Reactions with Alkenes + Addition Polymerisation + Alcohols Flashcards
Describe/outline the mechanism for the reaction between an alkene and a hydrogen halide molecule
electrophilic addition of hydrogen halide
HBr + ethene
overall equation
ethene + hbr -> bromoethane
The HBr has added to the ethene
This reaction is an example of electrophilic addition
The alkene in this reaction is ethene which is a symmetrical molecule
This means we can only make one possible product
However, if we carry out this reaction with an asymmetrical alkene such as but-1ene then we can make two possible products
H-Cl
H- Br
H - I
Halogen atoms are more electronegative than hydrogen
This means that hydrogen halide molecules have a permanent dipole (delta positive)
In the hydrogen halide molecules, the hydrogen atom has a small positive charge and the halogen atom has a small negative charge
HBr + ethene
The hydrogen bromide molecule approaches the alkene
The positive hydrogen on the HBr is attracted to the high electron density of the double bond. HBr is referred to as an electrophile
The +ve hydrogen atom in the HBr is acting as an electrophile
Now the positive charge on the hydrogen atom attracts the pair of electrons in the pi bond of the alkene
This pair of electrons move towards the hydrogen atom (curly arrow is used to show that pair of electrons moving)
A covalent bond is now forming towards the H atom in the HBr
This presents a problem as a H atom can only have one covalent bond
So at the same time, the pair of electrons in the covalent bond between the H and Br now move onto the Br atom
We show this as a curly arrow
When a covalent bond breaks like this, with both electrons going to one atom - this is called heterolytic fission
At the end of this stage we have two products
We have a positively charged intermediate molecule formed from the alkene. Called a carbocation intermediate
This contains a positively charged C atom
This atom is positively charged because it has lost its share of the electron pair that were in the pi bond
Also formed a negatively charged Bromide ion
This is negatively charged because both of the electrons that were in the covalent bond are now on the bromide ion
In the next stage, the electron pair on the bromide ion are attracted to the positive carbon atom in the carbocation
Represented with a curly arrow
This electron pair now forms a covalent bond to the carbocation
We formed our product which is Bromoethane
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Describe what is meant by an electrophile and heterolytic fission
double bond in alkenes consists of 4 electrons.
This means that the double bond is a region of high electron density
This region of high electron density determines how alkenes react
Alkenes react by a process called electrophilic addition
An electrophile is any positive ion or molecule which is attracted to a region of high electron density
draw curly arrow - make sure it starts exactly where the electrons are moving from and ends exactly where they are moving to
Use Markownikoff’s Rule to determine the products when a hydrogen halide adds to an assymetrical alkene
RULE - When a hydrogen halide reacts with an asymmetric alkene, the hydrogen atom of the hydrogen halide is more likely to bond to the carbon atom which is attached to the greater number of hydrogen atoms.
in but-1-ene - c1 is bonded to 2 H atoms whereas c2 is bonded to 1 H atom. So, the H of the HBr is more likely to bond with C1
Therefore, the major product is 2-bromobutane
When we add a hydrogen halide to a symmetric alkene, we can only make one product but-2-ene -> 2-bromobutane
when an asymmetric alkene reacts with a hydrogen halide molecule, we can form two different isomers
This depends on which carbon atom has the positive charge in the carbocation intermediate
e.g. in but-1-ene
if the positive charge is on carbon1 then we make 1-bromobutane
if positive charge is on Carbon2 we make 2-bromobutane
we dont make equal amounts of these two products (make more of 2-bromobutane than 1-bromobutane)
Explain why we get major and minor products when reacting asymmetric alkenes with hydrogen halides
we dont make equal amounts of these two products (make more of 2-bromobutane than 1-bromobutane)
2-bromobutane is the major product and 1-bromobutane is the minor product
A carbocation intermediate is an unstable molecule than only exists for a short period of time
The stability of the carbocation depends on how many alkyl groups are bonded to the positive charge
Alkyl group - a group containing C and H atoms e.g. methyl/ethyl group
If the positive carbon atom is bonded to only one alkyl group - called a primary carbocation
if its bonded to two alkyl groups - secondary carbocation
Secondary carbocations are more stable than a primary carbocation (more electron donating groups)
this is because the electrons in alkyl groups can shift towards the positive charge. The effect of this is to stabilise the positive charge. This is called the positive inductive effect
in primary C+, the positive charge is only stabilised by one alkyl group
while in secondary C+, the positive charge is stabilised by two alkyl groups
therefore this makes secondary c+ more stable than primary c+
Because a secondary carbocation is more stable - it exists for a longer period of time and is more likely to form the product (same for tertiary vs secondary C+)
explains why 2-bromobutane (forms secondary C+)is the major product and 1 bromobutane is the minor product (FORMS PRIMARY C+)
. This is because the alkyl groups donate electrons to the positive charge of the carbocation making it more stable.
