ChemistryCBSE Class 12 · Haloalkanes and Haloarenes

Reactions of Haloarenes: Class 12 Chemistry

Updated 15 October 2026 · Fact-checked

Haloarenes such as chlorobenzene resist nucleophilic substitution because resonance gives the C–Cl bond partial double-bond character and the carbon is $sp^2$. Electron-withdrawing nitro groups at ortho or para positions make substitution easier. In electrophilic substitution, the halogen deactivates the ring but directs incoming groups to ortho and para positions. With sodium they give Wurtz–Fittig and Fittig products.

Understand Reactions of Haloarenes

Haloarenes (aryl halides) have the halogen bonded directly to an $s p^{2}$ carbon of a benzene ring. They react quite differently from haloalkanes.

1. Nucleophilic substitution is difficult. NCERT gives four reasons. (a) Resonance: a lone pair on chlorine is delocalised into the ring, so the C–Cl bond gains partial double-bond character and becomes harder to break. (b) Hybridisation: the carbon is $s p^{2}$ , with more s-character than the $s p^{3}$ carbon of a haloalkane; it holds the bonding electrons more tightly, giving a shorter, stronger bond (C–Cl is 169 pm in chlorobenzene against 177 pm in chloromethane). (c) Instability of the phenyl cation: an SN1 route is ruled out because the phenyl cation is not stabilised by resonance. (d) Repulsion: an electron-rich nucleophile is repelled by the electron-rich π cloud of the ring, and backside attack (SN2) is blocked by the ring itself.

Forcing conditions. Chlorobenzene gives phenol only when heated with aqueous NaOH at 623 K and 300 atm, followed by acidification: $C_{6} H_{5} C l N a O H, 623 K, 300 atm C_{6} H_{5} O^{-} N a^{+} H^{+} C_{6} H_{5} O H$ .

Effect of electron-withdrawing groups. A nitro group at the ortho or para position makes the replacement much easier, and each extra nitro group lowers the temperature needed: 1-chloro-4-nitrobenzene reacts at 443 K, 1-chloro-2,4-dinitrobenzene at 368 K, and 2,4,6-trinitrochlorobenzene with warm water at 323 K. The nucleophile adds to the carbon bearing Cl to form a negatively charged intermediate; when $N O_{2}$ is ortho or para, resonance places that negative charge on the carbon bearing the nitro group, where it can be delocalised onto the nitro oxygens. A meta nitro group cannot do this, so it has little effect.

2. Electrophilic substitution. Chlorobenzene undergoes halogenation, nitration, sulphonation and Friedel–Crafts reactions, but more slowly than benzene. Chlorine's strong −I effect withdraws electron density and deactivates the ring. At the same time its +R effect (lone pair donated into the ring) increases electron density at the ortho and para positions more than at meta. So the halogen is deactivating but ortho, para-directing. The para product is usually the major one, because the ortho positions are crowded by the halogen.

Halogenation: $C l_{2}$ , anhydrous $F e C l_{3}$ → 1,2- and 1,4-dichlorobenzene.
Nitration: conc. $H N O_{3}$ + conc. $H_{2} S O_{4}$ → 1-chloro-2-nitrobenzene and 1-chloro-4-nitrobenzene.
Sulphonation: conc. $H_{2} S O_{4}$ → 2- and 4-chlorobenzenesulphonic acid.
Friedel–Crafts alkylation: $C H_{3} C l$ , anhydrous $A l C l_{3}$ → 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene.
Friedel–Crafts acylation: $C H_{3} C O C l$ , anhydrous $A l C l_{3}$ → 2- and 4-chloroacetophenone.

3. Reaction with metals. In the Wurtz–Fittig reaction, an aryl halide and an alkyl halide react with sodium in dry ether to give an alkylarene: $C_{6} H_{5} C l + C H_{3} C l + 2 N a \to C_{6} H_{5} C H_{3} + 2 N a C l$ . In the Fittig reaction, two aryl halides join: $2 C_{6} H_{5} C l + 2 N a \to C_{6} H_{5} - C_{6} H_{5} + 2 N a C l$ (biphenyl).

Para-dihalobenzenes melt higher than their ortho and meta isomers, because the symmetrical para molecules pack better in the crystal lattice.

