, p-methoxyphenol: $\mathrm{p}K_\mathrm{a} = 10.2$). Thus, one can expect that p-bromophenol should have a lower $\mathrm{p}K_\mathrm{a}$ than that of phenol (meaning more acidic than phenol) while p-aminophenol should have a higher $\mathrm{p}K_\mathrm{a}$ than that of phenol (meaning more basic than phenol). This is true in reality. Actual $\mathrm{p}K_\mathrm{a}$ values of p-bromophenol and p-aminophenol are $9.34$ and $10.3$, respectively ($9.34 \lt 9.95 \lt 10.3$).
Answer to this question is, as orthocresol pointed out, simpler than what you think. As you know, $\ce{-Br}$ is deactivate the ring (even though it is o,p-directing) while $\ce{-NH2}$ is great ring activator. What that means is regardless of the mesomeric effect, compared to $\ce{-H}$, $\ce{-Br}$ is overall electron withdrawing and $\ce{-NH2}$ is electron donating (remember, $\ce{-NH2}$ is a Lewis base).
Said that, I'd say basically, $\mathrm{pH}$ of a phenol ($\mathrm{p}K_\mathrm{a} = 9.95$) is depended on how stable phenolate ion is after it donated $\ce{H+}$. Electron withdrawing groups stabilize phenolate ion (e.g., $\ce{-NO2}$ group, p-nitrophenol: $\mathrm{p}K_\mathrm{a} = 7.1$) while electron donating groups destabilize it (e.g., $\ce{-OCH3}$ group, p-methoxyphenol: $\mathrm{p}K_\mathrm{a} = 10.2$). Thus, one can expect that p-bromophenol should have a lower $\mathrm{p}K_\mathrm{a}$ than that of phenol (meaning more acidic than phenol) while p-aminophenol should have a higher $\mathrm{p}K_\mathrm{a}$ than that of phenol (meaning more basic than phenol). This is true in reality. Actual $\mathrm{p}K_\mathrm{a}$ values of p-bromophenol and p-aminophenol are $9.34$ and $10.3$, respectively ($9.34 \lt 9.95 \lt 10.3$).
[sorry for posting this as an answer, but it isn't possible to include an image when posted as a comment]
Do we also need to consider hydrogen bonding effects in the conjugate base, since the measurement is in water? Presumably forming the phenolate would increase the basicity of the amine through the inductive effect, and this could be stabilized by hydrogen bonding to solvent. I have no idea the strength of this effect, but thought it worth considering.
[sorry for posting this as an answer, but it isn't possible to include an image when posted as a comment]
Do we also need to consider hydrogen bonding effects in the conjugate base, since the measurement is in water? Presumably forming the phenolate would increase the basicity of the amine through the inductive effect, and this could be stabilized by hydrogen bonding to solvent. I have no idea the strength of this effect, but thought it worth considering.
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In this situation, considering the distance between the $\ce{-OH}$ group and the substituents we can ignore inductive effects and focus on the mesomeric ones. $\ce{-Br}$ has a $+M_s$ effect, same as the $\ce{-NH2}$ group. To make an $\ce{-OH}$ group more acidic we need electron withdrawing groups as the slight positive charge built on the oxygen atom will facilitate the proton leaving. In order to find the more acidic molecule we look for the one with the most electron withdrawing substituent(or the least electron donating in this case). Consider the strength of the substituents' conjugation with the ring. Bromine is a huge atom, this difference in atomic radius compared to carbon gives their orbitals a hard time overlapping(I'm simplifying here for convenience's sake) and consequently a weaker $+M_s$ effect than $\ce{-NH2}$. Considering that, we can argue that 4-bromophenol has a lower pKa(is more acidic) than 4-aminophenol. And indeed, a quick google search validates my answer, the pKa of 4-bromophenol is 9.17 and the pKa of 4-aminophenol is 10.30 for the $\ce{-OH}$ group.
