Reactions of haloarenes
The C-X bond in haloarenes has partial double bond character, so nucleophilic substitution is hard unless -NO₂ groups sit ortho or para. Halogen is deactivating but ortho-para directing.
Why aryl halides resist nucleophilic substitution
- Resonance: a lone pair on the halogen is shared with the ring, so the C-X bond gets partial double bond character. It is harder to break than a single bond.
- Hybridisation: the carbon of C-X is sp². It holds the bonding electrons more tightly, so the bond is shorter and stronger than in a haloalkane.
- Unstable phenyl cation: if C-X broke on its own, the phenyl cation formed would not be stabilised by resonance, so the SN1 path is ruled out.
- Repulsion: the electron-rich ring repels an electron-rich nucleophile, so it is hard for the nucleophile to approach.
Nucleophilic substitution, and how -NO₂ helps
- 1★C₆H₅Cl + 2NaOH → C₆H₅ONa + NaCl + H₂O; then C₆H₅ONa + HCl → C₆H₅OH + NaClAqueous NaOH at 623 K and 300 atm, then acid. Your textbook gives this process no name (industry calls it Dow's process)
- 2O₂NC₆H₄Cl (Cl and NO₂ at 1 and 4) + 2NaOH → O₂NC₆H₄ONa + NaCl + H₂O; then H⁺ → 4-nitrophenolOnly 443 K is needed
- 31-Chloro-2,4-dinitrobenzene + aqueous NaOH, then H⁺ → 2,4-dinitrophenolOnly 368 K is needed
- 42-Chloro-1,3,5-trinitrobenzene + warm water → 2,4,6-trinitrophenol (picric acid) + HClJust warming with water is enough
- 5Mechanism, step 1: OH⁻ adds to the ring carbon that carries Cl, forming a negatively charged intermediate (a carbanion).
- 6★Mechanism, step 2: an -NO₂ group at the ortho or para position takes part of the negative charge by resonance, which stabilises the intermediate. Then Cl⁻ leaves and the aromatic ring is restored.A meta -NO₂ cannot take this charge by resonance, so your textbook says it has no effect on reactivity
★ key step
Why halogen is deactivating but ortho-para directing
The halogen's lone pairs are shared with the ring by resonance. This raises the electron density at the ortho and para positions more than at meta, so electrophiles go there. At the same time the halogen pulls electrons out of the ring by its −I effect (inductive effect). So the ring is less reactive than benzene: it is deactivated, and reactions need harsher conditions. The para product is usually major, because the large halogen crowds the ortho positions.
Electrophilic substitution of chlorobenzene
Each equation shows two isomeric products; each molecule of chlorobenzene gives one of them.
- 1C₆H₅Cl + Cl₂ → 1,2-dichlorobenzene (minor) + 1,4-dichlorobenzene (major), + HClAnhydrous FeCl₃
- 2★C₆H₅Cl + HNO₃ → 1-chloro-2-nitrobenzene (minor) + 1-chloro-4-nitrobenzene (major), + H₂OConc HNO₃ + conc H₂SO₄
- 3C₆H₅Cl + H₂SO₄ → 2-chlorobenzenesulphonic acid (minor) + 4-chlorobenzenesulphonic acid (major), + H₂OConc H₂SO₄, heat
- 4C₆H₅Cl + CH₃Cl → 1-chloro-2-methylbenzene (minor) + 1-chloro-4-methylbenzene (major), + HClFriedel-Crafts alkylation, anhydrous AlCl₃
- 5C₆H₅Cl + CH₃COCl → 2-chloroacetophenone (minor) + 4-chloroacetophenone (major), + HClFriedel-Crafts acylation, anhydrous AlCl₃
★ key step
Reaction with metals
- 1★C₆H₅Br + CH₃Br + 2Na → C₆H₅CH₃ + 2NaBrWurtz-Fittig reaction, dry ether: an aryl halide and an alkyl halide join to give an alkylarene (here toluene)
- 2★2C₆H₅Cl + 2Na → C₆H₅-C₆H₅ + 2NaClFittig reaction, dry ether: two aryl halides join to give a diaryl (here biphenyl)
★ key step
One industrial route to paracetamol, the common fever tablet, starts from chlorobenzene. Nitration gives 1-chloro-4-nitrobenzene, the -NO₂ group helps swap Cl for OH, and the 4-nitrophenol is reduced to 4-aminophenol, which is acetylated to give paracetamol.
3 marks: “Give three reasons why chlorobenzene is less reactive than chloromethane towards nucleophilic substitution.” Resonance gives C-Cl partial double bond character; the carbon is sp², so the bond is shorter and stronger; the phenyl cation is unstable (no SN1); the electron-rich ring repels the nucleophile.
2 marks: “Chlorobenzene is nitrated more slowly than benzene, yet the nitro group goes mainly to the ortho and para positions. Explain both facts.” Chlorine's −I effect withdraws electrons and deactivates the ring, so the reaction is slower. Its lone pairs, by resonance, raise electron density at ortho and para positions more than at meta, so electrophiles attack there.
2 marks: “Write the Wurtz-Fittig and Fittig reactions.” C₆H₅Br + CH₃Br + 2Na (dry ether) → C₆H₅CH₃ + 2NaBr. 2C₆H₅Cl + 2Na (dry ether) → C₆H₅-C₆H₅ + 2NaCl.
Ring holds X tight: resonance, sp², no cation, repulsion. NO₂ at ortho or para opens the door; at meta it stays shut.