Preparation of Haloarenes

Why Haloarenes Need Their Own Routes

Aryl halides are defined by a halogen bonded directly to the sp²-hybridised carbon of an aromatic ring. This is the dividing line from haloalkanes, where the halogen sits on an sp³ carbon. The distinction is not cosmetic: it dictates that the preparative chemistry of the two families barely overlaps.

Haloalkanes are conveniently built from alcohols, replacing the hydroxyl group with halogen using concentrated halogen acids, phosphorus halides, or thionyl chloride. NCERT is explicit that this strategy cannot be carried over to aryl systems. In phenols the carbon-oxygen bond has partial double-bond character — the oxygen lone pair is delocalised into the ring — making it stronger than an ordinary single bond and very difficult to cleave. The aromatic ring refuses to surrender its oxygen the way an alkyl chain surrenders its hydroxyl.

Haloarenes therefore arrive by two routes that have no counterpart in haloalkane synthesis: electrophilic substitution on the arene itself, and replacement of a diazonium group derived from an aromatic amine. The rest of this note dissects both.

Route 1 — Electrophilic Halogenation of Arenes

Aryl chlorides and bromides are made most directly by treating an arene with chlorine or bromine in the presence of a Lewis-acid catalyst such as iron filings or iron(III) halide. For benzene the canonical reactions are:

$$\ce{C6H6 + Cl2 ->[\text{anhydrous } FeCl3] C6H5Cl + HCl}$$ $$\ce{C6H6 + Br2 ->[\text{anhydrous } FeBr3] C6H5Br + HBr}$$

The iron(III) halide is the working catalyst whether you charge $\ce{FeCl3}$ directly or generate it in situ from iron filings and the halogen ($\ce{2Fe + 3Cl2 -> 2FeCl3}$). Its job is to convert the relatively unreactive halogen molecule into a far more electrophilic species capable of attacking the aromatic ring. NIOS §25.2.2 describes exactly this reaction and notes that "iron filings or iron(III) halide is used as the catalyst."

When the arene is a substituted benzene such as chlorobenzene, the products are ortho and para isomers — and these can be separated cleanly, because NCERT records that "the ortho and para isomers can be easily separated due to large difference in their melting points." The para isomer is notably higher melting because its symmetry packs better into the crystal lattice.

The Electrophilic Substitution Mechanism

The reaction is an electrophilic aromatic substitution, and it proceeds in the three conceptual stages drawn above. First, the Lewis acid abstracts a halide from the halogen molecule, leaving a strongly electrophilic halonium ion (or a tightly polarised halogen-Lewis-acid complex). Second, the π-electron cloud of the ring attacks this electrophile, breaking aromaticity to form a resonance-stabilised arenium ion (the σ-complex) in which one ring carbon has become sp³. Third, this intermediate loses a proton — the proton that the new halogen displaced — restoring the aromatic sextet and regenerating the catalyst.

Substitution, not addition, is the outcome precisely because the system pays a high price to break aromaticity in step two and is therefore strongly driven to recover it in step three. The same mechanistic skeleton governs nitration, sulphonation and Friedel-Crafts reactions of arenes, which is why mastering it here pays dividends across the whole aromatic-chemistry syllabus.

Orientation: Why ortho and para

When the arene already carries a halogen, where does the next substituent go? NCERT settles this clearly: the halogen atom, "besides being slightly deactivating is o,p-directing." The next electrophile therefore enters predominantly at the ortho and para positions relative to the existing halogen. The reasoning hinges on the dual personality of the halogen.

On one hand, the halogen withdraws electron density through its -I (inductive) effect, deactivating the ring overall and slowing every electrophilic substitution. On the other hand, the halogen lone pairs are donated into the ring by resonance, and the resonating structures place extra electron density specifically at the ortho and para positions. The inductive effect is the stronger of the two; it dominates reactivity and renders the ring net-deactivated. But for the question of where, the resonance contribution tips the balance toward ortho and para.

Reactivity is controlled by the stronger inductive effect; orientation is controlled by the resonance effect. This single sentence resolves the apparent paradox of a deactivating-yet-ortho/para-directing substituent.

