Aromatic Hydrocarbons: Benzene

Q: Why is benzene more stable than cyclohexa-1,3,5-triene?

Because its six pi electrons are delocalised over all six carbons instead of being locked into three separate double bonds. Delocalisation lowers the energy of the system by about 152 kJ/mol, the resonance energy, which shows up as a heat of hydrogenation of -208 kJ/mol instead of the calculated -360 kJ/mol.

Q: What exactly is Huckel's (4n+2) rule?

It states that a cyclic, planar, fully conjugated system is aromatic if it contains (4n+2) delocalised pi electrons, where n is a whole number. The allowed counts are 2, 6, 10, 14 and so on. All four conditions must hold together: cyclic, planar, complete conjugation and the correct electron count.

Q: Why does benzene prefer substitution over addition?

Substitution loses aromaticity only briefly, in the arenium ion, and gets it back when a proton leaves. Addition destroys the delocalised sextet permanently and costs about 152 kJ/mol. The substitution route also has the lower activation barrier, so it wins on both product stability and reaction rate.

Q: Is cyclooctatetraene aromatic or antiaromatic?

Neither. It is non-aromatic. With eight pi electrons it would be antiaromatic if it were planar, so it avoids that by puckering into a tub shape. In the tub the adjacent p orbitals are no longer parallel, delocalisation fails, and the molecule behaves like an ordinary polyene.

Q: What is the arenium ion and why does it matter?

The arenium ion, or sigma complex, is the carbocation formed when an electrophile bonds to a ring carbon. That carbon becomes sp3 and the positive charge is spread over the other five carbons. It matters because every electrophilic substitution of benzene, from nitration to acylation, passes through it, and its stability controls both the rate and the orientation of substitution.

Q: Why does Friedel-Crafts alkylation give rearranged products?

Because the electrophile is a free carbocation, which rearranges by hydride or alkyl shift to the most stable form before it meets the ring. A primary cation from n-propyl chloride shifts a hydride and becomes secondary, so the product is isopropylbenzene. To build a straight chain, acylate first and then reduce with Clemmensen or Wolff-Kishner.

Q: Why does tert-butylbenzene not give benzoic acid on oxidation?

Side-chain oxidation begins by removing a hydrogen from the benzylic carbon. In tert-butylbenzene that carbon is bonded to three methyl groups and the ring, so it has no hydrogen at all. With no benzylic hydrogen there is no way to start the oxidation, and the compound is unaffected even by hot acidified KMnO4.

Q: What are the bond lengths and bond angles in benzene?

All six carbon carbon bonds are equal at 139 pm, between a C-C single bond at 154 pm and a C=C double bond at 134 pm, which corresponds to a bond order of 1.5. The C-H bonds are 109 pm and every bond angle is 120 degrees. All twelve atoms lie in one plane.

Q: Why does benzene burn with a sooty flame?

Benzene is about 92 percent carbon by mass, a much higher proportion than any alkane of comparable size. In an ordinary flame there is not enough oxygen to burn all that carbon, so unburnt carbon particles glow and then escape as soot. The same high carbon content makes acetylene burn smokily.

ChemistryHydrocarbonsFor JEE aspirants

Aromatic hydrocarbons are cyclic, planar, fully conjugated hydrocarbons that are unusually stable and prefer substitution over addition. Benzene, $C_{6} H_{6}$ , is the parent aromatic hydrocarbon: a flat regular hexagon of six $s p^{2}$ carbons in which all six C-C bonds are equal at 139 pm and six $π$ electrons are delocalised above and below the ring. That delocalisation gives benzene a resonance energy of about 152 kJ/mol, which is exactly why benzene undergoes electrophilic substitution instead of addition. For JEE and NEET, the structure of benzene, Huckel's $(4 n + 2)$ rule and the arenium ion mechanism carry the most marks.

Key Formulas and Facts - Quick Reference

Benzene is $C_{6} H_{6}$ . A mononuclear arene follows $C_{n} H_{2 n - 6}$ with $n \geq 6$ .
Degree of unsaturation $D U = \frac{2 n _{C} + 2 - n _{H}}{2}$ . For benzene $D U = 4$ , that is three $π$ bonds plus one ring.
Huckel's rule: a species is aromatic if it is cyclic, planar, fully conjugated and holds $(4 n + 2)$ delocalised $π$ electrons, where $n = 0, 1, 2, 3, \dots$
Resonance energy of benzene $= 360 - 208 = 152$ kJ/mol (calculated minus observed heat of hydrogenation).
Geometry: every C-C bond $= 139$ pm, every C-H bond $= 109$ pm, every bond angle $= 12 0^{\circ}$ , bond order $= 1.5$ .
General electrophilic substitution: $C_{6} H_{6} + E^{+} ⟶ C_{6} H_{5} E + H^{+}$
Nitronium ion: $H N O_{3} + 2 H_{2} S O_{4} ⇌ N O_{2}^{+} + H_{3} O^{+} + 2 H S O_{4}^{-}$
Side-chain oxidation: $C_{6} H_{5} C H_{2} R K M n O_{4} / H^{+}, Δ C_{6} H_{5} C O O H$ , provided at least one benzylic hydrogen exists.
Combustion: $C_{6} H_{6} + \frac{15}{2} O_{2} ⟶ 6 C O_{2} + 3 H_{2} O$ , with a sooty luminous flame.

1. What Makes a Hydrocarbon Aromatic?

The word aromatic comes from the Greek aroma, meaning fragrance, and was first applied to compounds that happened to smell pleasant, long before anybody knew their structure. The word survives, but the meaning has completely changed. Today aromatic describes a class of compounds that show a characteristic extra stability despite being highly unsaturated.

Aromatic hydrocarbon: a cyclic, planar hydrocarbon in which a closed loop of $(4 n + 2)$ delocalised $π$ electrons produces stability far greater than any single Lewis structure would predict.

