Alkanes and free radical substitution

In a nutshell

Alkanes are hydrocarbons. Hydrocarbons contain the atoms carbon and hydrogen only. Hydrocarbons are used as fuels, as a lot of energy is released during complete combustion.

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Equations

These are the general word equations. You will need to be able to use these to write equations with specific substances.

​$$XY \rightarrow X^+ + \,Y^- \\XY \rightarrow X\cdot + \,Y\cdot$$​​

Properties of hydrocarbons

The shorter the length of the hydrocarbon:

• the less viscous they are
• the more volatile and flammable they are
• the lower the boiling point

Alkanes have induced dipole-dipole interactions. The longer the hydrocarbon chain, the greater number of electrons are able to interact giving rise to more induced dipole-dipole interactions. The more induced dipole-dipole interactions there are the more energy is required to break them, leading to properties such as a higher boiling point. 

Branched alkanes have a lower boiling point as they cannot pack closely together, reducing the number of induced dipole-dipole interactions.

Alkanes

Alkanes are the simplest hydrocarbons. Aliphatic alkanes have the general formula $$C_nH_{2n+2}$$. Whereas cyclic alkanes have the general formula $$C_nH_{2n}$$. They are saturated compounds as each carbon atom has four single bonds. Each carbon atom adopts a tetrahedral shape $$(T_d)$$ due to the presence of four pairs of bonding electrons. As bonding electrons repel equally a bond angle of $$109.5\degree$$ is adopted giving rise to a tetrahedral shape around the carbon.​

Examples

Reactions of alkanes

Alkanes can undergo bond fission. There are two types of bond fission:

1. Heterolytic fission
2. Homolytic fission

​​Heterolytic fission

Heterolytic fission is when one atom receives both electrons. Two different ions are formed, a cation and an anion. 

Homolytic fission

Homolytic fission is when both atoms in the bond receive one electron each. The species formed are radicals and they are uncharged and they have unpaired electrons. 

Free-radical substitution

Free radical substitution is when a free-radical replaces another atom or group of atoms in a molecule. Alkanes can react with halogens in photochemical reactions. Free-radical substitution have a three step mechanism. 

Example

​$$CH_4 \, + \, Cl_2 \xrightarrow{UV} CH_3Cl \, +\, HCl$$​​

Mechanism for the reaction between methane and chlorine:

Step 1 - Initiation: production of free radicals. This is done via homolytic fission where the energy for the breaking of the bond is provided by energy from sunlight (photodissociation). 

$$Cl_2 \xrightarrow{UV} 2\, Cl \cdot$$​

​Step 2 - Propagation: chain reaction of free radical formation and consumption.

The free radical produced is very reactive and goes onto react with a molecule of methane. The methyl radical can then go on to attack chlorine molecule forming another chlorine radical. And this process continues.

$$Cl\cdot + \,CH_4\rightarrow \cdot CH_3 \,+\, HCl \\\cdot CH_3 \, + \, Cl_2 \rightarrow CH_3 Cl +\, Cl \cdot$$​​

Step 3 - Termination: free radicals react together to form a stable molecule. There are many possible reactions for termination. For example,

$$Cl\cdot + \,\cdot CH_3 \rightarrow \,CH_3Cl\\\cdot CH_3 \, + \, \cdot CH_3 \rightarrow \,C_2H_6$$​​

A problem which arises is that a mixture of products is obtained. A chlorine radical may go on to react with chloromethane to form dichloromethane, which in turn may go on to form trichloro- and tetrachloromethane. This is not ideal as it requires time and energy to separate the mixture and obtain the desired product. Furthermore, for longer chain carbons, the free-radical attack may happen anywhere along the chain leading to isomers. 

Example

Reaction between butane and bromine can lead to the formation of $$1$$​-bromobutane and $$2$$​-bromobutane.