AQA A-Level Chemistry Paper 3, June 2018: Question 15

✅ Correct Answer

Correct Option: D

Structure:
H₂N—CH(COO⁻)—CH₂—COO⁻

At pH 12 (strongly alkaline), excess OH⁻ ions remove protons (H⁺) from both carboxylic acid groups present in aspartic acid, while the amino group remains unprotonated as —NH₂ .

💡 Key Knowledge

• Aspartic acid is an acidic amino acid containing two —COOH groups and one —NH₂ group.
• In strongly alkaline solution (pH 12–14):
  • Carboxylic acid groups lose H⁺:
    —COOH + OH⁻ → —COO⁻ + H₂O
  • Amine groups exist in neutral/free base form: —NH₂
• Total charge on the aspartate ion at pH 12 is -2.

🧠 Exam Technique: Systematic Elimination

1. Check the pH: pH 12 means high [OH⁻] (strongly basic).
2. Inspect the amine group: In base, amine groups cannot be positively charged. Any option with —NH₃⁺ is eliminated immediately ➔ Eliminates B and C.
3. Inspect ALL carboxylic acid groups: High pH will neutralise all acidic protons. Aspartic acid has two —COOH groups; both become —COO⁻ ➔ Option A only deprotonated one, which is wrong.
4. Confirm Option D: Both acid groups are —COO⁻ and the amine is neutral —NH₂ .

❌ Common Misconceptions & Errors

• Overlooking the side chain (Choosing A): Students frequently deprotonate the α-carboxylic acid group but forget that the side-chain —CH₂COOH is also a carboxylic acid that reacts with base.
• Confusing high and low pH (Choosing B): Structure B is the fully protonated species present in strongly acidic solutions ( pH 1 ).
• Assuming zwitterion dominates at pH 12 (Choosing C): The zwitterion predominantly exists around the isoelectric point ( pH ~3 for aspartic acid), not at pH 12.