OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2018: Question 22

18 marks · Hard difficulty · Extended Response

A multi-part question covering incomplete and complete combustion, intermolecular forces and boiling points, empirical and molecular formula calculations, organic synthesis equations involving carbonyl compounds and polyols, and structure elucidation using elemental analysis, mass spectrometry, and NMR spectroscopy.

Practise this question

Question

An exam question with multiple parts regarding fuels, combustion, boiling points, and structure determination. Table 22.1 lists boiling points and relative molecular masses for hexane, pentan-1-ol, and heptane. Part (a) asks for an incomplete combustion equation for heptane. Part (b) asks to explain differences in boiling points. Part (c) involves combustion data to find the molecular formula and structure of compound N, followed by an equation for synthesizing solketal from propane-1,2,3-triol and a carbonyl compound. Part (d) provides elemental analysis, a mass spectrum molecular ion peak, and an 1H NMR spectrum with D2O data to deduce the structure of an unknown fuel additive.
Question text

22 The relative molecular masses and boiling points of some fuels are shown in Table 22.1.

Fuel Relative molecular mass Boiling point / °C

hexane 86 69

pentan-1-ol 88 138

heptane 100 98

Table 22.1

(a) Write an equation for the incomplete combustion of heptane.

… [1]

(b) Explain the difference in the boiling points of the fuels in Table 22.1.

… [4]

(c) Fuel additives are often used to improve the combustion of a fuel.

(i) Compound N is a fuel additive containing carbon, hydrogen and oxygen only.

Complete combustion of 1.71 g of compound N produces 2.97 g of CO2 and 1.62 g of

H2O. The relative molecular mass of compound N is 76.0.

Calculate the molecular formula of N and suggest a possible structure for the compound.

compound N

[5]

(ii) Solketal has been investigated as a potential fuel additive.

HO O

O

solketal

Solketal is synthesised from propane-1,2,3-triol and a carbonyl compound.

Construct a balanced equation for this synthesis.

Show structures for the organic compounds in your equation.

[2]

(d)* A scientist is researching compounds that might be suitable as fuel additives.

One of the compounds gives the analytical results below.

Elemental analysis by mass:

C: 54.54%; H: 9.10%; O: 36.36%

Mass spectrum:

Molecular ion peak at m/z = 132.0

1H NMR spectrum in D O

54 3 2 1 0

chemical shift,/ppm

The numbers by the peaks are the relative peak areas.

When the spectrum is run without D2O, there are two additional peaks with the same relative

peak areas at 11.0 ppm and 3.6 ppm.

Use the information provided to suggest a structure for the compound.

Show all your reasoning. [6]

Additional answer space if required.

Mark scheme

Show the mark scheme The mark scheme provides answers and guidance for all parts of question 22. Part (a) shows incomplete combustion equations. Part (b) details marking points comparing London forces in heptane and hydrogen bonding in pentan-1-ol. Part (c)(i) outlines moles calculations, empirical ratio, molecular formula, and possible structures. Part (c)(ii) shows the equation for solketal formation. Part (d) uses a levels-of-response mark scheme for the 6-mark structure elucidation task, detailing empirical formula calculation, NMR peak analysis, and final chemical structures.

Question Answer Marks Guidance

22 (a) C H + 71/ O 7CO + 8H O 1 ALLOW multiples

7 16 2 2 2

OR IGNORE state symbols

C7H16 + 4O2 7C + 8H2O ALLOW equations for incomplete combustion that give

CO and/or C with CO2

e.g C7H16 + 9O2 4CO + 3CO2 + 8H2O

C7H16 + 6O2 4CO + 3C + 8H2O

(b) 4 ANNOTATE WITH TICKS AND CROSSES

Heptane compared to hexane ALLOW ORA throughout

heptane (has a longer chain so) has more points of contact

/ more surface interaction (between molecules)

ALLOW heptane has more electrons

heptane has stronger/more induced dipole(–dipole)

interactions IGNORE IDID

ALLOW stronger/more London forces

Pentan-1-ol compared to heptane and/or hexane IGNORE van der Waals’ forces/VDW for induced dipole–

dipole interactions (ambiguous as this term refers to both

pentan-1-ol has hydrogen bonds that are strong(er than permanent dipole–dipole interactions and induced

induced dipole–dipole interactions) dipole–dipole interactions)

OR

(alcohols have) hydrogen bonds and induced dipole(-dipole)

interactions/London forces IGNORE ‘pentan-1-ol can form hydrogen bonds with

water’

Energy required to break forces

More energy is required to break induced dipole(–dipole)

interactions in heptane than hexane ALLOW ‘more energy to break intermolecular forces’ if

OR intermolecular forces are not stated.

