OCR A-Level Chemistry AS Depth in chemistry (02), June 2018: Question 7

6 marks · Hard difficulty · Extended Response

Determine the structure of the trans stereoisomer compound F using its percentage composition, infrared spectrum, and mass spectrum data.

Practise this question

Question

An exam question showing the percentage composition of compound F (C 68.6%, H 8.6%, O 22.8%), an infrared spectrum with wavenumber from 4000 to 400 cm-1, and a mass spectrum showing m/z values up to 90. The question asks to determine the structure of compound F, explain reasoning, and show working.
Question text

Compound F is a trans stereoisomer which is a useful intermediate in organic synthesis.

The results of elemental and spectral analysis of compound F are shown below.

Percentage composition by mass: C, 68.6 %; H, 8.6 %; O, 22.8 %.

Infrared spectrum

transmittance

(%)

4000 3000 2000 1500 1000 500

wavenumber / cm–1

Mass spectrum

relative

intensity

10 20 30 40 50 60 70 80

m/z

In the mass spectrum, the peak with the greatest relative intensity is caused by the loss of a

functional group from the molecular ion of compound F.

Determine the structure of compound F.

Explain your reasoning and show your working.

… [6]

Mark scheme

Show the mark scheme A level-based mark scheme showing criteria for levels 1 to 3, with indicative scientific points including empirical formula calculation leading to C4H6O, IR absorption for carbonyl, molecular mass of 70, and identification of the trans stereoisomer structure.

Question Answer Marks Guidance

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

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

Level 3 (5-6 marks) Empirical formula

A comprehensive description including most of the evidence to justify empirical formula = C4H6O

the correct structure of F (accept cis or trans).

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

logically structured. The information presented is relevant and

substantiated.

IR and spectra and molecular formula

infrared absorption; 1630–1820 cm–1, due to

Level 2 (3–4 marks)

The candidate attempts all three scientific points, but explanations are C=O (aldehyde/ketone/carbonyl group)

molar mass = 70 g mol–1

incomplete.

OR (mass spectrum molecular ion peak m/z = 70)

Explains two scientific points thoroughly with few omissions. molecular formula = C4H6O

AND

an attempt at a feasible structure based on deduction from correct Functional groups, structure and stereochemistry

molecular formula alkene / C=C

aldehyde / –CHO (C H + fragment)

There is a line of reasoning presented with some structure. The 3 5

mass spectrum; peak at 41 due to C H + (loss of

information presented is relevant and supported by some evidence. 3 5

CHO)

Level 1 (1–2 marks) E/Z or cis-trans isomer: E/Z or cis-trans isomer:

The correct empirical formula

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

points.

OR

The candidate explains one scientific point thoroughly with few

omissions.

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

information is in the most part relevant.

cis trans (correct structure)

0 marks No response or no response worthy of credit.

Total 6

How to answer it

OCR AS Level Chemistry • Organic Analysis Exam Guide

Structural Determination of Compound F using Analytical Spectra

What this question tests

This 6-mark extended-response question assesses your ability to combine multiple analytical techniques to deduce an unknown organic structure. You must interpret elemental percentage composition to find the empirical formula, read an Infrared (IR) spectrum to identify functional groups, utilize a Mass Spectrum (MS) to find the molecular mass and key fragmentation patterns, and apply stereochemical clues (trans isomerism).

Question 7: Complete Structure Determination

Determine the structure of compound F. Explain your reasoning and show your working.

✅ Correct Answer & Final Structure

Compound F is trans-but-2-enal (or a trans stereoisomer drawing showing a carbon-carbon double bond with methyl and aldehyde groups arranged diagonally/trans across the double bond).

Structure layout description: H and CH₃ are on opposite sides of the C=C double bond, and H and CHO are on opposite sides.

💡 Key Knowledge Required

  • Combustion/Elemental Analysis: Converting % mass to empirical formula via moles and simplest ratios.
  • IR Spectroscopy: Identifying sharp absorption peaks around 1630–1820 cm⁻¹ as a carbonyl ( C=O ) group.
  • Mass Spectrometry: The molecular ion peak ( M⁺ ) gives the relative molecular mass ( m/z = 70 ). Fragmentation clues help piece the skeleton together.
  • Stereoisomerism: Understanding that trans means highest priority groups (or similar substituents) are positioned on opposite sides of the restricted-rotation C=C bond.

📐 Step-by-Step Calculations

  1. Assume 100 g sample:
    C = 68.6 g | H = 8.6 g | O = 22.8 g
  2. Divide by Relative Atomic Masses (Ar):
    C: 68.6 / 12.0 = 5.72 mol
    H: 8.6 / 1.0 = 8.60 mol
    O: 22.8 / 16.0 = 1.425 mol
  3. Find simplest ratio (divide by smallest, 1.425):
    C: 5.72 / 1.425 = 4
    H: 8.60 / 1.425 = 6
    O: 1.425 / 1.425 = 1
  4. Empirical Formula: C₄H₆O (Mr = 70.0)
  5. Molecular Formula Confirmation: The mass spectrum molecular ion peak ( m/z ) is at 70 , matching the empirical formula mass exactly. Therefore, Molecular Formula = Empirical Formula = C₄H₆O .

🧠 Exam Technique & Level of Response

This is a Level of Response question marked out of 6. To achieve Level 3 (5–6 marks), you must:

  • Show complete calculations for the empirical and molecular formula.
  • Explicitly link spectral data to functional groups (e.g., quote IR wavenumber range and attribute it to C=O).
  • Incorporate the fragmentation clue (loss of CHO fragment giving peak at m/z = 41 ).
  • Draw the explicit trans stereoisomer clearly.

❌ Common Errors & Pitfalls

  • Forgetting units or miscalculating ratios: Rounding moles too early, leading to incorrect empirical ratios like C₃H₅O.
  • Ignoring stereochemistry: Drawing a standard straight-chain skeletal structure or a cis isomer when the stem explicitly states compound F is a trans stereoisomer.
  • Vague spectral assignments: Stating "there is a peak" without stating the wavenumber range ( 1630–1820 cm⁻¹ ) or the specific bond ( C=O ).
  • Fragment confusion: Misinterpreting the mass spec peak at m/z = 41 ; the prompt notes a loss of a functional group ( −CHO mass = 29, so 70 − 29 = 41 , confirming an aldehyde).
Examiner Note: Top-tier responses systematically stepped through elemental analysis first to secure the molecular formula, used spectroscopic evidence to identify the aldehyde ( −CHO ) and alkene ( C=C ) groups, accounted for the m/z = 41 fragment via loss of CHO , and finally drew a structurally correct trans isomer layout.

Topics

Module 4: Core organic chemistry · Module 6: Organic chemistry and analysis · 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 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.