Chapter 9 · Hydrocarbons

9.4 Aromatic Hydrocarbons

Benzene, $\mathrm{C_{6}H_{6}}$, is so unsaturated on paper that it ought to add bromine at once — yet it does not. Naming and isomerism of arenes, the structure and resonance of benzene, aromaticity and the Hückel $(4n+2)$ rule, preparation, electrophilic substitution with mechanism, addition and combustion, the directing influence of substituents, and the toxicity of aromatic hydrocarbons.

In 1825 Michael Faraday isolated a sweet-smelling liquid from the oily residue of the gas used to light London's streets. Its formula, $\mathrm{C_{6}H_{6}}$, is missing eight hydrogen atoms compared with hexane — four rings or double bonds' worth of unsaturation — so it "should" react like a super-alkene. It does not. Benzene leaves bromine water and cold permanganate untouched; with bromine and a catalyst it substitutes a hydrogen instead of adding across a double bond, keeping its ring intact. The compounds built on this stubborn ring are called aromatic, after the fragrant oils and resins (oil of wintergreen, vanillin, cinnamon) from which many were first obtained.

This section explains why benzene behaves so differently. It covers the naming and isomerism of benzene derivatives; the Kekulé structures, resonance and delocalised π electrons of benzene; the general conditions for aromaticity (the Hückel $4n+2$ rule); how benzene is prepared; its characteristic reactions — electrophilic substitution (nitration, halogenation, sulphonation and the Friedel–Crafts reactions) with mechanism — and its few addition reactions and combustion; how a group already on the ring decides where and how fast the next one goes; and why benzene and its larger relatives are toxic and carcinogenic.

Naming, Isomerism and the Structure of Benzene

📘 Definition 9.15 — Aromatic hydrocarbons (arenes)
Aromatic hydrocarbons or arenes are hydrocarbons containing one or more benzene rings (benzenoid aromatics), such as benzene $\mathrm{C_{6}H_{6}}$, methylbenzene (toluene) $\mathrm{C_{6}H_{5}CH_{3}}$, naphthalene $\mathrm{C_{10}H_{8}}$ (two fused rings) and anthracene $\mathrm{C_{14}H_{10}}$ (three). A few aromatic compounds have no benzene ring at all (non-benzenoid aromatics, such as the tropylium ion below). The group $\mathrm{C_{6}H_{5}{-}}$ is called phenyl.
Substituted benzenes are named as derivatives of benzene: methylbenzene, ethylbenzene, chlorobenzene, nitrobenzene. For two substituents the positions are given by locants or by the prefixes ortho- (o-, 1,2), meta- (m-, 1,3) and para- (p-, 1,4): the three dimethylbenzenes (xylenes) are 1,2-, 1,3- and 1,4-dimethylbenzene (Chapter 8 gives the full rules).

Because the six positions of a benzene ring are equivalent, there is only one monosubstituted benzene $\mathrm{C_{6}H_{5}X}$, but three disubstituted ones $\mathrm{C_{6}H_{4}X_{2}}$ (o, m, p) — whether the two groups are the same or different — and three trisubstituted ones $\mathrm{C_{6}H_{3}X_{3}}$ (1,2,3-, 1,2,4- and 1,3,5-). Arenes also show chain/position isomerism through their side chains: ethylbenzene $\mathrm{C_{6}H_{5}C_{2}H_{5}}$ is an isomer of the three xylenes, $\mathrm{C_{8}H_{10}}$.

✏️ Example 1 — Aromatic isomers
(a) Draw and name the four aromatic isomers of $\mathrm{C_{8}H_{10}}$. (b) How many trimethylbenzenes are there? Name them. (c) How many aromatic compounds have the formula $\mathrm{C_{6}H_{4}BrCl}$? (d) Each xylene is nitrated so that one more H of the ring is replaced. Which xylene gives only one mononitro product?
(a) One ring ($\mathrm{C_{6}}$) plus two more carbons: as one ethyl group, ethylbenzene; or as two methyl groups, 1,2-, 1,3- and 1,4-dimethylbenzene (o-, m-, p-xylene). 4 isomers.
(b) 3: 1,2,3-, 1,2,4- and 1,3,5-trimethylbenzene (any other numbering is one of these, renumbered).
(c) 3: 1-bromo-2-chlorobenzene, 1-bromo-3-chlorobenzene, 1-bromo-4-chlorobenzene (o, m, p).
(d) Count the different ring H atoms. o-Xylene: two kinds (positions 3/6 and 4/5) → 2 products. m-Xylene: three kinds (2; 4/6; 5) → 3. p-Xylene: all four ring H equivalent → 1 product. p-Xylene.

The structure of benzene. The molecular formula, the existence of only one monosubstituted benzene, and the formation of cyclohexane ($\mathrm{C_{6}H_{12}}$) by adding three molecules of $\mathrm{H_{2}}$ led August Kekulé in 1865 to propose a six-membered ring of carbon atoms with alternating single and double bonds and one H on each carbon. Two facts did not fit a single fixed Kekulé structure. It predicts two different 1,2-disubstituted benzenes (with a C=C or a C–C between the substituents), but only one is known; Kekulé suggested that the double bonds keep shifting between two arrangements. And benzene does not show the addition reactions of alkenes. Modern measurements settle the matter: all six C–C bonds are equal, $139\ \mathrm{pm}$ — between C–C ($154$) and C=C ($134\ \mathrm{pm}$) — and the molecule is a flat regular hexagon.

