937275
Chemistry Unit 2- Alkenes and
Haloalkanes
- Alkenes
- These are hydrocarbons that have one or more
carbon-carbon double bond
- As you can see, the atoms around each
carbon form the trigonal planar shape and
there is a bond agngle of 120
- Their double bond means there is restricted rotation around
each carbon bond. In alkanes, the atoms have free rotatiomn
around each carbon atom because they have a single bond
- E/Z isomerism is where atoms have the same
molecular formula but a different 3-D spacial
arrangement around each atom
- It is also called geometrical isomerism
- It can only occur if there are 2 different groups attatched to each carbon
- When trying to work out if a molecule shows E/Z isomerism, you
work out where the two heaviest molecules are placed.
- If they're on the Zame Zide, then they show Z isomerism
- If they're on opposite sides, i.e. entgegen, then they show E isomerism
- Example, But-2-ene. The cis but-2-ene shows Z isomerism as the two heaviest molecules are on the same side
- The trans but-2-ene shows E isomerism as the two heaviest molecules are on different sides
- The trans but-2-ene shows E isomerism as the two heaviest molecules are on different sides
- If they're on the Zame Zide, then they show Z isomerism
- When trying to work out if a molecule shows E/Z isomerism, you
work out where the two heaviest molecules are placed.
- It can only occur if there are 2 different groups attatched to each carbon
- They don't show position, chain or functional group isomerism
- This is caused by the restricted rotation arround each carbon
- It is also called geometrical isomerism
- Reactivity/reactions
- Alkenes are more reactive than alkanes. Why?
- Alkanes are unsaturated, meaning not all bonds are
filled, they can be broken and more can be added to it
- They have a double bond, meaning there are 4 electrons within that region,
which will pull the 2 carbon atoms closer together and make the bond shorter.
- This then means that there is more negativity within a smaller space.
- The double bond is said to be a region of high electron density,
and this is what makes alkenes more reactive than alkanes.
- The double bond is said to be a region of high electron density,
and this is what makes alkenes more reactive than alkanes.
- This then means that there is more negativity within a smaller space.
- They have a double bond, meaning there are 4 electrons within that region,
which will pull the 2 carbon atoms closer together and make the bond shorter.
- Alkanes are unsaturated, meaning not all bonds are
filled, they can be broken and more can be added to it
- Electrophilic addition:
- An addition reaction is making 2 smaller molecules into 1 bigger molecule
- An electrophile is a species that is able to ACCEPT a pair of
electrons from an electron rich species
- An alkene is able to do electrophilic addition with 3 things
- H-X
- 1. The electrons in the double bond is given to
the hydrogen on the H-BR molecule
- The hydrogen now has 4 electrons, thats too many, so the Br will take
its original electron and also hydrogen's electron to become Br- ion
- The hydrogen is now bonded to one of the carbons and the
other carbon, has had one of its electrons take away from it.
It now has a positive charge and is now called a carbocation
- So the Br- will now become attracted to the carbocation
- A haloalkane will have formed
- A haloalkane will have formed
- So the Br- will now become attracted to the carbocation
- The hydrogen is now bonded to one of the carbons and the
other carbon, has had one of its electrons take away from it.
It now has a positive charge and is now called a carbocation
- The hydrogen now has 4 electrons, thats too many, so the Br will take
its original electron and also hydrogen's electron to become Br- ion
- There are two types of products that can be
formed: A major yield and a minor yeild
- You can determine which product will be your
major yeild and which will be your minor yeild
by looking at the carbocation
- The carbocation that is attatched to the most
carbon atoms is going to be the major yield
- Primary
- Attatched to only one other carbon atom
- Attatched to only one other carbon atom
- Secondary
- Attatched to 2 other carbon atoms
- Attatched to 2 other carbon atoms
- Tertiary
- Attatched to 3 other carbon atoms
- Attatched to 3 other carbon atoms
- You can't have a quaternary carbocation
because the carbon can only bond to 4 things:
3 other carbon, and the positive charge.
- Out of all carbocation, a tertiary carbocation is
the major product because its the most stable
- Primary
- The carbocation that is attatched to the most
carbon atoms is going to be the major yield
- This s only if you have an alkene that is not symmetrical
- If you have a symmetrical alkene, then
you'll get the same product
- If you have a symmetrical alkene, then
you'll get the same product
- You can determine which product will be your
major yeild and which will be your minor yeild
by looking at the carbocation
- 1. The electrons in the double bond is given to
the hydrogen on the H-BR molecule
- X2
- When you react an alkene with a halogen,
then you get a di-substituted haloalkane
- You dont get 2 products because the product formed is
symmetrical. The X2 doesnt have a permanent dipole
however it does have a temporary induced dipole
- This is because their
electronegativties are the
same
- This usually only works with I2 and Br2 because
they're big enough to have lasting dipoles
- This is because their
electronegativties are the
same
- You dont get 2 products because the product formed is
symmetrical. The X2 doesnt have a permanent dipole
however it does have a temporary induced dipole
- When you react an alkene with a halogen,
then you get a di-substituted haloalkane
- H2SO4
- When you react the alkene with the sulphuric acid,
then you'll get an alkyl hydrogen sulphate.
