Edexcel A-Level Chemistry Paper 2, June 2022: Question 6

16 marks · Medium difficulty · Short Open Response

Identify functional groups in carbonyl compounds, calculate the percentage yield of propanal from the oxidation of propan-1-ol, and explain boiling point and solubility differences using intermolecular forces.

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Question

Question text

6 Aldehydes and ketones are carbonyl compounds.

(a) Which of these compounds does not contain a ketone functional group?

(1)

A B

C D

O O

OH OH

(b) Which of these compounds has both an aldehyde functional group and a ketone

functional group?

(1)

A B

O O

C D

O O

(i) A student assembled the apparatus shown for this oxidation.

*P67094RA0924*water out

Liebig

condenser

water in

pear-shaped

flask

reaction mixture

anti-bumping

granules

heat

Explain why the use of this apparatus would give a very low yield of propanal.

(2)

(ii) The oxidising agent is acidified Na2Cr2O7 .

State the oxidation number of chromium in Na2Cr2O7 .

(1)

(iii) Complete the ionic half-equation for the oxidation of propan-1-ol.

(1)

CH CH CH OH → CH CH CHO + H+ + e–

32 2 3 2 …

(iv) State how the use of anti-bumping granules gives smoother boiling.

(1)

… 10

*P67094RA01024*

(v) Another student used the correct apparatus for this oxidation.

1.50g of propan-1-ol produced 0.609g of propanal.

Calculate the percentage yield of propanal by mass.

(3)

(d) The table contains data on propanone and ethanoic acid.

Substance Molar mass / g mol–1 Boiling temperature / °C Solubility in water

Propanone 58 56 completely miscible

Ethanoic acid 60 118 completely miscible

(i) Explain, by reference to the data and any intermolecular forces involved, the

difference in the boiling temperatures.

(4)

(ii) Explain, with the aid of a diagram, why propanone is completely miscible

with water. 11

*P67094RA01124* (2)

… *P67094RA01224*

(Total for Question 6 = 16 marks)

Mark scheme

Show the mark scheme

Question

Answer Mark

Number

6(a) (1)

The only correct answer is D ( )

A is not correct because there is a ketone group present

B is not correct because there is a ketone group present

C is not correct because there is a ketone group present

Question

Answer Mark

Number

6(b) (1)

The only correct answer is C ( )

A is not correct because there are two ketone groups but no aldehyde group

B is not correct because there are two ketone groups but no aldehyde group

D is not correct because there are two aldehyde groups but no ketone group

Question

Answer Additional Guidance Mark

Number

6(c)(i) An explanation that makes reference to Allow aldehyde for propanal (2)

• propanal is condensed back (to the pear-shaped Allow ‘apparatus is reflux’

flask) (1) Allow propanal is not being removed

/distilled off (from the oxidising agent)

• so propanal is (further) oxidised (to propanoic acid) Ignore just ‘reacts further’

propanal is more readily oxidised than propan-1-ol Do not award reference to propanal being

(1) completely oxidised

Question

Answer Additional Guidance Mark

Number

6(c)(ii) (1)

• (+)VI Allow (+) six / (+)6 / six (+) / 6(+)

Question

Answer Additional Guidance Mark

Number

6(c)(iii) Example of equation (1)

• balanced equation CH CH CH OH → CH CH CHO + 2H+ + 2e−

32 2 3 2

Question

Answer Additional Guidance Mark

Number

6(c)(iv) (1)

• provides a surface for bubbles to form / Allow distribution of heat more evenly / to

enables smaller bubbles to form / prevent superheating

provides nucleation sites for bubbles

or Ignore mixing / to stop bumping / spitting /

to prevent large bubbles forming explosion / liquid splashing out /

vigorous reaction / loss of reactants

Do not award reference to large gas

molecules

Question

Answer Additional Guidance Mark

Number

6(c)(v) Example of calculation (3)

