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Alkanes are saturated hydrocarbons that contain only single C–C and C–H bonds. Because these bonds are strong and non-polar (or only very slightly polar), alkanes are relatively unreactive. However, they do undergo one important reaction: free radical substitution with halogens in the presence of UV light.
In this lesson we will learn how a hydrogen atom on an alkane is replaced (substituted) by a halogen atom, and we will study the three-stage mechanism in detail.
Before we study the mechanism, we need to understand two ways in which a covalent bond can break.
In heterolytic fission, both electrons from the shared pair move to the same atom. This produces two ions: one positive (lost both bonding electrons) and one negative (gained both bonding electrons).
For example, if a Cl–Cl bond breaks heterolytically:
Cl–Cl → Cl− + Cl+
This type of fission is common in polar molecules and in solution, but it is not what happens in free radical substitution.
In homolytic fission, the shared pair of electrons is split equally: one electron goes to each atom. This produces two free radicals, each with an unpaired electron (shown by a dot).
Cl–Cl → Cl• + Cl•
Free radicals are highly reactive because of their unpaired electron. Homolytic fission is favoured in non-polar bonds and in the gas phase, which is why it occurs in the reaction of alkanes with halogens.
A free radical is a species with an unpaired electron. Free radicals are extremely reactive and are represented with a dot, e.g. Cl• or •CH3. A half curly arrow shows the movement of a single electron; a full curly arrow shows the movement of an electron pair.
| Feature | Detail |
|---|---|
| Reaction type | Free radical substitution |
| Reagents | Cl2(g) or Br2(g) |
| Conditions | UV light (or sunlight) |
| What happens | One H atom on the alkane is substituted by one halogen atom |
The overall equation for the reaction of methane with chlorine is:
CH4 + Cl2 → CH3Cl + HCl
Methane is converted into chloromethane (a haloalkane), and hydrogen chloride is produced as a by-product.
Free radical substitution proceeds through three stages: initiation, propagation, and termination.
UV light provides the energy to break the Cl–Cl bond by homolytic fission, producing two chlorine radicals:
Cl2 ⟶ 2Cl•
This step requires energy input (UV light) and generates the reactive species that start the chain reaction.
The propagation steps form a chain reaction. In each step, one radical is consumed but another is generated, so the reaction sustains itself.
Propagation step 1 (hydrogen abstraction): a chlorine radical attacks methane, removing a hydrogen atom to form HCl and a methyl radical:
CH4 + Cl• → •CH3 + HCl
Propagation step 2 (halogen addition): the methyl radical attacks a chlorine molecule, forming chloromethane and regenerating a chlorine radical:
•CH3 + Cl2 → CH3Cl + Cl•
The chlorine radical consumed in propagation step 1 is regenerated in propagation step 2. It acts as a catalyst for the chain reaction. This is why a single photon of UV light can lead to the formation of many product molecules.
The chain reaction ends when any two radicals combine, removing radicals from the system. There are three possible termination steps:
| Termination Step | Equation | Product |
|---|---|---|
| 1 | Cl• + Cl• → Cl2 | Chlorine |
| 2 | •CH3 + Cl• → CH3Cl | Chloromethane |
| 3 | •CH3 + •CH3 → C2H6 | Ethane |
Termination step 3 is particularly significant because it produces ethane (C2H6), a molecule that was not a reactant. The presence of trace amounts of ethane in the product mixture is evidence that the reaction proceeds via a free radical mechanism.
The same mechanism applies to other alkanes. For ethane reacting with chlorine:
CH3CH3 + Cl2 → CH3CH2Cl + HCl
The mechanism steps are:
Bromine can also be used instead of chlorine. For propane reacting with bromine:
CH3CH2CH3 + Br2 → CH3CHBrCH3 + HBr
The mechanism steps are:
With propane, the bromine radical preferentially attacks the secondary (middle) carbon because the secondary radical formed is more stable than a primary radical. This is why the major product is 2-bromopropane rather than 1-bromopropane.
If there is an excess of halogen, the substitution can continue. The haloalkane product itself has C–H bonds that can be substituted. For methane with excess chlorine, successive substitutions can occur:
| Step | Equation | Product |
|---|---|---|
| 1 | CH4 + Cl2 → CH3Cl + HCl | Chloromethane |
| 2 | CH3Cl + Cl2 → CH2Cl2 + HCl | Dichloromethane |
| 3 | CH2Cl2 + Cl2 → CHCl3 + HCl | Trichloromethane (chloroform) |
| 4 | CHCl3 + Cl2 → CCl4 + HCl | Tetrachloromethane |
This means it is difficult to obtain a pure sample of just one haloalkane from free radical substitution, because a mixture of mono-, di-, tri-, and tetra-substituted products is formed. This is a major limitation of this reaction.