Chemistry of Alkanes and Alkenes

Chemistry of Alkanes and Alkenes

Alkanes

• General formula for non-cyclic alkanes: $C<em>nH</em>{2n+2}$
• Saturated hydrocarbons with only single carbon-carbon bonds.
• Bonds are spaced tetrahedrally around carbon atoms.
• Isomerism: First example occurs with butane ($C<em>4H</em>{10}$), which has structural isomers like 2-methylpropane.
• Structural isomers have the same molecular formula but different structural formulas.
• Structural isomers possess different physical properties such as boiling point, melting point, and density.

Chemical Properties of Alkanes

• Alkanes are fairly unreactive due to strong, almost non-polar, single covalent bonds.
• They lack sites that encourage chemical attacks (old family name, paraffin, meant little reactivity).

Combustion

• Alkanes make useful fuels, especially lower members of the series.
• They react with oxygen in an exothermic reaction.
• Complete combustion:
  $CH<em>4(g) + 2O</em>2(g) \longrightarrow CO<em>2(g) + 2H</em>2O(l)$
• Incomplete combustion:
  $CH<em>4(g) + \frac{3}{2}O</em>2(g) \longrightarrow CO(g) + 2H_2O(l)$
• The greater the number of carbon atoms, the more energy produced.
• The greater the number of carbon atoms, the greater the amount of oxygen needed for complete combustion.
• Example of complete combustion of ethane:
  $C<em>2H</em>6(g) + \frac{7}{2}O<em>2(g) \longrightarrow 2CO</em>2(g) + 3H_2O(l)$
• Example of incomplete combustion of ethane:
  $C<em>2H</em>6(g) + \frac{5}{2}O<em>2(g) \longrightarrow 2CO(g) + 3H</em>2O(l)$

Free Radicals

• Reactive species (atoms or groups) which possess an unpaired electron.
• Their reactivity is due to them wanting to pair up the single electron.
• Formed by homolytic fission (homolysis) of covalent bonds.
• Formed during the reaction between chlorine and methane.
• Formed during thermal cracking.
• Involved in the reactions taking place in the ozone layer.

Chlorination of Methane

• Reagents: Chlorine and methane.
• Conditions: UV light or sunlight. Heat is an alternative energy source.
• Equations:
  • $CH<em>4(g) + Cl</em>2(g) \longrightarrow HCl(g) + CH_3Cl(g)$ (chloromethane)
  • $CH<em>3Cl(g) + Cl</em>2(g) \longrightarrow HCl(g) + CH<em>2Cl</em>2(l)$ (dichloromethane)
  • $CH<em>2Cl</em>2(l) + Cl<em>2(g) \longrightarrow HCl(g) + CHCl</em>3(l)$ (trichloromethane)
  • $CHCl<em>3(l) + Cl</em>2(g) \longrightarrow HCl(g) + CCl_4(l)$ (tetrachloromethane)
• Free radicals are very reactive; with sufficient chlorine, every hydrogen will eventually be replaced.
• Mechanism: Proceeds via free radical substitution because methane is attacked by free radicals, resulting in hydrogen atoms being substituted by chlorine atoms; this is a chain reaction.
• In the propagation step, one radical is produced for each one used.

Steps

• Initiation:
  $Cl_2 \longrightarrow 2Cl•$
  Radicals are created and the single dots represent unpaired electrons.
  The weakest bond (Cl-Cl) is broken as it requires less energy.
• Propagation:
  $Cl• + CH<em>4 \longrightarrow CH</em>3• + HCl$
  Radicals are used and then regenerated.
  $Cl<em>2 + CH</em>3• \longrightarrow CH_3Cl + Cl•$
  Free radicals are reactive because they want to pair up their single electron by abstracting a hydrogen atom from methane, forming a methyl radical.
  The methyl radical is very reactive and attacks a chlorine molecule.
• Termination: Radicals are removed.
  $Cl• + Cl• \longrightarrow Cl<em>2$$Cl• + CH</em>3• \longrightarrow CH<em>3Cl$$CH</em>3• + CH<em>3• \longrightarrow C</em>2H_6$
  Removing the reactive free radicals brings an end to the reaction, but this is not very likely at the start of the reaction because of their low concentration.

Overview Summary

• Alkanes need a very reactive species to react due to their lack of reactivity.
• Free radicals are formed by homolytic fission of covalent bonds, which is done by shining UV light on the mixture (heat could be used).
• Chlorine radicals are produced because the Cl-Cl bond is the weakest.
• Only one chlorine radical is needed to start things off.
• With excess chlorine, further substitution and a mixture of chlorinated products occur.

