Wurtz & Wurtz–Fittig Reactions

What the Wurtz Reaction Does

The Wurtz reaction is a preparation of alkanes in which an alkyl halide reacts with sodium metal in dry ether to build a higher alkane. Two molecules of the halide are consumed; their alkyl groups join through a fresh carbon–carbon bond, and the halogen atoms leave as sodium halide. The general equation is the single most important line to memorise:

$$\ce{2R-X + 2Na ->[dry\ ether] R-R + 2NaX}$$

The NIOS hydrocarbons unit lists this as one of the three routes from alkyl halides to alkanes, alongside direct reduction and the Grignard route. The textbook example couples bromoethane: two ethyl groups join to give the four-carbon chain, butane.

$$\ce{2CH3CH2Br + 2Na ->[dry\ ether] CH3CH2CH2CH3 + 2NaBr}$$

Three features deserve attention before the mechanism is unpacked. First, the reactivity of the halide follows the order R–I > R–Br > R–Cl, mirroring the ease with which the carbon–halogen bond breaks; alkyl bromides are the usual working compromise between reactivity and cost. Second, the stoichiometry is fixed at two halide molecules to two sodium atoms — one sodium reduces the halide to a reactive intermediate, the second mops up the liberated halide as sodium halide. Third, the carbon skeleton of each alkyl group is carried into the product intact, so the structure of the alkane can be read off directly by mentally welding two R fragments together. That predictability is exactly what makes the reaction a favourite for "identify-the-product" questions.

Because the reaction must be kept rigorously anhydrous, the solvent is dry (anhydrous) ether. Diethyl ether is chosen for several reasons at once: it is aprotic, so it cannot quench the carbanion-like intermediate; it is inert toward sodium; and it dissolves the organic halide while keeping the metal suspended at a clean surface. Any water present would destroy the sodium and protonate the reactive intermediate, returning the alkyl group as the simple alkane R–H rather than the coupled R–R — a low-yield, useless outcome the chemist takes pains to avoid.

Mechanism: Radical and Organosodium Routes

For NEET the mechanism is understood at two complementary levels, and the textbook does not commit to one alone. Both begin with sodium attacking the carbon–halogen bond, and both end at the same coupled alkane; they differ only in the character of the fleeting intermediate. Examiners reward a student who can state both pathways and, crucially, explain why the by-products that accompany the reaction point toward the radical route.

Organosodium (ionic) pathway

Sodium first reacts with one molecule of alkyl halide to form an organosodium intermediate (an alkyl sodium, R–Na), behaving like a carbanion-bearing nucleophile. The carbon now carries a high density of negative charge and is a strong base as well as a nucleophile. This species then attacks a second molecule of the alkyl halide in an S_N2-type displacement, expelling halide from the back face and forming the new C–C bond.

$$\ce{R-X + 2Na -> R-Na + NaX}$$

$$\ce{R-Na + R-X -> R-R + NaX}$$

Reading this pathway as an S_N2 step immediately explains the limitation that surfaces later: a bulky tertiary halide resists back-side attack, so the organosodium prefers to act as a base and strip off a β-hydrogen, giving an alkene. The same logic predicts that the cleanest couplings come from unhindered primary halides where the substitution geometry is unobstructed.

Free-radical pathway

Alternatively, single-electron transfer from sodium to the halide generates alkyl free radicals, which then combine to form the coupled alkane. The decisive experimental evidence for this route is the small but reproducible crop of side products: instead of only coupling, two radicals can disproportionate, one abstracting a hydrogen atom from the β-carbon of the other. This hands back one molecule of alkane (R–H) and one molecule of alkene, neither of which a purely ionic, clean S_N2 mechanism would predict.

$$\ce{R-X + Na -> R^{.} + NaX}$$

$$\ce{R^{.} + R^{.} -> R-R}$$

For ethyl radicals the disproportionation is written out explicitly: two ethyl radicals collide, one gives up a β-hydrogen, and the products are ethane and ethene rather than the desired butane.

$$\ce{2C2H5^{.} -> C2H6 + C2H4}$$

Both channels run in parallel; their relative weight depends on the halide and the conditions, but in every Wurtz preparation a chemist expects to see traces of the alkene and the lower alkane alongside the main coupled product. Recognising these by-products as the signature of competing radical chemistry is a frequent short-answer expectation.

Why Symmetrical, Even-Carbon Alkanes

When a single alkyl halide is used, the two groups that couple are identical. Joining two identical R groups produces a chain that is symmetrical about the new bond and contains exactly twice the carbon count of one group. Doubling any whole number gives an even number, so the Wurtz product from one halide always has an even number of carbon atoms.

This even-carbon habit dovetails with a physical-property fact the NIOS unit emphasises: among straight-chain alkanes, those with an even number of carbon atoms pack more closely in the solid state and accordingly show the higher melting points. The Wurtz reaction is the natural laboratory partner of that observation, since it delivers exactly those even-numbered chains.

Two corollaries are worth stating because they convert directly into marks. If a single halide is supplied, the product is not only even-numbered but symmetrical — it has a centre of symmetry at or between the two central carbons, because both halves are identical R groups. And the reverse reasoning is just as useful: given an alkane to be made by Wurtz, split the chain at its midpoint into two equal halves and the required halide is whichever group sits on either side. A clean midpoint split is possible only when the chain is symmetrical and even, which is precisely why the method is restricted to that class of targets.

