WJEC A-Level Chemistry Unit 4, June 2025: Question 8

✅ Correct Answer

Electrophilic addition

❌ Common Errors

Writing "nucleophilic addition" (confusing alkenes with carbonyls) or omitting "addition" and writing just "electrophilic".

✅ Correct Answer

The reaction proceeds via a secondary carbocation, which is more stable than the alternative primary carbocation (lower activation energy / formed at a faster rate).

💡 Key Knowledge

The secondary carbocation CH₃–C⁺H–COOH is stabilised by the inductive electron-releasing effect of the adjacent methyl group, compared to the less stable primary carbocation ⁺CH₂–CH₂–COOH .

✅ Correct Answer

CH₃CH(Br)COOH + 2NH₃ → CH₃CH(NH₂)COOH + NH₄Br

🧠 Exam Technique: Why 2 moles of NH₃?

The first NH₃ acts as a nucleophile displacing Br⁻. The second NH₃ acts as a base to remove a proton (H⁺), releasing the neutral amino acid and forming the ammonium salt NH₄Br .

✅ Correct Answer

H₃N⁺–CH(CH₃)–COOH

(Carboxylic acid group is uncharged –COOH , amine group is protonated –NH₃⁺ )

💡 Key Knowledge: pH vs. Isoelectric Point

At pH 4.0 (acidic conditions), the basic amine group acts as a base and accepts a proton to become –NH₃⁺ . The carboxylic acid group remains mostly un-ionised as –COOH .

✅ Correct Answer

RCH(NH₂)COOH

(Structural or displayed showing the –NH₂ and –COOH attached to the same carbon atom)

🧠 What does 'α' mean?

In an α-amino acid, both the amino ( –NH₂ ) group and the carboxyl ( –COOH ) group are bonded directly to the same central carbon atom (the α-carbon).

✅ Correct Answer (Either isomer accepted)

Ala–Gly: H₂N–CH(CH₃)–CO–NH–CH₂–COOH
or
Gly–Ala: H₂N–CH₂–CO–NH–CH(CH₃)–COOH

❌ Common Errors

Drawing the dipeptide as a zwitterion or forgetting to eliminate a water molecule ( H₂O ) when joining the amine and carboxylic acid groups.

✅ Correct Answer

The spatial arrangement/folding of the polypeptide backbone into an α-helix or β-pleated sheet, held together by hydrogen bonding between C=O and N–H groups.

❌ Don't Confuse with Tertiary Structure!

Secondary structure involves hydrogen bonds only within the peptide backbone. Mentioning interactions between R-groups (like disulfide bridges or ionic bonds) refers to the tertiary structure!

✅ Correct Answer

A chiral centre (or asymmetric carbon atom / carbon bonded to four different groups).

🧠 Identification

Locate the carbon with 4 different groups attached: –H , –NH₂ , –COOH , and the substituted benzyl group –CH₂–C₆H₃(OH)₂ .

✅ I. Advantage of Natural Extraction

• It is a renewable resource / plants can be regrown.
• Enzymes produce only the desired single enantiomer (L-dopa) with no unwanted D-enantiomer.
• Avoids toxic chemical reagents / avoids harmful synthetic by-products.

📐 II. Bean Mass Calculation

Step 1: Calculate total daily dose required
Daily dose = 3 × 250 mg = 750 mg = 0.750 g

Step 2: Calculate mass of beans needed (contains 6% L-dopa)
Mass of beans = 0.750 g / 0.06 = 12.5 g

✅ Completed Equation

C₆H₅CH₂COOH + 2NaOH → C₆H₅CH₃ + Na₂CO₃ + H₂O

Missing products: C₆H₅CH₃ (methylbenzene) and H₂O (water).

💡 What is Decarboxylation?

Decarboxylation removes a carboxyl group as carbon dioxide (forming carbonate in alkaline conditions, such as soda lime). Removing –COO– from phenylethanoic acid leaves methylbenzene ( C₆H₅CH₃ ).