1.6 Chlorination and bromination of alkanes, reaction mechanisms and the structure and uses of products

Part 1. ALKANES and the PETROCHEMICAL INDUSTRY - Doc Brown's Advanced A Level Organic Chemistry Revision Notes

The reactions of alkanes with halogens are important processes in the chemical industry for the production of a variety of useful products. The reaction between alkanes and halogens (chlorine and bromine) is described by reaction conditions, equations and the detailed mechanisms of halogenation, namely chlorination and bromination.

Sub-index for this page

1.6.1 Comments on lack of alkane reactivity

1.6.2 The reaction between chlorine and alkanes - uses of chloroalkanes

1.6.3 The reaction between bromine and alkanes - uses of bromoalkanes

1.6.4 Some footnotes on other examples of free radical chemistry

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1.6.1 The reactivity of alkanes - or lack of it!

Alkanes are not very reactive molecules. Most reactions of alkanes require some 'energetic' input to initiate a reaction e.g. high temperature and catalyst for cracking, uv light for chlorination or a spark to ignite them in combustion. In many cases this involves initiating free radical reactions.

A combination of two reasons account for the lack of reactivity of alkanes compared to most other homologous groups of organic molecules.

1. Bond Strength: Strong sigma (σ) bonds in alkane molecules

   • The single covalent C-C bonds (bond enthalpy 348 kJ mol^-1) and the C-H bonds (bond enthalpy 412 kJ mol^-1) in alkanes are very strong so bond fission does not readily happen.

   • The carbon and hydrogen atomic radii are small, giving a short and strong bonds.

   • Therefore, in terms of C-C and C-H bond fission, alkane reactions will tend to have high activation energies resulting in slow/no reaction.

2. Nature of bonding:

   • Carbon and hydrogen have similar electronegativities (C = 2.5, H = 2.1, Δ_elec = 0.4), so there is effectively no polar bond (non-polar bond) giving a slightly positive carbon (C^δ+) which can be attacked by electron pair donating nucleophiles. [e.g. contrasting with halogenoalkanes (^δ+C-Cl^δ-) or aldehydes/ketones (^δ+C=O^δ-)]

   • All the C-C and C-H bonds in alkanes are single covalent (saturated hydrocarbons) with no region of particularly high electron density like a pi (π) bond electron cloud susceptible to attack by electron pair accepting electrophiles. [e.g. like in the double C=C bond in unsaturated alkenes]

3. Free radicals are highly reactive species with an unpaired electron

   •  e.g. methyl radical CH₃▪, ethyl radical CH₃CH₂▪., the latter better shown as ▪CH₃ and  ▪CH₂CH₃ , because the ▪ dot represents the unpaired electron, which is in an orbital of a carbon atom.

   • These free radicals facilitate (propagate) rapid chain reactions e.g. in combustion!

   • Heat or uv light can generate free radicals by homolytically splitting halogen molecules such as chlorine and bromine into atoms that can then propagate a chain reaction to form substituted products known as halogenoalkanes (haloalkanes).

     • The bond energies of chlorine (Cl-Cl = 242 kJ/mol) and bromine (Br-Br = 193 kJ/mol) molecules are significantly lower than those of C-C and C-H bonds which is why in the ensuing free radical mechanism descriptions, the initiation step is the homolytic bond fission of the halogen molecule.

       • X-X  ===>  2X•  (where X = Cl or Br)

   • Although Cl▪ and Br▪ are individual atoms, they are highly reactive free radicals because they have an unpaired electron and seek another electron on another atom to 'pair up' and form a stable bond with this other atom.

1.6.2 The reaction of alkanes with chlorine

As pointed out already, alkanes are not very reactive unless burned and they will react with reactive chemicals like chlorine and bromine when heated or subjected to uv light to form chlorinated alkane hydrocarbons.

