OCR A-Level Chemistry AS Breadth in chemistry (01), June 2025: Question 25
9 marks · Medium difficulty · Structured Questions
Identify oxidation products, elimination behaviour, and calculate the volume of carbon dioxide produced from the combustion of isomeric alcohols using the ideal gas equation.
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Question text
25 This question is about alcohols.
Alcohols D, E, F and G, shown below, are structural isomers of C4H10O.
H H OH H H H CH3 H H H H
H C C C C H H3C C C OH H3C C OH HO C C C C H
H H H H CH3 H CH3 H H H H
Alcohol D Alcohol E Alcohol F Alcohol G
(a) Alcohols D, E, F and G are each heated under reflux with H+/ Cr O 2–.
Draw the structure of the organic product that is formed from each alcohol.
If there is no reaction, state ‘No reaction’.
Alcohol Organic product
D
E
F
G
[3]
(b) Alcohols D, E, F and G are each heated with concentrated H2SO4, which acts as an acid
catalyst.
Which alcohol forms a mixture of two alkenes that are structural isomers of one another?
Alcohol … [1]
(c) 25.9 g of alcohol F is completely combusted in oxygen at a temperature of 100°C and
1.50 × 105 Pa pressure.
The equation for this reaction is:
C4H10O + 6O2 4CO2 + 5H2O
Calculate the volume of CO (g) produced, in cm3.
Give your answer to an appropriate number of significant figures.
Volume of CO (g) = … cm3 [5]
Mark scheme
Show the mark scheme
Question Answer Mark Guidance
25 (a) 3 ALLOW any combination of skeletal OR structural OR
displayed formula as long as unambiguous
Alcohol Organic product
DO NOT ALLOW structure if H(s) are missing from
ONE structural formula
D
……. BUT ALLOW any further omissions as ECF
Take care with numbers of carbons, the branches
and the position of branching …. especially for E
E
IGNORE connectivity for methyl groups
BUT penalise incorrect connectivity for OH once
F No reaction –
e.g. HO
ALLOW structure of alcohol F for ‘no reaction’
G
ALL 4 responses correct → 3 marks
3 responses correct → 2 marks
2 responses correct → 1 marks
25 (b) D 1
25 (c) For CALCULATION, check the volume of CO2 on answer line23 ANNOTATE ANSWER WITH TICKS AND
MUST be derived from pV = nRT, 5 CROSSES
Award 5 marks for calculation for:
• answer = 2.89 104 (cm3) OR 28900 (cm3) Check entire response for credit worthy working
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Calculation of number of moles, n of CO2 produced If there is an alternative answer, check to see if
g 25.9 –1 there is any ECF credit possible using working
n(Alcohol F) = M = 74.0 = 0.35(0) (g mol ) ✓
n(CO2) = 4 0.35(0) = 1.4(0) mol ✓
Rearranging ideal gas equation to make V subject Marking point 3 can be awarded by direct substitution
nRT of values.
V = p ✓ ALLOW 150 (kPa) for 1.50 105 (conversion is
assessed in marking point 4)
Conversion of 100OC to 373K, and substituting all values into ALLOW use of 8.31 for R → 0.02892988
rearranged ideal gas equation DO NOT ALLOW use of 150 (kPa)
ALLOW 3 SF up to calculator value, correctly
1.40 × 8.314 × 373 3 rounded
V = 5 OR 0.02894380... (m )✓
1.50 10
V conversion of m3 → cm3 AND 3 SF (most appropriate)
64 3 3 (use of 8.31 for R gives same answer)
V = 0.02894 10 = 2.89 10 cm OR 28900 cm
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Note use of 373.15 → 29000
Fine for all 5 marks
Common errors:
No use of ratio x4
7240 → 4 marks
Not appropriate SF
28943.8 → 4 marks
No temp conversion
7760 → 4 marks
Use of 25.9 as n in pV=nRT
535000 → 3 marks
How to answer it
Reactions of Alcohols, Elimination & Ideal Gas Calculations
- Oxidation of alcohols: Identifying primary, secondary, and tertiary alcohols and predicting products formed under reflux with acidified dichromate (H⁺/Cr₂O₇²⁻).
- Acid-catalysed elimination (dehydration): Predicting isomeric alkene products formed from unsymmetrical alcohols.
- Stoichiometry & Ideal Gas Equation: Using pV = nRT to find gas volume, managing unit conversions (Pa, K, m³, cm³), and applying significant figure conventions.
Part (a): Oxidation of Isomeric Alcohols Under Reflux
Organic products of oxidation [3 Marks]
✅ Correct Products
| Alcohol | Classification | Product Formed |
|---|---|---|
| D (butan-2-ol) | Secondary (2°) | Butanone CH₃COCH₂CH₃ |
| E (2-methylpropan-1-ol) | Primary (1°) | 2-methylpropanoic acid (CH₃)₂CHCOOH |
| F (2-methylpropan-2-ol) | Tertiary (3°) | No reaction |
| G (butan-1-ol) | Primary (1°) | Butanoic acid CH₃CH₂CH₂COOH |
• All 4 correct = 3 marks
• 3 correct = 2 marks
• 2 correct = 1 mark
💡 Key Knowledge
- Primary alcohols heated under reflux with excess acidified dichromate oxidise fully to carboxylic acids ( -COOH ). If distilled, they form aldehydes.