Bromination of Methane

• Initiation:
  $Br_2 \longrightarrow 2Br•$
  Radicals are created.
• Propagation:
  $Br• + CH<em>4 \longrightarrow CH</em>3• + HBr$
  Radicals are used and then re-generated.
  $Br<em>2 + CH</em>3• \longrightarrow CH_3Br + Br•$
• Termination: Radicals are removed.
  $Br• + Br• \longrightarrow Br<em>2$$Br• + CH</em>3• \longrightarrow CH<em>3Br$$CH</em>3• + CH<em>3• \longrightarrow C</em>2H_6$
• Reaction:
  $Br<em>2 + CH</em>4 \longrightarrow$

Further Propagation

• If excess chlorine is present, further substitution takes place.
• Dichloromethane Formation:
  $Cl• + CH<em>3Cl \longrightarrow CH</em>2Cl• + HCl$
  $Cl<em>2 + CH</em>2Cl• \longrightarrow CH<em>2Cl</em>2 + Cl•$
• Trichloromethane Formation:
  $Cl• + CH<em>2Cl</em>2 \longrightarrow CHCl<em>2• + HCl$$Cl</em>2 + CHCl<em>2• \longrightarrow CHCl</em>3 + Cl•$
• Tetrachloromethane Formation:
  $Cl• + CHCl<em>3 \longrightarrow CCl</em>3• + HCl$
  $Cl<em>2 + CCl</em>3• \longrightarrow CCl_4 + Cl•$
• Mixtures: There will be a mixture of products due to the many possible reactions, and individual haloalkanes can be separated by fractional distillation.
• Dibromomethane Formation:
  $Br• + CH<em>3Br \longrightarrow CH</em>2Br• + HBr$
  $Br<em>2 + CH</em>2Br• \longrightarrow CH<em>2Br</em>2 + Br•$
• Tribromomethane Formation:
  $Br• + CH<em>2Br</em>2 \longrightarrow CHBr<em>2• + HBr$$Br</em>2 + CHBr<em>2• \longrightarrow CHBr</em>3 + Br•$
• Tetrabromomethane Formation:
  $Br• + CHBr<em>3 \longrightarrow CBr</em>3• + HBr$
  $Br<em>2 + CBr</em>3• \longrightarrow CBr_4 + Br•$

The Structure of Alkenes

• General formula for non-cyclic alkenes: $C<em>nH</em>{2n}$
• Hydrocarbons containing only C and H.
• Unsaturated: Atoms can be added to their formula.
• Contain a C=C double bond somewhere in their structure.
• Spatial arrangement around the C=C is planar, and the bond angles are 120°.

Structural Isomerism in Alkenes

• Different structures are possible due to:
  • Different positions for the double bond (e.g., pent-1-ene, pent-2-ene).
  • Branching (e.g., 3-methylbut-1-ene).

Geometrical Isomerism in Alkenes

• An example of stereoisomerism found in some, but not all, alkenes.
• Occurs due to the restricted rotation of C=C bonds.
• Cis: Groups/atoms are on the same side of the double bond.
• Trans: Groups/atoms are on opposite sides across the double bond.
• Isomers have different physical properties (e.g., boiling points, density) but similar chemical properties in most cases.
• Single covalent bonds can easily rotate; what appears to be a different structure is not.
• C=C bonds have restricted rotation, so the groups on either end of the bond are ‘frozen’ in one position; it is not easy to flip between the two.

How to tell if geometrical isomerism exists

• Two different atoms/groups attached to each end of the C=C.
• Once you get two similar atoms/groups attached to one end of a C=C, you cannot have geometrical isomerism.

Isomerism in Butene

• There are 3 structural isomers of $C<em>4H</em>8$ that are alkenes.
• Only one exhibits geometrical isomerism.
  • BUT-1-ENE
  • cis BUT-2-ENE
  • trans BUT-2-ENE
  • 2-METHYLPROPENE
• Alkanes with formula $C<em>4H</em>8$ can be obtained if the carbon atoms are in a ring.

Chemical Properties of Alkenes

• The main reaction of alkenes is addition.
• Because of the extra electron density in a C=C double bond, alkenes are attacked by species which ‘like’ electrons. These species are called electrophiles; they possess a positive or partial positive charge somewhere in their structure. Examples:
  • hydrogen halides
  • Concentrated $H<em>2SO</em>4$

Electrophilic Addition Mechanism

• The electrophile, having some positive character, is attracted to the alkene.
• The electrons in the pi bond come out to form a bond to the positive end.
• Because hydrogen can only have two electrons in its orbital, its other bond breaks heterolytically.
• The H attaches to one of the carbon atoms.
• A carbocation is formed.
• The species that left now has a lone pair. It acts as a nucleophile and attacks the carbocation using its lone pair to form a covalent bond.
• Overall, there is addition.

