OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2025: Question 22

6 marks · Medium difficulty · Structured Questions

Use TLC chromatogram data to determine Rf values and identify amino acids in a dipeptide, deduce dipeptide structures, and predict the products and functional group positioning of beta-keto acid decarboxylation.

Practise this question

Question

Question 22 consists of parts (a) and (b). Part (a) shows a TLC plate with a start line, solvent front, and two spots, alongside a reference table of four amino acid R groups with their corresponding Rf values: -H (0.18), -CH3 (0.25), -CH2CH(CH3)2 (0.64), and -CH(CH3)2 (0.77). Part (a)(i) asks for the Rf values of the two amino acids, and (a)(ii) asks to draw two possible structures for compound S, a dipeptide made from these amino acids. Part (b) introduces beta-keto acids with an example equation showing 3-oxobutanoic acid decomposing into propan-2-one and CO2. Part (b)(i) asks to draw the organic products formed from the decomposition of 2-methyl-3-oxopentanoic acid and 2-oxocyclopentanecarboxylic acid. Part (b)(ii) asks to suggest the structural difference between an alpha-keto acid and a beta-keto acid.
Question text

22 This question is about organic acids.

α-Amino acids have the general formula RCH(NH2)COOH.

(a) Compound S is made of two α-amino acids bonded together.

Compound S is hydrolysed into its component α-amino acids.

The resulting mixture is analysed using TLC to produce a chromatogram:

solvent front

start line

The R group and the Rf value is shown for some α-amino acids.

R group Rf value

–H 0.18

–CH3 0.25

–CH2CH(CH3)2 0.64

–CH(CH3)2 0.77

(i) Determine the Rf values of the two α-amino acids in the chromatogram.

[1]

(ii) Compound S has two possible structures.

Draw both of these structures.

[2]

Question 22(b) starts on Page 30

(b) This part is about compounds called β-keto acids.

The equation shows the decomposition of a β-keto acid.

O O O

+ CO2

OH

(i) Draw the structures of the organic products you would expect from the decomposition of the two

other β-keto acids below.

O O

OH

O O

OH

[2]

(ii) α-Amino acids have the NH2 group bonded to the carbon atom adjacent to the carboxylic acid

group.

Suggest the difference between an α-keto acid and a β-keto acid.

… [1]

Mark scheme

Show the mark scheme Mark scheme for Question 22: (a)(i) gives Rf values as 0.75 (range 0.71–0.79) and 0.25 (range 0.22–0.28) for 1 mark. (a)(ii) shows four accepted structures including linear dipeptides formed from alanine and valine (or cyclic/ester alternatives), awarding 2 marks for any two correct. (b)(i) gives the structures of pentan-3-one and cyclopentanone for 2 marks. (b)(ii) awards 1 mark for stating that in alpha-keto acids the C=O is adjacent to the COOH group, whereas in beta-keto acids the C=O is two carbons away from the COOH group.

Question Answer Mark Guidance

22 (a) (i) Calculation of Rf values 1

ALLOW ±1 mm on each measurement, i.e.

34.5 Rf = 0.71 – 0.79

Rf = 46 = 0.75

Rf = 0.22 – 0.28

AND

11.5 ALLOW values that round to 2dp to give a value in the

Rf = 46 = 0.25 ✓ acceptable range

Working is not required

22 (a) (ii) Any 2 from the following: ✓✓ 2 ALLOW any combination of skeletal OR structural OR

displayed formula as long as unambiguous

IGNORE connectivity

Check structures carefully to ensure they are different

from each other

ALLOW ECF from wrong Rf values in 22a(i)

R group should be consistent with Rf values however

ALLOW ECF if incorrect R group is repeated in second

structure

ALLOW ECF for one small slip e.g. missing H on NH group

22 (b) (i) Pentan-3-one 2 ALLOW any combination of skeletal OR structural OR

displayed formula as long as unambiguous

IGNORE non-organic products e.g. CO2, , H2O etc

✓

Cyclopentanone

✓

22 (b) (ii) Mark is for a clear description of the relative positions of 1 ALLOW labelled structures e.g.

