# 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 $$\gt$$ 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.

$$\ce{Phenol + acetone~enolate <=> phenoxide + acetone ~enol}$$

So, from the $$\mathrm{p}K_\mathrm{a}$$ 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 $$\mathrm{p}K_\mathrm{a}$$ 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.

This question was answered well enough 5-7 years ago, but there is something still lacking. The resonance diagrams drawn in brackets suggest that resonance is responsible for increasing the acidity of phenol; I find the wording misleading. Better would be that the resonance of the phenyl ring is responsible for the stabilization of phenolate anion. IMHO, drawing diagrams with two charges and switching them around the molecule amounts to creative imagination - sometimes useful, but only as a last resort.

Let's first picture what's going on with phenol, or more accurately, with the $$pi$$ electrons of benzene, the ultimate resonance molecule (Ref 1). Resonance, the amount of energy of hydrogenation (or combustion) less than calculated for a compound with the same number of C-H and C-C and C=C bonds, occurs because three localized C=C double bonds (which will have energy at the dotted line) interact and become more delocalized and stable: the totally interacting combination is significantly more stable than the three original double bonds (it contains 2 electrons; it is one molecular bond). Then because of symmetry, there are two more combinations of the three original orbitals, each with a node perpendicular to the ring, and degenerate (having the same energy). Each of these bonding combinations is a bond that is more stable than one of the original C=C double bonds.

The diagram in brackets in the original question suggests that a $$pi$$ lone pair from the more electronegative oxygen plops onto the phenyl ring, leaving oxygen with a positive charge. Not likely. However, a base can capture the phenolic proton, leaving oxygen with negative charge, more than it usually has. Too much, in fact. How does it adjust? By sharing the lone pair in an orbital perpendicular with the ring; remember, the oxygen is sp$$^2$$ hybridized, with 2 lone pairs now in the plane of the ring and a third perpendicular. The lone pair in an atomic orbital on oxygen interacts with one molecular orbital of the ring. That's my simplification. If we talk about molecular orbitals, we should include all the orbitals - but for simplicity, I'm going to discuss only a bond between the one oxygen atomic orbital and one of the phenyl ring molecular orbitals.

Which phenyl orbital should we consider? The three bonding orbitals are filled; in my simplification, they are tied up, out of consideration. However, there are three anti-bonding $$pi$$ orbitals, two of which have low enough energies to be somewhat compatible with the lone pair of oxygen - they could combine to give a spread-out molecular orbital, including the oxygen and the phenyl ring, reducing charge concentration on oxygen while increasing bonding somewhere else. Which orbital would be most effective?

Take the carbon at 3 o'clock to be the one connected to oxygen. The orbital labeled $$pi_5$$ has a node at that carbon and hence does not see the oxygen. However, $$pi_4$$ has significant density there (and so does $$pi_6$$, but it is too high energy to be concerned with) and so can unite with the oxygen AO to form a phenolic MO with electron density spread over the ring (phenol is then activated toward electrophilic attack).

This explains how phenol is more acidic than cyclohexanol.

There is another explanation of phenolic acidity, and I bring it up only to point out that I believe it is an oversimplification at least, and perhaps very inadequate (Ref 2):

It just smears everything together and gives no insight, and not even the creativity of drawing lines and arrows.

Connecting vinyl alcohol with phenol and cyclohexanol seems to be possible because its pK$$_a$$ was given as 10.9. Literature favors a different number: 19-20 (Ref 3). With this larger pK$$_a$$, there seems to be no close connection to discuss.

At room temperature, acetaldehyde (H$$_3$$CCH=O) is more stable than vinyl alcohol (H$$_2$$C=CHOH) by 42.7 kJ/mol (Ref 4). Vinyl alcohol can be stabilized by utilizing kinetic favorability of the deuterium kinetic isotope effect (k$$_{H+}$$/k$$_{D+}$$ = 4.75, k$$_{H2O}$$/k$$_{D2O}$$ = 12). The tautomerization process is significantly inhibited at ambient temperatures, and the half-life of the enol form can be increased to t$$_{1/2}$$ = 42 minutes. The difficulty of dealing with vinylic alcohols is a further disconnect from phenol and cyclohexane.

For reference, the empirical resonance energy of benzene is 143.1 kJ/mol (Ref 5). The stabilization of phenolate will be considerably smaller than this, and present only for the ion. If phenol were stabilized as much as the phenolate ion is, there would be no increased acidity compared to an alcohol. Comparing the resonance stabilization of benzene with ordinary bond energies shows a fairly small energy: a C-C bond has about 347 kJ/mol strength; a C=C bond about 614 kJ/mol, so a $$pi$$ bond has about 267 kJ/mole strength. The resonance energy of benzene is worth about a half of a $$pi$$ bond (Ref 6).