Haloalkanes and Haloarenes · Lesson 6 of 16
Preparing Haloarenes
“Prepare aryl halides through ring substitution or diazonium replacement and explain their different requirements.”
• Distinguish electrophilic ring halogenation from radical side-chain halogenation. • Write diazotisation and halide-replacement equations with correct conditions. • Explain when Cu(I) salts are used and when iodide replacement does not require them. • Plan aryl-halide routes without assuming ordinary phenol or alkyl substitution chemistry.
Arene halogenation preserves the aromatic skeleton
Benzene can replace a ring hydrogen by chlorine or bromine under suitable electrophilic substitution conditions. A Lewis acid such as FeCl₃ or FeBr₃ activates the halogen toward the aromatic π system. The ring temporarily loses aromaticity during electrophile attack and regains it when a proton is removed. The net result is substitution, C₆H₆+X₂→C₆H₅X+HX, not permanent addition of two halogens across one benzene double bond.
For toluene, the methyl group favours ortho and para ring substitution under ring-halogenation conditions. The products are o- and p-halotoluene rather than benzyl halide. Darkness and a suitable Lewis-acid catalyst distinguish this intended route from radical side-chain halogenation under UV light. Product separation may exploit differences in physical properties, but an ortho/para mixture is not automatically cleanly separated by simply writing two structures.
Problem
Contrast chlorination of toluene on the ring and on the side chain.
- 1.With suitable Cl₂/FeCl₃ ring-substitution conditions, the new C–Cl bond forms at an aromatic carbon, giving mainly ortho/para chlorotoluene products.
- 2.With controlled radical chlorination under light, replacement at benzylic CH₃ can give C₆H₅CH₂Cl; further side-chain chlorination may follow if not controlled.
- 3.The reagent halogen alone does not choose the site. State the activation conditions and recognise whether a radical or electrophilic intermediate is being invoked.
Direct iodination is often reversible and may be assisted by an oxidant that removes or oxidises the HI formed, helping the net substitution proceed. Direct fluorination is highly reactive and difficult to control; the chlorine/bromine procedure is not a general selective recipe for making aryl fluorides. Special fluorination routes belong to a different set of conditions rather than an unqualified replacement of Cl₂ by F₂.
An aromatic amine provides a different entry point
Aniline contains –NH₂ directly attached to a benzene ring. In cold aqueous mineral acid, sodium nitrite generates nitrous acid and related nitrosating species; the primary aromatic amino group can be converted into a diazonium group, –N₂⁺. For benzenediazonium chloride the balanced preparation is C₆H₅NH₂+NaNO₂+2HCl→[C₆H₅N₂]Cl+NaCl+2H₂O. The mixture is kept near 273–278 K during diazotisation. The positive charge belongs to the diazonium ion, with a counterion maintaining overall neutrality.
A diazonium group is a useful leaving arrangement because its replacement releases stable N₂ gas. This provides an aryl substitution route that does not require ordinary backside SN2 attack at an aromatic carbon. Aromatic diazonium salts are normally used freshly in solution under appropriate conditions; the lesson describes the transformations rather than treating isolated dry salts as ordinary stable storage substances.
Problem
Give the route and identify the role of CuCl.
- 1.First diazotise aniline with NaNO₂ and aqueous HCl at 273–278 K to form benzenediazonium chloride.
- 2.Then use CuCl in an appropriate acidic medium for Sandmeyer replacement: [C₆H₅N₂]Cl→C₆H₅Cl+N₂. Cu(I)-mediated chemistry assists replacement of the diazonium group.
- 3.The Cu salt is not an extra chlorine atom to add on top of the balanced formula indiscriminately. Write it over the arrow when using the compact net equation and preserve ring connectivity.
| Desired aryl product | Diazonium route | Key distinction |
|---|---|---|
| Chlorobenzene | [C₆H₅N₂]Cl —CuCl/HCl→ C₆H₅Cl+N₂ | Cu(I)-assisted Sandmeyer route |
| Bromobenzene | [C₆H₅N₂]Br —CuBr/HBr→ C₆H₅Br+N₂ | Use the appropriate bromide system |
| Iodobenzene | [C₆H₅N₂]Cl+KI→C₆H₅I+N₂+KCl | KI replacement does not require Cu(I) halide |
The arrow label supplies a reagent or catalyst condition rather than an additional product formula. In the iodide equation, both K and Cl appear explicitly because the salt-exchange bookkeeping is shown. A compact ionic form is C₆H₅N₂⁺+I⁻→C₆H₅I+N₂, with charge zero on both sides. It is useful to recognise which spectators have been omitted, not to change atom counts between representations.
Why this route differs from replacing phenolic OH
Simple HCl treatment does not generally turn phenol into chlorobenzene because an aryl C–O bond is not the same as an alkyl alcohol bond. Similarly, NaCl is not ordinarily able to displace a ring amino group from aniline directly. Diazotisation deliberately creates a new leaving group whose departure is coupled to N₂ formation. Route design changes the functional group first rather than assuming every heteroatom substituent leaves equally well.
Problem
Verify C₆H₅NH₂+NaNO₂+2HCl→[C₆H₅N₂]Cl+NaCl+2H₂O.
- 1.Carbon is six on each side. Nitrogen is one from aniline plus one from nitrite, matching the two in diazonium.
- 2.Oxygen is two from nitrite and appears in two water molecules. Hydrogen totals 7+2=9 on the left and 5+4=9 on the right.
- 3.Sodium and the two chlorines are preserved in the diazonium salt and NaCl. Balancing the net equation is a necessary check, although it is not a complete detailed nitrosation mechanism.
Quick Quiz
Quiz
What is the normal net reaction type in benzene plus Br₂/FeBr₃?
Explanation: The ring regains aromaticity after H is replaced by Br.
What temperature range is used for the stated aromatic diazotisation?
Explanation: Cold aqueous acid is used to form the diazonium solution.
Which reagent can replace the diazonium group by I without a Cu(I) halide?
Explanation: Iodide replacement releases N₂ and does not need the CuCl/CuBr Sandmeyer conditions.
Why does diazotisation help aniline become an aryl halide?
Explanation: The leaving arrangement changes before the final substitution.
What condition distinguishes intended benzylic radical halogenation from ring halogenation?
Explanation: The activation mode selects a different pathway and reactive site.
Practice Problems
- Write balanced diazotisation of aniline using NaNO₂/HCl.
- Give the aniline-to-bromobenzene sequence.
- Write an ionic equation for diazonium iodide replacement and check charge.
- Why is adding HCl to phenol not equivalent to the diazonium route?
- Predict the ring products when toluene is brominated under appropriate Lewis-acid conditions.
- Why should direct iodine and fluorine chemistry not be assumed identical to chlorine chemistry?
C₆H₅NH₂+NaNO₂+2HCl→[C₆H₅N₂]Cl+NaCl+2H₂O at about 273–278 K. Nitrite supplies the second N and the acid supports formation of the nitrosating system and diazonium counterion.
Key Takeaways
• Electrophilic arene halogenation replaces a ring H and restores aromaticity. • Lewis-acid ring conditions differ from radical side-chain conditions. • Cold diazotisation turns an aromatic amino group into a useful diazonium leaving group. • CuCl and CuBr enable the stated Sandmeyer chloride and bromide routes. • Iodide replacement of diazonium can proceed without a Cu(I) halide. • Stable N₂ formation helps explain the usefulness of diazonium replacement. • Balanced net equations and catalyst labels convey different parts of a reaction description.