Cambridge International AS & A Level Chemistry 9701 · A Level topic 30

Hydrocarbons (A Level)

What this chapter covers30.1

Benzene rings are everywhere in useful chemicals: in drugs such as paracetamol and procaine, in dyes, in explosives, in the monomers of polystyrene and Kevlar. Almost all of them are made by attaching new groups to a benzene ring, or by modifying a group already attached. This chapter covers the reactions that do it.

The previous chapter explained why the benzene ring is unusually stable: its six π electrons are delocalised, and the ring is 152 kJ mol⁻¹ more stable than a structure with three separate C=C bonds would be. That stability controls everything here. Benzene does not react like an alkene: it needs a powerful, positively charged electrophile, usually generated with a catalyst, and it reacts by substitution, so that the delocalised ring survives. The chapter covers five such substitutions, then two reactions that do not keep the ring intact or do not touch it at all — hydrogenation of the ring and oxidation of a side chain — and finally the rules that decide where on the ring a second group goes.

Benzene, methylbenzene and their reactions30.1.1

Benzene, C₆H₆, and methylbenzene, C₆H₅CH₃ (toluene), are colourless liquids that do not mix with water. Both burn with a very smoky flame, because the proportion of carbon in the molecule is high and combustion in air is incomplete. Methylbenzene is the syllabus example of an alkylbenzene: it has a benzene ring and a saturated side chain, and so it can react in two separate places — at the ring, like benzene, or at the CH₃ group, like an alkane.

All the reactions in this chapter are summarised in Figure 30.1. Five of them are electrophilic substitutions of a ring hydrogen. The other two are different in kind: hydrogenation adds hydrogen across the ring and destroys the aromatic system, and oxidation with potassium manganate(VII) attacks the side chain and leaves the ring untouched.

Nitration30.1.1(b), 30.1.2(a)

When benzene is warmed with a mixture of concentrated nitric acid and concentrated sulfuric acid, a pale yellow oil, nitrobenzene, is formed. A hydrogen atom on the ring has been replaced by a nitro group, –NO₂:

The electrophile: the nitronium ion

Nitric acid on its own does not nitrate benzene at a useful rate. Sulfuric acid is the stronger acid of the two, and in the mixture it protonates nitric acid, which then loses water to give the nitronium ion, NO₂⁺. The overall equation for its formation is usually written:

A simpler version, HNO₃ + H₂SO₄ → NO₂⁺ + H₂O + HSO₄⁻, is also correct. NO₂⁺ carries a full positive charge and is a powerful electrophile — powerful enough to attack the delocalised π system of benzene.

The mechanism

The reaction takes place in two steps (Figure 30.2).

1. Attack. Two electrons from the delocalised π system form a new bond to the nitrogen atom of NO₂⁺. The curly arrow starts inside the ring — at the π electrons — and ends at the N atom. The carbon that is attacked now has four σ bonds (to two ring carbons, to H and to NO₂) and becomes sp³. The ring has lost two of its six π electrons: the remaining four are delocalised over the other five carbon atoms, which share the positive charge. This intermediate is drawn as a horseshoe — a partial circle open at the sp³ carbon — with a + inside it.
2. Loss of H⁺. The C–H bond on the sp³ carbon breaks heterolytically. Its two electrons move back into the ring, restoring the complete delocalised π system, and H⁺ is released. The curly arrow starts on the C–H bond and ends inside the ring.

The H⁺ released combines with a hydrogensulfate ion, H⁺ + HSO₄⁻ → H₂SO₄, so the sulfuric acid is regenerated: it is a catalyst.

Why the temperature matters

The syllabus gives the temperature as between 25 °C and 60 °C. Above about 60 °C a second nitro group substitutes, giving mainly 1,3-dinitrobenzene, so the temperature is kept low to obtain the mononitro product. Methylbenzene reacts faster than benzene because the methyl group releases electron density into the ring (section 10), and it gives a mixture of 2-nitromethylbenzene and 4-nitromethylbenzene.

Halogenation of the ring30.1.1(a), 30.1.2(a)

Benzene does not react with bromine water or with bromine in the dark: the Br₂ molecule, with no permanent dipole, is not a strong enough electrophile. In the presence of a halogen carrier — anhydrous aluminium bromide, AlBr₃, or iron(III) bromide, FeBr₃ — bromine reacts at room temperature, the red-brown colour fades, and misty fumes of hydrogen bromide are given off:

Chlorine reacts in the same way with AlCl₃ or FeCl₃ to give chlorobenzene. The products are halogenoarenes (aryl halides), the subject of the next chapter.