Reactions of Haloarenes: formulas and key facts

Phenol from chlorobenzene (Dow-type conditions): $C_{6} H_{5} C l + 2 N a O H 623 K, 300 atm C_{6} H_{5} O N a + N a C l + H_{2} O; C_{6} H_{5} O N a H^{+} C_{6} H_{5} O H$; Forcing conditions show the low reactivity of aryl halides.
Effect of nitro groups (temperature needed): $4-nitro: 443 K; 2,4-dinitro: 368 K; 2,4,6-trinitro: 323 K (warm water)$; Only ortho and para nitro groups help.
Nitration of chlorobenzene: $C_{6} H_{5} C l + H N O_{3} co n c . H_{2} S O_{4} C l C_{6} H_{4} N O_{2} (o + p) + H_{2} O$; Para isomer is major.
Wurtz–Fittig reaction: $C_{6} H_{5} C l + C H_{3} C l + 2 N a dry ether C_{6} H_{5} C H_{3} + 2 N a C l$; 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 groups join (biphenyl).
Directing effect of halogen: $- I > + R \Rightarrow deactivating; + R \Rightarrow o, p -directing$; Explains slower but o/p substitution.

How to solve Reactions of Haloarenes questions

Expect reasoning ('give reasons'), product prediction ('complete the reaction') and conversions. Keep the three reaction types separate in your mind: nucleophilic, electrophilic, with metals.

1Identify the reagent type: nucleophile ( $O H^{-}$ ), electrophile source ( $C l_{2} / F e C l_{3}$ , $H N O_{3} / H_{2} S O_{4}$ , $R C l / A l C l_{3}$ ) or metal (Na).
2For nucleophilic substitution, check the ring: plain chlorobenzene needs 623 K and 300 atm; ortho/para nitro groups make it easier.
3For electrophilic substitution, put the new group ortho and para to the halogen; name both products and say para is major.
4For Wurtz–Fittig, combine the aryl and alkyl groups; for Fittig, join two aryl groups.
5For reasons, name the effect (resonance, $s p^{2}$ hybridisation, −I, +R, phenyl cation instability, repulsion) and connect it to the observation.
6For conversions, plan backwards from the target, choose the step that places groups in the right positions, and write conditions on each arrow.
7Write each equation with correct formulae and check that it is balanced (HCl or NaCl by-products).

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.

Ortho-para direction and separation of isomers

Suppose a small chemical unit chlorinates chlorobenzene with chlorine and anhydrous ferric chloride to make 1,4-dichlorobenzene for a customer. The crude product is a mixture.

Name the products formed and explain why both appear. Which one is major, and how could the difference in melting points help to obtain the para isomer?

Try it first, then show the solutionSolution

Chlorine on the ring is ortho, para-directing (+R effect), so the electrophile $C l^{+}$ enters at positions 2 and 4.
Products: 1,2-dichlorobenzene and 1,4-dichlorobenzene: $C_{6} H_{5} C l + C l_{2} \to C_{6} H_{4} C l_{2} + H C l$ .
The para isomer is major because the ortho positions are crowded next to the existing chlorine.
1,4-Dichlorobenzene is symmetrical and packs well in a crystal, so it has a much higher melting point than the ortho isomer and can be crystallised out on cooling.

Answer: 1,2- and 1,4-dichlorobenzene; the para isomer is major and, having the higher melting point, crystallises out on cooling.

Memory hook: Halogen on the ring points newcomers to ortho and para; the neat, symmetric para isomer freezes first.

Tricks to remember Reactions of Haloarenes

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

RHPR: four reasons aryl halides resist nucleophiles
Resonance, Hybridisation ( $s p^{2}$ ), Phenyl cation unstable, Repulsion by π electrons.
Use it when: 3-mark 'give reasons' questions.
Nitro at o/p, temperature drops
No nitro 623 K → one nitro 443 K → two 368 K → three 323 K.
Use it when: Ordering reactivity of nitrochlorobenzenes.
Halogen: slows the ring, steers to o/p
−I applies the brake (deactivating); +R turns the steering wheel to ortho and para.
Use it when: Explaining the directing effect.

