Alkane preparation methods

Alkanes are the least reactive family of organic compounds — saturated chains of $\text{C}_n\text{H}_{2n+2}$ held together by strong C–C and C–H single bonds. Precisely because they are so unreactive, chemists rarely start with one alkane and turn it into another; instead they build alkanes from more reactive starting materials: alkenes and alkynes, alkyl halides, and carboxylic acid salts. Each route works by the same underlying trick — replace a reactive site (a double bond, a halogen, a –COOH group) with hydrogen, or stitch two reactive fragments together into one saturated chain. This article collects the five standard laboratory and industrial methods a Class 11 student must know: catalytic hydrogenation, the Wurtz reaction, reduction of alkyl halides, soda-lime decarboxylation, and Kolbe's electrolysis — with the arithmetic of carbon counting worked through in full.

The core insight

Every standard alkane synthesis works by turning a reactive site into a C–H bond or joining two alkyl fragments into one C–C bond, so the finished alkane's carbon count is fixed entirely by how many carbons went into the starting material(s).

The core idea

1. Hydrogenation of unsaturated hydrocarbons. Alkenes and alkynes add hydrogen across their multiple bond(s) over a finely divided metal catalyst (Ni, Pd, or Pt), which weakens the $\pi$ bond enough for $\text{H}_2$ to add:

\text{C}_n\text{H}_{2n} + \text{H}_2 \xrightarrow{\text{Ni/Pd/Pt}} \text{C}_n\text{H}_{2n+2}

where: $n$ is the carbon count of the starting alkene, which is unchanged in the product — hydrogenation only saturates the multiple bond, it never alters the skeleton.

2. Wurtz reaction. Two molecules of an alkyl halide couple across a sodium metal surface in dry ether, forming a new C–C bond and losing sodium halide:

2\,R\text{–X} + 2\,Na \xrightarrow{\text{dry ether}} R\text{–}R + 2\,NaX

where: $R$ is the alkyl group and $X$ is a halogen (Cl, Br, or I). Because the reaction joins two identical alkyl fragments, it cleanly builds only symmetrical, even-carbon alkanes from a single halide; mixing two different halides gives a statistical jumble of three products (derived below).

3. Reduction of alkyl halides. The C–X bond is replaced directly by C–H using a reducing agent such as zinc and dilute HCl, or catalytic $\text{H}_2$ :

R\text{–X} + 2[H] \xrightarrow{Zn/HCl} R\text{–H} + HX

where: $[H]$ denotes hydrogen supplied by the reducing system; the carbon skeleton of $R$ is untouched, so this route (unlike Wurtz) gives an alkane with the same number of carbons as the starting halide, including odd numbers.

4. Soda-lime decarboxylation. Heating the sodium salt of a carboxylic acid with soda lime ( $\text{NaOH} + \text{CaO}$ ) strips out $\text{CO}_2$ and replaces the –COONa group with –H:

R\text{–COONa} + NaOH \xrightarrow[\Delta]{\text{CaO}} R\text{–H} + Na_2CO_3

where: the carboxyl carbon is lost as $\text{CO}_2$ (trapped by CaO as $\text{Na}_2\text{CO}_3$ ), so the alkane produced always has one carbon fewer than the parent acid.

5. Kolbe's electrolytic method (qualitative). Electrolysing an aqueous solution of a carboxylate salt liberates $\text{CO}_2$ at the anode and couples the resulting alkyl radicals into a higher, symmetrical alkane, with $\text{H}_2$ evolved at the cathode — mechanistically a radical cousin of the Wurtz coupling, but driven by current instead of sodium metal.

Figure

Five routes to an alkane converging from alkenes alkynes alkyl halides and carboxylate salts

Across all five routes the essential idea repeats: either add hydrogen across a multiple bond, replace a functional group (–X or –COOH) with –H, or couple two alkyl fragments — never anything more exotic than that.

Derivation ACarbon count and product mix in the Wurtz reaction

The goal is to show algebraically, using nothing beyond simple counting, why coupling two molecules of a single alkyl halide in the Wurtz reaction can only ever give an alkane with an even number of carbons, and why using two different alkyl halides together destroys the reaction's usefulness by producing a mixture of three distinct products rather than one clean alkane.

Step 1. Let the alkyl halide be $R\text{–X}$ , where $R$ has $m$ carbon atoms. The Wurtz reaction couples two $R$ fragments end to end: $2\,R\text{–X} + 2\,Na \rightarrow R\text{–}R + 2\,NaX$ . The product $R\text{–}R$ contains exactly $m + m = 2m$ carbon atoms, because no carbon is gained or lost in the reaction — only the two C–X bonds are broken and one new C–C bond is formed.

Step 2. Since $m$ is a whole number (a real alkyl group must have at least one carbon), $2m$ is always even. This is not a coincidence of any particular example; it is forced by the reaction joining two identical, whole fragments. So a single alkyl halide, run through the Wurtz reaction, can never deliver an alkane with an odd carbon count — the algebra rules it out before any lab work begins.

Step 3. Now suppose two different halides are used together: $R\text{–X}$ (with $m$ carbons) and $R'\text{–X}$ (with $p$ carbons), both reacting with sodium in the same flask. Sodium generates both alkyl fragments simultaneously, and they can combine in three ways: $R$ with $R$ (giving a $2m$ -carbon alkane), $R'$ with $R'$ (giving a $2p$ -carbon alkane), or $R$ with $R'$ (giving an $(m+p)$ -carbon alkane). All three couplings happen at once, in roughly statistical proportions, because sodium cannot tell the two fragments apart.

