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1. At room temperature and pressure:
- C1 - C4 are gases
- C5 - C15 are liquids
- C16 - are solids.
2. As the number of carbon atoms in a straight chain alkane increases, the boiling points rise in a uniform and predictable manner.
This is because each molecule is identical in terms of types of atom and bonds present, the only difference being the regular increase in mass.
3. Compounds with branched chains always have boiling points below that of the related straight chain compound.
hexane: boiling point = 69oC
2,2 dimethylbutane: boiling point = 49oC.
This is because the straight chain molecules have greater areas of contact between them and hence have stronger forces of attraction.
4. The melting point also increases with the number of carbon atoms in the chain. However, when plotted on a graph, two curves are obtained due to the fact that 'even' carbon molecules pack tighter together, and therefore van der Waal's forces are greater than 'odd' carbon molecules.
5. The alkanes are non-polar and are therefore immiscible in water and other polar solvents. Methane is the most soluble.
The C-C and C-H bonds in alkanes are very strong since they are non-polar and almost totally covalent in character. Therefore, alkanes are relatively inert with regard to most chemical reagents.
They have three main properties:
- Substitution reactions
- Catalytic (or thermal cracking)
The alkanes burn in a plentiful supply of oxygen to produce CO2 and H2O.
Example: 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g)
A gaseous mixture of an alkane and oxygen are extremely explosive. These reactions are used commercially when fuels such as natural gas, petrol and oil are burnt in air.
Substitution reactions involving chlorine and methane
A mixture of chlorine and methane:
- a) Does not react if kept in the dark at room temp.
- b) Does not react if kept in the dark at 300oC
- c) Reacts at room temperature if exposed to sunlight or U.V. light.
- d) Explodes if exposed to bright sunlight or sparked.
So, energy is required to initiate the reaction.
Four products are formed from this reaction:
These products are explained in terms of a stepwise reaction:
- CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
- CH3Cl(g) + Cl2(g) → CH2Cl2(g) + HCl(g)
- CH2Cl2(g) + Cl2(g) → CHCl3(g) + HCl(g)
- CHCl3(g) + Cl2(g) → CCl4(g) + HCl(g)
Each of these reactions is an example of a substitution reaction (a Cl atom is substituted for a H atom).
Mechanism: Free radical substitution
It has been shown that the reaction will proceed in the presence of a free radical.
There are 3 stages:
Light or heat will cause a few chlorine molecules to split homolytically into chlorine radicals having unpaired electrons.
Cl:Cl(g) → 2Clo(g) (chlorine radical)
The radicals are very reactive and will react with the first particle they meet - most probably a methane molecule because the formation of a H-Cl bond is more exothermic than the formation of a C-Cl bond.
- i) Clo(g) + CH4(g) → oCH3(g) + HCl(g) (formation of a methyl radical is more likely)
- Clo(g) + CH4(g) → CH3Cl(g) + Ho(g)
The methyl radical produced initiates further propagation steps:
- ii) oCH3(g) + Cl2(g) → CH3Cl(g) + Clo(g)
- iii) CH3Cl(g) + Clo(g) → oCH2Cl(g) + HCl(g)
- iv) oCH2Cl(g) + Cl2(g) → CH2Cl2(g) + Clo(g)
- v) CH2Cl2(g) + Clo(g) → oCHCl2(g) + HCl(g)
Some reactions occur in which atoms or radicals combine together to produce a molecule without a new radical being formed:
- Clo(g) + Clo(g) → Cl2(g)
- oCH3(g) + Clo(g) → CH3Cl(g)
- oCH2Cl(g) + Clo(g) → CH2Cl2(g)
Radicals are removed from the system thus preventing the chain reaction going to completion.
Bromination of methane occurs by a similar mechanism but requires more energy. Iodination does not take place.
Thermo-cracking is used to break down high molecular mass alkanes into low molecular mass alkanes as well as alkenes using heat and a catalyst. As bond breaking is a random process, a variety of products can be formed.