Electrophilic Addition of Hydrogen Bromide

• Reagent: Hydrogen bromide (HBr), which is electrophilic as the H is slightly positive.
• Condition: Room temperature.
• Equation:
  $C<em>2H</em>4(g) + HBr(g) \longrightarrow C<em>2H</em>5Br(l)$ (bromoethane)
• Mechanism:
  • Step 1: As the HBr nears the alkene, one of the carbon-carbon bonds breaks. The pair of electrons attaches to the slightly positive H end of H-Br. The HBr bond breaks to form a bromide ion. A carbocation is formed.
    $C<em>2H</em>4(g) + HBr(g) \longrightarrow [C<em>2H</em>5]^+(g) + Br^-(g)$
  • Step 2: The bromide ion behaves as a nucleophile and attacks the carbocation. Overall, there has been addition of HBr across the double bond.
    $[C<em>2H</em>5]^+(g) + Br^-(g) \longrightarrow C<em>2H</em>5Br(g)$

Electrophilic Addition of Hydrogen Chloride

• Mechanism:
• Equation:$C<em>2H</em>4(g) + HCl(g) \longrightarrow C<em>2H</em>5Cl(g)$
  • Step 1:
    $C<em>2H</em>4(g) + HCl(g) \longrightarrow [C<em>2H</em>5]^+(g) + Cl^-(g)$
  • Step 2:
    $[C<em>2H</em>5]^+(g) + Cl^-(g) \longrightarrow C<em>2H</em>5Cl(g)$

Electrophilic Addition of Bromine

• Reagent: Bromine (Neat liquid or dissolved in tetrachloromethane, CCl4).
• Condition: Room temperature. No catalyst or UV light required!
• Equation:
  $C<em>2H</em>4(g) + Br<em>2(l) \longrightarrow CH</em>2BrCH_2Br(l)$ (1,2 - dibromoethane)
• It is surprising that bromine should act as an electrophile as it is non-polar.
• Explanation: As a bromine molecule approaches an alkene, electrons in the pi bond of the alkene repel the electron pair in the bromine-bromine bond, thus inducing a dipole. The electron pair is now nearer one end, so the bromine molecule is polar and becomes electrophilic.

Electrophilic Addition of Sulphuric Acid

• Reagent: Concentrated sulphuric acid (85%).
• Conditions: 0°C
• Equation:
  $C<em>2H</em>4(g) + H<em>2SO</em>4(conc) \longrightarrow C<em>2H</em>5OSO_2OH(aq)$ (ethyl hydrogensulphate)
• Hydrolysis: The product can be converted to ethanol by boiling with water.
  $C<em>2H</em>5OSO<em>2OH(aq) + H</em>2O(l) \longrightarrow H<em>2SO</em>4(aq) + C<em>2H</em>5OH(l)$
• Phosphoric acid ($H<em>3PO</em>4$) and steam are used.
• Ethanol can also be made by fermentation.

Addition of Unsymmetrical Alkenes

• Addition of HBr to propene gives two isomeric brominated compounds.
• HBr is unsymmetrical and can add in two ways.
• Products are not formed to the same extent.
• The problem does not arise in ethene because it is symmetrical.

Markovnikoff’s Rule

• A Russian scientist, Markovnikoff, investigated the products of the addition of hydrogen halides to alkenes.
• He found that, when two products were formed, one was formed in a larger quantity.
• The modern version uses carbocation stability as a criterion for predicting the products.
• In the electrophilic addition to alkenes, the major product is formed via the more stable carbocation (carbonium ion).
• Carbocation Stability: Build up of charge in one place leads to instability. If it can be spread around or neutralised in some way, stability is increased. Alkyl groups are electron releasing and can “push” electrons towards the carbocations thus reducing the charge density.
• Carbocation Stability Order: methyl < primary (1°) < secondary (2°) < tertiary (3°)
• In the addition to propene, path A involves a 2° carbocation, path B a 1° carbocation. As the 2° ion is more stable, the major product (i.e., 2-bromopropane) is formed this way.

Markovnikoff’s Rule Example

• 
  $C<em>3H</em>6 + HCl \longrightarrow C<em>3H</em>7Cl$
• Path A:
  $CH<em>2CHCH</em>3 + Cl<em>2 \longrightarrow [CH</em>3CHCH<em>3]^+ + Cl^-$$[CH</em>3CHCH<em>3]^+ + Cl^- \longrightarrow CH</em>3CH<em>2ClCH</em>3$
• Path B:
  $CH<em>2CHCH</em>3 + Cl<em>2 \longrightarrow [CH</em>2CH<em>2CH</em>3]^+ + Cl^-$
  $[CH<em>2CH</em>2CH<em>3]^+ + Cl^- \longrightarrow CH</em>2ClCH<em>2CH</em>3$
• Major Product: Path A, Secondary Carbocation.
• Minor Product: Path B, Primary Carbocation.