OCR A-Level Chemistry AS Depth in chemistry (02), June 2022: Question 6

6 marks · Hard difficulty · Extended Response

Determine the structure of the unsaturated trans stereoisomer compound A using its percentage composition by mass, mass spectrum, and infrared spectrum.

Practise this question

Question

The question presents an organic compound A that is unsaturated and a trans stereoisomer, with percentage composition by mass C 55.8 percent, H 7.0 percent, O 37.2 percent. A mass spectrum shows peaks with a molecular ion peak at m/z 86 and relative intensity up to 100, and an infrared spectrum shows a broad absorption peak around 2500-3300 cm-1 and a peak around 1700 cm-1. Students are asked to use this information to determine the structure of compound A, explaining their reasoning and showing working.
Question text

The organic compound A is unsaturated and is a trans stereoisomer.

Compound A has the following composition by mass: C, 55.8%; H, 7.0%; O, 37.2%.

The mass spectrum and the infrared spectrum of compound A are shown below.

Mass spectrum

relative

intensity

10 20 30 40 50 60 70 80 90 100 110 120

m/z

Infrared spectrum

transmittance 50

(%)

4000 3000 2000 1500 1000 500

wavenumber / cm–1

Use the information to determine the structure of compound A.

Explain your reasoning and show your working. [6]

Additional answer space if required.

Mark scheme

Show the mark scheme The mark scheme outlines a 6-mark level-of-response answer detailing empirical formula calculation from percentage composition, interpretation of the mass spectrum for molecular mass and formula, identification of functional groups from IR peaks such as O-H and C=O for a carboxylic acid and C=C for an alkene, and deducing the correct trans-stereoisomer structure.

AO

Question Answer Marks Guidance

element

Please refer to the marking instructions on page 5 of the mark 6 AO3.1 LOOK AT THE SPECTRA for labelled peaks

scheme for guidance on how to mark this question. ×3 Indicative scientific points may include:

Level 3 (5-6 marks) AO3.2 Empirical formula

A comprehensive description including most of the evidence to ×3 • empirical formula = C2H3O

justify the correct structure of A (accept cis or trans). element %mass Ar moles ratio

C 55.8 12 4.65 2

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

H 7.0 1 7.0 3

logically structured. The information presented is relevant and

substantiated. O 37.2 16 2.325 1

Spectra and molecular formula

Level 2 (3–4 marks)

Mass spectrum

Explains two scientific points thoroughly with few omissions.

AND • (molecular ion peak m/z = 86)

• molar mass = 86 g mol–1

an attempt at a feasible structure with either a C=C OR COOH

• molecular formula = C4H6O2

There is a line of reasoning presented with some structure. The

information presented is relevant and supported by some Infrared absorption;

evidence. • broad peak at 2500–3300 cm–1, due to O–H in

carboxylic acid,

Level 1 (1–2 marks) • peak at 1630–1820 cm –1 due to C=O

The correct empirical formula • (peak at 1620–1680 cm –1 due to C=C)

AND a simple description based on at least one of the main

scientific points. Functional groups, structure and stereochemistry

OR • alkene / C=C

Some aspects from two scientific points are given

• carboxylic acid / –COOH

There is an attempt at a logical structure with a line of • mass spectrum; peak at 41 due to loss of COOH

reasoning. The information is in the most part relevant. • Correct structural formula: CH3CH=CHCOOH

i.e. cis OR trans

0 marks No response or no response worthy of credit.

• trans isomer indicates C=C bond with 2 different

groups attached to both double bonded carbons

• trans: common groups on opposite sides of double

bond

• Correct structure:

23 AO

element

NOTE: Correct trans assignment with justification

would be an example of a well-developed line of

reasoning that is substantiated.

