2019 H2 Chem P1 Q25 — Acid base equilibrium Solution

The answer is B.

A group is mostly deprotonated when pH \(>\) its p\(K_\text{a}\), and mostly protonated when pH \(<\) its p\(K_\text{a}\).
Assigning the p\(K_\text{a}\) values: the two \(-\text{CO}_2\text{H}\) groups are more acidic than \(-\text{NH}_3^+\). The \(-\text{CO}_2\text{H}\) next to the electron-withdrawing \(-\text{NH}_3^+\) group is the more acidic (its carboxylate is more stabilised), so p\(K_\text{a}\) 2.1 belongs to it, 4.1 to the side-chain \(-\text{CO}_2\text{H}\), and 9.5 to \(-\text{NH}_3^+\).
At pH 7:
• both \(-\text{CO}_2\text{H}\) groups are deprotonated to \(-\text{CO}_2^-\) (7 \(>\) 2.1 and 7 \(>\) 4.1);
• \(-\text{NH}_3^+\) stays protonated (7 \(<\) 9.5).
The major species is \(^-\text{O}_2\text{CCH}_2\text{CH}_2\text{CH(NH}_3^+\text{)CO}_2^-\) (overall charge 1-), which is structure B.

• A is wrong: it shows \(-\text{NH}_2\); the \(-\text{NH}_3^+\) group loses its proton only above pH 9.5.
• C is wrong: it keeps the side-chain \(-\text{CO}_2\text{H}\) (p\(K_\text{a}\) 4.1) protonated; that form is the major species only between pH 2.1 and 4.1.
• D is wrong: it is the fully protonated form, the major species only below pH 2.1.