Reactions of Haloarenes: solved examples

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

Example 1: Major product of chlorination · 1 mark · MCQBoard level

The major product when chlorobenzene reacts with $C l_{2}$ in the presence of anhydrous $F e C l_{3}$ is:
(a) 1,3-dichlorobenzene (b) 1,4-dichlorobenzene (c) benzene hexachloride (d) chlorocyclohexane

Try it first, then show the solutionSolution

Cl is ortho, para-directing; the para position is less hindered.

Answer: (b) 1,4-dichlorobenzene.

Example 2: Three chlorides compared · 2 marks · Very short answerBoard level

Arrange chlorobenzene, chloromethane and benzyl chloride ( $C_{6} H_{5} C H_{2} C l$ ) in increasing order of reactivity towards aqueous NaOH, giving one reason for the position of each.

Try it first, then show the solutionSolution

Order: chlorobenzene < chloromethane < benzyl chloride (½ mark).
Chlorobenzene: resonance gives C–Cl partial double-bond character and the carbon is $s p^{2}$ , so the bond is hard to break (½ mark).
Chloromethane: $s p^{3}$ carbon, unhindered, reacts readily by SN2 (½ mark).
Benzyl chloride: the C–Cl carbon is $s p^{3}$ and the benzylic carbocation is resonance-stabilised, so it reacts fastest (SN1 as well as SN2) (½ mark).

Answer: Chlorobenzene < chloromethane < benzyl chloride.

Example 3: Nitro groups and reactivity · 3 marks · Short answerHarder board level

Arrange in increasing order of reactivity towards aqueous NaOH, and explain why the ortho/para nitro group helps: chlorobenzene, 1-chloro-2,4-dinitrobenzene, 1-chloro-4-nitrobenzene, 2,4,6-trinitrochlorobenzene. Why does 1-chloro-3-nitrobenzene not show a similar increase?

Try it first, then show the solutionSolution

Order: chlorobenzene (623 K) < 1-chloro-4-nitrobenzene (443 K) < 1-chloro-2,4-dinitrobenzene (368 K) < 2,4,6-trinitrochlorobenzene (323 K) (1 mark).
When $O H^{-}$ adds to the carbon bearing Cl, a negatively charged intermediate forms. With $N O_{2}$ at ortho or para, one resonance form puts the negative charge on the carbon attached to $N O_{2}$ , so it spreads onto the nitro oxygens and the intermediate is stabilised (1½ marks).
A meta nitro group never carries this charge in any resonance form, so it gives little stabilisation (½ mark).

Answer: Chlorobenzene < 4-nitro < 2,4-dinitro < 2,4,6-trinitro; o/p nitro groups delocalise the negative charge of the intermediate, meta cannot.

Example 4: Deactivating yet o,p-directing · 3 marks · Short answerHarder board level

Explain why chlorine on a benzene ring is deactivating but ortho, para-directing. Write the equation for the nitration of chlorobenzene and name the products.

Try it first, then show the solutionSolution

Chlorine is strongly electronegative: its −I effect pulls electron density from the whole ring, making it less reactive than benzene (deactivating) (1 mark).
Its lone pairs are donated into the ring by resonance (+R), which raises electron density mainly at the ortho and para positions; the electrophile therefore attacks these positions (1 mark).
$C_{6} H_{5} C l + H N O_{3} co n c . H_{2} S O_{4} C l C_{6} H_{4} N O_{2} + H_{2} O$ : 1-chloro-2-nitrobenzene (minor) and 1-chloro-4-nitrobenzene (major) (1 mark).

Answer: −I outweighs +R for overall rate (deactivating); +R controls position (o, p). Products: 1-chloro-2-nitro and 1-chloro-4-nitrobenzene.

