I'm totally confused about the relationship between the polarity and acid strength of a given compoud.

Take the case of alcohols itself, it's said that primary alcohols have greater acid strength compared to secondary alcohols as the electron-releasing group in secondary alcohol decreases the polarity of $\ce{O-H}$ bond.

But why is this so? I mean, less polarity will help in liberating $\ce{H+}$ ions more easily so acidity is meant to be increased.

That means secondary alcohols are meant to have more acid strength.

Some help to solve this dilemme is appreciated.

  • $\begingroup$ Why will less polarity help in liberating $\ce{H+}$? $\endgroup$ – Avnish Kabaj Feb 17 '18 at 2:22
  • $\begingroup$ Greater polarity means they are more tightly attracted, right? $\endgroup$ – SlayerDiAngelo Feb 17 '18 at 2:31
  • $\begingroup$ I'm not sure why you even need to invoke "polarity" in this case. In fact, reduced polarity and high acidity seem to be separate cases arising from the root case of positive inductive effect of more carbon atoms towards the alcoholic group. $\endgroup$ – Gaurang Tandon Feb 17 '18 at 3:00
  • $\begingroup$ @Gaurang I didn't get how are both separate in this case aren't polarity and acidity linked if we are judging from the basis of inductive effects? $\endgroup$ – Avnish Kabaj Feb 17 '18 at 3:04
  • $\begingroup$ @AvatarShiny Yes, they are "linked" in the sense that they both arise from the same "cause". But, that doesn't mean the acidic strength is higher because of the reduced polarity. The relation is a correlation but not a causality. (video) $\endgroup$ – Gaurang Tandon Feb 17 '18 at 3:09

In analysing acid strength, bond polarity of the bond to the ionisable hydrogen atom is what should be considered, not the polarity of the molecule (i.e. the dipole moment). In the case of alcohols, the polarity of the $\ce {O-H}$ bond should be considered when assessing strength as an acid.

Why does the bond dipole matter? In the process of ionisation $\ce {ROH + H2O-> RO^{-} + H3O^{+}}$, the $\ce {O-H}$ bond is heterolytically cleaved. It is easy to observe that the greater the partial positive charge on the $\ce {H}$ atom, the more easy it is for it to be abstracted by the water molecule, due to stronger electrostatic attractions. Also, the more polar the bond, the lower the covalent character of the bond as the electron density shifts more towards the oxygen. Such a bond is also more easily cleaved. Recall that the covalent character of a covalent bond is highest when the atomic orbitals are of similar energies and that electron density is shared equally between the two atoms.

In the primary alcohol, there is one electron-donating alkyl group attached to the alpha carbon while in the secondary alcohol, there are two such groups attached to the alpha carbon. Due to the greater electron donation to the oxygen atom in the secondary alcohol, where $\ce {R = CHR'R''}$ , the oxygen atom actually withdraws less electron density from the $\ce {O-H}$ bond. We can think of it this way: The oxygen atom has a fixed demand for electron density. When this demand is met by an increased supply from the alkyl chain, the demand for electron density from the hydrogen atom would decrease. Thus, the $\ce {O-H}$ bond is less polar and less easily heterolytically cleaved.

Simarly, for the primary alcohol, where $\ce {R = CH2R'}$, there is also an increased supply of electron density from the alkyl chain R, as compared to $\ce {R = CH3}$. However, the increased supply from $\ce {R}$ is less so compared to that in the secondary alcohol. Thus, the primary alcohol would be a weaker acid than the methyl alcohol but a stronger acid than the secondary alcohol.

Conventionally, acid strength is assessed using conjugate base stability, rather than bond polarity. A brief overview of assessing conjugate base stability can be found here.

  • $\begingroup$ I'm so delighted to have a valid explanation. Can we apply the same reason in the case of acidity of phenol? $\endgroup$ – SlayerDiAngelo Feb 17 '18 at 3:45
  • $\begingroup$ Water abstracts $H^+$ ion, so do water have a negative character? Forgive me, if this sounds absurd. $\endgroup$ – SlayerDiAngelo Feb 17 '18 at 3:53
  • $\begingroup$ @SlayerDiAngelo This explanation is universal and works in every case. Of course, in different cases, there may be other factors which may have significant influences on the acidity as well. In the case of phenol, we would have to look at the delocalisation of the negative charge due to the pi system as well. $\endgroup$ – Tan Yong Boon Feb 17 '18 at 4:00
  • $\begingroup$ @SlayerDiAngelo The water molecule is able to abstract the hydrogen atom as it has a lone pair of electrons on the oxygen atom which can be donated to form a bond with the hydrogen atom. The abstraction of the hydrogen by water molecule is further enhanced by the electrostatic attraction between the partial charges on oxygen and hydrogen. $\endgroup$ – Tan Yong Boon Feb 17 '18 at 4:02

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