Aromatic hydrocarbons

What are aromatic hydrocarbons?

In plain words: benzene, C₆H₆, looks as if it has three double bonds, but it does not behave like an alkene. It does not add bromine as alkenes do, and it prefers to swap a hydrogen for another group (substitution) rather than add across a double bond. The reason is that its six π electrons are spread evenly round the whole ring, which makes the ring unusually stable; any reaction that keeps the ring intact is favoured.

Examples of arenes are benzene, toluene (methylbenzene), naphthalene and biphenyl (NCERT page 318).

Nomenclature and isomerism

The naming of benzene compounds was met in Unit 8, under substituted benzenes (NCERT Section 9.5.1, page 318):

• all six hydrogens of benzene are equivalent, so there is only one monosubstituted product of each kind, for example one methylbenzene (toluene);
• when two hydrogens are replaced, by the same or by different groups, three position isomers are possible: 1,2 (or 1,6) is ortho (o-), 1,3 (or 1,5) is meta (m-) and 1,4 is para (p-).

The dimethylbenzenes are 1,2-dimethylbenzene (o-xylene), 1,3-dimethylbenzene (m-xylene) and 1,4-dimethylbenzene (p-xylene).

Structure of benzene

Benzene was isolated by Michael Faraday in 1825. Its formula, C₆H₆, shows a high degree of unsaturation, yet it is a stable molecule (NCERT Section 9.5.2, pages 319–321). Three observations guided the search for its structure:

• it forms a triozonide, which points to three double bonds;
• it gives only one monosubstituted derivative, so all six carbons and all six hydrogens are alike;
• it gives only one ortho disubstituted product.

Kekulé’s structure. In 1865 August Kekulé proposed a ring of six carbons with alternate single and double bonds and one hydrogen on each carbon. This structure predicts two 1,2-dibromobenzenes, one with the bromines on a double bond and one with them on a single bond, but only one exists. Kekulé answered this with the idea that the double bonds oscillate between the two positions. Even so, his structure could not explain benzene’s unusual stability or its preference for substitution over addition.

Resonance. By valence bond theory, benzene is a resonance hybrid: the two Kekulé structures (A and B) are the main contributors, and the hybrid is drawn as a hexagon with a circle (or dotted circle) in it, the circle standing for six electrons delocalised over the six carbons.

Orbital picture. All six carbons are sp² hybridised. Two sp² orbitals of each carbon overlap with those of its neighbours to form six C–C σ bonds in the plane of the hexagon, and the third overlaps with the s orbital of a hydrogen to form a C–H σ bond. Each carbon keeps one unhybridised p orbital perpendicular to the ring. These could pair up as C1–C2, C3–C4, C5–C6 or as C2–C3, C4–C5, C6–C1, which are the two Kekulé structures (NCERT Fig. 9.7 (a) and (b)). But X-ray diffraction shows that all the C–C distances are equal, so each p orbital overlaps equally with both neighbours: the π electrons form two ring-shaped clouds, like doughnuts, one above and one below the plane of the ring (Fig. 9.7 (c), (d)).

The delocalised π electron cloud is held more strongly by the carbon nuclei than a cloud localised between two carbons, so benzene is more stable than the hypothetical cyclohexatriene. NCERT page 321 says the six π electrons move freely about the six carbon nuclei “instead of any two as shown in Fig. 9.6 (a) or (b)”; the figures meant are 9.7 (a) and (b), since Fig. 9.6 is the orbital picture of ethyne.

Bond lengths. X-ray data show that benzene is planar and that all six C–C bonds are the same length, 139 pm, between the C–C single bond (154 pm) and the C=C double bond (133 pm in NCERT here; 134 pm in Section 9.3.1). If structure A or B were right, there would be two different bond lengths. The absence of pure double bonds explains why benzene is reluctant to add reagents under normal conditions.

Aromaticity and Hückel’s rule

Benzene was taken as the parent “aromatic” compound. Today the name is given to any ring system, with or without a benzene ring, that has these characteristics (NCERT Section 9.5.3, page 321):

1. planarity;
2. complete delocalisation of the π electrons in the ring;
3. (4n + 2) π electrons in the ring, where n is an integer (n = 0, 1, 2, …). This is Hückel’s rule.

