ChemistryCBSE Class 11 · Hydrocarbons

Benzene: Aromaticity, Electrophilic Substitution and Directing Effects Class 11

Updated 17 October 2026 · Fact-checked

Benzene is aromatic: it is planar, cyclic and fully conjugated with (4n + 2) π electrons (six, for n = 1). The delocalised π cloud makes it stable, so it prefers electrophilic substitution (nitration, halogenation, sulphonation, Friedel–Crafts alkylation and acylation) to addition. A substituent already on the ring directs the next group: activating groups such as –CH₃ and –OH direct to ortho and para, and the deactivating –NO₂, –CHO and –COOH direct to meta; halogens are deactivating but ortho-para directing.

Understand Benzene: Aromaticity, Electrophilic Substitution and Directing Effects

Aromaticity. A compound is aromatic if it is planar, cyclic, completely conjugated and has $(4 n + 2) π$ electrons (Hückel's rule). Benzene has six, naphthalene ten, the cyclopentadienyl anion six; cyclobutadiene (four) and cyclooctatetraene (eight, non-planar) are not aromatic. Benzene is a resonance hybrid of two Kekulé structures; all six C–C bonds are equivalent and the carbons are $s p^{2}$ .

Electrophilic substitution (three stages). (1) The electrophile $E^{+}$ is generated. (2) It attacks the ring to form a non-aromatic carbocation (the arenium ion, stabilised by resonance). (3) A proton is lost and aromaticity is restored.

Nitration: conc. $H N O_{3}$ + conc. $H_{2} S O_{4}$ forms $N O_{2}^{+}$ (nitronium ion); product nitrobenzene.
Halogenation: $C l_{2}$ or $B r_{2}$ with $F e C l_{3}$ / $F e B r_{3}$ (Lewis acid) gives $C_{6} H_{5} C l$ or $C_{6} H_{5} B r$ .
Sulphonation: fuming $H_{2} S O_{4}$ (oleum), electrophile $S O_{3}$ , product benzenesulphonic acid.
Friedel–Crafts alkylation: $R - C l$ with anhydrous $A l C l_{3}$ gives alkylbenzene; acylation: $R C O C l$ (or an anhydride) with anhydrous $A l C l_{3}$ gives an aryl ketone, with acylium ion $R C^{+} = O$ . Neither works on rings strongly deactivated by groups such as $- N O_{2}$ .

Addition reactions (forced conditions). With $H_{2}$ /Ni on heating benzene gives cyclohexane. Chlorine in sunlight (UV) adds three molecules to give benzene hexachloride, $C_{6} H_{6} C l_{6}$ .

Directing effects. An electron-releasing group (–CH₃, –OH, –NH₂) activates the ring and directs the incoming group to the ortho and para positions. An electron-withdrawing group (–NO₂, –CN, –CHO, –COOH, –SO₃H) deactivates the ring and directs meta. Halogens withdraw electrons by induction (deactivating) but donate lone pairs by resonance (ortho/para directing).

Planning a synthesis. The order of steps decides the isomer: nitration then bromination gives the meta product, but bromination then nitration gives the para (and ortho) product.

Carcinogenicity. Benzene and polynuclear hydrocarbons such as benzopyrene are toxic and carcinogenic.

Benzene: Aromaticity, Electrophilic Substitution and Directing Effects: formulas and key facts

Hückel's rule: $number of π electrons = 4 n + 2$; n = 0, 1, 2, ...
Nitronium ion: $H N O_{3} + 2 H_{2} S O_{4} ⇌ N O_{2}^{+} + H_{3} O^{+} + 2 H S O_{4}^{-}$; Generated in the nitrating mixture.
Combustion of benzene: $C_{6} H_{6} + \frac{15}{2} O_{2} \to 6 C O_{2} + 3 H_{2} O$; Sooty flame because of the high carbon content.
Percentage yield: $% yield = \frac{actual}{theoretical} \times 100$; Use the mole ratio from the equation.

