Haloalkanes

Haloalkanes

Lessons

In this lesson, we will learn:
  • The definition of a haloalkane, their general formulae and some examples of them.
  • The major properties of haloalkanes and their differences to other types of organic compounds.
  • The different types of haloalkane, how they are produced and the reactions they are used in.
  • How to name haloalkanes according to IUPAC organic nomenclature.

Notes:
  • Haloalkanes (A.K.A alkyl halides or halogenoalkanes) are saturated organic compounds containing one or more carbon-halogen bonds.
    Halogen atoms only make one covalent bond, just like hydrogen, so haloalkanes have the same structure as alkanes except for the hydrogen atom(s) being replaced by halogens.

  • Because halogen atoms have the same valence as hydrogen, the general formula of haloalkanes is related to that of alkanes, only depending on the number of halogen atoms the compound has.
    • In haloalkanes with one halogen atom, it is CnH2n+1X, where X = F, Cl, Br, I.
    • In dihaloalkanes which have two halogen atoms it is CnH2nX2

    In any haloalkane, the sum of X and H in the formula will be 2n+2 just like in an alkane because its a saturated compound.
  • The properties of haloalkanes depends on what halogen atom they contain, but generally they are:
    • Very slightly soluble in water. The carbon-halogen bond is normally polar, so they are more water soluble than alkanes but much less soluble than alcohols.
    • More reactive than alkanes, except fluoroalkanes which are very unreactive. This is because haloalkanes react by breaking the carbon-halogen bond - the weaker this is, the more reactive the chemical is going to be. Carbon-fluorine bonds are amongst the strongest chemical bonds while the carbon-iodine bond is quite weak.
    • Relatively higher boiling and melting point compared to their alkane analogue. This is due to the stronger dipole-dipole interactive forces between haloalkane molecules than the London dispersion forces present in simple alkanes. In the examples, we talk about intermolecular forces being broken up when discussing chemicals dissolving in one another this is important when talking about melting and boiling points too!

  • As with alcohols, haloalkanes can be sub-categorized into primary, secondary and tertiary haloalkanes.
    • In primary haloalkanes, the halogen is bonded to a carbon with only one carbon-carbon bond. This would be the case when the halogen is bonded at the end of a carbon chain.
    • In secondary haloalkanes, the halogen is bonded to a carbon atom with two carbon-carbon bonds.
    • In tertiary haloalkanes, the halogen is bonded to a carbon atom with three carbon-carbon bonds – this would be the case when the halogen is bonded to the molecule at a branch in the carbon chain.e.

  • Haloalkanes can be produced by free radical substitution of alkanes with halogens.
    A free radical is a lone electron produced by homolytic fission of a covalent bond. This is where the bond splits evenly in half, so one electron of the covalent bond goes back to one atom and one back to the other. We call the lone one electron on the atom a radical and the resulting atom is very unstable it has lost its complete outer shell.

  • Free radical substitution is the substituting of H on an alkane with a halogen atom to make haloalkanes and other products, which happens by making alkyl and halogen radicals.
    • The free radical substitution of methane with chlorine starts with the homolytic fission (even breaking) of the Cl-Cl bond in Cl2 which is quite a weak bond. This can be broken by UV light, such as by the suns rays high in the atmosphere:

    • Cl2 \, \, 2 Cl \cdotp

      The above is the initiation step of the reaction. It is also sometimes called a photochemical (light chemical) reaction because light has caused a chemical change. These extremely reactive chlorine radicals cause further reactions in a chain reaction with alkanes like methane:

      Cl\cdotp + CH4 \, \, HCl + CH3\cdotp

      CH3 + Cl2 \, \, CH3Cl + Cl\cdotp

      The above are propagation (spreading) steps of the reaction, since we have reacted radicals but also produced more radicals in turn. In the second equation you can see chloromethane being produced and a chlorine radical, which will probably react with more methane to produce HCl in the first of the two above equations.

      Because these radicals are reactive, there is a tendency for them to react together to produce non-radical products. These are the termination steps of the free radical reaction:
      2 Cl \cdotp \, \, Cl2

      Cl\cdotp + CH3\cdotp \, \, CH3Cl

      CH3\cdotp + CH3\cdotp \, \, C2H6

      As you can see in the third equation above, it can produce ethane which can go on to react in the same way as methane, because they both still contain weak C-H bonds that can be reacted by halogen radicals. This is also true of the haloalkanes like CH3Cl, which can go through another free radical substitution and become CH2Cl2 or CHCl3. Some molecules could become chloroethane, dichloroethane, and the ethane could go on to become propane.

    It is extremely hard to predict the real distribution of products of these free radical substitutions. As long as there are halogen radicals being produced, the weak C-H and C-X (halogen) bonds can be broken and there is no hard limit to the number of substitutions that can take place.

