AQA A-Level Chemistry Paper 2, 2018: Question 8

✅ Mark Scheme Model Answer

• Mark 1: An electron-deficient H atom (or H that is δ⁺ / slightly positive).
• Mark 2: Attracts a lone pair / electron pair on the O atom (of a C=O group).

💡 Key Knowledge

• Nitrogen and oxygen are highly electronegative compared to hydrogen and carbon.
• The N–H bond is polar: N is δ⁻ and H is δ⁺.
• The C=O bond is polar: C is δ⁺ and O is δ⁻, leaving two lone pairs on oxygen.
• A hydrogen bond is an electrostatic attraction between the δ⁺ H atom and an O lone pair.

❌ Common Student Errors

• Describing it as a coordinate/dative bond: The mark scheme explicitly states "Penalise lone pair/electron pair donation". Hydrogen bonds are non-covalent electrostatic attractions, not dative covalent bonds.
• Vague polarity: Stating just "oxygen is negative" without mentioning the lone pair of electrons loses Mark 2.

🧠 Exam Technique

Always use precise bonding vocabulary for 2-mark hydrogen bond questions:

"Electrochemical attraction between the δ⁺ (electron-deficient) hydrogen on N–H and a lone pair of electrons on the electronegative oxygen atom of C=O."

✅ Correct Structure

The completed link between the two main peptide backbone carbons must show the full cysteine side chains and the disulfide bond:

💡 Key Knowledge

• Look up cysteine in your AQA Data Booklet: its side chain (R group) is —CH₂—SH .
• Two thiol ( —SH ) groups undergo oxidation to lose two H atoms, forming a covalent disulfide link ( —S—S— ).
• Unlike hydrogen bonds, disulfide bridges are strong, covalent bonds.

❌ Common Errors & Penalties

• Missing the —CH₂— groups: Directly linking the backbone —CH— to —S—S— scores 0 marks. Always check the Data Booklet for the full R group!
• Dashed or dotted lines: The mark scheme notes: "Penalise dashed/dotted S—S". Dashes imply intermolecular interactions; disulfide bonds are full single covalent bonds and must be drawn as solid lines.

✅ Acceptable Answers

• Tertiary (or 3°)
• Quaternary (or 4°)

💡 Hierarchy of Protein Structures

• Primary (1°): Sequence of amino acids joined by peptide bonds.
• Secondary (2°): Local folding (α-helices and β-pleated sheets) held only by hydrogen bonds between the peptide backbone ( C=O and N–H ).
• Tertiary (3°): Overall 3D folding of a single polypeptide chain maintained by interactions between R side chains (disulfide bridges, ionic bonds, H-bonds, van der Waals forces).
• Quaternary (4°): Arrangement and bonding between multiple polypeptide chains (subunits).

✅ Correct Answer (Either Isomer Accepted)

Option 1: Alanine–Serine (Ala-Ser)

Option 2: Serine–Alanine (Ser-Ala)

📐 Step-by-Step Construction

1. Find R-groups in Data Booklet:
   Alanine: R = —CH₃
   Serine: R = —CH₂OH
2. Link via a peptide (amide) bond:
   Join the carboxyl carbon of one amino acid to the amino nitrogen of the other: —C(=O)—NH— .
3. Add zwitterion charges:
   • Protonate the free amine group: —NH₂ → ⁺H₃N—
   • Deprotonate the free carboxylic acid group: —COOH → —COO⁻
4. Check neutrality:
   One positive charge (+1) and one negative charge (–1), giving a net charge of 0.

❌ Critical Error Warning (Consequential Error Rule)

"Incorrect peptide bond CE = 0"

If your peptide bond is drawn incorrectly (e.g. adding an extra oxygen atom, drawing —CO—O—NH— , or missing the NH group), this is a Chemical Error (CE) and you score 0/2 for the entire question, even if your zwitterion charges are correct.

🧠 Top Tip for Dipeptides

• The question specifies "a dipeptide" without setting the order. Either Ala-Ser or Ser-Ala is credited.
• Do not ionise the neutral —OH side-chain on serine. Serine's alcohol group does not form ions at neutral pH.
• Ensure the positive charge is shown on or next to the nitrogen atom ( ⁺H₃N— or H₃N⁺— ), not floating near carbon.