Chemical reactions of the alkanes

• The reasons for the lack of reactivity of the alkanes are:

  • C-C and C-H sigma bonds are strong
  • C-C bonds are non-polar
  • the electronegativity of carbon and hydrogen is so similar that the C-H bond can be considered non-polar
• The complete combustion of alkanes produces carbon dioxide and water
• Equation for balancing combustion of any alkane:

  $C_xH_y +$$(x + y/4) O_2 -> xCO_2 +$$y/2H_2O$
• As you descend the homologous series, each alkane molecule (each extra -CH2-) needs an extra 1 1/2 O2 molecules for complete combustion
• Incomplete combustion of alkanes produces carbon monoxide or soot (carbon) and water
• Carbon monoxide is a colourless, odourless, highly toxic gas. It combines irreversibly with haemoglobin in red blood cells to form carboxyhaemoglobin. This prevents the haemoglobin from transporting oxygen around the body.
• In the presence of sunlight, alkanes react with halogens. The high-energy ultraviolet radiation provides the initial energy for a reaction to take place. For example the bromination of methane:

  $CH_4(g)+$$Br_2(l) -> CH_3Br(g)+$$HBr(g)$ This is a substitution reaction as a hydrogen atom in the alkane has been substituted by a halogen atom.
• The bromination of methane is an example of radical substitution. The mechanism of radical substitution takes place in three stages:

  • initiation
  • propagation
  • termination
• INITIATION (radical substitution using bromine and methane as example)

  In the initiation stage, the reaction is started when the covalent bond in a bromine molecule is broken by homolytic fission. Each bromine atom takes one electron from the pair in the bond, forming two highly reactive bromine radicals. The energy for this bond fission is provided by UV radiation. The radical is shown with a single dot to represent the unpaired electron.
• PROPAGATION(radical substitution using methane and bromine as example)

  In the propagation stage, the reaction propagates through two propagation steps, a chain reaction.

  Propagation step 1: a bromine radical reacts with a C-H bond in the methane, forming a methyl radical (example of carbocation) and a molecule of hydrogen bromide

  Propagation step 2: each methyl radical reacts with another bromine molecule forming the organic product bromomethane together with a new bromine radical

  The new bromine radical then reacts with another CH4 molecule as in the first propagation step
• In theory the propagation steps could continue until. all the reactants have been used up. In practice propagation is terminated whenever two radicals collide. It has been estimated that up to a million propagation cycles take place before a termination step stops the reaction
• TERMINATION (radical substitution using methane and bromine as example)

  In the termination stage, two radicals collide, forming a molecule with all electrons paired. There are a number of termination steps with different radicals in the reaction mixture:

  $Br. +$$Br. -> Br_2$

  $CH_3. +$$CH_3. -> C_2H_6$

  $CH_3 . +$$Br. -> CH_3Br$

  When two radicals collide and react, both radicals are removed from the reaction mixture, stopping the reaction.
• In the mechanism of the radical substitution of methane and bromine, bromomethane (CH3Br) is formed in the second propagation step. Another bromine radical can collide with a bromomethane molecule, substituting a further hydrogen atom to for dibromoethane (CH2Br2). Further substitution can continue until all hydrogen atoms have been substituted. The result is a mixture of CH3Br, CH2Br2, CHBr3, and CBr4.
• If the carbon chain is longer than 2 carbons, reacting with a halogen will result in a mixture of monosubstituted isomers by substitution at different positions in the carbon chain. With further substitution there are even more possibilities.