Home >
Community >
Inductive vs resonance effects and the acidity of phenol
Upvote
28
Downvote
+ Resonance
+ Biochemistry
+ Phenols
Posted by
Nicole Williams
Inductive vs resonance effects and the acidity of phenol
The acidity of $\ce{A-H}$ is a measure of where the following equilibrium lies.
$$\ce{A-H <=> A^- + H+}$$
This means that we are comparing the relative stabilities of the products and reactants ($\ce{\Delta G=-RTlnK}$).
In the present example (assuming your $\mathrm{p}K_\mathrm{a}$ for propen-2-ol is really being measured in water, note too that in water the relative concentration of this enol of acetone is ~ $10^{-8}$, how can you measure the $\mathrm{p}K_\mathrm{a}$ of something so dilute?) we are comparing the relative stability between [phenol and phenoxide] to the relative stability between [propen-2-ol and the corresponding anion]. Just because the two systems have the same difference in relative stabilities does not mean that phenol and propen-2-ol have the same stability. In other words, just because the two systems have similar $\mathrm{p}K_\mathrm{a}$'s this cannot be interpreted to mean that the additional resonance structures in phenol and phenoxide are not significant contributors to both their stability and description.
EDIT: Jerepierre's comment as worked up by Dissenter (Addressing jerepierre's point)
in his edit Dissenter wrote,
Given this, we can conclude that the reactants are less stable in the acid-base sense than the products
I agree
Dissenter further posits,
acetone enolate is less stable than phenoxide. Phenol is also less stable than acetone enol.
That's where I disagree. We know that the reactants are less stable than the products, but I don't see how we know which reactant(s) is less stable than which of the products.
Let's say the reactants are $10\,\mathrm{kcal/mol}$ (made up number) less stable than the products. Is phenol $5\,\mathrm{kcal/mol}$ less stable than phenoxide and acetone enolate $5\,\mathrm{kcal/mol}$ less stable than acetone enol? Or is it $10$ and $0$ or $0$ and $10$, or $15$ and $-5$? How can you break down the overall reactant/product difference of $10\,\mathrm{kcal/mol}$ further?
The acidity of $\ce{A-H}$ is a measure of where the following equilibrium lies.
$$\ce{A-H <=> A^- + H+}$$
This means that we are comparing the relative stabilities of the products and reactants ($\ce{\Delta G=-RTlnK}$).
In the present example (assuming your $\mathrm{p}K_\mathrm{a}$ for propen-2-ol is really being measured in water, note too that in water the relative concentration of this enol of acetone is ~ $10^{-8}$, how can you measure the $\mathrm{p}K_\mathrm{a}$ of something so dilute?) we are comparing the relative stability between [phenol and phenoxide] to the relative stability between [propen-2-ol and the corresponding anion]. Just because the two systems have the same difference in relative stabilities does not mean that phenol and propen-2-ol have the same stability. In other words, just because the two systems have similar $\mathrm{p}K_\mathrm{a}$'s this cannot be interpreted to mean that the additional resonance structures in phenol and phenoxide are not significant contributors to both their stability and description.
EDIT: Jerepierre's comment as worked up by Dissenter (Addressing jerepierre's point)
in his edit Dissenter wrote,
Given this, we can conclude that the reactants are less stable in the acid-base sense than the products
I agree
Dissenter further posits,
acetone enolate is less stable than phenoxide. Phenol is also less stable than acetone enol.
That's where I disagree. We know that the reactants are less stable than the products, but I don't see how we know which reactant(s) is less stable than which of the products.
Let's say the reactants are $10\,\mathrm{kcal/mol}$ (made up number) less stable than the products. Is phenol $5\,\mathrm{kcal/mol}$ less stable than phenoxide and acetone enolate $5\,\mathrm{kcal/mol}$ less stable than acetone enol? Or is it $10$ and $0$ or $0$ and $10$, or $15$ and $-5$? How can you break down the overall reactant/product difference of $10\,\mathrm{kcal/mol}$ further?
@Dissenter I disagree. pKa can be used as a measure of which conjugate base is more stable. We can easily calculate the the equilibrium constant of any acid base reaction if we know the pKas of the acids. In this case, if acetone enolate is reacted with phenol, an equilibrium will be established with a ~10:1 ratio of phenolate to acetone enolate (setting aside protonation at the alpha-carbon). Doesnt that tell us that phenolate is more stable than acetone enolate? Essentially we are using the proton as the basis of comparison.More
@jerepierre What is it that you are disagreeing with?More
Upvote
VOTE
Downvote
more replies
That blog post you are referencing voices very strong opinions with which one can agree or disagree. I personally love one of the comments under the blog saying
I judge people by the units they use. Mr. Evans used kcal, so he lost.