📘 Definition 9.16 — Resonance and delocalisation in benzene
Benzene is a resonance hybrid of the two Kekulé structures (Section 4.1): neither exists on its own, and the real molecule, with six identical C–C bonds of bond order $1.5$, lies between them. It is drawn as a hexagon with a circle inside.
Orbital picture: each carbon is sp² hybridised and forms three σ bonds (two to neighbouring carbons, one to H) in the plane of the ring, at $120^{\circ}$. Each carbon keeps one unhybridised p orbital perpendicular to the ring. These six p orbitals overlap sideways all the way round the ring — each with both of its neighbours equally — giving two rings of π electron density, one above and one below the plane. The six π electrons are delocalised over all six carbons rather than localised in three double bonds.
Delocalisation lowers the energy: benzene is about $150\ \mathrm{kJ\,mol^{-1}}$ more stable than a hypothetical cyclohexa-1,3,5-triene with three localised double bonds (its resonance energy; Section 5.3 estimated about $160$ from bond enthalpies). Reactions that would destroy the delocalised ring (addition) are therefore unfavourable, while reactions that keep it (substitution) are favoured.
⚠️ The evidence of ozonolysis
Ozonolysis of 1,2-dimethylbenzene (o-xylene) gives three products — ethanedial $\mathrm{OHC{-}CHO}$, 2-oxopropanal $\mathrm{CH_{3}COCHO}$ and butanedione $\mathrm{CH_{3}COCOCH_{3}}$. The Kekulé structure with a C=C between the two methyl-bearing carbons would give only ethanedial and 2-oxopropanal; the one with a C–C there would give only ethanedial and butanedione. All three products appear, so both structures contribute to the real molecule.
📐 Rule 9.10 — Aromaticity: Hückel's rule
A molecule or ion is aromatic — it has a special stability like benzene — if:
  1. it is cyclic and planar;
  2. it is fully conjugated: every ring atom has a p orbital (is sp² or carries a lone pair or empty p orbital) so that the π electrons can be delocalised around the whole ring;
  3. it has $(4n+2)$ π electrons, where $n=0,1,2,\ldots$ — that is $2$, $6$, $10$, $14$, … (Erich Hückel, 1931).
Benzene: planar, six sp² carbons, 6 π electrons ($n=1$) — aromatic. Naphthalene: 10 π electrons ($n=2$) — aromatic. The cyclopentadienyl anion $\mathrm{C_{5}H_{5}^{-}}$ (two C=C plus a lone pair in a p orbital, 6 π) and the cycloheptatrienyl (tropylium) cation $\mathrm{C_{7}H_{7}^{+}}$ (three C=C plus an empty p orbital, 6 π) are aromatic. Cyclopentadiene itself is not: its $\mathrm{CH_{2}}$ carbon is sp³ and breaks the conjugation. Cyclobutadiene (4 π) and cyclooctatetraene (8 π, and tub-shaped rather than planar) are not aromatic.
✏️ Example 2 — Applying Hückel's rule
Decide whether each species is aromatic, giving the number of π electrons and the reason: (a) cyclopropenyl cation $\mathrm{C_{3}H_{3}^{+}}$; (b) cyclopentadienyl cation $\mathrm{C_{5}H_{5}^{+}}$; (c) cycloheptatriene $\mathrm{C_{7}H_{8}}$; (d) tropylium cation $\mathrm{C_{7}H_{7}^{+}}$; (e) pyridine $\mathrm{C_{5}H_{5}N}$ (a benzene ring with one CH replaced by N); (f) cyclooctatetraene dianion $\mathrm{C_{8}H_{8}^{2-}}$ (planar).
Method Check planarity and that every ring atom has a p orbital; then count π electrons: 2 per C=C, 2 for a carbanion lone pair, 0 for a carbocation's empty p orbital; test $4n+2$.
(a), (b) (a) One C=C + an empty p orbital: 2 π ($n=0$) — aromatic. (b) Two C=C + an empty p orbital: 4 π — not aromatic.
(c), (d) (c) The $\mathrm{CH_{2}}$ is sp³: not fully conjugated — not aromatic (6 π electrons, but not in a closed loop). (d) Remove $\mathrm{H^{-}}$ from that $\mathrm{CH_{2}}$: now sp², empty p orbital, 6 π in a planar ring — aromatic.
(e), (f) (e) Three C=C/C=N in a planar sp² ring: 6 π (the N lone pair lies in the ring plane and is not counted) — aromatic. (f) Four C=C + two extra electrons: 10 π ($n=2$), planar — aromatic.
Interactive 1 — Hückel 4n + 2 Checker

Choose a ring, or press ▶ Play to tour all of them. The drawing marks each ring atom's p orbital: one green dot per π electron (one for each carbon of a double bond, two for a carbanion lone pair), an empty circle for an empty p orbital, and a cross for an sp³ atom with no p orbital. The checklist tests the three conditions of Rule 9.10 in turn.

planar, cyclic?—
fully conjugated?—
π electrons—
verdict—

Preparing Benzene

Industrially, benzene comes from coal tar and, today mostly, from petroleum by catalytic reforming (aromatisation of hexane, Section 9.1). Three laboratory methods are:

✏️ Example 3 — Making benzene
(a) What mass of benzene can be obtained from $14.41\ \mathrm{g}$ of sodium benzoate? (b) What mass of zinc is needed to reduce $9.41\ \mathrm{g}$ of phenol? (c) Which salt would you heat with soda lime to obtain methylbenzene? ($M$: $\mathrm{C_{6}H_{5}COONa}$ $144.10$, $\mathrm{C_{6}H_{6}}$ $78.11$, $\mathrm{C_{6}H_{5}OH}$ $94.11$, Zn $65.38\ \mathrm{g\,mol^{-1}}$)
(a) $n=14.41/144.10=0.1000\ \mathrm{mol}$ → $0.1000\ \mathrm{mol}$ benzene → $7.81\ \mathrm{g}$.
(b) $n=9.41/94.11=0.1000\ \mathrm{mol}$; $1:1$ → $0.1000\times65.38=6.54\ \mathrm{g}$ of Zn.
(c) The $\mathrm{{-}COONa}$ must be on the ring where the H is wanted: a sodium methylbenzoate, e.g. sodium 4-methylbenzoate $\mathrm{CH_{3}C_{6}H_{4}COONa}$ (any of the o, m, p isomers gives methylbenzene). Sodium phenylethanoate $\mathrm{C_{6}H_{5}CH_{2}COONa}$ would also work.

