Halogen Compounds (A Level)

0. What Examiners Want

This guide covers Topic 31 of Cambridge International AS & A Level Chemistry 9701 (2025–27) and follows the learning outcomes below, word for word. The right-hand column shows where each one is taught. It builds on Topic 15 (halogenoalkanes), Topic 29 (naming arenes) and Topic 30 (the reactions of arenes). The reactivity of an acyl chloride compared with these two is the subject of Topic 33.

1. Halogenoarenes: Structure and Names

A halogenoarene (also called an aryl halide) is an arene in which one or more ring hydrogen atoms have been replaced by halogen atoms. The halogen is bonded directly to a carbon of the benzene ring. The simplest examples are chlorobenzene, C₆H₅Cl, and bromobenzene, C₆H₅Br.

The names follow the rules in Topic 29:

• A halogen on a benzene ring is a prefix: chlorobenzene, bromobenzene.
• If another group is present, number the ring so that the principal group (or the parent, such as methylbenzene) is carbon 1, in the direction that gives the lowest numbers. So the two isomers of C₇H₇Cl with chlorine on the ring are 2-chloromethylbenzene and 4-chloromethylbenzene. (They are also written as 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene, which mean the same.)
• A halogen on a side chain is named differently: C₆H₅CH₂Cl is (chloromethyl)benzene. The brackets show that the Cl is on the CH₃ carbon, not on the ring.

2. Making Halogenoarenes

Halogenoarenes are made by electrophilic substitution of an arene (the reaction in Topic 30). Chlorine or bromine reacts with the arene in the presence of a halogen carrier: anhydrous AlCl₃ for chlorination and AlBr₃ for bromination. Iron(III) chloride and bromide work in the same way.

Chlorination of benzene gives one product. Chlorination of methylbenzene gives a mixture of two isomers.

The mechanism. The halogen carrier makes the electrophile, which attacks the π ring. The method is the same as for bromination in Topic 30.

Chlorination of benzene. AlCl₃ accepts a lone pair from Cl₂, making Cl⁺ (the electrophile) and AlCl₄⁻. The catalyst is regenerated in the last step.

1. Make the electrophile: Cl₂ + AlCl₃ → Cl⁺ + AlCl₄⁻.
2. Attack: a pair of π electrons forms a bond to Cl⁺, giving the positively charged intermediate (the delocalisation is broken, and Cl and H are on the same carbon).
3. Loss of H⁺: the C–H electrons return to the ring. H⁺ + AlCl₄⁻ → HCl + AlCl₃.

Methylbenzene. The CH₃ group is 2,4-directing and activating (Topic 30), so methylbenzene reacts faster than benzene and chlorine enters at position 2 or position 4. The products are 2-chloromethylbenzene and 4-chloromethylbenzene. Bromination with Br₂ and AlBr₃ gives the 2- and 4-bromo equivalents.

Worked example. State the reagents and conditions, and the names of the products, for the chlorination of methylbenzene on the ring. [4]

Reagents: Cl₂ and AlCl₃ (or FeCl₃) as catalyst [1]; conditions: room temperature, anhydrous, in the dark (no UV light) [1]. Products: 2-chloromethylbenzene [1] and 4-chloromethylbenzene [1].

3. Recap: Reactivity of Halogenoalkanes

To explain why halogenoarenes are unreactive, you need to remember why halogenoalkanes are reactive (Topic 15).

• The C–X bond is polar: the carbon is δ+ because the halogen is more electronegative. A nucleophile (a lone-pair donor) attacks the δ+ carbon and replaces the halogen. This is nucleophilic substitution.
• Primary halogenoalkanes react by the SN2 mechanism: the nucleophile attacks the carbon from the back, in one step. Tertiary halogenoalkanes react by SN1: the C–X bond breaks first, giving a carbocation, which is then attacked.
• The rate of hydrolysis is controlled by the bond energy, not the polarity: C–I (240 kJ mol⁻¹) breaks faster than C–Br (280) and C–Cl (340).

The weaker the carbon–halogen bond, the faster the hydrolysis. The C–Cl bond in chlorobenzene is stronger than in chloroethane, and it does not hydrolyse.

The reactions of chloroethane that you must be able to compare are summarised below. The right-hand side of the figure shows the result of using the same reagents with chlorobenzene: none of them works.

Chloroethane undergoes substitution and elimination under the usual conditions. Chlorobenzene is inert to all four.

4. Why Halogenoarenes Are Unreactive

Chlorobenzene does not hydrolyse when warmed with aqueous sodium hydroxide. Industrially, a halogenoarene can be converted to a phenol only with aqueous alkali at very high temperature and pressure. The main reasons are as follows.

The two compounds side by side.

Reason 1: the C–Cl bond is stronger. A chlorine atom has lone pairs of electrons in p orbitals. In chlorobenzene one of these p orbitals overlaps sideways with the p orbitals of the ring carbons, so the lone pair is partly delocalised into the π system of the ring. This gives the C–Cl bond some double-bond character: it is shorter and stronger than the single bond in chloroethane, and so it is harder to break. (The carbon is also sp² hybridised, which holds the bonding electrons closer to the carbon than an sp³ carbon does.)