Example 5: Planning conversions · 5 marks · Long answerHOTS / competency

How will you bring about the following conversions? Write equations with conditions. (a) Chlorobenzene to toluene (b) Chlorobenzene to biphenyl (c) Chlorobenzene to 4-chloroacetophenone (d) Benzene to 4-nitrophenol

Try it first, then show the solutionSolution

(a) Wurtz–Fittig: $C_{6} H_{5} C l + C H_{3} C l + 2 N a dry ether C_{6} H_{5} C H_{3} + 2 N a C l$ (1 mark).
(b) Fittig: $2 C_{6} H_{5} C l + 2 N a dry ether C_{6} H_{5} - C_{6} H_{5} + 2 N a C l$ (1 mark).
(c) Friedel–Crafts acylation: $C_{6} H_{5} C l + C H_{3} C O C l anhyd. A l C l_{3} 4 - C l C_{6} H_{4} C O C H_{3} + H C l$ (with some 2-isomer) (1 mark).
(d) Benzene + $C l_{2}$ (anhyd. $F e C l_{3}$ ) → chlorobenzene; nitration (conc. $H N O_{3}$ / $H_{2} S O_{4}$ ) → 1-chloro-4-nitrobenzene (separate from the ortho isomer) (1 mark).
Then aqueous NaOH at 443 K followed by dilute HCl replaces Cl by OH, giving 4-nitrophenol; the para nitro group makes this step feasible (1 mark).

Answer: (a) Wurtz–Fittig with $C H_{3} C l$ /Na. (b) Fittig with Na. (c) $C H_{3} C O C l$ / $A l C l_{3}$ . (d) Chlorination → nitration → NaOH at 443 K, then $H^{+}$ .

Where students lose marks in Reactions of Haloarenes

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

Mistake: Putting the incoming group meta to chlorine
Do this: Halogens are ortho, para-directors; write both o and p products and mark para as major.
Usually costs: 1 mark per product.
Mistake: Giving phenol from chlorobenzene with dilute NaOH at room temperature
Do this: Write the conditions: 623 K, 300 atm, then acidify.
Usually costs: 1 mark.
Mistake: Mixing up Wurtz–Fittig and Fittig
Do this: Wurtz–Fittig: aryl + alkyl halide → alkylbenzene. Fittig: two aryl halides → diaryl.
Usually costs: 1 mark.
Mistake: Saying the halogen activates the ring because of +R
Do this: Net effect: −I > +R, so the ring is deactivated; +R only decides the position.
Usually costs: ½ to 1 mark.

Exam tips

Write reagents and conditions over the arrow (anhyd. $A l C l_{3}$ , dry ether, 623 K, 300 atm).
For o/p products, write both names and state which is major.
Use the four NCERT reasons by name in 'less reactive' questions.
In conversions, number each step and show intermediate names.
Remember the nitro temperatures (443 K, 368 K, 323 K) as evidence in explanations.

