AQA A-Level Chemistry Paper 2, 2017: Question 10

21 marks · Medium difficulty · State/Explain/Numerical

Deduce structures, spectra, combustion amounts, and polymer properties for several C6H10O2 isomers, including naming, ideal gas calculations, NMR, and polymer biodegradability.

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Question text

10 This question is about six isomers of C6H10O2

10.1 Give the full IUPAC name of isomer P.

[1 mark]

10.2 A sample of P was mixed with an excess of oxygen and the mixture ignited.

After cooling to the original temperature, the total volume of gas remaining was

335 cm3

When this gas mixture was passed through aqueous sodium hydroxide, the

carbon dioxide reacted and the volume of gas decreased to 155 cm3

Both gas volumes were measured at 25 °C and 105 kPa

Write an equation for the combustion of P in an excess of oxygen and calculate

the mass, in mg, of P used.

The gas constant R = 8.31 J K−1 mol−1

[5 marks]

21 ou

Mass of P used mg

10.3 Isomer Q (C6H10O2) is a cyclic compound. The infrared spectrum of Q is

shown in Figure 4 and the 13C NMR spectrum of Q is shown in Figure 5.

Figure 4

Figure 5

Use these spectra and Tables A and C in the Data Booklet to deduce the

structure of Q.

In your answer, state one piece of evidence you have used from each

spectrum.

[3 marks]

Structure of Q.

Evidence from Figure 4

Evidence from Figure 5

22 ou

10.4 Isomers R and S are shown.

R S

Although the 13C spectra of R and S both show the same number of peaks, the

spectra can be used to distinguish between the isomers.

Justify this statement using Table C from the Data Booklet.

Give the number of peaks for each isomer.

[3 marks]

Justification

Number of peaks 23 ou

10.5 Although the 1H spectra of R and S both show the same number of peaks, the

spectra can be used to distinguish between the isomers.

Justify this statement using the splitting patterns of the peaks.

Give the number of peaks for each isomer.

[3 marks]

Justification

Number of peaks 24 ou

10.6 The action of heat on 5-hydroxyhexanoic acid can lead to two different

products.

On gentle heating, 5-hydroxyhexanoic acid loses water to form a cyclic

compound, T (C6H10O2).

Under different conditions, 5-hydroxyhexanoic acid forms a polyester.

Draw the structure of T.

Draw the repeating unit of the polyester and name the type of polymerisation.

[3 marks]

Structure of T

Repeating unit of polyester

*23* Type of polymerisation 25 ou

10.7 Isomer U is shown.

The polymer formed by U and the polymer formed by 5-hydroxyhexanoic acid

in Question 10.6 both contain ester groups that can be hydrolysed.

Draw the repeating unit of the polymer formed by U.

Justify the statement that, although both polymer structures contain ester

groups, the polymer formed by U is not biodegradable.

[3 marks]

Repeating unit of polymer formed by U.

Justification

Mark scheme

Show the mark scheme

Question Answers Mark Additional Comments/Guidance

10.1 Z-2-methylpent-2-en (-1-) oic acid 1 Ignore missing hyphens or extra commas, spaces, hyphens

10.2 C6H10O2 + 7½ O2 6CO2 + 5H2O M1 Allow multiple

Volume of CO formed = 180 cm3 M2 If incorrect volume:155 gives 125mg / 335 gives 270mg

could score M1,M3,M4 – max 3

If incorrect volume from AE then penalise M2 and mark on

(Final answer is 0.806 x their volume)

Mol carbon dioxide = pV/RT = 105000 × (180 × 10-6) M3 If unit error in p, V or T lose M3 and M5

8.31 × 298 If incorrect rearrangement lose M3 and M5

= 7.632 × 10-3 If both errors seen then no further marks

Mol P, C H O used = 7.632 × 10-3 / 6 = 1.272 × 10-3 M4 M3 divided by 6 If wrong no further marks

6 10 2 – – –

Mass P used = 1.272 × 10-3 × 114(.0) g

M5 Mark for answer (allow ans to 2 sf)

= 145 mg

Check chemical equation before awarding final mark

H3C

COOH Mark independently

COOH Apply the list principle

OR H3C

Fig 4: IR OH (acid) peak (2500-3000cm-1) present M2 Ignore C=O signal at 1750 cm-1