Aromatic compounds split into two families:

Benzenoid: contains one or more benzene rings. This includes benzene itself, arenes (benzene rings carrying aliphatic side chains, such as toluene and xylene), polynuclear hydrocarbons in which rings are fused (naphthalene, anthracene, phenanthrene) and linked systems such as biphenyl.
Non-benzenoid: aromatic without containing a benzene ring at all. The tropylium cation, the cyclopentadienyl anion and azulene belong here.

Figure 1: Aromatic hydrocarbons are benzenoid (built on benzene rings) or non-benzenoid. Both families obey Huckel's

(4 n + 2)

rule.

2. Structure of Benzene: the Kekule Story

Benzene was first isolated in 1825 by Michael Faraday, from cylinders of compressed illuminating gas obtained by the pyrolysis of whale oil. Elemental analysis and molecular mass determination fixed the molecular formula as $C_{6} H_{6}$ . That formula gives a degree of unsaturation of 4, so chemists expected a substance that would behave like a very reactive alkene or alkyne.

Benzene refused to behave that way. It does not decolourise bromine water, it does not decolourise Baeyer's reagent, and instead of adding reagents across its multiple bonds it quietly substitutes one hydrogen for something else. Explaining that contradiction took forty years.

Kekule's ring, 1865

In 1865 Friedrich August Kekule proposed that the six carbons form a ring with alternating single and double bonds, a cyclohexa-1,3,5-triene. Rival structures were proposed by others over the following decades, but none of them survived experimental testing.

Figure 2: Rival structures for

C_{6} H_{6}

. Kekule's hexagon won because it alone predicts one monosubstituted product and three disubstituted isomers (ortho, meta, para).

Kekule's ring explained the substitution products beautifully: one monosubstituted product, three disubstituted isomers (ortho, meta and para) and three trisubstituted isomers. Every one of these counts matched experiment.

The objection, and Kekule's answer

One serious objection remained. If the double bonds are fixed in place, then two different ortho disubstituted products are possible: one in which the two substituted carbons are joined by a double bond, and one in which they are joined by a single bond. Bromination of benzene should therefore give two ortho dibromobenzenes. Experiment gives only one.

Figure 3: The classic objection to Kekule's structure. Two ortho isomers are predicted, one is observed. Resonance, not oscillation, is the modern answer.

Kekule answered by proposing that the double bonds oscillate rapidly back and forth between the two sets of positions, so the two forms are never separable. This became known as Kekule's dynamic formula. It was not quite right, but it pointed directly at the modern electronic picture: benzene is not flipping between two structures, it is a single structure that is a resonance hybrid of both.

Other evidence for the ring: benzene gives exactly one monosubstituted product, adds three molecules of $H_{2}$ to give cyclohexane, and on ozonolysis gives three molecules of glyoxal. All three results demand a six-membered ring with three alternate double bonds.

3. Modern Picture: Resonance and the Stability of Benzene

Benzene resists addition and readily undergoes substitution. That alone tells us benzene must be more stable than a hypothetical cyclohexa-1,3,5-triene with three isolated double bonds. The enthalpy of hydrogenation puts a number on how much more stable.

Enthalpy of hydrogenation is the enthalpy change when one mole of an unsaturated compound is hydrogenated. For disubstituted alkenes of the type $R - C H = C H - R$ it lies between 117 and 125 kJ/mol, so roughly 120 kJ/mol per double bond.

Compound	C=C bonds	Calculated $Δ H_{h y d}$ (kJ/mol)	Observed $Δ H_{h y d}$ (kJ/mol)	Difference
Cyclohexene	1	$- 120$	$- 120$	0
Cyclohexa-1,3-diene	2	$- 240$	$- 232$	8 kJ/mol (small, from conjugation)
Cyclohexa-1,3,5-triene (hypothetical)	3	$- 360$	not observed	-
Benzene (real)	3 formally	$- 360$	$- 208$	152 kJ/mol

Cyclohexene matches its prediction exactly. Cyclohexa-1,3-diene is 8 kJ/mol more stable than predicted, a modest bonus from conjugating two double bonds. Benzene is 152 kJ/mol more stable than predicted. That gap is far too large to be ordinary conjugation, and it is what we call the resonance energy of benzene.

Figure 4: Benzene releases

208

kJ/mol on hydrogenation instead of the calculated

360

kJ/mol. The

152

kJ/mol shortfall is its resonance energy.

What X-ray studies show

X-ray diffraction settles the structure. Benzene is a planar molecule. All six C-C bonds are identical at 139 pm, sitting between a pure C-C single bond (154 pm) and a pure C=C double bond (134 pm). All six carbons are $s p^{2}$ hybridised and all bond angles are exactly $12 0^{\circ}$ .

Figure 5: All six bonds in benzene are

139

pm, between

C = C

134

pm and

C - C

154

pm, giving a bond order of

1.5

If benzene really had three single and three double bonds, the ring would be a distorted hexagon with alternating long and short sides. It is not. Benzene is a resonance hybrid, with the two Kekule structures as the main contributing forms and three Dewar-type structures making a minor contribution.

Figure 6: The two Kekule forms are contributing structures, not real molecules. Real benzene is the single hybrid with a resonance energy of about

152

kJ/mol.

Remark: the two Kekule structures are not real molecules and benzene is not a mixture of them. Neither structure exists even for an instant. Only the hybrid is real, and it is more stable than either contributor.

Solved Example 1

The observed heat of hydrogenation of benzene is

- 208

kJ/mol. Given that a single isolated double bond in a ring releases about 120 kJ/mol on hydrogenation, calculate the resonance energy of benzene and explain what it measures.

Solution:

A hypothetical cyclohexa-1,3,5-triene has three isolated double bonds, so the calculated value is $3 \times (- 120) = - 360$ kJ/mol.

Resonance energy $=$ calculated $-$ observed $= 360 - 208 = 152$ kJ/mol.

