OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2017: Question 20

9 marks · Hard difficulty · Structured Questions

Determine the molecular formula, functional groups, and structure of a naturally occurring aromatic compound using percentage composition, mass spectrometry, qualitative test results, and carbon-13 NMR spectroscopy.

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Question

Question text

20 A chemist analyses a naturally occurring aromatic compound.

(a) The percentage composition and mass spectrum of the compound are shown below.

Percentage composition by mass: C, 70.58%; H, 5.92%; O, 23.50%.

Mass spectrum

relative 60

intensity

25 50 75 100 125 150

m/z

Determine the molecular formula of the compound.

Show your working.

molecular formula = … [3]

(b) Qualitative tests are carried out on the aromatic compound. The results are shown below.

Test Acidity Na2CO3(aq) 2,4-DNP Tollens’ reagent

No observable No observable

Observation pH = 5 Orange precipitate

change change

Determine the functional groups in the compound. Explain your reasoning.

Functional groups …

Explanation …

… [3]

200 180 160 140 120 100 80 60 40 20 0

chemical shift, δ/ppm

Using the spectrum and the results from (a) and (b), determine the structure of the compound.

Explain your reasoning.

Structure of compound

[3]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

20 (a) Empirical formula 3 ANNOTATE ANSWER WITH TICKS AND

CROSSES

Mole Ratio C : H : O = 5.88 : 5.92 : 1.47

70.58 5.92 23.50

ALLOW 12.0 : 1.0 : 16.0

Empirical formula = C4H4O

ALLOW 4:4:1 if linked to C:H:O

Molecular formula

Alternative method for 3 marks:

Molecular formula = C8H8O2 136 × 70.58/100

AND C: 12.0 = 8

Evidence of 136 in working or from labelled peak in 136 × 5.92/100

spectrum H: 1.0 = 8

136 × 23.50/100

O: = 2

16.0

(b) 3

Functional groups

Phenol AND ketone DO NOT ALLOW any other functional groups for first

marking point.

Explanation

ALLOW identity of functional groups in the

Links phenol to (weak) acidity explanation if not stated on functional group prompt

AND line.

no reaction with Na2CO3 (so not carboxylic acid)

Links 2,4-DNP(H) or Brady’s reagent observation to

carbonyl

AND ALLOW “aldehyde or ketone” in place of carbonyl

Tollens’ reagent observation (so not an aldehyde)

ALLOW peaks to be identified by:

Peaks between 110–160 ppm are the (four) aromatic

(carbon environments) Peaks labelled on spectrum

Compound contains a C=O between 190 - 200 ppm Peaks indicated on a chemical structure

AND

Compound contains a C-C at 20-30 ppm Peaks indicated from within text

Note: If identifying aromatic peaks from the

Structure O spectrum all four peaks should be indicated.

ALLOW any combination of skeletal OR structural

OR displayed formula as long as unambiguous

Total 9

How to answer it

Structure Determination of an Aromatic Compound

What this question tests

This multi-step synoptic organic question assesses your ability to combine analytical data from multiple sources to deduce an unknown organic structure. Key skills include calculating empirical and molecular formulae from percentage composition and mass spectrometry, interpreting functional group tests (acidity, 2,4-DNP, Tollens'), and correlating Carbon-13 NMR chemical shifts with molecular environments.

Part (a) - Empirical & Molecular Formula

Determining the Molecular Formula from % Composition and Mass Spec

📐 Step-by-Step Calculation

Find moles of each element:
C = 70.58 / 12.0 = 5.88 mol
H = 5.92 / 1.0 = 5.92 mol
O = 23.50 / 16.0 = 1.47 mol
Find the simplest whole-number ratio:
Divide by the smallest value (1.47):
C : H : O = 4 : 4 : 1
Determine Empirical Formula:
C₄H₄O (Relative mass = 68.0)
Use Mass Spectrum for Molecular Formula:
The molecular ion peak (highest m/z peak) is at m/z = 136 .
Since 136 is double 68 (136 / 68 = 2), multiply the empirical formula by 2.
Molecular formula = C₈H₈O₂

❌ Common Calculation Traps

Missing the Molecular Ion Peak: Failing to explicitly state or highlight 136 from the mass spectrum will cost you the final mark. Examiners require clear evidence of how you jumped from empirical to molecular formula.
Rounding Errors: Premature rounding of mole ratios can lead to incorrect empirical formulae (e.g., getting 3.99 instead of 4). Always keep full calculator values until the final ratio step.

Mark Scheme Breakdown (3 marks): 1 mark for correct mole ratios (C:H:O = 5.88 : 5.92 : 1.47 or equivalent) | 1 mark for empirical formula (C₄H₄O) | 1 mark for molecular formula (C₈H₈O₂) AND evidence of m/z = 136.

Part (b) - Functional Group Analysis

Deducing Functional Groups from Chemical Tests

✅ Correct Answer

Functional groups: Phenol AND Ketone

Explanation:

pH = 5 indicates weak acidity, and there is no observable change with Na₂CO₃(aq) . This rules out a carboxylic acid, confirming the weak acid is a phenol.
An orange precipitate with 2,4-DNP proves the presence of a carbonyl group (aldehyde or ketone).
No observable change with Tollens' reagent proves it is not an aldehyde, leaving a ketone as the only valid carbonyl option.

💡 Key Knowledge: Reagents & Observations

Na₂CO₃(aq): Reacts with strong/medium acids (like carboxylic acids) to release CO₂ gas (effervescence). Phenols are too weakly acidic to react with carbonates.
2,4-DNP (Brady's Reagent): Forms a yellow/orange crystalline precipitate with any aldehyde or ketone.
Tollens' Reagent: Forms a silver mirror specifically with aldehydes; ketones give no reaction.

Mark Scheme Breakdown (3 marks): 1 mark for identifying Phenol AND Ketone on the prompt line | 1 mark for linking phenol to weak acidity and no reaction with Na₂CO₃ (proving absence of -COOH) | 1 mark for linking 2,4-DNP positive result to carbonyl and Tollens' negative result to the absence of an aldehyde.

Part (c) - Carbon-13 NMR & Structure Determination

Synthesizing Data into a Final Structure

🧠 Exam Technique & NMR Analysis

Aromatic Region (110–160 ppm): There are four distinct peaks in this region, confirming a substituted benzene ring with non-equivalent carbon environments.
Carbonyl Carbon (190–200 ppm): A peak in this high shift region confirms a ketone carbonyl group.
Aliphatic Carbon (20–30 ppm): A peak here corresponds to an alkyl carbon next to a carbonyl (e.g., a -CH₃ group attached to C=O ).

✅ Final Correct Structure

The compound is 4-hydroxyacetophenone (or 1-(4-hydroxyphenyl)ethanone).

Structure description: A benzene ring substituted in a 1,4- (para) arrangement with a hydroxyl group ( -OH ) on one side and an ethanoyl group ( -COCH₃ ) directly attached on the opposite side.

Mark Scheme Breakdown (3 marks): 1 mark for identifying aromatic carbon environments (peaks between 110–160 ppm) | 1 mark for identifying C=O (190–200 ppm) and aliphatic C (20–30 ppm) | 1 mark for the fully correct drawn structure of 4-hydroxyacetophenone.

Topics

Module 4: Core organic chemistry · Module 6: Organic chemistry and analysis · Module 2: Foundations in chemistry · 6.1 Aromatic compounds, carbonyls and acids · 6.3 Analysis · 2.1 Atoms and reactions · 4.2 Alcohols, haloalkanes and analysis

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.