Methanol is slightly more acidic than water. Their $\mathrm{p}K_\mathrm{a}$ values, in water, are $15.5$ and $15.7$, respectively. All other aliphatic alcohols, however, are less acidic than water.

Is the following reasoning correct? This is my best rationalization; is there anything better or anything that can be added?

In both hydroxide and methoxide, we have an $\ce{O}$ bearing negative charge. In the former, we have a hydrogen attached to the $\ce{O}$. In the latter, we have a carbon attached to the $\ce{O}$. Carbon is more electronegative than hydrogen, therefore, carbon should be able to better withdraw electron density via induction from the $\ce{O}$. This is a stabilizing interaction.

We often think of methyl groups as inductively donating, but that's molecular profiling. They can inductively donate to a carbocation, partly because a carbocation is highly electronegative, so the carbocation pulls the electrons toward itself pretty well (so there's both pushing and pulling of electrons going on). When you have a methyl group attached to an $\ce{O}$, however, the methyl group is often inductively withdrawing. NMR data supports this.

Now the question is why aren't other aliphatic alcohols more acidic than water? In longer-chain aliphatic alkoxides, you don't just have a methyl group attached to the $\ce{O}$ bearing the negative charge - you have a bunch of methyl groups strung together. These lessen the amount of inductive withdrawal that the alpha carbon can do. Each$\ce{-CH2 -}$ unit attached to the alpha $\ce{-CH2 -}$ is somewhat inductive donating to the alpha $\ce{-CH2 -}$.

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    $\begingroup$ The $pK_a$ of water is 14.0 so it is more acidic than methanol. chemwiki.ucdavis.edu/Organic_Chemistry/Fundamentals/… $\endgroup$ – MaxW Dec 21 '15 at 6:34
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    $\begingroup$ @MaxW - well, that opens up a whole new can of worms because I do see the 15.7 value cited a lot. ;) Time for another question? $\endgroup$ – Dissenter Dec 21 '15 at 6:53
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    $\begingroup$ 14.0 is $-\log(\ce{[H+][OH^{-}]})$; 15.7 is $-\log{\ce{[H+][OH^{-}]\over[H2O]}}$. $\endgroup$ – Ivan Neretin Dec 21 '15 at 7:45
  • $\begingroup$ @IvanNeretin but can we say that the activity of water is its concentration? That is, we are determining the pka of water in water-so shouldn't we use the activity of water? $\endgroup$ – Aditya Anand Dec 21 '15 at 16:19
  • $\begingroup$ Some light in the darkness can be obtained from J. Chem. Educ. 2017, 94, 690−695 (DOI: 10.1021/acs.jchemed.6b00623) where the difference between the "rational" and the "conventional" pka (H2O) value is elegantly discussed. pubs.acs.org/doi/10.1021/acs.jchemed.6b00623 $\endgroup$ – JM Urbina Aug 25 '20 at 0:33

The relative acidities of these molecules in the aqueous solution$^{[1][2]}$ is:

$$\text{methanol > water > ethanol > isopropanol > t-butanol}$$

The relative acidities of these molecules in DMSO$^{[3]}$ is:

$$\text{methanol > ethanol > isopropanol > t-butanol > water}$$

The problem with only considering the acidities of these molecules in aqueous solution is that it assumes only one explanation for acidity. The inversion of the order of acidity in DMSO indicates that there are multiple factors at play.

The acidity trend in DMSO is explained by polarizability and the anionic hyperonjugation effect$^{[4]}$ of the conjugate bases of the molecules. The larger size of the alkyl substituents in the conjugate bases allows for increased distribution of negative charge over a larger volume. This reduces the charge density, and thereby the Coulombic repulsion between the conjugate base and the solvent. In DMSO, water is much less acidic than even methanol, which is also consistent with the relative polarizability of a methyl group and a hydrogen atom, but fails to explain the relative acidity of t-butanol and methanol.

The trend in alcohol acidity is explained by anionic hyperconjugation, which is weakened in larger alkyl substituents.

For methanol and the methoxide anion, the primary interaction occurs between a filled lone-pair ($n$) orbital on oxygen and an unoccupied $\pi^*_\ce{Me}$ orbital of the methyl group, resulting in a two-electron stabilizing effect.

$\hspace{3.8cm}$MO interactions

Deprotonation of the alcohol leads to an increase in the energy of the n orbitals on oxygen, and a decreased energy separation between the interacting orbitals $n$ and $\pi^*_\ce{Me}$, and hence a greater stabilizing interaction$^{[5]}$.

Although DMSO does solvate the solute to an extent, because is an aprotic solvent, it is incapable of hydrogen bonding with the solute. This allows DMSO acidities better reflect the intrinsic acidities of the molecules.

In aqueous solution, however, the conjugate bases are stabilized by hydrogen bonding. Because smaller ions are better solvated by water, water's acidity is increased more than methanol's, but because methanol is intrinsically more acidic due to its polarizability, it is still slightly more acidic than water, even in aqueous solution. As the size of the alkyl substituent increases, the alcohol becomes weaker due to the decreasing ability of water to solvate the conjugate base. This has a greater affect on the alcohol's acidity than the dispersion of the negative charge, and overrides the the intrinsic acidity, inverting the order of acidity.

$^{[1]}$ Evans $\mathrm{p}K_\mathrm{a}$ Table, Alcohols
$^{[2]}$ Wikipedia, Ethanol
$^{[3]}$ Bordwell $\mathrm{p}K_\mathrm{a}$ Table, Alcohols
$^{[4]}$ ACS, Acidities of Water and Methanol in Aqueous Solution and DMSO
$^{[5]}$ ACS, Does a Methyl Substituent Stabilize or Destabilize Anions?


If we compare the +I effect of methyl group and more substituted alkyl groups, then methyl group has less +I effect as compared to other alkyl groups. Since oxygen is more electronegative than both carbon and hydrogen.... So a methyl group cannot fulfill the electron needs of oxygen which is why oxygen takes electrons from Hydrogen and a hydrogen ion is released easily. If a more electron donating group was present it could have fulfilled the electron needs of oxygen and hydrogen ion wouldn't have been released easily thus making the compound less acidic


Methanol is more acidic than water because its conjugate base that is methoxide is weaket than waters conjugate base that is hydroxide and now we can say that weaker is the conjugate base more is its acidic character . or we can say that due to the presence of methyl group in CH3 bond between O and H become weak and it can easily release hydrogen where is in H2O bond between O and H is stronger so it will not easily release hydrogen and become a weak acid than methanol

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    $\begingroup$ Your answer is not very insightful. How does the presence of the methyl group cause the O-H bond in methanol to be weakened? $\endgroup$ – Tan Yong Boon Feb 17 '19 at 7:16

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