NCERT’s examples:

In the cyclopentadienyl anion, the lone pair on the negative carbon joins the four π electrons of the two double bonds; in the cycloheptatrienyl cation, the positive carbon has an empty p orbital, and the six π electrons of the three double bonds spread over all seven carbons. A ring that has a carbon with no p orbital in it (an sp³ CH₂, as in cyclopentadiene) cannot be completely delocalised, and a ring with 4 or 8 π electrons fails the 4n + 2 count (Example 3).

Preparation of benzene

Benzene is commercially isolated from coal tar. In the laboratory it can be prepared by the following methods (NCERT Section 9.5.4, pages 321–322):

• cyclic polymerisation of ethyne: 3HC≡CH → C₆H₆ (red hot iron tube, 873 K; NCERT Section 9.4.4);
• decarboxylation of an aromatic acid: sodium benzoate heated with soda lime, C₆H₅COONa + NaOH → C₆H₆ + Na₂CO₃ (CaO, heat), the same decarboxylation that gives alkanes from the sodium salts of acids (NCERT Section 9.2.2);
• reduction of phenol: phenol vapour passed over heated zinc dust, C₆H₅OH + Zn → C₆H₆ + ZnO.

Physical properties

Aromatic hydrocarbons are non-polar and are usually colourless liquids or solids with a characteristic aroma. Naphthalene balls are used in toilets and to preserve clothes because of their smell and their moth-repellent property. Arenes are immiscible with water but readily miscible with organic solvents, and they burn with a sooty flame (NCERT Section 9.5.5, page 322).

Electrophilic substitution

Arenes are characterised by electrophilic substitution reactions; under special conditions they also undergo addition and oxidation (NCERT page 322). In each substitution an electrophile, E⁺, replaces a hydrogen of the ring.

1. Nitration. Heating benzene with a mixture of concentrated nitric and sulphuric acids (the nitrating mixture) at 323–333 K introduces a nitro group: C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O (nitrobenzene; conc. H₂SO₄).

2. Halogenation. Arenes react with halogens in the presence of a Lewis acid such as anhydrous FeCl₃, FeBr₃ or AlCl₃ to give haloarenes: C₆H₆ + Cl₂ → C₆H₅Cl + HCl (chlorobenzene; anhydrous AlCl₃). With excess chlorine, in the dark and cold, all six hydrogens are replaced in turn: C₆H₆ + 6Cl₂ → C₆Cl₆ + 6HCl (hexachlorobenzene).

3. Sulphonation. Heating benzene with fuming sulphuric acid (oleum, H₂SO₄ containing SO₃) replaces a hydrogen by a sulphonic acid group: C₆H₆ + H₂SO₄ → C₆H₅SO₃H + H₂O (benzenesulphonic acid).

4. Friedel–Crafts alkylation. An alkyl halide with anhydrous aluminium chloride gives an alkylbenzene: C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl (toluene), and C₆H₆ + C₂H₅Cl → C₆H₅C₂H₅ + HCl (ethylbenzene). NCERT asks why 1-chloropropane gives isopropylbenzene, not n-propylbenzene: the primary carbocation CH₃CH₂CH₂⁺ rearranges, by a hydride shift, to the more stable secondary carbocation (CH₃)₂CH⁺, which then attacks the ring (Example 5).

5. Friedel–Crafts acylation. An acyl halide or acid anhydride with a Lewis acid (AlCl₃) gives an acylbenzene (a ketone):

• C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl (acetophenone)
• C₆H₆ + (CH₃CO)₂O → C₆H₅COCH₃ + CH₃COOH

NCERT spells the reaction “Friedel Craft’s” once on page 322; the names are those of Charles Friedel and James Crafts, so it is Friedel–Crafts.

The mechanism

Electrophilic substitution (S_E) is believed to go in three steps (NCERT pages 323–324):

(a) Generation of the electrophile. Anhydrous AlCl₃, a Lewis acid, takes a chloride from the reagent:

• Cl–Cl + AlCl₃ → Cl⁺ + [AlCl₄]⁻ (NCERT calls Cl⁺ the “chloronium ion”, but that name properly belongs to ions such as R₂Cl⁺; free Cl⁺ does not really form: AlCl₃ polarises Cl–Cl so that the outer chlorine acts as Cl⁺);
• CH₃–Cl + AlCl₃ → CH₃⁺ + [AlCl₄]⁻;
• CH₃COCl + AlCl₃ → CH₃CO⁺ + [AlCl₄]⁻ (the acylium ion);
• in sulphonation the electrophile is SO₃ from oleum; unlike the others, sulphonation is reversible (beyond the NCERT text).