Electrophilic Substitution, Addition and Combustion

Physical properties. Arenes are non-polar, colourless liquids or solids with a characteristic smell, immiscible with water but miscible with organic solvents. Benzene boils at $80\ ^{\circ}\mathrm{C}$ and burns with a very sooty flame (high C : H ratio).

The cloud of six delocalised π electrons makes benzene a source of electrons, so it is attacked by electrophiles (Chapter 8). But where an alkene would then add a nucleophile, benzene instead loses $\mathrm{H^{+}}$ and regains its aromatic ring.

📐 Rule 9.11 — Mechanism of electrophilic aromatic substitution
Every electrophilic substitution of benzene, $\mathrm{C_{6}H_{6}+E^{+}\rightarrow C_{6}H_{5}E+H^{+}}$, goes in three steps:
  1. Generation of the electrophile $\mathrm{E^{+}}$ from the reagents, usually with the help of a catalyst (an acid or a Lewis acid).
  2. Attack on the ring (slow): two π electrons of the ring form a σ bond to $\mathrm{E^{+}}$. The carbon attacked becomes sp³, and the ring is left with a positive charge spread over the other five carbons — the arenium ion (σ-complex), a carbocation stabilised by resonance (three contributing structures), but no longer aromatic.
  3. Loss of a proton (fast): a base removes $\mathrm{H^{+}}$ from the sp³ carbon; its electron pair returns to the ring and the aromatic sextet is restored.
📘 Definition 9.17 — The four electrophilic substitutions of benzene
  • Nitration: warming with the nitrating mixture (concentrated $\mathrm{HNO_{3}}$ + concentrated $\mathrm{H_{2}SO_{4}}$) at $323$–$333\ \mathrm{K}$ gives nitrobenzene. Electrophile: the nitronium ion, $\mathrm{NO_{2}^{+}}$: $$\begin{aligned}&\mathrm{HNO_{3}+2H_{2}SO_{4}}\\&\quad\mathrm{\rightleftharpoons NO_{2}^{+}+H_{3}O^{+}+2HSO_{4}^{-}}\end{aligned}$$
  • Halogenation: chlorine or bromine with a Lewis acid catalyst — anhydrous $\mathrm{FeCl_{3}}$, $\mathrm{FeBr_{3}}$ or $\mathrm{AlCl_{3}}$ (or iron filings) — in the dark gives chlorobenzene or bromobenzene: $\mathrm{Cl_{2}+FeCl_{3}\rightarrow Cl^{+}+FeCl_{4}^{-}}$. With excess chlorine the substitution goes on to hexachlorobenzene, $\mathrm{C_{6}Cl_{6}}$.
  • Sulphonation: heating with fuming sulphuric acid (oleum, $\mathrm{H_{2}SO_{4}}$ containing dissolved $\mathrm{SO_{3}}$) gives benzenesulphonic acid, $\mathrm{C_{6}H_{5}SO_{3}H}$; the electrophile is $\mathrm{SO_{3}}$ (or $\mathrm{HSO_{3}^{+}}$). Sulphonation is reversible.
  • Friedel–Crafts reactions (Charles Friedel and James Crafts, 1877), catalysed by anhydrous $\mathrm{AlCl_{3}}$. Alkylation: $\mathrm{C_{6}H_{6}+CH_{3}Cl\xrightarrow{\text{anhyd. }AlCl_{3}}C_{6}H_{5}CH_{3}+HCl}$, electrophile $\mathrm{CH_{3}^{+}}$ (from $\mathrm{CH_{3}Cl+AlCl_{3}\rightarrow CH_{3}^{+}+AlCl_{4}^{-}}$). Acylation: $\mathrm{C_{6}H_{6}+CH_{3}COCl\rightarrow C_{6}H_{5}COCH_{3}+HCl}$ (1-phenylethanone, acetophenone), electrophile the acylium ion $\mathrm{CH_{3}CO^{+}}$; ethanoic anhydride $\mathrm{(CH_{3}CO)_{2}O}$ can be used instead of the acid chloride.

Addition reactions happen only under forcing conditions, because they destroy the aromatic ring. With hydrogen over nickel at high temperature and pressure benzene gives cyclohexane, $\mathrm{C_{6}H_{6}+3H_{2}\rightarrow C_{6}H_{12}}$. With chlorine in ultraviolet light at about $500\ \mathrm{K}$ — no catalyst — it adds three molecules of $\mathrm{Cl_{2}}$, giving benzene hexachloride (BHC), $\mathrm{C_{6}H_{6}Cl_{6}}$ (1,2,3,4,5,6-hexachlorocyclohexane), once used as an insecticide. Combustion: $\mathrm{2C_{6}H_{6}(l)+15O_{2}(g)\rightarrow 12CO_{2}(g)+6H_{2}O(l)}$; for any hydrocarbon, $\mathrm{C_{x}H_{y}+(x+\tfrac{y}{4})O_{2}\rightarrow xCO_{2}+\tfrac{y}{2}H_{2}O}$.

