OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2022: Question 20
22 marks · Hard difficulty · Structured Questions
Answer questions on the chemistry of aromatic compounds including phenol testing, NMR spectroscopy, electrophilic substitution mechanisms, activating effects, oxidation, amine synthesis, reaction pathways, and esterification.
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Question text
20 This question is about the chemistry of aromatic compounds.
(a) Compounds J, K and L, shown below, are structural isomers.
CH3
CH2CH2OH OH H3C OH
CH3 CH3
Compound J Compound K Compound L
(i) What chemical test(s) could be used to confirm the presence of the phenol group in
compounds K and L?
… [1]
(ii) A student thought that 13C NMR spectroscopy could be used to distinguish between
compounds J, K and L.
Explain, with reasoning, whether the student is correct.
… [3]
(iii) Compound J is substituted at the 2- and 4- positions by chlorine in the presence of a
catalyst.
Outline the mechanism for the 4 substitution of compound J by chlorine in the presence
of a catalyst.
Show the role of the catalyst.
[4]
(b) Compounds K and L react with chlorine much more readily than compound J.
Explain why.
… [3]
(c) Compound J, C6H5CH2CH2OH, is reacted with acidified potassium dichromate(VI) under
reflux to form organic product M.
Write an equation for this reaction.
Use [O] to represent the oxidising agent and show the structure of M.
[2]
(d) A two-stage synthesis of an amine from compound J is shown below.
(i) Add the reagents for each stage of this synthesis.
CH2CH2OH
Compound J
CH2CH2OH CH2CH2OH
Reduction
NO2 NH2
[2]
(ii) Fill in the equation for the reduction stage of this synthesis.
CH2CH2OH + … CH2CH2OH + …
NO2 NH2
[1]
(e) 1-phenylethanol is a naturally occurring compound found in many vegetables and flowers.
1-phenylethanol can be synthesised from 2-phenylethanol in two stages.
H H OH H
Stage 1 Stage 2
C C OH Intermediate C C H
H H H H
2-phenylethanol 1-phenylethanol
Suggest reagents, conditions and equations for each stage in the synthesis.
Show structures for organic compounds.
Stage 1
reagents and conditions …
equation:
Stage 2
reagents and conditions …
equation:
[4]
(f) Acid anhydrides react in a similar way to acyl chlorides with phenols.
Benzoic anhydride is the acid anhydride of benzoic acid, C6H5COOH.
Benzoic anhydride reacts with butan-2-ol to form an ester.
Suggest an equation for this reaction. Show structures for organic compounds. Use C6H5 for
any phenyl groups.
[2]
Mark scheme
Show the mark scheme
AO
Question Answer Marks Guidance
element
20 (a) (i) Indicator AND observation of acidity 1 AO1.2 ALLOW
AND ×1 (Add) bromine AND white precipitate
No reaction with carbonate
ALLOW
(Add) FeCl3 AND violet/purple colour
20 (a) (ii) Compound J has 3 AO3.2 IGNORE any numbers shown on structures
6 peaks/environments/types of carbon ×3
IGNORE chemical shifts
Compound K has
5 peaks/environments/types of carbon
Compound L has
8 peaks/environments/types of carbon
20 (a) (iii) ANNOTATE ANSWER WITH TICKS AND CROSSES 4 AO1.2
×2
Action of catalyst 1 mark AO2.5 ALLOW use of FeCl3 or other halogen carriers
Formation of electrophile: Cl + AlCl → Cl+ + AlCl – ×2 (AlBr )
23 4 3
AND
Regeneration of catalyst: H+ + AlCl – → AlCl + HCl
---------------------------------------------------------
--------------------------------------------------------------------------- For curly arrows, ALLOW straight or snake-
Electrophilic attack 1 mark like arrows and small gaps (see examples):
---------------------------------------------------------
Curly arrow from π-bond to Cl+ 1st curly arrow must
• start from, OR close to circle of benzene
ring
AND
• go to Cl+
AO
element
Correct intermediate only 1 mark DO NOT ALLOW the following intermediate:
π-ring must cover more than half of benzene ring
AND
correct orientation, i.e. gap towards C with Cl
ALLOW + sign anywhere inside the ‘hexagon’ of
