Named reactions of haloarenes

Fittig reaction — aryl + aryl. Two molecules of an aryl halide react with sodium metal in dry ether. Sodium donates electrons that couple the two aromatic carbons directly, forming a new C–C bond between the rings and expelling the halide as sodium halide:

where: $X$ is a halogen (usually $Cl$ or $Br$), and the product $C_6H_5\text{-}C_6H_5$ is biphenyl, two benzene rings joined by a single bond. This is the aromatic sibling of the aliphatic Wurtz reaction, which couples two alkyl halides into a longer alkane; here the coupling partners are aromatic rings instead, so the product is a substituted biphenyl rather than an alkane.

Wurtz–Fittig reaction — aryl + alkyl. If an aryl halide and an alkyl halide are mixed together and treated with sodium in dry ether, the cross-coupling stitches the aryl group to the alkyl group, building an alkylbenzene:

where: $R$ is an alkyl group (e.g., $CH_3$, $C_2H_5$) and $C_6H_5$-$R$ is the resulting alkylbenzene. Because both an aryl and an alkyl halide are present in the same flask, sodium cannot be selective — it also couples aryl with aryl (giving some biphenyl, the Fittig side-reaction) and alkyl with alkyl (giving some of the corresponding alkane, the ordinary Wurtz side-reaction). The cross-coupled alkylbenzene is the major product only when the two halides are used in sensible proportions; the reaction mixture is never perfectly clean.

Reduction of a haloarene to benzene. A haloarene loses its halogen entirely on catalytic hydrogenolysis: hydrogen gas adds across the C–X bond in the presence of a metal catalyst, breaking that bond and replacing the halogen with hydrogen.

where: the catalyst is finely divided nickel or palladium and $\Delta$ denotes heating. The ring itself is untouched — only the C–X bond is broken — so this reaction is a clean way to delete a halogen substituent once it has served its purpose in a synthesis, without disturbing anything else on the ring.

Sodium metal is a strong single-electron donor. When it meets an aryl or alkyl halide, it transfers an electron to the C–X bond, which then breaks: the halogen leaves as halide ion (soon paired with $Na^+$ as $NaX$), and the carbon is left as a highly reactive fragment that seeks another such fragment to bond with. In a flask containing only one kind of aryl halide, the only fragments available are aryl fragments, so they pair up with each other — that is the Fittig reaction, and biphenyl is the only possible coupling product. The moment a second, different halide is added to the same flask, this argument changes completely. With an aryl halide and an alkyl halide present together, sodium generates both aryl and alkyl reactive fragments in the same pot. Any fragment can, in principle, pair with any other fragment present — aryl with aryl, alkyl with alkyl, or aryl with alkyl. All three pairings are chemically possible and none is forbidden, so all three products form: biphenyl (aryl-aryl), the alkane (alkyl-alkyl), and the alkylbenzene (aryl-alkyl, the desired Wurtz–Fittig product). Using an excess of one halide relative to the other biases the statistics toward more cross-coupling, but it can never make the self-coupled by-products vanish entirely — this is exactly why the Wurtz–Fittig reaction is described as "not perfectly clean" and why the crude product must be purified afterward. Catalytic reduction works by an entirely different mechanism. The metal catalyst ($Ni$ or $Pd$) adsorbs both the haloarene and hydrogen gas onto its surface, weakening the H–H bond and the C–X bond at the same time. A hydrogen atom is delivered to the ring carbon exactly where the halogen used to sit, and the halogen departs as $HCl$ (or the corresponding hydrogen halide). Because this is a surface-catalysed hydrogenolysis rather than an ionic substitution, it needs both the catalyst and heat, and it does not touch the aromatic ring itself — the ring's own carbon-carbon bonds are far too strong to be affected under these conditions.

Design a synthesis of toluene (methylbenzene) from bromobenzene and bromomethane, and identify the likely by-products of the reaction.

Step 1. Mix bromobenzene ($C_6H_5Br$) and bromomethane ($CH_3Br$) together in dry ether, then add sodium metal — the standard Wurtz–Fittig setup, an aryl halide plus an alkyl halide with sodium as the coupling agent.

Step 2. Write the desired cross-coupling. Sodium couples the aryl fragment from bromobenzene with the methyl fragment from bromomethane: $C_6H_5Br + CH_3Br + 2\,Na \xrightarrow{\text{dry ether}} C_6H_5CH_3 + 2\,NaBr$, giving toluene as the target product.

Step 3. Identify the two side-reactions that necessarily accompany it, because sodium cannot distinguish which fragments pair with which. Aryl-aryl coupling between two molecules of bromobenzene (the Fittig side-reaction) gives biphenyl, $C_6H_5$-$C_6H_5$. Alkyl-alkyl coupling between two molecules of bromomethane (the ordinary Wurtz side-reaction) gives ethane, $CH_3$-$CH_3$.

Step 4. Conclude. The crude product is a three-component mixture: toluene (the major, desired product when the halide proportions are chosen sensibly), biphenyl, and ethane. These must be separated — for example by fractional distillation, since a gas (ethane), a liquid (toluene), and a solid (biphenyl) have very different boiling points — to isolate pure toluene.

Competitive stretch

Starting from aniline, outline a route to benzene that passes through a haloarene intermediate and a reduction step. Step 1. Diazotise aniline: $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0\text{-}5°C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$. Step 2. Convert the diazonium salt to chlorobenzene via Sandmeyer's reaction: $C_6H_5N_2^+Cl^- + CuCl \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl + N_2\uparrow$. Step 3. Reduce chlorobenzene to benzene: $C_6H_5Cl + H_2 \xrightarrow{Ni/Pd,\ \Delta} C_6H_6 + HCl$. Final answer: aniline is converted to benzene in three steps — diazotisation, Sandmeyer substitution, then catalytic reduction — using the haloarene purely as a removable stepping-stone that ultimately deletes the amino group.

• Confusing the Fittig reaction (aryl + aryl, gives biphenyl) with the Wurtz–Fittig reaction (aryl + alkyl, gives an alkylbenzene) — both use $Na$/dry ether and sound alike, but the substrates and the product differ.
• Expecting the Wurtz–Fittig reaction to give one clean product — because two different halides are present, sodium always also forms some biphenyl and some of the corresponding alkane as by-products.
• Forgetting that reduction of a haloarene to benzene needs a metal catalyst ($Ni$/$Pd$) and hydrogen gas — it does not happen simply by heating the haloarene alone.
• Treating "Wurtz," "Fittig," and "Wurtz–Fittig" as three separate reactions to memorise rather than one family of sodium/dry-ether couplings that differ only in whether the halide(s) are alkyl, aryl, or a mixture.

Ask the tutor: want to practice predicting all the by-products of a given Wurtz–Fittig mixture, or trace through more multi-step syntheses that use a haloarene as an intermediate?