But of course, that doesn’t discredit the valuable contributions Mr Evans made especially to synthetic organic chemistry.
However, there is a flaw in the argument Mr Evans and the blog writer uses. Both are comparing the $\mathrm{p}K_\mathrm{a}$ of phenol with that of prop-1-en-2-ol. But the standard way of measuring $\mathrm{p}K_\mathrm{a}$s (see equation) is in aquaeous solution, and prop-1-en-2-ol is not stable in aqaeous solution and will rearrange to acetone! Ron mentioned that the relative concentration of prop-1-en-2-ol in aquaeous solution is $10^{-8}$ which, for a $1\,\mathrm{M}$ solution, is lower then the concentration of $\ce{H3O+}$ by autoprotonation of water.
$$\ce{HA + H2O <=> A- + H3O+}$$
As such, I wish to see exact measurement details before I accept both values as equivalent, and therefore I reject the comparison of $\mathrm{p}K_\mathrm{a}$ values made like this.
Instead, I would propose measuring the $\mathrm{p}K_\mathrm{a}$ values of these two related species which are guaranteed to be stable in the drawn form.
Let us continue this argument when these values are presented.
That blog post you are referencing voices very strong opinions with which one can agree or disagree. I personally love one of the comments under the blog saying
I judge people by the units they use. Mr. Evans used kcal, so he lost.
But of course, that doesn’t discredit the valuable contributions Mr Evans made especially to synthetic organic chemistry.
However, there is a flaw in the argument Mr Evans and the blog writer uses. Both are comparing the $\mathrm{p}K_\mathrm{a}$ of phenol with that of prop-1-en-2-ol. But the standard way of measuring $\mathrm{p}K_\mathrm{a}$s (see equation) is in aquaeous solution, and prop-1-en-2-ol is not stable in aqaeous solution and will rearrange to acetone! Ron mentioned that the relative concentration of prop-1-en-2-ol in aquaeous solution is $10^{-8}$ which, for a $1\,\mathrm{M}$ solution, is lower then the concentration of $\ce{H3O+}$ by autoprotonation of water.
$$\ce{HA + H2O <=> A- + H3O+}$$
As such, I wish to see exact measurement details before I accept both values as equivalent, and therefore I reject the comparison of $\mathrm{p}K_\mathrm{a}$ values made like this.
Instead, I would propose measuring the $\mathrm{p}K_\mathrm{a}$ values of these two related species which are guaranteed to be stable in the drawn form.
Let us continue this argument when these values are presented.
The acidity of $\ce{A-H}$ is a measure of where the following equilibrium lies.
$$\ce{A-H <=> A^- + H+}$$
This means that we are comparing the relative stabilities of the products and reactants ($\ce{\Delta G=-RTlnK}$).
In the present example (assuming your $\mathrm{p}K_\mathrm{a}$ for propen-2-ol is really being measured in water, note too that in water the relative concentration of this enol of acetone is ~ $10^{-8}$, how can you measure the $\mathrm{p}K_\mathrm{a}$ of something so dilute?) we are comparing the relative stability between [phenol and phenoxide] to the relative stability between [propen-2-ol and the corresponding anion]. Just because the two systems have the same difference in relative stabilities does not mean that phenol and propen-2-ol have the same stability. In other words, just because the two systems have similar $\mathrm{p}K_\mathrm{a}$'s this cannot be interpreted to mean that the additional resonance structures in phenol and phenoxide are not significant contributors to both their stability and description.
EDIT: Jerepierre's comment as worked up by Dissenter (Addressing jerepierre's point)
in his edit Dissenter wrote,
I agree
Dissenter further posits,
That's where I disagree. We know that the reactants are less stable than the products, but I don't see how we know which reactant(s) is less stable than which of the products.
Let's say the reactants are $10\,\mathrm{kcal/mol}$ (made up number) less stable than the products. Is phenol $5\,\mathrm{kcal/mol}$ less stable than phenoxide and acetone enolate $5\,\mathrm{kcal/mol}$ less stable than acetone enol? Or is it $10$ and $0$ or $0$ and $10$, or $15$ and $-5$? How can you break down the overall reactant/product difference of $10\,\mathrm{kcal/mol}$ further?