The Mixed-Halide Problem

The moment two different alkyl halides are mixed, the reaction loses its selectivity. Each halide can couple with its own kind or with the other, so three alkanes appear together: the two symmetrical products and the desired cross-coupled product.

$$\ce{R-X + R'-X + 2Na ->[dry\ ether] R-R + R'-R' + R-R'}$$

Concretely, treating a mixture of bromoethane and 1-bromopropane with sodium yields butane, hexane and pentane simultaneously:

$$\ce{CH3CH2Br + CH3CH2CH2Br + 2Na ->[dry\ ether] C4H10 + C6H14 + C5H12}$$

Because these alkanes have similar boiling points and no functional handle for separation, isolating the cross product in good yield is impractical. This is the reason a Wurtz reaction is reserved for symmetrical targets and is a poor choice for an unsymmetrical alkane.

It is worth pausing on the statistics, because the trap is sharper than "three products". If the two halides are present in equal amount and couple at comparable rates, the random pairing of fragments delivers the three alkanes in a roughly 1 : 2 : 1 ratio of R–R : R–R′ : R′–R′ — the cross product is the most abundant single component, yet it still amounts to only about half the mixture, and the two symmetrical alkanes together claim the rest. So even the "best case" leaves the wanted alkane diluted by two contaminants of similar volatility. Propane illustrates the dead end perfectly: it has three carbons, an odd number, so no single halide can give it, and the mixed route (methyl bromide + ethyl bromide) would hand back ethane, butane and propane all at once. The honest conclusion is that Wurtz is not a synthesis of choice for any unsymmetrical alkane, and certainly not for an odd-carbon one.

Limitations: Tertiary Halides & Methane

Two boundaries fence in the Wurtz reaction, and both are favourite NEET probes.

Tertiary halides fail

Tertiary alkyl halides are bulky and rich in β-hydrogens. In contact with sodium they prefer elimination over coupling, so the dominant products are alkenes rather than the coupled alkane. Yields fall in the order primary > secondary > tertiary, and for tertiary halides the coupling effectively does not work. For this reason the reaction is best confined to primary halides.

Methane is impossible

The Wurtz reaction always builds a new carbon–carbon bond, so its smallest product is ethane, formed from two methyl units. Methane has a single carbon and no C–C bond, so it lies below the floor of what coupling can construct and cannot be prepared this way.

The Wurtz–Fittig Reaction

Extending the coupling to aromatics gives the Wurtz–Fittig reaction: a mixture of an aryl halide and an alkyl halide, treated with sodium in dry ether, joins the aryl and alkyl fragments to give an alkyl-substituted benzene (an alkylbenzene). It is the metal-coupling counterpart to Friedel–Crafts alkylation, and a useful one because it sidesteps the carbocation rearrangements that plague the Friedel–Crafts route.

$$\ce{C6H5-Br + CH3CH2-Br + 2Na ->[dry\ ether] C6H5-CH2CH3 + 2NaBr}$$

Here chlorobenzene plus bromoethane and sodium delivers ethylbenzene. As with the aliphatic Wurtz, a competing symmetrical coupling of each partner accompanies the cross product, so the alkyl and aryl halides are chosen to make the desired arene the principal target. The three possible couplings are the aryl–aryl product (biphenyl, a Fittig side reaction), the alkyl–alkyl product (the symmetrical alkane) and the wanted aryl–alkyl arene; in practice the aryl–alkyl cross-coupling is favoured enough to make the method preparatively useful.

The real selling point of Wurtz–Fittig is what it avoids. Friedel–Crafts alkylation installs the same alkyl group but proceeds through a carbocation, which is prone to rearrangement — an attempt to attach an n-propyl group by Friedel–Crafts tends to deliver the isopropyl-substituted ring instead, because the primary cation rearranges to the more stable secondary one. The Wurtz–Fittig route runs through a radical or organosodium intermediate that does not rearrange, so a straight-chain alkyl group stays straight. It also works on aryl halides that resist Friedel–Crafts because their ring is deactivated. For these reasons the reaction is the textbook answer whenever a question stresses "without rearrangement" or "without using anhydrous aluminium chloride".

The Fittig Reaction

When two aryl halides are coupled with sodium in dry ether, the product is a biaryl — two aromatic rings joined by a single C–C bond. The simplest case unites two molecules of bromobenzene to give biphenyl.

$$\ce{2C6H5-Br + 2Na ->[dry\ ether] C6H5-C6H5 + 2NaBr}$$

Biphenyl is the parent of a large family of biaryls that matter well beyond the exam — they form the backbone of liquid crystals, polymers and many drug molecules — but for NEET the single fact to lock in is the bond formed: a direct ring-to-ring carbon–carbon bond between the two ipso carbons that previously bore the halogens. Substituted aryl halides give the corresponding substituted biaryls (for instance, two molecules of 4-bromotoluene give 4,4′-dimethylbiphenyl), and a mixture of two different aryl halides suffers the same selectivity problem as a mixed Wurtz, giving three biaryls together.

The three reactions therefore form a tidy family that differs only in which kind of halide is supplied: alkyl + alkyl (Wurtz), aryl + alkyl (Wurtz–Fittig) and aryl + aryl (Fittig). The new bond is alkyl–alkyl, aryl–alkyl and aryl–aryl respectively. The shared reagent set — sodium in dry ether — and the shared coupling logic make the trio a single concept wearing three names; once the halide pairing is read, the product follows automatically.

Coupling Map: Three Reactions at a Glance

Worked Synthesis Examples