1. 3) The final stage is called termination.
1.1. In termination, two free radicals together to form a molecule with no unpaired electrons.
1.1.1. This is now a stable molecyle and no longer takes part in the reaction.
1.2. There are three possible reactions in termination.
1.2.1. Br• + Br• → Br₂
1.2.1.1. Two bromine free radicals can form a bromine molecule.
1.2.2. CH₃• + CH₃• → C₂H₆
1.2.2.1. Two methyl free radicals can form a molecule of ethane.
1.2.3. CH₃• + Br• → CH₃Br
1.2.3.1. A methyl free radical and a bromine free radical can form a molecule of bromomethane.
2. There's one big problem with free radical substitution of alkanes.
2.1. We get a whole range of side products.
2.1.1. E.g. if a bromine free radical reacts with a molecule of bromomethane then we make dibromomethane.
2.1.1.1. we can get further reactions to form tribromomethane and tetrabromomethane.
3. At the end of this reaction we need to separate out our product molecules.
3.1. The best way of reducing the chance of these byproducts is to have an excess of methane.
3.1.1. This means there's a greater chance of a bromine radical colliding with a methane molecule and not a bromomethane molecule.
4. Another problem with free radical substitution is that it can take place at any point along the carbon chain, so a mixture of isomers can be formed.
5. Conditions for free radical substitution:
5.1. UV light - required to break the single covalent bond between the halogen by homolytic fission.
5.2. Occurs in the gaseous state.
6. Alkanes are unreactive
6.1. Alkanes are unreactive, non polar molecules - as carbon and hydrogen atoms have very similar electronegativity. The bonds are also strong and take a lot of energy to break.
7. One way alkanes can react is via free radicals
7.1. A free radical is any species with an unpaired electron. We show the unpaired electron as a dot.
7.2. Free radicals are highly reactive, so even though alkanes are unreactive, they can react with free radicals.
8. Free radical reaction between alkanes and halogens
8.1. Methane + Bromine --> Bromomethane + Hydrogen Bromide
8.1.1. CH4+Br2→ CH3Br+HBr
8.2. There's three stages in this reaction.
8.2.1. 1) Initiation
8.2.2. 2) Propagation
8.2.3. 3) Termination
9. At the start of the reaction, we have a mixture of methane and bromine molecules (Br₂ )
9.1. 1) In the first stage, initiation, we shine UV light onto the reaction mix. The energy of the UV light causes the single covalent bond between the two bromine atoms to break. When the bond breaks, one electron goes to each bromine atom. Because these now have an unpaired electron, they're now bromine free radicals.
9.1.1. Br₂ → (UV) 2Br•
9.1.2. When a covalent bond splits this way, it's called homolytic fission.
10. 2) The next stage is called propagation. It has two steps.
10.1. In the first step a bromine free radical reacts with a methane molecule.
10.1.1. Br• + CH₄ → HBr + CH₃•
10.1.2. To make an electron pair, the bromine free radical takes a hydrogen atom plus one electron from the methane molecule. This reaction produces hydrogen bromide and a methyl free radical.
10.2. In the second step, the methyl free radical now reacts with a bromine molecule. This produces our end product bromomethane, plus another free radical.
10.2.1. CH₃• + Br₂ → CH₃Br + Br•