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Is ethanoic acid most acidic in DMSO, water or methanol?

I suspect that the conjugate base would be most stabilized by a polar protic solvent (meaning it's weakest in DMSO), however, water and methanol are both protic and polar. Which causes ethanoic acid to be most acidic lowest $\mathrm{p}K_\mathrm{a}$? This was asked in an exam question where the following $\mathrm{p}K_\mathrm{a}$ values were meant to be assigned to the acid in each of the solvents: $9.5$, $12.6$, $4.8$.

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  • $\begingroup$ I agree with your reasoning regarding DMSO vs (water and methanol). Now, between the two, which would be best able to stabilize free protons? Alternatively, which is a stronger base, water or methanol, and thus which is more likely to form, H3O+ or CH3OH2+? $\endgroup$ – Jason Patterson May 19 '15 at 16:29
  • $\begingroup$ Is methanol more basic? inductive donation? $\endgroup$ – RobChem May 19 '15 at 16:39
  • $\begingroup$ According to Evans' $\mathrm{pK_a}$ table, ethanoic acid has $\mathrm{pK_a}$ values of 4.76 and 12.3 in water and DMSO respectively. This answers the question of the ordering. As to why I might have something to say but I need to think about it for a bit and do some reading. $\endgroup$ – bon May 19 '15 at 17:47
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Ethanoic acid has $\mathrm{pK_a}$ values of 4.76, 9.63 and 12.3 in water, methanol and DMSO respectively (source here and here). Comparing this with the $\mathrm{pK_a}$ values given in the question we can see that ethanoic acid is most acidic in water and least acidic in DMSO. $$\ce{pK_{a}(H2O) < pK_{a}(CH3OH) < pK_{a}(DMSO)}$$

In general the acidity of a species in different solvents is affected by three factors:

  1. The dielectric constant of the solvent. Solvents with larger dielectric constants favour charge separation because the force pulling the charges together as they are separated is lower. This is evidenced by the Born equation which estimates the Gibbs free energy of solvation of gaseous ions: $$\Delta G = -\frac{N_Az^2e^2}{8\pi\epsilon_0r_0}\Bigg(1-\frac{1}{\epsilon_r}\Bigg)$$
  2. The strength of solvation of the conjugate base of the acid.
  3. The strength of solvation of the conjugate acid of the solvent.

Dielectric constant: The dielectric constants of water, methanol and DMSO are 80.1, 32,7 and 46.7 respectively (source). This gives some support to the greater acidity in water but contradicts the observed acidity in methanol and DMSO. This suggests that dielectric constant may not be the major factor that determines the relatives acidities.

Anion solvation: As you suggested in the question, water and methanol are protic solvents but DMSO is aprotic and so the ethanoate ion will be more strongly solvated in water and methanol, shifting the acid-dissociation equilibrium in favour of dissociation. This is considered to be the main factor which explains the greater acidity of ethanoic acid in protic solvents as opposed to aprotic solvents. When comparing water and methanol the strength of anion solvation will be broadly similar because both form $\ce{O-H\bond{~}Anion}$ hydrogen bonds.

Cation solvation: The difference in the stability of the conjugate acid of the solvent may be the factor that causes the difference between water and methanol. The $\mathrm{pK_a}$ of $\ce{CH3OH2+}$ is approximately -2.5 (several sources, none giving exact answers) and the $\mathrm{pK_a}$ of $\ce{H3O+}$ is -1.7. Therefore protonated methanol is less stable than protonated water and so this disfavours dissociation of the acid in methanol. This page claims that this can be attributed to oxygen $p$-orbital donation into $\ce{C-H}~\sigma^*$ orbitals but I don't have enough knowledge of MO theory to verify this (maybe someone else can).

Overall the greater acidity of ethanoic acid in water than methanol seems to be down to the greater dielectric constant of water and the greater stability of the solvent cation formed.

This page says some good things about the subject (some of my ideas were developed from reading this).

Additional reading: This paper looks to have some detail on the anion solvation aspect but I can't access the full text so I'll leave it as a link.

