IB syllabus > organic (sl) > 11.1 

These notes were written for the old IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

IB syllabus for first examinations 2016

10.2 - Alkanes


10.2.1: Explain the low reactivity of alkanes in terms of bond enthalpies and bond polarity.


Bond strength

Alkanes are stuctures that contain only carbon-carbon and carbon-hydrogen covalent bonds.

Carbon - carbon bonds are strong in relative terms, having a bond energy of 342 kJ mol-1. Carbon - hydrogen bonds are even sgtronger, with a bond energy of 412 kJ mol-1.

This means that it is relatively difficult to break C-C and C-H bonds. Hence, alkanes are resistant to bond breaking, which is needed for reaction to occur.

Bond polarity

The polarity of a bond is determined by the difference in electronegativity between the two atoms that make up the bond.

Carbon - carbon bonds are totally non-polar (of course) while the electronegativities of carbon (2.5) and hydrogen (2.1) are close together, making the bond formed between carbon and hydrogen very weakly polar.

The consequence of this is that there are no regions of either reduced electron density (partial positive charge) or increased electron density (partial negative charge) that can be attacked by other reagents. This makes the alkanes difficult to react with and therefore unreactive, except towards free radicals (which are themselves highly reactive species).


10.2.2: Describe, using equations, the complete and incomplete combustion of alkanes.


Combustion is a free radical process in which a flammable compound reacts with oxygen. The products of combustion of hydrocarbons are carbon dioxide and water.

The carbon content of the organic molecule is not always oxidised in the process to carbon dioxide. If the supply of oxygen is limited then not all of the carbon turns to carbon dioxide and both carbon monoxide and carbon itself may be formed. This is called incomplete combustion .

There is not only one equation for incomplete combustion, whereas there can only be one equation for complete combustion.

Example - the complete combustion of propane:

C3H8 + 5O2 3CO2 + 4H2O

Example - the incomplete combustion of propane:

C3H8 + 4O2 CO2 + 2CO + 4H2O


10.2.3: Describe, using equations, the reactions of methane and ethane with chlorine and bromine.


Chlorine and bromine also undergo free radical reactions with alkanes in the presence of ultra-violet light.

CH4 + Cl2 CH3Cl + HCl

 


10.2.4: Explain the reactions of methane and ethane with chlorine and bromine in terms of a free-radical mechanism


This is because the ultra-violet light is energetic enough to break the weak halogen- halogen bond (Cl-Cl = 242 kJ mol-1) giving free halogen atoms (free radicals).

Cl2 2Cl•

These chlorine free radicals can then attack the alkane:

CH4 + Cl• CH3• + HCl

CH3• + Cl2 CH3Cl + Cl•

The products are chlorinated alkane (chloroalkanes). The reaction can continue as long as chlorine remains and a mixture of products is obtained. This is typical of a free radical process.

With bromine (and UV light) the reaction can be followed by the disappearance of the red/brown bromine colour and the formation of acidic fumes of hydrogen bromide.


Resources

Free radical substitution