AQA A-Level Chemistry Paper 2, 2019: Question 9

✅ Correct Answer

Concentrated hydrochloric acid (or conc. HCl )

💡 Key Knowledge

Peptide (amide) bonds are relatively stable and require harsh conditions to hydrolyse completely into individual amino acids:

• Standard lab conditions: 6 mol dm⁻³ HCl (conc HCl), heated under reflux (often ~110 °C for 24 hours).
• Concentrated aqueous strong bases (e.g., conc NaOH) are also acceptable alternatives.

❌ Common Errors

• Writing just HCl or hydrochloric acid without specifying concentrated or giving a concentration ≥ 5 mol dm⁻³. Dilute acid is too slow to achieve complete hydrolysis.
• Suggesting water alone ( H₂O ) — water is a reactant, but it is not a suitable laboratory reagent on its own as uncatalysed hydrolysis is extremely slow.

🧠 Exam Technique

When asked for reagents that hydrolyse proteins, polyesters, or polyamides, always be precise with concentration and conditions. Writing "conc." takes one second and prevents losing an easy mark.

✅ Correct Answer

Any one of the following:

• Using ninhydrin (spray/reagent)
• Using ultraviolet (UV) light
• Using iodine vapour (I₂)

💡 Key Knowledge

Amino acids are colourless compounds and are invisible on silica plates under ordinary visible light:

• Ninhydrin: Reacts with primary amino groups to form an intense purple/blue complex (Ruhemann's purple).
• UV Light: Many TLC plates contain a fluorescent indicator; spots appear dark where amino acids quench the fluorescence.

❌ Common Errors

• Vague answers like "shine a light on it" without specifying ultraviolet (UV).
• Confusing chemical tests: suggesting Biuret reagent (which tests for peptide bonds in unhydrolysed proteins, not free individual amino acids).

🧠 Exam Technique

Memorise "ninhydrin" specifically for amino acid TLC. It is the examiner's favourite answer and applies specifically to amino acid detection.

✅ Correct Answer

7 (or seven )

📐 How to Count (Figure 2 Analysis)

Count every separate final spot (grey circles) visible after running solvent 2:

• Row 1 (top level from solvent 1): 1 grey spot
• Row 2 (middle-upper level): 2 grey spots
• Row 3 (middle-lower level): 2 grey spots
• Bottom line (origin line of solvent 1): 2 grey spots (these amino acids were insoluble or had zero Rf in solvent 1, but moved across in solvent 2)
• Total distinct spots: 1 + 2 + 2 + 2 = 7 spots

❌ Common Errors

• Counting 4: Only counting the spots that moved in Solvent 1 (hollow spots) or counting rows.
• Counting 5: Forgetting to count the 2 grey spots along the baseline that only separated in Solvent 2.
• Counting 8: Including the original empty circle at the injection spot along with the 7 final grey spots.

🧠 Exam Technique

Look carefully at the key: hollow circles show intermediate spots after solvent 1; grey circles show the final positions after solvent 2. Each final resolved grey spot represents at least one distinct amino acid.

✅ Correct Answer

Either of the following points:

• Some of the amino acids did not separate (or did not dissolve) in the first / either solvent.
• Some amino acids have the same Rf value (or the same affinity for the stationary/mobile phase) in the first / either solvent.

💡 Key Knowledge

In standard 1D chromatography, two different amino acids with very similar side-chain polarities may have identical R_f values in Solvent 1 and travel together as a single spot.

By rotating 90° and running a second solvent with a different polarity, the relative solubilities and affinities change, allowing the co-eluting amino acids to separate.

❌ Common Errors (Examiner Trap!)

• Stating merely: "Amino acids have different Rf values in different solvents." — Zero marks! (Explicitly rejected on the mark scheme because it does not explain why two were needed).
• Saying "to make it more accurate" or "to get better results" without referencing separation, identical Rf values, or overlapping spots.

🧠 Exam Technique

Always state the problem that the second solvent solves: "In the first solvent, some amino acids have identical Rf values / do not separate." Framing your answer around solving incomplete separation ensures full credit.