Edexcel A-Level Chemistry Paper 2, June 2024: Question 8

16 marks · Hard difficulty · Open Response

Analyze amino acid properties, stereoisomers, dipeptide formation, zwitterions, synthetic pathways, and electrophoresis behavior.

Practise this question

Question

Question text

8 This question is about amino acids and related compounds.

The structures of two amino acids are shown.

CH2OH CH3

H2N C COOH H2N C COOH

H H

serine alanine

(a) Both amino acids contain a chiral carbon atom.

(i) State what is meant by the term ‘chiral carbon atom’.

(1)

(ii) Draw diagrams of the two stereoisomers of alanine, showing their

three‑dimensional shape.

(2)

(b) Draw the skeletal formulae of the two dipeptides that could form when serine

and alanine react.

(2)

What is the formula of this zwitterion?

(1)

CH2OH

A H N C CO–

CH2OH

B H3N C COOH

CH2OH

C H N C CO–

CH O–

D H3N C COOH

(d) Alanine can be synthesised as shown.

O O

H H H H H

C C C C

O C O

H3C CH3

HN O

CH3

Compound X

Step 1 C H O–Na+

–

O O

H H H H

C C C C

C O

H3C O CH +

3 + C H OH + Na

O 2 5

*P76896A02532*

Step 2 CH3Br

O O

H H CH3 H H

C + Br–

C C C

O C O

H3C CH3

HN O

CH3

Compound Y

Step 3

CH3

H2N C COOH + CH3COOH + 2CH3CH2OH + CO2

(i) Sodium ethoxide, C*P76896A02632*2H5ONa, needed in Step 1, is formed by the reaction of

sodium with ethanol.

Write the equation for this reaction.

State symbols are not required.

(1)

(ii) Name the two functional groups present in Compound Y.

(2)

(iii) Give the reagent needed and conditions required for Step 3.

(2)

(iv) Calculate the mass of Compound X required to make 15.0g of alanine,

assuming the overall yield of the synthesis is 55%.

(3)

(e) A mixture of amino acids can be separated using a technique called

paper electrophoresis. A simplified diagram is shown.

The movement of the amino acid on the paper depends on any charge on the

amino acid, and this is determined by the pH of the buffer solution.

positive electrode negative electrode

strip of filter paper concentrated spot of

soaked in a buffer amino acid mixture

solution of pH = 9 containing amino acid Z

Amino acidZ has the structure shown.

CH2COOH 27

*P76896A02732*H2NCCOOH

amino acid Z

Explain in which direction, if any, amino acid Z would move when a current flows

in the circuit with a buffer solution of pH = 9.

(2)

(Total for Question 8 = 16 marks)

Mark scheme

Show the mark scheme

Question

Answer Additional Guidance Mark

Number

8(a)(i) An answer that makes reference to the following point: (1)

• a carbon atom with four different atoms / groups (of Allow resulting molecule has non-superimposable

atoms) attached mirror images

Allow a carbon atom with four different functional

groups attached

Ignore a carbon atom with four different species

attached

Do not award ‘molecules’ for atoms / groups

Question

Answer Additional Guidance Mark

Number

8(a)(ii) Allow 1 mark for 2 tetrahedral mirror images (2)

• COOH HOOC shown with only lines and no wedges / dashed lines

Allow 1 mark for 2 tetrahedral mirror images

shown with two similar wedges on each structure

Accept dashed line for dashed wedges

Ignore connectivity

Question

Answer Additional Guidance Mark

Number

8(b) Allow 1 mark for 2 correct structures shown as structural (2)

• or displayed formulae

Ignore displayed NH2, NH, or OH groups

H2N

Penalise missing hydrogen atoms on amide group once

(1) only

Penalise NH group shown as part of chain once only

•

H2N

(1) Allow 1 if both dipeptides shown as correct polymer

repeat units

and

Question

Answer Mark

Number

8(c) The only correct answer is A (1)

