OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2017: Question 17

21 marks · Hard difficulty · Structured Questions

Outline the nitration of benzoic acid, calculate its percentage yield and purity, explain the relative ease of nitration of phenol, benzene and benzoic acid, and complete a multi-step organic synthesis of 3-bromophenylamine starting from nitrobenzene.

Practise this question

Question

A structured multi-part exam question about aromatic chemistry. Part (a) shows the nitration equation of benzoic acid to 3-nitrobenzoic acid, followed by a mechanism question and a 6-mark extended-response question on recrystallisation, percentage yield, and checking purity. Part (b) presents a table of conditions required for the nitration of phenol, benzene, and benzoic acid, followed by questions asking to state the trend in ease of nitration and explain it using bonding in arenes. Part (c) provides a reaction scheme for synthesizing 3-bromophenylamine from nitrobenzene, requiring the identification of an intermediate, reagents for bromination and reduction, and an explanation of directing effects when the order of steps is reversed.
Question text

17 This question is about the chemistry of aromatic compounds.

(a) Benzoic acid can be nitrated by concentrated nitric acid in the presence of concentrated

sulfuric acid as a catalyst, as shown in Equation 17.1.

The organic product of this reaction is 3-nitrobenzoic acid.

COOH COOH

H2SO4

+ HNO3 + H2O Equation 17.1

NO2

benzoic acid 3-nitrobenzoic acid

(i) Outline the mechanism for this nitration of benzoic acid.

Show how H2SO4 behaves as a catalyst.

[5]

(ii)* A chemist carries out the reaction in Equation 17.1 using 4.97 g of benzoic acid.

The chemist obtains 3-nitrobenzoic acid as an impure solid.

The chemist purifies the solid to obtain 4.85 g of 3-nitrobenzoic acid.

Describe a method to obtain a pure sample of 3-nitrobenzoic acid from the impure solid,

determine the percentage yield and check its purity.

… [6]

(b) A student investigates the relative ease of nitration of phenol, benzene, and benzoic acid.

OH COOH

phenol benzene benzoic acid

The student finds that the conditions required for the nitration of each compound are different,

as shown in Table 17.1.

Compound phenol benzene benzoic acid

Dilute HNO3 Concentrated HNO3 Concentrated HNO3

Conditions required

20 °C 55 °C 100 °C

for nitration

No catalyst H2SO4 catalyst H2SO4 catalyst

Table 17.1

(i) State the trend in the relative ease of nitration of phenol, benzene, and benzoic acid.

… [1]

(ii) Apply your knowledge of the bonding in arenes to explain the trend in part (b)(i).

… [3]

(c) A student synthesises 3-bromophenylamine, shown below, starting from nitrobenzene.

(i) Complete the flowchart showing the structure of the intermediate and the formulae of

the reagents for each stage.

NO2

bromination

reagents: …

reduction NH

reagents: …

Br

3-bromophenylamine

intermediate

[3]

(ii) Another student attempts the same synthesis but carries out reduction before bromination.

The student was surprised to find that two structural isomers of 3-bromophenylamine

had been formed instead of the desired organic product.

Explain this result and suggest the structures of the two isomers that formed.

Explanation …

Structures

[3]

Mark scheme

Show the mark scheme The mark scheme corresponding to question 17, detailing the marking points for the electrophilic substitution mechanism, the 3-level assessment grid for the recrystallisation and yield calculation, trends and explanations for ring activation/deactivation in phenol and benzoic acid, and structures and reagents for the organic synthesis pathway.

Question Answer Marks Guidance

17 (a) (i) Generation of electrophile 5 ANNOTATE ANSWER WITH TICKS AND CROSSES

HNO + H SO H O + HSO – + NO +

32 4 2 4 2 ALLOW HNO + 2H SO H O+ + 2HSO – +

32 4 3 4

NO +

Electrophilic substitution 2

Curly arrow from -bond to NO + + –

2 ALLOW HNO3 + H2SO4 H2NO3 + HSO4

then

COOH H NO + H O + NO +

23 2 2

ALLOW +NO OR NO +

NO + +

2 First curly arrow must come from the ring to NO2

---------------------------------------------------------------------

Correct intermediate DO NOT ALLOW the following intermediate:

Curly arrow back from C-H bond to reform -ring COOH

AND H+ as product

COOH COOH

+

NO2

H

+ + H+

NO2

-ring should cover approximately 4 of the 6 sides of

NO2

the benzene ring structure

H AND

the correct orientation, i.e. gap towards C with NO2

Regeneration of catalyst

H+ + HSO – H SO ALLOW + sign anywhere inside the ‘hexagon’ of

42 4

intermediate

(ii)* Please refer to the marking instructions on page 5 of 6 Indicative scientific points, with bulleted elements,

this mark scheme for guidance on how to mark this may include:

question.

