Wurtz Reaction: Mechanism, Limitations, Wurtz-Fittig and Fittig Reactions

ChemistryClass 11, HydrocarbonsFor JEE and NEET aspirants

The Wurtz reaction joins two alkyl groups. An alkyl halide treated with sodium metal in dry ether gives an alkane with twice as many carbon atoms: bromoethane gives butane. It is a good way to make a symmetrical alkane but a poor one for an unsymmetrical alkane, because two different halides give three alkanes. With aryl halides the same idea gives the Wurtz-Fittig reaction (aryl + alkyl) and the Fittig reaction (aryl + aryl). These reactions are in the NCERT chapters on hydrocarbons (Class 11) and haloalkanes and haloarenes (Class 12), and are asked in NEET, JEE Main and JEE Advanced.

At a glance

Wurtz:
$2C H_{3} C H_{2} Br + 2Na dry ether C H_{3} C H_{2} C H_{2} C H_{3} + 2NaBr$
Rule: R–X gives R–R, so a single alkyl monohalide gives a coupled alkane with an even number of carbons.
Two different alkyl groups: R–X + R′–X gives R–R, R–R′ and R′–R′.
Halide reactivity: $R - I$ $>$ $R - Br$ $>$ $R - Cl$ ; fluorides hardly react.
Wurtz-Fittig:
$C_{6} H_{5} Br + C H_{3} Br + 2Na dry ether C_{6} H_{5} C H_{3} + 2NaBr$
Fittig:
$2 C_{6} H_{5} Br + 2Na dry ether C_{6} H_{5} C_{6} H_{5} + 2NaBr$
Cannot make: methane (the smallest product is ethane, from a methyl halide).

1. Mechanism

Sodium is a strong one-electron donor. It gives an electron to the C–X bond, which breaks into an alkyl radical and a halide ion. Two paths then lead to the same alkane, and both are usually quoted.

Figure 1: Wurtz reaction of bromoethane. The fish-hook arrows (path 1) move one electron each; the full arrows in the SN2 step move an electron pair. Either path joins two ethyl groups, so the alkane has twice the carbons of the halide.

Electron transfer. $R - X$ + Na gives $R^{∙}$ + $N a^{+} X^{-}$ . This happens on the surface of the metal.
Radical path. Two alkyl radicals pair their electrons: $R^{∙}$ + $R^{∙}$ gives R–R.
Carbanion path. A second sodium atom turns $R^{∙}$ into the organosodium compound $R^{-} N a^{+}$ ; the carbanion attacks another molecule of R–X by SN2 and displaces $X^{-}$ .

Some radicals do not join: one radical takes a hydrogen atom from the carbon next to the radical centre of another (disproportionation), so ethyl radicals also give a little ethane and ethene. 2° halides give lower yields (some alkene and R–H form); with 3° halides elimination and disproportionation take over and almost no R–R forms.

Exam Trick

"Wurtz doubles." Count the carbons of the alkyl group and double them. Wurtz cannot make methane, or an odd-numbered alkane from a single alkyl monohalide; from two different halides an odd alkane comes only as one of three products.

2. What the Wurtz reaction can and cannot make

Halide(s) with Na, dry ether	Product	Comment
$C H_{3} Br$	Ethane	Methane cannot be made
$C H_{3} C H_{2} Br$	Butane	Twice the carbons
$(C H_{3})_{2} CHBr$	2,3-Dimethylbutane	One coupled product (2°, so a lower yield)
$C H_{3} Br$ + $C H_{3} C H_{2} Br$	Ethane, propane and butane	Three alkanes: poor for propane
$(C H_{3})_{3} CBr$	2-Methylpropene and 2-methylpropane; very little coupled alkane	3° halide: elimination and disproportionation win

Why dry ether. Sodium reacts with water, and the organosodium intermediate would take a proton from water to give the alkane RH instead of R–R. Ether dissolves the halide and does not react with sodium.

2.1 Wurtz-Fittig and Fittig reactions

An aryl halide and an alkyl halide with sodium in dry ether give an alkylbenzene (Wurtz-Fittig). Two aryl halides give a biaryl (Fittig): bromobenzene, chlorobenzene or iodobenzene gives biphenyl, $C_{6} H_{5} - C_{6} H_{5}$ , molar mass 154. Wurtz-Fittig works better than a Wurtz reaction with two different alkyl halides because the aryl and alkyl halides react at different rates, so the alkylbenzene is the main product; biphenyl and R–R form as minor by-products.

C_{6} H_{5} Cl + C_{2} H_{5} Cl + 2Na dry ether C_{6} H_{5} C_{2} H_{5} + 2NaCl

JEE Advanced

Rings from dihalides. When both halogens are in one molecule, the two ends can join: 1,3-dibromopropane with sodium (or zinc) gives cyclopropane. A 1,2-dihalide cannot close a ring: 1,2-dibromoethane with zinc or sodium loses both bromines and gives ethene.

Wurtz coupling in Grignard work. While RMgX is being made, some R–X couples with the reagent to give R–R. This side reaction is why the halide is added slowly to the magnesium.

Counting products. With n halides carrying different alkyl groups, at most $\frac{n ( n + 1 )}{2}$ coupled alkanes form: 3 for two. There are fewer when two pairings give the same alkane: $C H_{3} X$ , $C_{2} H_{5} X$ and n- $C_{3} H_{7} X$ give 5, because ethyl + ethyl and methyl + propyl both give butane.

3. Worked examples

Solved Example 1

Which alkyl halide gives 2,3-dimethylbutane in the Wurtz reaction?

Solution:

Split 2,3-dimethylbutane, $(C H_{3})_{2} CH - CH (C H_{3})_{2}$ , at its central C–C bond: two isopropyl groups. So 2-bromopropane (or 2-chloropropane) with sodium in dry ether.

Solved Example 2

A mixture of bromomethane and bromoethane is treated with sodium in dry ether. Name the products.

Solution:

Three alkanes: ethane (methyl + methyl), propane (methyl + ethyl) and butane (ethyl + ethyl). This is why the Wurtz reaction is not used to make propane.

Solved Example 3

Chlorobenzene is treated with sodium in dry ether. Give the product and its molar mass.

Solution:

Two phenyl groups join (Fittig reaction): biphenyl, $C_{12} H_{10}$ , molar mass $12 \times 12 + 10 = 154 g mo l^{- 1}$ .

4. Exam traps

Watch out

The product has twice the carbons of the alkyl group: $C H_{3} X$ gives ethane, so methane cannot be made.
Two halides with different alkyl groups give three alkanes, not one.
The solvent must be dry ether; water destroys sodium and the organosodium intermediate.
Tertiary halides give mainly the alkene and the alkane R–H, almost no R–R.
Aryl + alkyl halide is Wurtz-Fittig; aryl + aryl halide is Fittig. Both use Na in dry ether.
Wurtz joins two R groups; Kolbe electrolysis also gives R–R, but from carboxylate salts.
Sodium with an aryl halide alone gives biphenyl, a biaryl, not benzene.

Past-paper questions on Wurtz reaction

Our past papers: JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026. We found 10 questions in them that test this:

JEE Main 7 (2025-2026)
JEE Advanced 1 (2025)
NEET 2 (2018, 2023)

Each question was read and checked by hand (list checked 2026-10-16). A question that shows its structures only as pictures can be missed by our search. Each opens in a new tab, with the question to try and our step-by-step solution.

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