AQA A-Level Chemistry Paper 2, 2017: Question 8
12 marks · Medium difficulty · State/Explain/Describe
Write the generation equation and mechanism for the nitration of nitrobenzene to form 1,3-dinitrobenzene, and analyze TLC separation, Rf values, polarity, and solvent effects for isomeric dinitrobenzenes.
Practise this questionQuestion
Question text
08 This question is about nitrobenzenes.
08.1 Nitrobenzene reacts when heated with a mixture of concentrated nitric acid and
concentrated sulfuric acid to form a mixture of three isomeric dinitrobenzenes.
Write an equation for the reaction of concentrated nitric acid with concentrated
sulfuric acid to form the species that reacts with nitrobenzene.
[1 mark]
08.2 Name and outline a mechanism for the reaction of this species with
nitrobenzene to form 1,3-dinitrobenzene.
[4 marks]
Name of mechanism
Do
Mechanism 16 ou
08.3 The dinitrobenzenes shown were investigated by thin layer chromatography
(TLC).
In an experiment, carried out in a fume cupboard, a concentrated solution of
pure 1,4-dinitrobenzene was spotted on a TLC plate coated with a solid that
contains polar bonds. Hexane was used as the solvent in a beaker with a lid.
The start line, drawn in pencil, the final position of the spot and the final solvent
front are shown on the chromatogram in Figure 3
Figure 3
Use the chromatogram in Figure 3 to deduce the Rf value of
1,4-dinitrobenzene in this experiment.
Tick ( ) one box.
[1 mark]
A 0.41
B 0.46
C 0.52
Do
D 0.62 17 ou
08.4 State in general terms what determines the distance travelled by a spot in TLC.
[1 mark]
08.5 To obtain the chromatogram, the TLC plate was held by the edges and placed
in the solvent in the beaker in the fume cupboard. The lid was then replaced
on the beaker.
Give one other practical requirement when placing the plate in the beaker.
[1 mark]
08.6 A second TLC experiment was carried out using 1,2-dinitrobenzene and
1,4-dinitrobenzene. An identical plate to that in Question 8.3 was used under
the same conditions with the same solvent. In this experiment, the Rf value of
1,4-dinitrobenzene was found to be greater than that of 1,2-dinitrobenzene.
Deduce the relative polarities of the 1,2-dinitrobenzene and 1,4-dinitrobenzene
and explain why 1,4-dinitrobenzene has the greater Rf value.
[2 marks]
Relative polarities
Explanation
Do
18 ou
08.7 A third TLC experiment was carried out using 1,2-dinitrobenzene. An identical
plate to that in Question 8.3 was used under the same conditions, but the
solvent used contained a mixture of hexane and ethyl ethanoate.
A student stated that the Rf value of 1,2-dinitrobenzene in this third experiment
would be greater than that of 1,2-dinitrobenzene in the experiment in
Question 8.6
Is the student correct? Justify your answer.
*17* [2 marks]
Mark scheme
Show the mark scheme
Question Answers Mark Additional Comments/Guidance
Allow H SO + HNO → NO + + HSO – + H O
24 3 2 4 2
08.1 HNO + 2H SO → NO + + H O+ + 2HSO – 1 Allow a combination of equations which produce NO +
32 4 2 3 4 2
Penalise equations which produce SO 2–
Electrophilic substitution. 1 Ignore nitration
M1 Arrow from inside hexagon to N or + on N (Allow NO +)
M1 M3 2
O N O N H M2 Structure of intermediate
NO2
NO2 horseshoe centred on C1 and must not extend beyond
C2 and C6, but can be smaller
+ in intermediate not too close to C1 (allow on or “below”
M2 a line from C2 to C6)
08.2
3 M3 Arrow from bond into hexagon (Unless Kekule)
OR Kekule Allow M3 arrow independent of M2 structure
+ on H in intermediate loses M2 not M3
M1 M3
+ H
O2N O2N
NO2
NO2
M2
– – –
08.3 D 1
(Balance between) solubility in moving phase and retention by OR (relative) affinity for stationary/solid and
08.4 1
stationary phase mobile/liquid/solvent (phase)
08.5 Solvent depth must be below start line 1 Ignore safety
25 of 32
1,2- is more polar OR 1,4- is less polar 1
M2 dependent on correct M1
OR 1,2 is polar, 1,4- is non-polar
If M1 is blank then read explanation for possible M1 and M2
1,4- ( or Less/non polar is) less attracted to (polar) plate / 1
08.6
stationary phase / solid
OR (Less/non polar is) more attracted to / more soluble in
(non-polar) solvent / mobile phase / hexane Allow converse argument for 1,2
No CE = 0
Yes - mark on but there is NO MARK FOR YES
Mark independently following yes
08.7 Solvent (more) polar or ethyl ethanoate is polar 1
Polar isomer more attracted to / more soluble in / stronger 1 Penalise bonded to mobile phase in M2
affinity to the solvent (than before)
Total 12
How to answer it
Electrophilic Substitution & TLC of Dinitrobenzenes
This question assesses synthetic and analytical organic chemistry principles across aromatic chemistry and chromatography:
- Inorganic acid-base generation of electrophiles: Writing balanced equations for generating the nitronium ion (NO₂⁺).
- Aromatic reaction mechanisms: Drawing curly-arrow mechanisms for electrophilic substitution and accurately representing Wheland-type carbocation intermediates.
- Thin-layer chromatography (TLC) analysis: Estimating Rf values, explaining principles of separation (adsorption vs solubility), and practical setup rules.
- Intermolecular forces & polarity: Relating dipole moments to TLC retention and predicting the effect of changing solvent polarity.
