OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2022: Question 19

22 marks · Hard difficulty · Structured Questions

A structured multi-part organic chemistry question covering carboxylic acid reactions, esterification, hydrolysis, addition and condensation polymers, and a multi-step organic synthesis calculation.

Practise this question

Question

An exam question spanning multiple parts about carboxylic acids, polymers, and organic synthesis. Part (a) asks for full equations for reactions of propanoic acid with KOH, methanoic acid with magnesium, and aspartic acid with sodium carbonate. Part (b) presents a reaction flowchart involving 2-hydroxybutanoic acid, esterification, and reduction. Part (c) covers addition and condensation polymers from carboxylic acid monomers. Part (d) asks about a polymer formed from 2-aminopropanoic acid. Part (e) is a 6-mark synthesis and mass calculation question starting from 2-chloropropanoic acid to form an amino ester.
Question text

19 This question is about compounds that contain the carboxylic acid functional group.

(a) Carboxylic acids react with alkalis, metals and carbonates to form salts.

Write full equations for the following three reactions. Show structures for organic

compounds.

• the reaction of propanoic acid with aqueous potassium hydroxide:

• the reaction of aqueous methanoic acid with magnesium:

• the reaction of the α–amino acid, aspartic acid (R=CH2COOH), with an excess of

aqueous sodium carbonate, Na2CO3:

[4]

(b) The structure of 2-hydroxybutanoic acid is shown below.

H

CH3CH2 C COOH

OH

2-hydroxybutanoic acid

Fill in the flowchart for reactions involving 2-hydroxybutanoic acid.

C6H5COOH/H2SO4 (CH3)2CHOH/H2SO4

reflux reflux

H H

CH3CH2 C CN CH CH C COOH

OH OH

2-hydroxybutanoic acid

NaBH4

[4]

(c) This part is about polymers derived from carboxylic acid monomers.

(i) Poly(pent-3-enoic acid) is an addition polymer.

Draw the structure of pent-3-enoic acid and two repeat units of this polymer.

Pent-3-enoic acid

Two repeat units of

poly(pent-3-enoic acid)

[2]

(ii) Butanedicarboxylic acid and 1,4-dihydroxy-2-methylbenzene react to form a

condensation polymer.

Draw one repeat unit of this condensation polymer.

[2]

(iii) Three repeat units of a condensation polymer are shown below.

O O O

O O O

O O O

O O O

Draw the structure of the monomer required to form this polymer.

[1]

(d) A polymer is formed from 400 molecules of 2-aminopropanoic acid.

(i) Draw one repeat unit of this polymer.

[1]

(ii) What is the relative molecular mass, Mr, of the polymer?

Mr = … [2]

(e)* A student intends to synthesise compound I.

O

O

NH2

Compound I

Plan a synthesis to prepare 9.36 g of compound I starting from 2-chloropropanoic acid,

CH3CHClCOOH. The overall percentage yield of compound I from 2-chloropropanoic acid

is 64%.

In your answer, include starting mass of 2-chloropropanoic acid, reagents, conditions and

equations where appropriate. [6]

Additional answer space if required.

Mark scheme

Show the mark scheme The mark scheme provides expected chemical equations, structural drawings for organic products and polymers, molar mass calculations, and level-of-response guidance for the 6-mark synthesis question.

AO

Question Answer Marks Guidance

element

19 (a) C2H5COOH + KOH → C2H5COOK + H2O 4 AO2.6 ALLOW any combination of skeletal OR

×4 structural OR displayed formula as long as

unambiguous

IGNORE state symbols and use of equilibrium

sign

ALLOW KC2H5COO

DO NOT ALLOW a missing charge (e.g.

C H COO–K) the 1st time seen but IGNORE for

next equations.

2HCOOH + Mg → (HCOO)2Mg + H2 For salts,

ALLOW C H COO–K+ OR C H COO– + K+

25 2 5

DO NOT ALLOW –COO–K (covalent bond)

the 1st time seen but IGNORE for next

equations.