Sources: https://pubchem.ncbi.nlm.nih.gov/compound/4-bromophenol#section=Dissociation-Constants https://en.wikipedia.org/wiki/4-Aminophenol
In this situation, considering the distance between the $\ce{-OH}$ group and the substituents we can ignore inductive effects and focus on the mesomeric ones. $\ce{-Br}$ has a $+M_s$ effect, same as the $\ce{-NH2}$ group. To make an $\ce{-OH}$ group more acidic we need electron withdrawing groups as the slight positive charge built on the oxygen atom will facilitate the proton leaving. In order to find the more acidic molecule we look for the one with the most electron withdrawing substituent(or the least electron donating in this case). Consider the strength of the substituents' conjugation with the ring. Bromine is a huge atom, this difference in atomic radius compared to carbon gives their orbitals a hard time overlapping(I'm simplifying here for convenience's sake) and consequently a weaker $+M_s$ effect than $\ce{-NH2}$. Considering that, we can argue that 4-bromophenol has a lower pKa(is more acidic) than 4-aminophenol. And indeed, a quick google search validates my answer, the pKa of 4-bromophenol is 9.17 and the pKa of 4-aminophenol is 10.30 for the $\ce{-OH}$ group.
Sources:https://pubchem.ncbi.nlm.nih.gov/compound/4-bromophenol#section=Dissociation-Constantshttps://en.wikipedia.org/wiki/4-Aminophenol
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Answer to this question is, as orthocresol pointed out, simpler than what you think. As you know, $\ce{-Br}$ is deactivate the ring (even though it is o,p-directing) while $\ce{-NH2}$ is great ring activator. What that means is regardless of the mesomeric effect, compared to $\ce{-H}$, $\ce{-Br}$ is overall electron withdrawing and $\ce{-NH2}$ is electron donating (remember, $\ce{-NH2}$ is a Lewis base).
Said that, I'd say basically, $\mathrm{pH}$ of a phenol ($\mathrm{p}K_\mathrm{a} = 9.95$) is depended on how stable phenolate ion is after it donated $\ce{H+}$. Electron withdrawing groups stabilize phenolate ion (e.g., $\ce{-NO2}$ group, p-nitrophenol: $\mathrm{p}K_\mathrm{a} = 7.1$) while electron donating groups destabilize it (e.g., $\ce{-OCH3}$ group, p-methoxyphenol: $\mathrm{p}K_\mathrm{a} = 10.2$). Thus, one can expect that p-bromophenol should have a lower $\mathrm{p}K_\mathrm{a}$ than that of phenol (meaning more acidic than phenol) while p-aminophenol should have a higher $\mathrm{p}K_\mathrm{a}$ than that of phenol (meaning more basic than phenol). This is true in reality. Actual $\mathrm{p}K_\mathrm{a}$ values of p-bromophenol and p-aminophenol are $9.34$ and $10.3$, respectively ($9.34 \lt 9.95 \lt 10.3$).
Sources for $\mathrm{p}K_\mathrm{a}$ values: Evans pKa Table and Wikipedia.
Answer to this question is, as orthocresol pointed out, simpler than what you think. As you know, $\ce{-Br}$ is deactivate the ring (even though it is o,p-directing) while $\ce{-NH2}$ is great ring activator. What that means is regardless of the mesomeric effect, compared to $\ce{-H}$, $\ce{-Br}$ is overall electron withdrawing and $\ce{-NH2}$ is electron donating (remember, $\ce{-NH2}$ is a Lewis base).
Said that, I'd say basically, $\mathrm{pH}$ of a phenol ($\mathrm{p}K_\mathrm{a} = 9.95$) is depended on how stable phenolate ion is after it donated $\ce{H+}$. Electron withdrawing groups stabilize phenolate ion (e.g., $\ce{-NO2}$ group, p-nitrophenol: $\mathrm{p}K_\mathrm{a} = 7.1$) while electron donating groups destabilize it (e.g., $\ce{-OCH3}$ group, p-methoxyphenol: $\mathrm{p}K_\mathrm{a} = 10.2$). Thus, one can expect that p-bromophenol should have a lower $\mathrm{p}K_\mathrm{a}$ than that of phenol (meaning more acidic than phenol) while p-aminophenol should have a higher $\mathrm{p}K_\mathrm{a}$ than that of phenol (meaning more basic than phenol). This is true in reality. Actual $\mathrm{p}K_\mathrm{a}$ values of p-bromophenol and p-aminophenol are $9.34$ and $10.3$, respectively ($9.34 \lt 9.95 \lt 10.3$).
Sources for $\mathrm{p}K_\mathrm{a}$ values: Evans pKa Table and Wikipedia.
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