A Data Booklet may be used.
CommonWhat is the correct set of conditions for the conversion of benzene into nitrobenzene?
| acid | temperature | |
|---|---|---|
| A | dilute HNO3 | 100 °C |
| B | concentrated HNO3 | 0 °C |
| C | concentrated HNO3 and concentrated H2SO4 | 10 °C |
| D | concentrated HNO3 and concentrated H2SO4 | 120 °C |
Source: A Level 1999 P3 Q23 (November), via EJC Arenes Topical Quiz, Q6
What is the reason for adding concentrated sulfuric acid to concentrated nitric acid in the preparation of nitrobenzene? C6H6(l) + HNO3(l) → C6H5NO2(l) + H2O(l)
Source: A Level 1999 P3 Q23 (June), via EJC Arenes Topical Quiz, Q7
3-chlorobenzoic acid can be synthesised from benzene in three steps. Which of the following is the best method for this synthesis?
| step 1 | step 2 | step 3 | |
|---|---|---|---|
| A | Cl2, AlCl3 | CH3Cl, AlCl3 | KMnO4, H2SO4 |
| B | Cl2, AlCl3 | KMnO4, H2SO4 | CH3Cl, AlCl3 |
| C | CH3Cl, AlCl3 | KMnO4, H2SO4 | Cl2, AlCl3 |
| D | CH3Cl, AlCl3 | Cl2, AlCl3 | KMnO4, H2SO4 |
Source: CJC 2018 Prelim P1 Q18, via EJC Arenes Topical Revision, Q5
1-(chloromethyl)-4-nitrobenzene, ClCH2C6H4NO2, can be synthesised from benzene via a 3-step synthetic method. Which of the following sequences of reactions, when carried out, gives the best yield of 1-(chloromethyl)-4-nitrobenzene?
| step 1 | step 2 | step 3 | |
|---|---|---|---|
| A | alkylation | free radical substitution | nitration |
| B | nitration | alkylation | free radical substitution |
| C | alkylation | nitration | free radical substitution |
| D | free radical substitution | nitration | alkylation |
Source: RI 2019 Prelim P1 Q17, via EJC Arenes Topical Revision, Q8
CommonMethylbenzene can be used as an additive in unleaded petrol. Methylbenzene, C6H5CH3, is converted into C6H5CO2H by reaction I, into C6H5CH2Br by reaction II, and into A by reaction III. A:
Source: A Level 1998 P1 Q6(c), via EJC Arenes Topical Quiz, Q3(a) and (b)
Source: EJC 2022 JC1 Arenes Tutorial, Section B Q6
In Friedel–Crafts acylation, FeCl3 is employed to attach an acyl group to an aromatic ring via electrophilic substitution. For example, compound A can be prepared by the reaction of 2-nitroanisole with ethanoyl chloride, CH3COCl, in the presence of FeCl3.
→
Source: DHS 2017 Prelim P2 Q5(a)(i) and (iii) (adapted), via EJC Arenes Topical Revision, Section B Q1
ChallengeThe table below contains information about cyclohexene and naphthalene. Cyclohexene contains one carbon–carbon double bond and shows chemical properties common to the other alkenes. Naphthalene, C10H8, shows chemical properties common to aromatic compounds.
| conditions for reaction with hydrogen | product | enthalpy change of hydrogenation / kJ mol−1 | |
|---|---|---|---|
| cyclohexene | room temperature, nickel catalyst | cyclohexane | −118 |
| naphthalene | – | decalin | – |
Source: A Level 2017 P3 Q7 (modified), via EJC 2022 JC1 Arenes Tutorial, Section B Q1
Cconcentrated HNO3 and concentrated H2SO4, 10 °C[1]
Nitration of benzene requires the strong electrophile, NO2+, formed when nitric acid is protonated by the stronger sulfuric acid, followed by loss of water: HNO3 + 2H2SO4 → H2O + NO2+ + 2HSO4−
The nitration reaction is highly exothermic. Mononitration occurs below 55 °C, further nitration to dinitrobenzene can occur above 55 °C and trinitration occurs above 100 °C. Hence the reaction mixture must be cooled and maintained below 55 °C for mononitration.
CIt donates protons to the nitric acid, thus forming NO2+ ions.[1]
The function of sulfuric acid is to aid in the generation of the strong electrophile, NO2+, by first, protonation of nitric acid (sulfuric acid is a stronger Brønsted–Lowry acid).
CCH3Cl, AlCl3 → KMnO4, H2SO4 → Cl2, AlCl3[1]
Calkylation, nitration, free radical substitution[1]
Alkylation gives C6H5CH3. Nitration (–CH3 is 2,4-directing) gives 4-nitromethylbenzene. Free radical substitution then gives ClCH2C6H4NO2.
In B, –NO2 is 3-directing.
I reagent: KMnO4 with dilute H2SO4. Condition: heat under reflux. Type of reaction: side-chain oxidation[1]
II reagent: Br2, excess methylbenzene. Condition: uv light. Type of reaction: free radical (side-chain) substitution[1]
III reagent: Br2, anhydrous FeBr3 catalyst. Condition: room temperature. Type of reaction: electrophilic (aromatic) substitution[1]
The positional isomers are and [1]
4-bromomethylbenzene is more likely to be formed along with A.[1]
B reacts with Cl2 in the absence of uv light to give dense white fumes (of HCl) with a suitable catalyst ⇒ electrophilic substitution ⇒ presence of benzene ring (since it requires a catalyst to react with Cl2).
B can be oxidised by KMnO4 to form C ⇒ presence of alkyl side-chain on benzene.
C is exactly neutralised by one mole of NaOH ⇒ C only has 1 –CO2H group after side-chain oxidation ⇒ B only has 1 alkyl substituent.
B: [1]
C: [1]
C6H5CH2CH3 + 6[O] → C6H5CO2H + 2H2O + CO2[1]
The –NO2 group is a highly deactivating substituent. It reduces the electron density of the benzene ring such that it becomes less susceptible to an electrophilic attack by a rather unreactive electrophile like CH3CO+.[1]
The presence of a second, strongly activating group like –OCH3 increases the electron density of the benzene ring sufficiently for it to be susceptible to electrophilic attack by CH3CO+.[1]
Each of the 10 carbon atoms in naphthalene is sp2 hybridised, with one singly filled 2p orbital perpendicular to the plane of the molecule. A C–C σ bond is formed with neighbouring C atoms via sp2–sp2 overlap.[1]
All 10 p orbitals with one electron each overlap sideways, leading to a delocalised electron cloud of 10 π electrons covering all 10 C atoms. The bonds between the carbon atoms are in between a single C–C and a double C=C bond.[1]