Acids and Bases, Part 2: Resonance

Last time, in Acids and Bases, Part 1, we found a single rule that does most of the work: the stronger acid is the one that gives the more stable conjugate base, and the negative charge sits best on the more electronegative atom. Call it Factor 1.

Did you practise it? Good. Let's stretch it on a couple of problems first, then meet the factor you reach for when Factor 1 goes quiet.

Warm-up: let Factor 1 do its job

Picture a molecule that has both an O–H and an N–H. A base comes along. Which H does it take?

Pull the proton off oxygen and the negative charge lands on O. Pull it off nitrogen and the charge lands on N. So ask one question: where is the charge happier?

Oxygen is more electronegative than nitrogen. It holds the negative charge better. So the O–H is the more acidic one, and that's the H the base grabs. You decided that almost without trying.

One more. Suppose a molecule has both an S–H and an O–H. Which H is most acidic?

Hint: the most acidic H is the one that leaves behind the most stable conjugate base.

Sulfur sits below oxygen in the group. It's bigger, so it spreads that negative charge over a larger volume — the charge is more stable on S. So here the S–H is the most acidic. Factor 1, again, settles it.

When Factor 1 has nothing to say

Now the honest part. Sometimes Factor 1 just shrugs. Which is more acidic:

$CH_3COOH \quad \text{or} \quad CH_3OH$

Take a proton off either one and the negative charge ends up on an oxygen. Same atom, both times. Factor 1 can't pick a winner — the atoms tie.

So say hello to Factor 2: resonance.

How resonance changes stability

Here's the whole idea in one line:

More resonance = more stability.

A charge that's stuck in one place is fragile. A charge that's smeared over several atoms is calm. Resonance is nature's way of sharing the burden.

Look at the two conjugate bases. Methoxide has nowhere to send its charge:

But acetate spreads its charge across both oxygens. These two structures are equally real; the truth is the average of them:

That sharing is resonance stabilisation. Acetate is the calmer ion, so it forms more readily — which means $CH_3COOH$ is the stronger acid. Methanol can't compete.

Important: reach for resonance only after Factor 1 fails to decide. If one conjugate base lands its charge on a more electronegative atom, you're already done — don't overthink it.

One more, to make it stick

Imagine a molecule with two different O–H groups. Remove either H and the charge sits on oxygen both times — so Factor 1 ties again. Think resonance.

If pulling one particular H lets the resulting negative charge delocalise into a neighbouring $\pi$ system (a carbonyl, a ring, a double bond), that conjugate base is resonance-stabilised. The other one isn't. So that H is the more acidic one. The molecule that shares its charge always wins.

Your turn. Which is the stronger acid: ethanol ($CH_3CH_2OH$) or acetic acid ($CH_3COOH$)? Name the deciding factor.

Check: Acetic acid. Both conjugate bases put the charge on oxygen, so Factor 1 ties. But acetate is resonance-stabilised across two oxygens while ethoxide is not — Factor 2, resonance, decides it.

The two factors so far

• Factor 1 — the atom bearing the charge. More electronegative (or larger, down a group) holds the charge better. Try this first.
• Factor 2 — resonance. Use it only when Factor 1 ties. More resonance means more spread-out charge, a more stable conjugate base, a stronger acid.

Next we add Factor 3 — the inductive effect, for the cases where even resonance can't separate two acids. Stack the three in order and you can rank almost any pair on sight.