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Alkanes and their Reactions
13-Apr-15
Objective:
Be able to write equations for the combustion of alkanes
Be able to describe the free radical substitution of Alkanes
Alkanes
Hydrocarbon homologous series
Cn
H2n+2
Saturated compounds (all C-C bonds are single)
Relatively unreactive
Reactions of Alkanes
At room temperature the alkanes are very
unreactive.
Why?
Hint:
Electronegativity and
enthalpy
Alkanes are unreactive because:
The C-C and C-H bonds involve an equal sharing of
electrons. The electronegativities are similar and therefore
the bonds are not polar. This makes them less susceptible
to attack from many organic molecules.
1. Electronegativity
2. Bond Enthalpy
Alkanes are relatively simple molecules that contain only
single C-C and C-H bonds. These bonds have a relatively
high bond enthalpy. So they only react in the presence of a
strong energy source.
Heating Alkanes
In the absence of air
In a plentiful supply
of air
Cracking Combustion
A form of thermal
decomposition where large
molecules are broken down
into smaller more useful.
The burning of a
fuel to create
thermal energy.
Why is the combustion of alkanes so
important to our life?
1. Generating electricity
2. Fuel fires
3. Central heating
4. Cooking
5. transport
It is an exothermic reactions.
Energy is transferred from chemical energy
to thermal energy that we can use.
Alkanes are often used as fuels because they
release large amounts of energy on combustion.
• The products of the combustion of alkanes
contain relatively strong bonds (C=O and O-H)
that release a lot of energy on their formation.
Combustion
Which alkanes do we use for combustion?
Methane Propane Butane
• Methane 🡪 butane are colourless gases
• Propane and butane can be condensed under
pressure
• Alkanes containing 5-19 carbons are colourless
liquids
• Alkanes with 20+ carbons are colourless waxy
solids
1. Why do we use these alkanes for combustion?
2. Why is propane and butane used more than methane?
3. Suggest why we do not use ethane for combustion
Candles are made from alkanes that generally
contain 20-25 carbons
if this is true
why does the
whole candle
not combust?
complete Incomplete
Burns in a plentiful supply
of oxygen
Burns in a limited supply
of oxygen
Creates carbon dioxide (a
greenhouse gas)
Creates carbon monoxide
Very energy efficient Low energy efficiency
Creates soot (carbon)
CH4
+ 2O2
🡪 CO2
+ 2H2
O CH4
+ 1.5O2
🡪 CO + 2H2
O
CH4
+ O2
🡪 C + 2H2
O
Substitution Reaction
CH4
+ Br2
🡪 CH3
Br + HBr
1. UV light is needed to start a substitution
reaction
2. The bromine atom actually takes the place of
a Hydrogen atom.
3. The molecule formed is called
bromomethane
Why is this called an substitution reaction?
Recap from GCSE:
Bond fission: breaking bonds
When you break bonds there are 2 ways the
electrons can be shared out
1. Homolytic fission
2. Heterolytic fission
Same/destruction
different/destruction
Homolytic fission
• When the breaking of a covalent bond leads to
both atoms having an equal share of
electrons.
• Formation of atoms with unbonded electons=
free radicals
• Draw dot and cross diagrams to represent
what is happening
Cl-Cl 🡪 Cl◦ +Cl◦
Free radicals produced from homolytic
fission
1. They are not charged
2. But extremely reactive
Heterolytic fission
• Unequal sharing of electrons
• Results in two charged particles
• Covalent bond will already have a degree of
polarity
H-Cl 🡪 H+
+ Cl-
Summary Halogenation of Alkanes
• Requires ultraviolet (UV) radiation
• Requires 300o
C
• Bonds break by homolytic fission, which forms
free radicals – an atom with an unpaired
electron
• The radicals substitute
• There is a three stage mechanism for the
reaction
Step 1: Initiation
• A chlorine molecule, Cl-Cl is broken by
homolytic fission.
– UV radiation provides the energy for this to
happen
Cl-Cl 🡪 Cl
.
+ Cl
.
• The rest of the reaction requires no more
energy to be added, as the chlorine radical
attack the methane (or other alkane)
Step 2: Propagation
• Methane reacts with the chlorine radical
– A single C-H bond is broken by homolytic fission,
forming a
.
CH3
radical
– HCl is also formed
CH4
+ Cl
.
🡪 .
CH3
+ HCl
• The methyl radical then reacts with a chlorine
molecule, forming chloromethane and
another chlorine radical.
.
CH3
+ Cl2
🡪 CH3
Cl + Cl
.
• In halogenation of alkanes, the halogen undergoes
homolytic fission to produce two free radical chlorines
• These free radicals attack the C-H bonds in the alkane
• This produces a methyl free radical group which can
continue to react with chlorine
• Which produces another free radical. This recycling of free
radicals produces a self-propogating chain reaction
Step 2: Propagation
• Propagation reactions are rapid.