In nitration the electrophile is the nitronium ion, NO₂⁺, made by proton transfer from sulphuric acid to nitric acid:

• H₂SO₄ + HNO₃ ⇌ H₂O⁺–NO₂ + HSO₄⁻ (protonated nitric acid);
• H₂O⁺–NO₂ ⇌ H₂O + NO₂⁺.

It is interesting to note that in the process of generation of nitronium ion, sulphuric acid serves as an acid and nitric acid as a base.

NCERT Chemistry Part II (Class 11), Section 9.5.5, page 323

(b) Formation of the carbocation (arenium ion). The electrophile attacks the π cloud and bonds to one carbon, which becomes sp³ hybridised. The result is the σ-complex or arenium ion, stabilised by resonance over the other five carbons; but it has lost its aromatic character, because delocalisation stops at the sp³ carbon.

(c) Removal of a proton. To get the aromatic character back, the σ-complex loses H⁺ from the sp³ carbon to [AlCl₄]⁻ (in halogenation, alkylation and acylation) or to HSO₄⁻ (in nitration). This releases HCl and regenerates AlCl₃, or regenerates H₂SO₄, so both act as catalysts (in acylation the ketone product binds AlCl₃, so a full equivalent is used).

Addition, combustion and the general combustion equation

Under vigorous conditions benzene does add (NCERT page 324):

• hydrogen, at high temperature and/or pressure over nickel: C₆H₆ + 3H₂ → C₆H₁₂ (cyclohexane);
• chlorine, under ultraviolet light at 500 K: C₆H₆ + 3Cl₂ → C₆H₆Cl₆, benzene hexachloride (BHC), a mixture of stereoisomers; NCERT calls it gammaxane, a name that strictly belongs to its γ isomer (lindane).

Note the contrast: with Cl₂ and AlCl₃ in the dark, benzene gives C₆H₅Cl or C₆Cl₆ by substitution; with Cl₂ in UV light it gives C₆H₆Cl₆ by addition.

Combustion. Benzene burns in air with a sooty flame: C₆H₆ + 15/2 O₂ → 6CO₂ + 3H₂O. For any hydrocarbon (NCERT equation 9.83, page 324):

Each carbon takes one O₂ and every four hydrogens take one O₂. For an alkane C_nH_2n+2 this gives n + (2n + 2)/4 = (3n + 1)/2, the same as the alkane equation. NCERT’s equation 9.83 ends with a stray “n” after H₂O, which is a printing slip.

Directive influence of a group

When a monosubstituted benzene is substituted again, the three possible disubstituted products are not formed in equal amounts: either the ortho and para products, or the meta product, are formed mainly (NCERT Section 9.5.6, pages 324–325).

It has also been observed that this behaviour depends on the nature of the substituent already present in the benzene ring and not on the nature of the entering group.

NCERT Chemistry Part II (Class 11), Section 9.5.6, page 324

Ortho and para directing, activating groups. The –OH group of phenol donates a lone pair into the ring by resonance; the resonance structures put a negative charge on the ortho and para carbons, so the electron density is higher there and substitution takes place mainly at those positions. The –I effect of oxygen slightly reduces the electron density, but the overall effect of resonance is to raise it, so –OH activates the ring. Other activating groups: –NH₂, –NHR, –NHCOCH₃, –OCH₃, –CH₃, –C₂H₅ (alkyl groups have no lone pair; they release electrons by their +I effect and hyperconjugation).

Halogens: ortho and para directing but deactivating. In aryl halides the strong –I effect of the halogen lowers the overall electron density of the ring, so further substitution is harder; but resonance still makes the ortho and para positions richer than the meta position. So halogens are deactivating but ortho/para directing.

Meta directing, deactivating groups. –NO₂, –CN, –CHO, –COR, –COOH, –COOR and –SO₃H withdraw electrons. The nitro group, for example, lowers the electron density of the ring by its strong –I effect, and its resonance structures put positive charges on the ortho and para carbons. The meta positions are then the least electron-poor, and the electrophile attacks there.

