AQA A-Level Chemistry Paper 2, June 2025: Question 6

✅ Correct Answer

• D₂O rather than CCl₄: D₂O is polar (or CCl₄ is non-polar / D₂O dissolves lysine / D₂O is a good solvent for polar amino acids). [1 mark]
• D₂O rather than water (H₂O): D₂O has no hydrogen atoms (no ¹H protons) and therefore does not produce a solvent peak in the ¹H NMR spectrum. [1 mark]

💡 Key Knowledge

• Amino acids exist as ionic zwitterions or highly polar molecules containing -NH₂ and -COOH groups. They dissolve readily in polar solvents like water/D₂O, but are virtually insoluble in non-polar solvents like CCl₄.
• Deuterium (²H or D) has an even mass number and a nuclear spin that does not resonate in the frequency range scanned for ¹H NMR. Replacing ¹H with ²H makes the solvent "invisible" in ¹H NMR.

🧠 Exam Technique

Treat this as two separate, single-mark questions:

• Why not CCl₄? This is purely about solubility. Amino acids will not dissolve in non-polar tetrachloromethane.
• Why not H₂O? This is about spectral interference. Water contains immense numbers of ¹H atoms that would drown out the small sample signals.

❌ Common Errors

• Saying "D₂O is inert" or "CCl₄ reacts with lysine" — this is incorrect; it is strictly a solubility issue.
• Confusing ¹H and ²H by writing "D₂O has no protons" in a subatomic sense (deuterium contains 1 proton and 1 neutron). The mark scheme accepts "no H atoms" or "no ¹H atoms". To be precise and safe, state: "D₂O contains no ¹H nuclei / gives no peak in the ¹H NMR spectrum."

✅ Correct Answer

The neutral or zwitterionic structure of histidine:

Molecular formula: C₆H₉N₃O₂

💡 How to Dissect the Complex

• Identify the sequence: Gly-His-Lys.
  • Top residue: H₂N-CH₂-CO- (Glycine)
  • Middle residue: -NH-CH(CH₂-imidazole)-CO- (Histidine)
  • Bottom right residue: -NH-CH((CH₂)₄NH₂)-COOH (Lysine)
• Extract Histidine: Break the two peptide amide bonds on either side of the middle unit:
  ...-C(=O) | NH-CH(R)-C(=O) | NH-...
• Add -H to the nitrogen to reform the free amine -NH₂ .
• Add -OH to the carbonyl carbon to reform the carboxylic acid -COOH .

🧠 Exam Technique

• Remember every standard 2-amino acid has the core unit: H₂N-CH(R)-COOH .
• Once you locate the central alpha carbon attached to the heterocyclic ring via a -CH₂- group, simply construct the standard α-amino acid backbone around it.
• Both neutral form ( H₂N-CH(R)-COOH ) and zwitterion ( H₃N⁺-CH(R)-COO⁻ ) are fully credited.

❌ Common Errors

• Leaving the amino acid as an amide fragment (e.g. leaving off the -OH or writing it still bonded to Cu²⁺).
• Incorrect valency on the nitrogen atoms in the imidazole ring (ensure the ring matches Figure 6 with one =N- and one -NH-).
• Missing the -CH₂- spacer between the α-carbon and the 5-membered ring.

✅ Correct Answer

• Type of secondary structure before heating: α-helix (alpha helix) [1 mark]
• Type of secondary structure after heating: β-pleated sheet (beta-pleated sheet / beta sheet) [1 mark]

💡 Key Knowledge

• Secondary structure refers to the regular, repeating spatial arrangements of the polypeptide backbone held together by hydrogen bonds between C=O and N-H groups of peptide bonds.
• α-helix: A right-handed coiled ribbon structure with intra-chain hydrogen bonds running roughly parallel to the helical axis.
• β-pleated sheet: Flattened, pleated accordion-like ribbons where chains lie alongside each other with inter-strand hydrogen bonding.

🧠 Exam Technique

• You can use either the Greek letter or full English word: α-helix or alpha-helix, β-pleated sheet or beta-pleated sheet.
• Ensure you specify both parts of the term (do not just write "helix" or "sheet").

❌ Common Errors

• Confusing secondary structures with tertiary structures (e.g. writing "globular", "fibrous", or mentioning ionic bonds / disulfide bridges).
• Getting the order reversed: the spiral is the helix (before heating); the folded/pleated ribbon is the sheet (after heating).