A Level Chemistry: Halogen Compounds (Halogenoarenes) — Revision Notes

Condensed for revision. For the full explanation, use the Arenes and Halogenoarenes study guide, then test yourself with the practice questions. For nucleophilic substitution of halogenoalkanes (the AS background to this topic), see the AS Halogenoalkanes revision notes.

Syllabus: Cambridge International AS & A Level Chemistry 9701, 2025–2027, A Level content: subtopic 31.1 Halogen compounds.

Key words

Term	Meaning
Halogenoarene (aryl halide)	a compound with a halogen atom bonded directly to a benzene ring carbon, e.g. chlorobenzene, C₆H₅Cl
Halogenoalkane	a halogen atom bonded to an sp³ (saturated) carbon, e.g. chloroethane, C₂H₅Cl
Halogen carrier	the catalyst (AlCl₃ or AlBr₃) that polarises the halogen to make a stronger electrophile
Electrophilic substitution	an electrophile replaces an H atom on the ring; the delocalised ring is kept

Producing halogenoarenes (31.1.1)

Reagents and conditions

Arene	Reagent	Catalyst (halogen carrier)	Product
benzene	Cl₂	AlCl₃, anhydrous, room temperature	chlorobenzene
benzene	Br₂	AlBr₃, anhydrous, room temperature	bromobenzene
methylbenzene	Cl₂	AlCl₃, anhydrous, room temperature	2-chloromethylbenzene and 4-chloromethylbenzene

Equations:

C₆H₆ + Cl₂  → C₆H₅Cl + HCl            (AlCl₃ catalyst)
C₆H₆ + Br₂  → C₆H₅Br + HBr            (AlBr₃ catalyst)
C₆H₅CH₃ + Cl₂ → ClC₆H₄CH₃ + HCl       (AlCl₃ catalyst; 2- and 4-isomers)

The HCl or HBr is given off as steamy fumes. Only one H atom on the ring is replaced (mono-substitution) when one mole of halogen is used per mole of arene.

Role of the halogen carrier

The catalyst accepts a lone pair from the halogen molecule and generates the electrophile (Cl⁺, or a strongly polarised Cl–Cl–AlCl₃ complex). It is regenerated at the end:

Cl₂ + AlCl₃     → Cl⁺ + AlCl₄⁻        (electrophile formed)
C₆H₆ + Cl⁺      → C₆H₅Cl + H⁺         (via the intermediate cation)
H⁺ + AlCl₄⁻     → HCl + AlCl₃         (catalyst regenerated)

Adding the three steps gives the overall equation C₆H₆ + Cl₂ → C₆H₅Cl + HCl.

Mechanism in outline (full detail in topic 30.1): the π electrons of the ring attack Cl⁺; a positive intermediate forms in which the delocalised system is partly disrupted; the C–H bond then breaks, H⁺ is lost and the delocalised ring is restored. Substitution, not addition, happens because keeping the ring keeps the aromatic stabilisation.

Why methylbenzene gives two products

The –CH₃ group directs to the 2- and 4-positions (it is electron-donating), so chlorination of the ring gives a mixture of 2-chloromethylbenzene and 4-chloromethylbenzene. The –CH₃ group also makes the ring slightly more reactive than benzene.

Ring or side-chain? The conditions decide where substitution happens:

Conditions	Where Cl goes	Product
Cl₂, AlCl₃ catalyst, no UV light	ring (electrophilic substitution)	2- and 4-chloromethylbenzene
Cl₂, UV light (no catalyst)	side-chain (free-radical substitution)	(chloromethyl)benzene, C₆H₅CH₂Cl

C₆H₅CH₃ + Cl₂ → C₆H₅CH₂Cl + HCl       (UV light; side-chain)

Only the catalyst route makes a halogenoarene; the UV route makes a compound whose Cl is on an sp³ carbon, so it behaves like a halogenoalkane.