- The mechanism is the same: 1. The double bond
breaks and a pair of electrons are shared with the
hydrogen. Then the Hydrogen will give its orginial
electron to the O and it will become negatively charged.
This will then react, and bond with the carbocation.
- The alkyl refers to the side group
that is attatched to the hydrogen
sulphate.
- Again, you can get a major product and a minor
product from this. This will depend on whether
the carbocation is primary, seconday or tertiary.
- Alkyl hydrogen sulphates can be converted into alcohols,
by a hydrolysis reaction, i.e. the addition of water
- Notice how you get sulphuric acid at the
end aswell; it acts as a catalyst
- This type of reaction is called a hydration reaction
- This type of reaction is called a hydration reaction
- Notice how you get sulphuric acid at the
end aswell; it acts as a catalyst
- The mechanism is the same: 1. The double bond
breaks and a pair of electrons are shared with the
hydrogen. Then the Hydrogen will give its orginial
electron to the O and it will become negatively charged.
This will then react, and bond with the carbocation.
- When you react the alkene with the sulphuric acid,
then you'll get an alkyl hydrogen sulphate.
- H-X
- An addition reaction is making 2 smaller molecules into 1 bigger molecule
- Alkenes are more reactive than alkanes. Why?
- Addition polymerisation: This is where
alkenes join together in the presence of
high pressure and a catalyst
- The product will be a very long hydrocarbon chain.
- You can show any polymerisation reaction by opening
up the double bond and putting a bracket around it.
- These polyalkenes are saturated, and are very unreactive
- Polyethene: Plastic bags. Polypropene:
straw, biros, food containers.
- Polyethene: Plastic bags. Polypropene:
straw, biros, food containers.
- These polyalkenes are saturated, and are very unreactive
- You can show any polymerisation reaction by opening
up the double bond and putting a bracket around it.
- The product will be a very long hydrocarbon chain.
- These are hydrocarbons that have one or more
carbon-carbon double bond
- Haloalkanes
- Properties:
- Boiling points
- A very low, usually gaseous at room temperature. The
reason for this is because there are weak
intermolecular forces between them i.e. van der Waal's
- As you go down the homologous series, the boiling
points get higher because there are more van der Waal's
- So more energy is required to break these intermolecular forces.
- So more energy is required to break these intermolecular forces.
- As you go down the homologous series, the boiling
points get higher because there are more van der Waal's
- A very low, usually gaseous at room temperature. The
reason for this is because there are weak
intermolecular forces between them i.e. van der Waal's
- Solubility:
- Haloalkanes are non polar, and so they aren't
soluble in water and therefore are soluble in
hydrocarbon because it is also non polar
- Haloalkanes are non polar, and so they aren't
soluble in water and therefore are soluble in
hydrocarbon because it is also non polar
- Bond Strength
- Decreases as you go down the group 7
- The reason for this is because they get less
electronegative as you go down the group
- This is because shielding increases as well as distance. There is
weaker nuclear attraction and so the bond between the group 7 atom
and carbon is not as stong and gets easier to break as you go down.
- This is because shielding increases as well as distance. There is
weaker nuclear attraction and so the bond between the group 7 atom
and carbon is not as stong and gets easier to break as you go down.
- The reason for this is because they get less
electronegative as you go down the group
- Decreases as you go down the group 7
- Boiling points
- Reactivity/Reactions
- There is a difference in electronegativity between the
halogen and the carbon. This will create dipoles
- They go through Nucleophilic subsitution
- Substitution is replacing one atom with another
- A nucleophile is a species that is able to donate a
pair of electrons to an electron deficient species
- Hydroxide/OH- ion
- Produces an alcohol
- Also alcoholic Aqeuous
- Produces an alcohol
- Cyanide/CN- ion
- Produces a nitrile
- Produces a nitrile
- A molecule of ammonia
- Produces an amine
- Alcholic solvent
- Alcholic solvent
- Produces an amine
- Hydroxide/OH- ion
- Mechanism:
- The OH- will 'attack' the carbon as it
is a delta+ species.
- The carbon now has too many electrons so the Br atoms
takes its original electrons and one of the carbon's.
