Alkanes are generally unreactive, but they will react with halogens (such as chlorine or bromine) in the presence of ultraviolet (UV) light. This reaction is a photochemical chain reaction known as free radical substitution, where hydrogen atoms in the alkane are replaced by halogen atoms.
🔑 Key Principle
Free radical substitution is a chain reaction initiated by UV light, which breaks halogen bonds homolytically. The propagation steps generate the halogenoalkane product while regenerating the radical catalyst. The chain reaction is terminated when two radicals combine to form a stable molecule.
1. Homolytic Fission and Radicals
The reaction begins with the homolytic fission of a covalent bond. Fission means bond breaking, and it can occur in two ways:
- Heterolytic Fission: The bond breaks unevenly, with both electrons in the shared pair going to one of the bonding atoms, forming a cation and an anion.
- Homolytic Fission: The bond breaks evenly, with each bonding atom taking one electron from the shared pair, resulting in two highly reactive species with unpaired electrons.
A highly reactive chemical species with one or more unpaired electrons.
2. The Reaction Mechanism
The free radical substitution mechanism between methane and chlorine consists of three distinct stages: initiation, propagation, and termination.
The first step in a free radical reaction, in which radicals are generated from non-radical molecules by homolytic fission, typically triggered by UV light.
Stage 1: Initiation
UV light provides the energy required to break the covalent bond in the chlorine molecule (\(\text{Cl}_2\)). Since the \(\text{Cl}-\text{Cl}\) bond is weaker than the \(\text{C}-\text{H}\) bond, it breaks first: \[ \text{Cl}_2 \xrightarrow{\text{UV}} 2\text{Cl}^\bullet \]
This is homolytic fission, producing two chlorine free radicals. Each radical has an unpaired electron, represented by a dot (\(^\bullet\)).
A cyclic step in a free radical chain reaction, in which a radical reacts with a molecule to form a new radical and a new molecule, maintaining the radical concentration.
Stage 2: Propagation
Propagation is a chain reaction that proceeds in a two-step cycle:
- The chlorine radical (\(\text{Cl}^\bullet\)) reacts with a methane molecule (\(\text{CH}_4\)), abstracting a hydrogen atom to form hydrogen chloride (\(\text{HCl}\)) and a methyl radical (\(^\bullet\text{CH}_3\)): \[ \text{Cl}^\bullet + \text{CH}_4 \rightarrow \text{HCl} + ^\bullet\text{CH}_3 \]
- The highly reactive methyl radical (\(^\bullet\text{CH}_3\)) then reacts with a chlorine molecule (\(\text{Cl}_2\)), abstracting a chlorine atom to form chloromethane (\(\text{CH}_3\text{Cl}\)) and regenerating the chlorine radical (\(\text{Cl}^\bullet\)): \[ ^\bullet\text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^\bullet \]
Because a chlorine radical is regenerated in the second step, it can react with another methane molecule, repeating the cycle. This allows the reaction to continue as a self-sustaining chain reaction.
The final step in a free radical reaction, in which two free radicals collide and combine to form a stable, non-radical molecule, bringing the chain reaction to an end.
Stage 3: Termination
The chain reaction is brought to a stop when two free radicals collide and react to form a stable molecule. Because the concentration of radicals is low, these collisions are rare, but they eventually stop the reaction:
- Two chlorine radicals combine: \[ \text{Cl}^\bullet + \text{Cl}^\bullet \rightarrow \text{Cl}_2 \]
- A methyl radical and a chlorine radical combine: \[ ^\bullet\text{CH}_3 + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{Cl} \]
- Two methyl radicals combine: \[ ^\bullet\text{CH}_3 + ^\bullet\text{CH}_3 \rightarrow \text{C}_2\text{H}_6 \]
The formation of ethane (\(\text{C}_2\text{H}_6\)) in the last step provides strong evidence for the existence of methyl radicals in the reaction mixture.
When drawing mechanism diagrams, make sure the radical dot is placed on the specific atom containing the unpaired electron. For example, in a methyl radical, write it as \( ^\bullet\text{CH}_3 \) or \( \text{H}_3\text{C}^\bullet \), showing the dot on the carbon atom rather than the hydrogen atoms. Always use single-headed (fishhook) arrows to represent the movement of a single electron.