The acidity of $\ce{A-H}$ is a measure of where the following equilibrium lies.
$$\ce{A-H <=> A^- + H+}$$
This means that we are comparing the relative stabilities of the products and reactants ($\ce{\Delta G=-RTlnK}$).
In the present example (assuming your $\mathrm{p}K_\mathrm{a}$ for propen-2-ol is really being measured in water, note too that in water the relative concentration of this enol of acetone is ~ $10^{-8}$, how can you measure the $\mathrm{p}K_\mathrm{a}$ of something so dilute?) we are comparing the relative stability between [phenol and phenoxide] to the relative stability between [propen-2-ol and the corresponding anion]. Just because the two systems have the same difference in relative stabilities does not mean that phenol and propen-2-ol have the same stability. In other words, just because the two systems have similar $\mathrm{p}K_\mathrm{a}$'s this cannot be interpreted to mean that the additional resonance structures in phenol and phenoxide are not significant contributors to both their stability and description.
EDIT: Jerepierre's comment as worked up by Dissenter (Addressing jerepierre's point)
in his edit Dissenter wrote,
I agree
Dissenter further posits,
That's where I disagree. We know that the reactants are less stable than the products, but I don't see how we know which reactant(s) is less stable than which of the products.
Let's say the reactants are $10\,\mathrm{kcal/mol}$ (made up number) less stable than the products. Is phenol $5\,\mathrm{kcal/mol}$ less stable than phenoxide and acetone enolate $5\,\mathrm{kcal/mol}$ less stable than acetone enol? Or is it $10$ and $0$ or $0$ and $10$, or $15$ and $-5$? How can you break down the overall reactant/product difference of $10\,\mathrm{kcal/mol}$ further?
More
VOTE
VOTE
VOTE
VOTE
VOTE
VOTE
That blog post you are referencing voices very strong opinions with which one can agree or disagree. I personally love one of the comments under the blog saying
But of course, that doesn’t discredit the valuable contributions Mr Evans made especially to synthetic organic chemistry.
However, there is a flaw in the argument Mr Evans and the blog writer uses. Both are comparing the $\mathrm{p}K_\mathrm{a}$ of phenol with that of prop-1-en-2-ol. But the standard way of measuring $\mathrm{p}K_\mathrm{a}$s (see equation) is in aquaeous solution, and prop-1-en-2-ol is not stable in aqaeous solution and will rearrange to acetone! Ron mentioned that the relative concentration of prop-1-en-2-ol in aquaeous solution is $10^{-8}$ which, for a $1\,\mathrm{M}$ solution, is lower then the concentration of $\ce{H3O+}$ by autoprotonation of water.
$$\ce{HA + H2O <=> A- + H3O+}$$
As such, I wish to see exact measurement details before I accept both values as equivalent, and therefore I reject the comparison of $\mathrm{p}K_\mathrm{a}$ values made like this.
Instead, I would propose measuring the $\mathrm{p}K_\mathrm{a}$ values of these two related species which are guaranteed to be stable in the drawn form.
Let us continue this argument when these values are presented.
That blog post you are referencing voices very strong opinions with which one can agree or disagree. I personally love one of the comments under the blog saying
But of course, that doesn’t discredit the valuable contributions Mr Evans made especially to synthetic organic chemistry.
However, there is a flaw in the argument Mr Evans and the blog writer uses. Both are comparing the $\mathrm{p}K_\mathrm{a}$ of phenol with that of prop-1-en-2-ol. But the standard way of measuring $\mathrm{p}K_\mathrm{a}$s (see equation) is in aquaeous solution, and prop-1-en-2-ol is not stable in aqaeous solution and will rearrange to acetone! Ron mentioned that the relative concentration of prop-1-en-2-ol in aquaeous solution is $10^{-8}$ which, for a $1\,\mathrm{M}$ solution, is lower then the concentration of $\ce{H3O+}$ by autoprotonation of water.
$$\ce{HA + H2O <=> A- + H3O+}$$
As such, I wish to see exact measurement details before I accept both values as equivalent, and therefore I reject the comparison of $\mathrm{p}K_\mathrm{a}$ values made like this.
Instead, I would propose measuring the $\mathrm{p}K_\mathrm{a}$ values of these two related species which are guaranteed to be stable in the drawn form.
Let us continue this argument when these values are presented.
More
VOTE