Skip to content

Chemistry · Ch 13 — Hydrocarbons

Preparation

13.2.2

Preparation

Preparation of Alkanes

Petroleum and natural gas are the primary natural sources of alkanes. In the laboratory, alkanes can be prepared by several distinct methods, each useful for specific types of alkanes.

1. From Unsaturated Hydrocarbons: Hydrogenation

Alkenes and alkynes add dihydrogen gas (H2H_2) across their multiple bonds to form alkanes. This reaction requires a finely divided metal catalyst — typically platinum (PtPt), palladium (PdPd), or nickel (NiNi). The process is called hydrogenation.

The metal catalyst adsorbs H2H_2 molecules onto its surface, weakening the H−HH-H bond and making the hydrogen atoms highly reactive. Platinum and palladium are active enough to catalyse the reaction at room temperature. Nickel requires higher temperatures and pressures.

Examples:

CH2=CH2+H2→Pt/Pd/NiCH3−CH3(Ethene → Ethane)CH_2=CH_2 + H_2 \xrightarrow{Pt/Pd/Ni} CH_3-CH_3 \quad \text{(Ethene → Ethane)}

CH3−CH=CH2+H2→Pt/Pd/NiCH3−CH2−CH3(Propene → Propane)CH_3-CH=CH_2 + H_2 \xrightarrow{Pt/Pd/Ni} CH_3-CH_2-CH_3 \quad \text{(Propene → Propane)}

CH3−C≡C−H+2H2→Pt/Pd/NiCH3−CH2−CH3(Propyne → Propane)CH_3-C\equiv C-H + 2H_2 \xrightarrow{Pt/Pd/Ni} CH_3-CH_2-CH_3 \quad \text{(Propyne → Propane)}

Note

The hydrogenation of alkynes requires two moles of H2H_2 per mole of alkyne, as both the π\pi bonds are saturated. The reaction typically proceeds through an alkene intermediate, but under these conditions, the alkene is further hydrogenated to the alkane.

2. From Alkyl Halides

Two important methods use alkyl halides as starting materials.

(i) Reduction with Zinc and Dilute Acid

Alkyl halides (except fluorides) are reduced to alkanes when treated with zinc metal and dilute hydrochloric acid. The zinc provides electrons for the reduction, and the acid supplies protons.

R−X+H2→Zn,H+R−H+HXR-X + H_2 \xrightarrow{Zn, H^+} R-H + HX

Examples:

CH3−Cl+H2→Zn,H+CH4+HCl(Chloromethane → Methane)CH_3-Cl + H_2 \xrightarrow{Zn, H^+} CH_4 + HCl \quad \text{(Chloromethane → Methane)}

C2H5−Cl+H2→Zn,H+C2H6+HCl(Chloroethane → Ethane)C_2H_5-Cl + H_2 \xrightarrow{Zn, H^+} C_2H_6 + HCl \quad \text{(Chloroethane → Ethane)}

CH3CH2CH2−Cl+H2→Zn,H+CH3CH2CH3+HCl(1-Chloropropane → Propane)CH_3CH_2CH_2-Cl + H_2 \xrightarrow{Zn, H^+} CH_3CH_2CH_3 + HCl \quad \text{(1-Chloropropane → Propane)}

Tip

This method works well for preparing alkanes with the same number of carbon atoms as the starting alkyl halide. It is a straightforward reduction where the halogen atom is replaced by hydrogen.

(ii) Wurtz Reaction

When an alkyl halide is treated with sodium metal in dry ether (anhydrous conditions), a higher alkane is formed. This reaction is called the Wurtz reaction. Two alkyl groups couple together, and the halogen atoms are eliminated as sodium halide.

2R−X+2Na→dry etherR−R+2NaX2R-X + 2Na \xrightarrow{\text{dry ether}} R-R + 2NaX

The reaction is particularly useful for preparing alkanes with an even number of carbon atoms because two identical alkyl groups combine.

Examples:

2CH3−Br+2Na→dry etherCH3−CH3+2NaBr(Bromomethane → Ethane)2CH_3-Br + 2Na \xrightarrow{\text{dry ether}} CH_3-CH_3 + 2NaBr \quad \text{(Bromomethane → Ethane)}

2C2H5−Br+2Na→dry etherCH3−CH2−CH2−CH3+2NaBr(Bromoethane → n-Butane)2C_2H_5-Br + 2Na \xrightarrow{\text{dry ether}} CH_3-CH_2-CH_2-CH_3 + 2NaBr \quad \text{(Bromoethane → n-Butane)}

Watch out

If two different alkyl halides are used, a mixture of three alkanes is obtained. For example, reacting CH3BrCH_3Br and C2H5BrC_2H_5Br gives ethane (CH3−CH3CH_3-CH_3), butane (C2H5−C2H5C_2H_5-C_2H_5), and propane (CH3−C2H5CH_3-C_2H_5). This mixture is difficult to separate, so the Wurtz reaction is practical only when a single alkyl halide is used.

3. From Carboxylic Acids

Two methods use carboxylic acids as starting materials.

(i) Decarboxylation with Soda Lime

When the sodium salt of a carboxylic acid is heated with soda lime (a mixture of NaOHNaOH and CaOCaO), an alkane is produced. The alkane has one carbon atom fewer than the original carboxylic acid. This elimination of CO2CO_2 is called decarboxylation.

R−COONa+NaOH→CaO,ΔR−H+Na2CO3R-COONa + NaOH \xrightarrow{CaO, \Delta} R-H + Na_2CO_3

The calcium oxide in soda lime serves to keep the mixture porous and prevent the NaOHNaOH from fusing with the glass apparatus.

Example:

CH3COONa+NaOH→CaO,ΔCH4+Na2CO3(Sodium ethanoate → Methane)CH_3COONa + NaOH \xrightarrow{CaO, \Delta} CH_4 + Na_2CO_3 \quad \text{(Sodium ethanoate → Methane)}

Tip

Problem 9.6: To prepare propane (CH3CH2CH3CH_3CH_2CH_3), which has three carbon atoms, the starting carboxylic acid must have four carbon atoms — butanoic acid. The sodium salt required is sodium butanoate.

CH3CH2CH2COONa+NaOH→CaO,ΔCH3CH2CH3+Na2CO3CH_3CH_2CH_2COONa + NaOH \xrightarrow{CaO, \Delta} CH_3CH_2CH_3 + Na_2CO_3

(ii) Kolbe's Electrolytic Method

When an aqueous solution of the sodium or potassium salt of a carboxylic acid is electrolysed, an alkane is formed at the anode. The alkane produced contains an even number of carbon atoms.

2CH3COONa+2H2O→ElectrolysisCH3−CH3+2CO2+H2+2NaOH2CH_3COONa + 2H_2O \xrightarrow{\text{Electrolysis}} CH_3-CH_3 + 2CO_2 + H_2 + 2NaOH

The reaction proceeds through a free radical mechanism:

  1. Dissociation: The sodium salt dissociates in water. …
Start your 14-day free trial to unlock the full solution →
← Previous · 13.2.1Example 9.5 — Write structures for each of the following compounds. Why are the give…Next · 13.2.2 →Example 9.6 — Sodium salt of which acid will be needed for the preparation of propan…