H2O AND CO2 FOR CO2 + H2O ALLOW H2CO3

Correct formula of salt:

AO

Question Answer 22 Marks Guidance

element

19 (b) 4 AO2.5 ALLOW any combination of skeletal OR

×4 structural OR displayed formula as long as

unambiguous

ALLOW any vertical bond to the OH group

e.g. ALLOW

OR

OH HO

IGNORE connectivity of CH3CH2 group

IGNORE inorganic by-products

ALLOW HCl/H2O, H2SO4/H2O

IGNORE dilute

23 AO

element

19 (c) (i) Pent-3-enoic acid 2 ALLOW any combination of skeletal OR

structural OR displayed formula as long as

unambiguous

AO1.2 ALLOW either the E or Z isomer

2 repeat units of polymer

ALLOW ECF from pent-2-enoic acid

CH2COOH CH2COOH OR pent-4-enoic acid ONLY

H H

For repeat unit,

• ‘side bonds’ required on either side of

C C C C repeat unit from C atoms

• 2 repeat units required

CH3 H CH3 H AO2.5

IGNORE connectivity of CH2COOH in polymer

IGNORE brackets

IGNORE n

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

ALLOW any consistent repeat unit:

CH2COOH and CH3 groups can alternate or be

on opposite sides of chain

e.g.

CH2COOH CH2COOH

H H

C C C C

H CH3 CH3 H

AO

element

19 (c) (ii) CH3 2 end –O– may be at either side e.g.

CH3

O H H O

O H H O

C C C C O O O C C C C O

H H

H H

AO1.2

ester link

AO2.5

ALLOW CH3 to be on position 2 or 3 of the

ONE repeat units of correct polymer aromatic ring

‘End bonds’ MUST be shown (do not have to

be dotted)

IGNORE brackets

IGNORE n

19 (c) (iii) 1 AO3.2 ALLOW any combination of skeletal OR

structural OR displayed formula as long as

unambiguous

19 (d) (i) 1 AO2.5 end –N– may be at either side e.g.

H O H O

N C C C C N

H CH CH3 H

‘End bonds’ MUST be shown (do not have to

ONE repeat unit ONLY be dotted)

IGNORE brackets

IGNORE n

AO

element

19 (d) (ii) IF answer on answer line = 28418, AWARD 2 marks 2 AO2.2

IF answer on answer line = 28400, AWARD 1 mark ×2

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

Mr of 400 molecules = 400 × 89 = 35600 ALLOW ECF from incorrect repeat unit in 19di

Mr of polymer = 35600 – (399 × 18) = 28418 ALLOW ECF from incorrect Mr of 400 repeat

units

Alternative method based on repeat unit:

Mr of 400 repeat units = 400 × 71 = 28400

Mr of polymer = 28400 + 1 + 17 = 28418

19 (e)* Refer to marking instructions on page 5 of mark scheme for 6 AO3.3 Indicative scientific points may include:

guidance on marking this question. ×6

Calculation of mass of CH3CHClCOOCH3

Level 3 (5-6 marks) Using moles

Correct calculation of mass of CH3CHClCOOH. 9.36

AND • n(I) =

117.0

Planned synthesis includes substitution of –Cl and formation of = 0.08(00) (mol)

compound I (or its corresponding ammonium salt) with the 100

correct reagents and some conditions identified and equations • n(CH3CHClCOOC2H5) = 0.0800 ×

are mostly correct. = 0.125 (mol)

• Mass of CH3CHClCOOH = 108.5 × 0.125

There is a well-developed line of reasoning which is clear and = 13.5625 g

logically structured. The information presented is relevant and

substantiated.

AO

element

Level 2 (3-4 marks) Using mass

Calculation of mass of CH3CHClCOOH is correct 100

AND • Theoretical mass of I = 9.36 ×

Planned synthesis includes one step of the synthesis with the = 14.625 (g)

correct reagent and some conditions identified and equation is 14.625

mostly correct • Theoretical n(CH3CHClCOOH) =

117.0

OR = 0.125 (mol)

Calculation of mass of CH3CHClCOOH is partly correct • Mass of CH3CHClCOOH = 108.5 × 0.125

AND = 13.5625 g

Planned synthesis includes substitution of –Cl and formation of

compound I (or its corresponding ammonium salt) with the

correct reagents ALLOW slip/rounding errors such as errors in

OR Mr, e.g. use of 107.5 instead of 108.5 for

Attempts to calculate mass of CH3CHClCOOC2H5 but makes little CH3CHClCOOH → 13.4375

progress

AND

Planned synthesis includes substitution of –Cl and formation of ---------------------------------------------------------------

compound I (or its corresponding ammonium salt) with the Examples of partly correct calculations

correct reagents and some conditions identified and equations 64

are mostly correct Mass = 5.5552 g from 0.0800 × × 108.5

(% yield inverted)

There is a line of reasoning presented with some structure. The Mass = 8.68 g from 0.0800 × 108.5

information presented is relevant and supported by some (% yield omitted)

evidence.