Step 4. The result is a three-component mixture with no single dominant product, which is difficult and wasteful to separate by distillation for anything beyond illustrative small-scale work. This is precisely why the Wurtz reaction is taught as a symmetrical, single-halide method: it is algebraically guaranteed to work cleanly only when $R = R'$ , collapsing the three possible products into one. Check: with $R = \text{CH}_3$ ( $m=1$ ), $2m = 2$ gives ethane; with $R = \text{C}_2\text{H}_5$ ( $m=2$ ), $2m = 4$ gives butane — both even, matching Step 2.

Worked example

Problem. 2.0 mol of bromoethane ( $\text{CH}_3\text{CH}_2\text{Br}$ , molar mass $109\ \text{g/mol}$ ) is treated with excess sodium in dry ether. (a) Identify the alkane formed and explain why propane could not be made this way from bromoethane alone. (b) Calculate the mass of NaBr produced.

Step 1. Write the Wurtz equation for bromoethane: $2\,CH_3CH_2Br + 2\,Na \xrightarrow{\text{dry ether}} CH_3CH_2\text{–}CH_2CH_3 + 2\,NaBr$ . Each bromoethane contributes a 2-carbon ethyl fragment, so the coupled product has $2 + 2 = 4$ carbons.

Step 2. The product is butane ( $\text{C}_4\text{H}_{10}$ ). Using the Step 2 result from the derivation above, coupling two identical $m=2$ fragments can only give $2m = 4$ carbons — never an odd number like propane's 3. To get propane one would need to couple a 1-carbon and a 2-carbon fragment (bromomethane with bromoethane), but that mixed reaction also produces ethane and butane as by-products (Step 3 of the derivation), so propane cannot be isolated cleanly.

Step 3. Compute the mass of NaBr. From the balanced equation, 2 mol of $\text{CH}_3\text{CH}_2\text{Br}$ produce 2 mol of NaBr — a 1:1 mole ratio. So 2.0 mol bromoethane gives 2.0 mol NaBr. Molar mass of NaBr $= 23 + 80 = 103\ \text{g/mol}$ , so mass $= 2.0 \times 103 = 206\ \text{g}$ .

Answer: the product is butane ( $\text{C}_4\text{H}_{10}$ ), propane cannot be made cleanly from bromoethane alone, and 2.0 mol of bromoethane yields 206 g of NaBr.

Competitive stretch

Problem. Sodium propanoate ( $\text{CH}_3\text{CH}_2\text{COONa}$ ) is heated with soda lime. Identify the product and write the balanced equation.

Step 1. Soda-lime decarboxylation removes the carboxyl carbon as $\text{CO}_2$ (trapped as $\text{Na}_2\text{CO}_3$ ) and replaces –COONa with –H, shortening the chain by one carbon. Sodium propanoate has 3 carbons, so the alkane product has $3 - 1 = 2$ carbons.

Step 2. Balanced equation: $CH_3CH_2COONa + NaOH \xrightarrow[\Delta]{\text{CaO}} CH_3CH_3 + Na_2CO_3$ . The product is ethane ( $\text{CH}_3\text{CH}_3$ ).

Common traps

Forgetting that the Wurtz reaction only cleanly gives symmetrical, even-carbon alkanes — a single alkyl halide can never yield an odd-carbon product, by the $2m$ counting argument above.
Confusing the hydrogenation catalyst (Ni/Pd/Pt with $\text{H}_2$ gas, acting on an alkene/alkyne) with the reduction reagent for alkyl halides (Zn/HCl, acting on R–X) — different starting materials, different reagents.
Miscounting carbons in soda-lime decarboxylation — the product alkane always has one carbon fewer than the parent carboxylic acid salt, since the carboxyl carbon leaves as $\text{CO}_2$ .
Writing the Wurtz equation with only one molecule of $R$ –X — the correct stoichiometry needs two molecules of $R$ –X and two atoms of Na per molecule of coupled alkane.

Ask the tutor: if you want to practise predicting the alkane product (and its carbon count) from any of these five methods for a given starting material, ask the AI tutor to generate practice problems and check your equations.

Why does the Wurtz reaction need dry ether specifically?

Sodium metal reacts violently and uncontrollably with water or any protic solvent, so a dry, inert solvent is essential for the reaction to proceed cleanly. Dry ether is chosen because it dissolves the alkyl halide well, does not react with sodium itself, and provides an unreactive medium in which the alkyl fragments can meet at the sodium surface and couple. Any trace of moisture would consume the sodium in a side reaction and spoil the yield of alkane.

Can the Wurtz reaction be used to make methane?

No. Methane has just one carbon, and the Wurtz reaction always joins two identical alkyl fragments into a product with an even, doubled carbon count — the smallest possible Wurtz product is ethane ( $2 \times 1$ carbon), from two molecules of a 1-carbon halide like bromomethane. There is no way to halve a coupling reaction, so methane must be made by other routes, such as reducing a 1-carbon halide (Zn/HCl) or other methods outside this list.

Is Kolbe's electrolytic method just the Wurtz reaction done with electricity instead of sodium?

They are closely related in spirit but not identical. Both couple two identical alkyl fragments into a symmetrical, even-carbon alkane, and both fail cleanly if you mix two different starting materials, for the same statistical reason. The difference is the source of the reactive fragments and the starting material: Wurtz starts from an alkyl halide and uses sodium metal to generate alkyl radicals for coupling, while Kolbe's method starts from a carboxylate salt and uses electrolysis, releasing CO2 at the anode as the alkyl radical is generated and then coupled.

Alkane preparation methods

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads