ChemistryISC Class 12 · Haloalkanes and Haloarenes

Reactions of Haloarenes: Chlorobenzene for ISC Class 12

Q: Why are haloarenes less reactive than haloalkanes towards nucleophiles?

Resonance gives the C–X bond partial double-bond character, the carbon is sp², the phenyl cation is unstable, and the ring blocks backside attack.

Q: What is the Wurtz–Fittig reaction?

An aryl halide and an alkyl halide react with sodium in dry ether to give an alkylbenzene, for example chlorobenzene and chloromethane give toluene.

Q: Why is halogen deactivating but ortho-para directing?

Its −I effect lowers electron density in the whole ring (slower reaction), while its +R effect raises it at the ortho and para positions (where attack occurs).

Q: How is phenol obtained from chlorobenzene?

By heating with aqueous NaOH at 623 K and 300 atm to give sodium phenoxide, then acidifying (Dow's process).

Q: Why do nitro groups at ortho and para positions make substitution easier?

They withdraw electrons and delocalise the negative charge of the intermediate carbanion, stabilising it.

Updated 15 October 2026 · Fact-checked

Haloarenes resist nucleophilic substitution because the C–X bond has partial double-bond character from resonance, so chlorobenzene gives phenol only at 623 K and 300 atm. They undergo electrophilic substitution (chlorination, nitration, sulphonation) at ortho and para positions, more slowly than benzene, and couple with sodium in the Fittig and Wurtz–Fittig reactions.

Chlorobenzene gives phenol only under drastic conditions, nitrates at ortho and para positions, and couples with sodium to form biphenyl (Fittig).

Illustration drawn for QuestionBuddy.

Understand Reactions of Haloarenes

Preparation. (i) Electrophilic halogenation of benzene with Cl₂ or Br₂ in the presence of a Lewis acid (anhydrous FeCl₃, FeBr₃ or iron), in the dark: $C_{6} H_{6} + C l_{2} anhyd. FeCl_{3} C_{6} H_{5} C l + H C l$ . The Lewis acid polarises Cl₂ to give the electrophile Cl⁺. (ii) From benzenediazonium chloride (made from aniline with NaNO₂ and HCl at 273–278 K): with Cu₂Cl₂/HCl (Sandmeyer) or Cu powder/HCl (Gattermann), $C_{6} H_{5} N_{2} C l \to C_{6} H_{5} C l + N_{2}$ ; CuBr/HBr gives bromobenzene.

Physical properties. Haloarenes are colourless liquids or crystalline solids, heavier than water, insoluble in water but soluble in organic solvents. The para-dihalo isomer has a higher melting point than ortho and meta because its symmetrical molecules pack better in the crystal.

Low reactivity towards nucleophiles. Compared with chloroethane, chlorobenzene is far less reactive because: (a) resonance gives the C–Cl bond partial double-bond character, making it shorter and stronger; (b) the carbon is sp² hybridised, more electronegative than sp³, so it holds the bonding pair more tightly; (c) the phenyl cation is not stabilised by resonance, so SN1 does not occur; (d) the electron-rich ring repels the nucleophile and blocks backside attack, so SN2 cannot occur.

Nucleophilic substitution under drastic conditions.
• Replacement by –OH (Dow's process): chlorobenzene with aqueous NaOH at 623 K and 300 atm gives sodium phenoxide, which acid converts to phenol: $C_{6} H_{5} C l + 2 N a O H \to C_{6} H_{5} O N a + N a C l + H_{2} O$ ; $C_{6} H_{5} O N a + H C l \to C_{6} H_{5} O H + N a C l$ .
• Replacement by –NH₂: with ammonia and Cu₂O catalyst at high temperature and pressure, $C_{6} H_{5} C l + 2 N H_{3} \to C_{6} H_{5} N H_{2} + N H_{4} C l$ .
• Effect of electron-withdrawing groups. A nitro group at the ortho or para position makes substitution much easier, because it withdraws electron density and stabilises the negatively charged intermediate (carbanion) by resonance: 1-chloro-4-nitrobenzene reacts with NaOH at 443 K, 1-chloro-2,4-dinitrobenzene at 368 K, and 2,4,6-trinitrochlorobenzene with warm water (323 K). A meta nitro group has little effect, because the negative charge of the intermediate is never placed on the carbon carrying it. Mechanism (addition-elimination): OH⁻ adds to the carbon bearing Cl to give a resonance-stabilised carbanion, then Cl⁻ is lost.