10.3

Fig 5:13C NMR 4 peaks so 4 (non-equivalent) environments

Allow correct Fig 4 answers in Fig 5 and converse 28 of 32

Or Peak at 160-185 (show C=O) in (esters or) acids

Or Peak at 40-50 (show R-CO-CH) presence of carbonyl

Both M2 & M3 can be awarded on the spectra

R has 4 C next to C=O S has 2 C next to C=O M1 M1 for structural point

in range = 20-50 M2 M2 for resulting peak in spectra

R has two peaks and S only one peak in this range

Or R has more peaks (allowed if no numbers given)

10.4

S has a -C(H2)-C(H3) R does not M1

– – –

S has one peak in range = 5-40 R does not M2

/ lowest peak for S is lower than lowest for R

(Both have) three peaks

R Both singlets M1

M2 29 of 32

S has triplet and a quartet

R CH3/peak at 2.1-2.6 is a singlet M1

S CH3/peak at 0.7-1.2 is a triplet M2

10.5

R CH2/peak at 2.1-2.6 is a singlet M1

S CH2/peak at 2.1-2.6 is a quartet

(Both have) two peaks M3

H3C O O

CH3 O CH3 O

O C (CH2)3 C C (CH2)3 C O

10.6 H OR H

Must have trailing bonds

Ignore brackets and n

O CH3

C O C (CH2)3 – – –

OR H

condensation 1 Ignore esterification

COOCH2CH3 M1 Must have trailing bonds 30 of 32

Ignore brackets and n

CH2 C

CH3 M3 dependent on correct or close M2

Strong / non-polar C-C bonds (in the chain) M2

cannot be attacked by nucleophiles/acids/cannot be

hydrolysed. M3

10.7 OR

Only polar ester group M2

Can be attacked by nucleophiles/acids/can be hydrolysed M3

Allow 1 mark for in (polar) ester link in side chain/not in main

chain therefore polymer chain not broken

Total 21

How to answer it

Isomers of C₆H₁₀O₂: Structure, Spectroscopy & Polymers

AQA A-Level Chemistry • Advanced Organic Synthesis, Spectroscopy & Analysis (21 Marks Total)

What this question tests

This comprehensive question integrates multi-topic organic chemistry core themes across 6 isomers:

E/Z & IUPAC nomenclature: Applying Cahn-Ingold-Prelog (CIP) priority rules to substituted alkenoic acids.
Quantitative stoichiometry & ideal gas calculations: Using pV = nRT with combustion reactions and gas absorption by NaOH.
Structure elucidation: Interpreting infrared (IR) and ¹³C NMR spectra to deduce cyclic carboxylic acids.
Comparative spectroscopy: Distinguishing symmetrical diketones using ¹³C chemical shift environments and ¹H spin-spin splitting patterns ( n + 1 rule).
Polymerisation & cyclic ester formation: Intramolecular esterification (lactones) vs. condensation polyesters, addition polymers, and polymer biodegradability based on chain stability.

Part 10.1 • 1 Mark

IUPAC Nomenclature of Alkene Isomer P

✅ Correct Answer

(Z)-2-methylpent-2-enoic acid

Also accepted: Z-2-methylpent-2-en-1-oic acid.

💡 Key Knowledge

Longest Chain: 5 carbons containing both the C=C double bond and the principal functional group (-COOH) = pent-2-enoic acid.
CIP Priorities:
- At C2: -COOH (priority 1) > -CH₃ (priority 2)
- At C3: -CH₂CH₃ (priority 1) > -H (priority 2)
Since both high-priority groups are on the same side (top), it is assigned the Z stereodescriptor.

❌ Common Errors

Naming as E by confusing stereocentre priorities or confusing the ethyl group with methyl.
Forgetting to number the double bond position ( -2-en- ).
Numbering from the wrong end: the carbonyl carbon of -COOH must always be C1.

Mark scheme: 1 mark for the correct full name including 'Z'. Hyphens/commas/spaces are condoned if ordering and connectivity are unambiguous.

Part 10.2 • 5 Marks

Combustion & Ideal Gas Calculation

📐 Step-by-Step Calculation

Step 1: Write the balanced combustion equation [Mark 1]
C₆H₁₀O₂ + 7.5 O₂ → 6 CO₂ + 5 H₂O (or 2 C₆H₁₀O₂ + 15 O₂ → 12 CO₂ + 10 H₂O)

Step 2: Determine the volume of CO₂ gas produced [Mark 2]
NaOH absorbs acidic CO₂ gas. The remaining volume is unreacted excess O₂.
V(CO₂) = 335 cm³ − 155 cm³ = 180 cm³