✏️ Example 4 — Electrophiles and products
For each reaction of benzene, name the product and the electrophile, and write the equation for the formation of the electrophile: (a) conc. $\mathrm{HNO_{3}}$/conc. $\mathrm{H_{2}SO_{4}}$; (b) $\mathrm{Br_{2}}$/$\mathrm{FeBr_{3}}$; (c) $\mathrm{CH_{3}CH_{2}Cl}$/anhyd. $\mathrm{AlCl_{3}}$; (d) $\mathrm{CH_{3}COCl}$/anhyd. $\mathrm{AlCl_{3}}$. (e) Why must the $\mathrm{AlCl_{3}}$ be anhydrous? (f) Why does benzene react with $\mathrm{Cl_{2}}$/$\mathrm{FeCl_{3}}$ in the dark by substitution, but with $\mathrm{Cl_{2}}$ in UV light by addition?
(a) Nitrobenzene; $\mathrm{NO_{2}^{+}}$; $\mathrm{HNO_{3}+2H_{2}SO_{4}\rightleftharpoons NO_{2}^{+}+H_{3}O^{+}+2HSO_{4}^{-}}$.
(b) Bromobenzene; $\mathrm{Br^{+}}$; $\mathrm{Br_{2}+FeBr_{3}\rightarrow Br^{+}+FeBr_{4}^{-}}$.
(c) Ethylbenzene, $\mathrm{C_{6}H_{5}CH_{2}CH_{3}}$; $\mathrm{CH_{3}CH_{2}^{+}}$; $\mathrm{CH_{3}CH_{2}Cl+AlCl_{3}\rightarrow CH_{3}CH_{2}^{+}+AlCl_{4}^{-}}$.
(d) Acetophenone (1-phenylethanone), $\mathrm{C_{6}H_{5}COCH_{3}}$; $\mathrm{CH_{3}CO^{+}}$; $\mathrm{CH_{3}COCl+AlCl_{3}\rightarrow CH_{3}CO^{+}+AlCl_{4}^{-}}$.
(e) Water reacts with (hydrolyses) $\mathrm{AlCl_{3}}$, destroying the Lewis acid that generates the electrophile.
(f) $\mathrm{FeCl_{3}}$ produces the electrophile $\mathrm{Cl^{+}}$, and the ionic mechanism ends with loss of $\mathrm{H^{+}}$ (substitution, ring kept). UV light instead splits $\mathrm{Cl_{2}}$ into $\mathrm{Cl^{\bullet}}$ radicals, which add to the ring step by step — a radical addition giving $\mathrm{C_{6}H_{6}Cl_{6}}$.
✏️ Example 5 — Addition and combustion of benzene
(a) What mass of hydrogen is needed to convert $39.06\ \mathrm{g}$ of benzene into cyclohexane? (b) Write the balanced equation for the complete combustion of benzene and check the atom balance. (c) With $\Delta_{c}H^{\circ}(\mathrm{C_{6}H_{6},l})=-3268\ \mathrm{kJ\,mol^{-1}}$, find the heat released per gram. (d) Write the equation for the formation of BHC and give its molar mass. ($M$: $\mathrm{C_{6}H_{6}}$ $78.11$, $\mathrm{H_{2}}$ $2.016\ \mathrm{g\,mol^{-1}}$; Cl $35.45$)
(a) $n=39.06/78.11=0.5001\ \mathrm{mol}$; $3\ \mathrm{H_{2}}$ per mole → $1.500\ \mathrm{mol}\times2.016=3.02\ \mathrm{g}$.
(b) $x=6$, $y=6$: $\mathrm{C_{6}H_{6}+\tfrac{15}{2}O_{2}\rightarrow 6CO_{2}+3H_{2}O}$, i.e. $\mathrm{2C_{6}H_{6}(l)+15O_{2}(g)\rightarrow 12CO_{2}(g)+6H_{2}O(l)}$: C $12=12$, H $12=12$, O $30=24+6$ ✓.
(c) $3268/78.11=41.8\ \mathrm{kJ\,g^{-1}}$ — less than methane's $55.5$, as expected from the lower H content.
(d) $\mathrm{C_{6}H_{6}+3Cl_{2}\xrightarrow[500\,\text{K}]{\text{UV}}C_{6}H_{6}Cl_{6}}$; $M=6(12.01)+6(1.008)+6(35.45)=290.8\ \mathrm{g\,mol^{-1}}$.

Directive Influence of Substituents; Toxicity

When a benzene ring already carries a group, a second electrophile does not attack the five remaining positions at random. Nitrating methylbenzene gives mainly 2- and 4-nitromethylbenzene (o- and p-nitrotoluene), and faster than benzene; nitrating nitrobenzene gives mainly 1,3-dinitrobenzene, and much more slowly. The first group both directs the newcomer and activates or deactivates the ring.