• (M1) evaluation of number of moles of propan-1-ol (1) n(propan-1-ol) = (1.50 ÷ 60) = 0.025 (mol)

Method one using masses for percentage calculation n(propan-1-ol) = n(propanal)

• (M2) evaluation of maximum mass of propanal (1) max m(propanal) = (0.025 x 58)

= 1.45 (g)

• (M3) percentage yield (1) %Yield = ((0.609 ÷ 1.45) x 100) = 42 %

Method two using moles for percentage calculation

• (M2) evaluation of actual moles of propanal (1) n(propanal) = (0.609 ÷ 58) = 0.0105 (mol)

• (M3) percentage yield (1) %Yield =((0.0105 ÷ 0.025) x 100) = 42 %

Allow TE at each stage

Ignore SF except 1SF

Penalise incorrect Mr values once only

Correct answer without working scores (3)

Question

Answer Additional Guidance Mark

Number

6(d)(i) An explanation that makes reference to the following points: (4)

• similar molar masses so the number of electrons is Allow van der Waals’ forces /

similar/same resulting in similar London forces (1) dispersion forces / instantaneous

dipole-induced dipole forces

Ignore reference to ethanoic acid

having greater London forces

• propanone (and ethanoic acid) form Ignore reference to hydrogen bonding

permanent dipole(-dipole) forces (1) to water by propanone

Penalise abbreviation pd-d once only

• (only) ethanoic acid forms (intermolecular) hydrogen Ignore references to ethanoic acid

bonding (1) dimerization

• which is stronger so requires more energy to break Reference to energy must be linked to

(giving a higher boiling temperature) (1) the breaking of hydrogen bonds

Question

Answer Additional Guidance Mark

Number

6(d)(ii) An explanation that makes reference to the following (2)

points:

• forms hydrogen bonds with water (1) Allow H bonds for hydrogen bonds

• diagram of hydrogen bond (1)

Ignore bond angle and missing dipoles

and missing lone pair

Do not award incorrect dipoles

Do not award incorrect propanone

and/or water structure

Do not award if second hydrogen bond

drawn to the hydrogen of the CH3

(Total Question 6 = 16 marks)

Allow annotated equations to score these marks in both (i) and (ii)

Allow any unambiguous formulae for the organic molecules in both (i) and (ii) such as C2H5CN for CH3CH2CN

How to answer it

Aldehydes, Ketones, and Carboxylic Acids

What this question tests

This question assesses functional group identification in organic molecules, practical organic chemistry techniques (specifically reflux vs. distillation), redox half-equations and oxidation numbers, quantitative percentage yield calculations, and the analysis of intermolecular forces and hydrogen bonding affecting boiling points and solubility.

Question 6 (a)

Identifying Functional Groups (Ketones)

✅ Correct Answer

Compound D contains an ester linkage and a hydroxyl group, but does not contain a C=O double bond flanked by two carbon chains (a ketone).

❌ Common Errors

Selecting A, B, or C. All three of these molecules contain a distinct carbonyl peak forming part of a ring system that qualifies as a ketone group.

Question 6 (b)

Dual Functional Groups

✅ Correct Answer

Compound C contains both a terminal aldehyde group ( -CHO at the left end) and a ketone carbonyl group (in the body of the carbon chain).

💡 Key Knowledge

Aldehydes always sit at the end of a carbon chain (C=O with an attached H), whereas ketones are situated within the carbon skeleton between two alkyl groups.

Question 6 (c)(i)

Oxidation Practical Setup (Reflux vs Distillation)

✅ Correct Answer

Two distinct marks are available:

The apparatus shown is set up for reflux (vapours condense and drip back into the flask).
This causes the propanal to be further oxidised into propanoic acid, or prevents propanal from being distilled off as it forms.

🧠 Exam Technique

To prepare an aldehyde from a primary alcohol, you must use distillation with immediate boil-off. Using reflux forces complete oxidation to the carboxylic acid.

