Organic chemistry Part 10: Reactivity and reaction mechanisms

10.2 Some reaction mechanisms of ALKANES - lack of reactivity and halogenation of alkanes by a free radical substitution mechanism

[Author ©  Dr Phil Brown PhD: Doc Brown's exam revision notes suitable for A level chemistry students of advanced pre-university/college level organic chemistry courses:  halogenation of alkanes  [page updated April 14th 2026 *]

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Part 10.2 ALKANES - introduction to the reaction mechanisms of alkanes.

A Level Revision Notes PART 10 Summary of organic reaction mechanisms - A mechanistic introduction to organic chemistry and explanations of different types of organic reactions The reaction mechanisms of alkanes

Free radical chlorination/bromination to give halogenoalkanes (haloalkanes, alkyl halides).

On other pages ...

Free radical thermal cracking to give shorter alkanes and ionic catalytic cracking to give shorter alkanes and alkenes.

Revision notes include full diagrams and explanation of the mechanisms of alkanes and the 'molecular' equation and reaction conditions and other con-current reaction pathways and products are also explained.

• Alkanes are not very reactive molecules. Most reactions require some energy input to initiate a reaction e.g. high temperature and catalyst for cracking, uv light for chlorination or a spark to ignite them (initiating free radical reactions).

• A combination of two main reasons account for this lack of reactivity compared to most other homologous groups of organic molecules.

  1. C-C and C-H bond Strength:

     • The single covalent C-C (bond enthalpy 348 kJ mol^-1) and C-H (bond enthalpy 412 kJ mol^-1) bonds are very strong so bond fission does not readily happen. The carbon atom radius is small, giving a short and strong bond with other small atoms. Therefore the reactions will tend to have high activation energies resulting in slow/no reaction.

  2. Polarity nature of the C-H bond:

     • Carbon and hydrogen have similar electronegativities, so there is no polar bond giving a slightly positive carbon (C^δ+) which can be attacked by electron pair donating nucleophiles. [e.g. see halogenoalkanes (^δ+C-Cl^δ-) or aldehydes/ketones (^δ+C=O^δ-)]

     • All the C-C and C-H bonds are single covalent and no region of particularly high electron density susceptible to attack by electron pair accepting electrophiles. [e.g. like a double C=C bond see alkenes which are highly reactive despite the fact the C=C double bond is not polar]

• Hydrocarbon molecules such as saturated alkanes are reacted with chlorine or bromine to manufacture valuable chloroalkanes or bromoalkanes e.g.

  • (i) ethane + bromine ==> bromoethane + hydrogen bromide

  • + Br₂ ==> + HBr

  • (ii) propane + chlorine ==> 1-chlorobutane/2-chlorobutane + hydrogen chloride

  • + Cl₂ ==> + HCl

  • + Cl₂ ==> + HCl

  • Note that there are two possible monosubstitution products possible via this free radical substitution mechanism described and explained below.

  • This is an important organic synthesis process to manufacture organochlorine and organobromine compounds.

• What is the reaction mechanism of chlorine reacting with alkanes like methane and ethane etc.

• Under the influence of high temperature (heat) or uv light, alkanes will react with chlorine or bromine via a free radical substitution reaction mechanism.

• The basic reaction is: R₃C-H + Cl₂ ==heat/uv==> R₃C-Cl + HCl [mechanism 6]

  • R = H or alkyl e.g. CH₃, CH₃CH₂ etc. or aryl e.g. C₆H₅, CH₃C₆H₄ etc.

• The reaction is initiated by higher temperatures e.g. 250-400^oC or uv light at room temperature.

• If other hydrogen atoms are available on the original hydrocarbon then polysubstituted chloroalkanes will be formed

  • e.g. methane => chloromethane => dichloromethane => trichloromethane => tetrachloromethane

    • CH₄ ==> CH₃Cl ==> CH₂Cl₂ ==> CHCl₃ ==> CCl₄

  • propane can form initially 1-chloropropane or 2-chloropropane

    • and then 1,1- or 1,2- or 1,3- or 2,2-dichloropropanes etc. etc.!

    • i.e. CH₃CH₂CH₃ ==> CH₃CH₂CH₂Cl or CH₃CHClCH₃ 

      • => CH₃CH₂CHCl₂ or CH₃CHClCH₂Cl or ClCH₂CH₂CH₂Cl or CH₃CCl₂CH₃  

    • etc. and ultimately CCl₃CCl₂CCl₃

  • You can write a similar set of equations for ethane, butane and pentane etc...

    • ... and the mechanisms can similarly constructed

mechanism 6 - free radical chlorination of an alkane, R = H, alkyl or aryl.

If R = H,  CR₃H = methane, and the mechanism shows the monosubstitution of an alkane.

You can write a similar mechanism for the mono-bromo substitution of an alkane in which the reaction can be initiated with uv light or an elevated reactor temperature.

• Step (1) is the initiation step when the chlorine molecule is split into two chlorine atoms/radicals by homolytic bond fission by the impact-absorption of the ultraviolet photon. Its quantum of energy, E=hv, must be great enough to break the Cl-Cl bond.

  • Homolytic bond fission means the original pair of (Cl-Cl) bonding electrons is split between the two radicals formed.

  • Step (1) illustrates how to use half-arrows to show a homolytic bond fission step

    • Not all exam boards demand half-arrows, so the 'style' is deliberately varied in the diagram.

  • The red dots represent the unpaired electron on the free radical and the half-arrows show the individual electron 'shifts'.

  • The breaking of the Cl-Cl bond in the chlorine molecules begins the reaction because it is the weakest of the bonds of any reactant molecule involved. Bond enthalpies/kJmol^-1:

  • Cl-Cl = 242, C-C = 348, C-H = 412, and even the new bond formed, C-Cl, is 338.

  • Free radicals are highly reactive species with an unpaired electron and tend to form a new bond as soon as it is possible by e.g. in this case by ...

    • abstracting another atom from another molecule e.g. propagation steps (2) H abstracted, and (3) chlorine abstracted or by pairing up with another radical e.g. termination steps (4) to (6).

• Steps (2) and (3) are chain propagation steps, because as well as producing one of the reaction products, a new free radical is also produced to continue the reaction, which is why such reactions are sometimes referred to as 'chain reactions'.

  • Step (2) Illustrates how to use half-arrows in a chain propagation step where an attacking radical abstracts an atom from a stable and complete molecule and another radical is formed in the process.

• Steps (4) to (6) are three possible chain termination steps which remove the highly reactive free radicals as two unpaired electrons form a new bond, in this case single C-C covalent bonds.

  • Step (5) illustrates how to use half-arrows to indicate a termination step where the unpaired electrons of the two radicals pair up to form a new bond, in this case a C-Cl bond.

• FURTHER COMMENTS

  • The mechanism for bromination is similar.

  • When the alkane is methane, traces of ethane are found in the final mixture of products.

    • This provides evidence for a mechanism involving a methyl radical.

    • It would be formed from combining two methyl radicals: H₃C• + •CH₃ ==> H₃C-CH₃ 

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10.2.3 The free radical thermal cracking of alkanes On separate page now

10.2.4 An ionic mechanism for catalytic cracking On separate page now

Other important associated organic chemistry links