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.


Addressing jerepierre's point

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

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.
The acidity of $\ce{A-H}$ is a measure of where the following equilibrium lies.

$$\ce{A-H &lt;=&gt; A^- + H+}$$

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?