It measures how much more stable real benzene is than the imaginary molecule with three localised double bonds. This is the energy barrier any reaction must pay if it wants to destroy the aromatic sextet, which is precisely why benzene substitutes rather than adds.

4. Orbital Structure of Benzene

Each carbon in benzene is $s p^{2}$ hybridised. It uses its three $s p^{2}$ orbitals to form two sigma bonds to neighbouring carbons and one sigma bond to hydrogen. This gives a flat hexagonal framework with all twelve atoms in one plane and every angle $12 0^{\circ}$ .

That leaves one unhybridised $2 p$ orbital on every carbon, standing perpendicular to the ring plane. All six are parallel, so each overlaps equally with both of its neighbours. The result is not three separate $π$ bonds but a single delocalised system: two doughnut-shaped electron clouds, one above and one below the ring, holding six electrons shared by all six carbons.

Figure 7: Each carbon contributes one unhybridised

2 p

orbital. Sideways overlap of all six gives two

π

clouds holding six delocalised electrons.

When a bonding orbital is spread over more than two atoms it is said to be delocalised. Delocalisation lowers the energy of the system, and that is the physical origin of the 152 kJ/mol resonance energy.

5. Aromaticity and Huckel's $(4 n + 2)$ Rule

The modern theory of aromaticity was set out by Erich Huckel in 1931. Huckel's insight was that aromaticity is a function of electronic structure alone. Any ring, any polynuclear system, any heterocycle and any cyclic ion can be aromatic if it meets the electronic conditions.

The four conditions

Cyclic. The conjugated system must close on itself.
Planar. Complete delocalisation of the $π$ cloud is only possible if all the $p$ orbitals are parallel, and that requires a flat ring. This is why benzene is aromatic and cyclooctatetraene is not: cyclooctatetraene puckers into a tub shape.
Complete conjugation. Every atom in the ring must carry a $p$ orbital. A single $s p^{3}$ carbon anywhere in the ring breaks the loop and kills aromaticity.
$(4 n + 2)$ $π$ electrons, where $n = 0, 1, 2, 3, \dots$ This gives the magic numbers 2, 6, 10, 14, 18. A count of $4 n$ electrons (4, 8, 12) in a planar conjugated ring makes the system antiaromatic, which is worse than having no aromaticity at all.

Figure 8: Apply the four tests in order. Only a cyclic, planar, fully conjugated ring with

(4 n + 2)

π

electrons is aromatic.

Benzene has 6, naphthalene 10, and anthracene and phenanthrene 14 delocalised $π$ electrons in a conjugated cyclic system, so all four are aromatic. Cyclopentadiene and cyclohepta-1,3,5-triene each contain an $s p^{3}$ $C H_{2}$ group that breaks the conjugated loop, so both are non-aromatic. Cyclooctatetraene has eight $π$ electrons, a $4 n$ count, but it escapes antiaromaticity by puckering into a non-planar tub, leaving it simply non-aromatic.

Figure 9: The same

(4 n + 2)

count runs through rings of every size. Charge changes the electron count, and one

s p^{3}

carbon or a puckered ring removes aromaticity altogether.

Species	Ring size	$π$ electrons	Verdict	Reason
Benzene	6	6 $(n = 1)$	Aromatic	Planar, cyclic, fully conjugated
Naphthalene	10 (fused)	10 $(n = 2)$	Aromatic	Planar fused system
Cyclopropenyl cation	3	2 $(n = 0)$	Aromatic	Smallest aromatic ion
Cyclopentadienyl anion	5	6 $(n = 1)$	Aromatic	Lone pair joins the $π$ system
Tropylium cation	7	6 $(n = 1)$	Aromatic	Empty $p$ orbital completes the loop
Cyclobutadiene	4	4	Antiaromatic	$4 n$ count in a planar ring
Cyclopentadienyl cation	5	4	Antiaromatic	$4 n$ count
Cyclooctatetraene	8	8	Non-aromatic	Tub shaped, not planar
Cyclopentadiene	5	4	Non-aromatic	$s p^{3}$ $C H_{2}$ breaks conjugation
Cyclohepta-1,3,5-triene	7	6	Non-aromatic	$s p^{3}$ $C H_{2}$ breaks conjugation

Solved Example 2

Using Huckel's rule, predict which of the following will show aromatic stabilisation: (a) tropylium cation, (b) cyclooctatetraene, (c) cyclopentadiene, (d) cyclohepta-1,3,5-triene.

Solution:

Only (a), the tropylium cation.

(a) The seven-membered ring carries three C=C bonds, that is 6 $π$ electrons, and the positively charged carbon supplies an empty $p$ orbital that completes the conjugated loop. With $6 = 4 (1) + 2$ the ion is aromatic.

(b) Cyclooctatetraene has 8 $π$ electrons, a $4 n$ count, and it is tub shaped rather than planar. Non-aromatic.

(d) Cyclohepta-1,3,5-triene has the right count of 6 $π$ electrons but also an $s p^{3}$ $C H_{2}$ group, so conjugation is interrupted. Non-aromatic.

Solved Example 3

Which of the following are aromatic, and why? (i) cyclopentadienyl anion, (ii) cyclopentadienyl cation, (iii) tropylium cation, (iv) cyclohepta-1,3,5-triene.

Solution:

(i) and (iii) are aromatic.

(i) The anion has two C=C bonds (4 electrons) plus the lone pair on the negatively charged carbon, which occupies a $p$ orbital in the ring plane system. Total $= 6 = 4 n + 2$ with $n = 1$ . Aromatic.

(ii) The cation has only the two C=C bonds, so 4 $π$ electrons. This is a $4 n$ system, which makes it antiaromatic and unusually unstable.

(iii) Six $π$ electrons in a planar, fully conjugated seven-membered ring. Aromatic.