In chloroethane the C–Cl bond is a simple σ bond. In chlorobenzene the chlorine lone pair overlaps with the ring π system.

Reason 2: the ring repels nucleophiles. A nucleophile such as OH⁻ is negatively charged. The benzene ring has a cloud of π electrons above and below the plane, so it repels the incoming ion. Also, the SN2 route needs the nucleophile to attack the carbon from the side opposite the chlorine; in chlorobenzene that side is inside the ring, so the attack is blocked.

OH⁻ can reach the δ+ carbon of chloroethane from behind. In chlorobenzene the ring is in the way and its electron cloud repels the ion.

Reason 3: no stable carbocation for SN1. The SN1 route would need a phenyl cation, C₆H₅⁺. The empty orbital lies in the plane of the ring, so it is not stabilised by the π system, and the cation is very unstable. (Tertiary halogenoalkanes can do SN1 because their carbocation is stabilised by three alkyl groups.)

Worked example. Explain why chlorobenzene is much less reactive than chloroethane towards nucleophiles. [3]

(1) In chlorobenzene a lone pair on chlorine overlaps with (is delocalised into) the π system of the ring [1]. (2) The C–Cl bond therefore has partial double-bond character, so it is stronger and harder to break than the C–Cl bond in chloroethane [1]. (3) The electron-rich ring repels the nucleophile (OH⁻), so the carbon cannot be attacked [1].

5. The Silver Nitrate Test

A halogenoalkane is hydrolysed to release its halide ion, and the halide ion is then identified with silver nitrate. A halogenoarene gives no halide ion, so there is no precipitate.

Add NaOH(aq) and warm; cool and acidify with dilute nitric acid; add silver nitrate. Only chloroethane gives a precipitate.

1. Warm with aqueous NaOH. CH₃CH₂Cl + OH⁻ → CH₃CH₂OH + Cl⁻.
2. Cool and acidify with dilute HNO₃. This neutralises the excess OH⁻, which would otherwise form a brown precipitate of silver oxide with Ag⁺, hiding the result.
3. Add AgNO₃(aq). Ag⁺(aq) + Cl⁻(aq) → AgCl(s), a white precipitate. A bromoalkane gives a cream precipitate (AgBr) and an iodoalkane a pale yellow one (AgI).

For chlorobenzene the C–Cl bond is not hydrolysed, so there are no Cl⁻ ions in step 3 and no precipitate forms, even after prolonged heating.

Worked example. 0.645 g of chloroethane (M_r = 64.5) is hydrolysed completely and the mixture is treated with excess acidified silver nitrate. Calculate the mass of silver chloride formed (M_r = 143.3). [2]

Amount of C₂H₅Cl = 0.645 / 64.5 = 0.0100 mol [1]. Each mole gives one mole of Cl⁻ and so one mole of AgCl: mass = 0.0100 × 143.3 = 1.43 g [1]. (If chlorobenzene were present in the same flask it would add nothing to this mass.)

6. Ring or Side Chain?

The formula C₇H₇Cl fits three isomers that you should know, and their reactivity depends on where the chlorine is.

Ring chlorine does not hydrolyse; side-chain chlorine does.

• Cl on the ring (2- and 4-chloromethylbenzene): made with Cl₂ and AlCl₃ in the dark. It is a halogenoarene and does not hydrolyse.
• Cl on the side chain ((chloromethyl)benzene): made with Cl₂ and UV light (free-radical substitution, Topic 30). It is a primary halogenoalkane carrying a phenyl group, so it does hydrolyse with warm NaOH(aq) and gives a white precipitate in the silver nitrate test.

This gives a simple way to tell the isomers apart: warm each with NaOH(aq), acidify with HNO₃ and add AgNO₃(aq). Only the side-chain isomer gives a precipitate.

Worked example. Compounds M and N both have the formula C₇H₇Cl. M has Cl on the ring (4-position) and N has Cl on the CH₃ carbon. Describe a test that distinguishes them. [3]

Warm each with aqueous NaOH [1], cool and acidify with dilute HNO₃, then add AgNO₃(aq) [1]. N gives a white precipitate; M gives no precipitate [1].

7. Quick Sheet and Checklist

Checklist. Before the exam you should be able to:

• [ ] define a halogenoarene and name chlorobenzene, 2-chloromethylbenzene, 4-chloromethylbenzene and (chloromethyl)benzene
• [ ] give the reagents and conditions for making a halogenoarene from benzene and from methylbenzene, and state the products
• [ ] write the equation for forming the electrophile (Cl⁺ or Br⁺) and describe the role of the halogen carrier
• [ ] distinguish ring substitution (AlCl₃, dark) from side-chain substitution (UV light)
• [ ] recall why halogenoalkanes are reactive and how bond energy controls the rate
• [ ] explain, in three linked points, why chlorobenzene is much less reactive than chloroethane
• [ ] describe the silver nitrate test, give the purpose of each step, and state the result for each compound
• [ ] distinguish C₇H₇Cl isomers with the chlorine on the ring and on the side chain