OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2018: Question 18

9 marks · Medium difficulty · Structured Questions

Explain whether 13C NMR can distinguish between nitrophenol isomers, explain the reactivity of phenol compared to benzene, and complete the electrophilic substitution mechanism for methylbenzene reacting with sulfur trioxide.

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Question

Question text

18 This question is about aromatic compounds.

(a) Phenol undergoes nitration more readily than benzene.

(i) A student carries out the nitration of phenol with dilute nitric acid to produce 2-nitrophenol

and 4-nitrophenol.

A small amount of 3-nitrophenol is also produced.

OH OH

NO2

2-nitrophenol 3-nitrophenol 4-nitrophenol

The student thought that 13C NMR spectroscopy could be used to distinguish between

these three nitrophenols.

Explain whether the student is correct.

… [3]

(ii) Explain why phenol is nitrated more readily than benzene.

… [3]

(b) Methylbenzene reacts with sulfur trioxide, SO3, to form D, shown below.

CH3

O S O

O–

The electrophile in this reaction is SO3.

Complete the mechanism for the formation of D.

Show curly arrows and the structure of the intermediate.

CH3

O O

intermediate

S δ+

δ–

CH3

+ H+

O S O

O–

[3]

Mark scheme

Show the mark scheme

18 (a) (i) 3

Number of peaks 2 marks

IGNORE any numbers shown on structures

2-nitrophenol AND 3-nitrophenol have six

peaks/environments/types of carbon ALLOW 1 mark only IF a response identifies that all the

compounds have 6 peaks/environments/types of C

4-nitrophenol has four peaks/environments/types of OR all the compounds have 4 peaks/environments/

carbon types of carbon

Statement 1 mark IGNORE chemical shifts

4-nitrophenol can be distinguished

2-nitrophenol and 3-nitrophenol cannot be distinguished

DO NOT ALLOW ECF from an incorrect number of

peaks/environments/types of carbon

(ii) (In phenol) a (lone) pair of electrons on O is(partially) 3 ALLOW the electron pair in the p-orbitals of the O atom

delocalised/donated into the -system / ring becomes part of the -system / ring

ALLOW diagram to show movement of lone pair into ring

ALLOW lone pair of electrons on O is (partially)

drawn/attracted/pulled/ into -system / ring

IGNORE activating

Electron density increases/is higher (than benzene)

ORA IGNORE charge density

IGNORE electronegativity

(phenol) is more susceptible to electrophilic attack

OR IGNORE phenol reacts more readily (no reference to

(phenol) attracts/accepts electrophile/HNO3 more electrophile)

(phenol) polarises electrophile/HNO more ALLOW NO + for electrophile

ORA

Question Answer 14 Marks Guidance

(b) 3 ANNOTATE WITH TICKS AND CROSSES

NOTE: curly arrows can be straight, snake-like, etc.

but NOT double headed or half headed arrows

Curly arrow from -bond to S in SO3 1st curly arrow must

AND go to the S of SO3

curly arrow from the S=O bond to O atom AND

CH3 start from, OR close to circle of benzene ring

O O

S + 2nd curly arrow must start from, OR be traced back to,

any part of S=O bond and go to O

ALLOW 2nd curly arrow from S=O to any O in SO3

Intermediate must have correct SO – structure fully

displayed

Correct intermediate DO NOT ALLOW the following intermediate:

CH3

Curly arrow from C-H bond to reform -ring

CH3 +

+ O S O

O -

H S -ring must cover more than half of the benzene ring

structure

O O -

AND

the correct orientation, i.e. gap towards C with SO –

ALLOW + sign anywhere inside the ‘hexagon’ of the

intermediate.

DO NOT ALLOW mark for intermediate if CH3 is missing

curly arrow must start from, OR be traced back to, any

part of C-H bond and go inside the ‘hexagon’

Total 9

How to answer it

Aromatic Chemistry: Phenol Reactivity & Substitution Mechanisms

📌 What this question tests

This question examines core concepts in aromatic and analytical organic chemistry from OCR Chemistry A:

¹³C NMR Spectroscopy of Disubstituted Benzenes: Determining carbon environments using plane-of-symmetry arguments to distinguish isomers.
Relative Reactivity of Arenes: Explaining why phenol is activated towards electrophilic attack compared to benzene (orbital overlap, electron density, and polarisation).
Electrophilic Aromatic Substitution Mechanism: Adapting standard benzene mechanism steps to an unfamiliar neutral electrophile (SO₃) attacking methylbenzene.

Part (a)(i) — Distinguishing Nitrophenol Isomers using ¹³C NMR

Symmetry and Number of Carbon Environments [3 Marks]

✅ Model Answer & Mark Scheme

2-nitrophenol has 6 carbon environments / peaks [1 mark]
3-nitrophenol has 6 carbon environments / peaks [combined with above]
4-nitrophenol has 4 carbon environments / peaks [1 mark]
Conclusion: The student is incorrect (or only partially correct) because 4-nitrophenol can be distinguished, but 2-nitrophenol and 3-nitrophenol cannot be distinguished from each other by number of peaks alone [1 mark] .

💡 Key Knowledge: Symmetry in Benzene Rings

4-nitrophenol (1,4-disubstituted): Contains a vertical plane of symmetry passing directly through C1 (bearing -OH) and C4 (bearing -NO₂). This makes C2 ≡ C6 and C3 ≡ C5, giving exactly 4 unique carbon environments.
2-nitrophenol (1,2-) & 3-nitrophenol (1,3-): Neither molecule possesses a plane of symmetry across the ring because the two substituents (-OH and -NO₂) are different. All 6 carbon atoms are chemically distinct, resulting in 6 peaks each.