• This is a chain reaction, as each time a
chloromethane molecule is formed, another
chlorine radical is formed to react with
another methane molecule
• Therefore the reaction will continue until no
chlorine is left
Step 3: Termination
• Two radicals combine to form a molecule
• There are three possibilities for the reaction
between chlorine and methane:
Cl
.
+ Cl
.
🡪 Cl2
.
CH3
+ Cl
.
🡪 CH3
Cl
.
CH3
+ .
CH3
🡪 C2
H6
• This stage removes radicals, thus terminating the
reaction.
• The propagation step would have happened
Further Reactions of Chloromethane
• Chloromethane can undergo further reactions
with chlorine radicals, forming:
–
–
–
Initiation-Propagation-Termination
Task
• Show the mechanism for the bromination of
methane.
– Include all three steps!
Which stage of substitution reactions
do the following describe?
Two free radicals collide and produce
a molecule
UV light provides the energy needed for
homolytic fission of a Chlorine molecule
A free radical reacts with molecule to
crate another free radical
What order do the stages of the
mechanism?
Which stage of substitution reactions
do the following describe?
Two free radicals collide and produce
a molecule
UV light provides the energy needed for
homolytic fission of a Chlorine molecule
A free radical reacts with molecule to
crate another free radical
What order do the stages of the
mechanism?
Draw curly arrow and dot and cross
diagrams for the following reactions
H-Br 🡪 hydrogen and bromine ions
Br-Br 🡪 bromine free radicals
Quiz time!
•
CH3
+ •
CH3
🡪 C2
H6
What step is this equation in the
mechanism of free radical substitution?
termination
•
CH3
+ Cl2
🡪 CH3
Cl + Cl•
What step is this equation in the
mechanism of free radical substitution?
propagation
Cl2
🡪 2Cl•
What step is this equation in the
mechanism of free radical substitution?
initiation
Cl•
+ C3
H8
🡪 •
C3
H8
+ HCl
What step is this equation in the
mechanism of free radical substitution?
propagation
2Cl•
🡪 Cl2
What step is this equation in the
mechanism of free radical substitution?
termination
C6
H12
+ Br•
🡪 HBr + •
C6
H11
What step is this equation in the
mechanism of free radical substitution?
propagation
I2
🡪 2I•
What step is this equation in the
mechanism of free radical substitution?
initiation
•
C4
H9
+ Br2
🡪 C4
H9
Br + Br•
What step is this equation in the
mechanism of free radical substitution?
propagation
2Br•
🡪 Br2
What step is this equation in the
mechanism of free radical substitution?
termination
•
C4
H9
+ •
C2
H5
🡪 C6
H14
What step is this equation in the
mechanism of free radical substitution?
propagation

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Reactions of Alkanes.pptx.pdf

  • 1. Alkanes and their Reactions 13-Apr-15 Objective: Be able to write equations for the combustion of alkanes Be able to describe the free radical substitution of Alkanes
  • 2. Alkanes Hydrocarbon homologous series Cn H2n+2 Saturated compounds (all C-C bonds are single) Relatively unreactive
  • 3. Reactions of Alkanes At room temperature the alkanes are very unreactive. Why? Hint: Electronegativity and enthalpy
  • 4. Alkanes are unreactive because: The C-C and C-H bonds involve an equal sharing of electrons. The electronegativities are similar and therefore the bonds are not polar. This makes them less susceptible to attack from many organic molecules. 1. Electronegativity 2. Bond Enthalpy Alkanes are relatively simple molecules that contain only single C-C and C-H bonds. These bonds have a relatively high bond enthalpy. So they only react in the presence of a strong energy source.
  • 5. Heating Alkanes In the absence of air In a plentiful supply of air Cracking Combustion A form of thermal decomposition where large molecules are broken down into smaller more useful. The burning of a fuel to create thermal energy.
  • 6. Why is the combustion of alkanes so important to our life? 1. Generating electricity 2. Fuel fires 3. Central heating 4. Cooking 5. transport It is an exothermic reactions. Energy is transferred from chemical energy to thermal energy that we can use.
  • 7. Alkanes are often used as fuels because they release large amounts of energy on combustion. • The products of the combustion of alkanes contain relatively strong bonds (C=O and O-H) that release a lot of energy on their formation. Combustion
  • 8. Which alkanes do we use for combustion? Methane Propane Butane • Methane 🡪 butane are colourless gases • Propane and butane can be condensed under pressure • Alkanes containing 5-19 carbons are colourless liquids • Alkanes with 20+ carbons are colourless waxy solids 1. Why do we use these alkanes for combustion? 2. Why is propane and butane used more than methane? 3. Suggest why we do not use ethane for combustion
  • 9. Candles are made from alkanes that generally contain 20-25 carbons if this is true why does the whole candle not combust?
