Halogenation of Alkanes (DP IB Chemistry): Revision Note

Halogenation of alkanes

Stability of alkanes

• Alkanes are relatively stable / unreactive due to the strengths of the C–C and C–H bonds and their non-polar nature

Strength of bonds

• Alkanes consist of carbon and hydrogen atoms which are bonded together by single bonds

• Unless a lot of heat is supplied, it is difficult to break these strong C-C and C-H covalent bonds

• This decreases the alkanes’ reactivities in chemical reactions

Lack of polarity

• The electronegativity of the carbon and hydrogen atoms in alkanes are almost the same

• This means that both atoms share the electrons in the covalent bond almost equally

 Pauling electronegativity values for the elements

• As a result of this, alkanes are nonpolar molecules and have no partial positive or negative charges (δ⁺ and δ^–  respectively)

Structural formula of ethane showing bond polarities

• Alkanes, therefore, do not react with polar reagents

  • They have no electron-deficient areas to attract nucleophiles

  • And also lack electron-rich areas to attract electrophiles

• Alkanes only react in combustion reactions and undergo substitution by radicals

Free-radical substitution of alkanes

• Alkanes can undergo free-radical substitution in which a hydrogen atom gets substituted by a halogen (chlorine/bromine)

• Since alkanes are very unreactive, ultraviolet light (sunlight) is needed for this substitution reaction to occur

Proving that energy from UV light is required for radical reactions with halogens

• The free-radical substitution reaction consists of three steps

Initiation step

• In the initiation step, the halogen bond (Cl-Cl or Br-Br) is broken by UV energy to form two radicals

• The covalent Cl-Cl bond is broken by energy from the UV light

• Each atom takes one electron from the covalent bond

• This produces two radicals in a homolytic fission reaction

Cl–Cl 2Cl^•

• For more information about the initiation step, see our revision note about homolytic fission

Propagation step

• The halogen free radicals are very reactive and will attack the unreactive alkanes

• One of the methane C-H bond breaks homolytically to produce an alkyl radical

CH₄ + Cl^• → ^•CH₃ + HCl

• The alkyl radical can attack another chlorine molecule to form a halogenoalkane

• This also regenerates the chlorine free radical

^•CH₃ + Cl₂ → CH₃Cl + Cl^• 

• The regenerated chlorine free radical can then repeat the cycle

• For example, the chlorination of ethane is:

ethane + chlorine radical → ethyl radical + hydrogen chloride

CH₃CH₃ + Cl^• → ^•CH₂CH₃ + HCl

ethyl radical + chlorine molecule → chloroethane + regenerated chlorine radical

^•CH₂CH₃ + Cl₂ → CH₃CH₂Cl + Cl^•

• This reaction is not very suitable for preparing specific halogenoalkanes as a mixture of substitution products is formed

• If there is enough halogen present, all the hydrogens in the alkane will eventually get substituted

• For example, the chlorination of ethane could continue:

chloroethane + chlorine radical → radical + hydrogen chloride

CH₃CH₂Cl + Cl^• → ^•CH₂CH₂Cl + HCl

radical + chlorine molecule → 1,2-dichloroethane + regenerated chlorine radical

^•CH₂CH₂Cl + Cl₂ → CH₂Cl₂ + Cl^•

• This process can repeat until hexachloroethane, C₂Cl₆, is formed

Termination step

• The termination step is when the chain reaction terminates (stops) due to two free radicals reacting together and forming a single unreactive molecule

  • Multiple products are possible

• For example, the single substitution of ethane by chlorine can form: 

ethyl radical + chlorine radical → chloroethane

^•CH₂CH₃ + Cl^• → CH₃CH₂Cl

ethyl radical + ethyl radical → butane

^•CH₂CH₃ + ^•CH₂CH₃ → CH₃CH₂CH₂CH₃

chlorine radical + chlorine radical → chlorine molecule

Cl^• + Cl^• → Cl₂