B is not correct because the CO2H group has not lost a proton

C is not correct because the NH2 group has not gained a proton

D is not correct because the CH2OH group has lost a proton

Question

Answer Additional Guidance Mark

Number

8(d)(i) Examples (1)

• balanced equation C2H5OH + Na → C2H5ONa + ½H2

C H OH + Na → C H O−Na+ + ½H

25 2 5 2

C2H6O + Na → C2H5ONa + ½H2

C H O + Na → C H O−Na+ + ½H

26 2 5 2

Allow multiples

Allow displayed / skeletal / molecular formulae

Ignore state symbols even if incorrect

Do not award C H OH + Na → C H Oδ−Naδ+ + ½H

25 2 5 2

Question

Answer Additional Guidance Mark

Number

8(d)(ii) An answer that makes reference to the following point: (2)

• ester (1) If 3 answers given and 2 are correct, award 1 mark

• (N-substituted) amide (1) If 4 or more answers given and 2 are correct award 0

marks

Ignore structures, even if incorrect

Question

Answer Additional Guidance Mark

Number

8(d)(iii) An answer that makes reference to the following points: Ignore references to concentrated / dilute (2)

• hydrochloric acid / HCl(aq) (1) Allow HCl

Allow correct name or formulae of any strong acid e.g.

HNO3, H2SO4,

Allow phosphoric acid / H3PO4

Allow NaOH followed by any identified strong acid

Ignore H+ / H O+ / catalyst

• (heat under) reflux (1)

Mark independently

Question

Answer Additional Guidance Mark

Number

8(d)(iv) Example of calculation (3)

• calculation of moles of alanine (1) 15.0 ÷ 89.0 = 0.16854 (mol)

• scale to 100% (1) 0.16854 × (100 ÷ 55) = 0.30644 (mol)

• calculation of mass of compound X (1) 0.30644 × 217 = 66.496 = 66.5 g

• scale mass of alanine to 100% (1) 15.0 × (100 ÷ 55) = 27.273 (g)

• calculation of moles of alanine (1) 27.273 ÷ 89.0 = 0.30644 (mol)

• calculation of mass of compound X (1) 0.30644 × 217 = 66.496 = 66.5 g

(1) 15.0 ÷ 89.0 = 0.16854 (mol)

• calculation of moles of alanine

(1) 0.16854 × 217 = 36.573

• calculate mass of X

(1) 36.573 × (100 ÷ 55) = 66.496 = 66.5 g

• scale to 100 %

Correct answer no working scores 3

Ignore SF except 1 SF

Allow TE throughout

If no mark scored allow 1 mark for 2 correct Mr

values, 89 and 217

Question

Answer Additional Guidance Mark

Number

8(e) An explanation that makes reference to the following points: Note (2)

The negative charge on the ion and loss of both

hydrogen ions from the COOH groups can both be

awarded from a correct diagram of the ion

• would move towards the positive electrode / to the

left (1) Do not award zwitterion

and

as it will form a negative ion

• (as basic pH means) the acid groups will both lose a (1) Allow the acid groups will both lose a hydrogen

proton / has two COO– groups (ion)

M2 can be awarded from diagram / amended

structure in question

(Total for Question 8 = 16 marks)

TOTAL FOR PAPER = 90 MARKS

How to answer it

Amino Acids and Related Compounds Study Guide

What this question tests

This question covers key organic chemistry concepts relating to amino acids: optical isomerism (chiral centres and 3D stereoisomers), peptide bond formation (dipeptides), zwitterion structures, multi-step organic synthesis (ester/amide chemistry, reagents, and percentage yield calculations), and paper electrophoresis separation based on amino acid charge at specific pH values.

Part (a)(i)

Definition of a Chiral Carbon Atom

✅ Correct Answer

A carbon atom with four different atoms or groups of atoms attached to it.

❌ Common Errors

Students often lose this mark by stating "four different molecules" or "four different elements" instead of atoms or groups of atoms. Avoid referring to "molecules" for attached substituents.

