Preparation of haloarenes

1. Direct electrophilic substitution on benzene. Benzene reacts with chlorine or bromine in the presence of a Lewis acid catalyst (a "halogen carrier") such as anhydrous $FeCl_3$, $FeBr_3$, or $AlCl_3$:

where: $FeCl_3$ is the halogen-carrier catalyst (an $AlCl_3$ or $FeBr_3$ analogue works identically for the corresponding halogen), and $HCl$ is the by-product that carries away the displaced ring hydrogen. The catalyst polarises the halogen molecule, generating an electrophilic, $Cl^{\delta+}$-like species reactive enough to attack the electron-rich benzene ring. This route works cleanly for chlorination and bromination, but not for iodination (see point 4) or fluorination (too vigorous and uncontrollable to run safely on benzene).

2. From diazonium salts — Sandmeyer's reaction. Aniline is first converted to a diazonium salt by diazotisation at low temperature:

Treating the diazonium salt with cuprous chloride or cuprous bromide, in the presence of the matching halogen acid, then replaces the diazonium group with $Cl$ or $Br$, releasing nitrogen gas:

3. From diazonium salts — Gattermann's reaction. A closely related method swaps the pre-formed cuprous halide for freshly precipitated copper powder used directly with the matching halogen acid:

Both reactions achieve the identical net transformation — diazonium group out, halogen in, nitrogen gas released — and differ only in the form of the copper reagent: a pre-made cuprous salt for Sandmeyer, bare copper powder for Gattermann.

4. Why direct iodination fails. Molecular iodine does add to benzene under forcing conditions, but the reaction is reversible: the $HI$ produced is a strong reducing agent that reduces the newly formed iodobenzene straight back to benzene, so the equilibrium sits unfavourably unless an oxidising agent is added to mop up the $HI$. Because that extra oxidising step is awkward to run cleanly, aryl iodides are made instead via the diazonium route, most simply by treating the diazonium salt directly with potassium iodide (see the worked example).

There is no equation to derive here — the syllabus asks for reaction routes, not a quantitative law — so the reasoning is about mechanism availability rather than a formula. An ordinary haloalkane is made from an alcohol or by direct nucleophilic substitution because the reacting carbon is $sp^3$ and open to backside attack ($S_N2$) or to forming a carbocation ($S_N1$). Neither pathway is available on an aromatic ring: the ring carbon is $sp^2$, locked in the planar, delocalised π system, and backside attack is blocked by the ring geometry, while a simple aryl carbocation is far too unstable to form under ordinary conditions. So substitution at the ring carbon has to be electrophilic, not nucleophilic — the ring's own electron density is the reactive site, which is exactly what a Lewis-acid-activated halogen exploits directly on benzene. The diazonium route sidesteps the problem entirely rather than solving it: instead of trying to displace something already fixed to the ring, chemists first attach a group, $-N_2^+$, that is an excellent leaving group precisely because it leaves as stable, neutral $N_2$ gas. Once that leaving group is in place, the diazonium carbon becomes reactive enough (via a radical pathway, catalysed by copper) for the halide or iodide ion to take its place. This is why Sandmeyer's and Gattermann's reactions always start from aniline (or another primary aromatic amine) rather than from benzene directly — the amine is what lets you build the diazonium leaving group in the first place. The clearest test case is iodine. Direct iodination attempted as an equilibrium, $C_6H_6 + I_2 \rightleftharpoons C_6H_5I + HI$, does not go cleanly to completion, because $HI$ is a strong enough reducing agent to reduce iodobenzene back to benzene and regenerate $I_2$, pulling the equilibrium backward. Adding an oxidising agent (such as $HIO_3$) to mop up the $HI$ would drag the equilibrium forward in principle, but it is experimentally awkward compared with a route that has no reverse reaction at all. The diazonium route sidesteps the problem structurally rather than fighting an equilibrium: $C_6H_5N_2^+Cl^- + KI \rightarrow C_6H_5I + N_2\uparrow + KCl$ releases nitrogen gas as an irreversible driving force, so the reaction runs cleanly to completion — which is why this, not direct iodination, is the standard method for making iodobenzene.

Starting from aniline, prepare iodobenzene, and explain why the diazonium route is the appropriate method here rather than direct electrophilic substitution on benzene.

Step 1. Diazotise aniline at low temperature: $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0\text{–}5\,°C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$. The 0–5°C range is essential — above it the diazonium salt decomposes before it can be used.

Step 2. Treat the diazonium salt directly with potassium iodide solution. No copper catalyst is required for this step, since iodide is a good enough nucleophile on its own: $C_6H_5N_2^+Cl^- + KI \rightarrow C_6H_5I + N_2\uparrow + KCl$.

Step 3. Justify the route choice: direct electrophilic iodination of benzene with $I_2$ is reversible because the $HI$ by-product reduces iodobenzene back to benzene, so without an added oxidising agent the reaction barely proceeds. The diazonium pathway instead loses nitrogen gas irreversibly, so it runs to completion cleanly.

Answer: iodobenzene, $C_6H_5I$, obtained via aniline in two steps — diazotisation, then treatment with KI — because this is the reliable, irreversible route, unlike direct iodination.

Competitive stretch

If the question instead asked for bromobenzene from aniline, the same diazonium salt could be treated with cuprous bromide and hydrobromic acid (true Sandmeyer conditions), $C_6H_5N_2^+Cl^- + CuBr \xrightarrow{Cu_2Br_2/HBr} C_6H_5Br + N_2\uparrow$ — but since direct electrophilic substitution on benzene with $Br_2/FeBr_3$ already gives bromobenzene in a single, cheaper step, an exam answer should note the diazonium route is unnecessary for Br (and Cl); it is really needed only for I.

• Assuming benzene can be iodinated as easily as it is chlorinated or brominated — the $HI$ by-product makes direct iodination reversible and low-yielding, which is exactly why the diazonium route exists for aryl iodides.
• Confusing halogen-carrier-catalysed electrophilic substitution (needs $FeCl_3$/$FeBr_3$/$AlCl_3$, acts directly on benzene) with the Sandmeyer/Gattermann routes (need a pre-formed diazonium salt, work by displacing $N_2$) — these are two mechanistically unrelated families of reaction that happen to give the same product class.
• Mixing up Sandmeyer's reaction (cuprous halide, $Cu_2Cl_2$ or $Cu_2Br_2$) with Gattermann's reaction (copper powder used directly with the halogen acid) — both convert a diazonium salt to an aryl halide with loss of nitrogen gas, differing only in the copper reagent's form.
• Forgetting the 0–5°C temperature condition for diazotisation, or forgetting that the diazonium salt is unstable and should be used promptly rather than stored.

Ask the tutor: want to see how the same diazonium salt is used to install $-CN$ or $-OH$ groups instead of a halogen, or work through more Sandmeyer/Gattermann practice equations?