Generation of the Nitrating Electrophile
1 Mark
✅ Correct Equation
HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻
OR (acceptable alternative):
HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O
💡 Key Knowledge
Sulfuric acid acts as a Brønsted-Lowry acid (proton donor) and protonates nitric acid, which acts as a base:
H₂SO₄ + HNO₃ ⇌ H₂NO₃⁺ + HSO₄⁻
The protonated nitric acid subsequently loses water to form the active electrophile: H₂NO₃⁺ → NO₂⁺ + H₂O.
❌ Common Errors
- Writing equations that produce sulfate ions ( SO₄²⁻ ) rather than hydrogensulfate ions ( HSO₄⁻ ). Sulfuric acid is not strong enough to fully deprotonate twice in this non-aqueous mixture.
- Forgetting the positive charge on the nitronium ion ( NO₂⁺ ).
Mechanism: Formation of 1,3-Dinitrobenzene
4 Marks
✅ Correct Answer & Mechanism Breakdown
Name of mechanism: Electrophilic substitution (1 mark)
- Arrow 1 (M1): Curly arrow starts from inside the delocalised benzene π-ring and points directly to the N atom (or the + charge on N) of NO₂⁺. The attack occurs at position 3 (meta) relative to the existing -NO₂ group.
- Intermediate (M2): A six-membered ring with a tetrahedral carbon at C3 bearing both -H and -NO₂. A horseshoe/horseshoe-shaped broken ring covers C2 through to C6 (5 carbons), open towards C3. A single + charge is located in the centre of the ring (or below the C2–C6 line, not directly on C3).
- Arrow 2 (M3): Curly arrow starts on the C–H bond at C3 and points directly into the broken π-system to restore aromaticity, releasing H⁺.
🧠 Exam Technique: Drawing the Horseshoe
- The opening of the horseshoe must face the sp³ carbon holding the new group (C3).
- The ends of the horseshoe must span from C2 around to C6. Do not extend the ends past C2 or C6!
- Never place the + charge on the tetrahedral carbon; it must be delocalised inside the ring.
- Naming the reaction "nitration" scores 0 for the mechanism name—"electrophilic substitution" is required.
Deducing Rf from TLC Plate
1 Mark
✅ Correct Option
D (0.62)
📐 Step-by-Step Calculation
- Measure from the pencil start line to the solvent front (total solvent distance, Lsolvent). On the exam paper, this is approximately 50 mm.
- Measure from the pencil start line to the centre of the spot (spot distance, Lspot ≈ 31 mm).
- Calculate the ratio:
Rf = 31 mm / 50 mm = 0.62
❌ Common Errors
Measuring from the very bottom of the plate instead of the pencil start line. Doing this gives an artificially low ratio (~0.52 or ~0.46), leading directly to distractor options B or C.
Factors Determining TLC Travel Distance
1 Mark
✅ Correct Answer
The balance between solubility in the moving/mobile phase and retention by (affinity for / adsorption onto) the stationary phase.
💡 Key Knowledge
- Mobile phase (solvent): Moves up the plate. Substances that are more soluble in the mobile phase spend more time dissolved and travel further.
- Stationary phase (silica/alumina): Retains compounds through intermolecular attractions (e.g. hydrogen bonding, dipole-dipole). Substances strongly adsorbed travel slower.
Practical Setup Requirement for TLC
1 Mark
✅ Correct Answer
The solvent depth must be below the pencil start line (sample spots).
🧠 Why This Is Essential
If the solvent level is above the start line, the spotted sample will dissolve directly into the solvent pool in the bottom of the beaker rather than travelling up the plate, ruining the chromatogram.
Comparing Polarities and Rf of Isomers
2 Marks
✅ Correct Answer
Relative polarities (Mark 1):
1,2-dinitrobenzene is more polar (OR 1,4-dinitrobenzene is non-polar / less polar).
Explanation (Mark 2):
1,4-dinitrobenzene is less attracted to the polar stationary phase / plate
OR
1,4-dinitrobenzene is more soluble in the non-polar mobile phase (hexane).
💡 Understanding Dipole Cancellation
- In 1,4-dinitrobenzene, the two nitro groups are opposite (para, 180°). The bond dipoles are symmetrical and cancel out, making the molecule non-polar.
- In 1,2-dinitrobenzene, the dipoles act in similar directions (ortho), giving a large net molecular dipole.
- The silica plate is polar (contains polar Si–O and O–H bonds). The more polar 1,2-isomer binds tightly to the plate, resulting in a lower Rf. The non-polar 1,4-isomer interacts weakly with the plate and dissolves well in non-polar hexane, giving a higher Rf.
Effect of Adding a Polar Solvent (Ethyl Ethanoate)
2 Marks
✅ Correct Answer
Is the student correct? Yes (Consequential marking applies; no mark awarded for just stating 'Yes', but stating 'No' scores 0 overall).
- Mark 1: The new solvent mixture is more polar (or ethyl ethanoate is polar).
- Mark 2: The polar isomer (1,2-dinitrobenzene) is more attracted to / more soluble in this solvent (has a stronger affinity for the mobile phase than before).
❌ Common Errors & Traps
- Contradiction Error (CE): Saying "No" instantly awards 0/2.
- Imprecise language: Stating the isomer "bonds to the mobile phase". Penalised! Use terms like attracted to, soluble in, or greater affinity for.
- Failing to mention that ethyl ethanoate increases the overall polarity of the mobile phase.
Topics
Organic Chemistry · 3.3.10 Aromatic Chemistry · 3.3.16 Chromatography
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.