Transcript 10.2 Alkanes

Alkanes
IB Chemistry Topic 10.2
10.2 Alkanes Asmt. Stmts
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10.2.1 Explain the low reactivity of alkanes in terms
of bond enthalpies and bond polarity.
10.2.2 Describe, using equations, the complete and
incomplete combustion of alkanes.
10.2.3 Describe, using equations, the reactions of
methane and ethane with chlorine and bromine.
10.2.4 Explain the reactions of methane and ethane
with chlorine and bromine in terms of a free-radical
mechanism.
10.2.1
Explain the low reactivity of alkanes
in terms of bond enthalpies and
bond polarity.
10.2.1
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Have low reactivity
 Bond enthalpies are relatively strong
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Low polarity
Only readily undergo combustion reactions
with oxygen (very flammable) and
substitution reactions with halogens in UV
light
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348 kJ mol-1 to break a C-C bond
412 kJ mol-1 to break a C-H bond
10.2.2
Describe, using equations, the
complete and incomplete
combustion of alkanes.
Complete vs. Incomplete Combustion
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Hydrocarbons (only contain C & H)
Complete combustion
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Alkanes burn in an excess supply of oxygen to
form carbon dioxide and water:
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Example:
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C8H18 (g) + 12 ½ O2 (g) → 8 CO2 (g) + 9 H2O (l)
exothermic (-∆H)
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Incomplete combustion
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If oxygen supply is limited, the gas carbon
monoxide and carbon is formed
C8H18 (l) + O2 (g) → C (s) + CO (g) + CO2 (g) + H2O (l)
(notice left over carbon (black soot) and dangerous CO)
10.2.3
10.2.3 Explain the reactions of
methane and ethane with chlorine
and bromine in terms of a freeradical mechanism.
10.2.3 Free Radical
Substitutions
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Many organic molecules undergo substitution
reactions.
In a substitution reaction one atom or group
of atoms is removed from a molecule and
replaced with a different atom or group.
Example:
Cl2 + CH4  CH3Cl + HCl
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10.2.3 Reactions of Alkanes:
with Halogens
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Alkanes do not react with halogens in the
dark at room temperature, but will react in the
presence of sunlight (UV).
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A substitution reaction will occur where
some or all of the hydrogens will be
replaced with a halogen
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C2H6 (g) + Br2 (g) → C2H5Br (l) + HBr (g)
Cl2 + CH4  CH3Cl + HCl
Think electronegativity!!!
10.2.3 Reactions of Alkanes:
with Halogens
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Energy absorbed from light allows homolytic
fission: each resulting atom receives one
unpaired electron, known as free radicals
Cl-Cl → Cl• + Cl•
H
H
H
+
Cl Cl
H
h
Cl
H
H
H
methane
chloromethane
h
H
Cl
C
H
dichlorome
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this happens by a process know as free
radical substitution that happens in 3
steps
1.
2.
3.
Initiation
Propagation
Termination
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initiation
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Cl2  Cl* + Cl*
propagation
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initiated by UV light breaking a chlorine molecule into
two free radicals by a process called homolytical
fission (* = unpaired electron)
each resulting atom receives one
unpaired electron, known as free
radicals that have lots of energy
keeps the chain going (radical in reactants and products)
CH4 + Cl*  CH3* + HCl
CH3* + Cl2  CH3Cl + Cl*
termination
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this removes free radicals (*) from the system without
replacing them by new ones
Cl* + Cl*  Cl2
CH3* + Cl *  CH3Cl
CH3* + CH3*  CH3CH3
each resulting atom
receives one unpaired
electron, known as free
radicals
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Homolytic fission:
http://www.youtube.com/watch?v=XAWWf86
TJ0c
http://www.youtube.com/watch?v=MvWQOY
Lm1f4
Free radical damage:
http://www.youtube.com/watch?v=JV9bORH
bSZU
10.2.4 Free Radical
Mechanism-The Initiation Step
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The ultraviolet light is a source of energy that
causes the chlorine molecule to break apart into
2 chlorine atoms, each of which has an unpaired
electron
The energies in UV are exactly right to break the
bonds in chlorine molecules to produce chlorine
atoms.
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10.2.4 Homolytic Fission
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Free radicals are formed if a bond splits
evenly - each atom getting one of the two
electrons. The name given to this is
homolytic fission.
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10.2.4 Free Radical
Propagation
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The productive collision happens if a chlorine radical
hits a methane molecule.
The chlorine radical removes a hydrogen atom from the
methane. That hydrogen atom only needs to bring one
electron with it to form a new bond to the chlorine, and
so one electron is left behind on the carbon atom. A new
free radical is formed - this time a methyl radical, CH3 .
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10.2.4 Free Radical
Propagation II
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If a methyl radical collides with a chlorine
molecule the following occurs:
CH3. + Cl2  CH3Cl + Cl.
The methyl radical takes one of the chlorine
atoms to form chloromethane
In the process generates another chlorine free
radical.
This new chlorine radical can now go through
the whole sequence again, It will produce yet
another chlorine radical - and so on and so on.
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10.2.4 Termination Steps
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The free radical propagation does not go on
for ever.
If two free radicals collide the reaction is
terminated.
2Cl.  Cl2
CH3. + Cl .  CH3Cl
CH3 . + CH3.  CH3CH3
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10.2.4 Exercise
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Write the steps in the free radical mechanism for the
reaction of chlorine with methyl benzene. The overall
reaction is shown below. The methyl group is the part
of methyl benzene that undergoes attack.
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10.2.4 Solution
Initiation
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Cl2  2Cl.
Propagation
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Termination
2Cl.  Cl2
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10.2.4 Reactions of Alkanes:
with Halogens
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2.
3.
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Substitution of an alkane with a halogen has
3 steps:
Initiation
Propagation
Termination
See pg. 193 in IBCC for detailed diagrams
of these steps
Rate of reaction: Cl2 > Br2 > I2 … Why???