Edexcel A-Level Chemistry AS Paper 2, June 2019: Question 7

7 marks · Hard difficulty · Open Response

Calculate the percentage composition by mass and deduce the displayed formulae of three isomers of C4H8O using IR spectroscopy and structural data.

Practise this question

Question

Exam question 7 about compounds with molecular formula C4H8O. Part (a) is a multiple-choice question asking for the percentage by mass of each element in C4H8O, with a table of four options A, B, C, D. Part (b) provides infrared spectroscopy and structural clues for three different isomers A, B, and C with formula C4H8O, and asks to deduce their displayed formulae with justification.
Question text

7 This question is about compounds with the molecular formula C4H8O.

(a) What is the percentage by mass of each element in C4H8O?

(1)

Percentage carbon Percentage hydrogen Percentage oxygen

A 66.67 11.11 22.22

B 60.00 20.00 20.00

C 41.38 3.45 55.17

D 30.77 61.54 7.69

*(b) Three different compounds, A, B and C, have the molecular formula C4H8O.

Information about these three compounds includes:

Ɣ all three compounds have infrared absorptions at about 3500 cm–1

Ɣ the infrared spectra of A and B each contain a peak at about 1650 cm–1,

while that of C does not

Ɣ only A has a branched carbon chain

Ɣ B is the E-isomer of a pair of stereoisomers.

Deduce a possible displayed formula for each of the compounds A, B and C.

You must use all the information and the Data Booklet to fully justify each of your

structures.

(6)

… *P55611A02028*

(Total for Question 7 = 7 marks)

Mark scheme

Show the mark scheme Mark scheme for question 7, indicating option A is correct for part (a). Part (b) is marked using a 6-mark levels-of-response rubric assessing indicative chemical points (IR absorptions, C=C presence, E-isomerism, cyclic structures) alongside structure and lines of reasoning.

How to answer it

Structural Isomerism, Infrared Spectroscopy & Stereoisomerism (C₄H₈O)

Edexcel AS Level Chemistry

What this question tests

This question assesses your ability to calculate percentage composition by mass, interpret infrared (IR) spectroscopy absorption data using the Data Booklet, recognize E-Z stereoisomerism in alkenes, apply structural constraints (branched vs straight chains, cyclic vs acyclic), and draw fully displayed formulae with clear logic and sustained reasoning.

Part (a): Percentage by Mass Calculation

Question Part (a) - 1 Mark

✅ Correct Answer

A (66.67% C, 11.11% H, 22.22% O)

Awarded 1 mark for selecting option A.

📐 Step-by-Step Calculation

  1. Find total Mᵣ of C₄H₈O:
    (4 × 12.0) + (8 × 1.0) + (16.0) = 72.0 g mol⁻¹
  2. Percentage Carbon: (48.0 / 72.0) × 100 = 66.67%
  3. Percentage Hydrogen: (8.0 / 72.0) × 100 = 11.11%
  4. Percentage Oxygen: (16.0 / 72.0) × 100 = 22.22%

❌ Common Errors

  • Distractor B: Uses atomic numbers instead of relative atomic masses.
  • Distractor C: Ignores atomic mass multipliers and uses raw number of each atom present.
  • Distractor D: Uses only atom counts and completely omits atomic masses.

Part (b): Deducing Displayed Formulae from Spectra and Clues

Question Part (b) - 6 Marks (Level of Response + Indicative Content)

💡 Key Knowledge & Data Booklet Clues

  • 3500 cm⁻¹ peak: Indicates an O–H stretch (alcohol or carboxylic acid, but formula has only 1 oxygen and 8 hydrogens, pointing to an alcohol or enol).
  • 1650 cm⁻¹ peak: Indicates a C=C double bond (present in A and B, absent in C).
  • E-isomerism: Compound B must contain a carbon-carbon double bond with two different groups attached to each carbon of the C=C bond, arranged across the double bond as an *E*-stereoisomer.
  • Cyclic structure: Since compound C has no C=C double bond (no 1650 cm⁻¹ peak) and no other unsaturation, it must be cyclic to account for the ring degree of unsaturation.

✅ Correct Structures (Displayed Formulae)

  • Compound A (Branched chain with C=C and O–H):
    2-methylprop-2-en-1-ol or 2-methyl-1-propen-1-ol displayed formula showing branched carbon skeleton, C=C bond, and –OH group.
  • Compound B (E-isomer with straight/unbranched chain, C=C and O–H):
    *(E)*-but-2-ene-1-ol displayed formula drawn clearly showing the *E* configuration across the double bond (groups on opposite sides).
  • Compound C (Cyclic structure with O–H, no C=C):
    Cyclobutanol or methylcyclopropanol isomers showing a closed ring and an –OH group.
6 marks total: Up to 4 marks for Indicative Content (IP1–IP6) and up to 2 marks for Quality of Written Communication, structure, and sustained reasoning.

🧠 Exam Technique & Level of Response Strategy

  • Link your clues: Explicitly state *why* a peak leads to a deduction (e.g., "Absorption at 3500 cm⁻¹ shows an O–H bond is present").
  • Draw fully displayed formulae: Ensure every single atom and every single bond is explicitly drawn out. Do not use condensed groups like CH₃ or CH₂OH unless the question permits (here, displayed formula requires all bonds shown).
  • Top-level responses: Seamlessly integrate all 6 indicative points with clear structural linkages rather than bullet-pointing disconnected facts.

Topics

Physical Chemistry · Organic Chemistry · Topic 5: Formulae, Equations and Amounts of Substance · Topic 6: Organic Chemistry I · Topic 7: Modern Analytical Techniques I

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.