Substitution Reactions of Alkanes
This lesson covers:
- How halogens undergo radical substitution reactions with alkanes
- The reaction mechanism for the halogenation of alkanes
Photochemical halogenation of alkanes
Halogens react with alkanes in special light-induced reactions called photochemical reactions.
For the reaction to occur:
- Ultraviolet (UV) light must be present.
- This UV light provides the activation energy to start the reaction.
The overall reaction is a substitution, where a hydrogen atom in the alkane molecule is replaced by a halogen atom like chlorine or bromine.
Free radical substitution mechanism
Photochemical halogenation of alkanes follows a three-step free radical substitution mechanism:
- Initiation - UV light produces reactive radicals
- Propagation - Radicals react in a chain reaction
- Termination - Radicals join to form stable molecules
Methane reacts vigorously with chlorine gas in the presence of UV light, as represented by the equation:
CH4 + Cl2 ➔ CH3Cl + HCl
The three-stage mechanism for this reaction is detailed below.
Stage 1 - Initiation
- UV light breaks the Cl-Cl bond in chlorine via homolytic fission
- This gives two chlorine radicals (Cl•):
Cl2 ➔ 2Cl•
- The unpaired electron makes Cl• highly reactive.
Stage 2 - Propagation
- The chlorine radical attacks a methane molecule in a substitution reaction:
Cl• + CH4 ➔ CH3• + HCl
- The methyl radical (CH3•) attacks another chlorine molecule:
CH3• + Cl2 ➔ CH3Cl + Cl•
- This propagation cycle continues until reagents are used up.
Stage 3 - Termination
- Two radicals join to form a stable covalent bond:
Cl• + CH3• ➔ CH3Cl
- Other combinations like 2CH3• ➔ C2H6 or 2Cl• ➔ Cl2 are also possible.
The end products of the termination step depend on which reagent is in excess:
- Excess chlorine - Further substitution on products like chloromethane occurs, producing a mixture of products such as CH3Cl, CH2Cl2, CHCl3, and CCl4.
- Excess methane - Predominantly single substitution occurs to form chloromethane.