Describe the electrophilic addition of a halogen molecule to an alkene
ethene + bromine
ethene + bromine ->1,2 dibromoethane
in this reaction, the halogen molecule adds across the double bond
example of electrophilic addition
mechanism different to that of Hydrogen halide. This difference is due to the fact that halogen molecules do not have a permanent dipole
In the first stage of the reaction, the bromine molecule approaches the ethene molecule
Bromine molecule does not have a permanent dipole
However, the double bond of the alkene is a region of high electron density
This high electron density repels the electron pair of the covalent bond in the bromine molecule
This means that the bromine molecule now has an induced dipole
In stage 2, the pair of electrons in the pi bond of the alkene are attracted to the positive bromine
The positive bromine is acting as an electrophile
The electron pair now forms a covalent bond to this bromine atom
At the same time, the covalent bond in the bromine molecule now breaks and the pair of electrons move onto the other br atom
When a covalent bond breaks like this, with both electrons going to the same atom - called heterolytic fission
At the end of stage 2, we have a carbocation intermediate and a Br ion
In stage 3, the electron pair on the bromide ion are attracted to the positive c atom in the carbocation intermediate
This electron pair now forms a covalent bond and we have our final product.
Product is 1,2-dibromoethane
The halogen molecule adds across the double bond
means that the two halogen atoms end up on two adjacent carbon atoms
we cannot get both halogen atoms on the same c atom
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Differences between adding a halogen to an alkene and adding a hydrogen halide to an alkene
Adding a hydrogen halide
Hydrogen halide molecules has a permanent dipole
whereas in halogen molecules, the dipole is induced
When we add a hydrogen halide to an asymmetric alkene, we make a major and a minor product
However, when we add a halogen to an asymmetric alkene, we only make one product
e.g. adding bromine to but-1-ene, we can only make 1,2-dibromobutane
Describe how to test for (the presence) unsaturated molecules using bromine water
We can use the reaction with a halogen to test for the presence of an unsaturated molecule such as an alkene
To do this, we use bromine water which has an orange/brown colour
To test if a substance is unsaturated, we add drops of bromine water and gently shake the test tube
If our substance is unsaturated then the bromine will add across the double bond and the product of the reaction will be colourless
We will see the orange bromine water decolourise
However, if our test substance is saturated, then the bromine will not react an the bromine water will remain orange
describe the hydration of alkenes using steam and an acid catalyst
electrophilic addition of water to alkenes to make alcohols - hydration reaction
ethene + water (g) ⇌ ethanol
in hydration, the water is in the form of steam
phosphoric acid catalyst
300 degrees C
60 atm
In the first stage of the reaction, the pair of e- in the pi bond of the ethene are attracted to one of the positive hydrogen atoms in the phosphoric acid
The positive hydrogen atom is acting as an electrophile
Now the pair of electrons in the pi bond form a covalent bond to the positive hydrogen atom
At the same time, the covalent bond between the hydrogen and oxygen breaks. The pair of electrons now move completely onto the oxygen atom
when a covalent bond breaks like this, with both electrons going to the same atom - called heterolytic fission
At the end of stage one - we have a carbocation intermediate with a positively charged carbon atom
Also have a dihydrogen phosphate ion with a negatively charged oxygen atom
Phosphoric acid is a catalyst. Therefore the phosphoric acid will be regenerated in a later stage
In stage 2, the carbocation intermediate reacts with a molecule of water in the form of steam
The oxygen atom in the water molecule has two lone pairs of electrons
One of these lone pairs now forms a covalent bond between the oxygen and the positive carbon atom in the carbocation intermediate
At the end of stage 2, we now have an intermediate molecule containing a positive oxygen atom
This oxygen is positive because its lone pair of electrons have formed a covalent bond
The final sage involves the dihydrogen phosphate ion - made earlier
The dihydrogen phosphate ion forms a covalent bond to a hydrogen in the intermediate molecule.
