Edexcel A-Level Chemistry Paper 2, June 2023: Question 9

10 marks · Hard difficulty · Calculations

Identify organic structures from reaction pathways involving halogenoalkanes and Grignard reagents, and calculate the percentage of nitrogen in a compound using back-titration data.

Practise this question

Question

Question 9 presents two parts about organic compounds. Part (a) outlines reaction steps starting from compound A (C3H7Cl) reacting with aqueous NaOH to form B (C3H8O), which oxidises to C (C3H6O), and separately reacting A with magnesium in dry ether to form D (C3H7MgCl), which reacts with carbon dioxide then acidification to form E (C4H8O2), requiring the identification of structures A to E for 5 marks. Part (b) describes a 1.19g sample of nitrogen-containing compound Q heated with NaOH solution, passing the resulting ammonia into 100.0 cm3 of 0.225 mol dm-3 HCl, and back-titrating 25.0 cm3 portions of unreacted acid with 15.5 cm3 of 0.100 mol dm-3 NaOH, requiring the calculation of the percentage of nitrogen in Q for 5 marks.
Question text

9 This question is about the analysis of some organic compounds.

(a) A compound A (C3H7Cl) reacts with dilute aqueous sodium hydroxide to

produce B (C3H8O). B can be oxidised to C (C3H6O), which cannot be oxidised

any further.

A reacts with magnesium in dry ether to give D (C3H7MgCl). When carbon dioxide

is passed through the solution of D, followed by acidification, E (C4H8O2) is formed.

Identify the structures of A to E.

(5)

(b) An organic compound, Q, contains carbon, hydrogen and nitrogen only.

When a 1.19g sample of the compound was heated with sodium hydroxide

solution, all of the nitrogen was converted into ammonia. The ammonia was

passed into 100.0 cm3 of 0.225 mol dm–3 hydrochloric acid.

NH3(g) + HCl(aq) → NH4Cl(aq)

25.0 cm3 portions of the resulting solution containing unreacted hydrochloric acid

required a mean titre of 15.5 cm3 of 0.100 mol dm–3 sodium hydroxide

for neutralisation.

Calculate the percentage of nitrogen in Q.

(5)

*P71913A02428*

(Total for Question 9 = 10 marks)

Mark scheme

Show the mark scheme Mark scheme for question 9 detailing the expected answers and alternative guidance. Part (a) awards marks for structures A (2-chloropropane), B (propan-2-ol), C (propanone), D (2-propylmagnesium chloride), and E (2-methylpropanoic acid). Part (b) outlines the step-by-step titration calculation involving moles of NaOH, moles of HCl in 100 cm3, initial moles of HCl, moles of NH3 reacted, and finally the percentage of nitrogen in the compound.

(Total for Question 8 = 14 marks)

Question

Answer Additional Guidance Mark

Number

9(a) An answer that makes reference to the following points Allow displayed formulae, any combination of structural (5)

and displayed formulae or skeletal formulae

Ignore names even if incorrect

• structure of A as 2-chloropropane (1) CH3CHClCH3

• structure of B as propan-2-ol (1) CH3CH(OH)CH3

Allow structure of propan-1-ol if A is 1-chloropropane

• structure of C as propanone (1) CH3COCH3

Allow structure of propanal if A is 1-chloropropane

Do not allow propanoic acid as the formula is incorrect.

• structure of D as 2-propylmagnesium chloride (1) CH3CH(MgCl)CH3

Allow CH3CH(ClMg)CH3

• structure of E as 2-methylpropanoic acid (1) CH3CH(CH3)COOH

TE throughout

Question

Answer Additional Guidance Mark

Number

9(b) Example of calculation (5)

• calculation of mol of NaOH (1) mol of NaOH = 15.5 × 0.100 = 1.55 × 10−3 / 0.00155 (mol)

1000

3 mol HCl = 4 × 1.55 × 10−3 = 6.2 × 10−3 / 0.0062 (mol)

• calculation of mol HCl in 100 cm (1)

mol HCl at start = 100 × 0.225 = 2.25 × 10−2 / 0.0225 (mol)

• calculation of mol HCl at start (1)

1000

mol NH = 0.0225 – 0.0062 = 0.0163 / 1.63 × 10−2 (mol)