  • 10. complete Incomplete Burns in a plentiful supply of oxygen Burns in a limited supply of oxygen Creates carbon dioxide (a greenhouse gas) Creates carbon monoxide Very energy efficient Low energy efficiency Creates soot (carbon) CH4 + 2O2 🡪 CO2 + 2H2 O CH4 + 1.5O2 🡪 CO + 2H2 O CH4 + O2 🡪 C + 2H2 O
  • 11. Substitution Reaction CH4 + Br2 🡪 CH3 Br + HBr 1. UV light is needed to start a substitution reaction 2. The bromine atom actually takes the place of a Hydrogen atom. 3. The molecule formed is called bromomethane Why is this called an substitution reaction? Recap from GCSE:
  • 12. Bond fission: breaking bonds When you break bonds there are 2 ways the electrons can be shared out 1. Homolytic fission 2. Heterolytic fission Same/destruction different/destruction
  • 13. Homolytic fission • When the breaking of a covalent bond leads to both atoms having an equal share of electrons. • Formation of atoms with unbonded electons= free radicals • Draw dot and cross diagrams to represent what is happening Cl-Cl 🡪 Cl◦ +Cl◦
  • 14. Free radicals produced from homolytic fission 1. They are not charged 2. But extremely reactive
  • 15. Heterolytic fission • Unequal sharing of electrons • Results in two charged particles • Covalent bond will already have a degree of polarity H-Cl 🡪 H+ + Cl-
  • 16. Summary Halogenation of Alkanes • Requires ultraviolet (UV) radiation • Requires 300o C • Bonds break by homolytic fission, which forms free radicals – an atom with an unpaired electron • The radicals substitute • There is a three stage mechanism for the reaction
  • 17. Step 1: Initiation • A chlorine molecule, Cl-Cl is broken by homolytic fission. – UV radiation provides the energy for this to happen Cl-Cl 🡪 Cl . + Cl . • The rest of the reaction requires no more energy to be added, as the chlorine radical attack the methane (or other alkane)
  • 18. Step 2: Propagation • Methane reacts with the chlorine radical – A single C-H bond is broken by homolytic fission, forming a . CH3 radical – HCl is also formed CH4 + Cl . 🡪 . CH3 + HCl • The methyl radical then reacts with a chlorine molecule, forming chloromethane and another chlorine radical. . CH3 + Cl2 🡪 CH3 Cl + Cl .
  • 19. • In halogenation of alkanes, the halogen undergoes homolytic fission to produce two free radical chlorines • These free radicals attack the C-H bonds in the alkane • This produces a methyl free radical group which can continue to react with chlorine • Which produces another free radical. This recycling of free radicals produces a self-propogating chain reaction Step 2: Propagation
  • 20. • Propagation reactions are rapid. • This is a chain reaction, as each time a chloromethane molecule is formed, another chlorine radical is formed to react with another methane molecule • Therefore the reaction will continue until no chlorine is left
  • 21. Step 3: Termination • Two radicals combine to form a molecule • There are three possibilities for the reaction between chlorine and methane: Cl . + Cl . 🡪 Cl2 . CH3 + Cl . 🡪 CH3 Cl . CH3 + . CH3 🡪 C2 H6 • This stage removes radicals, thus terminating the reaction. • The propagation step would have happened
  • 22. Further Reactions of Chloromethane • Chloromethane can undergo further reactions with chlorine radicals, forming: – – –
  • 24. Task • Show the mechanism for the bromination of methane. – Include all three steps!
  • 25. Which stage of substitution reactions do the following describe? Two free radicals collide and produce a molecule UV light provides the energy needed for homolytic fission of a Chlorine molecule A free radical reacts with molecule to crate another free radical What order do the stages of the mechanism?
  • 26.
  • 27. Which stage of substitution reactions do the following describe? Two free radicals collide and produce a molecule UV light provides the energy needed for homolytic fission of a Chlorine molecule A free radical reacts with molecule to crate another free radical What order do the stages of the mechanism?
  • 28. Draw curly arrow and dot and cross diagrams for the following reactions H-Br 🡪 hydrogen and bromine ions Br-Br 🡪 bromine free radicals
  • 30. • CH3 + • CH3 🡪 C2 H6 What step is this equation in the mechanism of free radical substitution? termination
  • 31. • CH3 + Cl2 🡪 CH3 Cl + Cl• What step is this equation in the mechanism of free radical substitution? propagation
  • 32. Cl2 🡪 2Cl• What step is this equation in the mechanism of free radical substitution? initiation
  • 33. Cl• + C3 H8 🡪 • C3 H8 + HCl What step is this equation in the mechanism of free radical substitution? propagation
  • 34. 2Cl• 🡪 Cl2 What step is this equation in the mechanism of free radical substitution? termination
  • 35. C6 H12 + Br• 🡪 HBr + • C6 H11 What step is this equation in the mechanism of free radical substitution? propagation
  • 36. I2 🡪 2I• What step is this equation in the mechanism of free radical substitution? initiation
  • 37. • C4 H9 + Br2 🡪 C4 H9 Br + Br• What step is this equation in the mechanism of free radical substitution? propagation
  • 38. 2Br• 🡪 Br2 What step is this equation in the mechanism of free radical substitution? termination
  • 39. • C4 H9 + • C2 H5 🡪 C6 H14 What step is this equation in the mechanism of free radical substitution? propagation