At the same time, the covalent bond in the intermediate now breaks by heterolytic fission and the pair of electrons in the bond now move completely onto the oxygen
So at the end of this stage, we have made our product molecule ethanol and we have regenerated our phosphoric acid catalyst
hydration of ethene produces ethanol
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State the major and minor products from the hydration of asymmetric alkenes
propene + steam
When we hydrate an asymmetric alkene e.g. propene
How do we determine which carbon atom will bond to the hydrogen atom of the water and which will bond to the OH group
Use the rule
the hydrogen is more likely to bond to the carbon atom which is already bonded to the greater number of hydrogen atoms
In propene, C1 is bonded to 2 H atoms, while C2 is bonded to 1 H ATOM
This means that the hydrogen atom in the water is more likely to bond with C1 in propene
This makes propan-2-ol our major product and propan-1-ol our minor product
Describe the reaction between alkenes and sulfuric acid
Due to their double bond which is a region of high electron density
This means that alkenes react by electrophilic addition
e.g. with hydrogen haldies or halogens
electrophilic addition of
alkenes + conc. sulfuric acid
H2SO4 contains two hydrogen atoms covalently bonded to two O atoms
Oxygen is a highly electronegative element
Because of this, the oxygen atoms have a partial negative charge and the H atoms have a partial +ve charge
The pair of electrons in the pi bond of the ethene are attracted to one of the positive hydrogen atom in the sulfuric acid
The positive hydrogen atom is acting as an electrophile
The pair of electrons in the pi bond form a covalent bond to the positive hydrogen atom
At the same time, the covalent bond between the hydrogen and oxygen breaks
And the pair of electrons now move completely onto the oxygen atom
When a covalent bond breaks like this with both electrons going to same atom - called heterolytic fission
At the end of stage 1, we have a carbocation intermediate with a positively charged C atom
We also have a hydrogensulfate ion with a negatively charged oxygen atom
In stage 2, the lone pair of electrons on the (oxygen of the) hydrogensulfate ion are attracted to the positive Carbon atom on the carbocation intermediate
The lone pair of electrons now forms a covalent bond to this carbon atom
We have made our product which is ethylhydrogensulfate
This reaction takes place by electrophilic addition
If we add water to ethylhydrogen sulfate, we make the alcohol ethanol and we also reform the sulfuric acid
So this reaction can be used to form alcohols from alkenes.
Overall equation ethene + water -> ethanol
->- SULFURIC ACID
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if we use an asymmetric alkene for the electrophilic addition of sulfuric acid and then water to make an oxygen
e.g. with propene
Using the rule ,we know that the hydrogen from the sulfuric acid is more likely to bond to the carbon atom which is already bonded to the greater number of hydrogen atoms
in propene, carbon 1 is bonded to two hydrogen atoms whereas c2 is bonded to one h atom
So in this case, the hydrogen in the sulfuric acid is more likely to bond to c1 and the hydrogen sulfate bonds to carbon 2
propylhydrogensulfate
When we add water, the OH of the water molecule takes the place of the hydrogensulfate
This means that our major product is propan-2-ol
The minor product is propan-1-ol but only a small amount of this forms
categories of polymers
Addition polymers
Condensation polymers
Describe what is meant by a monomer
Polymers are large molecules
We form polymers by joining together thousands of small identical molecules called monomers
in a process called polymerisation
When we form addition polymers, the monomers are alkenes
What does polymerisation require and what happens during polymerisation
Polymerisation requires high temperature and pressure as well as a catalyst
When alkenes form a polymer, the double bond in the alkene opens up and joins one monomer to another
When we polymerise ethene, we make poly(ethene) - used to make plastic bags and bottles
The product polymer has no double bonds in the carbon backbone
This means that despite being forms from an alkene addition polymers are actually alkanes
Addition polymers contain a large number of carbon-hydrogen and carbon-carbon bonds
These bonds are both non-polar and relatively strong which make them difficult to break
So because of that, addition polymers are unreactive molecules
This lack of reactivity means that addition polymers can exist in the environment for a very long time
It appears the C atoms at either end only have three covalent bonds
However during polymerisation, other molecules are added to cap the ends of the polymer chain
Describe what is meant by an addition polymer
Describe what is meant by a repeating unit
write and draw the polymerisation reaction for ethene to poly(ethene)
The repeating unit shows the arrangement of atoms that are repeated in the polymer chain
to identify the repeating unit
take any two adjacent carbon atoms on the main chain
Draw those two carbon atoms plus any atoms above and below
Square brackets
covalent bond extending through square brackets
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monomer -> repeating unit
