Electron withdrawing groups and acidity

Question

I know that a strong electron withdrawing group reduces electron density and makes a molecule more acidic. Also, a meta substituent will have less effect than ortho and para substituence and the para isomer will be more acidic than the ortho one. The point that I wonder: Does the acidity always increase with the number of withdrawing group as well?

Some group, let's say, phenol has one strong electron withdrawing group like nitro, carboxyl, chloro etc. If phenol would have two more of that same withdrawing group attached, would it be simply more acidic? Or, should another factors be considered to determine the acidity? I thought that maybe it would be more reliable to look $\mathrm{p}K_\mathrm{a}$ values but couldn't find any molecule that fits that situation.

Answer 1

Yes and no, with more electron withdrawing groups at ortho and para positions would certainly increase acidity but other factors do play a role. For example, In trinitrobenzoic acid

here the two $\ce{NO2}$ at ortho positions are very bulky and due to steric repulsion rotate the $\ce{COOH}$ group thus making the compound non planar due to which there will be no more resonance thus, reducing acidity drastically. This is the most important factor that plays role in reducing the acidity.

Answer 2

Let’s look at some data:

• phenol: $\mathrm{p}K_\mathrm{a} = 9.86$
• 3-nitrophenol: $\mathrm{p}K_\mathrm{a} = 8.34$
• 4-nitrophenol: $\mathrm{p}K_\mathrm{a} = 7.23$
• 3,4-dinitrophenol: $\mathrm{p}K_\mathrm{a} =5.42$

$\Delta \mathrm{p}K_\mathrm{a}(\text{P/3-NP}) = 1.52$
$\Delta \mathrm{p}K_\mathrm{a}(\text{4-NP/3,4-dNP}) = 1.81$

$\Delta \mathrm{p}K_\mathrm{a}(\text{P/4-NP}) = 2.63$
$\Delta \mathrm{p}K_\mathrm{a}(\text{3-NP/3,4-dNP}) = 2.92$

As you can see, the differences in $\mathrm{p}K_\mathrm{a}$ are not strictly additive but enfore each other; one could almost say in a cooperative manner.