Learners' AlleyDesigned by Khow Willard

Arenes

Set AAnswer key

From the lesson

13 marks

(a)

Br₂, anhydrous FeBr₃, room temperature, dark[1]

(b)

CH₃CH₂Cl, anhydrous AlCl₃, room temperature[1]

Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻[1]

(a)

Br₂, anhydrous FeBr₃, room temperature, in the dark[1]

Products: 2- and 4-bromomethylbenzene[1]

(b)

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]

Practice

23 marks

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]

1Benzene has a simple molecular structure and does not conduct electricity. 2Benzene undergoes substitution reactions easily as a result of its resonance stability. 3The bond order of all C–C bonds in benzene is 1.5, due to resonance, which arises from the continuous overlap of unhybridised p orbitals. 4Hydrogenation of benzene and of hypothetical cyclohexa-1,3,5-triene gives the same product, cyclohexane. Benzene is more stable than cyclohexa-1,3,5-triene due to resonance, so it exists at a lower energy level, resulting in a less exothermic enthalpy of hydrogenation.

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

(a)

(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]

(b)

NO₂⁺[1]

(c)

(i) [1]

(ii) [1]

(a)

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]

(b)

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]

(a)

Br is less electronegative than Cl, hence will react with AlCl₃ to form the electrophile Br⁺, which undergoes electrophilic substitution with benzene.[1]

(b)

Electrophilic substitution. BrCl + AlCl₃ ⇌ Br⁺ + AlCl₄⁻

[2]

NoteThree correct steps score 2 marks, and two correct steps score 1 mark.

(a)

C₆H₅CH₃ + 9O₂ → 7CO₂ + 4H₂O[1]

(b)

CH₃CH₂CH₂CH₂CH₂CH₂CH₃ → + 4H₂[1]

(c)

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]

(d)

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]

(e)

CH₃C₆H₄CH₃ + 6[O] → HO₂CC₆H₄CO₂H + 2H₂O[1]