How the halogen carrier works

Aluminium in AlBr₃ has only six electrons in its outer shell and can accept a lone pair. It accepts one from a bromine molecule, which polarises and then breaks the Br–Br bond heterolytically, releasing an electrophile that behaves as Br⁺:

The mechanism is then identical to nitration: attack on Br⁺ by the π electrons, a positive intermediate, and loss of H⁺. The H⁺ reacts with AlBr₄⁻ to regenerate the catalyst and form HBr (Figure 30.3).

Friedel–Crafts alkylation and acylation30.1.1(c)(d)

The same kind of catalyst can generate a carbon electrophile, which forms a new C–C bond to the ring. These are the Friedel–Crafts reactions, the most important way of attaching a carbon chain to a benzene ring.

Alkylation

Benzene heated with a chloroalkane and AlCl₃ gives an alkylbenzene. With chloromethane the product is methylbenzene:

AlCl₃ accepts a lone pair from the chlorine of the chloroalkane and removes it as AlCl₄⁻, leaving a carbocation — the electrophile:

Other halogenoalkanes work in the same way: chloroethane gives ethylbenzene and 2-chloropropane gives (1-methylethyl)benzene, known as cumene. An alkyl group releases electrons into the ring, so the product is more reactive than benzene and can be alkylated again; an excess of benzene is used to limit this.

Acylation

With an acyl chloride instead of a chloroalkane, the electrophile is an acylium ion, RCO⁺, and the product is an aromatic ketone. Ethanoyl chloride gives phenylethanone:

In the mechanism the ring attacks the carbon atom of the acylium ion, which carries the positive charge. The –COCH₃ group withdraws electrons from the ring, so the product is less reactive than benzene and a second acylation does not occur (Figure 30.4).

Drawing the mechanism correctly30.1.2(a)(b)

Every electrophilic substitution in this chapter follows the same pattern, whatever the electrophile E⁺. It is worth learning as a single mechanism with a variable part, and it is examined as one: the syllabus names nitration and bromination as the examples, but questions regularly ask for the mechanism with an electrophile you have not seen, such as Cl–C⁺=O or a chlorine atom attacking pyridine.

The electron-rich and electron-poor partners

The ring's delocalised π system is the source of electrons. The electrophile has a vacant orbital or a positive charge on one particular atom — N in NO₂⁺, C in CH₃⁺ and CH₃CO⁺, the halogen in Cl⁺ and Br⁺ — and that atom is the one that forms the new bond.

Step 1: formation of the intermediate

• Arrow 1 starts inside the hexagon, at the delocalised π electrons, and ends at the electron-deficient atom of E⁺. It does not start at a carbon atom or at a corner of the hexagon.
• Bond formed: a new C–E σ bond, using two electrons from the π system.
• Intermediate: the attacked carbon is now sp³ and carries both H and E. The remaining four π electrons are delocalised over the other five carbons. Draw the partial ring as a horseshoe that does not pass through the sp³ carbon, and put the + inside the horseshoe — not on the sp³ carbon and not on E.

Step 2: loss of H⁺

• Arrow 2 starts on the C–H bond of the sp³ carbon (not on the H atom) and ends inside the ring.
• Bond broken: C–H, heterolytically; both electrons return to the ring, which regains its six delocalised π electrons.
• Products: the substituted arene and H⁺. Always show the H⁺; if the question asks, show it combining with HSO₄⁻ or AlX₄⁻ to regenerate the catalyst.

Why substitution wins over addition30.1.2(b)

After step 1 the intermediate has a choice. It could complete an addition, as the carbocation in alkene chemistry does, by bonding to a negative ion. Or it could lose H⁺ and complete a substitution. With benzene, substitution happens every time. The reason is the delocalisation energy.

The addition product would be a cyclohexadiene: two localised C=C bonds and no delocalised ring. Forming it would throw away the stabilisation of about 150 kJ mol⁻¹ that the aromatic ring provides. Losing H⁺, on the other hand, returns the two electrons of the C–H bond to the ring and restores the full delocalised π system. The substitution product keeps the aromatic stability; the addition product would not.

The same idea explains the energy profile of the reaction (Figure 30.5). Step 1 destroys the complete delocalisation, so it has a high activation energy and is the slow, rate-determining step; that is why a strong electrophile is needed at all. Step 2 restores the delocalisation and is fast, with a small activation energy.