(iv) Six $π$ electrons, so the count is right, but the ring contains an $s p^{3}$ carbon and is therefore devoid of complete conjugation. Non-aromatic.

Solved Example 4

Which is more stable, the cyclopentadienyl anion or the cyclopentadienyl cation? Which is more stable, the tropylium cation or the tropylium anion?

Solution:

In the five-membered ring the anion is far more stable, because losing a proton from cyclopentadiene creates a 6 $π$ electron aromatic system. This is why cyclopentadiene is an unusually acidic hydrocarbon, with $p K_{a}$ around 16.

In the seven-membered ring the cation is more stable. Removing a hydride from cycloheptatriene leaves an empty $p$ orbital and a 6 $π$ electron aromatic system. The anion would have 8 $π$ electrons, a $4 n$ count, and would be antiaromatic.

The lesson worth carrying into the exam: the same ring can be aromatic or antiaromatic depending only on its charge, because charge changes the electron count.

6. Preparation of Benzene

Benzene was first synthesised by Berthelot, who passed acetylene through a red-hot tube. In the laboratory it was first made by heating benzoic acid or phthalic acid with calcium oxide. Four routes matter for the exam.

Cyclic polymerisation of acetylene. Three molecules of ethyne passed through a red-hot iron tube at about 873 K trimerise to benzene. The same reaction on propyne gives 1,3,5-trimethylbenzene (mesitylene).
$3 C H \equiv C H red-hot Fe tube 873 K C_{6} H_{6}$
Decarboxylation of sodium benzoate. Heating sodium benzoate with soda lime (NaOH and CaO) strips off the carboxyl group as carbonate.
$C_{6} H_{5} C O O N a + N a O H C a O, Δ C_{6} H_{6} + N a_{2} C O_{3}$
From phenol. Phenol vapour passed over heated zinc dust is reduced to benzene.
$C_{6} H_{5} O H + Z n Δ C_{6} H_{6} + Z n O$
Aromatisation of n-hexane. The industrial route, also called catalytic reforming or hydroforming.
$n - C_{6} H_{14} C r_{2} O_{3} - A l_{2} O_{3} 773 K, 10 - 20 a t m C_{6} H_{6} + 4 H_{2}$

Figure 10: Benzene from acetylene, from sodium benzoate, from phenol and from

n

-hexane. Aromatisation of

n

-hexane is the industrial route.

Also worth knowing: benzenesulphonic acid hydrolysed with superheated steam gives back benzene, because sulphonation is reversible. And aromatisation of $n$ -heptane by the same $C r_{2} O_{3} - A l_{2} O_{3}$ catalyst gives toluene, not benzene, since the seven carbons cyclise to a ring plus a methyl group.

Solved Example 5

(a)

n

-heptane is heated for a prolonged period with a mixture of

C r_{2} O_{3}

and

A l_{2} O_{3}

at about 873 K. What is the main product? (b) Are divinylacetylene and benzene isomers?

Solution:

(a) Methylbenzene (toluene). Aromatisation cyclises six of the seven carbons into a ring and leaves the seventh as a methyl substituent, with loss of four molecules of $H_{2}$ .

(b) Yes. Divinylacetylene is $C H_{2} = C H - C \equiv C - C H = C H_{2}$ , which is $C_{6} H_{6}$ , the same molecular formula as benzene. Both have $D U = 4$ . They are structural isomers, and this is a favourite true-or-false question.

7. Physical Properties of Benzene

A colourless, mobile liquid with a characteristic aromatic odour.
Boiling point 353 K ( $8 0^{\circ} C$ ); melting point 278.6 K ( $5. 5^{\circ} C$ ), so it freezes on a cold day.
Density about $0.879 g c m^{- 3}$ , so it is lighter than water and floats.
Immiscible with water, freely miscible with organic solvents such as ether, alcohol and $C C l_{4}$ . Benzene is itself an excellent solvent for fats, resins and iodine.
Inflammable, burning with a sooty luminous flame because of its high carbon content.
The ring resists oxidation. Acidified permanganate attacks benzene only slowly and under vigorous conditions, eventually degrading it to $C O_{2}$ and $H_{2} O$ .
Benzene is carcinogenic and its vapour is toxic, which is why it has been replaced by toluene in most laboratory work.

8. Electrophilic Substitution: the Master Mechanism

The major reactions of the benzene ring are electrophilic substitution reactions. Before looking at individual reagents it pays to understand why substitution wins, because one argument covers every reaction in this chapter.

Why substitution and not addition?

The $π$ cloud of benzene is electron rich, so an electrophile is certainly attracted to it. The question is what happens after the electrophile attaches.

In addition, one weak $π$ bond breaks and two strong $σ$ bonds form. Energetically that looks favourable for an ordinary alkene. But for benzene the price is the destruction of the delocalised sextet, which costs about 152 kJ/mol of resonance energy. The product is no longer aromatic.
In substitution, the ring temporarily loses aromaticity to form the intermediate, then gets it straight back when a proton leaves. Nothing is permanently lost.

Figure 11: Addition would cost benzene its resonance energy of about

152

kJ/mol, so the substitution route wins on both barrier height and product stability.

The three-step mechanism

Every electrophilic substitution of benzene, whichever reagent you use, follows the same three steps and passes through the same intermediate.

Figure 12: Every ring substitution of benzene runs through the same arenium ion. Attack is slow, loss of

H^{+}

is fast, and aromaticity is restored at the end.

Arenium ion (sigma complex, benzenonium ion): the carbocation intermediate formed when an electrophile bonds to a ring carbon. The attacked carbon changes its hybridisation from trigonal $s p^{2}$ to tetrahedral $s p^{3}$ , and the positive charge is delocalised over the remaining five carbons, principally the two ortho and the one para position.

Rate-determining step: attack of the $π$ cloud on the electrophile (step 2). Loss of the proton in step 3 is fast, because it is driven by the huge energy gain of restoring aromaticity.