• Despite the reactivity of chlorine you still need something extra to initiate the reaction like uv light at room temperature or relatively high temperature (250^oC - 400^oC) in this case.
  • So homolytic bond fission occurs via uv photo absorption of a high thermal energy collision.
• A substitution reaction occurs and a chloro–alkane is formed e.g. a hydrogen is swapped for a chlorine and the hydrogen combines with a 2nd chlorine atom e.g.
• The free radical substitution reaction between methane and chlorine
  • methane + chlorine ===> chloromethane + hydrogen chloride
    • CH₄ + Cl₂ ===> CH₃Cl + HCl
    • + Cl₂ ===> + HCl

  • 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

      • The further sequential substitution equations are easy to work out:

      • CH₃Cl  +  Cl₂  ===>  CH₂Cl₂  +  HCl

      • CH₂Cl₂  +  Cl₂  ===>  CHCl₃  +  HCl

      • CHCl₃  +  Cl₂  ===>  CCl₄  +  HCl

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

      • i.e. CH₃CH₂CH₃  +  Cl₂  ===>   {CH₃CH₂CH₂Cl or CH₃CHClCH₃ }  +  HCl

      • and then 1,1- or 1,2- or 1,3- or 2,2-dichloropropanes etc. etc.! (just shown schematically)

      • ===> CH₃CH₂CHCl₂  or  CH₃CHClCH₂Cl  or  ClCH₂CH₂CH₂Cl  or  CH₃CCl₂CH₃  + HCl

      • etc. and ultimately CCl₃CCl₂CCl₃ !!!

      • Because it is quite difficult to control this reaction when chlorine (or bromine later) is in excess, its not easy to control the yield of the desired halogenoalkane.

      • With substitution occurring in different places on the same carbon atom, or different carbon atoms, the variety of possible products  places quite a limitation on this reaction as a means of chemical synthesis. Fractional distillation can be used to separate the products.

  • The basic schematic mechanism for the free radical chain substitution reaction between an alkane and chlorine is given in the mechanism 6 diagram below.
  • 
  • Mechanism 6. R = H or alkyl in various combinations, when all the R groups s are H, CR₃H = methane

  • 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 - hence each radical has an unpaired electron.

    • 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.

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

      • abstracting another atom from another molecule e.g. step (2) H abstracted, and step (3) chlorine abstracted in the propagation steps or by pairing up with another radical e.g. termination steps (4) to (6). The terms propagation step and termination step are explained below.

  • 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 free radical '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.

    • Termination steps break the chain i.e. they prohibit a possible chain propagation step.

    • 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.

• The methane - chlorine free radical substitution reaction in detail

  • 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 

    • Similarly if the alkane was ethane, traces of butane are formed, similarly, for other alkanes you will traces of an alkane with twice as many carbon atoms as the original reactant alkane.

    • Mechanism 48 the free radical mechanism of chlorine reacting with methane to form chloromethane.

    • Initiation step - homolytic bond fission of the chlorine molecule

      • Cl₂  ===>  2Cl•

    • Propagation steps - two needed to form the monosubstituted product

      • CH₄  +  Cl•  ===>  •CH₃  +  HCl

      • •CH₃  +  Cl₂  ===>  CH₃Cl  +  Cl•

    • Termination steps - the collision of any two free radicals

      • •Cl  +  •Cl  ===>  Cl₂

      • •CH₃  +  •Cl  ===>  CH₃Cl

      • •CH₃  +  •CH₃  ===>  CH₃CH₃

    • Chlorine atoms produced in the initiation step (via uv or heat).
    • In the two propagation steps a chlorine atom abstracts hydrogen atom from methane to form HCl. The resulting methyl radical abstracts a chlorine atom from a chlorine molecule to form the product. BUT, you are left with a chlorine atom/radical that continues the chain reaction via.
    • Termination occurs when two radicals meet and combine. This has a very low probability because the free radical concentrations are very small in the first place. This allows the propagation steps to occur several hundred times before termination occurs.

  • My original mechanism 48 sketch is shown below, but in the text version above, I've shown the methyl radical more correctly with the unpaired electron on the carbon atom.