🧠 Exam Technique

Always state the explicit number of peaks for every isomer first, then address the prompt directly ("Explain whether the student is correct"). Never simply say "they have different peaks"; specify the exact numbers (6, 6, 4).

❌ Common Errors

Assuming 3-nitrophenol has symmetry: students often confuse 1,3-disubstituted benzenes with symmetrical compounds—remember the substituents are different!
Relying on chemical shifts: the mark scheme explicitly states IGNORE chemical shifts; the differentiation is based purely on peak count.
No Error Carried Forward (ECF): if peak counts are wrong, the final conclusion mark cannot be awarded.

Mark Breakdown: 2 marks for all three peak counts correct (6, 6, and 4) • 1 mark for the logical deduction that 4-nitrophenol is distinguishable but the 2- and 3-isomers cannot be distinguished from each other.

Part (a)(ii) — Reactivity: Phenol vs Benzene

Activating Effect of the -OH Group [3 Marks]

✅ Model Answer & Mark Scheme

A lone pair of electrons on the oxygen atom of the -OH group is delocalised / donated into the π-system (ring) of phenol [1 mark] .
This causes the electron density of the ring to increase (or be higher than in benzene) [1 mark] .
Therefore, phenol polarises the electrophile more / is more susceptible to electrophilic attack (attracts the electrophile more strongly) [1 mark] .

💡 The 3-Step "Activating Group" Template

Origin: Mention lone pair on the heteroatom (O) donating into the delocalised π-system.
Ring State: Ring has higher electron density.
Action: Polarises the electrophile more effectively (no halogen carrier needed).

🧠 Top-Grade Precision

Use comparative language throughout: "electron density is higher than in benzene" and "polarises the electrophile more". State clearly that it is the lone pair in an oxygen p-orbital that overlaps with the π-system, not just the "oxygen atom".

❌ Common Errors

Saying "charge density" or "electronegativity" instead of electron density (examiners actively ignore charge density).
Failing to mention the electrophile in mark 3 (e.g. stating simply "phenol reacts faster" without mentioning electrophile attraction or polarisation).
Stating that oxygen's high electronegativity pulls electrons away—while inductive withdrawal occurs, the positive mesomeric (+M) resonance donation dominates.

Mark Breakdown: 1 mark for lone pair on O delocalised into π-system • 1 mark for increased electron density • 1 mark for greater attraction/polarisation of electrophile (or NO₂⁺ / HNO₃).

Part (b) — Mechanism: Electrophilic Substitution with SO₃

Electrophilic Attack on Methylbenzene [3 Marks]

✅ Mechanism Requirements

Step 1 Curly Arrows:
- Arrow 1: From the π-electron ring of methylbenzene to the partially positive sulfur atom (S^δ+) in SO₃.
- Arrow 2: From one of the S=O double bonds onto that oxygen atom (forming O⁻).
[1 mark]
Intermediate Structure: Correctly drawn arenium intermediate containing:
- The methyl group (-CH₃) remaining at C1.
- At C4 (para position): a single bond to -H and a single bond to -SO₃⁻ (fully displayed: -S(=O)₂O⁻).
- Partially delocalised π-system: an open horseshoe covering at least 5 ring carbons (C2 through C6), with opening facing C4.
- A positive charge (+) located inside the partial ring hexagon.
[1 mark]
Step 2 Curly Arrow: From the C-H bond at C4 back into the partial π-ring to regenerate aromaticity [1 mark] .

🧠 Diagram Drawing Guide

How to draw the intermediate correctly:
1. Draw the hexagon with the top vertex bonded to -CH₃.
2. At the bottom vertex (C4), draw two outward bonds: one to -H and one to -SO₃⁻ (showing S(=O)₂O⁻).
3. Draw the horseshoe opening: the open ends must reach C2 and C6 (covering 5 ring carbons). The gap must strictly face C4.
4. Put a + sign inside the center of the broken ring.
5. Draw a curly arrow starting directly from the C-H single bond line pointing cleanly inside the horseshoe ring.

❌ Common Mechanism Penalties

Missing arrow on SO₃: Forgetting to break an S=O double bond to O. Sulfur cannot accept electrons from the ring without breaking a π-bond!
Horseshoe too small: The broken ring in the intermediate must cover more than half the ring (5 carbon atoms). If it only covers 3 carbons, mark is lost.
Wrong arrow origin: The regenerating arrow must start on the C-H bond, not on the H atom itself.
Forgotten methyl group: Omitting the -CH₃ group on the intermediate loses the intermediate mark immediately.

💡 Why SO₃ is an Electrophile

Sulfur trioxide (SO₃) is a neutral electrophile. The three strongly electronegative oxygen atoms pull electron density away from sulfur, leaving sulfur highly electron-deficient with a significant partial positive charge (S^δ+).

When the ring attacks S, a pair of electrons in an S=O double bond shifts onto an oxygen, producing a sulfonate intermediate with a formal negative charge on that oxygen (-SO₃⁻).

Mark Breakdown: 1 mark for both first curly arrows (ring to S AND S=O to O) • 1 mark for correct intermediate with correct horseshoe, + charge, and attached -SO₃⁻ • 1 mark for curly arrow from C-H bond into ring.

Topics

Module 6: Organic chemistry and analysis · 6.1 Aromatic compounds, carbonyls and acids · 6.3 Analysis

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.