Arenes & phenol: A Level Chemistry questions with answers
Free Arenes & phenol questions for A Level Chemistry (9701), part of Inorganic & organic. Each question has the answer and the marking points an examiner would look for.
electrophilic substitution B1 generation of CH3+B1 curly arrow from ring to electrophile M1 intermediate with broken delocalisation A1 loss of H+ restoring ring A1
5
(b)
lone pair on O delocalised into ring, increasing electron density; 2,4,6-tribromophenol (white precipitate)
O lone pair overlaps with π system B1 ring electron density higher, attracts electrophile B1 2,4,6-tribromophenol B1
3
(c)
ethanol < phenol < ethanoic acid; phenoxide ion stabilised by delocalisation of charge into ring (less than carboxylate)
correct order B1 phenoxide charge delocalised into ring B1 less delocalisation than carboxylate (two O atoms) B1
3
Question 2Structured · 8 marksA* challenge
Phenol reacts with bromine water at room temperature but benzene needs a catalyst to react with bromine.
(a)
Describe what is seen when bromine water is added to phenol, and name the product.
[2]
(b)
Explain why phenol reacts more readily than benzene.
[3]
(c)
Write an equation for the formation of the electrophile in the nitration of benzene.
[1]
(d)
Explain why phenol is a stronger acid than ethanol.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
orange → colourless, white ppt; 2,4,6-tribromophenol
orange to colourless and a white precipitate B1 2,4,6-tribromophenol B1
2
(b)
O lone pair delocalised into ring
a lone pair on O overlaps with the π system M1 increasing the electron density of the ring M1 so the ring polarises Br₂ and attracts electrophiles more strongly M1
CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄⁻ B1 curly arrow from ring to CH₃CO⁺ M1 intermediate with positive charge and broken delocalisation A1 loss of H⁺ restores ring, AlCl₃ + HCl regenerated A1
4
Question 4Structured · 13 marks
A student nitrates 6.90 g of methylbenzene, C₆H₅CH₃, using a mixture of concentrated nitric and sulfuric acids at 30 °C. She obtains 5.65 g of methyl-4-nitrobenzene, CH₃C₆H₄NO₂.
(a)
Write an equation for the formation of the electrophile from the acids.
[2]
(b)
Calculate the percentage yield of methyl-4-nitrobenzene. (Mr: C₆H₅CH₃ = 92; CH₃C₆H₄NO₂ = 137)
[3]
(c)
Name and describe the mechanism of this reaction. State where each curly arrow starts and ends.
[3]
(d)
Explain why methylbenzene is nitrated at a lower temperature than benzene.
[2]
(e)
Suggest why the yield is less than 100%, other than loss during purification.
[1]
(f)
Explain why benzene undergoes substitution rather than addition reactions.
n(methylbenzene) = 6.90/92 = 0.0750 mol C1 theoretical mass = 10.3 g C1 55.0 % A1
3
(c)
electrophilic substitution; arrow ring → N of NO₂⁺; arrow C–H → ring, H⁺ lost
electrophilic substitution B1 arrow from the delocalised ring to the N of NO₂+, forming a positively charged intermediate B1 arrow from the C–H bond into the ring, H+ lost and the delocalised ring restored B1
3
(d)
CH₃ donates electrons; ring more electron-rich; more reactive
the methyl group is electron-donating B1 increasing the electron density of the ring, so it is more susceptible to electrophilic attack B1
2
(e)
2- and 4- isomers / further nitration
a mixture of isomers (2- and 4-nitro) forms / some dinitration occurs B1
1
(f)
delocalised ring stable; addition would destroy delocalisation
the delocalised π system gives extra stability B1 substitution keeps the delocalised ring, addition would destroy it B1
2
Question 5Structured · 7 marks
Benzene undergoes electrophilic substitution.
(a)
Describe the mechanism of the nitration of benzene, including the formation of the electrophile.
[5]
(b)
Explain why benzene does not undergo addition reactions easily.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
NO₂⁺ from HNO₃ + H₂SO₄; attack; intermediate; loss of H⁺
HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻ B1 curly arrow from ring to NO₂⁺ B1 intermediate with broken delocalisation B1 loss of H⁺ B1 aromatic ring restored B1
5
(b)
delocalised stability
delocalised π system is very stable B1 addition would destroy it B1
2
Question 6Structured · 10 marks
This question is about phenol and its derivatives.
(a)
Write an equation for the reaction of phenol with aqueous sodium hydroxide.
[1]
(b)
Explain why phenol does not react with sodium carbonate while ethanoic acid does.
[2]
(c)
Phenol reacts with dilute nitric acid at room temperature. State the products and explain why nitration is easier than for benzene.
[3]
(d)
Describe how phenylamine is converted into an azo dye using phenol, giving reagents and conditions for each step.
[4]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
C₆H₅OH + NaOH → C₆H₅ONa + H₂O
C₆H₅OH + NaOH → C₆H₅O⁻Na⁺ + H₂O B1
1
(b)
phenol weaker acid than carbonic acid; ethanoic stronger
phenol is too weak an acid to protonate carbonate (weaker than carbonic acid) B1 ethanoic acid is stronger and releases CO₂ B1
2
(c)
2- and 4-nitrophenol; OH lone pair delocalised into ring; activated
2-nitrophenol and 4-nitrophenol B1 the oxygen lone pair is delocalised into the ring B1 the higher electron density makes the ring more reactive towards electrophiles B1
3
(d)
NaNO₂ + HCl below 10 °C → benzenediazonium chloride; add to alkaline phenol → coloured azo compound
NaNO₂ and dilute HCl B1 below 10 °C (to prevent decomposition of the diazonium ion) B1 benzenediazonium chloride formed B1 coupled with phenol in alkaline solution giving a yellow-orange azo dye B1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in