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Alkanes: Preparation

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Alkanes: Preparation - Lesson Summary

Alkanes Preparation
Alkanes are the major components of petroleum and natural gas. However, alkanes can be obtained in a laboratory, either from unsaturated hydrocarbons, alkyl halides or carboxylic acid.
Hydrogenation of unsaturated hydrocarbon ethene in presence of finely divided catalysts like platinum, palladium or nickel will produce ethane.

Hydrogenation is the chemical reaction that results from the addition of hydrogen to a molecule.
Ethene (C₂H₄) contains a double bond between the two carbon atoms.

Hydrogenation of Propyne requires two molecules of hydrogen to obtain Propane. Alkanes obtained from alkyl halides by the reduction method or the Wurtz reaction. The reduction method involves the addition of hydrogen to a molecule.

Chloromethane reduced to methane, using the zinc as reducing agent.  The function of zinc is to deliver hydrogen atoms to the chloromethane molecule. When the hydrogen molecule reaches the chloromethane molecule, it breaks the carbon to chlorine bond. One hydrogen atom replaces chlorine, while the other hydrogen atom attaches itself with the newly released chloride ion to form hydrogen chloride.
Wurtz reaction:
In this reaction the alkyl halides are treated with sodium metal in an anhydrous ethereal solution.

Sodium transfers one electron to the halogen to produce a sodium halide and an alkyl radical. The alkyl radical then accepts an electron from another sodium atom to form an alkyl anion, and the sodium becomes cationic. And alkyl anion then displaces the halide and forms a new carbon-carbon covalent bond.

        R - X       +    2Na   + X - R        Dry Ether →           R - R + 2NaX
   Alkyl Halide                  Alkyl Halide                          Alkane

In the preparation of ethane from bromomethane, two molecules of bromomethane react with two atoms of sodium. Thus, two molecules of bromomethane and two atoms of sodium form one molecule of ethane and two molecules of sodium bromide.

CH 3 - Br           +  2Na   +          Br - CH 3     Dry Ether →     CH 3 - CH 3 + 2NaBr          
Bromomethane                      Bromomethane                     Ethane

Similarly two molecules of bromoethane react with two atoms of sodium to form n- butane.

C 2H 5 - Br           +  2Na   +          Br - C 2H 5     Dry Ether →     C 2H 5 - C 2H 5 + 2NaBr          

One of the sodium atoms transfers one electron to the bromine of one of the bromoethane molecules, forming one sodium bromide and one ethyl radical. This ethyl radical then accepts an electron from the second sodium atom to form ethyl anion and sodium cation.

This ethyl anion then displaces the bromine from the second molecule of bromoethane to form a new carbon-carbon covalent bond forming an n-butane molecule. Meanwhile, the sodium cation attaches with bromine to form sodium bromide. Thus one molecule of bromoethane and two atoms of sodium form one n-butane molecule and two molecules of sodium bromide.

In both the reactions, it is to be notice that the number of carbon atoms in the product molecule is always twice that of the reactant. Thus, the Wurtz reaction is limited to the synthesis of symmetric alkanes.
Preparation of alkanes from carboxylic acids:
Remove the carboxyl group from the carboxylic acids in the form of carbon dioxide to get alkanes. Hence this reaction is known as decarboxylation. As the carboxyl group contains a carbon atom, the alkane produced has one carbon atom less than the total number of carbon atoms present in the compound.

Ex: To obtain methane from sodium ethanoate heat the compound with a mixture of sodium hydroxide and calcium oxide, called soda lime.
On heating in the presence of a catalyst, calcium oxide, the -COONa group from sodium ethanoate is replaced by the hydrogen atom from sodium hydroxide, forming methane and sodium hydroxide gets converted into sodium carbonate.
Ex: preparation of propane from sodium butanoate.

On heating in the presence of catalyst calcium oxide, the –COONa group is replaced by the hydrogen atom from sodium hydroxide, forming propane and sodium hydroxide gets converted into sodium carbonate.

Preparation of alkanes by Kolbe’s method:

2CH 3COO Na   +       2H 2O     Electrolysis →     CH 3 - CH 3 + 2CO 2 + H 2 + 2NaOH         

In this reaction, two molecules of sodium acetate and two molecules of water, when electrolyzed, yield one molecule of ethane, two molecules of carbon dioxide, one molecule of hydrogen and two molecules of sodium hydroxide.

At cathode, water molecules break to form hydrogen and hydroxyl ions. Two hydrogen ions combine to form a hydrogen molecule that is released as gas. At anode first sodium acetate molecules form acetate and sodium ions.

Then two acetate ions accept two electrons to form two acetate free radicals. These, on further electrolysis, form two methyl free radicals and two molecules of carbon dioxide are released as gas.

The two methyl free radicals then combine to form ethane. As two free radicals combine to form an alkane, this method can only be used to obtain an alkane with an even number of carbon atoms.


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