- The carbon now has too many electrons so the Br atoms
takes its original electrons and one of the carbon's.
- Equation: C2H5Br + NaOH ------> C2H5OH + NaBr
- Equation: C2H5Br + NaOH ------> C2H5OH + NaBr
- The carbon now has too many electrons so the Br atoms
takes its original electrons and one of the carbon's.
- The carbon now has too many electrons so the Br atoms
takes its original electrons and one of the carbon's.
- The CN- will 'attack' the carbon as it
is a delta+ species
- The carbon now has too many electrons
so the Br atoms takes its original
electrons and one of the carbon's.
- It becomes a Br- ion and reacts
with the Na+ or K+ that's just sitting
there in the solution
- It becomes a Br- ion and reacts
with the Na+ or K+ that's just sitting
there in the solution
- We don't use HCN because
its a poisonous gas
- Notice how the lone pair on the CN- is on the carbon
- Equation: C2H5Br + KCN -----> C2H5CN + KBr
- C2H5CN: Propanitrile
- C2H5CN: Propanitrile
- Equation: C2H5Br + KCN -----> C2H5CN + KBr
- The carbon now has too many electrons
so the Br atoms takes its original
electrons and one of the carbon's.
- The lone pair on the nitrogen of the
ammonia molecule will 'attack' the
carbon as it's a delta+ species
- The carbon now has too many electrons so the
Br atoms takes its original electrons and one of
the carbon's.
- It forms a Br- ions
- There is a positive charge on the carbon
because the nitrogen is bonded to 4 things
- In order to make it stable, we need to
'deprotonate' it. In other words, you take of a
proton or H+ ions because it's the same thing
- To do this, you need another ammonia molecule which will dative
covalently bond with the H+ ions to form ammonium (NH4+)
- This will then bond with the Br- ions
that was formed before
- Equation: C2H5Br + 2NH3 -------> C2H5NH2 + NH4Br
- C3H5NH2-- Ethylamine
- C3H5NH2-- Ethylamine
- Equation: C2H5Br + 2NH3 -------> C2H5NH2 + NH4Br
- This will then bond with the Br- ions
that was formed before
- To do this, you need another ammonia molecule which will dative
covalently bond with the H+ ions to form ammonium (NH4+)
- In order to make it stable, we need to
'deprotonate' it. In other words, you take of a
proton or H+ ions because it's the same thing
- It forms a Br- ions
- Conditions with ammonia:
- Sealed container as NH3 is gaseous
- High pressure: Higher rate of reaction
- Alcoholic solution for the haloalkane
- Alcoholic solution for the haloalkane
- High pressure: Higher rate of reaction
- Sealed container as NH3 is gaseous
- The carbon now has too many electrons so the
Br atoms takes its original electrons and one of
the carbon's.
- Conditions for OH- to react:
- 1. NaOH or the reagent has to be in a polar solvent e.g. water
- 2. The haloalkane needs to be in a non-polar solvent e.g. alcohol
- Therefore, an alcoholic aqueous solution is required
- The reaction needs to take place in either room temperature or be warm, but NOT heated
- The reaction needs to take place in either room temperature or be warm, but NOT heated
- Therefore, an alcoholic aqueous solution is required
- 2. The haloalkane needs to be in a non-polar solvent e.g. alcohol
- 1. NaOH or the reagent has to be in a polar solvent e.g. water
- The OH- will 'attack' the carbon as it
is a delta+ species.
- Substitution is replacing one atom with another
- We know that the reagents are CN-, OH- and
ammonia. However, you can't just go to a store
and get and OH- or CN- ion of the shelf.
- It's got to be ionically bonded to something else, e.g.
NaOH, NaCN, KOH, KCN etc. in order to stabilise it.
- We include the reagent in the chemical equation but
not the mechanism
- The Na+, or K+ ion will act as the spectator ion
- We include the reagent in the chemical equation but
not the mechanism
- It's got to be ionically bonded to something else, e.g.
NaOH, NaCN, KOH, KCN etc. in order to stabilise it.
- There is a difference in electronegativity between the
halogen and the carbon. This will create dipoles
- Production
- 1. Electrophilic addition
- With H-X
- With X2
- With H-X
- Free Radical Substitution
- This uses an Alkane and a halogen
- A free radical is a species with an
unpaired electron. It is very unstable
and will react with almost anything
- Conditions:
- A sealed container
- UV light, this provided the energy to break the halogen
bond which is extremely hard to break on its own
because they have similar electronegativitys
- UV light, this provided the energy to break the halogen
bond which is extremely hard to break on its own
because they have similar electronegativitys
- A sealed container
- Equation: CH4 + Cl2 ------> CH3Cl + HCl
- Mechanism:
- 1. Initiation: This is where the UV light will break the
bond between the Cl2. The two electrons that are
shared now separates and each Cl gets its own
electron and forms a free radical
- This is called homolytic fission: where a
bond breaks and the electrons go seperate
ways to form a free radical
- 2. Propagation I: The chlorine free radical reacts with
the methane and forms a methane free radical and a
HCl molecule
- Propagation II: The methane free radical will react with a
Cl2 molecule to form chloromethane and a Cl free radical
- This should in theory start the cycle again because
you're back to where you started.