A heterocyclic example

Pyridine, C₅H₅N, has the same delocalised six-electron π system as benzene (chapter 29), and it undergoes the same electrophilic substitution: with Cl₂ and AlCl₃ it gives 3-chloropyridine and HCl. The mechanism is drawn exactly as for benzene, with the horseshoe of the intermediate spread over the four carbons and the nitrogen that are not attacked. Only one positive charge is shown, inside the partial ring.

Past-paper practice · Set 30B · The mechanism of electrophilic substitution

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 30B.1[4]

Answer and marking guidance

(i) HBr ✔. (ii) curly arrow from inside the hexagon to Br⁺ AND curly arrow from the C–H bond back into the ring ✔; correct intermediate — partial ring over five carbons, + inside it, H and Br on the sp³ carbon ✔. (iii) electrophilic substitution ✔. Examiner insight: frequent errors in (ii): drawing the partial delocalised ring through the sp³ carbon; placing the + on the sp³ carbon instead of inside the partial ring; starting the second arrow on the H atom instead of on or near the C–H bond.

Question 30B.2[4]

Answer and marking guidance

(i) Cl₂ + AlCl₃ → Cl⁺ + AlCl₄⁻ ✔. (ii) M1 curly arrow from inside the ring to Cl⁺ ✔; M2 intermediate ✔; M3 curly arrow from the C–H bond into the ring AND H⁺ shown ✔. Examiner insight: (i) was done well — the commonest error was forming a chloride ion. In (ii) many good answers were seen; difficulties were curly arrows in the wrong direction, poorly drawn intermediates and leaving out the H⁺.

Question 30B.3[5]

Part (c) of this question, on chlorobenzene and chloroethane, is in the chapter on halogen compounds.

Answer and marking guidance

(a) 120° AND sp² ✔. (b)(i) C₂H₅Cl + AlCl₃ → CH₃CH₂⁺ + AlCl₄⁻ ✔. (ii) arrow from the ring to the CH₂⁺ carbon AND arrow from the C–H bond into the ring ✔; correct intermediate ✔; products C₆H₅C₂H₅ AND H⁺ ✔. Examiner insight: (a) and (b)(i) were usually correct. In (b)(ii) the common errors were an intermediate whose partial ring included the sp³ carbon or carried the + on that carbon, the second arrow starting on the H rather than the C–H bond, and no H⁺ at the end.

Question 30B.4[4]

This part follows 30A.1 in the same question: cumene is (1-methylethyl)benzene.

Answer and marking guidance

Attacking species NO₂⁺ ✔; curly arrow starting inside the hexagon and going to the N of NO₂⁺ ✔; correct intermediate (partial ring, + inside, H and NO₂ on the attacked carbon, which is para to the side chain in the mark scheme's drawing) ✔; second curly arrow from the C–H bond into the ring ✔. Examiner insight: quite discriminating; many accurate intermediates and curly arrows were seen. Some candidates did not use the space next to the cumene structure to show the first attack, by a pair of electrons from the ring, on NO₂⁺.

Question 30B.5[3]

The pyridine structure is given in part (d) of the same question (29A.3).

Answer and marking guidance

M1 curly arrow from the ring (delocalised π system) to Cl⁺ ✔; M2 correct intermediate — partial ring over the other five ring atoms (four C and the N), + inside, H and Cl on the attacked carbon ✔; M3 curly arrow from C–H into the ring AND H⁺ lost ✔. Examiner insight: many clear answers. Errors included an intermediate with two + charges — one in the ring and one on the H — and showing HCl as the leaving group without a Cl⁻ having been formed.

Question 30B.6[3]

Answer and marking guidance

M1 curly arrow from inside the hexagon to the C of Cl–C⁺=O ✔; M2 intermediate: partial ring with +, H and COCl on the sp³ carbon ✔; M3 curly arrow from the C–H bond into the ring AND H⁺ formed ✔. Examiner insight: accessible to those who had learned the mechanism. Errors: the first arrow not going to the carbon atom; a + in the ring and a + on the –COCl group; the second arrow starting on the H rather than on the C–H bond; H⁺ omitted.