intermediate
DO NOT ALLOW intermediates substituted at
positions 3 or 5
IGNORE intermediates substituted at position 2
OR di-substituted at positions 2,4
------------------------------------------------------------------------- ---------------------------------------------------------------
Reforming benzene ring 1 mark Curly arrow must start from, OR be traced back
to, any part of C–H bond and go inside the
Curly arrow from C–H bond to reform π-ring ‘hexagon’
30 AO
element
20 (b) (In phenols) a (lone) pair of electrons on O is (partially) 3 AO1.1 ALLOW the electron pair in the p-orbitals of the O
delocalised/donated into the ring / π-system × 3 atom becomes part of the ring / π-system
ALLOW diagram to show movement of lone pair
into ring
ALLOW lone pair of electrons on O is (partially)
drawn/attracted/pulled/ into ring / π-system
ALLOW lone pair on O
DO NOT ALLOW (two) lone pairs are
delocalised/donated into the ring / π-system
Electron density increases/is higher (than benzene) IGNORE activating
ORA
IGNORE charge density
IGNORE electronegativity
(phenols) are more susceptible to electrophilic attack
OR IGNORE phenols react more readily with
(phenols) attract/accept electrophile/Cl2 more electrophiles/Cl2 (given in question)
OR
(phenols) polarise electrophile/Cl2 more ALLOW Cl+ for electrophile
ORA IGNORE Cl for electrophile
20 (c) 2 ALLOW any combination of skeletal OR structural
OR displayed formula as long as unambiguous
AO2.5
ALLOW C6H5 for phenyl group
AO2.6
organic product
Correct balanced equation
20 (d) (i) 2 AO1.2
×2 IGNORE references to concentration
IGNORE ‘dilute’ for HCl
IGNORE H2
IGNORE NaOH if seen as a reagent to convert
nitro group into amine
e.g ‘Sn/(concentrated) HCl then NaOH’ scores the
mark
20 (d) (ii) 1 AO2.6
20 (e) Stage 1 4 ALLOW any combination of skeletal OR structural
Reagents: H2SO4 AO3.1 OR displayed formula as long as unambiguous
ALLOW H+ OR HCl OR H PO
DO NOT ALLOW other named acids
IGNORE concentration/pressure
AO2.6 IGNORE water/steam
Stage 2
Reagents: Steam/H O(g) AND acid/H+ (catalyst) AO3.1
For steam,
ALLOW H2O with temperature ≥100ºC
ALLOW use of H3PO4/H2SO4 as catalyst
AO2.6 DO NOT ALLOW HCl
IGNORE pressure
20 (f) 2 ALLOW any combination of skeletal OR structural
Structure of ester product 32 OR displayed formula as long as unambiguous
AO3.1
Correct balanced equation AO3.2 ALLOW
Total 22
How to answer it
Chemistry of Aromatic Compounds Study Guide
What this question tests
This multi-part synthesis question assesses your mastery of aromatic chemistry. Key skills and knowledge tested include: functional group tests (phenol acidity vs. carboxylic acids), 13C NMR spectroscopy interpretation via carbon environments, electrophilic substitution mechanisms (halogenation of arenes), activating effects of electron-donating groups, oxidation of primary alcohols to carboxylic acids, multi-step synthetic routes involving nitration and reduction, acid-catalyzed hydration/dehydration reactions, and the esterification of phenols using acid anhydrides.
Question 20(a) — Structural Isomers & Mechanisms
Part (a)(i)
✅ Correct Answer
Use an acid-base indicator (or pH probe) to show acidity, combined with Na₂CO₃ (sodium carbonate) showing no reaction.
Alternative valid tests: Bromine water (forms white precipitate) or aqueous iron(III) chloride ( FeCl₃ ) yielding a violet/purple solution/complex.
❌ Common Errors
Failing to specify that carbonate must give no reaction. Phenols are weakly acidic and do not react with weak bases like carbonates (unlike carboxylic acids), which is a key distinction.
Part (a)(ii)
✅ Correct Answer
Compound J: 6 peaks (environments/types of carbon)
Compound K: 5 peaks
Compound L: 8 peaks
💡 Key Knowledge
13C NMR spectroscopy counts non-equivalent carbon environments. Symmetry planes in substituted benzenes (like the symmetrical methyl groups in Compound K) reduce the number of unique carbon environments.