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
Chlorobenzene and chloromethane heated with sodium in dry ether give mainly:
(a) biphenyl (b) ethane (c) toluene (d) benzene
Hint
Wurtz–Fittig reaction.
Answer
(c) toluene.
Q2 · 2 marks · Very short answerBoard level
Complete the reactions: (a) $C_{6} H_{5} C l + C H_{3} C O C l anhyd. A l C l_{3}$ (b) $C_{6} H_{5} C l + conc. H_{2} S O_{4} Δ$
Hint
Ortho and para products.
Answer
(a) 2-Chloroacetophenone (minor) + 4-chloroacetophenone (major) + HCl. (b) 2-Chlorobenzenesulphonic acid (minor) + 4-chlorobenzenesulphonic acid (major) + $H_{2} O$ .
Q3 · 3 marks · Short answerHarder board level
Under what conditions is chlorobenzene converted into phenol? Write the equations and explain why such conditions are needed.
Hint
High temperature and pressure.
Answer
Heat with aqueous NaOH at 623 K and 300 atm: $C_{6} H_{5} C l + 2 N a O H \to C_{6} H_{5} O N a + N a C l + H_{2} O$ ; then acidify: $C_{6} H_{5} O N a + H C l \to C_{6} H_{5} O H + N a C l$ . Harsh conditions are needed because the C–Cl bond has partial double-bond character (resonance) and the $s p^{2}$ carbon holds it strongly.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Chlorobenzene undergoes electrophilic substitution more slowly than benzene.
Reason (R): The −I effect of chlorine reduces the electron density 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
Deactivating group.
Answer
(a).
Q5 · 2 marks · Very short answerHarder board level
Why does 1,4-dichlorobenzene have a higher melting point than 1,2- and 1,3-dichlorobenzene, although their boiling points are similar?
Hint
Think of crystal packing.
Answer
The para isomer is symmetrical, so its molecules fit closely in the crystal lattice and more energy is needed to melt it. Boiling points depend on intermolecular forces in the liquid, which are similar for the isomers.
Q6 · 2 marks · Very short answerHarder board level
Which is more reactive towards nucleophilic substitution, 1-chloro-4-nitrobenzene or 1-chloro-3-nitrobenzene? Explain with reference to the intermediate formed.
Hint
Where can the negative charge go?
Answer
1-Chloro-4-nitrobenzene. In the intermediate formed by attack of $O H^{-}$ , the negative charge can reach the carbon bonded to the para $N O_{2}$ and be delocalised onto its oxygens; with meta $N O_{2}$ this is not possible.
Q7 · 3 marks · Short answerHOTS / competency
Explain why the C–Cl bond in chlorobenzene (169 pm) is shorter than in chloromethane (177 pm), and why an SN1 mechanism is not possible for chlorobenzene.
Hint
Two effects shorten the bond.
Answer
The carbon in chlorobenzene is $s p^{2}$ (33% s-character) rather than $s p^{3}$ (25%), so it is more electronegative and holds the bonding pair closer; resonance also gives the bond partial double-bond character. SN1 would require a phenyl cation, which is very unstable because its empty $s p^{2}$ orbital is in the plane of the ring and cannot be stabilised by the π system.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student treats chlorobenzene with three reagents in separate experiments: (i) aqueous NaOH at 373 K, (ii) $C H_{3} C l$ with anhydrous $A l C l_{3}$ , (iii) $C H_{3} C l$ with sodium in dry ether. In (i) almost nothing happens; in (ii) and (iii) she obtains different products, each containing a methyl group.
(a) Why is there no reaction in (i)? (1 mark)
(b) Name the reaction type and products in (ii). (1 mark)
(c) Name the reaction and product in (iii). (1 mark)
(d) Why do the products of (ii) still contain chlorine while the product of (iii) does not? (1 mark)
Hint
One reaction substitutes a ring H, the other replaces Cl.
Answer
(a) The C–Cl bond is strengthened by resonance and $s p^{2}$ hybridisation; replacement needs about 623 K and 300 atm. (b) Friedel–Crafts alkylation (electrophilic substitution): 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene. (c) Wurtz–Fittig reaction: toluene. (d) In (ii) the electrophile replaces a ring hydrogen and Cl stays as a directing group; in (iii) sodium removes the halogens and joins the aryl and methyl groups.

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 to convert chlorobenzene to phenol?Show answer
Aqueous NaOH, 623 K, 300 atm, then acidification.
2.Product of the Fittig reaction of chlorobenzene?Show answer
Biphenyl.
3.Is Cl on benzene activating or deactivating?Show answer
Deactivating (but o, p-directing).
4.Temperature for replacing Cl in 1-chloro-2,4-dinitrobenzene?Show answer
368 K.
5.C–Cl bond length in chlorobenzene?Show answer
169 pm.
6.Major product of nitration of chlorobenzene?Show answer
1-Chloro-4-nitrobenzene.

Reactions of Haloarenes: frequently asked questions

Why are haloarenes less reactive than haloalkanes towards nucleophilic substitution?

Because of resonance (partial double-bond character in C–X), the sp^2 carbon (shorter, stronger bond), the instability of the phenyl cation, and repulsion between the nucleophile and the ring's π electrons.

Why is chlorine ortho and para directing?

Its lone pairs are donated into the ring by resonance, increasing electron density at the ortho and para positions, where electrophiles attack.

What is the Wurtz–Fittig reaction?

A reaction of an aryl halide and an alkyl halide with sodium in dry ether that joins the aryl and alkyl groups, for example chlorobenzene and chloromethane give toluene.

How do nitro groups affect nucleophilic substitution in haloarenes?

Nitro groups at ortho or para positions withdraw electrons and stabilise the negatively charged intermediate, so substitution happens under much milder conditions.

Why is the para isomer the major product in electrophilic substitution of chlorobenzene?

The ortho positions are next to the chlorine atom and are sterically crowded, so attack at the para position is favoured.