📐 Rule 9.12 — Ortho/para and meta directing groups
  • Ortho- and para-directing, activating groups release electrons to the ring: $\mathrm{{-}NH_{2}}$, $\mathrm{{-}NHR}$, $\mathrm{{-}NHCOCH_{3}}$, $\mathrm{{-}OH}$, $\mathrm{{-}OCH_{3}}$ (by the $+R$ effect of a lone pair on the atom joined to the ring), and alkyl groups $\mathrm{{-}CH_{3}}$, $\mathrm{{-}C_{2}H_{5}}$ (by $+I$ and hyperconjugation). The resonance structures of, for example, phenol place a negative charge at the two ortho positions and the para position, so the electron density there is highest; the ring as a whole is richer in electrons than benzene, so it reacts faster.
  • Meta-directing, deactivating groups withdraw electrons from the ring: $\mathrm{{-}NO_{2}}$, $\mathrm{{-}CN}$, $\mathrm{{-}CHO}$, $\mathrm{{-}COR}$, $\mathrm{{-}COOH}$, $\mathrm{{-}COOR}$, $\mathrm{{-}SO_{3}H}$ (by $-R$ and $-I$). Their resonance structures place a positive charge at the ortho and para positions, so those positions lose the most electron density; the meta positions are the least depleted and are attacked, and the whole reaction is slower than for benzene.
  • Halogens ($\mathrm{{-}F}$, $\mathrm{{-}Cl}$, $\mathrm{{-}Br}$, $\mathrm{{-}I}$) are ortho/para-directing but deactivating: their strong $-I$ effect withdraws electron density from the whole ring (slower reaction), while their $+R$ effect still makes the ortho and para positions the richer ones.
Usually the para product predominates over the ortho, partly because the ortho positions are crowded by the first group.
✏️ Example 6 — Predicting where the second group goes
Name the main product(s): (a) nitration of methylbenzene; (b) nitration of nitrobenzene; (c) chlorination ($\mathrm{Cl_{2}/FeCl_{3}}$) of chlorobenzene; (d) bromination of benzoic acid; (e) nitration of methoxybenzene (anisole). (f) Arrange benzene, nitrobenzene, methylbenzene, chlorobenzene and methoxybenzene in order of increasing rate of nitration. (g) To make 1-bromo-3-nitrobenzene from benzene, should you nitrate first or brominate first?
(a) $\mathrm{{-}CH_{3}}$ is o/p-directing: 1-methyl-2-nitrobenzene and 1-methyl-4-nitrobenzene (o- and p-nitrotoluene).
(b) $\mathrm{{-}NO_{2}}$ is m-directing: 1,3-dinitrobenzene (m-dinitrobenzene); the reaction needs stronger conditions.
(c) Cl is o/p-directing (though deactivating): 1,2- and 1,4-dichlorobenzene, mainly the 1,4-isomer.
(d) $\mathrm{{-}COOH}$ is m-directing: 3-bromobenzoic acid.
(e) $\mathrm{{-}OCH_{3}}$ is strongly activating, o/p: 2- and 4-nitromethoxybenzene (1-methoxy-2-nitro- and 1-methoxy-4-nitrobenzene).
(f) Nitrobenzene $\lt$ chlorobenzene $\lt$ benzene $\lt$ methylbenzene $\lt$ methoxybenzene (strong withdrawer; weak withdrawer by $-I$; none; weak donor by $+I$/hyperconjugation; strong donor by $+R$).
(g) The two groups are meta to each other, so the meta director must already be on the ring: nitrate first (benzene → nitrobenzene), then brominate ($\mathrm{Br_{2}/FeBr_{3}}$) — the $\mathrm{{-}NO_{2}}$ sends Br to the meta position. Brominating first would give o- and p-bromonitrobenzene.
Interactive 2 — Electrophilic Substitution Director

Choose the group already on the ring and the reaction, or press ▶ Play. The ring shows where the group's resonance effect concentrates (δ−) or removes (δ+) electron density; green dots mark the positions the electrophile attacks. The rate gauge compares the reaction with that of benzene itself, and the readouts name the product(s).

group type—
directs to—
rate vs benzene—
main product(s)—
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — limits of the Friedel–Crafts reactions

Not required by the CBSE syllabus. Friedel–Crafts alkylation and acylation are listed by JEE Main, NEET and JEE Advanced; their limits are a standard part of exam questions on them.

(1) Friedel–Crafts reactions fail on strongly deactivated rings (nitrobenzene is even used as a solvent for them) and on rings carrying $\mathrm{{-}NH_{2}}$, whose lone pair binds the $\mathrm{AlCl_{3}}$. (2) In alkylation the carbocation can rearrange: 1-chloropropane with $\mathrm{AlCl_{3}}$ gives mainly (1-methylethyl)benzene (cumene), because the primary $\mathrm{CH_{3}CH_{2}CH_{2}^{+}}$ shifts a hydride to become the more stable secondary $\mathrm{(CH_{3})_{2}CH^{+}}$. (3) Alkylation over-alkylates: the alkylbenzene formed is more reactive than benzene, so di- and tri-alkylbenzenes form too. Acylation avoids (2) and (3) — the acylium ion does not rearrange, and the $\mathrm{{-}COR}$ group deactivates the ring — so a straight chain is best introduced as an acyl group and then reduced (Class 12).

Carcinogenicity and toxicity. Benzene and many aromatic hydrocarbons with several fused rings (polynuclear hydrocarbons) are toxic and carcinogenic. Long exposure to benzene vapour damages the bone marrow and can cause leukaemia, which is why its content in petrol is strictly limited and it is no longer used as a general laboratory solvent (methylbenzene is used instead). Polynuclear hydrocarbons with four or more fused rings, such as benzo[a]pyrene, are formed when organic matter — tobacco, coal, wood, petroleum and grilled food — burns incompletely; in the body they are converted into reactive species that bind to DNA and can trigger cancer. They are major carcinogens in cigarette smoke and urban air pollution.