Electrophilic substitution. The halogen deactivates the ring (its −I effect withdraws electrons, so reactions are slower than with benzene), but it is ortho-para directing, because its +R effect (lone pair donated into the ring) raises electron density at the ortho and para positions. The para product usually predominates because the ortho positions are crowded by the halogen.
• Chlorination: $C_{6} H_{5} C l + C l_{2} anhyd. FeCl_{3} C_{6} H_{4} C l_{2} + H C l$ (1,2- and 1,4-dichlorobenzene).
• Nitration: with conc. HNO₃ and conc. H₂SO₄, $C_{6} H_{5} C l + H N O_{3} \to C_{6} H_{4} C l N O_{2} + H_{2} O$ (1-chloro-2-nitro- and 1-chloro-4-nitrobenzene, para major). The electrophile is the nitronium ion: $H N O_{3} + 2 H_{2} S O_{4} \to N O_{2}^{+} + H_{3} O^{+} + 2 H S O_{4}^{-}$ .
• Sulphonation: with conc. H₂SO₄ on heating, $C_{6} H_{5} C l + H_{2} S O_{4} \to C_{6} H_{4} C l (S O_{3} H) + H_{2} O$ (2- and 4-chlorobenzenesulphonic acids).
• Friedel–Crafts alkylation and acylation with anhydrous AlCl₃ give 1-chloro-2- and 1-chloro-4-methylbenzene, or the corresponding chloro-acetophenones.
General mechanism: (1) generation of the electrophile E⁺; (2) attack of the π electrons on E⁺ to form a resonance-stabilised arenium ion (σ complex); (3) loss of H⁺ restores aromaticity.

Reactions with metals. Wurtz–Fittig: aryl halide + alkyl halide + sodium in dry ether gives an alkylbenzene: $C_{6} H_{5} C l + 2 N a + C H_{3} C l \to C_{6} H_{5} C H_{3} + 2 N a C l$ . Fittig: two aryl halides give a diaryl: $2 C_{6} H_{5} C l + 2 N a \to C_{6} H_{5} C_{6} H_{5} + 2 N a C l$ (biphenyl). Grignard reagent: $C_{6} H_{5} B r + M g dry ether C_{6} H_{5} M g B r$ ; with water it gives benzene, and with CO₂ followed by acid it gives benzoic acid. Reduction: nascent hydrogen (for example Ni–Al alloy with NaOH) converts chlorobenzene to benzene.

Reactions of Haloarenes: formulas and key facts

Dow's process: $C_{6} H_{5} C l + 2 N a O H 623 K, 300 atm C_{6} H_{5} O N a + N a C l + H_{2} O$; Acidify to get phenol.
Nitronium ion (nitration electrophile): $H N O_{3} + 2 H_{2} S O_{4} \to N O_{2}^{+} + H_{3} O^{+} + 2 H S O_{4}^{-}$; Conc. H₂SO₄ protonates HNO₃ and removes water.
Wurtz–Fittig reaction: $C_{6} H_{5} C l + 2 N a + C H_{3} C l dry ether C_{6} H_{5} C H_{3} + 2 N a C l$; Aryl + alkyl halide.
Fittig reaction: $2 C_{6} H_{5} C l + 2 N a dry ether C_{6} H_{5} C_{6} H_{5} + 2 N a C l$; Two aryl halides give a diaryl (biphenyl).
Directing effect of halogen: $- I (deactivates) > + R (directs o, p)$; Slower than benzene, but ortho-para directing.
Activation by nitro groups (NaOH): $C_{6} H_{5} C l (623 K) < p - N O_{2} (443 K) < 2, 4 - (N O_{2})_{2} (368 K) < 2, 4, 6 - (N O_{2})_{3} (323 K)$; Only ortho and para nitro groups help.

How to solve Reactions of Haloarenes questions

ISC haloarene questions are conversions, reasons and mechanisms. For each conversion, decide whether you need an electrophile (ring substitution), a nucleophile under drastic conditions (replacing X) or a metal (coupling).

1To put a halogen on the ring: halogen with anhydrous FeCl₃/FeBr₃; or from aniline via the diazonium salt (Sandmeyer or Gattermann).
2To add a group to the ring of a haloarene: electrophilic substitution; give both ortho and para products and say para is major.
3To replace the halogen: use NaOH (623 K, 300 atm) for –OH or NH₃/Cu₂O (high T and P) for –NH₂; mention that o/p nitro groups make it easier.
4To join carbon chains: Fittig (aryl + aryl) or Wurtz–Fittig (aryl + alkyl) with sodium in dry ether.
5For mechanisms, write the electrophile-generating step, the arenium ion with resonance, and loss of H⁺.
6Balance every equation and write conditions over the arrow.

Reactions of Haloarenes 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.

Para isomers pack better

Suppose a student finds that moth-repellent cakes sold for wardrobes are made of 1,4-dichlorobenzene, a white solid that slowly sublimes, whereas a bottle of 1,2-dichlorobenzene in the laboratory is a liquid at room temperature.

Explain why the para isomer is a solid and the ortho isomer a liquid, and write the equation for making the dichlorobenzenes from chlorobenzene.

Try it first, then show the solutionSolution

Chlorination of chlorobenzene with Cl₂ and anhydrous FeCl₃ gives both isomers: $C_{6} H_{5} C l + C l_{2} \to C_{6} H_{4} C l_{2} + H C l$ , para major.
The two isomers have the same molar mass and similar intermolecular forces.
The para isomer is symmetrical, so its molecules fit closely in a crystal lattice; more energy is needed to break the lattice, giving a much higher melting point.
The less symmetrical ortho isomer packs poorly and melts below room temperature.

Answer: Symmetrical 1,4-dichlorobenzene packs better in the crystal, so it is a solid; 1,2-dichlorobenzene packs poorly and is a liquid. Both come from chlorination of chlorobenzene.

Memory hook: Symmetry packs tight: para melts higher.

Tricks to remember Reactions of Haloarenes

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

Halogen: slows down, still points o/p
Think of the halogen as a slow tour guide: it slows the group (−I, deactivating) but still points to the ortho and para seats (+R).
Use it when: Reason questions on reactivity and orientation.
Fittig fits two rings
Fittig joins two aryl rings (biphenyl); Wurtz–Fittig adds a Wurtz-style alkyl to one ring.
Use it when: Naming coupling reactions.
623-300 for plain chlorobenzene
No nitro: 623 K and 300 atm. Each extra o/p nitro group cuts the temperature: 443 K, 368 K, then warm water.
Use it when: Effect of nitro groups on nucleophilic substitution.

Reactions of Haloarenes: solved examples

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

Example 1: Major nitration product · 1 mark · MCQBoard level

The major product of nitration of chlorobenzene with a mixture of conc. HNO₃ and conc. H₂SO₄ is:
(a) 1-chloro-3-nitrobenzene
(b) 1-chloro-4-nitrobenzene
(c) 1,3-dinitrobenzene
(d) nitrobenzene

Try it first, then show the solutionSolution

Cl is ortho-para directing (+R effect).
The para position is less crowded than ortho, so the para isomer is major.

Answer: (b) 1-chloro-4-nitrobenzene.

Example 2: Aniline to chlorobenzene · 2 marks · Short answerBoard level

How will you convert aniline into chlorobenzene? Write equations and conditions.

Try it first, then show the solutionSolution

Diazotisation at 273–278 K: $C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l \to C_{6} H_{5} N_{2} C l + N a C l + 2 H_{2} O$ (1 mark).
Sandmeyer reaction with Cu₂Cl₂/HCl: $C_{6} H_{5} N_{2} C l \to C_{6} H_{5} C l + N_{2}$ (1 mark).

Answer: Diazotise aniline (NaNO₂/HCl, 273–278 K), then treat with Cu₂Cl₂/HCl.

Example 3: Mechanism of nitration of chlorobenzene · 3 marks · Short answerHarder board level

Write the mechanism of nitration of chlorobenzene to give 1-chloro-4-nitrobenzene, showing how the electrophile is formed.

Try it first, then show the solutionSolution

Electrophile: $H N O_{3} + 2 H_{2} S O_{4} \to N O_{2}^{+} + H_{3} O^{+} + 2 H S O_{4}^{-}$ (1 mark).
The π electrons of chlorobenzene attack NO₂⁺ at the para position, forming an arenium ion (σ complex) whose positive charge is delocalised over the ring and, through the lone pair of Cl, onto chlorine (1 mark).
HSO₄⁻ removes H⁺ from the carbon carrying NO₂, restoring the aromatic ring: 1-chloro-4-nitrobenzene + H₂SO₄ (1 mark).

Answer: NO₂⁺ formed from HNO₃ and H₂SO₄ attacks para to Cl; the arenium ion loses H⁺ to give 1-chloro-4-nitrobenzene.

Example 4: Replacing chlorine by –OH and –NH₂ · 3 marks · Short answerHarder board level

Write equations, with conditions, for converting chlorobenzene into (a) phenol and (b) aniline. Why are such severe conditions needed?

Try it first, then show the solutionSolution

(a) $C_{6} H_{5} C l + 2 N a O H 623 K, 300 a t m C_{6} H_{5} O N a + N a C l + H_{2} O$ , then $C_{6} H_{5} O N a + H C l \to C_{6} H_{5} O H + N a C l$ (1 mark).
(b) $C_{6} H_{5} C l + 2 N H_{3} C u_{2} O, hi g h T, P C_{6} H_{5} N H_{2} + N H_{4} C l$ (1 mark).
Resonance gives the C–Cl bond partial double-bond character, the carbon is sp², and neither SN1 (unstable phenyl cation) nor SN2 (backside blocked by the ring) is possible (1 mark).

Answer: (a) Dow's process: NaOH at 623 K and 300 atm, then acid. (b) NH₃ with Cu₂O at high T and P. The C–Cl bond is strengthened by resonance.

Example 5: A bromobenzene reaction map · 5 marks · Long answerHOTS / competency

Benzene is treated with Br₂ and anhydrous FeBr₃ to give A. A reacts with Mg in dry ether to give B. B gives C with water, and D with CO₂ followed by dilute acid. A heated with sodium in dry ether gives E.
(a) Identify A to E. [2½]
(b) Write equations for the formation of B and E, and name the reaction that gives E. [2½]

Try it first, then show the solutionSolution

(a) A = bromobenzene, C₆H₅Br; B = phenylmagnesium bromide, C₆H₅MgBr; C = benzene; D = benzoic acid, C₆H₅COOH; E = biphenyl, C₆H₅C₆H₅ (½ mark each).
(b) $C_{6} H_{5} B r + M g dry ether C_{6} H_{5} M g B r$ (1 mark).
$2 C_{6} H_{5} B r + 2 N a dry ether C_{6} H_{5} C_{6} H_{5} + 2 N a B r$ (1 mark).
This is the Fittig reaction (½ mark).

Answer: A C₆H₅Br, B C₆H₅MgBr, C C₆H₆, D C₆H₅COOH, E biphenyl (Fittig reaction).

