AQA A-Level Chemistry Paper 2, 2017: Question 6

✅ Correct Answer

Draw phenylalanine showing both formal charges (+ and −) on the molecule:

Mark scheme: 1 mark for correct dipolar structure. Either ⁺H₃N— or H₃N⁺— is accepted. Either —COO⁻ or displayed —C(=O)O⁻ is accepted.

💡 Key Knowledge

• An amino acid has both a basic amino group and an acidic carboxyl group.
• In a zwitterion, the proton transfers from the carboxylic acid to the amine:
  -COOH + -NH₂ → -COO⁻ + -NH₃⁺
• The side chain for phenylalanine from the Data Booklet is -CH₂C₆H₅ (benzyl group). It is non-polar and does not ionise.

🧠 Exam Technique

Always consult the Data Booklet for the exact side chain ( R group). Be careful when placing the positive charge: place it clearly on or next to the Nitrogen atom, never on the carbon backbone.

❌ Common Errors

• Forgetting to show the charges entirely (drawing unionised H₂N-CH(R)-COOH ).
• Ionising only one end (giving a net +1 or -1 ion instead of a neutral zwitterion).
• Drawing the side chain incorrectly (e.g. omitting the -CH₂- spacer between the α-carbon and the benzene ring).

✅ Correct Answer

At high pH (alkaline conditions, excess OH⁻), the amino acid acts as an acid and loses a proton from the carboxyl group:

Mark scheme: 1 mark for the deprotonated carboxylate anion with an unprotonated amine group. (Alcohol side chain -CH₂OH remains intact).

💡 Key Knowledge

• High pH (alkaline): OH⁻ ions remove protons → -COOH becomes -COO⁻ , while -NH₂ remains unionised.
• Low pH (acidic): H⁺ ions protonate basic groups → -NH₂ becomes -NH₃⁺ , while -COOH remains neutral.
• Aliphatic alcohol groups (like the -CH₂OH in serine) are far too weakly acidic to deprotonate in aqueous alkali (unlike phenolic -OH ).

🧠 Exam Technique

Think: "High pH = Base = Wants to remove H⁺". The most acidic proton is on the -COOH group. The product must carry a net negative charge ( -1 ).

❌ Common Errors

• Leaving the nitrogen protonated as -NH₃⁺ in alkali.
• Deprotonating the alcoholic -OH group on serine to -O⁻ (alcohols do not deprotonate at normal high pH).

🧠 Exam Technique: "Show All Atoms and Bonds"

When the question states: "show all the atoms and bonds in the amide link", writing condensed -CONH- will score 0 marks for that link! You must explicitly draw:

• The carbon-to-oxygen double bond: C=O
• The carbon-to-nitrogen single bond: C—N
• The nitrogen-to-hydrogen single bond: N—H

❌ Common Errors

• Drawing two identical dipeptides instead of swapping the N-terminus and C-terminus.
• Drawing a condensed peptide link -CONH- and losing the mark despite correct connectivity.
• Inverting connectivity (e.g. connecting side-chain -OH instead of the α-amino or α-carboxyl groups).

✅ 1. Name of Mechanism [1 Mark, M1]

(nucleophilic) addition-elimination

Examiner note: "Electrophilic addition-elimination" is penalized and scores 0.

✅ 2. IUPAC Name of Organic Product [1 Mark, M6]

N-ethylpropanamide

Examiner note: Must have the capital N- prefix to show that the ethyl group is substituted onto nitrogen.

📐 Mechanism Breakdown [4 Marks, M2–M5]

Step 1: Nucleophilic Attack

• M2: Curly arrow from the lone pair on N of CH₃CH₂NH₂ to the carbonyl carbon atom ( C=O ).
• M3: Curly arrow from the C=O double bond onto the oxygen atom.

Step 2: Tetrahedral Intermediate Structure

• M4: Draw correct intermediate:
             :O:⁻
             |
  CH₃CH₂ — C — Cl
             |
           ⁺NH₂—CH₂CH₃
  Must clearly show the negative charge on Oxygen (O⁻) and the positive charge on Nitrogen (N⁺).

Step 3: Elimination (3 Curly Arrows)

• M5: Three simultaneous/sequential arrows:
  1. Arrow from lone pair on O:⁻ back to reform the C=O double bond.
  2. Arrow from the C—Cl bond to the chlorine atom (expelling Cl⁻ ).
  3. Arrow from an N—H bond to the N⁺ atom (loss of proton). Note: A second amine molecule :NH₂CH₂CH₃ removing the H⁺ is accepted; using Cl⁻ to remove H⁺ is penalised.