More energy is required to break hydrogen bonds

IGNORE it is harder to break the intermolecular forces

no reference to energy)

IGNORE more energy needed to separate molecules

IGNORE more energy is needed to break bonds

Question Answer 31 Marks Guidance

(c) (i) 5 Consult your team leader if an alternative creditworthy

approach is seen

n(CO2) = 2.97/44 = 0.0675 (mol)

n(H2O) = 1.62/18 = 0.0900 (mol)

IGNORE ratio of CO2 to H2O is 3:4

Ratio of C : H ALLOW this mark from the correct molecular formula

3 : 8 OR a correct structure if not shown in working

Molecular formula

C3H8O2 DO NOT ALLOW an incorrect molecular formula

Structure

any correct structure of C3H8O2

Mark independently from molecular formula but structure

e.g. MUST contain 3C, 8H and 2O

H H H

ALLOW any combination of skeletal OR structural OR

HO C C C OH displayed formula as long as unambiguous

H H H ALLOW any vertical bond to the OH group

e.g. ALLOW

OR

OR

H H H OH HO

H C O C O C H DO NOT ALLOW OH–

H H H etc

(c) (ii) 2 ALLOW any combination of skeletal OR structural OR

OH O HO O displayed formula as long as unambiguous

+ + H2O

HO OH

O

Carbonyl compound identified as propanone

Rest of equation

(d)* Please refer to the marking instructions on page 5 of this 6 Indicative scientific points:

mark scheme for guidance on how to mark this question.

Empirical and Molecular Formula

Level 3 (5–6 marks) C : H : O = 54.54/12 : 9.10/1 : 36.36/16

Compound is a structure of C6H12O3 that is consistent with 4.545 : 9.10 : 2.273

splitting pattern and chemical shifts in NMR spectrum33.

2 : 4 : 1

AND

Comprehensive reasoning with most of the data analysed. Empirical formula = C H O

There is a well-developed line of reasoning which is clear uses m/z = 132.0 to determine molecular formula as

and logically structured. The information presented is C H O

relevant and substantiated. 6 12 3

1H NMR analysis

Level 2 (3–4 marks)

Compound has a feasible chemical structure that is Spectrum:

consistent with the splitting pattern in NMR spectrum but = 4.0 ppm, quartet, 1H, CH3–CH–O

may have incorrect molecular formula. = 1.3 ppm, singlet, 6H, (CH3)2–C

AND = 1.2 ppm, doublet, 3H, CH3–CH–

Reasoning provided with some of the data analysed.

Without D2O:

Peak at 11.0 ppm COOH or OH

There is a line of reasoning presented with some structure.

The information presented is relevant and supported by peak at 3.6 ppm OH

some evidence.

Note: Data Sheet shows O-H chemical shift can occur

Level 1 (1–2 marks) around 11.0 ppm

Correct determination of empirical formula and/or molecular

formula.

OR

Analyses most of the NMR data.

OR Structure

Attempts to determine empirical and/or molecular formula

AND analyses some of the NMR data. ALLOW any combination of skeletal OR structural OR

displayed formula as long as unambiguous

There is an attempt at a logical structure with a line of

reasoning. The information is in the most part relevant. Contains

region that gives doublet and quartet

0 marks

e.g.

No response or no response worthy of credit.

H O C

H C C C C

H H C

region that gives singlet

e.g.

CH3

C C C

CH3

Examples of structures consistent with splitting and

chemical shift in NMR

OH CH3 O

H3C C C C

H CH OH

OH O CH3

H3C C C C OH

H CH3

H

H3C O C CH3

C C OH

H3C OH

CH3 OH

H3C C C OH

O C H

CH3

Note: there may be other possible structures that are

consistent with the splitting pattern and chemical shifts in

NMR – if an alternative structure is seen, please contact

your team leader

Total 18

How to answer it

Fuel Additives, Intermolecular Forces & Structure Determination

OCR A-Level Chemistry • Organic Analysis & Synthesis

What this question tests

This comprehensive multi-part question tests your knowledge of alkane combustion equations, intermolecular forces (London forces vs hydrogen bonding), combustion elemental analysis calculations, organic synthetic equations (diol condensation with carbonyl compounds), and advanced structural elucidation using mass spectrometry, empirical formulas, and high-resolution proton (¹H) NMR spectroscopy.

Part (a) — Incomplete Combustion of Heptane

1 Mark

✅ Correct Answer

C₇H₁₆ + 7.5 O₂ ➔ 7 CO + 8 H₂O

OR C₇H₁₆ + 4 O₂ ➔ 7 C + 8 H₂O (or valid multiples)

❌ Common Errors

  • Producing carbon dioxide (CO₂) instead of carbon monoxide (CO) or solid carbon (C).
  • Balancing errors with oxygen atoms when using fractional coefficients.
Mark: 1 mark for a balanced equation showing incomplete combustion products (CO or C alongside H₂O).

Part (b) — Explaining Boiling Point Differences

4 Marks

💡 Key Knowledge

Boiling points depend on the energy required to overcome intermolecular forces between molecules, not covalent bonds.

🧠 Exam Technique

Structure your answer in two clear comparisons: (1) Heptane vs. hexane (size/surface area), and (2) Pentan-1-ol vs. heptane/hexane (functional group/hydrogen bonding).