  • 
  • Next, considering a few details of the subsequent polysubstitution reactions of methane and chlorine by looking at the propagation steps that need to further substitution products.
  • 
  • Diagram mechanism 52 (above): I've sketched out the propagation steps for the polysubstitution of methane.
    • Note I've correctly drawn the unpaired electron on the carbon atom of the carbon-based free radicals.
  • 
  • The dot and cross outer electron diagrams for the  •CH₃,  •CH₂Cl,  •CHCl₂ and •CCl₃ free radicals.
  • All the carbon based free radicals involved in the sequential chlorination of methane are pyramidal in shape.
    • The pyramid shape arises from the mutual repulsion of four lots of electrons (best appreciated in the dot & cross diagrams), three bonding pairs and one lone electron (not involved in chemical bonding).
    • The four electron groups are arranged in a tetrahedral arrangement around the central carbon atom, so the bond angles should be around 109^o (as in methane, NOT 120^o, these free radicals are NOT trigonal planar in shape).
• The ethane - chlorine substitution reaction
  • ethane + chlorine ==> chloroethane + hydrogen chloride
  • CH₃CH₃  +  Cl₂  ===>  CH₃CH₂Cl  +  HCl
  •   +  Cl₂ ===> +   HCl
  • 
  • Diagram mechanism 49: My original sketch for the mechanism for the monosubstitution of ethane with chlorine to give chloroethane - text version below with the unpaired electron, more correctly on the carbon atom.

  • Initiation step - homolytic bond fission of the chlorine molecule

    • Cl₂  ===>  2Cl•

  • Propagation steps - two needed to form the monosubstituted product

    • CH₃CH₃  +  Cl•  ===>  •CH₂CH₃  +  HCl

    • •CH₂CH₃  +  Cl₂  ===>  CH₃CH₂Cl  +  Cl•

  • Termination steps - the collision of any two free radicals

    • •Cl  +  •Cl  ===>  Cl₂

    • •CH₂CH₃  +  •Cl  ===>  CH₃CH₂Cl

    • •CH₂CH₃  +  •CH₂CH₃  ===>  CH₃CH₂CH₂CH₃

  • You should find a trace of butane in the final reaction mixture of products - evidence supporting this mechanism.

The butane - chlorine substitution reaction

butane  +  chlorine  ===>  {1-chlorobutane  or  2-chlorobutane}  +  hydrogen chloride

to give two possible isomeric (C₄H₉Cl) monosubstitution products.

+  Cl₂ {  or  } +  HCl

+ Cl₂ or   +  HCl

Diagram mechanism 50: Sketch of the mechanism for the monosubstitution of butane with chlorine. Note the possibility of different abstractions at different positions on the carbon chain of butane ultimately giving a mixture of two structural isomer products - the structural isomers 1-chlorobutane and 2-chlorobutane.

The big dots represent the unpaired electron of the halogen atom or alkyl radical.

Technically, you can expect traces of octane, 3,4-dimethylhexane and 3-methyheptane in the final reaction mixture.

The presence of CH₃(CH₂)₆CH₃, CH₃CH₂CH(CH₃)CH(CH₃)CH₂CH₃ and

(CH₃)₂(CH₂)₅CH₃  from the termination steps support the mechanism described above.

The other two possible termination steps will give 1-chlorobutane and 2-chlorobutane.

 

The cyclohexane - chlorine substitution reaction

cyclohexane  +  chlorine  ===>  chlorocyclohexane  + hydrogen chloride

+ Cl₂  +  HCl

 

In all these examples you can substitute bromine for chlorine to predict the mechanism steps and the bromoalkane products (see section 1.6.3 below).

The uses of chloroalkanes

Chloromethane and chloroethane are gases at room temperature, but bigger chloro–alkane molecules are liquids and useful solvents in the laboratory or industry. However, they are still quite volatile and chlorohydrocarbon vapours can be harmful if breathed in. Halogenoalkanes are used as refrigerants.

Long-chain chloroalkanes have a low solubility in water and are used as flame retardants in rubber, paint, leathers and sealing compound formulations.

1.6.3 The reaction of alkanes with bromine

The bromine - ethane reaction

e.g. monosubstitution: ethane  +  bromine  ===>  bromoethane  +  hydrogen bromide

Initiation step - homolytic bond fission of the bromine molecule

Br₂  ===>  2Br•

Propagation steps - two needed to form the product

CH₃CH₃  +  Br•  ===>  •CH₂CH₃  +  HBr

•CH₂CH₃  +  Br₂  ===>  CH₃CH₂Br  +  Br•

Termination steps - the meeting of any two radicals

•Br  +  •Br  ===>  Br₂

•CH₂CH₃  +  •Br  ===>  CH₃CH₂Br

•CH₂CH₃  +  •CH₂CH₃  ===>  CH₃CH₂CH₂CH₃

You should find a trace of butane in the final reaction mixture of products - evidence supporting this mechanism.