Step 3: Convert units and calculate moles of CO₂ via pV = nRT [Mark 3]
• p = 105 kPa = 105,000 Pa (N m⁻²)
• V = 180 cm³ = 180 × 10⁻⁶ m³
• T = 25 °C = 25 + 273 = 298 K
• R = 8.31 J K⁻¹ mol⁻¹
n(CO₂) = (p × V) / (R × T) = (105,000 × 180 × 10⁻⁶) / (8.31 × 298) = 7.632 × 10⁻³ mol

Step 4: Use reacting mole ratio to find moles of P [Mark 4]
Ratio of P : CO₂ is 1 : 6
n(P) = (7.632 × 10⁻³) / 6 = 1.272 × 10⁻³ mol

Step 5: Calculate mass of P in mg [Mark 5]
Mᵣ(C₆H₁₀O₂) = (6 × 12.0) + (10 × 1.0) + (2 × 16.0) = 114.0 g mol⁻¹
Mass of P = 1.272 × 10⁻³ mol × 114.0 g mol⁻¹ = 0.145 g
Mass in mg = 0.145 × 1000 = 145 mg (allow 150 mg for 2 sig figs)

❌ Common Traps

Unit conversion errors: cm³ must be multiplied by 10⁻⁶ to convert to m³. kPa must be multiplied by 10³ to convert to Pa.
Volume mix-up: Using 155 cm³ or 335 cm³ instead of finding the difference (180 cm³).
Final mass unit: Forgetting to multiply grams by 1000 to express the final answer in mg.

Mark allocation: M1 (Equation), M2 (V = 180 cm³), M3 (Moles of CO₂ via pV=nRT), M4 (M3 ÷ 6), M5 (Mass = 145 mg).

Part 10.3 • 3 Marks

Structure Deduction from IR & ¹³C NMR (Isomer Q)

✅ Deductions & Structure

Structure of Q: Cyclopentanecarboxylic acid

Structure Drawing: A 5-membered cyclopentane carbon ring with a single -COOH substituent group attached to one of the ring carbons.

Figure 4 (IR) Evidence: Very broad absorption band at 2500–3000 cm⁻¹ confirming an O–H (carboxylic acid) group.

Figure 5 (¹³C NMR) Evidence: Exactly 4 peaks indicating 4 non-equivalent carbon environments (due to the plane of symmetry in the ring).

🧠 Exam Technique: Symmetry Analysis

A formula of C₆H₁₀O₂ has 2 degrees of unsaturation (Rings + π-bonds = 6 + 1 − 10/2 = 2).
The IR broad band (2500–3000 cm⁻¹) and sharp band (~1710 cm⁻¹) prove a -COOH group (1 C=O double bond).
Since Q is cyclic, the remaining 5 carbons form a ring (1 ring).
In cyclopentanecarboxylic acid, symmetry means:
• C1 (attached to -COOH): 1 type
• C2 & C5: identical (1 type)
• C3 & C4: identical (1 type)
• -COOH carbon: 1 type (δ ~180 ppm)
Total = 4 peaks!

Mark scheme: 1 mark for correct drawn structure; 1 mark for IR citation (O-H acid at 2500-3000 cm⁻¹); 1 mark for NMR citation (4 peaks = 4 environments OR peak at δ 160-185 ppm for carboxylic acid C=O).

Part 10.4 • 3 Marks

Distinguishing Isomers R and S using ¹³C NMR

💡 Structures Overview

R: Hexane-2,5-dione: CH₃-CO-CH₂-CH₂-CO-CH₃

S: Hexane-3,4-dione: CH₃-CH₂-CO-CO-CH₂-CH₃

Both molecules possess internal symmetry and will therefore display the same total number of peaks: 3 peaks.

✅ Justification from Data Booklet (Table C)

Structural difference: In R, all 4 non-carbonyl carbons are bonded directly next to a C=O group ( -CH₂-C=O and CH₃-C=O ). In S, only 2 carbons are next to a C=O group ( -CH₂-C=O ); the two CH₃ carbons are attached to CH₂ .
Chemical shift difference:
• Carbons adjacent to C=O ( R-CH₂-C=O or R-CO-CH₃ ) appear in the range δ = 20–50 ppm.
• R has two peaks in the δ = 20–50 ppm range, whereas S has only one peak in this range.
(Alternative): S has an alkyl methyl carbon bonded to an alkyl group ( R-CH₃ ), giving a peak at δ = 5–40 ppm (lower chemical shift) which R does not have.
Number of peaks: 3

Mark scheme: M1 (Structural distinction regarding carbons adjacent to C=O), M2 (Corresponding peak shift range difference from Table C), M3 (State: 3 peaks for each isomer).

