AQA A-Level Chemistry Paper 2, 2018: Question 3

16 marks · Medium difficulty · State/Explain/Numerical

Answer a series of questions on the oxidation of propan-1-ol, testing for propanoic acid, calorimetry calculation of combustion of ethanol, mechanism of dehydration of pentan-2-ol, and stereoisomerism.

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Question

Question text

03 The oxidation of propan-1-ol can form propanal and propanoic acid.

The boiling points of these compounds are shown in Table 1.

Table 1

Compound Boiling point / °C

propan-1-ol 97

propanal 49

propanoic acid 141

In a preparation of propanal, propan-1-ol is added dropwise to the oxidising agent and

the aldehyde is separated from the reaction mixture by distillation.

03.1 Explain, with reference to intermolecular forces, why distillation allows propanal to be

separated from the other organic compounds in this reaction mixture.

[3 marks]

03.2 Give two ways of maximising the yield of propanal obtained by distillation of the

reaction mixture.

[2 marks]

03.3 Describe how you would carry out a simple test-tube reaction to confirm that the

sample of propanal obtained by distillation does not contain any propanoic acid.

[2 marks]

03.4 A student carried out an experiment to determine the enthalpy of combustion of

ethanol.

Combustion of 457 mg of ethanol increased the temperature of 150 g of water from

25.1 °C to 40.2 °C

Calculate a value, in kJ mol–1, for the enthalpy of combustion of ethanol in this

experiment.

Give your answer to the appropriate number of significant figures.

(The specific heat capacity of water is 4.18 J K–1 g–1)

[3 marks]

Enthalpy of combustion9 kJ mol–1

03.5 A mixture of isomeric alkenes is produced when pentan-2-ol is dehydrated in the

presence of hot concentrated sulfuric acid. Pent-1-ene is one of the isomers

produced.

Name and outline a mechanism for the reaction producing pent-1-ene.

[4 marks]

Name of mechanism

Mechanism

03.6 A pair of stereoisomers is also formed in the reaction in Question 03.5.

Name the less polar stereoisomer formed.

Explain how this type of stereoisomerism arises.

[2 marks]

Name

Explanation

Mark scheme

Show the mark scheme

Question Answers Mark Additional Comments/Guidance

Aldehyde/propanal has dipole-dipole forces (between molecules) M1

If any ‘covalent bonds broken’ CE=0 for clip.

Ignore Van der Waal forces

Alcohol/propan-1-ol AND Carboxylic acid/ propanoic acid have

hydrogen bonding (between molecules). M2 Ignore reference to energy

03.1

The forces between the molecules in aldehyde are weaker (than M3 M3 only awarded following correct M1 OR M2

those in alcohol and acid so it will evaporate first.)

Allow converse for M3

M1 Allow temperature in range 49-96 inclusive

Keep the temperature of the reaction mixture below the boiling point

of propan-1-ol/below 97°C

Allow description of cooling the vessel

03.2

Cool the distillate / collecting vessel

M2 Ignore reference to oxidising agents

Penalise lid / sealed container

Add named carbonate/hydrogencarbonate OR magnesium to a M1 Incorrect chemical CE=0

Allow formula (mark on for incorrect formula)

sample of the distillate.

Allow blue litmus or correct named indicator

03.3 Blue litmus turns red confirms acid present or

Effervescence/fizz/bubbles would confirm presence of acid or

converse

Allow gas/CO2 produced which turns lime water

M2 cloudy OR gas/H2 produced which burns with a

–squeaky pop – –

(Temperature difference = 15.1 oC)

If T wrong – AE mark on otherwise can only award M2

q = 150 × 4.18 × 15.1 or 9467.7 J or 9.4677kJ M1 If use 457 in M1, can only score M2

If use 457 in M2 can score 2 for - 0.953 kJ mol-1

amount ethanol burned = 0.457/46.0 = 9.93 × 10-3 mol M2

Heat change per mole = (M1/1000)/M2 = 952.99 kJ mol-1

03.4

ΔH = – 953 kJ mol–1 must be 3sfs and must be negative M3 BEWARE if they miss conversion to kJ and also miss