Exercises

I
Naming, isomerism and structure. Atomic masses: C $12.01$, H $1.008$.
  • 1 Name: (a) $\mathrm{C_{6}H_{5}CH_{2}CH_{3}}$; (b) $\mathrm{C_{6}H_{5}Cl}$; (c) $\mathrm{C_{6}H_{4}(CH_{3})_{2}}$ with the groups at 1 and 4; (d) $\mathrm{C_{6}H_{5}NO_{2}}$.
  • 2 Draw and name the three isomers of dichlorobenzene, using both locants and o/m/p.
  • 3 How many isomers of trichlorobenzene, $\mathrm{C_{6}H_{3}Cl_{3}}$, are there? Name them.
  • 4 Give two pieces of evidence that a single Kekulé structure does not describe benzene.
  • 5 Describe the orbital picture of benzene: hybridisation, σ framework, and the π electron cloud.
  • 6 What is meant by the resonance energy of benzene, and why does it make benzene prefer substitution to addition?
  • 7 Find the molar mass of naphthalene, $\mathrm{C_{10}H_{8}}$, and the number of its π electrons. Which value of $n$ in $4n+2$ does it fit?
  • 8 How many different mononitro products (ring substitution) can ethylbenzene give? And p-xylene?
1 (a) ethylbenzene; (b) chlorobenzene; (c) 1,4-dimethylbenzene (p-xylene); (d) nitrobenzene.
2 1,2-dichlorobenzene (o-), 1,3-dichlorobenzene (m-), 1,4-dichlorobenzene (p-).
3 3: 1,2,3-, 1,2,4- and 1,3,5-trichlorobenzene.
4 All six C–C bonds are the same length ($139\ \mathrm{pm}$), not alternately 154 and 134; only one 1,2-disubstituted isomer exists (one Kekulé form would give two); o-xylene gives three ozonolysis products; benzene does not readily add bromine (any two).
5 Six sp² carbons form a planar hexagon of C–C σ bonds plus six C–H σ bonds at $120^{\circ}$. Each carbon has one p orbital perpendicular to the ring; the six overlap sideways all round, giving delocalised π clouds above and below the ring that hold six electrons.
6 The extra stability (about $150\ \mathrm{kJ\,mol^{-1}}$) of real benzene compared with a structure with three localised C=C bonds. Addition would destroy the delocalised ring and lose this stability; substitution keeps it.
7 $10(12.01)+8(1.008)=128.2\ \mathrm{g\,mol^{-1}}$; five double bonds in the Kekulé structure → 10 π electrons $=4(2)+2$, $n=2$.
8 Ethylbenzene: 3 (2-, 3- and 4-nitro). p-Xylene: 1 (all four ring H are equivalent).
II
Aromaticity and preparation. $M$: $\mathrm{C_{2}H_{2}}$ $26.04$, $\mathrm{C_{6}H_{6}}$ $78.11\ \mathrm{g\,mol^{-1}}$.
  • 1 State Hückel's rule, with all three conditions.
  • 2 Cyclobutadiene is planar and fully conjugated. Is it aromatic? Explain.
  • 3 Explain why the cyclopentadienyl anion is aromatic but cyclopentadiene is not.
  • 4 How many π electrons does the cycloheptatrienyl (tropylium) cation have? Is it aromatic?
  • 5 Cyclooctatetraene has 8 π electrons and a tub shape. Give two reasons why it is not aromatic.
  • 6 Write the equation for the preparation of benzene from sodium benzoate and name the reagent.
  • 7 What mass of benzene could, at most, be obtained from $2.60\ \mathrm{g}$ of ethyne?
  • 8 How is phenol converted into benzene? Write the equation.
1 A cyclic, planar, fully conjugated molecule or ion (a p orbital on every ring atom) with $(4n+2)$ π electrons, $n=0,1,2,\ldots$, is aromatic.
2 No: it has 4 π electrons, which is not $4n+2$ for any whole number $n$.
3 In the anion every carbon is sp² and the lone pair occupies a p orbital: 6 π electrons delocalised round a planar ring. In cyclopentadiene the $\mathrm{CH_{2}}$ carbon is sp³ with no p orbital, so the ring is not fully conjugated.
4 6 (three C=C; the positive carbon has an empty p orbital); planar and fully conjugated — aromatic.
5 It is not planar, so the p orbitals cannot overlap all round the ring; and 8 is not a $4n+2$ number.
6 $\mathrm{C_{6}H_{5}COONa+NaOH\xrightarrow[\Delta]{CaO}C_{6}H_{6}+Na_{2}CO_{3}}$; soda lime.
7 $\mathrm{3C_{2}H_{2}\rightarrow C_{6}H_{6}}$ conserves mass: $2.60\ \mathrm{g}$ ($0.0998\ \mathrm{mol}$ ethyne → $0.0333\ \mathrm{mol}\times78.11=2.60\ \mathrm{g}$).
8 Phenol vapour is passed over heated zinc dust: $\mathrm{C_{6}H_{5}OH+Zn\rightarrow C_{6}H_{6}+ZnO}$.
III
Electrophilic substitution, addition and combustion. $M$: $\mathrm{C_{6}H_{6}}$ $78.11$, $\mathrm{C_{6}H_{12}}$ $84.16\ \mathrm{g\,mol^{-1}}$.
  • 1 What is the nitrating mixture? Write the equation for the formation of the electrophile.
  • 2 What is the role of anhydrous $\mathrm{FeCl_{3}}$ in the chlorination of benzene?
