Edexcel A-Level Chemistry Paper 2, June 2023: Question 4

9 marks · Medium difficulty · Short Open Response

Calculate the empirical formula of a hydrocarbon from combustion data, define an electrophile, and explain the relative resistance of benzene compared to ethene to bromination.

Practise this question

Question

Question 4 asks about hydrocarbons. Part (a) asks to calculate the empirical formula of a hydrocarbon given that a 2.50 g sample gives 7.59 g of carbon dioxide on complete oxidation, worth 4 marks. Part (b)(i) is a multiple-choice question defining an electrophile with options A to D, worth 1 mark. Part (b)(ii) asks to explain why benzene is resistant to bromination but ethene reacts readily with bromine at room temperature, worth 4 marks.
Question text

4 This question is about some hydrocarbons.

(a) A 2.50g sample of a hydrocarbon gave 7.59g of carbon dioxide on

complete oxidation.

Calculate the empirical formula of the hydrocarbon.

(4)

(b) Benzene and ethene react with bromine under different conditions but both

reactions involve an electrophile.

(i) An electrophile is a substance that

(1)

A accepts a pair of electrons

B accepts an unpaired electron

C donates a pair of electrons

D donates an unpaired electron

(ii) Explain why benzene is resistant to bromination but ethene reacts readily

with bromine at room temperature.

(4)

… 8

… *P71913A0828*

(Total for Question 4 = 9 marks)

Mark scheme

Show the mark scheme The mark scheme provides detailed answers for Question 4. Part (a) lists marks for calculating moles of CO2, carbon and hydrogen masses, mole ratio, and the final empirical formula C2H5 with example calculations. Part (b)(i) indicates option A is the correct answer. Part (b)(ii) awards marks for mentioning benzene's delocalised electrons, its stability or high activation energy, ethene's localised electron density, and susceptibility to electrophilic attack.

Question

Answer Additional Guidance Mark

Number

4(a) Example of calculation (4)

• calculation of mol of CO (1) mol CO = 7.59 = 1.725 × 10−1 / 0.1725 (mol)

• calculation of masses of C and H (1) mass C = 0.1725 × 12 = 2.07 (g)

or

12 x 7.59 = 2.07 (g) subsumes M1 as well

and

mass H = 2.50 – 2.07 = 0.43 (g)

• calculation of mol and ratio of C : H (1) C H

mol 2.07/12 0.43 /1

= 0.1725 = 0.43

ratio 0.1725/0.1725 0.43/0.1725

= 1 = 2.4928

• empirical formula (1) C2H5

Correct answer with no working scores 1

Correct answer with some working scores 4

Question Answer Mark

number

4(b)(i) (1)

The only correct answer is A (accepts a pair of electrons)

B is incorrect because electrophiles involve a pair of electrons, not an unpaired electron

C is incorrect because nucleophiles donate a pair of electrons

D is incorrect because nucleophiles donate a pair of electrons, not an unpaired electron

Question

Answer Additional Guidance Mark

Number

4(b)(ii) An explanation that makes reference to the following points: (4)

• benzene (is resistant to bromination because it) has (1)

delocalisation of electrons / delocalised electrons

(in π bonds)

• benzene is (kinetically) stable Allow more energy is required to break up the structure of

or benzene

the activation energy for the reaction is high (1) Allow a (Friedel-Crafts) catalyst / halogen carrier /AlCl3,

FeBr3 is needed

• ethene (reacts readily because it) has localised electron Allow ethene has a C=C / (carbon – carbon) double bond

density (in one π bond) / does not have delocalised electrons

or

increased / high electron density (of the double / π bond

compared to benzene) (1)

• which makes it more susceptible to / Allow benzene is less susceptible to electrophilic attack

easier to undergo electrophilic attack (than (than ethene)

benzene) (1) Allow the π / double bond in ethene is weaker (than in

benzene)

Allow ethene is a better nucleophile (than benzene)

Allow the Br2 / Br-Br is more easily polarised (by ethene)

Do not award electrophilic attack by Br+

If no other mark scored allow (1) for ‘benzene undergoes

substitution and ethene undergoes addition’

(Total for Question 4 = 9 marks)

How to answer it

Hydrocarbons, Combustion Analysis & Arenes

📋 What this question tests

This question assesses core organic chemistry and quantitative analysis skills: calculating empirical formulas from combustion data, defining fundamental reaction terminology (electrophiles), and explaining chemical stability and reactivity differences between aromatic compounds (benzene) and alkenes (ethene) using electron density and delocalisation concepts.