(allow range -953 to -954) conversion to g, they get answer = - 953 which scores 1

+953 can score M1 and M2

Allow -950 or -960 for rounding to 2sf

Elimination M1 Penalise base elimination

Mechanism : Either (E1)

3 M2 for protonation of alcohol, i.e. lp plus arrow to H+

H H

CH3CH2CH2 C CH3 CH CH CH C CH or to H of H O in H2SO4 and from H-O bond to O

32 2 3

OH OH

H+

M3 for protonated alcohol plus arrow showing loss of

03.5 water

H H

CH3CH2CH2 C CH3 CH3CH2CH2 C CH3

OH2 M4 for arrow showing loss of H+

H H From correct carbocation (E1)

CH CH CH C CH CH3CH2CH2 C CH2

32 2 2

H wrong alcohol used / alkene formed loses M4

– – –

OR (E2) +

M2 for protonation of alcohol, i.e. lp plus arrow to H

H H

or to H of H O in H2SO4 and from H-O bond to O

CH3CH2CH2 C CH3 CH CH CH C CH

32 2 3

OH OH

H+ M3 for protonated alcohol plus arrow showing loss of

water

H H H

CH3CH2CH2 C CH CH3CH2CH2 C CH2 +

2 M4 for arrow showing simultaneous loss of H

OH2

wrong alcohol used / alkene formed loses M4

E-pent-2-ene M1 Allow trans

03.6 C=C bond cannot rotate and

Allow (two) different groups on each/either side of

Each carbon in the double bond has (2) different groups attached. M2 the double bond.

Total 16

How to answer it

Organic Synthesis, Calorimetry & Elimination Mechanisms

WHAT THIS QUESTION TESTS

This question comprehensively assesses core AS and A2 organic and physical chemistry topics:

Intermolecular Forces & Distillation: Distinguishing dipole-dipole forces from hydrogen bonding to explain relative boiling points and purification.
Practical Organic Techniques: Practical parameters used to maximise aldehyde yield and standard chemical tests for distinguishing functional groups.
Calorimetry Calculations: Using q = mcΔT , converting mass units (mg to g), and calculating molar enthalpy of combustion with correct signs and significant figures.
Organic Mechanisms: Step-by-step acid-catalysed dehydration (elimination) of a secondary alcohol to form a specific terminal alkene.
Stereoisomerism: Geometric (E/Z) isomerism, polarity differences, and structural criteria for restricted rotation.

QUESTION 03.1 • 3 MARKS

Intermolecular Forces and Distillation

Explaining separation of propanal from propan-1-ol and propanoic acid

✅ Model Answer

Mark 1: Propanal has dipole–dipole forces (or permanent dipole–dipole forces) between molecules.
Mark 2: Propan-1-ol and propanoic acid both have hydrogen bonding between molecules.
Mark 3: The intermolecular forces in propanal are weaker than the hydrogen bonding in propan-1-ol and propanoic acid, so propanal has the lowest boiling point and distils off first.

💡 Key Knowledge

Aldehydes: Contain a polar carbonyl group (C=O) but no H atom directly attached to oxygen, so they cannot form intermolecular hydrogen bonds with themselves.
Alcohols & Carboxylic Acids: Contain O–H groups, producing strong intermolecular hydrogen bonding.
Boiling point order: propanal (49 °C) < propan-1-ol (97 °C) < propanoic acid (141 °C) .

🧠 Exam Technique

Always state the specific type of force in both the substance being removed and the remaining substances. Never just say "weaker forces" without explicitly naming them.

❌ Common Errors & Examiner Warnings

Contradiction Error (CE = 0): Stating that covalent bonds break when boiling results in 0 marks immediately for the entire question part!
Forgetting to mention propanoic acid alongside propan-1-ol when identifying hydrogen bonding.
Vaguely writing "van der Waals forces" without specifying dipole-dipole attractions.