Solved Example 6

Benzene has double bonds, yet it normally gives electrophilic substitution rather than electrophilic addition. Explain why.

Solution:

In electrophilic substitution, one $π$ bond is broken and a new $σ$ bond is formed between a ring carbon and the electrophile. The intermediate arenium ion then loses a proton and the aromatic character is restored. The overall energy change is small, and the stable aromatic product survives.

In electrophilic addition, the delocalised sextet would be permanently destroyed. Aromatic character is lost and stability is decreased by roughly 152 kJ/mol. The product is a non-aromatic cyclohexadiene, which is much less stable than benzene.

Because substitution preserves the resonance energy and addition throws it away, benzene substitutes.

9. The Electrophilic Substitution Reactions of Benzene

Reaction	Reagent and conditions	Attacking electrophile	Product
Nitration	conc. $H N O_{3}$ + conc. $H_{2} S O_{4}$ , 323-333 K	$N O_{2}^{+}$ (nitronium ion)	Nitrobenzene
Sulphonation	Fuming $H_{2} S O_{4}$ (oleum), $Δ$	$S O_{3}$ or $^{+} S O_{3} H$	Benzenesulphonic acid
Halogenation	$C l_{2}$ or $B r_{2}$ with $F e C l_{3}$ , $F e B r_{3}$ or $A l C l_{3}$ , dark	$C l^{+}$ or $B r^{+}$	Chlorobenzene, bromobenzene
Friedel-Crafts alkylation	$R - X$ with anhydrous $A l C l_{3}$	$R^{+}$ (carbocation)	Alkylbenzene
Friedel-Crafts acylation	$R C O C l$ or $(R C O)_{2} O$ with anhydrous $A l C l_{3}$	$R C O^{+}$ (acylium ion)	Aryl ketone

Nitration

The nitrating mixture is concentrated nitric acid plus concentrated sulphuric acid. The role of the sulphuric acid is not to provide the electrophile directly but to protonate nitric acid, which then loses water to give the nitronium ion. Note the unusual reversal of roles: here nitric acid behaves as the base and sulphuric acid as the acid.

Figure 13: Nitration of benzene. The attacking species is the linear nitronium ion

N O_{2}^{+}

, generated by sulphuric acid acting on nitric acid.

Sulphonation and halogenation

Sulphonation with fuming sulphuric acid gives benzenesulphonic acid. It is the one electrophilic substitution that is readily reversible: superheated steam pushes the equilibrium back to benzene, which makes the sulphonic acid group a useful temporary blocking group in synthesis.

Halogenation needs a Lewis acid carrier such as $F e C l_{3}$ , $F e B r_{3}$ or anhydrous $A l C l_{3}$ , which polarises the halogen molecule and delivers $C l^{+}$ or $B r^{+}$ to the ring. Light must be excluded, or the reaction switches to a free radical pathway.

Solved Example 7

Why can benzene not be iodinated directly with

I_{2}

? Show how benzene can nevertheless be converted to iodobenzene using

I_{2}

Solution:

Iodination is reversible. The reaction $C_{6} H_{6} + I_{2} ⇌ C_{6} H_{5} I + H I$ produces hydrogen iodide, which is a strong reducing agent and immediately reduces iodobenzene back to benzene. The equilibrium therefore sits on the left and no useful product accumulates. Iodine is also the least reactive halogen towards the ring.

Fix: carry out the reaction in the presence of an oxidising agent that destroys $H I$ as it forms, such as $H N O_{3}$ , $H I O_{3}$ or $H g O$ .

$C_{6} H_{6} + I_{2} H N O_{3} (oxidising agent) C_{6} H_{5} I + H_{2} O$

By removing $H I$ , the oxidising agent shifts the equilibrium to the right, in line with Le Chatelier's principle.

10. Friedel-Crafts Reactions

Friedel-Crafts alkylation

This reaction introduces an alkyl group into the benzene ring in the presence of a catalyst. The alkylating agent can be an alkyl halide $R - X$ , an alcohol $R O H$ , or an alkene. The catalyst is a Lewis acid, usually anhydrous $A l C l_{3}$ . All three routes converge on the same electrophile: a carbocation.

Figure 14: Friedel-Crafts alkylation. The electrophile is a free carbocation, which is why the product is so often the rearranged one, and why the reaction seldom stops at one alkyl group.

Because the electrophile is a free carbocation, the reaction inherits every carbocation problem you already know. The three limitations below are heavily examined.

Rearrangement. The carbocation rearranges to the most stable form before it reaches the ring. Benzene with $C H_{3} C H_{2} C H_{2} C l$ and $A l C l_{3}$ gives mainly isopropylbenzene (cumene), not $n$ -propylbenzene, because the primary cation shifts a hydride to become secondary.
Polyalkylation. An alkyl group is electron releasing, so the product ring is more reactive than benzene itself and reacts again. Using a large excess of benzene limits this.
Deactivated rings fail. Friedel-Crafts alkylation does not work on nitrobenzene or on any strongly deactivated ring, and it fails on aniline because the $- N H_{2}$ lone pair complexes the $A l C l_{3}$ . Aryl and vinyl halides cannot be used as the alkylating agent.

Intramolecular version: if the side chain already attached to the ring carries a halogen at the far end and contains four or five carbons, the carbocation can attack its own ring. This intramolecular Friedel-Crafts alkylation closes a new fused ring, giving systems such as indane and tetralin.

Friedel-Crafts acylation

Acylation replaces a ring hydrogen with an acyl group $R - C O -$ , using a carboxylic acid, an ester, an acid chloride or an acid anhydride as the acylating agent, again with a Lewis acid catalyst. The electrophile is the acylium ion.

Figure 15: Friedel-Crafts acylation. The acylium ion is resonance stabilised with a complete octet on every atom, so it never rearranges.

