A Data Booklet may be used.
CommonGive the reagents and conditions to make each compound from benzene.
Source: Learners' Alley, Arenes slides
Source: Learners' Alley, Arenes slides
CommonMethylbenzene reacts with Br2. Give the conditions and organic product for
Source: Learners' Alley, Arenes slides
Source: Learners' Alley, Arenes slides
Source: Learners' Alley, Arenes slides
Source: Learners' Alley, Arenes slides
Source: Learners' Alley, Arenes slides
CommonIn the reaction represented by the equation C6H6 + Cl2 → C6H5Cl + HCl, what type of reaction has benzene undergone?
Source: A Level 1993 P4 Q21, via EJC Arenes Topical Quiz, Q5
The molecule of benzene, C6H6, is a regular hexagon in which the π electrons are described as delocalised. Which of the following statements support this structure?
Source: SRJC 2017 Prelim P1 Q22, via EJC Arenes Topical Revision, Q4
A comparison is made of the nitration reactions of some aromatic compounds. How will the reaction temperatures required compare?
| lowest | highest | ||
|---|---|---|---|
| A | benzoic acid | benzene | phenylamine |
| B | phenylamine | benzene | ethylbenzene |
| C | phenylamine | ethylbenzene | benzaldehyde |
| D | benzoic acid | benzaldehyde | ethylbenzene |
Source: DHS 2017 Prelim P1 Q18, via EJC Arenes Topical Revision, Q9
Source: A Level 2004 P4 Q5, via EJC Arenes Topical Quiz, Q2
Describe a simple chemical test to distinguish the following pairs of compounds. State your observations and give balanced equations for the reactions involved.
Source: EJC 2022 JC1 Arenes Tutorial, Section B Q5(a) and (b)
CommonBrCl reacts with benzene in the presence of AlCl3 in an electrophilic substitution reaction to produce bromobenzene.
Source: 2025 NYJC J2 Prelim P3 QA1(c)(i) and (ii)
ChallengeMethylbenzene, C6H5CH3, is required on a large scale as an industrially important organic compound, and is also added to high grade unleaded petrol.
Source: A Level 1997 P2 Q5 (a) to (d) and (f), via EJC Arenes Topical Quiz, Q4
Br2 + FeBr3 → Br+ + FeBr4−[1]
Br2, anhydrous FeBr3, room temperature, in the dark[1]
Products: 2- and 4-bromomethylbenzene[1]
Br2, UV light[1]
Product: (bromomethyl)benzene, C6H5CH2Br[1]
Substitution keeps the delocalised π system and its extra stability: addition would destroy it.[1]
NO2+[1]
HNO3 + 2H2SO4 → NO2+ + 2HSO4− + H3O+[1]
–CH3 is electron-donating: it increases the ring's electron density[1]
and stabilises the carbocation intermediate.[1]
It has no benzylic H.[1]
Belectrophilic substitution[1]
The reaction involves a Cl substituting for one of the H in benzene. Benzene is nucleophilic due to the π electron cloud with high electron density. Hence, benzene attracts electrophilic reagents.
B3 and 4[1]
Cphenylamine, ethylbenzene, benzaldehyde[1]
Electron-donating groups increase the electron density of the benzene ring and are said to activate the ring. The more electron-rich the benzene ring, the more nucleophilic the ring, and hence the lower the reaction temperature.
Electron-donating ability: –NH2 > –C2H5 > –H > –CHO > –CO2H
(i) reagent: Cl2, anhydrous FeCl3, benzene. Condition: room temperature[1]
(ii) reagent: Br2, anhydrous FeBr3, benzene. Condition: room temperature[1]
(iii) reagent: concentrated HNO3, concentrated H2SO4, benzene. Condition: temperature below 55 °C[1]
(i) [1]
(ii) [1]
Test: to separate samples of each compound, add hot KMnO4 in dilute H2SO4. Observations: for benzene, the purple solution remains. For methylbenzene, the purple solution is decolourised, and a white precipitate is observed on cooling.[1]
Equation: C6H5CH3 + 3[O] → C6H5CO2H + H2O[1]
Test: to separate samples of each compound, add Br2 with FeBr3, in the dark. Observations: for cyclohexane, no colour change is observed and no gas is evolved. For benzene, reddish brown Br2 is decolourised, with the evolution of white fumes of HBr.[1]
Equation: C6H6 + Br2 → C6H5Br + HBr (FeBr3)[1]
Electrophilic substitution. BrCl + AlCl3 ⇌ Br+ + AlCl4−
[2]
NoteThree correct steps score 2 marks, and two correct steps score 1 mark.
High temperature.[1]
By Le Chatelier's Principle, the system would attempt to counter the high temperature by absorbing heat, hence favouring the forward reaction, which is endothermic.[1]
Since the forward reaction leads to the formation of more gaseous particles, a low pressure would favour a high equilibrium yield.[1]
By Le Chatelier's Principle, the system would attempt to counter the low pressure by producing more gaseous particles, hence favouring the forward reaction.[1]