Ch 1 · Arenes · AS & A Level Chemistry 9701

Topic 1 of 1 in Hydrocarbons — AS & A Level Chemistry 9701 — 67 lessons.

Introduction

Benzene's delocalised ring makes arenes react quite differently from alkenes, favouring substitution reactions that keep the ring intact. This lesson covers the full range of reactions expected for benzene and methylbenzene, the mechanism common to all of them, and how the presence of a substituent already on the ring, or the conditions used, changes both where and how a reaction occurs.

Key ideas — Halogenation of benzene and methylbenzene

Benzene reacts with chlorine or bromine only in the presence of a halogen carrier catalyst such as AlCl_3, FeBr_3 or iron filings (which react in situ to form the halogen carrier). Under these conditions, at room temperature and in the dark, a hydrogen atom on the ring is substituted: C_6H_6 + Cl_2 xrightarrowAlCl_3 C_6H_5Cl + HCl, forming chlorobenzene. Methylbenzene reacts in the same way under the same conditions, substituting a ring hydrogen to give a mixture of 2-chloromethylbenzene and 4-chloromethylbenzene, since the methyl group directs substitution to these positions.

🧑‍🏫 Teacher's explanation

Think of benzene as a very stable, high-security club. The electrons in the benzene ring aren't stuck in single or double bonds; they are 'delocalised' in a cloud above and below the ring. This makes the ring incredibly stable. If you try to add something to it (addition), you break that beautiful, stable cloud, which the molecule hates. Instead, it prefers 'substitution'—swapping one hydrogen for a halogen. It's like a club member swapping a hat for a cap; the member stays in the club, and the club's structure remains intact.

📖 Definitions

🔍 Worked example — Easy

Calculate the molar mass of chlorobenzene, C_6H_5Cl. (Ar: C=12, H=1, Cl=35.5)

🔍 Worked example — Standard

Write the balanced chemical equation for the bromination of benzene in the presence of FeBr_3.

🔍 Worked example — Exam-style

Predict the major products formed when methylbenzene reacts with Cl_2 in the presence of AlCl_3 at room temperature.

⚠️ Common errors

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🎯 Exam question

(a) State the type of reaction that occurs. [1] (b) Draw the structure of the two main organic products formed. [2] (c) Explain why the methyl group directs the bromine to the 2 and 4 positions rather than the 3 position. [2] [5 marks]

Key ideas — Nitration and Friedel–Crafts reactions

Nitration: benzene is warmed with a mixture of concentrated nitric acid and concentrated sulfuric acid at about 50 °C, substituting a ring hydrogen with a nitro group: C_6H_6 + HNO_3 xrightarrowH_2SO_4 C_6H_5NO_2 + H_2O, forming nitrobenzene.

🧑‍🏫 Teacher's explanation

Think of benzene as a very stable, high-security club. The six carbon atoms share a 'cloud' of electrons above and below the ring, which acts like a protective shield. Because these electrons are delocalised, the ring is extremely stable and doesn't want to break that shield by doing an 'addition' reaction (which would destroy the stability). Instead, it prefers 'substitution'—where one hydrogen is kicked out and replaced by something else, leaving the protective electron cloud perfectly intact.

📖 Definitions

🔍 Worked example — Easy

Calculate the molar mass of nitrobenzene (C_6H_5NO_2) given C=12, H=1, N=14, O=16.

🔍 Worked example — Standard

Benzene reacts with CH_3Cl in the presence of AlCl_3 to form methylbenzene and HCl. Write the balanced equation for this Friedel-Crafts alkylation.

🔍 Worked example — Exam-style

A sample of benzene (C_6H_6) is reacted with excess nitric acid and sulfuric acid to produce nitrobenzene (C_6H_5NO_2). If 78.0 g of benzene is used, calculate the mass of nitrobenzene produced. (C=12, H=1, N=14, O=16)

⚠️ Common errors

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🎯 Exam question

Key ideas — Side-chain oxidation and ring hydrogenation

The alkyl side-chain of an alkylbenzene, but not the ring itself, is oxidised by hot, acidified potassium manganate(VII); regardless of the length of the side-chain, it is shortened all the way back to the ring, converting it to a carboxylic acid group directly attached to the ring, e.g. methylbenzene is oxidised to benzoic acid, C_6H_5COOH. In contrast, the ring itself resists this oxidation, and can instead be hydrogenated: benzene reacts with hydrogen gas at high pressure over a nickel catalyst and at around 150–200 °C, adding three moles of H_2 across the ring to give cyclohexane, C_6H_6 + 3H_2 xrightarrowNi mathrmC_6H_12. This requires far more forcing conditions than the hydrogenation of an isolated alkene double bond, reflecting the ring's delocalisation stability.