Oxidising the side chain30.1.1(e)

The benzene ring is not oxidised by potassium manganate(VII), but an alkyl side chain attached to it is. When methylbenzene is heated under reflux with alkaline potassium manganate(VII), the purple colour fades and a brown precipitate of manganese(IV) oxide forms. The methyl group is oxidised to a carboxylate group, and acidifying the mixture with dilute acid then gives benzoic acid, which crystallises as the solution cools:

The oxidation is complete: it is the carbon atom joined to the ring that ends up as the –COOH carbon, and the rest of any longer side chain is lost. Ethylbenzene, propylbenzene and (1-methylethyl)benzene all give benzoic acid, not phenylethanoic acid or any longer acid. Each alkyl group on the ring is oxidised separately, so 1,3-dimethylbenzene gives benzene-1,3-dicarboxylic acid. Groups that are not alkyl — halogens, –NO₂ — are unaffected: 1-methyl-4-nitrobenzene gives 4-nitrobenzoic acid.

Hydrogenation of the ring30.1.1(f)

Although benzene resists addition, it can be made to add hydrogen. Heated with hydrogen and a platinum or nickel catalyst, benzene is converted to cyclohexane, and methylbenzene to methylcyclohexane:

Three molecules of hydrogen add, one for each pair of π electrons. The conditions are harsher than those for hydrogenating an alkene, and the reaction goes all the way to the cyclohexane ring: once the first molecule of H₂ has added, the delocalisation is lost and the remaining C=C bonds add hydrogen easily. This is the reaction whose enthalpy change, −208 kJ mol⁻¹, was used in chapter 29 as evidence for the delocalisation energy of benzene. An alkyl side chain is unchanged: cumene gives (1-methylethyl)cyclohexane.

Halogenation: ring or side chain?30.1.3

Methylbenzene has hydrogen atoms in two very different places — on the ring and on the methyl group — and chlorine can replace either. Which one is replaced depends entirely on the conditions, because the conditions decide the mechanism (Figure 30.6).

• In the ring: chlorine with an AlCl₃ (or FeCl₃) halogen carrier, at room temperature and in the dark. The catalyst generates Cl⁺, which attacks the electron-rich ring by electrophilic substitution. The methyl group directs the chlorine to the 2- and 4-positions, giving 2-chloromethylbenzene and 4-chloromethylbenzene.
• In the side chain: chlorine in ultraviolet light (or sunlight, or at a high temperature), with no catalyst. UV light splits Cl₂ homolytically into chlorine radicals, which substitute into the CH₃ group by free-radical substitution, exactly as they do with an alkane. The first product is (chloromethyl)benzene, C₆H₅CH₂Cl; with excess chlorine, further substitution gives C₆H₅CHCl₂ and C₆H₅CCl₃.

With a longer side chain there may be more than one side-chain product. (1-methylethyl)benzene, cumene, has two kinds of side-chain hydrogen — the single H on the carbon attached to the ring, and the six equivalent H atoms of the two methyl groups — so free-radical chlorination gives two monochloro products, C₆H₅CCl(CH₃)₂ and C₆H₅CH(CH₃)CH₂Cl. Putting the chlorine on "one methyl group or the other" does not give two different compounds: the two methyl groups are equivalent.

Past-paper practice · Set 30A · Reactions, reagents and conditions

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 30A.1[9]

Answer and marking guidance

(a)(i) D = 2-chloropropane, CH₃CHClCH₃ ✔; E = hydrogen chloride, HCl ✔. (ii) (Friedel–Crafts) alkylation ✔. (b)(i) AlCl₃ (or FeCl₃) as catalyst — a halogen carrier, anhydrous ✔. (ii) the 2- and 4-substituted products: 1-chloro-2-(1-methylethyl)benzene and 1-chloro-4-(1-methylethyl)benzene ✔. (iii) sunlight / UV light, or a temperature above about 100 °C ✔. (iv) C₆H₅CCl(CH₃)₂ and C₆H₅CH(CH₃)CH₂Cl ✔. (c) hot KMnO₄(aq): benzoic acid, C₆H₅COOH ✔; H₂ + Ni, high pressure: (1-methylethyl)cyclohexane — the ring is hydrogenated ✔. Examiner insight: (a)(i) was usually right; common errors for D were 1-chloropropane and invented names. In (a)(ii) many wrote "Friedel–Crafts" without "alkylation", which was not credited. In (b)(i) answers containing water ("Cl₂(aq) + AlCl₃") were not credited because AlCl₃ would be hydrolysed. In (b)(iv) many drew the same molecule twice — chlorine on the "left" and "right" methyl groups is one compound. The reduction in (c) was better known than the oxidation.

Question 30A.2[5]

Part (a) of this question is 30B.1; part (b)(iv), on carbon-13 NMR, belongs to the analytical chapter.