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Thermodynamically, solvation (including (de)protonation) relies on an energy difference: energy required to make a "hole" in the solvent for $\ce{CH_3COOH}$, as well as the energy required to make two holes - one for $\ce{CH_3COO^-}$ and one for $\ce{H^+}$ - compared to the energy released when stabilizing interactions form (dipole-dipole interactions, charge-dipole, cation-pi interactions, etc.) between solvent and solute molecules.

NOTE: When we say "holes" in the solvent, we literally mean the energy one would need to push aside the solvent molecules from each other to make a little hole big enough to put a single ethanoic acid molecule.

  1. If the energy required to make a "hole" in the solvent is larger than the energy released by making all the stabilizing interactions possible, the molecule will not be effectively solvated (e.g. hexanes in water: hexane cannot form lasting stabilizing interactions with water.)
  2. If the energy required to make a "hole" in the solvent is less than the energy released forming stabilizing interactions, the molecule will be solvated.(ethanoic acid in water)

It's about the ability of a solvent to stabilize the neutral and charged species.

Water: $\ce{CH_3COOH}$ and its acid and conjugate base are all polar molecules/ions. Water stabilizes $\ce{CH_3COOH}$ by forming excellent dipole dipole and dipole-charge interactions (and highly symmetrically because water is symmetric). The $\ce{H^+}$ from the acid forms $\ce{H_3O^+}$ in water (also highly symmetric) and is also stabilized by $\ce{H_2O}$, so often you read about "proton hopping" in acidic aqueous solution, which also happens in methanol, but to much less extent. You can also say water's lone pairs are larger and less hindered than methanol's, or you could say that the sigma-like MO for water overlaps better (and more symmetrically) with incoming molecules than methanol's. High symmetry allows much better orbital mixing (via more degenerate MOs), lowering the energy of water and hydronium significantly.

Methanol: Methanol is polar and has a $\mathrm{p}K_\mathrm{a}$ (so I've heard) around water's. Methanol $\ce{O}$ atom can accommodate a proton and when it does, it is stabilized partly by hyperconjugation (a type of orbital mixing; see below for visual) but is not stabilized to the extent water is in part because methanol is not as symmetric (neutral or protonated) as water is = less MO mixing. Yet, methanol forms good dipole-dipole and dipole-charge stabilizing interactions.

DMSO: DMSO is polar, but aprotic, suggesting it does not hold on to protons very well. It can form dipole-dipole interactions, but the sulfoxide $\ce{S=O}$ part likely doesn't overlap very well in a "sigma way" (head-to-head), but in a "pi-way" resulting in weaker overlap.

A note to illustrate hyperconjugation:

If protonated methanol is stabilized, it is likely stabilized intramolecularly by hyperconjugation, which is electron overlap between the exposed (empty) $\mathrm{p}$-orbital on oxygen and the $\ce{C-H}$ sigma bond.

My attempt at hyperconjugative orbital mixing in protonated methanol

It will also be stabilized intermolecularly if surrounded by other polar electron-rich species (methanol, $\ce{CH_3COO^-}$ or $\ce{CH3COOH}$): These molecules will also temporarily donate electron density to the empty $\mathrm{p}$-orbital on the protonated methanol oxygen.

These effects combine to allow for methanol to harbour a proton, making it a little better base than one would think.

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  • $\begingroup$ Can you address more of the issues that are asked about in the question? $\endgroup$ – jonsca May 20 '15 at 0:13
  • $\begingroup$ Great job, that's perfect! $\endgroup$ – jonsca May 20 '15 at 7:09
  • $\begingroup$ @user3158761 The oxygen p-orbital in $\ce{CH3OH2+}$ isn't empty though. It contains 2 electrons. $\endgroup$ – bon May 20 '15 at 9:56
  • $\begingroup$ @bon p orb forms as a result of symmetry change about the O from bent to trigonal planar upon protonation. It contains some e- density, due to donation and mixing with other MOs, but is definitely relatively +charged because it represents part of the LUMO of $CH_3OH_2^+$, so is mostly "empty". I apologize for considering the p-orb apart from a MO. Thanks! $\endgroup$ – Matthew MacLennan May 20 '15 at 15:16

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