  • 3 Name the product of benzene with (a) $\mathrm{CH_{3}Cl}$/anhyd. $\mathrm{AlCl_{3}}$, (b) ethanoic anhydride/anhyd. $\mathrm{AlCl_{3}}$.
  • 4 Name the reagent and the product of the sulphonation of benzene.
  • 5 Write the three steps of the mechanism for the bromination of benzene ($\mathrm{Br_{2}/FeBr_{3}}$).
  • 6 Why does the arenium ion lose $\mathrm{H^{+}}$ instead of adding $\mathrm{Br^{-}}$, as a carbocation from an alkene would?
  • 7 What mass of cyclohexane is formed when $7.81\ \mathrm{g}$ of benzene is completely hydrogenated?
  • 8 Write the balanced equation for the complete combustion of methylbenzene.
1 Concentrated nitric acid and concentrated sulphuric acid. $\mathrm{HNO_{3}+2H_{2}SO_{4}\rightleftharpoons NO_{2}^{+}+H_{3}O^{+}+2HSO_{4}^{-}}$.
2 It is a Lewis acid: it accepts a chloride ion from $\mathrm{Cl_{2}}$, generating the electrophile, $\mathrm{Cl_{2}+FeCl_{3}\rightarrow Cl^{+}+FeCl_{4}^{-}}$, and is regenerated when $\mathrm{FeCl_{4}^{-}}$ takes the $\mathrm{H^{+}}$ ($\rightarrow$ HCl + $\mathrm{FeCl_{3}}$).
3 (a) Methylbenzene (toluene). (b) Acetophenone (1-phenylethanone), $\mathrm{C_{6}H_{5}COCH_{3}}$, with ethanoic acid.
4 Fuming sulphuric acid (oleum); benzenesulphonic acid, $\mathrm{C_{6}H_{5}SO_{3}H}$.
5 (i) $\mathrm{Br_{2}+FeBr_{3}\rightarrow Br^{+}+FeBr_{4}^{-}}$. (ii) $\mathrm{C_{6}H_{6}+Br^{+}\rightarrow [C_{6}H_{6}Br]^{+}}$ (arenium ion, slow). (iii) $\mathrm{[C_{6}H_{6}Br]^{+}+FeBr_{4}^{-}\rightarrow C_{6}H_{5}Br+HBr+FeBr_{3}}$.
6 Losing $\mathrm{H^{+}}$ restores the aromatic sextet and its resonance energy (about $150\ \mathrm{kJ\,mol^{-1}}$); adding $\mathrm{Br^{-}}$ would give a non-aromatic diene. The substitution product is far more stable.
7 $n=7.81/78.11=0.1000\ \mathrm{mol}$ → $0.1000\times84.16=8.42\ \mathrm{g}$.
8 $\mathrm{C_{7}H_{8}(l)+9O_{2}(g)\rightarrow 7CO_{2}(g)+4H_{2}O(l)}$ ($x+\tfrac{y}{4}=7+2=9$).
IV
Directive influence and toxicity.
  • 1 Classify as ortho/para- or meta-directing: $\mathrm{{-}OH}$, $\mathrm{{-}NO_{2}}$, $\mathrm{{-}CH_{3}}$, $\mathrm{{-}CHO}$, $\mathrm{{-}Cl}$, $\mathrm{{-}COOH}$, $\mathrm{{-}NH_{2}}$, $\mathrm{{-}CN}$.
  • 2 Name the main products of the nitration of chlorobenzene.
  • 3 Name the main product of the bromination ($\mathrm{Br_{2}/FeBr_{3}}$) of nitrobenzene.
  • 4 Use resonance to explain why the $\mathrm{{-}OH}$ group activates the ring and directs to the ortho and para positions.
  • 5 Explain why chlorine is deactivating but ortho/para-directing.
  • 6 Arrange in order of increasing reactivity towards nitration: methylbenzene, benzene, nitrobenzene, phenol.
  • 7 Plan a two-step synthesis of 1-chloro-4-nitrobenzene from benzene, giving the order of steps and the reason.
  • 8 Name two sources of carcinogenic polynuclear aromatic hydrocarbons in everyday life, and say why benzene is no longer used as a school laboratory solvent.
1 o/p: $\mathrm{{-}OH}$, $\mathrm{{-}CH_{3}}$, $\mathrm{{-}Cl}$, $\mathrm{{-}NH_{2}}$. m: $\mathrm{{-}NO_{2}}$, $\mathrm{{-}CHO}$, $\mathrm{{-}COOH}$, $\mathrm{{-}CN}$.
2 1-chloro-2-nitrobenzene and 1-chloro-4-nitrobenzene (mainly the 4-isomer).
3 1-bromo-3-nitrobenzene (m-bromonitrobenzene).
4 A lone pair on O is delocalised into the ring ($+R$); in the resonance structures a negative charge appears on C2, C4 and C6. These positions become electron-rich and attract the electrophile, and the ring overall has more π electron density than benzene, so it reacts faster.
5 Cl is very electronegative: its $-I$ effect withdraws electron density from the whole ring, making it less reactive than benzene. But its lone pairs still donate by $+R$, which puts extra density at the ortho and para positions relative to meta, so those are attacked.
6 Nitrobenzene $\lt$ benzene $\lt$ methylbenzene $\lt$ phenol.
7 Chlorinate first ($\mathrm{Cl_{2}/FeCl_{3}}$ → chlorobenzene), then nitrate (conc. $\mathrm{HNO_{3}}$/$\mathrm{H_{2}SO_{4}}$): Cl directs o/p, and the para product is separated from the ortho. Nitrating first would put Cl meta.
8 Tobacco smoke, vehicle exhaust, burning coal or wood, charred (grilled) food. Benzene is carcinogenic (it can cause leukaemia) and toxic to the bone marrow; methylbenzene is used instead.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET Which species is aromatic? (A) cyclooctatetraene (B) cyclopentadienyl anion (C) cyclobutadiene (D) cyclopentadiene

