Preparation of Haloarenes: Electrophilic Halogenation and Sandmeyer Reaction
Chemistry · Haloalkanes And Haloarenes · NEET
Haloarenes (like chlorobenzene) are made in two main ways. First, direct halogenation: benzene reacts with Cl2 or Br2 using a Lewis acid catalyst (anhydrous FeCl3, AlCl3, or FeBr3) to give the aryl halide. Second, the Sandmeyer reaction: a benzene diazonium salt is treated with CuCl or CuBr to put Cl or Br on the ring. Memory hook: "Direct for Cl and Br, Diazonium for I" — because iodine will not go straight onto the ring, so iodobenzene needs a diazonium salt with KI.
Two NEET routes to haloarenes: direct electrophilic halogenation (benzene + Cl2/FeCl3) works for Cl and Br; the Sandmeyer reaction converts a diazonium salt to any aryl halide, and is the only way to get iodobenzene (with KI).
Your doubts, answered
How is chlorobenzene prepared from benzene?
Pass Cl2 through benzene with a Lewis acid catalyst (anhydrous FeCl3, AlCl3 or the metal iron). The catalyst makes the electrophile Cl+, which replaces one hydrogen on the ring. This is electrophilic aromatic substitution: C6H6 + Cl2 --(anhyd. FeCl3)--> C6H5Cl + HCl. The reaction happens in the dark and without heat, or you get side-chain products instead.
Why can iodine NOT be added directly to the benzene ring?
The direct reaction of benzene with I2 is very slow and reversible, so almost no iodobenzene forms. Also, the HI produced is a strong reducing agent that turns the product back into benzene. So NEET expects you to know: iodobenzene is NOT made by direct halogenation. Instead you make it from a benzene diazonium salt plus KI.
What is the Sandmeyer reaction and how do you use it to make haloarenes?
Start from aniline (C6H5NH2). Treat it with NaNO2 and HCl at 273-278 K (0-5 degrees C) to form benzene diazonium chloride, C6H5N2+Cl-. Then warm this with cuprous halide: with CuCl/HCl you get chlorobenzene, and with CuBr/HBr you get bromobenzene. The diazonium group (N2+) leaves as N2 gas and the halogen takes its place. Cu(I) is the key catalyst.
What is the difference between the Sandmeyer and Gattermann reactions?
Both replace the diazonium group with Cl or Br. Sandmeyer uses cuprous halide (CuCl or CuBr) with the matching HX. Gattermann uses copper powder with HCl or HBr instead of the cuprous salt. Sandmeyer gives a better yield, so it is the one NEET usually asks about.
How is iodobenzene prepared from aniline?
Make benzene diazonium chloride from aniline first (NaNO2/HCl at 273-278 K). Then just add potassium iodide (KI) and warm gently: C6H5N2+Cl- + KI --> C6H5I + N2 + KCl. No copper catalyst is needed for iodine. This is the standard route because iodine will not go directly onto the ring.
Is halogenation of benzene electrophilic or free-radical?
Ring halogenation (making chlorobenzene or bromobenzene) with a Lewis acid catalyst is ELECTROPHILIC substitution. But if you use UV light and no catalyst, Cl2 ADDS to benzene by a free-radical path to give hexachlorocyclohexane (BHC), which is a different product. And with toluene under UV light, Cl2 attacks the side chain by free radicals, not the ring.
⚠️ The NEET trap ✗ Thinking any diazonium reaction or any Cl2 reaction with benzene is electrophilic substitution. ✓ Only benzene + Cl2 with a Lewis acid (FeCl3/AlCl3) is electrophilic substitution. The Sandmeyer/diazonium reaction is an ionic substitution of the N2+ group, NOT electrophilic aromatic substitution. 🧠 Catalyst = Lewis acid means electrophilic; catalyst = Cu(I) on a diazonium salt means Sandmeyer, not electrophilic.
Real NEET questions
NEET 2019
Among the following, the reaction that proceeds through an electrophilic substitution is:
A · C6H5N2+I- + Cu2Cl2 -> C6H5I + N2
B · C6H6 + Cl2 --(AlCl3)--> C6H5Cl + HCl ✓
C · C6H6 + 3Cl2 --(hv)--> C6H6Cl6
D · CH4 + Cl2 --(heat)--> CH3Cl + HCl
Solution: Chlorination of benzene with Cl2 over the Lewis acid AlCl3 makes the electrophile Cl+, which replaces a ring hydrogen. This is classic electrophilic aromatic substitution giving chlorobenzene. Option A is a Sandmeyer-type reaction of a diazonium salt (ionic, not electrophilic). Option C is free-radical addition of Cl2 under UV giving BHC. Option D is free-radical substitution on methane. Only B is electrophilic substitution.
NEET 2022
Which sequence of reagents is suitable to synthesize chlorobenzene from benzene?
A · Benzene, Cl2, anhydrous FeCl3 ✓
B · Phenol, NaNO2, HCl, CuCl
C · Aniline (C6H5NH2), HCl
D · Aniline (C6H5NH2), HCl, heating
Solution: Direct electrophilic halogenation gives chlorobenzene: C6H6 + Cl2 --(anhyd. FeCl3)--> C6H5Cl + HCl. The Lewis acid FeCl3 makes the Cl+ electrophile. Option B starts from phenol, which has no -NH2 to diazotise, so no diazonium forms. Options C and D can form a diazonium/anilinium salt but have no CuCl, so no chlorine can be delivered to the ring.
NEET 2025
Statement I: Benzenediazonium salt is prepared by the reaction of aniline with acid at 273-278 K. It decomposes easily in the dry state. Statement II: Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI. Choose the most appropriate answer.
A · Statement I is correct but Statement II is incorrect
B · Statement I is incorrect but Statement II is correct
C · Both Statement I and Statement II are correct ✓
D · Both Statement I and Statement II are incorrect
Solution: Statement I is correct: aniline with NaNO2/HCl at 273-278 K gives benzene diazonium chloride, which is unstable and decomposes in the dry state, so it is kept in solution. Statement II is also correct: iodine will not go directly onto the ring, so iodobenzene is made from the diazonium salt with KI: C6H5N2+ + KI -> C6H5I + N2 + KCl. Both statements are correct (NCERT-based answer C).
Solved Haloalkanes And Haloarenes NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What are the two main methods to prepare haloarenes for NEET?
(1) Electrophilic halogenation: benzene + Cl2 or Br2 with a Lewis acid catalyst (FeCl3, AlCl3, FeBr3) in the dark. (2) Sandmeyer reaction: benzene diazonium salt + CuCl or CuBr gives chloro- or bromobenzene. Iodobenzene is a special case made from the diazonium salt with KI.
Why is anhydrous FeCl3 or AlCl3 needed in ring halogenation?
These are Lewis acids. They polarise the Cl-Cl bond and generate the electrophile Cl+ (or a Cl+ carrier), which is needed to attack the stable benzene ring. Without the catalyst, benzene will not react with Cl2 in the dark.
Can fluoroarenes be made by direct halogenation?
No. Direct fluorination of benzene is too violent and hard to control, and direct iodination does not work either. Both fluoro- and iodo-arenes are usually made from diazonium salts (fluorobenzene by the Balz-Schiemann route, iodobenzene with KI).
Does the halogen in the ring direct further substitution?
Yes. Halogens are ortho/para directing but deactivating. So a second electrophile enters mainly at the ortho and para positions, which links to the topic Electrophilic Substitution in Haloarenes.
What is the next topic after haloarene preparation?
Physical Properties of Haloalkanes, which explains boiling point, density and solubility trends. Understanding preparation first makes those property comparisons easier to remember.