the C=O/ketone and carboxylic acid/COOH in alpha and

beta keto acids

Alpha-keto acids:

ketone/C=O is adjacent/next/bonded to carboxylic acid/COOH

AND

Alpha-keto acids:

Beta-keto acids:

ALLOW ketone/C=O is on the adjacent C to the carboxylic

ketone/C=O two carbons away from carboxylic acid/COOH ✓

acid//COOH (matches alpha amino acid description)

OWTTE

ALLOW Bonded to the same C

Beta-keto acids:

ALLOW ketone/C=O on 3rd carbon (from COOH)

ALLOW has carbon in-between functional groups

IGNORE bonded to different C (must be specific)

Alternative approach – comparison of decomposition

ALLOW alpha keto acids decompose to form an aldehyde

and beta keto acids form a ketone OWTTE

How to answer it

Analysis of Organic Acids: TLC, Dipeptides & Decarboxylation

📋 What This Question Tests

This question assesses core organic synthesis and analytical skills:

  • Thin-Layer Chromatography (TLC): Accurate measurement and calculation of Rf values to identify unknown amino acids from reference data.
  • Peptide Chemistry: Deducing the two possible isomeric dipeptide structures formed by condensation of two distinct α-amino acids.
  • Reaction Mechanism & Pattern Recognition: Predicting products of thermal decarboxylation in substituted β-keto acids.
  • Chemical Nomenclature: Using contextual definitions to distinguish between α- and β-positioning of functional groups relative to carboxylic acids.

Part (a)(i): Calculating Rf Values from the Chromatogram

1 Mark • Assessment: Analytical Chromatography

📐 Step-by-Step Calculation

  1. Measure solvent front distance: From start line to solvent front = 46 mm (approx. 4.6 cm).
  2. Measure lower spot distance: From start line to centre of spot = 11.5 mm.
    Rf(lower) = 11.5 / 46 = 0.25
  3. Measure upper spot distance: From start line to centre of spot = 34.5 mm.
    Rf(upper) = 34.5 / 46 = 0.75

✅ Correct Answer

Rf values: 0.25 and 0.75

Mark Scheme:
• Lower spot: 0.22 – 0.28 (matches –CH₃, Rf = 0.25)
• Upper spot: 0.71 – 0.79 (matches –CH(CH₃)₂, Rf = 0.77)
[1 mark for both correct]

🧠 Exam Technique

  • Always measure to the centre of the spot, never the top or bottom edge.
  • Measure strictly from the pencil start line, not the bottom of the plate.
  • Rf has no units and is always a value between 0 and 1. Give your answers to 2 decimal places.

❌ Common Errors

  • Measuring from the bottom edge of the TLC plate instead of the baseline.
  • Inverting the equation (e.g. dividing distance to solvent front by spot distance).

Part (a)(ii): Structures of Dipeptide Compound S

2 Marks • Assessment: Peptide Bond Formation & Isomerism

💡 Key Knowledge: Identifying the Monomers

From the calculated Rf values, match the spots to the table:

  • Rf ≈ 0.25 corresponds to R = –CH₃ (Alanine: H₂N–CH(CH₃)–COOH)
  • Rf ≈ 0.75 corresponds to R = –CH(CH₃)₂ (Valine: H₂N–CH(CH(CH₃)₂)–COOH)

When two different amino acids form a dipeptide, two constitutional isomers can form depending on which amino acid retains its free –NH₂ group (N-terminus) and which retains its free –COOH group (C-terminus).

✅ Correct Structures (Any 2 of the following)

Structure 1 (Ala-Val):
H₂N–CH(CH₃)–CONH–CH(CH(CH₃)₂)–COOH
(Alanine at N-terminus, Valine at C-terminus)

Structure 2 (Val-Ala):
H₂N–CH(CH(CH₃)₂)–CONH–CH(CH₃)–COOH
(Valine at N-terminus, Alanine at C-terminus)

Mark Scheme: 1 mark per correct unique structure. Skeletal, displayed, or structural formulas accepted if unambiguous.

