# Why is acetone less acidic than cyclohexanol?

The $\mathrm{p}K_\mathrm{a}$ of acetone is 19, whereas the $\mathrm{p}K_\mathrm{a}$ of cyclohexanol is 16.

However, acetone is resonance stabilized and its major resonance structure displays the negative charge on an oxygen atom, like cyclohexanol. Is there any qualitative reason to explain why acetone is less acidic than cyclohexanol?

Thank you very much.

• Acetone is a CH-acid, and cyclohexanol is an alcohol. If anything, one should be surprised that the difference in their acidities is not much greater than it actually is. – Ivan Neretin May 23 '18 at 8:41
• @IvanNeretin I suppose what you're saying is that in acetone it is a C-H bond that is broken, which is much less polar than an O-H bond? – makamoe May 23 '18 at 16:23
• Yes, exactly. $\!$ – Ivan Neretin May 23 '18 at 16:38
• The average C-H bond energy is $\approx 410$ and the O-H $\approx 460$ kJ/mol so the result is surprising; a 1000 times difference in equilibrium constant but it seems it is the wrong way round? – porphyrin May 24 '18 at 7:44
• Bond dissociation energies are usually for homolytic bond cleavage, which might explain the difference. – a-cyclohexane-molecule Jul 22 '18 at 23:57

if you consider the structure of acetone, it has 2 $\ce{CH3}$ groups and a $\ce{C}$ double bond $\ce{O}$ ($\ce{C=O}$). Therefore, for the $\ce{H}$ to disassociate, it must be removed from the $\ce{CH3}$ group which leaves now a $\ce{C=O}$, $\ce{CH2}$, and a $\ce{CH3}$ group where the negative charge lies on $\ce{C}$ a part of the $\ce{CH2}$ group. Conversely, the cyclohexanol would have the negative charge lying on the $\ce{O}$ because the $\ce{H}$ a part of the $\ce{OH}$ group would be removed. Just by simple electron negativity rules, $\ce{O}$ is much more favored to possess a (partial) negative charge than $\ce{C}$. Thus, cyclohexanol is a much more stable acid than acetone which is reflected in its $\mathrm{p}K_\mathrm{a}$ value.