I got this question in my exam:

Arrange the following in increasing order of acidic strength in gaseous phase

(i) $\ce{CH3CH2COOH}$ (ii) $\ce{CH3COOH}$

I felt that increasing order must be (i) $<$ (ii). But answer given is (ii) $<$ (i). Why will the order change in gaseous phase?

  • $\begingroup$ Just guessing that tendency of creation of a dimer may be stronger at acetic acid. $\endgroup$
    – Poutnik
    Nov 23 '19 at 15:15
  • $\begingroup$ It's in the gaseous state: IMO dimerization is more likely in aqueous medium, where the molecules are much closer to each other. $\endgroup$ Nov 23 '19 at 15:25
  • 1
    $\begingroup$ @AniruddhaDeb In aqueous medium, water competes as hydrogen bonding partner (and is present at higher concentration). In the gas phase, the only available hydrogen bonding partner is another acetic acid molecule. $\endgroup$ Nov 24 '19 at 3:35
  • $\begingroup$ @Aniruddha Deb The acid dimerization is typical for the gaseous state and nonpolar solvents like benzene. As the result, Acetic acid distribution constant for benzen/water is highly dependent on concentration, as higher concentration of acid in benzene supports the dimerization. ( I remember doing it in past as a physical chemistry lab experiment ) $\endgroup$
    – Poutnik
    Nov 24 '19 at 7:21

Acidity or acid strength can be expressed either in terms of the proton affinity (PA) of the conjugate base, e.g., PA($\ce{A-}$), or in terms of the acid dissociation constant, e.g., $\mathrm{p}K_{\mathrm{a}(\ce{HA})}$. The former expression is often favored among gas-phase chemists, whereas the latter expression is most familiar to solution chemists.

In aqueous solution, $\mathrm{p}K_{\mathrm{a}(\ce{CH3CO2H})} = 4.76$ while $\mathrm{p}K_{\mathrm{a}(\ce{CH3CH2CO2H})} = 4.87$, thus acidity of acetic acid is greater than that of propanoic acid. Keep in mind that solvent plays a huge role in $\mathrm{p}K_{\mathrm{a}}$ measurements. For example, compare $\mathrm{p}K_{\mathrm{a}}$ measurements in $\ce{H2O}$ and in $\ce{DMSO}$:

$$ \begin{array}{c|cc} \hline \ce{HA} & \mathrm{p}K_{\mathrm{a}} \text{ in } \ce{H2O} & \mathrm{p}K_{\mathrm{a}} \text{ in } \ce{DMSO} \\ \hline \ce{CH3CO2H} & 4.76 & 12.3 \\ \ce{H2O} & 15.7 & 31.2 \\ \ce{CH3OH} & 15.5 & 27.9 \\ \ce{C6H5OH} & 9.95 & 18.0 \\ \hline \end{array} $$

However, the proton affinity (PA) measurements of the conjugate bases of these acids in gas phase tell a different story. Proton affinity is a thermodynamic measurement conducted in the gas phase. These measurements give a pure view of base/acid strength, without complications from solvent effects such as salvation, thus take the solvent out of the equation.

Cardwell, et al. (Ref.1) has measured the proton affinities of various $\ce{R-CO2H}$ acids, including ethanoic and propanoic acids ($\ce{R}=\ce{CH3}$ and $\ce{CH3CH2}$, respectively). They conclude that:

The acidity increment (with the size of $\ce{R}$-group) is relatively small and can be attributed to the stabilizing effect of the polarizability of the alkyl group on the negative ion.


  1. Gary Caldwell, Richard Renneboog, Paul Kebarle, “Gas phase acidities of aliphatic carboxylic acids, based on measurements of proton transfer equilibria,” Canadian Journal of Chemistry 1989, 67(9), 611-618 (https://doi.org/10.1139/v89-092).
  2. Also read: Diethard K. Bohme, Edward Lee-Ruff, L. Brewster Young, “Acidity order of selected Brønsted acids in the gas phase of $\pu{300 K}$,” Journal of American Chemical Society 1972, 94(15), 5153-5159 (https://doi.org/10.1021/ja00770a002) and references there in.

One possible explanation for this is hyperconjugation due to the attached alkyl chain, as shown below: enter image description here

since the acetate ion has one more $\alpha$-H atom than the propionate ion, the acetate ion is more stable than the propionate ion in the gaseous state and has more tendency to donate the hydrogen atom, making it more acidic.

  • 1
    $\begingroup$ Please cite the source, so that I may believe it $\endgroup$
    – Zenix
    Nov 23 '19 at 16:06

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