How to solve Benzene: Aromaticity, Electrophilic Substitution and Directing Effects questions

Decide the electrophile and the directing effect of what is already on the ring, then write the position of the new group.

1Identify the reagent and write the electrophile it generates.
2Look at the substituent on the ring: activating (o,p) or deactivating (m).
3Place the new group, remembering that halogens are o,p directing.
4For a multi-step synthesis, introduce the group that orients the next step first.
5For numericals use the balanced equation, convert grams to moles and apply the yield.

Benzene: Aromaticity, Electrophilic Substitution and Directing Effects in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Burning a hydrocarbon fuel sample

Suppose a lab burns 7.8 g of benzene (78 g mol⁻¹) completely in air (22.7 L mol⁻¹ at STP).

Find the volume of CO₂ at STP and the mass of water formed.

Try it first, then show the solutionSolution

Moles of C₆H₆ = 7.8/78 = 0.10 mol; CO₂ = 0.60 mol = 0.60 × 22.7 = 13.6 L.
Water = 0.30 mol = 0.30 × 18 = 5.4 g.

Answer: 13.6 L of CO₂ and 5.4 g of water.

Memory hook: Six C and six H atoms give 6 CO₂ and 3 H₂O.

Dye intermediates

Suppose a chemist plans to make a nitro-substituted bromobenzene as an intermediate and must choose the order of nitration and bromination.

Which order gives the meta isomer?

Try it first, then show the solutionSolution

Nitro is a meta director, so nitrate first to give nitrobenzene.
Then brominate with Br₂/FeBr₃ to place bromine at the meta position.

Answer: Nitration first, then bromination.

Memory hook: The first group steers the second.

Tricks to remember Benzene: Aromaticity, Electrophilic Substitution and Directing Effects

Short memory aids. Say each one aloud once, then use it on a question the same day.

Six, ten, fourteen
4n + 2 gives 2, 6, 10, 14 π electrons for n = 0, 1, 2, 3.
Use it when: Hückel's rule questions.
Activators are o,p; deactivators are m
Electron-donating groups direct o,p; electron-withdrawing groups direct m; halogens are the exception (o,p but deactivating).
Use it when: Orientation.
Electrophile list
NO₂⁺ for nitration, SO₃ for sulphonation, Cl⁺/Br⁺ (polarised halogen) for halogenation, R⁺ or RCO⁺ for Friedel–Crafts.
Use it when: Mechanism questions.
Dry catalyst
Friedel–Crafts needs anhydrous AlCl₃, and fails with strongly deactivated rings.
Use it when: Conditions.

Benzene: Aromaticity, Electrophilic Substitution and Directing Effects: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Hückel's rule · 2 marks · Very short answerBoard level

Decide whether naphthalene and the planar, fully conjugated cycloheptatrienyl (tropylium) cation, C₇H₇⁺, are aromatic, using Hückel's rule.

Try it first, then show the solutionSolution

Naphthalene is planar and fully conjugated with ten π electrons = 4(2) + 2, so it is aromatic. (1)
In C₇H₇⁺ the seven carbons share three double bonds (six π electrons) and the positive carbon contributes an empty p orbital; 6 = 4(1) + 2, so the ion is aromatic. (1)

Answer: Both are aromatic: ten and six π electrons respectively.

Example 2: Friedel–Crafts acylation · 3 marks · Short answerHarder board level

Write the equation for the reaction of benzene with ethanoyl chloride in the presence of anhydrous AlCl₃. Name the electrophile and the product, and say what product benzene gives with chloromethane under the same catalyst.

Try it first, then show the solutionSolution

CH₃COCl + AlCl₃ → CH₃C⁺=O + AlCl₄⁻; the electrophile is the acylium ion. (1)
C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl (anhydrous AlCl₃), the product being acetophenone (methyl phenyl ketone). (1)
C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl (toluene, by alkylation). (1)

Answer: Acetophenone via the acylium ion; toluene with CH₃Cl.

