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The Wurtz family of reactions all do one thing: sodium removes two halogen atoms and the two carbon pieces left behind join up. Once you see that, predicting products, by-products and quantities becomes bookkeeping. This post covers the aryl versions, Wurtz-Fittig and Fittig, with worked examples that go past the textbook equations.
What does NCERT say, and which exams include it?
The NCERT Class 12 chapter Haloalkanes and Haloarenes (Reprint 2026-27) describes three reactions with sodium in dry ether:
Wurtz reaction: alkyl halides give hydrocarbons "containing double the number of carbon atoms present in the halide".
Wurtz-Fittig reaction: "A mixture of an alkyl halide and aryl halide gives an alkylarene when treated with sodium in dry ether".
Fittig reaction: aryl halides give compounds "in which two aryl groups are joined together"; the example product is diphenyl (biphenyl).
The JEE (Advanced) 2026 brochure names "Fittig, Wurtz-Fittig" under haloarenes. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus do not name the reactions, but list "general methods of preparation, properties and reactions" for halogen compounds, so the NCERT text is the guide. The CBSE Chemistry curriculum for 2026-27 includes the chemical reactions of haloarenes. Check the syllabus for your exam year.
The reaction pair is old. Tollens and Fittig published "Ueber die Synthese der Kohlenwasserstoffe der Benzolreihe" (on the synthesis of hydrocarbons of the benzene series) in Liebigs Annalen in 1864.
How do you write the equation?
Ar–X + R–X + 2Na → Ar–R + 2NaX (Wurtz-Fittig)
2Ar–X + 2Na → Ar–Ar + 2NaX (Fittig)
Every new C–C bond uses two halogen atoms and two sodium atoms. Dry ether matters because sodium reacts with water.
Worked example 1: making ethylbenzene
Target: C₆H₅–CH₂CH₃. Cut the bond between the ring and the side chain. The pieces are C₆H₅– and –CH₂CH₃, so use bromobenzene and bromoethane with sodium in dry ether.
Possible coupled products:
C₆H₅–C₂H₅, ethylbenzene (the target).
C₂H₅–C₂H₅, butane (from two bromoethane molecules, a Wurtz by-product).
C₆H₅–C₆H₅, biphenyl (from two bromobenzene molecules, a Fittig by-product).
Worked example 2: how much sodium, how much product?
Take 15.7 g of bromobenzene (157.0 g/mol) and an equal number of moles of bromoethane.
Moles of C₆H₅Br = 15.7/157.0 = 0.100 mol.
The equation needs 2 mol of Na per mol of C₆H₅Br: 0.200 mol × 23.0 g/mol = 4.60 g of sodium.
Maximum ethylbenzene (106.2 g/mol), if every bromobenzene ended up in the target: 0.100 × 106.2 = 10.6 g.
The real yield is lower, partly because some of each halide is used up in the by-products listed above.
Worked example 3: biphenyl from benzene
NCERT Exercise 6.11 asks how to convert benzene to biphenyl. Two steps:
Benzene with Cl₂ and a Lewis acid catalyst such as iron(III) chloride gives chlorobenzene. NCERT's preparation section gives this route for aryl chlorides.
Chlorobenzene with sodium in dry ether gives biphenyl by the Fittig reaction.
Numbers: 22.5 g of chlorobenzene (112.6 g/mol) is 0.200 mol. Two molecules make one biphenyl, so the maximum is 0.100 mol × 154.2 g/mol = 15.4 g of biphenyl, using 0.200 mol (4.60 g) of sodium.
What if both halides are aryl halides?
Chlorobenzene and 4-chlorotoluene with sodium give three biaryls: biphenyl, 4,4′-dimethylbiphenyl and the cross product 4-methylbiphenyl. This is the same three-way pairing as a Wurtz reaction with two different alkyl halides, which our post on preparing alkanes works through for ethane, propane and butane.
Which mistakes cost marks?
Losing a carbon or a halogen. Write the formula of each piece before you join them.
Using 1 Na per coupling. Each new C–C bond needs 2 Na.
Naming the wrong reaction. Two alkyl halides: Wurtz. One alkyl and one aryl: Wurtz-Fittig. Two aryl: Fittig.
Forgetting by-products when a question asks for all possible products.
Writing wet ether or aqueous conditions. Sodium needs dry ether.
Haloarenes also undergo substitution reactions, which behave very differently from haloalkanes; our post on SN1 and SN2 in haloalkanes covers the alkyl side.
A practice routine
For five alkylbenzenes, cut the ring–chain bond and name the two halides you would use.
For each, list all three coupled products.
Solve two quantity problems: sodium needed and maximum product mass.
Answer NCERT Exercises 6.11 (x) and 6.19 (xviii), benzene to biphenyl, in full.
If you can write the equations but cannot plan a route from a target, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Tonight, plan the synthesis of propylbenzene by Wurtz-Fittig and list every coupled product it could give.