Answer and marking guidance

(i) reagent: chloroethane (or bromoethane / iodoethane) ✔; catalyst: AlCl₃ or FeCl₃ ✔. (ii) A is benzoic acid, C₆H₅COOH (C₆H₅COONa, the salt formed in alkaline solution, also allowed) ✔ — the whole ethyl side chain is oxidised to –COOH. (iii) step 3: LiAlH₄ (reduces COOH to CH₂OH) ✔; step 4: H₂ with a Pt (or Ni) catalyst ✔. Examiner insight: (i) was generally well answered; the commonest error was writing chloroethane as CH₂CH₃Cl. (ii) was found challenging: a common wrong answer was C₆H₅CH₂COOH, from oxidising only the end carbon. (iii) was answered well.

Question 30A.3[3]

Part (b)(iii), the equation for step 2, is in the chapter on acyl chlorides; part (b)(iv) is 30B.6.

Answer and marking guidance

(i) L is benzene-1,3-dicarboxylic acid — both CH₃ groups oxidised to –COOH ✔. (ii) step 1: heat / reflux with acidified or alkaline KMnO₄ (then acidify) ✔; step 2: PCl₅ or SOCl₂ (or heat with PCl₃) ✔. Examiner insight: (i) was usually correct. In (ii) common errors were missing one component of step 1 (acidic or alkaline, or heat) and suggesting HCl or Cl₂ + AlCl₃ for step 2.

Question 30A.4[3]

Fig. 7.1 of this question, showing the whole procaine synthesis, is reproduced with 29A.4 in the previous chapter.

Answer and marking guidance

(i) W is 4-nitromethylbenzene (1-methyl-4-nitrobenzene) ✔ — the ring is nitrated first, while the CH₃ group is still there to direct to the 4-position. (ii) step 1: concentrated HNO₃ and concentrated H₂SO₄ ✔; step 2: hot (alkaline) KMnO₄, followed by acidification ✔. Examiner insight: (i) was usually answered well; a common wrong answer was benzoic acid — oxidising first would give a –COOH group, which directs to the 3-position, not the 4-position. The use of alkaline KMnO₄ to oxidise the side chain was often seen.

Directing effects30.1.4

When a benzene ring already carries one group, a second electrophile does not attack all the remaining positions equally. Some groups send it mainly to the positions next to them (2 and 6) and opposite them (4); others send it to the positions one further away (3 and 5). The syllabus limits the groups to six (Figure 30.7 and Table 30.1).

The same electronic effect changes the rate. Groups that release electrons into the ring make it more attractive to electrophiles, so the ring reacts faster than benzene — dramatically so for phenol and phenylamine, which react with bromine water with no catalyst at all (the chapters on hydroxy compounds and nitrogen compounds). Groups that withdraw electrons make the ring react more slowly than benzene.

Why an alkyl group directs to the 2- and 4-positions

The directing effect can be traced to the stability of the positive intermediate. In the intermediate, the positive charge is spread over three of the five carbons of the horseshoe: those at the 2-, 4- and 6-positions relative to the carbon that was attacked. If the attack is at C2 or C4 of methylbenzene, one of those three carbons is C1 — the carbon carrying the methyl group. A positive charge on C1 is a tertiary carbocation, stabilised by electron donation from the attached alkyl group, so this intermediate is lower in energy and forms faster. If the attack is at C3, the charge never reaches C1 and every contributing carbocation is secondary (Figure 30.8).

Planning the order of steps

Directing effects decide the order in which groups must be introduced when two groups are needed on a ring. The rule is simple: the group already on the ring decides where the next one goes, so the first group must be the one that directs to the position required.

Past-paper practice · Set 30C · Directing effects and planning routes

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 30C.1[3]

Answer and marking guidance

(i) catalyst / halogen carrier ✔. (ii) the –OH group directs to the 2- and 4-positions, and both 2-positions are already occupied, so only the 4-position is available ✔; the –COOH groups direct to their 3-positions, and the position shown is the only such position free ✔ — both effects point to the same carbon. Examiner insight: about half the candidates gave only one reason (the effect of OH or of COOH). Weaker answers described the directing effect of the incoming chlorine, which plays no part.