3. Limits of Free Radical Substitution
Free radical substitution is a poor method for preparing pure halogenoalkanes due to two major limitations: further substitution and structural isomerism.
Further Substitution
The products of the reaction are themselves molecules containing C-H bonds. As the concentration of the halogenoalkane (e.g. \(\text{CH}_3\text{Cl}\)) increases, chlorine radicals will collide with it instead of methane, leading to further substitution:
- Step 1: \[ \text{Cl}^\bullet + \text{CH}_3\text{Cl} \rightarrow \text{HCl} + ^\bullet\text{CH}_2\text{Cl} \]
- Step 2: \[ ^\bullet\text{CH}_2\text{Cl} + \text{Cl}_2 \rightarrow \text{CH}_2\text{Cl}_2 + \text{Cl}^\bullet \]
This cycle can repeat until all hydrogen atoms are replaced, producing a mixture of chloromethane (\(\text{CH}_3\text{Cl}\)), dichloromethane (\(\text{CH}_2\text{Cl}_2\)), trichloromethane (\(\text{CHCl}_3\)), and tetrachloromethane (\(\text{CCl}_4\)). These compounds must be separated by fractional distillation, which is costly and inefficient.
Controlling the product mixture:
- To favour the mono-substituted product (\(\text{CH}_3\text{Cl}\)), a large excess of methane is used. This makes collisions between chlorine radicals and methane molecules far more likely than collisions with chloromethane molecules.
- To favour the multi-substituted product (\(\text{CCl}_4\)), a large excess of chlorine is used.
Positional Isomers
For alkanes with three or more carbon atoms, substitution can occur at different carbon atoms along the chain. This leads to a mixture of positional isomers, which also reduces the yield of a specific desired product. For example, chlorinating propane will yield a mixture of 1-chloropropane and 2-chloropropane.
If an exam question asks how to synthesise a pure sample of chloromethane or bromoethane, always state that you would use a large excess of the starting alkane rather than the halogen. This reduces further substitution and increases the yield of the mono-substituted halogenoalkane.
Solution:
- Initiation: Homolytic fission of the bromine molecule by UV light: \[ \text{Br}_2 \xrightarrow{\text{UV}} 2\text{Br}^\bullet \]
- Propagation Step 1: Bromine radical abstracts a hydrogen from ethane, forming a ethyl radical and hydrogen bromide: \[ \text{Br}^\bullet + \text{C}_2\text{H}_6 \rightarrow \text{HBr} + ^\bullet\text{C}_2\text{H}_5 \]
- Propagation Step 2: Ethyl radical reacts with a bromine molecule, forming bromoethane and regenerating the bromine radical: \[ ^\bullet\text{C}_2\text{H}_5 + \text{Br}_2 \rightarrow \text{C}_2\text{H}_5\text{Br} + \text{Br}^\bullet \]
- Termination (any of the following are acceptable):
- Two bromine radicals combine: \[ \text{Br}^\bullet + \text{Br}^\bullet \rightarrow \text{Br}_2 \]
- An ethyl radical and a bromine radical combine: \[ ^\bullet\text{C}_2\text{H}_5 + \text{Br}^\bullet \rightarrow \text{C}_2\text{H}_5\text{Br} \]
- Two ethyl radicals combine to form butane: \[ ^\bullet\text{C}_2\text{H}_5 + ^\bullet\text{C}_2\text{H}_5 \rightarrow \text{C}_4\text{H}_{10} \]
Solution:
- Dichloromethane is formed by further substitution of the initial product, chloromethane (\(\text{CH}_3\text{Cl}\)).
- A chlorine radical reacts with chloromethane, abstracting a hydrogen atom to form a chloromethyl radical and hydrogen chloride: \[ \text{Cl}^\bullet + \text{CH}_3\text{Cl} \rightarrow \text{HCl} + ^\bullet\text{CH}_2\text{Cl} \]
- The chloromethyl radical then reacts with a chlorine molecule to produce dichloromethane and regenerate the chlorine radical: \[ ^\bullet\text{CH}_2\text{Cl} + \text{Cl}_2 \rightarrow \text{CH}_2\text{Cl}_2 + \text{Cl}^\bullet \]
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