Total 6

How to answer it

Elucidating Organic Structure from Spectra and Composition

What this question tests

This synoptic Level of Response question assesses your ability to combine analytical data (percentage composition, mass spectrometry, and infrared spectroscopy) to deduce an unknown organic structure. You must calculate empirical and molecular formulas, identify characteristic functional groups from IR absorptions, interpret fragmentation peaks, and apply stereochemical knowledge (trans isomers).

Complete Examination Breakdown & Worked Solution

Question 6 (6 Marks total - Level of Response)

✅ Correct Final Answer

Compound A is a trans stereoisomer of but-2-enoic acid (also known as crotonic acid).

CH₃CH=CHCOOH

Drawn as a skeletal or structural formula clearly showing the trans (E) geometry across the carbon-carbon double bond.

💡 Key Knowledge Required

  • Empirical Formula: Converting % mass to moles using relative atomic masses (Aᵣ).
  • Molecular Formula: Using the molecular ion peak (M⁺) from the mass spectrum to scale the empirical formula.
  • IR Spectroscopy: Identifying O-H in carboxylic acids (broad 2500–3300 cm⁻¹) and C=O (~1680–1750 cm⁻¹), plus C=C alkene stretch (~1620–1680 cm⁻¹).
  • Stereochemistry: Recognising that a trans isomer requires different groups on both ends of the C=C double bond arranged on opposite sides.

🧠 Exam Technique (Level of Response)

This is a 6-mark banded question. To hit Level 3 (5–6 marks), you must present a comprehensive, logically structured description covering all lines of evidence: empirical formula, molecular mass, functional groups from IR/mass spec, and justification of the trans stereochemistry.

❌ Common Errors & Pitfalls

  • Forgetting to scale up the empirical formula using the molecular ion peak ( m/z = 86 ), leaving it as C₂H₃O .
  • Misassigning the broad IR absorption to an alcohol O-H rather than a carboxylic acid O-H.
  • Drawing a cis isomer when the question explicitly states compound A is a trans stereoisomer.

📐 Step-by-Step Calculation & Analysis

  1. Step 1: Empirical Formula Calculation

    Assume 100 g of compound A:

    • Moles of C = 55.8 / 12.0 = 4.65 mol
    • Moles of H = 7.0 / 1.0 = 7.00 mol
    • Moles of O = 37.2 / 16.0 = 2.325 mol

    Divide by the smallest value (2.325):

    • C : H : O = 2 : 3 : 1

    Empirical Formula = C₂H₃O (relative mass = 43.0)

  2. Step 2: Molecular Formula Determination

    The mass spectrum shows a molecular ion peak (M⁺) at m/z = 86 . Since the empirical mass (43.0) goes into 86 twice, multiply the empirical formula by 2.

    Molecular Formula = C₄H₆O₂

  3. Step 3: Infrared (IR) Spectrum Interpretation
    • Broad absorption peak between 2500 and 3300 cm⁻¹ indicates an O-H stretch in a carboxylic acid ( -COOH ).
    • Sharp absorption peak around 1680–1710 cm⁻¹ indicates a C=O carbonyl stretch.
    • Absorption peak around 1640 cm⁻¹ indicates a C=C alkene stretch (confirming unsaturation).
  4. Step 4: Putting Structure and Stereochemistry Together
    • Functional groups present: an alkene ( C=C ) and a carboxylic acid ( -COOH ).
    • The mass spectrum peak at m/z = 41 corresponds to the loss of a carboxylic acid group ( M⁺ - 45 ).
    • The question states compound A is a trans stereoisomer, meaning the carbon chain continues across the double bond with the main substituent groups positioned on opposite sides.
Mark Scheme Guidance: Level 3 is awarded for a fully justified structure linking quantitative data (formula) and qualitative data (IR/MS + trans geometry) into a clear logical sequence.

Topics

Module 2: Foundations in chemistry · Module 6: Organic chemistry and analysis · Module 4: Core organic chemistry · 2.1 Atoms and reactions · 4.1 Basic concepts and hydrocarbons · 6.3 Analysis

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