Questions on Alkanes
1. State the balanced equations for the complete combustion (under standard conditions) of
2-methylbutane, C5H12 (boiling point: 28oC), 3-methylpentane, C6H14, and octane, C8H18.
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2. If the ratio of air to gasoline (petrol) entering a car engine is reduced then the amount of air pollution
caused by oxides of nitrogen emitted in the exhaust gas is lowered. However, the concentration of two
other air pollutants is increased. Identify these two pollutants and explain how they arise.
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3. The first step in the reaction between methane and chlorine in the presence of ultraviolet light is the
homolytic fission of the chlorine to chlorine single bond, Cl-Cl(g), in gaseous chlorine.
(a) Explain the meaning of the term homolytic fission.
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(b) Explain why homolytic fission occurs with the Cl-Cl bond in chlorine and not the C-H bond in
methane.
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(c) State the name of the product formed when the Cl-Cl bond is broken homolytically and state its
electron configuration.
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(d) Explain why only a few homolytic fission reactions involving chlorine need to be successful in order
to bring about the complete reaction between chlorine and methane to form chloromethane and
hydrogen chloride.
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4. The reaction between bromine and ethane occurs in the presence of ultraviolet light.
(a) Explain why ultraviolet light is necessary for the reaction to proceed.
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(b) Describe, using equations, the stepwise mechanism of the reaction between one mol of bromine and
one mol of ethane to form one mol of bromoethane and one mol of hydrogen bromide.
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(c) Describe how you could make pure 1,2-dibromoethane from bromine and ethane.
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Answers
1. C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l)
2C6H14(l) + 19O2(g) → 12CO2(g) + 14H2O(l)
2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l)
2. The two pollutants are carbon monoxide and particles of solid carbon (particulates).
They are formed as there is insufficient oxygen (present in the air inside the combustion chamber in the
car) to bring about complete combustion of the gasoline.
3. (a) During homolytic fission the bond breaks symmetrically so that one electron forming the bond
between two atoms becomes attached to one of the atoms and the other electron becomes attached to
the other atom resulting in the formation of two radicals.
(b) The ultraviolet light provides the energy to break the Cl-Cl bond homolytically. This energy, 242 kJ
mol-1, is much less than the 414 kJ mol-1 of energy required to break a C-H bond.
(c) The product is a chlorine (free) radical. The electron configuration is 1s22s22p63s23p5.
(d) Once a chlorine free radical is formed it reacts with a methane molecule to produce hydrogen
chloride and a methyl radical. This methyl radical can react with another chlorine molecule to form
chloromethane and generate a new chlorine radical which can then repeat the process. This is a
propagation step. The formation of new radicals will only stop when a termination reaction occurs. (A
termination step may be between two radicals or between a radical and an impurity or the walls of the
reaction vessel).
4. (a) The ultraviolet light provides the energy to break the Br-Br bond in bromine homolytically.
(b) Initiation: Br2(g) + ultraviolet light → 2Br.(g)
Propagation: Br.(g) + C2H6(g) → C2H5.(g) + HBr(g)
Propagation: C2H5.(g) + Br2(g) → C2H5Br(g) + Br.(g)
Termination: C2H5.(g) + Br.(g) → C2H5Br(g) (there are other possible termination reactions)
(c) React ethane with excess bromine in the presence of ultraviolet light. This would give a mixture of
brominated products. For example, 1,1–dibromoethane, 1,2– dibromoethane, 1,1,1- tribromoethane
etc. The desired product, 1,2-dibromoethane, would need to be separated from the mixture using
fractional distillation (as they all have different boiling points) or by using some form of
chromatography, e.g. HPLC, GC or GLC, (as they will all have different retention times).