Solved examples

Common mistakes to avoid

Quick revision summary

• Arenes: benzene and related rings; benzenoid (with a benzene ring) and non-benzenoid. One monosubstituted benzene; three disubstituted (o 1,2, m 1,3, p 1,4).
• Structure: Kekulé (1865) with oscillating double bonds; now a resonance hybrid; six sp² carbons, delocalised π cloud above and below the ring; all C–C 139 pm (C–C 154, C=C 133 pm); planar.
• Hückel: planar, completely delocalised, (4n + 2) π electrons: benzene 6, C₅H₅⁻ 6, C₇H₇⁺ 6, naphthalene 10, anthracene 14.
• Preparation: 3C₂H₂ (873 K); C₆H₅COONa + NaOH (CaO); phenol + Zn dust. Commercially from coal tar.
• Substitution: nitration (conc. HNO₃ + H₂SO₄, NO₂⁺), halogenation (X₂ + FeX₃/AlCl₃), sulphonation (oleum), Friedel–Crafts alkylation and acylation (AlCl₃); mechanism: E⁺, arenium ion, loss of H⁺.
• Addition: 3H₂/Ni → cyclohexane; 3Cl₂/UV, 500 K → BHC. Combustion with a sooty flame.
• Directing: –OH, –NH₂, –OCH₃, –CH₃: activating, o/p; halogens: deactivating, o/p; –NO₂, –CN, –CHO, –COR, –COOH, –SO₃H: deactivating, meta.

Questions students ask

What is the difference between Friedel–Crafts alkylation and acylation?

Friedel–Crafts alkylation puts an alkyl group on a benzene ring, from an alkyl halide, while acylation puts on an acyl group, RCO–, from an acyl chloride or anhydride and gives a ketone; both need a Lewis acid, usually anhydrous AlCl₃. Acylation is usually cleaner: the acylium ion, RC≡O⁺, is stabilised by resonance and does not rearrange, and the C=O group deactivates the ring, so the reaction stops after one substitution. In alkylation the carbocation can rearrange and the alkyl group activates the ring, so di- and trialkylbenzenes form too. So butylbenzene is made by acylation with butanoyl chloride, then reduction of the C=O to CH₂ (Clemmensen, Class 12).

Why does aniline not undergo the Friedel–Crafts reaction?

Aniline does not undergo Friedel–Crafts alkylation or acylation because its –NH₂ group is a Lewis base: the nitrogen lone pair bonds to the catalyst, the Lewis acid AlCl₃, forming a salt-like adduct, C₆H₅NH₂·AlCl₃, in which the nitrogen carries a positive charge. That positive nitrogen withdraws electrons strongly, so the ring is deactivated much as in nitrobenzene, and the catalyst is used up. NCERT meets this case in Class 12, under amines; phenol gives poor yields for a similar reason, as its –OH oxygen also binds AlCl₃.

What is the difference between electrophilic addition and electrophilic substitution?

Electrophilic addition and electrophilic substitution both begin with an electrophile attacking π electrons to form a positively charged intermediate, but in addition that intermediate takes up a nucleophile, so two groups add across a double bond, while in substitution it loses H⁺, so a hydrogen is replaced. Cyclohexene adds bromine at room temperature with no catalyst, giving 1,2-dibromocyclohexane. Benzene needs a Lewis acid such as FeBr₃ and gives bromobenzene, C₆H₅Br, and HBr, keeping its aromatic ring, so on its own it does not decolourise bromine in CCl₄ as cyclohexene does.

What is the difference between benzene and cyclohexatriene?

Cyclohexatriene is the hypothetical molecule of a single Kekulé structure, a C₆H₆ ring with three fixed C=C bonds alternating with C–C bonds, while real benzene has six equal C–C bonds of 139 pm and is about 150 kJ mol^⁻¹ more stable. The figure comes from enthalpies of hydrogenation: cyclohexene, with one C=C, releases about 120 kJ mol^⁻¹ on hydrogenation to cyclohexane, so three C=C bonds should release about 360 kJ mol^⁻¹, but benzene releases only about 208 kJ mol^⁻¹. The difference is benzene’s resonance energy, the extra stability that makes it prefer substitution to addition.

What is the difference between aromatic, antiaromatic and non-aromatic compounds?

