OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2024: Question 22

6 marks · Medium difficulty · Structured Questions

Calculate the molar mass of a polymer formed from 500 molecules of a given alpha-amino acid, and draw 2 repeat units of addition and condensation polymers for a bifunctional monomer.

Practise this question

Question

Question 22 states alpha-amino acids have the general formula RCH(NH2)COOH, where R contains C and H only with a molar mass of 91 g mol^-1. Part (a) asks to determine the molar mass of a polymer formed from 500 molecules of this alpha-amino acid to the nearest whole number. Part (b) shows the structure of an amino acid containing both a C=C double bond and a carboxylic acid/amine group, and asks to draw 2 repeat units for both its addition polymer and condensation polymer in provided boxes.
Question text

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

The R group in an α-amino acid contains C and H only.

This R group has a molar mass of 91 g mol –1.

(a) A polymer is formed from 500 molecules of this α-amino acid.

Determine the molar mass of this polymer.

Give your answer to the nearest whole number.

molar mass of polymer = … g mol –1 [3]

(b) The amino acid below can form addition and condensation polymers.

H H

C C O

H CH C

H2N OH

Draw 2 repeat units of these polymers.

Display the sections linking the monomers together.

addition polymer (2 repeat units)

condensation polymer (2 repeat units)

[3]

Mark scheme

Show the mark scheme The mark scheme for question 22(a) awards 3 marks for determining the molar mass of the polymer as 73518, detailing steps for finding the amino acid Mr (165), multiplying by 500, and subtracting water molecules for condensation. Part (b) awards 3 marks for correct structures of two repeat units for both the addition polymer (showing a carbon-carbon backbone with side chains) and the condensation polymer (showing an amide link backbone with C=C side chains).

Question Answer Marks Guidance

22 (a) IF answer on answer line = 73518 AWARD 3 marks 3

IF answer on answer line = 73500 AWARD 2 marks

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Mr of amino acid = 165 ✓

Mr of 500 molecules = 500 165 = 82500 ✓ ALLOW ECF from incorrect Mr of amino acid

Mr of polymer = 82500 – (499 18) = 73518 ✓ Alternative method:

(final answer must be given to nearest whole number) Mr of repeat unit = 147 ✓

147 x 500 = 73500 ✓

73500 + 18 = 73518 ✓

Common error for 2 marks

36518 Use of Mr 91

82500 Not shown 165 in working

Common error for 1 mark

45500 Use of Mr 91

22 (b) Addition polymer 27 3 For BOTH structures,

ALLOW any combination of skeletal OR structural OR

H H H H displayed formula as long as unambiguous

C C C C ‘End bonds’ MUST be shown (with either a solid or dashed

line)

H H BUT ALLOW ECF IF end bonds omitted in both structures

H2N CH HC NH2

DO NOT ALLOW more than 2 repeat units

COOH BUT ALLOW ECF in subsequent structure

COOH ✓

IGNORE connectivity of side groups in both diagrams

------------------------------------------------------------

Condensation polymer CARE: ALLOW any consistent repeat unit:

side groups can alternate or be on opposite sides of chain

ALLOW NH in amide link i.e. without bond shown

ALLOW –NH– at either end

IGNORE brackets IGNORE n or subscript numbers

ALLOW C2H3 as side chain for condensation polymer

ALLOW 1 mark if correct structures given by wrong way

Amide link ✓ round

2 repeat units of correct polymer ✓

How to answer it

Amino Acids, Molar Mass Calculations, and Polymerisation

What this question tests

This question tests your understanding of amino acid structure (specifically calculating molar masses of whole molecules and condensation polymers accounting for condensation water losses) alongside the dual polymerisation capabilities of bifunctional molecules capable of both addition polymerisation (via alkene C=C bonds) and condensation polymerisation (via amine and carboxylic acid groups).

Question Part (a)

Polymer Molar Mass Calculation

✅ Correct Answer

Final Answer: 73518 g mol⁻¹

Awarded 3 marks for the correct final answer on the answer line.

📐 Step-by-Step Calculation

  1. Find the Mᵣ of the single amino acid:
    General formula: RCH(NH₂)COOH
    Mᵣ = R-group (91) + CH (13) + NH₂ (16) + COOH (45) = 165
  2. Find the unadjusted mass of 500 molecules:
    500 × 165 = 82500
  3. Account for condensation water loss:
    Forming a polymer chain of 500 units creates 499 peptide bonds, releasing 499 molecules of H₂O.
    Mass loss = 499 × 18 = 8982
    Polymer Mᵣ = 82500 - 8982 = 73518

❌ Common Errors & Traps

  • The 500 vs 499 trap: Subtracting 500 × 18 instead of 499 × 18 gives 73500 (awards 2 marks).
  • Forgetting the amino acid backbone: Using just the R-group mass of 91 gives 45500 (1 mark) or 36518 (2 marks). Always remember the amino acid formula includes the amine, carboxylic acid, and central CH carbon!
Question Part (b)

Drawing Addition and Condensation Polymers

💡 Key Knowledge

The given monomer is multifunctional:

  • It contains a C=C double bond which undergoes addition polymerisation (polyalkene formation, backbone is purely carbon-carbon).
  • It contains an -NH₂ and a -COOH group which react together to form a condensation polymer (polyamide/peptide link, backbone contains nitrogen and carbonyl carbons).

✅ Correct Structures & Marking Points

  • Addition Polymer: Backbone must be a continuous chain of 4 carbons ( -C-C-C-C- ) with single bonds, showing exactly 2 repeat units, correct side groups attached, and clear trailing end bonds.
  • Condensation Polymer: Must feature the amide link ( -NH-CH-C(=O)- ), showing 2 repeat units and trailing end bonds indicating polymerisation continuity.
3 marks total: 1 mark for correct addition polymer, 1 mark for correct amide link in condensation polymer, 1 mark for exactly 2 repeat units of the correct polymer.

🧠 Exam Technique & Guidance

  • End bonds are crucial: Always extend single bonds past your brackets or outside your terminal atoms to show the polymer continues.
  • Repeat unit count: Strictly draw exactly 2 repeat units as requested by the command word. Do not draw 1 or 3+.
  • Side-chain flexibility: Side groups can alternate or be drawn on opposite sides of the carbon chain, but connectivity to the main chain must make chemical sense.

Topics

Module 6: Organic chemistry and analysis · 6.2 Nitrogen compounds, polymers and synthesis

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.