Number of repeating units must be the same as the number of monomers to balance the equation
Work out the monomer and repeating unit for a given polymer or vice versa
polymer to monomer -
monomer always has a double bond between the two c atoms
repeating unit never has a double bond between the two carbon atoms
if 14 c atoms in the chain then n = 7 in repeating unit
other groups arund c=c are referred to as side groups
Describe the role of plasticisers
pvc is a rigid polymer - used to make plastic pipes and other products
if we add a chemical called a plasticiser, we can produce flexible pvc
flexible pvc is relatively soft and is used to make flooring and the insulation on electrical cables
in pvc, the polymer chains are attracted to each other by intermolecular forces - both permanent dipole-dipole interactions and vdW forces
A plasticiser is a small molecule that fits between the polymer chains
This causes the chains to move further apart
This weakens the intermolecular forces between the chains
So the effect of the plasticiser is to allow the polymer chains to move over each other, making the polymer flexible
Alcohols
all alcohols have the alcohol functional group
alcohol functional group - also called hydroxyl group
use a number to show which c atom is bonded to the hydroxyl group
if an alcohol molecule contains 2 hydroxyl groups it is called a diol
ethan-1,2-diol
alcohols with 3 hydroxyl groups - called triols
propane-1,2,3-triol
2-chloropropan-1-ol
Aldehydes, ketones and carboxylic acids (these functional groups) have naming priority over the alcohol functional group
3-hydroxypropanal
1-hydroxypropan-2-one
molecule is named based on the higher priority functional group
The alcohol group is shown by the prefix hydroxy
Classification of alcohols
Primary, secondary and tertiary alcohols
In primary alcohols, the carbon atom bonded to the hydroxyl group is bonded to one other carbon atom
e.g. ethanol, propan-1-ol, methanol (doesnt fit the definition but still considered as a primary alcohol)
In secondary alcohols, the carbon atom bonded to the hydroxyl group is bonded to two other carbon atoms
e.g. propan-2-ol
Finally in tertiary alcohols, the carbon atom bonded to the hydroxyl group is bonded to three other carbon atoms
2-methylbutan-2-ol
Whether an alcohol is a primary, secondary or tertiary alcohol can affect how it will react
A hydroxyl group (OH) requires a single bond to the carbon. If a carbon were to also be bonded to four other carbon atoms, it would have five bonds total, violating the octet rule (which states that a carbon atom can have at most four bonds).
Describe and explain the properties of alcohol molecules - volatility
Alcohols have a higher BP than the alkane with the same number of carbon atoms
Alkanes are non-polar molecules
Because of this lack of polarity, only vDW forces are acting between alkane molecules
vDW forces are weak and do not take a lot of energy to break
Because of this they have low BP
However, alcohols are polar molecules. Due to the alcohol functional group.
Oxygen atoms are much more electronegative than hydrogen atoms
Because of this, the oxygen atom in the alcohol functional group has an negative charge and the H atom has a positive charge
This means that alcohol molecules can form both VdW forces and hydrogen bonds to each other
Hydrogen bonds are relatively strong intermolecular forces requiring a relatively large amount of energy to break
Because alochol molecules have both VdW forces and hydrogen bonds they have BP’s than alkanes with the same number of c atoms
The volatility of a molecule tells us how readily a molecule turns into a gas
Because alcohols have higher BP than alkanes this means that alcohols are less volatile than alkanes with the same no. of c atoms
As we increase the number of carbon atoms, the difference in the boiling points between the alcohols and the corresponding alkanes reduces
Explain why.
In alkanes, there are only vDW forces
In alcohols, there are both vDW and hydrogen bonds
In an alcohol with a short carbon chain such as ethanol, the major intermolecular force is hydrogen bonding due to the alcohol functional group
Whereas VdW forces play a much less significant role
So this means that short chain alcohols have a much greater boiling point than the corresponding alkane
However in alcohols with long carbon chains, e.g. decan-1-ol, the contribution of vDW forces increases and the relative importance of hydrogen bonding is reduced
This means that the BP of long chain alcohols are only slightly greater than the corresponding alkanes
Describe and explain the properties of alcohol molecules - solubility in water
Hydrogen bonding explains another property of alcohol molecules
Alcohols are highly soluble in water
That is because the alcohol functional group can form hydrogen bonds with water molecules
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Describe and explain how solubility changes as we increase the length of the carbon chain of alcohols
As we increase the length of the carbon chain, alcohols become less soluble in water
This is because the non-polar carbon chain cannot form hydrogen bonds.
So as we increase the length of the carbon chain, a greater part of the molecule is unable to hydrogen bond to water molecules
This makes long chain alochols less water-soluble than short chain alcohols