Reactivity: halogenoalkane vs halogenoarene (31.1.2)

What is observed

Test	Chloroethane, C₂H₅Cl	Chlorobenzene, C₆H₅Cl
Heat under reflux with NaOH(aq)	hydrolysed to ethanol	no reaction
Warm with aqueous ethanolic AgNO₃	white precipitate of AgCl forms slowly (Cl⁻ released)	no precipitate

C₂H₅Cl + NaOH  → C₂H₅OH + NaCl         (reflux)
C₆H₅Cl + NaOH  → no reaction under the same conditions

Why chlorobenzene is so unreactive

1. p-orbital overlap. A lone pair on the chlorine is in a p orbital that overlaps with the π system of the ring. The lone pair becomes partly delocalised into the ring.
2. Partial double-bond character. This overlap gives the C–Cl bond some double-bond character, so it is shorter and stronger than the C–Cl bond in chloroethane. More energy is needed to break it.
3. The ring repels nucleophiles. The electron-rich π cloud repels an approaching nucleophile such as OH⁻, and the ring blocks attack from the side opposite the chlorine.
4. Aromatic stability. Substituting at a ring carbon would disrupt the stable delocalised system.

In chloroethane none of this applies: the C–Cl carbon is sp³, the bond is polar (Cδ+–Clδ−), the lone pairs stay on Cl, and OH⁻ can attack the δ+ carbon.

Relative ease of hydrolysis (links to topic 33.3): acyl chlorides (fast, cold water) > alkyl chlorides (need heating with NaOH(aq)) > aryl chlorides (no reaction under these conditions).

Exam traps

• The catalyst is AlCl₃ for Cl₂ and AlBr₃ for Br₂. It must be anhydrous; say “halogen carrier” and explain that it generates the electrophile.
• Do not use UV light when asked for a halogenoarene from methylbenzene: UV gives side-chain substitution.
• Methylbenzene gives 2- and **4-**chloromethylbenzene, not the 3-isomer.
• Explain unreactivity with the p-orbital / π-system overlap and the stronger C–Cl bond. “Benzene is stable” on its own is not enough.
• Chlorine is still more electronegative than carbon in chlorobenzene; the answer is bond strength and repulsion of the nucleophile, not “the bond is non-polar”.
• Each substitution gives one HCl per Cl added: balance H and Cl.

Self-test

1. State the reagent and catalyst to convert benzene into bromobenzene, and write the equation.
2. Write equations to show how AlCl₃ generates the electrophile and how it is regenerated.
3. Name the two main products of reacting methylbenzene with Cl₂ in the presence of AlCl₃.
4. How would you change the conditions to chlorinate the methyl group instead? Name the product.
5. Chloroethane and chlorobenzene are each heated under reflux with NaOH(aq). What happens in each case?
6. Explain, in terms of bonding, why the C–Cl bond in chlorobenzene is stronger than in chloroethane.
7. Give one reason, apart from bond strength, why OH⁻ does not substitute chlorobenzene.
8. Put these in order of increasing ease of hydrolysis: chlorobenzene, ethanoyl chloride, chloroethane.
9. (Illustrative yield.) 7.80 g of benzene is chlorinated with Cl₂ and AlCl₃. The yield of chlorobenzene is 60.0%. Calculate the mass of chlorobenzene obtained. (Ar: H = 1.0, C = 12.0, Cl = 35.5)

Answers:

1. Br₂ with AlBr₃ catalyst (anhydrous); C₆H₆ + Br₂ → C₆H₅Br + HBr.
2. Cl₂ + AlCl₃ → Cl⁺ + AlCl₄⁻; H⁺ + AlCl₄⁻ → HCl + AlCl₃.
3. 2-chloromethylbenzene and 4-chloromethylbenzene.
4. Use UV light and no catalyst (free-radical substitution); (chloromethyl)benzene, C₆H₅CH₂Cl.
5. Chloroethane is hydrolysed to ethanol (C₂H₅Cl + NaOH → C₂H₅OH + NaCl); chlorobenzene does not react.
6. A lone pair in a p orbital on Cl overlaps with the ring’s π system and is partly delocalised, giving the C–Cl bond partial double-bond character, so it is shorter and stronger.
7. The electron-rich π cloud of the ring repels the nucleophile (or: attack would disrupt the stable delocalised ring; the ring blocks attack from the far side).
8. Chlorobenzene < chloroethane < ethanoyl chloride.
9. Mr(C₆H₆) = 6(12.0) + 6(1.0) = 78.0, so n = 7.80 / 78.0 = 0.100 mol. Mr(C₆H₅Cl) = 6(12.0) + 5(1.0) + 35.5 = 112.5. 1 : 1 ratio, so theoretical mass = 0.100 × 112.5 = 11.25 g. Actual mass = 0.600 × 11.25 = 6.75 g.

These are original notes written for revision. The percentage yield in question 9 is illustrative. Check the full syllabus wording in the official 9701 syllabus.