- No more UV light is required
- 3. Termination: This is where 2 free radicals react with
each other and the reaction then stops as you no longer
have a free radical
- 3. Termination: This is where 2 free radicals react with
each other and the reaction then stops as you no longer
have a free radical
- No more UV light is required
- You could get other possible propagation steps that
lead to different substituted haloalkanes e.g.
dichloromethane or even tetrachloromethane
- These can be separated using fractional distillation as
they will have slightl different boiling points
- E.g. you could get a chloromethane react with a
cl. free radical. It will form HCl and then will form a
chloromethane free radical. This will then react
with a chlorine to form dichloromethane
- These can be separated using fractional distillation as
they will have slightl different boiling points
- This should in theory start the cycle again because
you're back to where you started.
- Propagation II: The methane free radical will react with a
Cl2 molecule to form chloromethane and a Cl free radical
- This is called homolytic fission: where a
bond breaks and the electrons go seperate
ways to form a free radical
- 1. Initiation: This is where the UV light will break the
bond between the Cl2. The two electrons that are
shared now separates and each Cl gets its own
electron and forms a free radical
- Depletion of the Ozone layer
- This occurs by free radical substitution
- The ozone layer is made up of O3 and it protects
the earth/land from the UV rays from the sun
- CFC's: These are chlorofluorocarbons. They used
to be used as refrigerants or aerosols
- However they are now banned because they were
contributing to the destruction of the ozone layer
- They are harmless on land, but when they get up in the
atmosphere, the bonds start to break from the UV light
found in the atmosphere to form free radicals.
- These free radicals will then react with the ozone layer
- Cl. + O3 ------> ClO. + O2
- ClO. + O3-----> 2O2 + Cl.
- In total : 2O3------> 3O2
- Although this isn't harmful, the ozone layer has been destroyed
- Although this isn't harmful, the ozone layer has been destroyed
- The cycle will start again
- In total : 2O3------> 3O2
- ClO. + O3-----> 2O2 + Cl.
- Cl. + O3 ------> ClO. + O2
- These free radicals will then react with the ozone layer
- They are harmless on land, but when they get up in the
atmosphere, the bonds start to break from the UV light
found in the atmosphere to form free radicals.
- However they are now banned because they were
contributing to the destruction of the ozone layer
- This occurs by free radical substitution
- This uses an Alkane and a halogen
- 1. Electrophilic addition
- Elimination:
- This is where you turn a haloalkane back into an alkene
- In order to do this, we need to break 2 bonds.
- In a haloalkane, the easiest bond to break is the H-X bond because they
have a greater difference in electronegativity than a C-H bond does
- The second bond to break has to be a C-H bond
- You have to get rid of a H that is ADJACENT to the carbon that
you've just pulled off a halogen from in order to form a double bond
- To do this, we need a base: a hydroxide ion
- This will pull off the H+ ions as a base is a proton acceptor
- The H+ will obviously take no electrons with it
- So the carbon has one extra electron, and it will have a negative charge on it
- Both the carbons form another bond (+ and -) and the double bond has formed
- Both the carbons form another bond (+ and -) and the double bond has formed
- So the carbon has one extra electron, and it will have a negative charge on it
- The H+ will obviously take no electrons with it
- This will pull off the H+ ions as a base is a proton acceptor
- To do this, we need a base: a hydroxide ion
- You have to get rid of a H that is ADJACENT to the carbon that
you've just pulled off a halogen from in order to form a double bond
- It will take both it's electrons with it, so there
will be a plus charge on the carbon
- The second bond to break has to be a C-H bond
- In a haloalkane, the easiest bond to break is the H-X bond because they
have a greater difference in electronegativity than a C-H bond does
- Mechanism:
- In order to do this, we need to break 2 bonds.
- Conditions:
- The right conditions are required otherwise
the OH will act as a nucleophile and not a base
- 1. Alcoholic solution/solvent only
- Heat, not warm
- Heat, not warm
- The right conditions are required otherwise
the OH will act as a nucleophile and not a base
- This is where you turn a haloalkane back into an alkene
- Properties:
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