Where students lose marks in Reactions of Haloarenes

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

Mistake: Starting a conversion from the wrong compound or missing the catalyst
ISC examiners noted that in 'benzene to biphenyl' many started from chlorobenzene and omitted the anhydrous AlCl₃/FeBr₃ step (2024).
Do this: Start from the compound given: benzene → bromobenzene (Br₂, FeBr₃) → biphenyl (Na, dry ether).
Usually costs: 1 mark of 2.
Mistake: Calling FeCl₃ a nucleophile generator
In a 2025 assertion-reason item many candidates accepted that FeCl₃ prepares a nucleophile.
Do this: FeCl₃ is a Lewis acid that generates the electrophile Cl⁺; halogenation of benzene is electrophilic substitution.
Usually costs: 1 mark.
Mistake: Unbalanced coupling equations
Students forget that two aryl halide molecules and two sodium atoms are needed.
Wrong working
$C_{6} H_{5} C l + N a \to C_{6} H_{5} C_{6} H_{5}$
Right working
$2 C_{6} H_{5} C l + 2 N a \to C_{6} H_{5} C_{6} H_{5} + 2 N a C l$
Do this: Count sodium and halide on both sides; two halogen atoms need two sodium atoms.
Usually costs: ½ to 1 mark.
Mistake: Saying chlorine activates the ring
Students confuse 'ortho-para directing' with 'activating'.
Do this: Halogens are deactivating (−I) but ortho-para directing (+R).
Usually costs: 1 mark.

Exam tips

In every conversion, write reagents and conditions over the arrow (anhyd. FeCl₃, 623 K, 300 atm, dry ether).
Name the reaction where one exists (Sandmeyer, Gattermann, Fittig, Wurtz–Fittig, Dow's process).
For electrophilic substitution, give both o- and p-products and mark the para isomer as major.
Learn the four reasons for low reactivity of haloarenes as a set: resonance, sp² carbon, unstable phenyl cation, repulsion by the ring.
Use the nitro-group temperature ladder (623, 443, 368, 323 K) to support reasoning about activation.

Practice questions on Reactions of Haloarenes

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
In the chlorination of benzene, anhydrous FeCl₃ acts as:
(a) a nucleophile
(b) a Lewis acid that helps generate the electrophile Cl⁺
(c) a reducing agent
(d) a free-radical initiator
Hint
It accepts a lone pair from chlorine.
Answer
(b). $C l_{2} + F e C l_{3} \to C l^{+} + F e C l_{4}^{-}$ ; the electrophile, not a nucleophile, is generated.
Q2 · 2 marks · Short answerBoard level
Write the equations for the Fittig and Wurtz–Fittig reactions using bromobenzene, and name the organic product of each.
Hint
Use bromomethane for the alkyl halide.
Answer
Fittig: $2 C_{6} H_{5} B r + 2 N a \to C_{6} H_{5} C_{6} H_{5} + 2 N a B r$ , biphenyl. Wurtz–Fittig: $C_{6} H_{5} B r + 2 N a + C H_{3} B r \to C_{6} H_{5} C H_{3} + 2 N a B r$ , toluene (methylbenzene). Both in dry ether.
Q3 · 2 marks · Short answerHarder board level
Chlorobenzene gives both ortho and para products in electrophilic substitution, but the para product is usually major. Explain both facts.
Hint
Think of the +R effect and of crowding.
Answer
The lone pairs on Cl are donated into the ring (+R effect), increasing electron density at the ortho and para positions, so the electrophile attacks there. The ortho positions are next to the large Cl atom, so attack there is sterically hindered, and the para product predominates.
Q4 · 3 marks · Short answerHarder board level
Write the equation for the sulphonation of chlorobenzene, name the products and identify the electrophile.
Hint
Concentrated sulphuric acid, heat.
Answer
$C_{6} H_{5} C l + H_{2} S O_{4} \to C_{6} H_{4} C l (S O_{3} H) + H_{2} O$ (conc. H₂SO₄, heat). Products: 2-chlorobenzenesulphonic acid (minor) and 4-chlorobenzenesulphonic acid (major). The electrophile is SO₃ (sulphur trioxide, from concentrated H₂SO₄).
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Chlorobenzene is ortho-para directing in electrophilic substitution.
Reason (R): The +R effect of chlorine increases the electron density at the ortho and para positions of the ring.
(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
Resonance structures place negative charge at o and p.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
How will you bring about the following conversions? (a) Aniline to bromobenzene. (b) Chlorobenzene to 4-chloroacetophenone. (c) Chlorobenzene to 4-chlorotoluene.
Hint
One Sandmeyer and two Friedel–Crafts reactions.
Answer
(a) NaNO₂/HCl at 273–278 K, then CuBr/HBr: $C_{6} H_{5} N_{2} C l + H B r C u B r C_{6} H_{5} B r + N_{2} + H C l$ . (b) CH₃COCl with anhydrous AlCl₃ (Friedel–Crafts acylation): $C_{6} H_{5} C l + C H_{3} C O C l \to C_{6} H_{4} C l (C O C H_{3}) + H C l$ , para isomer major. (c) CH₃Cl with anhydrous AlCl₃ (Friedel–Crafts alkylation): $C_{6} H_{5} C l + C H_{3} C l \to C_{6} H_{4} C l (C H_{3}) + H C l$ , separate the para isomer.
Q7 · 3 marks · Short answerHOTS / competency
Arrange in increasing order of reactivity towards aqueous NaOH, and explain the position of the meta compound: chlorobenzene, 1-chloro-3-nitrobenzene, 1-chloro-4-nitrobenzene, 1-chloro-2,4-dinitrobenzene.
Hint
Where does the negative charge of the intermediate sit?
Answer
Chlorobenzene ≈ 1-chloro-3-nitrobenzene < 1-chloro-4-nitrobenzene < 1-chloro-2,4-dinitrobenzene. Addition of OH⁻ gives a carbanion whose negative charge is delocalised to the carbons ortho and para to the site of attack. A nitro group at those positions withdraws and stabilises the charge (more nitro groups, more stabilisation). A meta nitro group is never at a carbon bearing the charge, so it gives little help.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a dye chemist needs 4-nitrophenol and has only chlorobenzene. She nitrates chlorobenzene, separates the para isomer, then heats it with aqueous NaOH at about 443 K and finally acidifies.
(i) Why does she nitrate first rather than make phenol first? [1]
(ii) Write the equation for the reaction with NaOH. [1]
(iii) Why can she use 443 K instead of the 623 K and 300 atm needed for chlorobenzene itself? [1]
Hint
A para nitro group helps nucleophilic substitution.
Answer
(i) With chlorine on the ring, nitration gives mainly the para isomer, and the nitro group then makes Cl easy to replace; making phenol first would need 623 K and 300 atm, and phenol nitrates to mixtures and can be oxidised. (ii) $C_{6} H_{4} C l N O_{2} + 2 N a O H \to C_{6} H_{4} (N O_{2}) O N a + N a C l + H_{2} O$ , then acidify to 4-nitrophenol. (iii) The para nitro group withdraws electrons and stabilises the carbanion intermediate by resonance, so the substitution needs much milder conditions.