Part 10.5 • 3 Marks

Distinguishing Isomers R and S using ¹H NMR Splitting

✅ Justification & Peak Splitting Patterns

Isomer R (Hexane-2,5-dione): Both proton signals appear as singlets.
• The -CH₃ groups have no adjacent protons (isolated by C=O) → singlet.
• The central -CH₂-CH₂- protons are equivalent and isolated from other protons → singlet.
Isomer S (Hexane-3,4-dione): Shows a triplet and a quartet (classic ethyl group pattern).
• The -CH₃ protons are adjacent to 2 protons ( -CH₂- ) → triplet (n+1 = 3).
• The -CH₂- protons are adjacent to 3 protons ( -CH₃ ) → quartet (n+1 = 4).
Total number of peaks: 2 for each isomer.

🧠 Top Tip for NMR Questions

Always cite both isomers clearly:

State what happens for R (two singlets).
State what happens for S (one triplet and one quartet).
Do not confuse the number of peaks/environments (2) with the splitting multiplicity (singlet, triplet, quartet).

Mark scheme: M1 (R: both singlets), M2 (S: triplet and quartet), M3 (State: 2 peaks for each isomer).

Part 10.6 • 3 Marks

Cyclisation & Condensation Polymerisation of 5-Hydroxyhexanoic Acid

✅ Structure of Cyclic Compound T (Lactone)

Structure: A 6-membered ring containing 5 carbon atoms and 1 oxygen atom.
• The carbon adjacent to the ring oxygen on one side has a =O (carbonyl) group.
• The carbon adjacent to the ring oxygen on the other side bears a methyl group ( -CH₃ ).
(6-methyltetrahydro-2H-pyran-2-one / 6-methyl-ε-caprolactone family)

✅ Repeating Unit & Polymer Type

Polyester repeating unit:
—O—CH(CH₃)—CH₂—CH₂—CH₂—CO—
or drawn showing open trailing bonds on both ends passing through brackets:
—[—O—CH(CH₃)—(CH₂)₃—C(=O)—]—

Type of polymerisation: Condensation polymerisation.

❌ Common Errors

Counting ring atoms incorrectly: Intramolecular reaction between C5-OH and C1-COOH forms a 6-membered ring (5 carbons + 1 oxygen). Students often draw a 5-membered ring by omitting one CH₂.
Missing trailing bonds: Marks are lost if open trailing bonds at the ends of the polymer repeating unit are omitted.
Writing 'esterification': The question specifically asks for the type of polymerisation, so the answer must be condensation (or condensation polymerisation).

Mark scheme: 1 mark for cyclic ester T; 1 mark for polyester repeating unit with trailing bonds; 1 mark for 'condensation'.

Part 10.7 • 3 Marks

Addition Polymerisation of Ethyl Methacrylate & Biodegradability

✅ Repeating Unit of Polymer Formed by U

The C=C double bond breaks to form the main carbon-carbon chain:

       COOCH₂CH₃
       |
—[—CH₂—C—]—
       |
       CH₃

Ensure open trailing bonds extend outside brackets.

💡 Justification for Non-Biodegradability

Main carbon chain backbone: The polymer backbone consists entirely of strong, non-polar C–C bonds.
Inertness to hydrolysis: These C–C bonds cannot be attacked by nucleophiles (e.g. water/OH⁻) or enzymes, so the main chain cannot be broken down/hydrolysed.
Side chain distinction: Although the side group contains a polar ester link, hydrolysing it only removes the side ester group; it does not break the polymer backbone.

❌ Common Misconceptions

Confusing addition with condensation polymers: Simply stating "it contains ester groups so it hydrolyses and is biodegradable". The examiner requires you to recognise that the ester is merely a pendant group; the backbone is an addition chain of C–C bonds!
Failing to state that C–C bonds in the backbone are non-polar and strong.

Mark scheme: M1 (Repeating unit with trailing bonds), M2 (Strong / non-polar C-C bonds in the main chain), M3 (Cannot be attacked by nucleophiles/acids / cannot be hydrolysed / ester is only in side chain so polymer backbone is not broken).

Topics

Physical Chemistry · Organic Chemistry · 3.1.2 Amount of Substance · 3.3.1 Introduction to Organic Chemistry · 3.3.4 Alkenes · 3.3.6 Organic Analysis · 3.3.9 Carboxylic Acids and Derivatives · 3.3.12 Polymers · 3.3.15 Nuclear Magnetic Resonance Spectroscopy

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.