1. Purification

Level 3 (5–6 marks) Recrystallisation

Dissolve impure solid in minimum volume of hot

Outlines the main steps of recrystallisation to produce water/solvent

a pure sample of 3-nitrobenzoic acid from the impure Cool solution and filter solid

solid. Wash with cold water/solvent and dry

AND

Calculates correct percentage yield of 3-nitrobenzoic 2. Percentage yield

acid. 4.97

AND n(benzoic acid) used = 122 = 0.0407 (mol)

Method of checking purity to include comparison to 4.85

relevant data. n(3-nitrobenzoic acid) made = 167 = 0.0290 (mol)

0.0290

A well-structured response with the steps for percentage yield = 0.0407 100 = 71.3 (%)

recrystallisation and the determination of purity being

given in the correct order. Correct use of terminology ALLOW 71 to calculator value of 71.29001554

throughout. correctly rounded.

Level 2 (3–4 marks) CHECK for extent of errors by ECF

Attempts all three scientific points but explanations

may be incomplete. Alternative correct calculation may calculate theoretical

OR mass of 3-nitrobenzoic acid that can be produced as

Explains two scientific points thoroughly with very few 0.0407 167 = 6.80 (g) followed by:

omissions.

4.85

percentage yield = 6.80 100 = 71.3 (%)

The description of checking for purity or

recrystallisation is clear and any calculations

structured. Key terminology used appropriately. Calculation must attempt to calculate n(benzoic acid)

in mol.

Level 1 (1–2 marks)

A simple explanation based on at least two of the

main scientific points.

OR 3. Checking purity

Explains one scientific point thoroughly with few Obtain melting point

omissions. Compare to known values

Pure sample will have a (sharp) melting point very

There is an attempt at a logical structure. The close to data book value

description of the practical techniques provides some

detail but may not be in the correct order. ALLOW alternative approach based on spectroscopy

or TLC

Purification step is unclear with few scientific

terms and little detail, e.g. just ‘recrystallise’. Spectroscopy

Calculation is difficult to follow, may just include a Run an NMR/IR spectrum

calculation of moles of reactants and/or products. Compare to (spectral) database

Purity check specifies a method but this is unclear Spectrum of pure sample will contain same peaks

with little detail, e.g. take melting point. and not others

0 marks TLC

No response or no response worthy of credit. Run a TLC

Compare (Rf value) to known data

Pure sample will have a very similar Rf

(b) (i) Phenol is the most easily nitrated/ 1 Response must give rank order of reactivity

most reactive

AND e.g. nitration becomes more difficult from phenol (to

Benzoic acid is the least easily nitrated benzene) to benzoic acid

/least reactive OR

nitration becomes easier from right to left in the table

(ii) Reactivity of phenol 3 ANNOTATE ANSWER WITH TICKS AND CROSSES

a (lone) pair of electrons on O is (partially) 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

Reactivity of benzoic acid

IGNORE activating and deactivating.

The –COOH group on benzoic acid is an electron

withdrawing group

Links electron density in -bond to reactivity

In phenol electron density is higher ALLOW the following alternatives for susceptibility to

AND attack:

The ring is more susceptible to attack

phenol attracts electrophiles / NO + more

OR

phenol polarises electrophiles / NO + more

In benzoic acid electron density is lower

AND

The ring is less susceptible to attack

benzoic acid attracts electrophiles / NO + less

benzoic acid polarises electrophiles / NO + less

(c) (i) Bromination: Br2 AND AlBr3/FeBr3/Fe 3 ALLOW any combination of skeletal OR structural OR

displayed formula as long as unambiguous

Intermediate

ALLOW any suitable halogen carrier catalyst

NO2

ALLOW Kekulé structure

IGNORE names (question asks for formulae)

Br IGNORE reaction conditions even if incorrect

IGNORE ‘dilute’ for HCl

Reduction: Sn AND (concentrated) 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

(ii) 3

NH2 is 2,4 directing IGNORE references to electron donating/withdrawing

groups

Products (1 mark for each):

NH2 NH2 ALLOW –NH2 activates the ring causing the new group

to join at positions 2 and 4.