Synthesis: Either order for 2 stages

Substitution of –Cl → amine:

• Reagents: (excess) NH3

• Condition: ethanol

• Equation: CH3CHClCOOH + 2NH3 →

CH3CHNH2COOH + NH4Cl

OR

CH3CHClCOOH + NH3 →

CH3CHNH2COOH + HCl

AO

element

Level 1 (1-2 marks) Esterification of amine → compound I

Calculation of mass of CH3CHClCOOH is partly correct • Reagents: CH3CH2OH

OR • Conditions: acid (catalyst), e.g. H2SO4

Planned synthesis includes both steps with some of the reagents (reflux/heat)

and conditions identified • Equation:

OR CH3CHNH2COOH + CH3CH2OH →

Attempts equations for both steps but these may contain errors CH3CHNH2COOCH2CH3 + H2O

OR OR --------------------------------------------------------

Describes one step of the synthesis with reagents, conditions and Esterification of carboxylic acid → ester

equation mostly correct

• Reagents: CH3CH2OH

• Conditions: acid (catalyst), e.g. H2SO4

There is an attempt at a logical structure with a line of reasoning.

(reflux/heat)

The information is in the most part relevant.

• Equation:

0 marks CH3CHClCOOH + CH3CH2OH →

No response or no response worthy of credit. CH3CHClCOOCH2CH3 + H2O

Substitution of –Cl → amine:

• Reagents: (excess) NH3

• Condition: ethanol

• Equation: e.g

CH3CHClCOOCH2CH3 + 2NH3 →

CH3CHNH2COOCH2CH3 + NH4Cl

OR

CH3CHClCOOCH2CH3 + NH3 →

CH3CHNH2COOCH2CH3 + HCl

OR

CH3CHClCOOCH2CH3 + NH3 →

CH3CHNH3ClCOOCH2CH3

(ammonium salt)

Total 22

How to answer it

Carboxylic Acids, Polymers & Multi-Step Organic Synthesis

What this question tests

This synoptic exam question tests core organic chemistry reactions of carboxylic acids and their derivatives, functional group interconversions (reduction, esterification, nucleophilic substitution), polymer chemistry (addition and condensation polymers with repeat units), and a complex multi-step synthetic route calculation involving percentage yield and moles.

Part (a): Reactions of Carboxylic Acids

Chemical equations and salt formation

✅ Correct Answers

  • Propanoic acid + KOH: C₂H₅COOH + KOH → C₂H₅COOK + H₂O
  • Methanoic acid + Mg: 2HCOOH + Mg → (HCOO)₂Mg + H₂
  • Aspartic acid + Na₂CO₃ (excess): Salt structure showing disodium aspartate: H₂N-CH(CH₂COONa)-COONa + H₂O + CO₂

💡 Key Knowledge

  • Carboxylic acids react with alkalis to form carboxylate salts and water.
  • Acids react with reactive metals (like Mg) to form carboxylate salts and hydrogen gas.
  • Acids react with carbonates to form carboxylate salts, water, and carbon dioxide gas. Remember aspartate has two carboxylic acid groups, so both react when sodium carbonate is in excess!

❌ Common Errors

  • Omitting charges on carboxylate ions (e.g., writing C₂H₅COO K instead of ionic or correctly bonded structures).
  • Drawing incorrect covalent bonds between oxygen and potassium ( --COO-K ).
  • Forgetting that aspartic acid contains two carboxylic acid functional groups that both react with excess carbonate.

🧠 Exam Technique

  • State symbols are generally ignored unless specifically requested, but balancing stoichiometry (especially for the Mg and carbonate equations) is strictly required for the marks.

Part (b): Flowchart of 2-Hydroxybutanoic Acid Reactions

Functional group transformations and reagents

✅ Correct Answers

  • Top-Left Box (Esterification): Structure of the ester formed from C₆H₅COOH and 2-hydroxybutanoic acid (ester linkage attached to the secondary alcohol group).
  • Top-Right Box (Esterification with alcohol): Structure of the ester formed with (CH₃)₂CHOH reacting at the -COOH group.
  • Reagent Box ( C₆H₅COOH/H₂SO₄ arrow): H⁺ / H₂O or H⁺(aq) .
  • Bottom Box (Reduction): Structure of the diol resulting from the reduction of the carboxylic acid group to a primary alcohol ( CH₃CH₂-CH(OH)-CH₂OH ).

💡 Key Knowledge

  • NaBH₄ selectively reduces aldehydes and ketones, but here NaBH₄ reduces the -COOH group (note: standard A-Level context often uses LiAlH₄ for carboxylic acid reduction, but follow the specific reaction scheme provided).
  • Esterification requires a carboxylic acid and an alcohol in the presence of an acid catalyst ( H₂SO₄ ) under reflux.