Part (a)(iii) — Mechanism for Chlorination of Compound J
🧠 Exam Technique: Catalyst & Generation
- Catalyst equation: Cl₂ + AlCl₃ → Cl⁺ + AlCl₄⁻
- Regeneration equation: H⁺ + AlCl₄⁻ → AlCl₃ + HCl
- Curly arrows: First arrow must start from the benzene ring (pi-bond) and point directly to the incoming electrophile ( Cl⁺ ).
❌ Common Errors
Drawing the intermediate horseshoe positive charge covering less than half of the ring, or placing the positive charge outside the ring hexagon. Ensure the intermediate shows the H and Cl attached to the same carbon with a broken ring inside.
Question 20(b) — Reactivity of Phenols vs Arenes
✅ Correct Answer
1. Lone pair of electrons on the oxygen atom is (partially) delocalized/donated into the benzene ring / pi-system.
2. Electron density in the ring increases (higher than in benzene).
3. Therefore, phenols are more susceptible to electrophilic attack / polarize electrophiles ( Cl₂ ) more effectively.
💡 Key Knowledge
Activating groups donate electron density into the ring through p-orbital overlap, making the ring more attractive to incoming electrophiles compared to unsubstituted benzene.
Question 20(c) — Oxidation of Alcohols
✅ Correct Answer
Equation: C₆H₅CH₂CH₂OH + 2[O] → C₆H₅CH₂COOH + H₂O
Structure of M: Phenylacetic acid ( C₆H₅CH₂COOH or structural formula).
🧠 Exam Technique
Always use 2[O] for the oxidation of a primary alcohol straight to a carboxylic acid under reflux conditions. Ensure balanced side products ( H₂O ).
Question 20(d) — Multi-Stage Synthesis (Nitration & Reduction)
Part (d)(i) — Reagents
✅ Correct Answer
Stage 1 (Nitration): Concentrated HNO₃ AND concentrated H₂SO₄ .
Stage 2 (Reduction): Sn (tin) AND concentrated HCl (followed by NaOH ).
Part (d)(ii) — Reduction Equation
✅ Correct Answer
Nitro compound + 6[H] → Amine compound + 2H₂O
❌ Common Errors
Balancing reduction equations using H₂ instead of the standard reduction notation [H] or miscounting water molecules (2 H₂O are produced per nitro group reduced).
Question 20(e) — Isomer Synthesis & Acid-Catalyzed Hydration
✅ Correct Answer
Stage 1 (Dehydration to alkene):
• Reagents/Conditions: Concentrated H₂SO₄ (or H₃PO₄ ) and heat.
• Equation: C₆H₅CH₂CH₂OH → C₆H₅CH=CH₂ + H₂O
Stage 2 (Hydration to secondary alcohol):
• Reagents/Conditions: Steam ( H₂O(g) ) with acid catalyst ( H₃PO₄ or H₂SO₄ ) at temperature ≥ 100°C.
• Equation: C₆H₅CH=CH₂ + H₂O → C₆H₅CH(OH)CH₃
💡 Key Knowledge
Markovnikov's rule dictates that addition of water across the unsymmetrical double bond in Stage 2 yields 1-phenylethanol as the major product because the hydrogen adds to the carbon with more hydrogens, stabilizing the intermediate carbocation.
Question 20(f) — Esterification with Acid Anhydrides
✅ Correct Answer
Ester Structure: Benzoic anhydride reacting with butan-2-ol forms 1-methylpropyl benzoate (or C₆H₅COOCH(CH₃)CH₂CH₃ ).
Balanced Equation:
(C₆H₅CO)₂O + CH₃CH(OH)CH₂CH₃ → C₆H₅COOCH(CH₃)CH₂CH₃ + C₆H₅COOH
🧠 Exam Technique
When writing equations involving acid anhydrides and alcohols, remember that one half of the anhydride forms the ester while the remaining half forms a carboxylic acid byproduct (unlike acyl chlorides which produce HCl ).
Topics
Module 6: Organic chemistry and analysis · 6.1 Aromatic compounds, carbonyls and acids · 6.2 Nitrogen compounds, polymers and synthesis · 6.3 Analysis
Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.