What you are about to read may be a very mind-boggling paragraph but please do not regard it as nonsense. Please think through it thoroughly.

In Chapter 6 of Organic Chemistry (4th ed.) by Maitland Jones Jr. and Steven A. Fleming, the following is written (p. 237):

In 1970, Professor John Brauman (b. 1937) and his co-workers at Stanford university showed that in the gas phase the opposite acidity order obtained. The intrinsic acidity of the four alcohols of Table 6.6 is exactly opposite to that found in solution. The acidity order measured in solution reflects a powerful effect of the solvent, not the natural acidities of the alcohols themselves. Organic ions are almost all unstable species, and the formation of the alkoxide anions depends critically on how easy it is to stabilize them through interaction with solvent molecules, a process called solvation. tert-Butyl alcohol is a weaker acid in solution than methyl alcohol because the large tert-butyl alkoxide ion is difficult to solvate. The more alkyl groups, the more difficult it is for the stabilizing solvent molecules to approach (Fig. 6.22). Of course, in the gas phase where solvation is impossible, the natural acidity order is observed.

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Let me provide some context. The author is discussing the acidity of alcohols in the gas phase and in the aqueous phase. In Table 6.6 as shown above, the pKa values are provided to show the decreasing acidity of the alcohols as the alkyl portion increases in size. The author explains this by saying that it is not really due to the electron-donating effect of the alkyl groups, concentrating the negative charge on the oxygen atom. Rather, he argues that it is due to the decreased extent of solvation due to the steric hindrance provided by the bulkiness of the alkyl groups. This is a reasonable rationalisation.

However, the authors then go on to discuss the acidity of the alcohols in the gas phase and mention that the acidity trend is reversed in this situation. The alkyl groups now increase the acidity of the alcohol, rather than decrease it. The authors then state that it is due to the electron-withdrawing nature of the alkyl substituents. They even provide a qualitative molecular orbital picture of this "electron-withdrawing" interaction, as shown in the following paragraph (p. 238):

Alkyl groups have both filled and empty molecular orbitals (see the problems at the end of Chapter 1 for several examples). A pair of electrons adjacent to an alkyl group can be stabilized through overlap with the LUMO of the alkyl group. (Similarly, an alkyl group stabilizes an adjacent empty orbital through overlap with the alkyl HOMO.)

I would like to ask if anyone has ever heard of such an unconventional rationalisation and whether there is a sound theoretical basis for the postulations made above by the authors. If the postulations are indeed inconsistent with current chemical understanding, please provide a clear response detailing the areas of conflict.

  • $\begingroup$ What's supposed to be "unconventional"? That one may have, say -CH3 < -H < X in one situation and -H < -CH3 < -X in another? Or some hyperconjugation? $\endgroup$
    – Mithoron
    Oct 23, 2018 at 19:35
  • 1
    $\begingroup$ @Mithoron The unconventional part is that alkyl groups which are normally said to be electron-donating are electron-withdrawing here. $\endgroup$ Oct 23, 2018 at 22:51
  • 1
    $\begingroup$ Ah, seems you didn't got it, all this +/- I has nothing to do with actual electron density transfer. H is arbitrarily put as "0" and substituents making stronger acid as "withdrawing" but actually C-H bonds are polarised to carbon and C-C can be symmetrical. All this is hardly "real" science, just a little something for highschoolers to have any relations for acidity etc. $\endgroup$
    – Mithoron
    Oct 23, 2018 at 23:30
  • $\begingroup$ Check out hyperconjugation, or the anomeric effect $\endgroup$
    – Blaise
    Nov 2, 2018 at 12:22
  • $\begingroup$ Related: chemistry.stackexchange.com/questions/42535/… $\endgroup$ Feb 17, 2019 at 7:35

3 Answers 3


I wonder if the pKa for water (15.7) is a typo. Usually we think it to be 14.0 (at 25C, anyway). And I wonder how much the gas phase acidity changes over this series.

Acidity in the gas phase means ROH must dissociate to RO- and H+ but doesn't put the proton onto anything else. Since the proton isn't solvated, it must have been knocked loose, say in a mass spectrometer, where the fragmentation potential can be measured. That just seems quite different from our usual idea of "acidity". But OK, it's an observation of a characteristic of a series of related compounds.