The Iodination Problem and Its Oxidant Fix

Direct halogenation works smoothly for chlorine and bromine, but iodine and fluorine are the two exceptions you must know for NEET. Iodine is sluggish and, worse, its reaction with benzene is reversible:

$$\ce{C6H6 + I2 <=> C6H5I + HI}$$

The trouble is the by-product. NIOS states it plainly: "the HI produced reduces the aryl iodide back to the aromatic hydrocarbon." Left alone, the equilibrium drifts backward and no iodobenzene survives. The cure is to add an oxidising agent — nitric acid, iodic acid ($\ce{HIO3}$), periodic acid ($\ce{HIO4}$) or mercuric oxide — which destroys the HI as fast as it forms. NCERT and NIOS both give the representative oxidation:

$$\ce{5HI + HIO3 -> 3I2 + 3H2O}$$

By consuming HI, the oxidant pulls the equilibrium forward (Le Chatelier's principle), and iodobenzene can finally be isolated. Fluorine sits at the opposite extreme: its reactivity is so high that direct fluorination "is very violent and cannot be controlled," so fluoroarenes are never made this way. Both exceptions are handled, instead, by the diazonium route below.

Route 2 — From Amines via Diazonium Salts

The second great route begins not from an arene but from a primary aromatic amine such as aniline. When the amine, dissolved or suspended in cold aqueous mineral acid, is treated with sodium nitrite, a diazonium salt is formed. This step — diazotisation — must be run cold, at 273–278 K, because diazonium salts are thermally fragile:

$$\ce{C6H5NH2 + NaNO2 + 2HCl ->[273\text{-}278\,K] C6H5N2^{+}Cl^{-} + NaCl + 2H2O}$$

The diazonium ion, $\ce{C6H5N2+}$, is the launchpad. Its $\ce{-N2+}$ group is an outstanding leaving group: it departs as inert nitrogen gas, and in doing so it can be replaced by a wide range of substituents — including each of the four halogens. This is what makes the diazonium route so general, and it is the only practical entry to fluoroarenes and (conveniently) iodoarenes, the two halogens that direct halogenation handles poorly.

Sandmeyer, Gattermann and the Iodo Route

For chlorine and bromine, two named reactions convert the diazonium salt into the haloarene. In the Sandmeyer reaction, mixing the freshly prepared diazonium salt with cuprous chloride or cuprous bromide replaces the diazonium group by –Cl or –Br:

$$\ce{C6H5N2^{+}Cl^{-} ->[CuCl] C6H5Cl + N2}$$ $$\ce{C6H5N2^{+}Cl^{-} ->[CuBr] C6H5Br + N2}$$

The Gattermann reaction reaches the same haloarenes but swaps the reagent: it uses copper powder together with the corresponding halogen acid (Cu/HCl or Cu/HBr) rather than a pre-formed cuprous halide. NIOS describes it as "reacting benzene diazonium chloride with copper powder in presence of corresponding halogen acid." The trade-off is practical — Gattermann sidesteps making the cuprous salt but typically returns lower yields than Sandmeyer.

Iodine is the easiest case of all. NCERT notes that "replacement of the diazonium group by iodine does not require the presence of cuprous halide and is done simply by shaking the diazonium salt with potassium iodide":

$$\ce{C6H5N2^{+}Cl^{-} + KI -> C6H5I + KCl + N2}$$

This is precisely why iodoarenes are made from amines and not by direct ring iodination — no oxidant, no cuprous reagent, just KI.

Fluoroarenes: The Balz-Schiemann Route

Since direct fluorination is uncontrollable, fluoroarenes are made exclusively from diazonium salts by the Balz-Schiemann route. The diazonium salt is treated with fluoroboric acid ($\ce{HBF4}$), which precipitates a sparingly soluble diazonium fluoroborate. Gentle heating of this dry salt then expels nitrogen and boron trifluoride, leaving the fluoroarene:

$$\ce{C6H5N2^{+}Cl^{-} + HBF4 -> C6H5N2^{+}BF4^{-} + HCl}$$ $$\ce{C6H5N2^{+}BF4^{-} ->[\Delta] C6H5F + N2 + BF3}$$

The thermal decomposition step is the signature of this method: the solid fluoroborate is isolated, dried, and then warmed to release the products cleanly. This is the standard NEET answer for "how do you get fluorobenzene," and it is worth pairing in memory with the KI route for iodobenzene, since both fill the gaps left by direct halogenation.

Choosing a Route: A Decision Table

With both routes laid out, the practical question is which to use for a target haloarene. The table consolidates the reagents, conditions and source-grounded reasoning so the choice becomes mechanical in an exam.