The acylium ion has two resonance structures. In the first the positive charge sits on carbon, which then has an incomplete octet. In the second the charge sits on oxygen and a carbon-oxygen triple bond forms, so every atom has a complete octet. The second structure is therefore the major contributor, and this is exactly why the acylium ion is so stable and never rearranges.

Feature	Friedel-Crafts alkylation	Friedel-Crafts acylation
Electrophile	Carbocation $R^{+}$	Acylium ion $R C O^{+}$
Rearrangement	Common	Never
Multiple substitution	Yes, product is activated	No, the ketone deactivates the ring
Catalyst required	Catalytic $A l C l_{3}$	More than one equivalent, the ketone complexes it
Product	Alkylbenzene	Aryl ketone

Synthetic trick: to attach a straight alkyl chain without rearrangement, acylate first and then reduce the ketone by the Clemmensen ( $Z n - H g$ , $H C l$ ) or Wolff-Kishner ( $N H_{2} N H_{2}$ , $K O H$ ) method. Acylation supplies the correct skeleton and reduction removes the oxygen.

Solved Example 8

Starting with benzene, outline a synthesis of 1-bromo-2-(trichloromethyl)benzene and of 1-bromo-3-(trichloromethyl)benzene.

Solution:

The trick is the order of the steps, because $- C H_{3}$ is an ortho, para director while $- C C l_{3}$ is a meta director.

For the 1,2 (ortho) isomer:

$C_6H_6 \xrightarrow{CH_3Cl/AlCl_3} C_6H_5CH_3 \xrightarrow{Br_2/Fe} \text{o-bromotoluene} \xrightarrow{3Cl_2/\text{light}} \text{1-bromo-2-(trichloromethyl)benzene}$

Brominate while the methyl group is still there, so the methyl directs bromine to the ortho position. Only then chlorinate the side chain under light.

For the 1,3 (meta) isomer:

$C_6H_6 \xrightarrow{CH_3Cl/AlCl_3} C_6H_5CH_3 \xrightarrow{3Cl_2/\text{light}} C_6H_5CCl_3 \xrightarrow{Br_2/Fe} \text{1-bromo-3-(trichloromethyl)benzene}$

Convert the methyl into $- C C l_{3}$ first. Being strongly electron withdrawing, it now directs the incoming bromine to the meta position.

Solved Example 9

What is the product of the reaction of benzene with ethylene oxide (epoxide) in the presence of anhydrous

A l C l_{3}

, followed by aqueous work-up?

Solution:

2-phenylethanol, $C_{6} H_{5} C H_{2} C H_{2} O H$ .

$A l C l_{3}$ coordinates to the epoxide oxygen and opens the strained three-membered ring, generating a carbon centre that is electrophilic enough to be attacked by the benzene $π$ cloud. This is a Friedel-Crafts alkylation in which the leaving group stays attached as an alkoxide. Hydrolysis on work-up then delivers the primary alcohol.

11. Addition Reactions of Benzene

Benzene resists addition, but it does not refuse it outright. Under forcing conditions, and usually with no Lewis acid present, the ring will add.

Catalytic hydrogenation. Three molecules of hydrogen add over a nickel catalyst at 473-573 K to give cyclohexane.
$C_{6} H_{6} + 3 H_{2} N i 473 - 573 K C_{6} H_{12}$
Photochemical chlorination. In bright sunlight or UV light, and with the Lewis acid deliberately left out, chlorine adds to give benzene hexachloride (BHC, gammexane, lindane), once used as an insecticide.
$C_{6} H_{6} + 3 C l_{2} h ν C_{6} H_{6} C l_{6}$
Ozonolysis. Ozone adds across all three double bonds to give a triozonide, which on hydrolysis with $H_{2} O$ and $Z n$ gives three molecules of glyoxal. This is the classical structural proof of three alternate double bonds.
$C_{6} H_{6} (i) O_{3}; (ii) H_{2} O / Z n 3 O H C - C H O$
Combustion. Benzene burns in air with a sooty flame because of its high carbon percentage.
$C_{6} H_{6} + \frac{15}{2} O_{2} ⟶ 6 C O_{2} + 3 H_{2} O$

Figure 16: Benzene adds

H_{2}

C l_{2}

and

O_{3}

only under force. The three glyoxal molecules from ozonolysis confirm three alternate double bonds.

Industrial oxidation: although the ring resists ordinary oxidising agents, benzene vapour passed over $V_{2} O_{5}$ in air at about 773 K is oxidised to maleic anhydride. This is the one oxidation of the benzene ring itself worth remembering.

12. Reactions of the Side Chain

When benzene carries an alkyl side chain, two different sites are available: the aromatic ring and the side chain. Which one reacts is decided entirely by the conditions, and that is the single most examined idea in this part of the chapter.

Halogenation: ring or side chain?

With a Lewis acid, in the dark: electrophilic substitution on the ring, at the ortho and para positions, because an alkyl group is ortho, para directing.
With light or high temperature and no Lewis acid: free radical substitution in the side chain.

Figure 17: Lewis acid and darkness send chlorine to the ring; light or heat send it to the side chain. This fork is a standard JEE question.

Toluene chlorinated in light gives benzyl chloride first, then benzal chloride ( $C_{6} H_{5} C H C l_{2}$ ) and finally benzotrichloride ( $C_{6} H_{5} C C l_{3}$ ) as more chlorine is supplied.

Why the benzylic position?

The ease of abstracting a hydrogen atom follows the stability of the radical that is left behind. The benzylic radical is resonance stabilised over four positions: the benzylic carbon plus the two ortho and the one para carbon of the ring.

Stability of free radicals: benzyl $\approx$ allyl $>$ $3^{\circ}$ $>$ $2^{\circ}$ $>$ $1^{\circ}$ $>$ methyl $>$ vinyl and aryl.

Figure 18: The benzyl radical is resonance stabilised over four carbons, which is why free radical halogenation of toluene always hits the side chain first.