 

The propane - bromine substitution reaction

propane  +  bromine   ===>   {1-bromopropane or 2-bromopropane}  +  hydrogen bromide

word equation followed by displayed formula equation, abbreviated structural formula equation and skeletal formula equation

  + Br₂    or    + HBr

  + Br₂  {  or }   + HBr

  + Br₂    or    + HBr

Note the formation of two isomeric (C4H9Br) monosubstituted products, 1-bromopropane and 2-bromopropane.

Diagram mechanism 51 (above): The mechanism for the monosubstitution of propane with bromine.

The big dots represent the unpaired electron of the atom or alkyl radical on a chlorine atom (Cl•) or a carbon atom of the carbon based free radical (e.g. •CH₂CH₂CH₃).

Again, note the possibility of two structural positional isomers (C₃H₇Br), 1-bromopropane and 2-bromopropane.

The mechanism is identical to that for chlorination of alkanes.

The cyclohexane - bromine substitution reaction

cyclohexane  +  bromine  ===>  bromocyclohexane C₆H₁₁Br

+ Br₂  +  HBr

There is only one possible monosubstitution product.

If excess bromine present, then other disubstituted products can form

Further substitution can take place to give three isomeric disubstitution products of C₆H₁₀Br₂  e.g.

+ Br₂ ,  ,   +  HBr

overall ...

cyclohexane  +  bromine  ===>  {1,2-   &   1,3-   &  1,4-dibromocyclohexane}  + hydrogen bromide

+ 2Br₂ ,  ,   +  2HBr

Another example of a multiple substitution of an alkane by a halogen

Note: I don't think iodine is reactive enough to produce iodoalkanes by this method.

Uses of bromoalkanes

Bromine compounds are used as flame retardant additives in polymer compositions and fire extinguishers.

Due to their toxicity, bromine compounds are also used as fumigants e.g. bromomethane has been used to control insects, weeds and rodents, but this application is controversial and is being phased out.

1.6.4 Some footnotes on free radical chemistry

(a) First encounters with free radical

In your advanced level chemistry course you usually first come across free radical substitution reactions when looking at the reaction between hydrogen and chlorine (one of the simplest free radical mechanisms, below) or the methane - chlorine reaction (already described in detail above on this page).

(i) initiation step: Cl₂  ===>  2Cl^.

Homolytic bond fission by heat or light to give two chlorine free radicals.

(ii) propagation steps: Cl^.  +  H₂  ===>  HCl  +  H^.   followed by    H^.  +  Cl₂ ===>  HCl  +  Cl^.

Two steps giving the product and a free radical to continue the chain reaction

(iii) termination steps: H^.  +  Cl^.  ===>  HCl   or   2H^.  ===>  H₂   or   2Cl^.  ===>  Cl₂

The three possible ways of ending a free radical reaction chain sequence.

You can see it is very similar to the methane - chlorine reaction, a H radical instead of CH3 radical, but no complications due to further substitution.

(b) Use of CFCs and destruction of the ozone layer

The chemistry of ozone formation and its destruction by CFCs involves free radical chemical reactions.

Ozone, CFC's and halogen organic chemistry links

(c) Polymerisation

Ethene can be polymerised to poly(ethene) using oxygen or organic peroxide catalysts which generate free radicals which facilitate polymerisation.

Free radical polymerisation to give poly(alkene) polymers e.g. ethene ==> poly(ethene)

(d) Biochemistry

Many reactions in living cells quite naturally generate free radicals as a by-product, which, because of their high reactivity, need to be removed before causing unwanted chemical reactions and cell damage.

Free radicals are formed in the body's normal metabolic activity e.g. respiration and they are important in the production of enzymes and hormones.

Free radicals can damage cell membranes and the genetic molecules of DNA.

Excess of free radicals are linked to premature aging and various chronic illnesses.

Smoking, sunbathing and air pollution from road vehicles also exposes the body to free radicals.

Note that uv radiation in bright sunlight can directly produce free radicals in skin cells!

However, the body has a natural defence mechanism in the form of molecules called anti-oxidants which counteract excess free radicals and prevent them harming cell function.

Two vitamins are known to be part of this defence mechanism.  Vitamin C in fresh fruit and vegetables and Vitamin E in fats, oils, nuts, egg yolk, liver and green vegetables.

 

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