Consider this problem:

Which is more acidic, $\ce{MeOH}$ or $\ce{MeSH}$?

According to my teacher:

To determine which is more acidic, figure out which forms the more stable conjugate base. In this case, $\ce{MeS-}$ is more stable than $\ce{MeO-}$. In general stability reduces in the order: $$\ce{RS- > RP- > RO- > RN- > RC-}$$

Why is the compound with $\ce{S-}$ or $\ce{P-}$ more stable than that with $\ce{O-}$? I know that oxygen is more electronegative than either sulphur or phosphorus. Shouldn't the descending order start with $\ce{RO-}$?

  • $\begingroup$ Consider the length of the RS-H bond vs the RO-H bond, this makes the RS-H bond weaker than the RO-H bond $\endgroup$
    – Waylander
    Mar 21, 2021 at 10:27
  • $\begingroup$ @Waylander I'm guessing RS-H is longer, but I don't know why. I saw your edit: why is RS-H longer than RO-H? $\endgroup$ Mar 21, 2021 at 10:30
  • 1
    $\begingroup$ Sulfur is a larger atom, the outermost electron shell is further from the nucelus $\endgroup$
    – Waylander
    Mar 21, 2021 at 10:50
  • $\begingroup$ @Waylander Isn't the outermost electron shell even farther in phosphorus? $\endgroup$ Mar 21, 2021 at 11:09
  • 1
    $\begingroup$ For phosphorus, it's a clear mistake, AFAICT - phosphine would probably be a bit more acidic then amine, but not even close to alcohol. $\endgroup$
    – Mithoron
    Mar 21, 2021 at 16:59

1 Answer 1


There is more to it than meets the electronegative eye.

Get into the habit of transferring attributes into energy. In this case the relevant energies are the electron affinity of the base money ($\ce{RO}$ vs $\ce{RS}$) versus the bond energy of the moiety to hydrogen. A strong anionic Bronsted-Lowry base will have a low value for the first versus a high value for the latter.

Surprise! Sulfur has a higher electron affinity than oxygen! Add the fact that the oxygen-hydrogen bond is stronger than the hydrogen-sulfur bond and by both measures, $\ce{RO^-}$ appears as the stronger base.

The higher electron affinity of sulfur and even selenium, compared with oxygen, has another curious effect. We know about the aromaticity of certain ring ketones in which the electronegative oxygen atom imparts a positive charge to a three- or seven-membered carbon ring. It turns out that despite what an elctronegativity table might suggest, the higher electron affinity of sulfur and selenium versus oxygen causes this characteristic to carry over to the corresponding $\ce{C=S}$ and $\ce{C=Se}$ compounds.

  • $\begingroup$ Your explanation makes sense for $\ce{RS- > RO- }$, but is $\ce{RP-}$ more stable than $\ce{RO-}$ (I realize I'm asking a slightly different question now)? The figures in the electron affinity table you linked and the fact that phosphorus is larger than sulphur, would suggest that $\ce{RP-}$ is the stronger base. $\endgroup$ Mar 21, 2021 at 11:26
  • $\begingroup$ My bond-energy table does not include phospgorus-hydrigen, but if the latter bond is weaker than sulfur-hydrogen it would compete against your arguments. $\endgroup$ Mar 21, 2021 at 11:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.