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Homework Help: Methanol vs water acidity

  1. Sep 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Soooo...looking at Ka values, methanol is more acidic than water. Ethanol is less acidic.

    I understand that the reason ethanol is less acidic than water is because the ethanol ion has an inductive effect where more electrons are donated to the oxygen, so it being more negative it is more likely to pick up a hydrogen again. Why then is methanol more acidic than water?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Sep 23, 2012 #2


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    Methanol suffers less inductive effect from Methyl group than Ethanol suffers from Ethyl group.

    Since the acidic strength is determined in aqueous solution itself, by the tendency of the molecule to deprotonate itself due to its acidic H atom being pulled out due to H-Bonds from water.

    The strength of H-Bond between H-atom of water and O-atom of water is somewhat uniform (Water Water everywhere) but there is the difference in methanol (due to different groups attached to O atom in methanol). This might be the cause why Methanol is more acidic than water.
  4. Sep 29, 2012 #3


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    I think this sort of explanation (but which maybe is, surely was, in the textbooks and teaching) has to be treated with caution.

    Of one thing I am sure, you must never limit yourself to looking only at the molecules ROH, and RO- you have to look at them in the surrounding solvent.

    Your explanation actually is applicable to the gas phase for the equilibrium ROH ⇔ RO- + H+ where it turns out however to be wrong.

    The trend in the alcohols, I quote
    "was thought at one time to be due to a molecular electronic effect. The larger and more branched alkyl groups... were believed to have increasing electron-donating inductive effects which increasingly destabilise the alkoxide conjugate bases.

    The gas phase activities are in the opposite order however."

    (Tert-butanol is a stronger acid than methanol in fact)
    (CH3)3COH > (CH3)2CHOH > C2H5OH > CH3OH

    "The gas phase results, which reflect the inherent acidities, invalidate the earlier molecular electronic interpretation and establish that the inversion in water must be due to solvation. But is it nor clear whether we are seeing and enthalpy or entropy solvation effect without separate ΔAHH0 and ΔAHS0 effects."

    I am surprised that these quantities were not available at the time of writing (1984). The quotes come from "The physical basis of organic Chemistry" by Howard Maskill. Electronic explanations do seem to work for some things however, or rather I guess, solvent effects although always important, can be constant enough in certain comparisons.

    But from the several pages of the above book (all relevant I've got :blushing:) about the factors determining acid/base strengths it would be quite hard for a student to distil much predictive ability IMO. So I ask whether there is any more up to date and also more didactic text or run-down - surely the chemical education community has been active on this since it has taught a bit wrong previously?

    The question is most clearly formulated as: why is the equilibrium

    CH3O- + C2H5OH ⇔ CH3OH + C2H5O-

    to the left? IOW why, in water, does negative the charge prefer to be next the methyl rather than the ethyl group?

    The non-polar group can be thought of as a hole in the water so it is surrounded by structured water with bonds between water molecules as in a water-air surface.

    Then my amateur explanation would be putting the charge involves breaking up a bigger surface and more bonds with the bigger nonpolar groups, so the charge prefers the smaller groups. *

    But I would very much like to hear what is the accepted picture.

    *Even then not sure I can account for why the pKAH of water (15.74) is between that of methanol and ethanol, which Maskill says is anomalous and 'must be due to solvation effects'
    Last edited: Sep 30, 2012
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