Mark breakdown: M1: Dipole–dipole for propanal. M2: Hydrogen bonding for alcohol AND carboxylic acid. M3: Comparison showing aldehyde forces are weaker so it evaporates first (dependent on M1 or M2).

QUESTION 03.2 • 2 MARKS

Maximising Distillation Yield

Practical steps to obtain maximum propanal and minimise loss

✅ Model Answer

Method 1: Keep the reaction temperature below the boiling point of propan-1-ol (i.e. below 97 °C / maintain temperature between 49 °C and 96 °C).
Method 2: Cool the collection vessel / receiver flask in an ice bath.

🧠 Exam Technique

Consider the process from start to finish:

In the reaction flask: Boil off propanal (b.p. 49 °C) immediately as it forms, but keep below 97 °C so unreacted alcohol doesn't boil over.
In the receiving flask: Propanal is volatile (b.p. 49 °C); cooling the distillate in an ice bath prevents evaporation loss into the lab.

❌ Common Errors

Sealing the apparatus: Suggesting a stopper or lid on the collection flask is penalised because heating a closed system creates a dangerous pressure build-up.
Mentioning reagents (e.g. "use excess dichromate" or "add acid") — the question specifically asks for ways to maximise yield by distillation.

Mark breakdown: M1: Control reaction temperature below 97 °C (range 49–96 °C). M2: Cool the distillate / collection vessel (e.g., using an ice bath).

QUESTION 03.3 • 2 MARKS

Chemical Test for Carboxylic Acid Impurity

Distinguishing propanoic acid from propanal

✅ Model Answer

Reagent: Add named metal carbonate or hydrogencarbonate (e.g. sodium hydrogencarbonate / NaHCO₃ / sodium carbonate / Na₂CO₃) OR magnesium ribbon (Mg).

Observation: No effervescence / no bubbles observed (confirming acid is absent).

Alternative valid approach: Add blue litmus paper → paper remains blue / does not turn red.

❌ Common Errors

Giving the positive test observation only: The question asks to confirm that it does not contain acid. State that effervescence/fizzing is not seen.
Using Tollens' reagent or Fehling's solution: These test for the aldehyde functional group, NOT the presence/absence of a carboxylic acid impurity!
Incorrect chemical formulae (e.g. NaCO₃ instead of Na₂CO₃) results in Chemical Error (CE = 0).

Mark breakdown: M1: Suitable named reagent (carbonate, hydrogencarbonate, Mg, or blue litmus). M2: Correct expected negative observation (no fizzing / remains blue).

QUESTION 03.4 • 3 MARKS

Calorimetry: Enthalpy of Combustion

Calculating standard enthalpy change of combustion from experimental data

📐 Step-by-Step Calculation

Step 1: Calculate heat energy released (q)
ΔT = 40.2 − 25.1 = 15.1 °C
m(water) = 150 g
q = m × c × ΔT = 150 × 4.18 × 15.1 = 9467.7 J = 9.4677 kJ

[Awarded M1]

Step 2: Calculate amount (moles) of ethanol burned
Mᵣ(C₂H₅OH) = (2 × 12.0) + (6 × 1.0) + 16.0 = 46.0 g mol⁻¹
Convert mg to g: 457 mg = 0.457 g
n = mass / Mᵣ = 0.457 / 46.0 = 9.9348 × 10⁻³ mol

[Awarded M2]

Step 3: Calculate ΔH and apply correct sign and significant figures
ΔH = − q / n = − 9.4677 kJ / (9.9348 × 10⁻³ mol) = −952.98 kJ mol⁻¹
Round to 3 significant figures (matching data precision: 457 mg, 150 g, 15.1 °C):
Enthalpy of combustion = −953 kJ mol⁻¹ (acceptable range: −953 to −954)

[Awarded M3]

❌ Common Calculation Pitfalls

Mass mix-up: Using 457 mg in q = mcΔT instead of the mass of water (150 g). The heated substance is water!
Unit Trap (mg to g): Forgetting to divide 457 mg by 1000 gives 457 g, leading to an answer off by a factor of 1000.
Missing Negative Sign: Combustion is an exothermic process. Omitting the minus sign loses M3!
Significant Figures: Failing to round to 3 significant figures loses the final mark.