Bromine is more selective than chlorine. Bromination is more endothermic in the hydrogen abstraction step, so its transition state resembles the radical more closely and the stability difference between benzylic and other positions matters more. Bromination therefore gives essentially pure benzylic product, while chlorination gives mixtures.

Oxidation of the side chain

The benzene ring is very resistant to oxidation, so the side chain is always the part that is attacked. Whatever the length of the side chain, the ultimate oxidation product is benzoic acid, because oxidation begins at the benzylic carbon and eats the chain back to a single carboxyl carbon.

Figure 19: The benzene ring resists oxidation, so the side chain is attacked. Benzoic acid is the endpoint whenever a benzylic hydrogen exists.

At least one benzylic hydrogen is compulsory. If the carbon attached to the ring carries no hydrogen, as in tert-butylbenzene, no oxidation occurs.
Two side chains oxidise together. o-xylene gives phthalic acid, m-xylene gives isophthalic acid, p-xylene gives terephthalic acid.
With an electron withdrawing group ( $- I$ or $- R$ ) on the ring, the ring is even more stable, and oxidation cleanly gives the substituted benzoic acid.
With $- O H$ or $- N H_{2}$ on the ring, the ring becomes so electron rich that it is itself broken down by any oxidising agent. Ring rupture is prevented by protecting the group first, usually by acetylation.

Solved Example 10

Give the product of oxidation with hot acidified

K M n O_{4}

for: (a) n-propylbenzene, (b) tert-butylbenzene, (c) o-xylene, (d) p-nitrotoluene.

Solution:

(a) Benzoic acid, $C_{6} H_{5} C O O H$ . The whole three-carbon chain is degraded, because the benzylic $C H_{2}$ carries hydrogens. Chain length makes no difference.

(b) No reaction. The benzylic carbon in $C_{6} H_{5} C (C H_{3})_{3}$ has no hydrogen, so there is nothing for the oxidising agent to remove.

(d) p-nitrobenzoic acid. The nitro group is electron withdrawing, so it stabilises the ring against oxidation and only the methyl group is attacked.

Solved Example 11

Arrange the three isomeric xylenes in increasing order of polarity.

Solution:

p-xylene $<$ m-xylene $<$ o-xylene.

Each methyl group contributes a bond dipole pointing away from the ring. The molecular dipole is the vector sum of the two.

In para-xylene the two dipoles are at $18 0^{\circ}$ and cancel exactly, so $μ = 0$ . In meta-xylene they are at $12 0^{\circ}$ , giving a resultant equal to one bond moment. In ortho-xylene they are at $6 0^{\circ}$ , giving the largest resultant, about $3$ times one bond moment.

13. Alkenyl Benzenes

Alkenyl benzenes carry a carbon-carbon double bond in the side chain. The two you must know are styrene (vinylbenzene, $C_{6} H_{5} C H = C H_{2}$ ) and 1-phenylpropene ( $C_{6} H_{5} C H = C H - C H_{3}$ ).

Preparation

Dehydrogenation of ethylbenzene over $C r_{2} O_{3} - A l_{2} O_{3}$ at about 870 K. This is the industrial route to styrene.
Dehydration of 1-phenylethanol with $H_{2} S O_{4}$ or $A l_{2} O_{3}$ on heating.
Dehydrohalogenation of (1-bromoethyl)benzene with alcoholic $K O H$ .

Reactivity

The side-chain double bond is more reactive than the benzene ring towards electrophilic reagents, so much milder conditions are needed for addition across the double bond than for substitution on the ring. The reason is the intermediate: the first step generates either a carbocation or a free radical, and in an alkenyl benzene that intermediate can be benzylic, which is stabilised by conjugation with the ring. That extra stability makes an alkenyl benzene more reactive than a simple alkene.

Figure 20: In styrene the alkene side chain is more reactive than the ring. Both the ionic and the radical route pass through a benzylic intermediate, and that is what decides where the bromine lands.

Solved Example 12

Predict the major product when styrene reacts with

H B r

(a) in the absence of peroxide and (b) in the presence of benzoyl peroxide.

Solution:

(a) 1-bromo-1-phenylethane, $C_{6} H_{5} C H B r - C H_{3}$ . The proton adds first. Adding it to the terminal $C H_{2}$ leaves the positive charge on the carbon next to the ring, giving the resonance stabilised benzylic carbocation. Bromide then attacks there. This is Markovnikov addition.

(b) 2-bromo-1-phenylethane, $C_{6} H_{5} C H_{2} - C H_{2} B r$ . Peroxide generates $B r \cdot$ , which adds first. Adding bromine to the terminal carbon leaves the odd electron on the benzylic carbon, again the stabilised intermediate. The result is the anti-Markovnikov (peroxide effect) product.

Notice that both routes pass through a benzylic intermediate. What changes is which atom arrives first, and that flips the regiochemistry.

Common Mistakes to Avoid

Watch out

Treating benzene like an alkene. Benzene does not decolourise bromine water and does not decolourise Baeyer's reagent. Any answer that says otherwise is wrong, and this is a standard trap in NEET assertion-reason questions.
Counting $π$ electrons carelessly. Count only the electrons that are actually part of the cyclic conjugated loop. A lone pair counts only if it sits in a $p$ orbital that is parallel to the rest of the ring system, as in the cyclopentadienyl anion.
Calling cyclooctatetraene antiaromatic. It has $4 n$ electrons, but it escapes by puckering into a tub. A non-planar ring cannot be antiaromatic, so cyclooctatetraene is non-aromatic.
Applying Huckel's rule to non-cyclic or non-planar systems. The electron count is the last test, not the first. Check cyclic, planar and fully conjugated first.
Forgetting the $s p^{3}$ carbon. Cyclohepta-1,3,5-triene has 6 $π$ electrons and still is not aromatic, because one $C H_{2}$ breaks the loop.
Writing the unrearranged Friedel-Crafts product. $n$ -propyl chloride plus benzene and $A l C l_{3}$ gives isopropylbenzene as the major product, not $n$ -propylbenzene.
Attempting Friedel-Crafts on a deactivated ring. Neither alkylation nor acylation works on nitrobenzene, and aniline fails because its lone pair ties up the $A l C l_{3}$ .
Using catalytic $A l C l_{3}$ in acylation. The ketone product complexes the catalyst, so more than one full equivalent is required.
Oxidising a chain with no benzylic hydrogen. tert-butylbenzene gives no benzoic acid, however harsh the conditions.
Mixing up the halogenation conditions. Lewis acid plus darkness sends the halogen to the ring; light or heat with no catalyst sends it to the side chain.
Calling the resonance energy an experimental enthalpy. It is a difference: calculated minus observed. Nothing measures it directly.