Question 6 (c)(ii)

Oxidation Number of Chromium

✅ Correct Answer

+VI (or +6 )

💡 Key Knowledge

In dichromate(VI) ions ( Cr₂O₇²⁻ ), oxygen has an oxidation state of -2. Therefore, 2Cr + 7(-2) = -2, meaning 2Cr = +12, so each Cr is +6.

Question 6 (c)(iii)

Ionic Half-Equation

✅ Correct Answer

CH₃CH₂CH₂OH → CH₃CH₂CHO + 2H⁺ + 2e⁻

🧠 Exam Technique

Balance oxygen atoms using H₂O if needed, balance hydrogens using H⁺ ions, and balance overall charge by adding electrons ( e⁻ ) to the product side.

Question 6 (c)(iv)

Role of Anti-Bumping Granules

✅ Correct Answer

They provide a rough surface for nucleation/bubbles to form, preventing violent, large bubble formation (bumping).

❌ Common Errors

Do not state that they "speed up the reaction" or "absorb gas molecules"—they are purely physical catalysts for smooth boiling.

Question 6 (c)(v)

Percentage Yield Calculation

📐 Step-by-Step Calculation

Molar mass of propan-1-ol (C₃H₈O): (3 × 12.0) + (8 × 1.0) + 16.0 = 60.0 g mol⁻¹
Moles of propan-1-ol used: 1.50 g / 60.0 g mol⁻¹ = 0.025 mol
Theoretical maximum moles/mass of propanal: Since the molar ratio is 1:1, max moles of propanal = 0.025 mol.
Molar mass of propanal (C₃H₆O) = 58.0 g mol⁻¹.
Max mass = 0.025 × 58.0 = 1.45 g.
Percentage Yield:
(Actual mass / Theoretical mass) × 100
= (0.609 g / 1.45 g) × 100 = 42%

❌ Calculation Traps

Be careful not to mix up the molar masses of propan-1-ol (60) and propanal (58). Always evaluate maximum theoretical yield using stoichiometry before finding the percentage.

Question 6 (d)(i)

Boiling Temperature Comparison

✅ Correct Answer

Propanone and ethanoic acid have similar molar masses, meaning they have similar London (dispersion) forces.
Both molecules also exhibit permanent dipole-dipole forces due to the polar C=O bond.
However, ethanoic acid can form intermolecular hydrogen bonds, whereas propanone cannot form hydrogen bonds with itself.
Therefore, more energy is required to break the stronger hydrogen bonds in ethanoic acid, resulting in a higher boiling temperature.

🧠 Exam Technique

For 4 marks, structure your answer systematically: (1) state similarity in induced dipole-dipole forces, (2) note permanent dipoles, (3) identify the crucial difference (hydrogen bonding in the acid), and (4) explicitly link this to energy required.

Question 6 (d)(ii)

Miscibility and Hydrogen Bonding with Water

✅ Correct Answer

Propanone is completely miscible because it can form hydrogen bonds with water.
Diagram description: Show a hydrogen bond clearly indicated by a dotted line ( ... ) between the lone pair on the oxygen of propanone (C=O) and a hydrogen atom attached to oxygen in a water molecule ( O-H...O=C ). Partial charges ( δ+ and δ- ) should be visible.

❌ Common Errors

Do not draw hydrogen bonds originating from hydrogen atoms bonded to carbon chains (e.g., from the methyl groups of propanone)—hydrogen bonds must involve the electronegative oxygen atom of the carbonyl group and water's O-H.

Topics

Organic Chemistry · Physical Chemistry · Core Practicals · Core Practical 5: Investigate the oxidation of ethanol · Topic 6: Organic Chemistry I · Topic 17: Organic Chemistry II · Topic 2: Bonding and Structure · Topic 5: Formulae, Equations and Amounts of Substance · Topic 3: Redox I

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.