Alkanes IB Chemistry Topic 10.2

10.2 Alkanes Asmt. Stmts 10.2.1 Explain the low reactivity of alkanes in terms of bond enthalpies and bond polarity. 10.2.2 Describe, using equations, the complete and incomplete combustion of alkanes. 10.2.3 Describe, using equations, the reactions of methane and ethane with chlorine and bromine. 10.2.4 Explain the reactions of methane and ethane with chlorine and bromine in terms of a free-radical mechanism.

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

10.2.1 Have low reactivity Bond enthalpies are relatively strong 348 kJ mol-1 to break a C-C bond 412 kJ mol-1 to break a C-H bond Low polarity Only readily undergo combustion reactions with oxygen (very flammable) and substitution reactions with halogens in UV light

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

Complete vs. Incomplete Combustion Hydrocarbons (only contain C & H) Complete combustion Alkanes burn in an excess supply of oxygen to form carbon dioxide and water: Example: C8H18 (g) + 12 ½ O2 (g) → 8 CO2 (g) + 9 H2O (l) exothermic (-∆H) Incomplete combustion If oxygen supply is limited, the gas carbon monoxide and carbon is formed C8H18 (l) + O2 (g) → C (s) + CO (g) + CO2 (g) + H2O (l) (notice left over carbon (black soot) and dangerous CO)

10.2.3 10.2.3 Explain the reactions of methane and ethane with chlorine and bromine in terms of a free-radical mechanism.

10.2.3 Free Radical Substitutions Many organic molecules undergo substitution reactions. In a substitution reaction one atom or group of atoms is removed from a molecule and replaced with a different atom or group. Example: Cl2 + CH4  CH3Cl + HCl 8

10.2.3 Reactions of Alkanes: with Halogens Alkanes do not react with halogens in the dark at room temperature, but will react in the presence of sunlight (UV). A substitution reaction will occur where some or all of the hydrogens will be replaced with a halogen C2H6 (g) + Br2 (g) → C2H5Br (l) + HBr (g) Cl2 + CH4  CH3Cl + HCl Think electronegativity!!!

10.2.3 Reactions of Alkanes: with Halogens Energy absorbed from light allows homolytic fission: each resulting atom receives one unpaired electron, known as free radicals Cl-Cl → Cl• + Cl•

this happens by a process know as free radical substitution that happens in 3 steps Initiation Propagation Termination

Cl2  Cl* + Cl* CH4 + Cl*  CH3* + HCl Cl* + Cl*  Cl2 Initiation Initiated by UV light breaking a chlorine molecule into two free radicals by a process called homolytical fission (* = unpaired electron) Cl2  Cl* + Cl* Propagation Keeps the chain going (radical in reactants and products) CH4  +  Cl*  CH3*  +  HCl CH3*  +  Cl2  CH3Cl  +  Cl* Termination This removes free radicals (*) from the system without replacing them by new ones Cl* + Cl*  Cl2 CH3* + Cl *  CH3Cl CH3* + CH3*  CH3CH3 each resulting atom receives one unpaired electron, known as free radicals that have lots of energy

10.2.4 Free Radical Mechanism-The Initiation Step The ultraviolet light is a source of energy that causes the chlorine molecule to break apart into 2 chlorine atoms, each of which has an unpaired electron The energies in UV are exactly right to break the bonds in chlorine molecules to produce chlorine atoms. 13

10.2.4 Homolytic Fission Free radicals are formed if a bond splits evenly - each atom getting one of the two electrons. The name given to this is homolytic fission. 14

10.2.4 Free Radical Propagation The productive collision happens if a chlorine radical hits a methane molecule. The chlorine radical removes a hydrogen atom from the methane. That hydrogen atom only needs to bring one electron with it to form a new bond to the chlorine, and so one electron is left behind on the carbon atom. A new free radical is formed - this time a methyl radical, CH3 . 15

10.2.4 Free Radical Propagation II If a methyl radical collides with a chlorine molecule the following occurs: CH3.  +  Cl2  CH3Cl  +  Cl. The methyl radical takes one of the chlorine atoms to form chloromethane In the process generates another chlorine free radical. This new chlorine radical can now go through the whole sequence again, It will produce yet another chlorine radical - and so on and so on. 16

10.2.4 Termination Steps The free radical propagation does not go on for ever. If two free radicals collide the reaction is terminated. 2Cl.  Cl2 CH3.  +  Cl .  CH3Cl CH3 .  +  CH3.  CH3CH3 17

each resulting atom receives one unpaired electron, known as free radicals

Homolytic fission: http://www.youtube.com/watch?v=XAWWf86TJ0c http://www.youtube.com/watch?v=MvWQOYLm1f4 Free radical damage: http://www.youtube.com/watch?v=JV9bORHbSZU

10.2.4 Exercise Write the steps in the free radical mechanism for the reaction of chlorine with methyl benzene. The overall reaction is shown below. The methyl group is the part of methyl benzene that undergoes attack. 20

10.2.4 Solution Cl2  2Cl. Termination 2Cl.  Cl2 Initiation Propagation Termination 2Cl.  Cl2 21

10.2.4 Reactions of Alkanes: with Halogens Substitution of an alkane with a halogen has 3 steps: Initiation Propagation Termination See pg. 193 in IBCC for detailed diagrams of these steps Rate of reaction: Cl2 > Br2 > I2 … Why??? Electronegativity values!!!