Ch 1 · Halogen compounds · AS & A Level Chemistry 9701

Topic 1 of 1 in Halogen compounds — AS & A Level Chemistry 9701 — 48 lessons.

Introduction

A halogenoarene has its halogen atom bonded directly to a benzene ring rather than to a saturated carbon chain. Although chloroethane and chlorobenzene both contain a C-Cl bond, they behave completely differently towards nucleophiles, and the reason lies entirely in how the ring's delocalised electrons interact with the halogen's lone pairs.

Key ideas — Producing a halogenoarene from an arene

A halogenoarene is produced by the electrophilic substitution of an arene with a halogen in the presence of a halogen carrier catalyst, exactly as met for benzene itself: C_6H_6 + Cl_2 xrightarrowAlCl_3 C_6H_5Cl + HCl, forming chlorobenzene. This is the only route to a halogenoarene in the syllabus; unlike a halogenoalkane, a halogenoarene cannot be made from an "arenol" by direct substitution of -OH, since the strength of the C-O bond attached directly to the ring makes that route impractical at this level.

🧑‍🏫 Teacher's explanation

Think of benzene as a highly stable, circular 'fortress' of electrons. While a halogenoalkane is like a house with a single, weak window (the C-X bond) that is easy to break into, benzene is like a reinforced bunker where the electrons are shared across the whole structure. Because these electrons are busy holding the ring together, they don't want to leave to bond with a nucleophile.

📖 Definitions

🔍 Worked example — Easy

Write a balanced chemical equation for the reaction between benzene and bromine to produce bromobenzene, including the necessary catalyst and the byproduct.

🔍 Worked example — Standard

Explain why chlorobenzene is significantly less reactive towards nucleophilic substitution than chloroethane.

🔍 Worked example — Exam-style

A student attempts to react chlorobenzene with aqueous NaOH to produce phenol. Predict the outcome and explain why this reaction is difficult compared to reacting chloroethane with NaOH.

⚠️ Common errors

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🎯 Exam question

Key ideas — Delocalisation raises the C–Cl bond strength in chlorobenzene

In chloroethane, the chlorine atom is bonded to an ordinary sp^3 carbon and simply withdraws electron density inductively, leaving the C-Cl bond as an isolated, fully single, polar bond. In chlorobenzene, one of chlorine's lone pairs occupies a p orbital that is correctly aligned to overlap sideways with the ring's delocalised π system; this lone pair partially delocalises into the ring, so that the C-Cl bond gains partial double-bond character. A bond with partial double-bond character is both shorter and, crucially, stronger than a pure single bond, since more electron density is shared between the carbon and chlorine nuclei than in an isolated C-Cl single bond.

🧑‍🏫 Teacher's explanation

Imagine a single-lane road versus a multi-lane highway. In chloroethane, the chlorine atom is like a car on a single-lane road; it's just there, and the bond is a simple, single connection. However, in chlorobenzene, the chlorine atom is like a car that has access to an extra lane because it's parked next to a massive, flowing highway of electrons (the benzene ring).

📖 Definitions

🔍 Worked example — Easy

Compare the bond enthalpy (strength) of the C-Cl bond in CH_3 extCH_2 extCl and C_6 extH_5 extCl.

🔍 Worked example — Standard

Explain why chlorobenzene is less reactive towards nucleophilic substitution than chloroethane.

🔍 Worked example — Exam-style

A student claims that the C-Cl bond in chlorobenzene is longer than the C-Cl bond in chloroethane. Is this correct? Justify your answer.

⚠️ Common errors

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🎯 Exam question

Key ideas — Effect on reactivity in nucleophilic substitution

Because bond strength (not polarity) governs how readily a bond breaks, the stronger, partially double C-Cl bond in chlorobenzene is far more resistant to heterolytic fission than the weaker, purely single C-Cl bond in chloroethane. Warming chloroethane with aqueous sodium hydroxide readily substitutes -Cl with -OH to give ethanol, CH_3CH_2Cl + OH^- arrow CH_3CH_2OH + Cl^-; chlorobenzene under the same conditions shows no observable reaction, since the delocalisation of the chlorine lone pair into the ring, together with the ring's own electron density, makes the ring carbon far less attractive to nucleophilic attack and the C-Cl bond far harder to break.

🧑‍🏫 Teacher's explanation

Imagine you are trying to pull a single, loose thread out of a sweater. This is like a halogenoalkane (e.g., chloroethane). The bond is a simple single bond, and the electron density is concentrated, making it easy for a nucleophile to 'attack' and push the chlorine out.

📖 Definitions

🔍 Worked example — Easy

Predict the products and state symbols for the reaction of chloroethane with aqueous sodium hydroxide: CH_3CH_2Cl + NaOH(aq) arrow ?

🔍 Worked example — Standard

Explain why chlorobenzene does not undergo nucleophilic substitution with aqueous NaOH under the same conditions as chloroethane.

🔍 Worked example — Exam-style

Compare the bond enthalpy of the C-Cl bond in CH_3CH_2Cl and C_6H_5Cl. Which is higher and why?

⚠️ Common errors

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🎯 Exam question

Key ideas — Two combined effects of the ring on the C–X bond

The benzene ring affects the carbon–halogen bond in two related ways. First, delocalisation of a halogen lone pair into the ring gives the bond partial double-bond character, shortening and strengthening it, as described above. Second, the ring carbon bearing the halogen is sp^2 hybridised, so the electron pair in the sigma bond is held closer to the carbon nucleus than at an sp^3 carbon, further shortening the bond. Both effects act in the same direction, making a halogenoarene's carbon–halogen bond consistently shorter and stronger than the equivalent bond in a comparable halogenoalkane, which is why halogenoarenes are essentially inert to the nucleophilic substitution conditions that readily convert halogenoalkanes to alcohols, nitriles or amines.

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Halogen compounds