Nitrogen Compounds: Question 9

Worked solution

Step 1: Predicting the predominant form at pH 1

At its isoelectric point (around pH 6 for alanine), the amino acid exists predominantly as the neutral zwitterion, $\text{CH}_3\text{CH(NH}_3^+\text{)COO}^-$, with the amine group protonated and the acid group deprotonated. At pH 1, the solution is well below this isoelectric point, meaning there is a large excess of $\text{H}^+$ ions available.

This excess of $\text{H}^+$ has two effects:

• It keeps the amine group protonated as $-\text{NH}_3^+$ (as it already was in the zwitterion).
• It also protonates the carboxylate group, converting $-\text{COO}^-$ back into undissociated $-\text{COOH}$, since the large excess of $\text{H}^+$ suppresses the acid’s own ionisation.

So the predominant form is the fully protonated cation, $\text{CH}_3\text{CH(NH}_3^+\text{)COOH}$, which carries an overall charge of $+1$.

Step 2: Predicting the direction of migration

In electrophoresis, a species migrates towards the electrode of opposite charge. A cation (overall positive charge) is attracted to, and migrates towards, the negative electrode, the cathode.

Step 3: Why the other options are wrong

• A is the neutral zwitterion, the form that dominates only near the isoelectric point (around pH 6), not at pH 1; it also incorrectly claims no migration for what would, in any case, be the wrong species here.
• C shows the anionic form (with -COO- but -NH2 unprotonated), which dominates well above the isoelectric point, not below it, and would migrate towards the anode, not the species present here.
• D shows the fully neutral, unprotonated structure, which does not exist as the major species at any pH for a simple amino acid like alanine, and is not present in solution as the “does not migrate” case at pH 1.

Final answer

B, $\text{CH}_3\text{CH(NH}_3^+\text{)COOH}$, the fully protonated cation, migrates towards the cathode.