• calculation of mol NH3 reacted with HCl (1) 3

• calculation of percentage of N in compound (1) mass N = 0.0163 × 14 = 0.2282 (g)

and

% N = 0.2282 × 100 = 19.176 (%)

1.19

Allow TE at each stage but M5 must be <100%

Correct answer scores 5

Ignore SF except 1 SF

(Total for Question 9 = 10 marks)

How to answer it

Analysis of Organic Compounds & Back-Titration Calculations

📌 What this question tests

This multi-step question assesses your knowledge of functional group chemistry (halogenoalkanes, alcohols, ketones, and carboxylic acids), Grignard reagent formation and chain extension with carbon dioxide, and quantitative chemistry via a complex back-titration to determine the percentage composition of nitrogen in an unknown organic compound.

Part (a): Organic Synthesis & Functional Group Interconversion

Identifying Compounds A to E (5 Marks)

✅ Correct Answers

  • A: 2-chloropropane ( CH₃CHClCH₃ )
  • B: propan-2-ol ( CH₃CH(OH)CH₃ )
  • C: propanone ( CH₃COCH₃ )
  • D: 2-propylmagnesium chloride ( CH₃CH(MgCl)CH₃ )
  • E: 2-methylpropanoic acid ( CH₃CH(CH₃)COOH )

💡 Key Knowledge

  • Nucleophilic substitution: Halogenoalkanes react with aqueous NaOH to form alcohols.
  • Oxidation: Secondary alcohols oxidize to ketones using acidified potassium dichromate( VI ). Ketones resist further oxidation.
  • Grignard Reactions: Haloalkanes react with Mg in dry ether to form organomagnesium compounds. Reaction with CO₂ followed by acid gives a carboxylic acid with a carbon chain extension.

🧠 Exam Technique

  • Deduce the structure of A step-by-step from the carbon skeleton ( C₃ ). Since B oxidizes to a ketone (C), B must be a secondary alcohol (propan-2-ol), making A 2-chloropropane.
  • Compound E contains 4 carbons ( C₄H₈O₂ ), confirming carbon dioxide insertion into the Grignard reagent.

❌ Common Errors

  • Confusing primary and secondary halogenoalkanes, leading to incorrect alcohol or aldehyde/carboxylic acid products.
  • Failing to account for the extra carbon atom introduced when Grignard reagents react with CO₂ .

Part (b): Back-Titration & Percentage Composition

Calculating the Percentage of Nitrogen in Compound Q (5 Marks)

📐 Step-by-Step Calculation

  1. Moles of NaOH used in titration:
    15.5 × 0.100 / 1000 = 1.55 × 10⁻³ mol (in 25 cm³)
  2. Moles of unreacted HCl in 100 cm³:
    Since 25 cm³ portions were titrated, scale up by multiplying by 4:
    4 × 1.55 × 10⁻³ = 6.20 × 10⁻³ mol
  3. Initial moles of HCl added:
    100 × 0.225 / 1000 = 2.25 × 10⁻² mol
  4. Moles of NH₃ reacted with HCl:
    2.25 × 10⁻² - 6.20 × 10⁻³ = 1.63 × 10⁻² mol (equal to moles of N in Q)
  5. Percentage of Nitrogen:
    Mass of N = 1.63 × 10⁻² × 14.0 = 0.2282 g
    % N = (0.2282 / 1.19) × 100 = 19.2%

🧠 Exam Technique & Marking Points

  • M1: Calculate moles of NaOH .
  • M2: Scale up to find unreacted HCl in the full 100 cm³ flask (factor of 4).
  • M3: Calculate initial moles of HCl .
  • M4: Subtract to find moles of NH₃ reacted.
  • M5: Convert to mass of N and find percentage using the sample mass ( 1.19 g ).

❌ Common Calculation Traps

  • Forgetting to multiply the titration moles by 4 to account for the 25 cm³ aliquots out of 100 cm³ total solution.
  • Subtracting initial moles from unreacted moles the wrong way around, resulting in negative values.

Topics

Organic Chemistry · Physical Chemistry · Topic 6: Organic Chemistry I · Topic 17: Organic Chemistry II · Topic 5: Formulae, Equations and Amounts of Substance

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.