A carboxylic acid, an inductive or resonance effect, Part 2

I want to add to what I said earlier. I was reading an article discussing resonance. I can see that those who disagreed with me would be following that line of thinking.

In the example below, I compare the acidity of cyclohexanol (pKa ~16) with phenol (pKa 10). The difference between these two structures is the double bonds in the phenol. The effect of the double bonds is to interact with the electrons on the oxygen atom. This is a resonance effect. The anion is stabilized. Below, I have a carboxylic acid in which both structures possess unsaturated carbons enabling a resonance effect. A carboxylic acid is more acidic (pKa ~4.5). Because both structures are capable of showing a resonance effect, the difference between them must be due to another factor. In this case it is an inductive effect.

As I wrote the above, I also realized an intuitive disbelief. In my earlier blog, I started with a need to explain how the electrons should be pulled away from the proton in the OH bond. This is the inverse square law. The force between a proton and the electron to which it is attracted must be reduced. If the distance is increased, the force will be reduced. I don’t believe I can increase acidity without using it. If this is so, I need to explain how electrons can be pulled away form the oxygen atom. Resonance does not accomplish this. Unsaturation does.

I have an example of how HCl might ionize in the presence of water. The model shows a transition state in which a proton is mutually attracted to electrons of a chlorine atom and oxygen atom. If the force (F7) is greater for the electrons of oxygen, the proton will remain with it. The forces (F6 and F7) should following an inverse square law. The smaller the separation of a proton to an electron pair, the stronger the force.

I also noted in the prior post that given the bond length of HCl and H2O, the repulsive forces between the proton and chlorine or oxygen nucleus can be compared. Water (96 pm) would have 75% of the repulsive force of HCl (121 pm) toward a proton.

If a similar model is applied to cyclohexanol, phenol, or a carboxylic acid, then an increase in the proton-electron pair distance must precede the loss of the proton. Given the electron withdrawing character of an sp2-carbon, its effect to pull electrons away from the OH bond cannot be ignored. In fact, it provides ample evidence for an increase in acidity without resorting to resonance.

The “R” represents a nucleus of a group that exerts an electromagnetic force to pull electrons toward it. By doing so, it results in an increase in the proton-electron pair distance. That increase in distance results in a reduction in the force and an increase in acidity. I find it very difficult to argue for any other force.

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