The tendency to designate specific places for electrons (LUMO, HOMO) is understandable, but sometimes this description is uncertain and unnecessary. In a gas phase RO- ion, the negative charge is not 100% around the nucleus of the oxygen atom. Surely it will become unsymmetrical and drift where it might find a little positivity, like on the carbons and the hydrogens. Alkyl groups are polarizable and stabilize both anions and cations (https://onlinelibrary.wiley.com/doi/abs/10.1002/ejoc.200700004).

It's simpler to explain the acidity changes in water and in the gas phase by referring to the size-dependent polarizability of the alkyl groups, but using molecular orbitals may have seemed better for publication.

  • $\begingroup$ I never knew that the effect is stabilising for both the cation and the anion. I guess it is not right to say that alkyl groups are electron-releasing all the time. Thanks for the insights. I think I'll just award you the points. $\endgroup$ Nov 8, 2018 at 10:29

Note that the table compares very small acidities, and the difference for different alkyl substituents is small (although the table for the gas phase data is missing), so the contribution of the alkyl groups to the total electron withdrawing effect (stabilizing the loss of a proton) is small. One doesn't think of carbon as electron withdrawing relative to oxygen, but it does have a higher electron affinity than H. The point, in any case, is to explain relative acidities rather than absolute acidities.

Consider then the general gas phase dissociation reaction

$ X-O-H \ce{->} X-O^- $ + $ H^+$

or alternatively the electron affinity of $X-O$ radicals:

$ X-O^. + e^- \ce{->} X-O^- $

A simple but intuitive physical model that suggests why $[tBuO^-]/[tBuO^.]$ should be larger than $[MetO^-]/[MetO^.]$ in the gas phase is the particle in a box model. If you assume the charge density is uniform in the alkyl groups (reasonable for hydrocarbons), then the extra electron has more room to move about (a bigger "box") in the tBu ion. Since the energy levels are inversely proportional to the size of the box, then the energy of the electron will be lower (more stable) in the larger tBu ion. The greater delocalization in the $tBuO^-$ ion lowers the energy of the electron relative to the case of the methoxy ion, so that the ionization energy is lower and acidity highest for tBuOH in vaccuum.

Arguing in term of LUMOs might be more accurate but a simple delocalization model is sufficient in this case.

It would also be interesting to compare acidities in an organic solvent. See e.g. this article.

  • $\begingroup$ That is an interesting perspective. Could you find a source to support yourself? $\endgroup$ Nov 4, 2018 at 23:46
  • $\begingroup$ What would facilitate this delocalisation of charge, considering that the carbon and hydrogen atoms are less electronegative? $\endgroup$ Nov 5, 2018 at 7:56
  • $\begingroup$ Molecules are composed of electrons and positively charged nuclei. While electrons repel the lone electron on the oxygen, the nuclei attract it. The electrons are happy to buzz about over a larger volume, this lowers their energy, particularly if there are some attractive charges such as protons. This is a variation on the theme that atoms can bond to form molecules despite individual atoms being neutral. $\endgroup$
    – Buck Thorn
    Nov 5, 2018 at 9:00
  • $\begingroup$ You might consider comparing electron affinities of other compounds, See e.g. pubs.acs.org/doi/abs/10.1021/j100021a060?journalCode=jpchax . Or consider even atomic EAs: en.wikipedia.org/wiki/Electron_affinity. Despite the larger size, larger atoms with identical nuclear charge often share comparable EAs, which can be explained as being due to the size of the larger atoms (delocalization). $\endgroup$
    – Buck Thorn
    Nov 5, 2018 at 9:23
  • $\begingroup$ Perhaps use of the term "delocalization" is a bit loose, since this typically refers to conjugated systems. But the idea is that (i) we are comparing similar compounds, and (ii) electrons are not localized, so when we speak of more electronegative elements attracting electrons keep in mind that the electrons are still buzzing about over as large a volume as possible where there are attractive interactions. $\endgroup$
    – Buck Thorn
    Nov 5, 2018 at 9:51

There is absolutely nothing unconventional about this rationalisation, although it appears to not be widely known.

Carbocations are stabilized by hyperconjugation. This is an orbital overlap that can be quantified. The question is whether hyperconjugation forms part of the polarizability (redistribution of electrons). Polarizability seems (to me) to be quite a nebulous term.

Carbanions are also stabilized by orbital overlap, which could also potentially be covered under the same term.

Electronegativity is a poor argument to use to explain the extent and direction of stabilization.


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