Check (A) 8 π, non-planar; (C) 4 π; (D) sp³ $\mathrm{CH_{2}}$ breaks conjugation; (B) planar, fully conjugated, 6 π ($n=1$).
Answer (B).

E2. JEE Main In the Friedel–Crafts acylation of benzene with $\mathrm{CH_{3}COCl}$ and anhydrous $\mathrm{AlCl_{3}}$, the electrophile is (A) $\mathrm{CH_{3}^{+}}$ (B) $\mathrm{CH_{3}CO^{+}}$ (C) $\mathrm{Cl^{+}}$ (D) $\mathrm{AlCl_{4}^{-}}$

Idea $\mathrm{CH_{3}COCl+AlCl_{3}\rightarrow CH_{3}CO^{+}+AlCl_{4}^{-}}$: the Lewis acid removes chloride, leaving the acylium ion. $\mathrm{AlCl_{4}^{-}}$ is an anion, not an electrophile.
Answer (B).

E3. Board · 3 marks Write the mechanism of the nitration of benzene.

1 mark Electrophile: $\mathrm{HNO_{3}+2H_{2}SO_{4}\rightleftharpoons NO_{2}^{+}+H_{3}O^{+}+2HSO_{4}^{-}}$.
1 mark Slow step: the π electrons attack $\mathrm{NO_{2}^{+}}$, giving the resonance-stabilised arenium ion $\mathrm{[C_{6}H_{6}NO_{2}]^{+}}$ (one sp³ carbon carrying H and $\mathrm{NO_{2}}$).
1 mark Fast step: $\mathrm{HSO_{4}^{-}}$ removes $\mathrm{H^{+}}$, restoring the aromatic ring: $\mathrm{[C_{6}H_{6}NO_{2}]^{+}+HSO_{4}^{-}\rightarrow C_{6}H_{5}NO_{2}+H_{2}SO_{4}}$.

E4. JEE Main Numerical answer. How many of these groups are meta-directing in electrophilic aromatic substitution? $\mathrm{{-}NO_{2}}$, $\mathrm{{-}OH}$, $\mathrm{{-}CHO}$, $\mathrm{{-}Cl}$, $\mathrm{{-}CN}$, $\mathrm{{-}CH_{3}}$, $\mathrm{{-}COOH}$, $\mathrm{{-}NHCOCH_{3}}$, $\mathrm{{-}SO_{3}H}$

Sort Electron-withdrawing by $-R$ (multiple bond to an electronegative atom at the attached atom): $\mathrm{{-}NO_{2}}$, $\mathrm{{-}CHO}$, $\mathrm{{-}CN}$, $\mathrm{{-}COOH}$, $\mathrm{{-}SO_{3}H}$ → meta. $\mathrm{{-}OH}$, $\mathrm{{-}NHCOCH_{3}}$, $\mathrm{{-}CH_{3}}$, $\mathrm{{-}Cl}$ → o/p.
Answer 5.

E5. JEE Advanced One or more options correct. (A) All the C–C bonds in benzene have the same length. (B) Benzene decolourises bromine water at room temperature. (C) Benzene adds three molecules of chlorine in UV light to give $\mathrm{C_{6}H_{6}Cl_{6}}$. (D) Chlorobenzene is nitrated faster than benzene.

(A) True: $139\ \mathrm{pm}$ (delocalisation).
(B) False: benzene does not add bromine without a catalyst.
(C) True: radical addition gives benzene hexachloride.
(D) False: Cl is deactivating ($-I$), though o/p-directing.
Answer (A), (C).

E6. Board · Case study Methylbenzene (toluene) has replaced benzene as an industrial solvent and is the starting material for the explosive TNT. (i) Show how methylbenzene is made from benzene, naming the catalyst and the electrophile. (2 marks) (ii) Name the main products of the mononitration of methylbenzene. (1 mark) (iii) Why is methylbenzene preferred to benzene as a solvent? (1 mark)

(i) Friedel–Crafts alkylation: $\mathrm{C_{6}H_{6}+CH_{3}Cl\xrightarrow{\text{anhyd. }AlCl_{3}}C_{6}H_{5}CH_{3}+HCl}$; catalyst anhydrous $\mathrm{AlCl_{3}}$; electrophile $\mathrm{CH_{3}^{+}}$.
(ii) 1-Methyl-2-nitrobenzene and 1-methyl-4-nitrobenzene (o- and p-nitrotoluene), since $\mathrm{{-}CH_{3}}$ is o/p-directing.
(iii) Benzene is carcinogenic (linked to leukaemia); methylbenzene, though harmful, is much less toxic in this respect.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How much more stable is benzene than "cyclohexatriene"? Can the resonance energy be measured, rather than just estimated from bond enthalpies?

Why it matters. The stability of aromatic rings explains why they survive in drugs, dyes and polymers — and why they persist in the environment. Calorimetry of hydrogenation, pioneered by George Kistiakowsky in the 1930s, gave the first direct measurement; modern computational chemistry still calibrates against such data.

What to do.

  1. Approximate enthalpies of hydrogenation to cyclohexane are: cyclohexene $-120$, cyclohexa-1,3-diene $-232$, benzene $-208\ \mathrm{kJ\,mol^{-1}}$. Draw an enthalpy-level diagram with cyclohexane at the bottom.
  2. If double bonds did not interact, what would you expect for cyclohexa-1,3-diene and for a cyclohexatriene? Find the stabilisation in each case.
  3. Compare the benzene value with the bond-enthalpy estimate in Section 5.3 (about $161\ \mathrm{kJ\,mol^{-1}}$). Why might the two methods differ?
  4. Benzene's hydrogenation needs high temperature and pressure while cyclohexene's goes at room temperature. Link this to your diagram.

Tools: Hess's law (Section 5.3), Definition 9.16, a thermochemistry database for checking the data.

Hand in: The labelled diagram, both stabilisation energies, and a 150-word explanation.

Going further. Hydrogenating benzene to cyclohexa-1,3-diene (adding one $\mathrm{H_{2}}$) is endothermic. Use the data to find its enthalpy change and explain why the reaction does not stop there.

Instructor note: expected diene $2(-120)=-240$, measured $-232$ → conjugation stabilisation about $8\ \mathrm{kJ\,mol^{-1}}$; expected triene $3(-120)=-360$, measured $-208$ → resonance energy about $152\ \mathrm{kJ\,mol^{-1}}$. Bond-enthalpy estimates use mean values and gas-phase data, so differ by ~10 kJ. Going further: benzene + $\mathrm{H_{2}}$ → diene: $-208-(-232)=+24\ \mathrm{kJ\,mol^{-1}}$ (endothermic); once formed, the diene hydrogenates much faster and exothermically, so it never accumulates.

🧪 Self-check quiz

0/5 correct

Q1. Which of these obeys Hückel's rule and is aromatic?

Q2. The electrophile in the nitration of benzene is

Q3. Which group directs an incoming electrophile to the meta position?

Q4. The catalyst for the Friedel–Crafts methylation of benzene with chloromethane is

Q5. How many structural isomers of dimethylbenzene are there?