Example 3: Orientation of the incoming group · 3 marks · Short answerHarder board level

Give the major product(s) and the reason in each case: (a) nitration of chlorobenzene, (b) bromination of nitrobenzene with Br₂/FeBr₃. (c) Explain why chlorobenzene reacts more slowly than benzene yet gives o- and p-products.

Try it first, then show the solutionSolution

(a) o-Chloronitrobenzene and p-chloronitrobenzene (chlorine is an o,p director). (1)
(b) m-Bromonitrobenzene, because the nitro group is a meta director and deactivates the ring. (1)
(c) Chlorine withdraws electrons through the σ bond (−I effect), making the ring less reactive; its lone pairs donate by resonance and enrich the o and p positions. (1)

Answer: (a) o- and p-chloronitrobenzene; (b) m-bromonitrobenzene; (c) −I deactivates, +R directs o,p.

Example 4: Order of steps in a synthesis · 4 marks · Long answerHarder board level

Starting from benzene, outline the preparation of (a) m-bromonitrobenzene and (b) p-bromonitrobenzene, giving the reagents. Explain why the order is different.

Try it first, then show the solutionSolution

(a) Benzene with conc. HNO₃/conc. H₂SO₄ gives nitrobenzene; then Br₂/FeBr₃ gives m-bromonitrobenzene. (1)
The nitro group is a meta director, so bromine enters the meta position. (1)
(b) Benzene with Br₂/FeBr₃ gives bromobenzene; then conc. HNO₃/conc. H₂SO₄ gives p-bromonitrobenzene (with some ortho isomer, separated). (1)
Bromine is an ortho-para director, so nitration goes mainly to the para position. (1)

Answer: (a) nitrate then brominate; (b) brominate then nitrate; the first group sets the position.

Example 5: Identifying an aromatic hydrocarbon · 4 marks · Long answerHOTS / competency

A hydrocarbon Y has 92.3% carbon and a vapour density of 39. It does not decolourise bromine water or cold dilute KMnO₄, burns with a sooty flame, and gives one mononitro product. (a) Find its molecular formula. (b) Identify Y and give the reason for the lack of reaction with bromine water. (c) Find the volume of CO₂ at STP (22.7 L mol⁻¹) from the combustion of 3.9 g of Y.

Try it first, then show the solutionSolution

(a) Molar mass = 2 × 39 = 78; C = 0.923 × 78 = 72, so six C atoms, and H = 78 − 72 = 6: C₆H₆. (1)
(b) Y is benzene (one mononitro product, sooty flame). The π electrons are delocalised over the ring, so benzene does not add bromine like an alkene; it substitutes only with a Lewis acid catalyst. (1)
(c) Moles of Y = 3.9/78 = 0.050 mol; CO₂ = 6 × 0.050 = 0.30 mol. (1)
Volume = 0.30 × 22.7 = 6.8 L. (1)

Answer: C₆H₆ (benzene); resonance stabilisation; 6.8 L of CO₂.

Where students lose marks in Benzene: Aromaticity, Electrophilic Substitution and Directing Effects

The slips examiners see most often, and what to write instead.

Mistake: Writing o,p-products for the nitration of nitrobenzene or the bromination of benzoic acid.
Ignoring the deactivating, meta-directing group.
Do this: −NO₂, −CHO, −COOH and −SO₃H direct to the meta position.
Usually costs: Product marks.
Mistake: Saying chlorine is a meta director because it is deactivating.
Mixing up activity with orientation.
Do this: Halogens deactivate by −I but direct o,p by +R.
Usually costs: Concept marks.
Mistake: Omitting the Lewis acid catalyst or writing aqueous AlCl₃ in Friedel–Crafts.
Incomplete conditions.
Do this: Write anhydrous AlCl₃ (and FeBr₃ or FeCl₃ for halogenation).
Usually costs: Condition marks.
Mistake: Counting 8 π electrons for cyclooctatetraene and calling it aromatic.
Hückel's rule needs 4n + 2, and the ring is non-planar.
Do this: Check planarity and the electron count.
Usually costs: Conceptual marks.

