2017 H2 Chem P1 Q26 — Nitrogen Compound Solution

The answer is A (proline, with a protonated ring N and \(\text{CO}_2\text{H}\)).
Hot 6 mol \(\text{dm}^{-3}\) HCl hydrolyses every peptide (amide) linkage, releasing the five amino acids. In this strongly acidic medium every amine group is protonated and every carboxyl group stays as \(\text{CO}_2\text{H}\). The five products are:
• valine: \(\text{H}_3\text{N}^+\text{CH(CH(CH}_3)_2)\text{CO}_2\text{H}\)
• tyrosine: \(\text{H}_3\text{N}^+\text{CH(CH}_2\text{C}_6\text{H}_4\text{OH)CO}_2\text{H}\)
• proline: the ring secondary amine is protonated to \(\text{NH}_2^+\), with \(\text{CO}_2\text{H}\) on the ring carbon next to N
• aspartic acid: \(\text{H}_3\text{N}^+\text{CH(CH}_2\text{CO}_2\text{H)CO}_2\text{H}\)
• alanine: \(\text{H}_3\text{N}^+\text{CH(CH}_3)\text{CO}_2\text{H}\)
Option A is the protonated proline, formed from the ring residue in the middle of the chain.
• B is wrong: the valine is drawn as a zwitterion with \(\text{CO}_2^-\). In 6 mol \(\text{dm}^{-3}\) HCl the carboxyl group is present as \(\text{CO}_2\text{H}\).
• C is wrong: the aspartic acid side chain is drawn as \(\text{CO}_2^-\). In the acidic medium it is present as \(\text{CO}_2\text{H}\).
• D is wrong: the side chain has two \(\text{CH}_2\) groups between the ring and the chiral carbon. The tyrosine residue in the chain has only one \(\text{CH}_2\) group, so this compound is not formed.