Br

ALLOW ortho and para directing for 2,4 directing

IGNORE 6-directing

ALLOW Kekulé structure

Br

IGNORE names

Total 21

How to answer it

Chemistry of Aromatic Compounds Study Guide

What this question tests

This question assesses your understanding of electrophilic substitution mechanisms involving arenes, catalyst generation, practical chemistry techniques (recrystallization, percentage yield, melting point determination), activating and deactivating ring substituents, and directing effects in multi-step organic synthesis.

Question 17 (a)(i)

Mechanism of Nitration and Catalyst Role

✅ Correct Answer & Mark Scheme

  • Generation of electrophile: HNO₃ + H₂SO₄ → H₂O + HSO₄⁻ + NO₂⁺
  • Electrophilic attack: Curly arrow originating from the benzene π-ring pointing directly to the NO₂⁺ ion.
  • Intermediate: Horseshoe-shaped intermediate with a positive charge located inside the incomplete circle, and both -COOH , -NO₂ , and -H shown on the carbon atom.
  • Loss of proton: Curly arrow from the C-H bond back into the delocalized ring.
  • Regeneration of catalyst: H⁺ + HSO₄⁻ → H₂SO₄

🧠 Exam Technique

Mechanism arrows must be drawn with precision. The first curly arrow must start from the delocalized π-ring (not from a carbon atom). The horseshoe intermediate must show the positive charge clearly contained within the broken ring structure, centred around the top and side carbons.

Question 17 (a)(ii) - Level of Response (6 Marks)

Practical Purification, Percentage Yield, and Purity Check

📐 Step-by-Step Calculation

  1. Moles of benzoic acid: Molar mass = 122 g mol⁻¹. 4.97 / 122 = 0.0407 mol
  2. Moles of 3-nitrobenzoic acid (actual): Molar mass = 167 g mol⁻¹. 4.85 / 167 = 0.0290 mol
  3. Percentage yield: (0.0290 / 0.0407) × 100 = 71.3% (Accept 71.29%).

💡 Key Knowledge (Recrystallization & Purity)

  • Dissolve the impure solid in a minimum volume of hot solvent/water.
  • Cool the solution to recrystallize, then filter under reduced pressure (suction filtration).
  • Wash with cold solvent and dry.
  • Purity Check: Measure the melting point using melting point apparatus and compare with literature values. A pure sample will have a sharp melting point very close to the data book value.

❌ Common Errors

Students often lose marks by stating "dissolve in hot water" without specifying a minimum volume, or by failing to mention that a sharp melting point matching literature values confirms high purity.

Question 17 (b)(i) & (ii)

Relative Ease of Nitration & Ring Bonding

✅ Correct Answer

(i) Trend: Phenol is the most reactive (easiest to nitrate), benzene is intermediate, and benzoic acid is the least reactive (most difficult to nitrate).

(ii) Explanation: The -OH group in phenol has a lone pair of electrons on oxygen that is partially delocalized/donated into the π-system, increasing electron density and making it more susceptible to electrophilic attack. Conversely, the -COOH group in benzoic acid is electron-withdrawing, decreasing ring electron density and making it less susceptible to attack.

🧠 Exam Technique

When comparing reactivities of aromatic compounds, always link the substituent group explicitly to its effect on electron density within the benzene ring and how that affects attraction toward the incoming electrophile ( NO₂⁺ ).

Question 17 (c)(i) & (ii)

Multi-Step Synthesis & Directing Effects

💡 Key Knowledge & Answers

  • (i) Bromination Reagents: Br₂ with a halogen carrier catalyst such as AlBr₃ , FeBr₃ , or Fe .
  • Intermediate Structure: 1-bromo-3-nitrobenzene (bromine meta to the nitro group).
  • Reduction Reagents: Sn (tin) and concentrated HCl , followed by NaOH .
  • (ii) Wrong Order Effect: If reduction happens first, the -NH₂ group formed is 2,4-directing (ortho/para directing). Subsequent bromination yields a mixture of 2-bromophenylamine and 4-bromophenylamine instead of the desired 3-isomer.

❌ Common Errors

Writing general terms like "dilute HCl" instead of concentrated HCl during reduction will lose marks. Also, ensure structural or skeletal formulas clearly show the relative positions (ortho, meta, para) of substituents on the benzene ring.

Topics

Module 6: Organic chemistry and analysis · Practical Activity Groups · 6.1 Aromatic compounds, carbonyls and acids · 6.2 Nitrogen compounds, polymers and synthesis · PAG 6: Synthesis of an organic solid

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.