Part (c): Polymers from Carboxylic Acid Monomers

Addition and Condensation Polymerisation

✅ Correct Answers

  • (c)(i) Pent-3-enoic acid monomer: CH₃-CH=CH-CH₂-COOH
  • (c)(i) Two repeat units: An addition polymer backbone -C-C-C-C- with correct side-chains and open bonds at both ends showing 2 repeat units.
  • (c)(ii) Condensation polymer unit: Shows alternating ester/amide links formed from butanedicarboxylic acid and 1,4-dihydroxy-2-methylbenzene.
  • (c)(iii) Monomer from repeat units: Dicarboxylic acid/diol structure showing the exact precursor molecule.

🧠 Exam Technique

  • For addition polymers, break the C=C double bond and show single bonds extending out of the square brackets or ending with dashed continuation bonds.
  • For condensation polymers, ensure the ester linkage ( -COO- ) is clearly drawn showing correct atom connectivity (e.g. oxygen bonded to the correct carbon on the aromatic ring).

Part (d): Amino Acid Polymers (Proteins/Polypeptides)

Repeat units and relative molecular mass (Mr) calculations

✅ Correct Answers

  • (d)(i) Repeat unit: -HN-CH(CH₃)-CO- with continuation bonds at both ends.
  • (d)(ii) Mr of polymer: 28418 (Accept 28400 if calculated via standard unit multiplication without water loss adjustments, but 28418 is fully rigorous).

📐 Calculation Breakdown (Part d-ii)

  1. Molar mass of monomer (2-aminopropanoic acid / alanine): C₃H₇NO₂ = 89 g mol⁻¹
  2. Mass of polymer chain: 400 molecules reacting means 400 repeat units minus the loss of water molecules during condensation (399 water molecules lost).
  3. Mr = (400 × 89) - (399 × 18) = 35600 - 7182 = 28418

Part (e): Multi-Step Synthesis & Percentage Yield Calculation

Planning a synthesis and calculating starting mass

💡 Reaction Pathway Overview

To convert 2-chloropropanoic acid ( CH₃CHClCOOH ) into Compound I (an ethyl ester with an amine/ammonium substitution):

  • Stage 1 (Nucleophilic Substitution): React CH₃CHClCOOH with excess NH₃ to substitute the chlorine atom with an amine group ( -NH₂ ), forming 2-aminopropanoic acid (or its ammonium salt).
  • Stage 2 (Esterification): React the resulting amino acid with ethanol ( CH₃CH₂OH ) in the presence of a concentrated H₂SO₄ catalyst under reflux to form the ethyl ester (Compound I).

📐 Step-by-Step Mass Calculation

  1. Find Mr of Compound I ( C₅H₁₁NO₂ ): (5 × 12.0) + (11 × 1.0) + (14.0) + (2 × 16.0) = 117.0 g mol⁻¹
  2. Calculate moles of Compound I desired:
    Moles = Mass / Mr = 9.36 g / 117.0 g mol⁻¹ = 0.0800 mol
  3. Account for 64% Percentage Yield:
    Theoretical moles required = 0.0800 × (100 / 64) = 0.125 mol
  4. Find Mr of starting material (2-chloropropanoic acid, CH₃CHClCOOH ):
    C₃H₅ClO₂ = (3 × 12.0) + (5 × 1.0) + 35.5 + (2 × 16.0) = 108.5 g mol⁻¹
  5. Calculate starting mass:
    Mass = Moles × Mr = 0.125 mol × 108.5 g mol⁻¹ = 13.5625 g (round to appropriate sig figs, e.g., 13.6 g ).

❌ Common Calculation Traps

  • Inverting the percentage yield: Multiplying by 64/100 instead of 100/64 . Always ask yourself: do you need more or less starting material when yield is less than 100%? You always need more!
  • Using incorrect molecular formulas or molar masses for the starting material or final product.

🧠 Top-Level Examiner Tips

  • Clearly state reagents and conditions for both steps to secure full Level 3 marks.
  • Show working clearly with units at each stage so error-carried-forward (ECF) marks can be awarded if an intermediate Mr has a minor arithmetic slip.

Topics

Module 4: Core organic chemistry · Module 6: Organic chemistry and analysis · 6.1 Aromatic compounds, carbonyls and acids · 6.2 Nitrogen compounds, polymers and synthesis · 4.1 Basic concepts and hydrocarbons · 4.2 Alcohols, haloalkanes and analysis

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.