# Why is ortho-hydroxybenzoic acid more acidic than its para-isomer?

Why is ortho-hydroxybenzoic acid $$(\mathrm{p}K_\mathrm{a} = 2.98)$$ more acidic than its para-isomer $$(\mathrm{p}K_\mathrm{a} = 4.58)$$?

March's Advanced Organic Chemistry (7th ed) gives the reason to be intramolecular hydrogen bonding between the $$\ce{OH}$$ and $$\ce{COO-}$$ groups of the conjugate base.

However, note that intermolecular hydrogen bonding (between the $$\ce{OH}$$ and $$\ce{COO-}$$ of two different molecules) is possible in the para-isomer. Interestingly, this question also states that intermolecular hydrogen bonding is relatively stronger than intramolecular one. So, shouldn't the para-isomer be more acidic?

I don't think that in the first link it is stated that the intermolecular $$\ce{H}$$-bonding is stronger than the intramolecular one. On the contrary, the accepted answer starts as:

"I don't think there is much of a difference between the strengths of intramolecular or intermolecular hydrogen bonds." With which I agree.

In that post they discuss the effect of intermolecular and intramolecular $$\ce{H}$$-bonding on the melting and boiling points of the compounds.

The answer by @AllYouCanEat86 that follows in the same post I think is more relevant to your question.

Therefore in our case there are intermolecular $$\ce{H}$$-bonds in p-hydroxybenzoic acid but they don't last for long due to the thermal motion and higher degrees of freedom of the $$\ce{COO-}$$ and $$\ce{OH}$$ groups. In contrast, in the ortho-isomer the $$\ce{COO-}$$ and $$\ce{OH}$$ groups are locked in close proximity due to the aromatic ring and therefore are forced to maintain the $$\ce{H}$$-bond for much longer. This could be thought as an equivalent to the chelation effect in metal complexes. Lastly, one could also argue that in the case of o-hydroxybenzoinc acid, there is both the option of intermolecular and intramolecular $$\ce{H}$$-bonding compared to the para where there is only intermolecular $$\ce{H}$$-bonding.

• Also, usually the pKa of a substance is determined in dilute solution, isn't it? So just due to dilution, there won't be many / any inter-molecular H bonds, but there can still be plenty of intra-molecular ones. – Curt F. Feb 21 '18 at 20:06
• Just curious, does steric inhibition of resonance also play a role in increasing the acidic strength of Ortho? – FreakyLearner Apr 6 '18 at 8:18

The intramolecular hydrogen bond in ortho-hydroxybenzoic acid between the $$\ce{COO-}$$ and the -$$\ce{OH}$$ groups forms a six membered ring. Since this structure is very stable, the deprotonation of the acid is favoured and the compound is more acidic.

In the case of the para-isomer, there is hydrogen bonding between different molecules which is much less stable. Therefore it's acidity is lower than that in the ortho-isomer.

In the link you have provided, they explain the lower melting and boiling points of the ortho-isomer. That is again because of the hydrogen bonding, but it does not give an indication is the relative strength of the intra and intermolecular types.

To melt a compound, the intermolecular forces that pack them together must be eliminated. In the ortho-isomer, an intramolecular hydrogen bond is formed. But this bond does not maintain the molecules together. When the molecules are separated, this hydrogen bond goes with each molecule in which it occurs. So it does not matter if this bond is strong or weak. It has not to be broken.

On the other hand, in the para-isomer, the intermolecular hydrogen bonds maintain the different molecules together, so more energy is required to separate them. The hydrogen bonds must be broken, and as a consequence, the melting point is higher.

This is the case in ortho- and para-hydroxybenzoic acid as well. The former has a melting point of $$\pu{158 ^\circ C}$$ and the latter has $$\pu{215 ^\circ C}$$.