Mark breakdown: M1: Correct calculation of q (9467.7 J or 9.468 kJ). M2: Correct calculation of moles of ethanol (9.93 × 10⁻³ mol). M3: Answer −953 kJ mol⁻¹ with negative sign and to 3 s.f.

QUESTION 03.5 • 4 MARKS

Dehydration Mechanism: Pentan-2-ol to Pent-1-ene

Acid-catalysed elimination of a secondary alcohol

✅ Model Answer

Name of mechanism: Elimination (or acid-catalysed elimination / dehydration).

Mechanism Description (E1 pathway):

Step 1: Protonation
Draw pentan-2-ol: CH₃CH₂CH₂CH(OH)CH₃ .
Curly arrow from a lone pair on the oxygen atom of the –OH group to an H⁺ ion (or H of H–OSO₃H , with arrow from H–O bond to O).
Step 2: Loss of Water Molecule
Structure has a protonated oxygen: –O⁺H₂ .
Curly arrow from the C–O⁺ bond to the O⁺ atom, releasing H₂O and forming the secondary carbocation: CH₃CH₂CH₂C⁺HCH₃ .
Step 3: Deprotonation to form Pent-1-ene
Show a C–H bond on the terminal carbon: CH₃CH₂CH₂C⁺H–CH₂–H .
Curly arrow from the terminal C–H bond into the adjacent C–C bond to form the C=C double bond, ejecting H⁺ and forming CH₃CH₂CH₂CH=CH₂ .

❌ Critical Examiner Watch-outs

Wrong Mechanism Name: Writing "Base elimination" or "Electrophilic elimination" is penalised. Just write Elimination.
Wrong Proton Lost: The question specifically demands pent-1-ene! If you remove the proton from carbon-3, you form pent-2-ene instead and lose the final mechanism mark.
Arrow Origins: The arrow for deprotonation must come cleanly from the C–H bond, not from the hydrogen atom itself.

Mark breakdown: M1: "Elimination". M2: Lone pair on OH attacks H⁺. M3: C–O⁺ bond breaks to release H₂O. M4: Arrow from C1–H bond forming C=C double bond in pent-1-ene.

QUESTION 03.6 • 2 MARKS

Stereoisomerism in Pent-2-ene

Identifying the less polar isomer and explaining origin of E/Z isomerism

✅ Model Answer

Name of less polar stereoisomer:
E-pent-2-ene (or trans-pent-2-ene).

Explanation (both points needed for M2):

There is restricted rotation around the C=C double bond.
Each carbon atom in the double bond is attached to two different groups/atoms (on C2: –H and –CH₃; on C3: –H and –CH₂CH₃).

💡 Why is E-pent-2-ene less polar?

In the E-isomer, the two alkyl groups (propyl-like ethyl and methyl) are on opposite sides of the double bond.
Alkyl groups are electron-releasing (+I inductive effect). When positioned opposite one another, their small bond dipoles partially oppose and cancel each other out, giving a lower net molecular dipole.

❌ Common Errors

Stating "no rotation around the molecule" — you must specify the C=C double bond.
Stating "four different groups attached to the double bond" — that describes optical isomerism (chiral centres). For E/Z isomerism, each C of the C=C must have two different groups attached to it.

Mark breakdown: M1: E-pent-2-ene (or trans). M2: Restricted rotation about C=C AND two different groups attached to each carbon of the double bond.

Topics

Physical Chemistry · Organic Chemistry · Required Practicals · 3.1.3 Bonding · 3.1.4 Energetics · 3.3.1 Introduction to Organic Chemistry · 3.3.5 Alcohols · 3.3.6 Organic Analysis · Required Practical 2: Measurement of an enthalpy change · Required Practical 5: Distillation of a product from a reaction · Required Practical 6: Tests for alcohol, aldehyde, alkene and carboxylic acid

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.