An aromatic compound has a cyclic, planar, fully conjugated ring with (4n + 2) π electrons, n = 0, 1, 2 … (not the number of rings), and is unusually stable; an antiaromatic one is also cyclic, planar and fully conjugated but has 4n π electrons, n = 1, 2 …, which makes it unusually unstable; a non-aromatic one lacks at least one of the three features. The cyclopropenyl cation, C₃H₃⁺, with 2 π electrons (n = 0), is aromatic; cyclobutadiene, C₄H₄, with 4, is antiaromatic and highly unstable; cyclohexa-1,3-diene, with two sp³ CH₂ groups in its ring, is non-aromatic. Antiaromaticity is beyond the NCERT text.

What is the difference between benzene, phenyl and benzyl?

Benzene is the compound C₆H₆, while phenyl and benzyl are groups: phenyl, C₆H₅–, is benzene less one hydrogen, joined through a ring carbon, and benzyl, C₆H₅CH₂–, is joined through a CH₂ outside the ring. So chlorobenzene, C₆H₅Cl, has its chlorine on the ring, but benzyl chloride, C₆H₅CH₂Cl, has it on the side chain and, unlike chlorobenzene, undergoes nucleophilic substitution readily, like an alkyl halide. Any arene group joined through a ring carbon is an aryl group, Ar–.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Why does benzene give only one monosubstituted product but three disubstituted products?

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   All six hydrogens are equivalent, so replacing any one gives the same compound. With two substituents, their relative positions can be 1,2 (ortho), 1,3 (meta) or 1,4 (para).

2. Q2. Why is benzene extraordinarily stable though it appears to contain three double bonds (NCERT Exercise 9.10)?

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   Its six π electrons are delocalised over all six carbons, in clouds above and below the ring; the delocalised cloud is held more strongly by the nuclei than localised double bonds would be. Benzene is a resonance hybrid, more stable than either Kekulé structure or the hypothetical cyclohexatriene.

3. Q3. What are the necessary conditions for a system to be aromatic (NCERT Exercise 9.11)?

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   It must be a planar ring, its π electrons must be completely delocalised round the ring, and it must contain (4n + 2) π electrons, where n = 0, 1, 2, … (Hückel’s rule).

4. Q4. Write the equations for the preparation of benzene from (a) sodium benzoate and (b) phenol.

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   (a) C₆H₅COONa + NaOH → C₆H₆ + Na₂CO₃ (soda lime, heat). (b) C₆H₅OH + Zn → C₆H₆ + ZnO (heated zinc dust).

5. Q5. Name the electrophile in (a) nitration, (b) chlorination with AlCl₃, (c) Friedel–Crafts acylation with CH₃COCl.

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   (a) The nitronium ion, NO₂⁺. (b) Cl⁺ (NCERT: chloronium ion), with [AlCl₄]⁻. (c) The acylium ion, CH₃CO⁺.

6. Q6. Why does the arenium ion lose a proton instead of adding a nucleophile?

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   The arenium ion has lost its aromatic character, because delocalisation stops at the sp³ carbon. Losing H⁺ from that carbon restores the aromatic ring; adding a nucleophile would not.

7. Q7. Out of benzene, m-dinitrobenzene and toluene, which undergoes nitration most easily, and why (NCERT Exercise 9.23)?

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   Toluene: the –CH₃ group releases electrons and activates the ring. Benzene comes next; m-dinitrobenzene, with two strongly deactivating –NO₂ groups, reacts least easily.

8. Q8. Arrange in decreasing order of reactivity towards an electrophile: chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene (NCERT Exercise 9.22 (a)).

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   Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene. Each –NO₂ group withdraws more electron density from the ring.

9. Q9. What is formed when benzene reacts with chlorine (a) with anhydrous AlCl₃ in the dark and (b) in ultraviolet light at 500 K?

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   (a) Substitution: chlorobenzene, C₆H₅Cl (and, with excess chlorine, hexachlorobenzene, C₆Cl₆). (b) Addition: benzene hexachloride, C₆H₆Cl₆ (BHC).

10. Q10. Write the balanced equation for the combustion of toluene (NCERT Exercise 9.8 (iv)).

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    C₇H₈ + 9O₂ → 7CO₂ + 4H₂O; from the general equation, x + y/4 = 7 + 2 = 9.