Halogen compounds

The single subtopic here is printed under "A Level subject content" in the 9701 syllabus and both of its objectives carry the tier "A Level". None of it is AS, and it is a separate topic from the AS topic 15 of the same name, which covers halogenoalkanes. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper and the whole 9701 bank on this site comes from it, so no practice here is tagged to this topic.

This is one of the two shortest topics in the A Level course, with two objectives. Both concern halogenoarenes, and the second is really a single explanation, worth understanding once properly rather than memorising.

Syllabus points

31.1 Halogen compounds

• Recall the reactions by which halogenoarenes can be produced: substitution of an arene with Cl₂ or Br₂ in the presence of a catalyst, AlCl₃ or AlBr₃, to form a halogenoarene, exemplified by benzene to form chlorobenzene, and methylbenzene to form 2-chloromethylbenzene and 4-chloromethylbenzene.
• Explain the difference in reactivity between a halogenoalkane and a halogenoarene, as exemplified by chloroethane and chlorobenzene.

Making a halogenoarene

The reaction is the electrophilic substitution from topic 30, and the conditions are the ones that matter:

C₆H₆ + Cl₂ → C₆H₅Cl + HCl

with an AlCl₃ catalyst, acting as a halogen carrier, and no ultraviolet light. The equivalent bromination uses Br₂ with AlBr₃.

With methylbenzene, the methyl group is electron donating and therefore 2,4-directing, so the products are 2-chloromethylbenzene and 4-chloromethylbenzene.

The contrast with the side-chain reaction is worth restating here because it is the point at which the two halves of the halogen chemistry meet. Under ultraviolet light and with no catalyst, methylbenzene reacts by a free-radical mechanism in the side-chain, giving (chloromethyl)benzene, which is a halogenoalkane attached to a ring and behaves like a halogenoalkane. Under a halogen carrier the halogen goes into the ring and the product is a halogenoarene, which does not.

So the same two reagents give products of completely different reactivity depending on the conditions, and that is the practical reason the next section matters.

Why a halogenoarene is so much less reactive

Chloroethane, CH₃CH₂Cl, undergoes nucleophilic substitution readily. Warmed with aqueous sodium hydroxide it gives ethanol; with aqueous silver nitrate in ethanol it gives a white precipitate of silver chloride within minutes.

Chlorobenzene, C₆H₅Cl, does not. It does not react with aqueous sodium hydroxide under those conditions, and it gives no precipitate with aqueous silver nitrate, because no chloride ion is released. Substituting the chlorine requires extreme conditions, of the order of 300 °C and 200 atmospheres.

The explanation has three parts and all three earn marks:

1. The lone pair on the chlorine overlaps with the delocalised pi system of the ring. The chlorine's p orbital lies parallel to the ring's p orbitals, so its lone pair is partially delocalised into the ring.
2. This gives the carbon-to-chlorine bond partial double bond character, so the bond is shorter and stronger than the C-Cl bond in chloroethane, and much harder to break.
3. The ring's high electron density repels the incoming nucleophile. A nucleophile is electron rich by definition, so it is repelled by the pi system rather than attracted, and it cannot approach the carbon atom it would need to attack.

Any one of those three alone is a partial answer. The bond strength argument is the one most often given on its own, and the repulsion argument is the one most often left out.

A fourth point, worth having ready if the question asks about the mechanism rather than the reactivity: the carbon bearing the chlorine in chlorobenzene is sp² hybridised and part of a planar ring, so there is no accessible site behind it for a nucleophile to attack, and the sort of backside approach that a primary halogenoalkane allows is geometrically impossible.

The comparison, in a sentence. Chloroethane reacts because its C-Cl bond is a plain polar single bond with an electron-deficient carbon that a nucleophile can reach; chlorobenzene does not because its C-Cl bond is strengthened by delocalisation and its carbon is shielded by an electron-rich ring that repels nucleophiles.

Where this leads

Two consequences are useful in later synthesis questions.

• A halogenoarene is a dead end for nucleophilic substitution. If a route needs an OH group on a benzene ring, it cannot be made by hydrolysing chlorobenzene. Phenol is made instead from a diazonium salt, which is topic 32.
• A halogen attached to a side-chain is not. (Chloromethyl)benzene, C₆H₅CH₂Cl, has an ordinary halogenoalkane C-Cl bond, so it hydrolyses readily and can be converted to an alcohol, a nitrile or an amine in the usual ways. Deciding whether a chlorine in a structure is on the ring or on a side-chain therefore decides whether the whole reaction happens at all.

The same argument extends to acyl chlorides in topic 33, where the order of hydrolysis rates is acyl chloride, then halogenoalkane, then halogenoarene, from very fast to effectively unreactive.

Common mistakes

• Predicting ring substitution under ultraviolet light, or side-chain substitution with a halogen carrier.
• Forgetting that methylbenzene gives two ring products, at positions 2 and 4.
• Explaining the low reactivity of chlorobenzene only by bond strength, and leaving out the repulsion of the nucleophile by the ring.
• Saying the C-Cl bond in chlorobenzene is stronger because chlorine is more electronegative there. The electronegativity is the same; the difference is the delocalisation of the lone pair into the ring.
• Saying chlorobenzene gives a precipitate slowly with silver nitrate. It gives none under those conditions.
• Treating (chloromethyl)benzene as unreactive because it contains a benzene ring. Its C-Cl bond is on the side-chain and behaves normally.