- Secondary alcohols oxidise to ketones ( >C=O ). They cannot oxidise further without breaking the carbon skeleton.
- Tertiary alcohols do not have a hydrogen atom on the carbon carrying the -OH group, so they resist oxidation under these conditions.
🧠 Exam Technique: Drawing Structures
- Watch the carbonyl oxygen: Ensure double bonds to oxygen are clearly shown ( C=O ).
- Connectivity matters: Draw hydroxyl bonds as C-O-H , not C-H-O . The bond from carbon must go directly to oxygen.
- Check your carbon count: For alcohol E, students frequently lose a branch. It must be a 3-carbon carboxylic acid chain with a methyl group on carbon 2: (CH₃)₂CHCOOH .
❌ Common Errors
- Stopping at the aldehyde for E and G. The question specifies reflux, which forces oxidation completely to the carboxylic acid.
- Attempting to oxidise tertiary alcohol F into a ketone or alkene. The correct response is strictly "No reaction".
- Missing hydrogen atoms in displayed structures (penalised across the question).
Part (b): Acid-Catalysed Dehydration / Elimination
Formation of structural isomers [1 Mark]
✅ Correct Answer
Alcohol D
💡 Key Knowledge
Elimination of water (dehydration) removes the -OH group and an -H from an adjacent carbon atom:
- Alcohol D (butan-2-ol): The -OH is on C2. Hydrogen can be removed from C1 to give but-1-ene, or from C3 to give but-2-ene. These are two structural (position) isomers.
- Alcohol E: Can only eliminate towards C1, giving 2-methylpropene.
- Alcohol F: All three adjacent methyl groups are equivalent, giving only 2-methylpropene.
- Alcohol G (butan-1-ol): Hydrogen can only be lost from C2, forming but-1-ene only.
Part (c): Combustion & Ideal Gas Equation Calculation
Calculating volume of CO₂(g) produced [5 Marks]
📐 Step-by-Step Calculation
Step 1: Calculate molar mass and moles of Alcohol F
Molar mass of C₄H₁₀O = (4 × 12.0) + (10 × 1.0) + 16.0 = 74.0 g mol⁻¹
n(C₄H₁₀O) = mass / Mᵣ = 25.9 / 74.0 = 0.350 mol [Mark 1]
Step 2: Use stoichiometric ratio to find moles of CO₂
From equation: 1 mol C₄H₁₀O produces 4 mol CO₂.
n(CO₂) = 4 × 0.350 = 1.40 mol [Mark 2]
Step 3: Rearrange ideal gas equation for volume
pV = nRT ⇒ V = nRT / p [Mark 3]
Step 4: Convert units and substitute into formula
- Temperature: T = 100 + 273 = 373 K (or 373.15 K)
- Pressure: p = 1.50 × 10⁵ Pa
- Gas constant: R = 8.314 J mol⁻¹ K⁻¹
V = (1.40 × 8.314 × 373) / (1.50 × 10⁵) = 0.02894 m³ [Mark 4]
Step 5: Convert m³ to cm³ and apply significant figures
To convert from m³ to cm³ , multiply by 10⁶ :
V = 0.02894 × 10⁶ = 28940 cm³
Values in question ( 25.9 g , 100 °C , 1.50 × 10⁵ Pa ) are given to 3 significant figures:
Volume of CO₂ = 2.89 × 10⁴ cm³ (or 28900 cm³) [Mark 5]
❌ Common Calculation Traps
- Forgetting the mole ratio (×4): Gives an answer of 7240 cm³ (maximum 4 marks). Always check the balanced chemical equation!
- Temperature not converted to Kelvin: Using 100 instead of 373 K gives 7760 cm³ (maximum 4 marks).
- Incorrect volume conversion: Multiplying by 1000 (converting to dm³) instead of 10⁶ (to cm³).
- Inappropriate significant figures: Writing unrounded calculator displays like 28943.8 cm³ loses the final mark.
- Substituting mass directly into pV = nRT: Using 25.9 as n gives 535000 cm³ (maximum 3 marks).
🧠 Exam Tip: Fast Unit Conversion Check
Always write down units beside your variables before calculating:
- p must be in Pa (not kPa).
- V calculated from pV = nRT is always in m³.
- 1 m³ = 1000 dm³ = 1 000 000 cm³ .
- T must always be in K ( °C + 273 ).
Topics
Module 4: Core organic chemistry · Module 2: Foundations in chemistry · 4.2 Alcohols, haloalkanes and analysis · 2.1 Atoms and reactions
Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.