Acidity order of nitrophenols

Question

As in other posts I have seen and learnt that, if one is given four compounds, namely para-nitrophenol, ortho-nitrophenol, meta-nitrophenol and phenol and is told to arrange them in order of acidity, then one has to take into account the $\text{-R}$ and $\text{-I}$ effect.

Now, for the inductive effect, the greatest intensity is observed at the ortho-position, followed by meta-, and then para-.

For the resonating effect, the intensity is greatest for para and lowest for ortho, while meta has no such effect. Now keeping in mind, that $\text{-R}$ and $\text{-I}$ both tend to withdraw electrons away from the phenol group (and thus increases the acidity), and the $\text{R}$ effect has more contribution than the $\text{I}$ effect, we can thus deduce that para-nitrophenol is most acidic, followed by ortho-nitrophenol, then meta-nitrophenol and since phenol has no ring activating group, it comes out to be least acidic in the given compounds.

The facts are the proof of my theory, where the $\mathrm{p}K_\mathrm{a}$ for: $$ \begin{array}{c|c} \text{Compound} & \mathrm{p}K_\mathrm{a} \\ \hline para\text{-nitrophenol} & 7.16\\ ortho\text{-nitrophenol} &7.2\\ meta\text{-nitrophenol} & 9.3\\ \text{phenol} & 9.95 \end{array} $$

So far so good.

But one of my colleagues, when I asked the problem, tried to give the theory in different way. He said, in para-nitrophenol, hydrogen bonding is strongest, so it can't be most acidic, which in his view port should be least. I knew this is not the case, but I just can't explain or point out where he's been wrong. Can anyone of you make it little easy for me to understand it?

Answer 1

You have correctly identified that m-nitrophenol should be the weakest acid among the isomers as the resulting anion cannot be stabilised by the $\text{-M}$ effect of nitro group owing to its meta location.

The acidity order of the remaining two nitrophenols can be explained in this way:

1. Hydrogen bonding is possible in both of them. But note that:

$$\text{H-bonding} \propto \delta^+ \text{charge on $\ce{H}$ atom}$$ Inductive effect (here, $\text{-I}$) is distance dependent. So $\delta^+$ charge on the $\ce{H}$ of ortho-nitrophenol's -$\ce{OH}$ is going to be more than the $\delta^+$ charge the $\ce{H}$ atom of para-nitrophenol's -$\ce{OH}$.

2. Also, quoting from this comment:

Usually the $\mathrm{p}K_\mathrm{a}$ of a substance is determined in dilute solution. 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.

3. The answer above the comment relates the strength of these two types of hydrogen bonds to kinetic motion of molecules:

Inter-molecular interaction, such as inter-molecular hydrogen bonding only take place for a short time when two molecules are in each other's vicinity. Shortly after, the molecules separate and the interaction does no longer exist.

Intra-molecular interactions on the other hand are much stronger since the two atoms forming the hydrogen bond remain in each others proximity due to structure of the molecule. In other words, they exhibit strain.

These three factors help us conclude that:

Strength of intra-molecular hydrogen bonding in o-nitrophenol $\gt$ Strength of intermolecular hydrogen bonding in p-nitrophenol

Thus, p-nitrobenzoic acid is more acidic than the o-isomer.

...and since phenol has no ring activating group, it comes out to be least acidic in the given compounds.

This reasoning is incorrect. Phenol has no group that stabilises the negative charge on the resulting phenoxide ion through $\text{-M}$ or $\text{-I}$ effect (a deactivating group). Therefore, it's the weakest acid of all under investigation.

Answer 2

Due to $\text{-R}$ effect of nitro group at ortho- and para-positions, the conjugate base gets stabilised more than that at meta-position. Now at the ortho-position, H-bonding is present, which makes removal of $\ce{H+}$ ion difficult. Therefore correct acidic order of nitrophenols are para- $\gt$ ortho- $\gt$ meta-.

Answer 3

In both ortho- and para-nitrophenol, there is one more resonance structure. So due to resonance effects both ortho- and para-nitrophenol are more acidic than the meta-nitrophenol.

Now we compare the acidity of ortho- and para-nitrophenol. In ortho-nitrophenol, interamolecular H-bonding takes place because of attachment with adjacent carbon atom. But in para-nitrophenol, there is no interamolecular H-bonding taking place. So that's why para-nitrophenol is more acidic than ortho-nitrophenol. So, by increasing order of acidity,

para-nitrophenol $\gt$ ortho-nitrophenol $\gt$ meta-nitrophenol.

Answer 4

More the stability of phenoxide after removal of $\ce{H}$ more will be acidity.

1. ortho-Nitrophenol phenoxide ion is stabilised by -R and -I effect where both -I and -R effect are maximum.
2. meta-Nitrophenol phenoxide ion is stabilised by -I effect only.
3. para-Nitrophenol phenoxide ion is stabilised by -R and -I effect where both -I and -R effect are minimum.

Thus according to stability of anions o-derivative will be more acidic than p-derivative which will be more acidic than m-derivative.

But due to H-bonding in ortho-nitrophenol it is reluctant to remove hydrogen due to which it's acidity decreases and order becomes:

p-derivative $\gt$ o-derivative $\gt$ m-derivative $\gt$ phenol (acidity in decreasing order)

Source- Jagdamba Singh: A logical approach to Modern Organic Chemistry. Revised edition 2009. Pg 151