Question 4(a)

Empirical Formula Calculation from Combustion Data

📐 Step-by-Step Calculation

  1. Find moles of CO₂ produced:
    Moles = 7.59 / 44.0 = 0.1725 mol
  2. Determine moles and mass of Carbon:
    Moles of C = Moles of CO₂ = 0.1725 mol
    Mass of C = 0.1725 × 12.0 = 2.07 g
  3. Determine mass and moles of Hydrogen:
    Mass of H = Total sample mass - Mass of C = 2.50 - 2.07 = 0.43 g
    Moles of H = 0.43 / 1.0 = 0.43 mol
  4. Find the simplest whole-number ratio:
    C : H ratio = 0.1725 / 0.1725 : 0.43 / 0.1725
    Ratio = 1 : 2.4928 (rounds to 1 : 2.5 )
  5. Scale up to integers:
    Multiply by 2 to get whole numbers: C₂H₅

❌ Common Errors & Calculation Traps

  • Forgetting to scale: Leaving the ratio as C1H2.5 instead of doubling to C₂H₅.
  • Hydrogen mass confusion: Trying to use water mass (which isn't given) instead of subtracting the carbon mass from the total hydrocarbon mass.
  • No working shown: A correct answer with zero working only scores 1 mark out of 4 under Edexcel rules. Always show intermediate moles and masses!
Mark Breakdown (4 marks total): (1) calculation of moles of CO₂; (2) calculation of masses of C and H; (3) calculation of mol ratio C:H; (4) final empirical formula C₂H₅ .
Question 4(i)

Definition of an Electrophile

✅ Correct Answer

A – accepts a pair of electrons

💡 Key Knowledge

  • Electrophile ("electron-loving"): An electron-pair acceptor.
  • Nucleophile ("nucleus-loving"): An electron-pair donor (options C).
  • Distinguish carefully between pairs of electrons and unpaired electrons (radicals).
Mark Breakdown (1 mark): 1 mark for selecting option A.
Question 4(ii)

Comparing the Reactivity of Benzene and Ethene

💡 Key Knowledge & Marking Points

  • Benzene structure: Pk-electrons are delocalised across the ring system, giving it extra thermodynamic and kinetic stability (high activation energy for reactions).
  • Ethene structure: Contains a localised π-bond between two carbon atoms with a high, concentrated electron density.
  • Reactivity outcome: Ethene reacts readily via electrophilic addition at room temperature because its localised electrons readily polarise approaching molecules, whereas benzene requires a halogen carrier catalyst (e.g., FeBr₃ or AlCl₃) to polarise the electrophile enough to break its stable delocalised ring.

🧠 Exam Technique & Examiner Guidance

  • To secure all 4 marks, you must explicitly contrast both molecules. Mentioning benzene alone is insufficient.
  • Use precise terminology: write "delocalised electrons" for benzene and "localised electron density / π-bond" for ethene.
  • Examiners accept alternative phrasing such as "ethene is a better nucleophile" or "ethene has a weaker π-bond compared to the delocalised system of benzene."
Mark Breakdown (4 marks total): (1) Benzene has delocalised electrons / π-system; (2) Benzene is stable / has a high activation energy; (3) Ethene has localised electron density in its double bond; (4) Ethene is more susceptible to electrophilic attack than benzene.

Topics

Physical Chemistry · Organic Chemistry · Topic 5: Formulae, Equations and Amounts of Substance · Topic 6: Organic Chemistry I · Topic 18: Organic Chemistry III

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.