Exam tips

Write the electrophile of each reaction before the equation.
Show the three stages: electrophile, arenium ion, loss of H⁺.
Name catalysts exactly: conc. H₂SO₄, anhydrous AlCl₃, FeBr₃.
In synthesis questions state why the order of steps matters.
Quote the Hückel count with the value of n.

Practice questions on Benzene: Aromaticity, Electrophilic Substitution and Directing Effects

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The electrophile in the sulphonation of benzene with fuming sulphuric acid is
(a) SO₃ (b) HSO₄⁻ (c) H₂SO₄ (d) SO₂
Hint
Oleum contains dissolved SO₃.
Answer
(a).
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Nitrobenzene does not undergo Friedel–Crafts alkylation.
Reason (R): The nitro group strongly deactivates the ring towards electrophilic attack.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
The ring must be electron-rich enough to attack the carbocation.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
Write the equations for (a) the chlorination of benzene in the dark with FeCl₃ and (b) the reaction of benzene with excess chlorine in bright sunlight. Name the products.
Hint
One is substitution, the other is addition.
Answer
(a) C₆H₆ + Cl₂ → C₆H₅Cl + HCl (chlorobenzene). (b) C₆H₆ + 3Cl₂ → C₆H₆Cl₆ (benzene hexachloride, 1,2,3,4,5,6-hexachlorocyclohexane).
Q4 · 3 marks · Short answerHarder board level
Write the major mononitration product of (a) toluene, (b) benzaldehyde and (c) benzoic acid brominated with Br₂/FeBr₃ (for part c give the bromination product). State the directing effect in each case.
Hint
–CH₃ activates; –CHO and –COOH withdraw electrons.
Answer
(a) o- and p-nitrotoluene (−CH₃ is an o,p director). (b) 3-nitrobenzaldehyde (−CHO is a meta director). (c) 3-bromobenzoic acid (−COOH is a meta director).
Q5 · 4 marks · Long answerHOTS / competency
15.6 g of benzene is nitrated with excess nitrating mixture and the yield of nitrobenzene (123 g mol⁻¹) is 60%. (a) Write the equation and the electrophile. (b) Find the theoretical and actual mass of nitrobenzene. (c) What further product appears if conditions are made more severe?
Hint
Moles of benzene = 15.6/78.
Answer
(a) C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O (conc. H₂SO₄); electrophile NO₂⁺. (b) Moles of benzene = 0.20 mol, theoretical mass = 0.20 × 123 = 24.6 g; actual = 0.60 × 24.6 = 14.8 g. (c) m-Dinitrobenzene, because the first nitro group directs the second to the meta position.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Hückel's rule?Show answer
(4n + 2) π electrons.
2.Electrophile in nitration?Show answer
The nitronium ion NO₂⁺.
3.Catalyst for Friedel–Crafts reactions?Show answer
Anhydrous AlCl₃.
4.Directing effect of −NO₂?Show answer
Meta (deactivating).
5.Product of benzene with H₂/Ni?Show answer
Cyclohexane.
6.Hybridisation of carbon in benzene?Show answer
sp².

Benzene: Aromaticity, Electrophilic Substitution and Directing Effects: frequently asked questions

Why does benzene prefer substitution to addition?

Substitution keeps the delocalised π system (aromatic stability), while addition would destroy it.

Why is a Lewis acid needed for halogenation?

It polarises the halogen molecule to give a stronger electrophile.

Why does Friedel–Crafts alkylation give polyalkylation?

The alkyl group activates the ring, so the product reacts faster than benzene itself.

Is benzene harmful?

Benzene is toxic and carcinogenic, so handling needs care.

What is the arenium ion?

The resonance-stabilised cyclohexadienyl carbocation formed when an electrophile attacks the ring.