🧑‍🏫 Teacher's explanation

Think of the benzene ring as a highly stable, reinforced circular fortress. The electrons are not stuck in one place; they are 'delocalised' in a cloud above and below the ring, making the ring itself incredibly resistant to change. It is like a circular wall that is much harder to break than a single straight fence.

📖 Definitions

🔍 Worked example — Standard

Ethylbenzene is reacted with hot, acidified KMnO_4. Write the balanced equation for the oxidation of the side-chain to form the corresponding carboxylic acid.

🔍 Worked example — Exam-style

Benzene is hydrogenated to cyclohexane. Calculate the volume of H_2 gas required to react completely with 0.50 mol of benzene at RTP, assuming V_m = 24 dm^3mol^-1.

⚠️ Common errors

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🎯 Exam question

(a) Identify the organic product formed. [1] (b) Write the balanced chemical equation for this reaction. [2] (c) State the conditions required to convert benzene into cyclohexane. [2] [5 marks]

Key ideas — Mechanism of electrophilic substitution

Every reaction above proceeds by the same two-step electrophilic substitution mechanism. First, a powerful electrophile (generated using the catalyst) is attracted to the ring's delocalised π electron density and forms a new sigma bond to one ring carbon, using two of the six delocalised electrons; this produces a positively charged, non-aromatic cyclohexadienyl (arenium) intermediate, in which the remaining four π electrons are delocalised over only five carbons. Second, a base (such as AlCl_4^- or HSO_4^-) removes the proton from the carbon now bonded to both the electrophile and a hydrogen atom, restoring full delocalisation across all six ring carbons.

🧑‍🏫 Teacher's explanation

Think of benzene as a group of friends standing in a circle, all holding hands to share a single, large, communal blanket (the delocalised π electron cloud). This 'blanket' makes the ring incredibly stable. If a stranger (an electrophile) tries to pull one person away to form a private bond, the 'blanket' is broken, and the group becomes unstable and unhappy (the non-aromatic intermediate). To fix this, the group quickly kicks out a hydrogen atom to allow the 'blanket' to be shared perfectly among all six members again. This is why benzene prefers substitution over addition: addition would permanently tear the blanket, while substitution just temporarily disrupts it before it's restored.

📖 Definitions

🔍 Worked example — Easy

Identify the electrophile in the nitration of benzene using concentrated H_2SO_4 and HNO_3.

🔍 Worked example — Standard

In the bromination of benzene, why is a Lewis acid catalyst like FeBr_3 required?

🔍 Worked example — Exam-style

Compare the stability of benzene to the cyclohexadienyl intermediate. Explain why substitution is preferred over addition.

⚠️ Common errors

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🎯 Exam question

Key ideas — Why substitution predominates over addition

An electrophile could, in principle, add across one ring double bond instead of substituting. Addition would use up two of the six delocalised π electrons permanently and remove two carbons from the delocalised system, destroying most of the ring's delocalisation (resonance) energy. Losing a proton to regenerate the intermediate as a fully delocalised aromatic ring is energetically far more favourable than retaining an addition product with a broken, localised ring, so substitution, which restores aromaticity, dominates overwhelmingly over addition.

🧑‍🏫 Teacher's explanation

Imagine a group of six friends holding hands in a circle, constantly passing a single ball around the circle very quickly. This 'ball' represents the π electrons. Because the ball is moving so fast and is shared by everyone, the group is incredibly stable and energetic. This stability is what we call 'aromaticity'.

📖 Definitions

🔍 Worked example — Easy

Explain why benzene reacts with Br_2 to form bromobenzene and HBr rather than 1,2-dibromocyclohexadiene.

🔍 Worked example — Standard

Calculate the molar mass of benzene, C_6H_6, and explain why it is more stable than a hypothetical cyclohexatriene.

🔍 Worked example — Exam-style

A benzene ring undergoes nitration with HNO_3 in the presence of H_2SO_4. Predict the main product and state the role of the NO_2^+ ion.

⚠️ Common errors

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🎯 Exam question

Key ideas — Side-chain versus ring halogenation, and directing effects

Whether halogenation occurs at the ring or the side-chain of an alkylbenzene depends entirely on the conditions: a halogen carrier catalyst in the dark or at room temperature favours electrophilic substitution on the ring, whereas ultraviolet light and heat, with no halogen carrier present, favours free-radical substitution in the side-chain, exactly as for an alkane, since UV light homolyses the halogen to generate radicals rather than polarising it into an electrophile.

🧑‍🏫 Teacher's explanation

Think of the benzene ring as a highly stable, circular cloud of electrons. Because these electrons are shared across the whole ring (delocalised), the ring is very happy and doesn't want to break that circle to undergo addition reactions like an alkene would. Instead, it prefers 'substitution'—swapping one atom for another to keep the stable ring intact.

All chapters in Hydrocarbons

Arenes