  • Haloalkanes can also be prepared by reacting alcohols with strong hydrohalic acids (HX, where X is a halogen). You can think of the -OH hydroxyl group being swapped out for the X- on the acid reacting with it.
    For example, ethanol with hydrochloric acid and propanol with hydrobromic acid are shown below:


  • CH3CH2OH + HCl \, \, CH3CH2Cl + H2O

    CH3CH2CH2OH + HBr \, \, CH3CH2CH2Br + H2O

  • You can do a test tube reaction on haloalkanes to test for the halide present in your compound. This is done by heating gently with sodium hydroxide, then adding acidified silver nitrate (AgNO3) followed by ammonia (NH3) solution. The colour of the precipitate (insoluble solid) formed will show the halide present:

    Ion present

    Test observation

    F-

    No precipitate

    Cl-

    White precipitate (AgNO3) which redissolves with added NH3.

    Br-

    Pale cream precipitate which redissolves with added conc. NH3

    I-

    Yellow precipitate which does not redissolve with added conc. NH3



  • Haloalkanes like chlorofluorocarbons (CFCs) were very useful as refrigerants but they are banned in many countries today because they destroy ozone in the stratosphere. This happens by free radical substitution as described earlier.
    • A common CFC was dichlorodifluoromethane, CCl2F2 which could be broken up by UV light like halogen molecules:

    CCl2F2 \, \, \cdotp CClF2 + Cl\cdotp

    This Cl\cdotp radical would now react with ozone:

    O3 + Cl\cdotp \, \, ClO\cdotp + O2

    ClO\cdotp + O3 \, \, 2 O2 + Cl\cdotp

    The Cl\cdotp radical produced can now just go through the chain reaction again to break down even more ozone.
    This is the major problem of free radical substitution reactions: one radical atom/molecule could react hundreds or thousands of reactant molecules because it gets regenerated. When CFCs were used in refrigerators, even tiny amounts of the CFCs used persisted in the atmosphere and caused serious damage to the ozone layer.

  • Haloalkanes can be reacted to produce a number of other chemicals including:
    • Amines, which are made by a substitution reaction in the following equation (using 1-chloropropane as an example)
      CH3CH2CH2Cl + 2NH3 CH3CH2CH2NH2 + NH4Cl
    • Alcohols in a substitution reaction with sodium or potassium hydroxide in the following equation:
      CH3CH2CH2Cl + NaOH CH3CH2CH2OH + NaCL
    • Alkenes in an elimination reaction with sodium or potassium hydroxide in the following equation:
      CH3CH2CH2Cl + NaOH CH3CH=CH2 + NaCL + H2
      You should notice that the reaction with sodium or potassium hydroxide can produce alkenes and alcohols. Which product forms normally depends on a few conditions and the type of haloalkane used – this will be explored in another lesson!

  • Haloalkanes can be named according to IUPAC systematic rules already learned, where halogens have equal priority to alkyl branches. This has some basic consequences when naming organic compounds with halogen atoms in them (see the examples below):
    • Alcohol groups and alkene double bonds have priority over halogen atoms when numbering/ordering the carbon chain.
    • When naming the alkyl and halogen substituents in a molecule, name them in alphabetical order, even if this leads to 'dipping' between numbers or naming alkyl and halogen substituents.
  • Introduction
    Haloalkanes: Introduction
    a)
    What is a haloalkane?

    b)
    Properties of haloalkanes.

    c)
    Types of haloalkanes

    d)
    Reactions of haloalkanes

    e)
    Naming haloalkanes (IUPAC organic nomenclature)


  • 1.
    Compare and explain the properties of the haloalkanes compared to alkanes and alcohols.
    a)
    Explain why methane's (CH4) boiling point is around -160°C, while dichloromethane's (CH2Cl2) boiling point is around 40°C. Use ideas about intermolecular forces.

    b)
    Explain why dichloromethane (CH2Cl2) is not soluble in water but methanol is soluble in water. Use ideas about intermolecular forces.


  • 2.
    Apply the rules of IUPAC organic nomenclature to draw structural and skeletal formula.
    Draw the structures of the following molecules given by their IUPAC names.
    a)
    3-bromo-3-ethyl-2-iodopentane

    b)
    1,3-difluoro-3-methylhexane

    c)
    3,4-dibromobut-1-ene


  • 3.
    Apply knowledge of IUPAC organic nomenclature to identify and correct systematic naming.
    There are mistakes in the IUPAC systematic names of the chemicals below. Identify and correct the mistakes and give the true IUPAC name of the compounds.
    a)
    1-chloro-2,2-dibromopropane

    b)
    2-bromo-3-chloromethylpropane

    c)
    2-methyl-2-iodobutane