Question 30C.2[2]

Answer and marking guidance

A tertiary carbocation is more stable than a secondary one ✔; the CH₃ group donates electron density (positive inductive effect) to the adjacent positive carbon, stabilising it ✔. For attack at C2 or C4 the positive charge of the intermediate is shared by C1, which carries the CH₃ group, giving a tertiary cation; for attack at C3 it never reaches C1. Examiner insight: most found this challenging. Many recognised the methyl group as electron-donating, but few went on to discuss the relative stability of the carbocations.

Question 30C.3[5]

Answer and marking guidance

(i) X = (1-methylethyl)benzene, C₆H₅CH(CH₃)₂ ✔; Y = 1-(1-methylethyl)-4-nitrobenzene (NO₂ para to the alkyl group) ✔. (ii) step 1: (CH₃)₂CHBr with FeBr₃ or AlBr₃ ✔; step 2: concentrated HNO₃ and concentrated H₂SO₄ ✔; step 3: Sn and concentrated HCl ✔. Examiner insight: (i) was answered well, though some gave X as nitrobenzene and Y as the 4-alkyl nitro compound — this cannot work, because NO₂ directs to the 3-position. Sn is a reactant, not a catalyst, in the reduction. In (ii) some used C₂H₅CH₂Cl instead of CH₃CHClCH₃, or forgot that the acids must be concentrated.

Question 30C.4[4]

Answer and marking guidance

(i) HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻ (or HNO₃ + H₂SO₄ → NO₂⁺ + H₂O + HSO₄⁻) ✔. (ii) M1 curly arrow from the ring to the N of NO₂⁺ ✔; M2 correct intermediate AND curly arrow from C–H back into the ring ✔. (iii) electrophilic substitution ✔. Examiner insight: in (i) some gave SO₄⁻, NO₂ or H₃SO₄⁺, or unbalanced charges. In (ii) the errors were careless: the first arrow must start inside the ring; the partial ring must be drawn carefully; the + belongs inside the partial ring, not on the sp³ carbon; the second arrow starts on or near the C–H bond. (iii) wrong names included electrophilic addition and nucleophilic substitution.

Reagents and conditions at a glance30.1

Misconceptions and how the topic is assessed30.1

Self-test30.1

Twelve questions on the whole chapter. Each gives its reason once you answer.

Definitions to learn30.1

Summary

• Benzene and methylbenzene react with strong electrophiles by electrophilic substitution: nitration (conc. HNO₃/conc. H₂SO₄, 25–60 °C, NO₂⁺), halogenation (X₂ + AlX₃, X⁺), Friedel–Crafts alkylation (CH₃Cl + AlCl₃, heat, CH₃⁺) and acylation (CH₃COCl + AlCl₃, heat, CH₃CO⁺).
• Each catalyst generates the electrophile and is regenerated when H⁺ is lost: H⁺ + HSO₄⁻ → H₂SO₄; H⁺ + AlX₄⁻ → AlX₃ + HX.
• Mechanism: arrow from the π system to E⁺; intermediate with a horseshoe over five carbons and + inside; arrow from C–H into the ring; H⁺ released. Step 1 is slow and rate-determining.
• Substitution predominates because it restores the delocalised π system and its stabilisation; addition would destroy it.
• Hot alkaline KMnO₄, then dilute acid, oxidises any alkyl side chain completely to –COOH. H₂ with Pt or Ni and heat hydrogenates the ring to a cyclohexane.
• Halogen with AlX₃ in the dark substitutes the ring (electrophilic); halogen in UV light substitutes the side chain (free radical).
• –NH₂, –OH, –R direct to 2,4 (and 6) and activate the ring; –NO₂, –COOH, –COR direct to 3 and deactivate it. The order of steps in a synthesis is chosen so the right group is present to direct each new one.

Examination checklist

• Can I give complete reagents and conditions for all six arene reactions and for side-chain halogenation?
• Can I write balanced equations for generating NO₂⁺, Cl⁺, Br⁺, CH₃⁺ and CH₃CO⁺, and for regenerating each catalyst?
• Can I draw the mechanism with both curly arrows correctly placed, the correct intermediate, and H⁺ — including for an unfamiliar electrophile or ring?
• Can I explain why benzene undergoes substitution rather than addition?
• Can I predict the product of oxidising any alkylbenzene, and of hydrogenating any arene?
• Can I choose conditions for ring or side-chain halogenation and draw the distinct products of each?
• Can I state the directing effect of –NH₂, –OH, –R, –NO₂, –COOH and –COR, and use it to decide the order of steps?
• Can I explain why an alkyl group directs to the 2- and 4-positions in terms of carbocation stability?

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