✅ Model Explanation

  • Heptane vs. Hexane: Heptane has a longer carbon chain, providing a larger surface area and more points of contact between molecules. This results in stronger induced dipole-dipole interactions (London forces) requiring more energy to overcome.
  • Pentan-1-ol vs. Alkanes: Pentan-1-ol contains an -OH group, allowing it to form hydrogen bonds between molecules. Hydrogen bonds are significantly stronger than the London forces present in alkanes, hence pentan-1-ol has a much higher boiling point (138 °C).
Marks: 4 marks total. 2 marks for comparing the two alkanes (surface area/chain length & London force strength), and 2 marks for pentan-1-ol (hydrogen bonding & energy required to break them).

Part (c)(i) — Combustion Analysis & Molecular Formula

5 Marks

📐 Step-by-Step Calculation

  1. Find moles of CO₂ produced:
    n(CO₂) = mass / M_r = 2.97 / 44.0 = 0.0675 mol
    Therefore, moles of Carbon = 0.0675 mol
  2. Find moles of H₂O produced:
    n(H₂O) = 1.62 / 18.0 = 0.0900 mol
    Moles of Hydrogen = 0.0900 × 2 = 0.1800 mol
  3. Determine masses of C and H:
    Mass of C = 0.0675 × 12.0 = 0.810 g
    Mass of H = 0.1800 × 1.0 = 0.180 g
  4. Find mass and moles of Oxygen in Compound N:
    Total mass = 1.71 g. Mass of O = 1.71 - (0.810 + 0.180) = 0.720 g
    Moles of O = 0.720 / 16.0 = 0.0450 mol
  5. Establish C : H : O Ratio:
    C : H : O = 0.0675 : 0.1800 : 0.0450
    Divide by smallest (0.0450) ➔ 1.5 : 4 : 1
    Multiply by 2 to get whole numbers ➔ 3 : 8 : 2. Empirical formula: C₃H₈O₂ .
  6. Molecular Formula & Structure:
    The relative molecular mass is given as 76.0. The relative mass of C₃H₈O₂ is (3×12) + (8×1) + (2×16) = 76.0. Therefore, the molecular formula is C₃H₈O₂ .
    Suggested Structure: Propane-1,3-diol ( HO-CH₂-CH₂-CH₂-OH ) or Propane-1,2-diol.
Marks: 5 marks (1 for n(CO₂)/n(H₂O), 1 for element ratio, 1 for molecular formula C₃H₈O₂, 2 for valid structure).

Part (c)(ii) — Synthesis of Solketal

2 Marks

✅ Correct Equation

Propane-1,2,3-triol + Propanone ➔ Solketal + Water

C₃H₈O₃ + C₃H₆O ➔ C₆H₁₂O₃ + H₂O

💡 Key Knowledge

This is a condensation/acetal formation reaction where a diol reacts with a carbonyl compound (a ketone, specifically propanone) in the presence of an acid catalyst to form a cyclic acetal ring and eliminate water.

Marks: 2 marks (1 mark for correctly identifying propanone as the carbonyl reactant, 1 mark for balanced equation and correct structures).

Part (d) — Structure Elucidation from Analytical Data

6 Marks (Level of Response)

🧠 Strategy for Top-Level (Level 3) Answers

Systematically process each piece of analytical data in order: Elemental analysis ➔ Empirical formula ➔ Molecular formula (using M+ peak) ➔ Functional groups (D₂O shake & NMR shifts) ➔ Connectivity (splitting patterns).

💡 Analytical Breakdown

  • Elemental Analysis: C = 54.54%, H = 9.10%, O = 36.36%. Dividing by atomic masses gives an empirical formula of C₃H₆O .
  • Mass Spec: Molecular ion peak (M⁺) at m/z = 132.0. The molar mass of C₃H₆O is 58.0. Since 132 / 58 = 2.2 (or using full formula derivation), the molecular formula is C₃H₁₂O₃ (or correct multiple confirming exact molecular formula C₆H₁₂O₃ based on data integration).
  • ¹H NMR & D₂O Exchange: Peaks at 11.0 ppm and 3.6 ppm disappear or shift upon adding D₂O, confirming the presence of exchangeable protons: a carboxylic acid group ( -COOH ) or alcohol groups ( -OH ).
  • Splitting & Integration: Quartet at 4.0 ppm (area 1, CH-O), singlet at 1.3 ppm (area 6, two equivalent CH₃ groups attached to carbon), doublet at 1.2 ppm (area 3, CH₃-CH).

✅ Possible Structure

A structure consistent with the spectroscopic data features branching methyl groups and oxygen heteroatoms, such as:

2-hydroxy-3,3-dimethylbutanoic acid derivatives or related isomers with quaternary centers producing singlets and coupling patterns matching the integration values (6 : 3 : 1).

Marks: 6 marks awarded via Level of Response criteria (Level 3 requires correct molecular formula, detailed NMR analysis of splitting/shifts, and a fully consistent structure).

Topics

Module 4: Core organic chemistry · Module 6: Organic chemistry and analysis · Module 2: Foundations in chemistry · 6.1 Aromatic compounds, carbonyls and acids · 6.2 Nitrogen compounds, polymers and synthesis · 6.3 Analysis · 2.1 Atoms and reactions · 4.1 Basic concepts and hydrocarbons · 4.2 Alcohols, haloalkanes and analysis

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.