# Inductive vs resonance effects and the acidity of phenol

A lot of people rationalize the acidity of phenol by saying that resonance is responsible for much of phenol's acidity as opposed to aliphatic alcohols.

However, this image suggests that in fact the inductive effect is responsible for a great deal of phenol's additional acidity. As we can see in the picture immediately above, it seems that the sp$^2$ carbon neighboring to the oxygen is pulling most of the weight here; that the resonance contributors with charges on carbons are minor contributors. The additional resonance contributors for phenol seem to be doing almost nothing here.

So, who's right? Resonance>inductive effect in phenol or the other way around?

EDIT: The argument I have summarized above is presented here at this website entitled "I Judge People By Their Grammar and Knowledge of Phenol"

While the chemistry in the paper was good, as The Chem Blog has noted, the authors’ lack of attention to detail was borderline disrespectful. I expected more from one of the most storied labs in synthetic organic chemistry.

[...]

It turns out that an inductive effect—not a resonance effect—is the predominant reason for the increased acidity of phenol relative to aliphatic alcohols.

[...]

For those interested, these data come from Evans’ Chem 206 lecture notes (Lecture 20, restricted access), where the point is hammered home in glorious detail. Professor Evans’ PowerPoint slides should be framed and displayed in the Smithsonian.

$\ce{Phenol + acetone~enolate \leftrightharpoons phenoxide + acetone ~enol}$

So, from the pKa values we know that the above equilibrium lies mostly to the right (according to jerepierre). Given this, we can conclude that the reactants are less stable in the acid-base sense than the products; acetone enolate is less stable than phenoxide.

Phenol is also less stable than acetone enol.

But I don't see how this contradict's the blog article. It does contradict ron's point about being unable to use the provided pKa values to compare the relative stabilities of the acids.

• Are your 3 pKa values all measured in the gas phase (I can't imagine that someone measured the pKa of propen-2-ol in solution)? – ron Dec 3 '14 at 2:10
• You seem to forget that the $\ce{C=C}$ double bond in the enol is able to provide resonance stabilization as well and I would think that in both cases - for phenol and the enol - the resonance stabilization has the major effect on the acidity not the hybridization. – Philipp Dec 3 '14 at 2:10
• @Philipp True, but that makes for what? Two contributing resonance structures for the enol? A carbocation and an oxide ion? Phenoxide however has four ... the oxide ion and three carbocations. It seems that the two remaining carbocation contributing structures give rise to less than an order of magnitude of pKa difference. The presence of an sp2 carbon itself gives rise to much, much more than an order of magnitude of pKa difference. – Dissenter Dec 3 '14 at 4:16
• @ron - why wouldn't one be able to measure the pKa of an unstable enol in solution? Also, how does one get pKa in the gas phase? Doesn't Ka and therefore pKa imply the presence of water? – Dissenter Dec 3 '14 at 4:17
• @Dissenter The number of resonance structures is not always the key to stability. The effect of providing any possibility for delocalization might give the major contribution (I don't know) and the possibility to spread the charge over an even larger "area" in Phenol might simply not make that much of a difference. Coulomb's law goes as $1/r^{2}$ so the difference between having the electrons very close together and moderately far apart is much more pronounced than between going from having electrons moderately far apart and very far apart. But my argument might be wrong. – Philipp Dec 3 '14 at 16:32

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?

• So in your opinion the author of the blog article is wrong in his reasoning? I was this close to actually showing the article to my professors because they, like most, stick with resonance in explaining the acidity of phenol. – Dissenter Dec 3 '14 at 16:42
• Yes, in my opinion. Do you see any flaws in my argument? Do you see the flaw I'm pointing out in his? – ron Dec 3 '14 at 18:05
• @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 pKa's 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). Doesn't that tell us that phenolate is more stable than acetone enolate? Essentially we are using the proton as the basis of comparison. – jerepierre Dec 3 '14 at 18:46
• @jerepierre What is it that you are disagreeing with? – ron Dec 3 '14 at 18:56
• @jerepierre I don't follow, could you expand your comment into an answer? – ron Dec 3 '14 at 19:12

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.