🧠 Structural Representation Details

To ensure maximum marks when drawing:

  • Show the peptide link clearly: –C(=O)–NH– .
  • Keep the backbone clear: N–C–C–N–C–C .
  • Check the R groups: one carbon must bear a methyl group –CH₃ , and the other must bear an isopropyl group –CH(CH₃)₂ .
  • Leave one end as a primary amine ( –NH₂ ) and the other as a carboxylic acid ( –COOH ).

❌ Common Traps & Where Marks Were Lost

  • Drawing the same molecule twice: Flipping the molecule horizontally without changing connectivity produces the identical structure. Always verify N-terminus vs C-terminus.
  • Missing hydrogens on the amide nitrogen: Drawing –C(=O)–N– without the H loses marks unless strictly using skeletal formula.
  • Connectivity errors: Attaching R groups to the carbonyl carbon instead of the α-carbon.

Part (b)(i): Decarboxylation of β-Keto Acids

2 Marks • Assessment: Pattern Recognition & Reaction Products

💡 Decarboxylation Pattern

The question provides an example decomposition equation:

CH₃–CO–CH₂–COOH ➔ CH₃–CO–CH₃ + CO₂

Reaction Rule: The carboxylic acid group ( –COOH ) is removed as CO₂ . The remaining bond is satisfied by a hydrogen atom (the carbon that held the –COOH becomes protonated).

✅ Reaction 1 Product

Starting material: 2-methyl-3-oxopentanoic acid
CH₃–CH₂–CO–CH(CH₃)–COOH

Product: Pentan-3-one

CH₃–CH₂–CO–CH₂–CH₃

Structure to draw: A 5-carbon chain with a C=O double bond at carbon-3.

1 Mark for correct skeletal or structural formula.

✅ Reaction 2 Product

Starting material: 2-oxocyclopentanecarboxylic acid
(5-membered ring with adjacent C=O and –COOH)

Product: Cyclopentanone

Structure to draw: A cyclopentane ring with one ketone group ( =O ) on any ring carbon.

1 Mark for correct ring ketone structure.

❌ Common Errors

  • Over-counting carbons in product 1: Forgetting that the branch carbon ( –CH₃ ) and the –CH– become part of an ethyl group: –CH(CH₃)– + H becomes –CH₂CH₃ . Drawing a branched ketone like 2-methylbutan-3-one is incorrect.
  • Opening the ring in product 2: The ring remains intact during decarboxylation; only the exocyclic –COOH group is lost.

Part (b)(ii): α-Keto Acid vs β-Keto Acid

1 Mark • Assessment: Chemical Nomenclature & Functional Group Positioning

✅ Correct Explanation

  • α-keto acid: The ketone group ( C=O ) is on the carbon atom adjacent to (directly bonded to) the carboxylic acid group ( –COOH ).
  • β-keto acid: The ketone group ( C=O ) is two carbons away from the carboxylic acid group (i.e. on the 3rd carbon along, separated by a –CH₂– or –CHR– group).
Mark Scheme: 1 mark for clearly stating both relative positions. Both α and β descriptions must be correct.

🧠 Using Prompt Clues

The question states: "α-Amino acids have the NH₂ group bonded to the carbon atom adjacent to the carboxylic acid group."

Top students simply substitute "ketone / C=O group" in place of "NH₂ group":

  • If α-amino = NH₂ on the adjacent carbon, then α-keto = C=O on the adjacent carbon.
  • By extension, β-keto = C=O on the second carbon away from the COOH group.

Topics

Module 1: Development of practical skills in chemistry · Module 6: Organic chemistry and analysis · 1.1 Practical skills assessed in a written examination · 6.1 Aromatic compounds, carbonyls and acids · 6.2 Nitrogen compounds, polymers and synthesis · 6.3 Analysis

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.