Part (a)(ii)

Stereoisomers of Alanine

✅ Correct Answer

Two tetrahedral mirror images showing 3D spatial arrangement. Use wedges and dashed lines for bonds going in/out of the plane. The central C connects to COOH, NH₂, H, and CH₃ in opposite mirror configurations.

🧠 Exam Technique

When drawing 3D stereoisomers, make sure to show tetrahedral geometry clearly using wedges (coming towards you) and dashes (going away). Examiners accept dashed lines for dashed wedges, but clear tetrahedral orientation is essential.

Part (b)

Skeletal Formulae of Dipeptides

✅ Correct Answer

Two distinct skeletal formulae representing the combination of serine and alanine:
1. Serine-alanine dipeptide (peptide link formed between serine's amino group and alanine's carboxylic acid group, or vice versa).
2. Alanine-serine dipeptide (the alternative sequence).

❌ Common Errors

Do not draw condensed structural units incorrectly inside skeletal frameworks. Ensure the amide functional group (-CONH-) is clearly visible without breaking bonding rules.

Part (c)

Zwitterion Structure of Serine

✅ Correct Answer

A is the correct option.

H₃N⁺-CH(CH₂OH)-COW⁻

💡 Key Knowledge

A zwitterion forms via internal proton transfer from the carboxylic acid group (-COOH becomes -COO⁻) to the basic amine group (-NH₂ becomes -NH₃⁺).

Part (d)(i)

Formation of Sodium Ethoxide

✅ Correct Answer

C₂H₅OH + Na → C₂H₅ONa + 0.5 H₂

(Multiples like 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂ are also fully correct). State symbols are not required.

Part (ii)

Functional Groups in Compound Y

✅ Correct Answer

Ester (-COO-)
(N-substituted) amide (-CONH-)

❌ Common Errors

If you list more than two functional groups and any are incorrect, you lose marks. Only list the exact functional groups requested.

Part (iii)

Reagents and Conditions for Step 3

✅ Correct Answer

Reagent: Hydrochloric acid ( HCl(aq) ) or any named strong acid (e.g., H₂SO₄ ).
Condition: Heat under reflux.

🧠 Exam Technique

Step 3 hydrolyses ester and amide bonds back into carboxylic acids and amines. Always pair acid or base hydrolysis reagents with "heat under reflux" to ensure the reaction goes to completion.

Part (iv)

Synthesis Mass Calculation

📐 Step-by-Step Calculation

Calculate moles of alanine produced:
M(alanine, C₃H₇NO₂) = (3×12.0) + (7×1.0) + 14.0 + (2×16.0) = 89.0 g mol⁻¹.
Moles = 15.0 g / 89.0 g mol⁻¹ = 0.1685 mol.
Adjust for 55% percentage yield:
Theoretical moles needed = 0.1685 × (100 / 55) = 0.3064 mol.
Calculate mass of Compound X:
M(Compound X, C₁₀H₁₉NO₄) = 217 g mol⁻¹.
Mass = Moles × M = 0.3064 mol × 217 g mol⁻¹ = 66.5 g .

❌ Common Calculation Traps

Forgetting to scale up for the incomplete yield (dividing instead of multiplying by 100/55) is the most common error. Always check whether your theoretical starting mass is larger or smaller than your final isolated mass!

Part (e)

Paper Electrophoresis of Amino Acid Z

✅ Correct Answer

Direction: Moves towards the positive electrode (to the left).
Reason: At basic pH (pH = 9), both carboxylic acid groups lose protons to form a negative ion with two carboxylate groups ( -COO⁻ ), making the amino acid overall negatively charged. Oppositely charged ions are attracted to the positive anode.

💡 Key Knowledge

In paper electrophoresis, the buffer solution pH determines the protonation state. High pH (alkaline/basic) removes protons from acidic sites (creating negative ions that migrate to the positive electrode), while low pH (acidic) protonates basic sites (creating positive ions that migrate to the negative electrode).

Topics

Organic Chemistry · Topic 17: Organic Chemistry II · Topic 18: Organic Chemistry III

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.