Frequently Asked Questions

Why is benzene more stable than cyclohexa-1,3,5-triene?

Because its six $π$ electrons are delocalised over all six carbons instead of being locked into three separate double bonds. Delocalisation lowers the energy of the system by about 152 kJ/mol, the resonance energy, which shows up as a heat of hydrogenation of $- 208$ kJ/mol instead of the calculated $- 360$ kJ/mol.

What exactly is Huckel's $(4 n + 2)$ rule?

It states that a cyclic, planar, fully conjugated system is aromatic if it contains $(4 n + 2)$ delocalised $π$ electrons, where $n$ is a whole number. The allowed counts are 2, 6, 10, 14 and so on. All four conditions must hold together: cyclic, planar, complete conjugation and the correct electron count.

Why does benzene prefer substitution over addition?

Substitution loses aromaticity only briefly, in the arenium ion, and gets it back when a proton leaves. Addition destroys the delocalised sextet permanently and costs about 152 kJ/mol. The substitution route also has the lower activation barrier, so it wins on both product stability and reaction rate.

Is cyclooctatetraene aromatic or antiaromatic?

Neither. It is non-aromatic. With eight $π$ electrons it would be antiaromatic if it were planar, so it avoids that by puckering into a tub shape. In the tub the adjacent $p$ orbitals are no longer parallel, delocalisation fails, and the molecule behaves like an ordinary polyene.

What is the arenium ion and why does it matter?

The arenium ion, or sigma complex, is the carbocation formed when an electrophile bonds to a ring carbon. That carbon becomes $s p^{3}$ and the positive charge is spread over the other five carbons. It matters because every electrophilic substitution of benzene, from nitration to acylation, passes through it, and its stability controls both the rate and the orientation of substitution.

Why does Friedel-Crafts alkylation give rearranged products?

Because the electrophile is a free carbocation, which rearranges by hydride or alkyl shift to the most stable form before it meets the ring. A primary cation from $n$ -propyl chloride shifts a hydride and becomes secondary, so the product is isopropylbenzene. To build a straight chain, acylate first and then reduce with Clemmensen or Wolff-Kishner.

Why does tert-butylbenzene not give benzoic acid on oxidation?

Side-chain oxidation begins by removing a hydrogen from the benzylic carbon. In tert-butylbenzene that carbon is bonded to three methyl groups and the ring, so it has no hydrogen at all. With no benzylic hydrogen there is no way to start the oxidation, and the compound is unaffected even by hot acidified $K M n O_{4}$ .

What are the bond lengths and bond angles in benzene?

All six C-C bonds are equal at 139 pm, between a C-C single bond at 154 pm and a C=C double bond at 134 pm, which corresponds to a bond order of 1.5. The C-H bonds are 109 pm and every bond angle is $12 0^{\circ}$ . All twelve atoms lie in one plane.

Why does benzene burn with a sooty flame?

Benzene is about 92 percent carbon by mass, a much higher proportion than any alkane of comparable size. In an ordinary flame there is not enough oxygen to burn all that carbon, so unburnt carbon particles glow and then escape as soot. The same high carbon content makes acetylene burn smokily.

Previous year questions on Aromatic Hydrocarbons: Benzene

9 questions from past papers, each with a step-by-step solution.

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1. What Makes a Hydrocarbon Aromatic?

2. Structure of Benzene: the Kekule Story

Kekule's ring, 1865

The objection, and Kekule's answer

3. Modern Picture: Resonance and the Stability of Benzene

What X-ray studies show

4. Orbital Structure of Benzene

5. Aromaticity and Huckel's (4n+2) Rule

The four conditions

6. Preparation of Benzene

7. Physical Properties of Benzene

8. Electrophilic Substitution: the Master Mechanism

Why substitution and not addition?

The three-step mechanism

9. The Electrophilic Substitution Reactions of Benzene

Nitration

Sulphonation and halogenation

10. Friedel-Crafts Reactions

Friedel-Crafts alkylation

Friedel-Crafts acylation

11. Addition Reactions of Benzene

12. Reactions of the Side Chain

Halogenation: ring or side chain?

Why the benzylic position?

Oxidation of the side chain

13. Alkenyl Benzenes

Preparation

Reactivity

Common Mistakes to Avoid

Frequently Asked Questions

Why is benzene more stable than cyclohexa-1,3,5-triene?

What exactly is Huckel's (4n+2) rule?

Why does benzene prefer substitution over addition?

Is cyclooctatetraene aromatic or antiaromatic?

What is the arenium ion and why does it matter?

Why does Friedel-Crafts alkylation give rearranged products?

Why does tert-butylbenzene not give benzoic acid on oxidation?

What are the bond lengths and bond angles in benzene?

Why does benzene burn with a sooty flame?

Previous year questions on Aromatic Hydrocarbons: Benzene

Ready to master Hydrocarbons?

5. Aromaticity and Huckel's $(4 n + 2)$ Rule

What exactly is Huckel's $(4 n + 2)$ rule?