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.Conditions for Dow's process?Show answer
Aqueous NaOH at 623 K and 300 atm, then acidification.
2.Electrophile in nitration?Show answer
The nitronium ion, NO₂⁺.
3.Product of the Fittig reaction of chlorobenzene?Show answer
Biphenyl.
4.Why is chlorobenzene ortho-para directing?Show answer
The +R effect of chlorine increases electron density at the ortho and para positions.
5.Reagent in the Gattermann reaction for chlorobenzene?Show answer
Copper powder with HCl on benzenediazonium chloride.
6.Which nitro position does not activate chlorobenzene towards NaOH?Show answer
Meta.

Reactions of Haloarenes: frequently asked questions

Why are haloarenes less reactive than haloalkanes towards nucleophiles?

Resonance gives the C–X bond partial double-bond character, the carbon is sp², the phenyl cation is unstable, and the ring blocks backside attack.

What is the Wurtz–Fittig reaction?

An aryl halide and an alkyl halide react with sodium in dry ether to give an alkylbenzene, for example chlorobenzene and chloromethane give toluene.

Why is halogen deactivating but ortho-para directing?

Its −I effect lowers electron density in the whole ring (slower reaction), while its +R effect raises it at the ortho and para positions (where attack occurs).

How is phenol obtained from chlorobenzene?

By heating with aqueous NaOH at 623 K and 300 atm to give sodium phenoxide, then acidifying (Dow's process).

Why do nitro groups at ortho and para positions make substitution easier?

They withdraw electrons and delocalise the negative charge of the intermediate carbanion, stabilising it.