1
Q
What is the order of priority of functional groups?
A
carboxyl - oic acid amide - amide / carbamoyl carbonyl (aldehyde) - al / formyl carbonyl (ketone) - one / oxo hydroxyl (alcohol) - ol / hydroxy amino (amine) - amine / amino alkene (ene) alkyne (yne) halo
2
Q
Saturated
A
Single bonds between C atoms (alkanes)
3
Q
Unsaturated
A
Double/triple bonds between C atoms (alkenes, alkynes)
4
Q
The reaction of Alkenes with hydrogen
A
Hydrogenation
- Readily react in the presence of a Nickel catalyst to form an alkane
- Double bond ‘opens up’ to accept the H atoms via an addition reaction
- Results in the corresponding alkane forming (Ethene = Ethane)
5
Q
The reaction of Alkenes with halogens
A
Halogenation
- no catalyst
- Double bond ‘opens up’ to accept the halogen atoms via an addition reaction
6
Q
The reaction of Alkenes with hydrogen halides
A
- Hydrogen halides are molecules that contain 1 H and 1 halogen atom
- no catalyst
- addition reaction
7
Q
The reaction of Alkenes with water
A
Hydration
- Requires heat and a catalyst (temp of about 300°C)
- Alkenes form corresponding alkanol (ethene = ethanol)
- Phosphoric acid or sulfuric acid used as catalyst
8
Q
Markovnikov
A
The substituent (the halide in the hydrogen halide) adds to the more substituted carbon (the carbon with more H atoms already attached to it), the H will add to the other open bond on the carbon. (rich get richer).
9
Q
Reactions of alkanes - substitution
A
- Because they are saturated they have to make room, for other atoms to substitute onto the molecule and replace existing atoms
- Requires UV light
10
Q
What is polymerisation
A
It is a chemical reaction in which many identical small molecules called monomers, combine together to form 1 large molecule.
Occurs by the eq:
nH2C=CH2 → [―CH2CH2―]n
11
Q
Explain why boiling point increases with each member in a homologous series.
A
- With each member in a homologous series, there is added molecular mass
- The strength of dispersion forces increases with molecular mass, as there are more electrons added, thus with each added member the strength of intermolecular forces increases
- This means more energy must be input to overcome these forces to turn these molecules into a gaseous phase
12
Q
Explain how packing can impact the boiling point of an organic compound.
A
- Packing increases the boiling point of an organic compound
- Efficient packing resulting from a linear structure allows molecules to align more closely to one another, resulting in stronger intermolecular forces being experienced, thus raising boiling point
13
Q
Identify the relationship between branching and boiling point
A
- Chain branching decreases boiling point
- Adding chains disrupts the shape of the molecule meaning multiple molecules cannot align closely with one another, thus decreasing IMFs experienced and boiling point.
14
Q
Why do methane and ethane have higher melting points than propane?
A
- Methane and Ethane are smaller than propane allowing for a more ordered packing structure in the solid-state compared to propane
- This increases the overall IMF strength between the molecules and thus increases melting points
15
Q
Why do hydrocarbons not dissolve in water?
A
- Hydrocarbons cannot dissolve in water since the adhesive forces they form with water are not able to overcome the cohesive forces within water.
- The cohesive forces within water is a strong network H- bonding between each water molecule
- The adhesive force formed between water and hydrocarbons are only relatively weak dispersion forces, hence this is not enough to overcome the cohesive force the hydrocarbon is not able to dissolve in water.
16
Q
Water is less volatile than hexane. Account for this difference in terms of their intermolecular forces.
A
Water has stronger intermolecular forces compared to hexane, as it has H-bonding whereas hexane has weak dispersion forces. This means that hexanes intermolecular forces require less energy to overcome, allowing it to become a gas more easily. In contrast waters intermolecular forces require more energy to overcome, thus making it less volatile than hexane.
17
Q
Would hexane dissolve into water? If not, what would we observe with the layers of solution formed? Justify your answer.
A
Hexane would not dissolve in water as the weak dispersion adhesive forces between the molecules would not be able to overcome the Waters strong h-bonding cohesive forces. Hexane would sit above water if the solutions were added together, as water is more dense than hexane due to its increased strength of intermolecular forces of h-bonding which reduce the distance between its molecules allowing it to have a higher measure of mass per unit of volume.
18
Q
Which compound would have a higher viscosity, Hexane or water. Explain your prediction.
A
Water would have a higher viscosity as it’s molecules are more attracted to each other due to its strong intermolecular forces of h-bonding, this increased attraction also allows for increased resistance to fluid flow. Hexane has only weak dispersion forces between its molecules, thus decreasing it’s resistance to fluid flow.
19
Q
Explain why amines have a lower bp than carboxylic acids despite also being able to form multiple hydrogen bonds.
A
- Amines have a lower bp as the hydrogen bonds formed are to N which is not as electronegative as O
- The H-bonds are thus weaker and can be overcome more easily
- Carboxylic acids are arranged in a dimer structure, which strengthens their H-bonds, further raising bp.
20
Q
Explain why small amines, amides and carboxylic acids dissolve completely in water.
A
- All have highly polar functional groups
- The adhesive force of H-bonding between water and the polar functional groups on these compounds is strong enough to overcome the similarly strong H-bond cohesive forces and thus dissolution occurs.
21
Q
Explain why longer amines, amides and carboxylic acids do not dissolve well in water.
A
- As chain increases, non-polar section increases
- This section cannot form adhesive forces with water, rather due to their increasing molecular mass, which provides them with additional electrons, they strengthen the cohesive dispersion forces of the organic solute, making adhesive forces too weak to overcome these cohesive forces, decreasing solubility.
22
Q
Between Ethanoic acid and Dibromoethanoic acid, which acid will have a lower pH?
A
- Dibromoethanoic acid will have a lower pH
- This is due to the presence of two highly electronegative Br atoms, which, due to the inductive effect, draw electronegativity away from the O- in the hydroxyl group, weakening the O-H bond, allowing the H to be lost easier. (making it a stronger acid)
23
Q
Condensation reaction
A
A reaction where two or more molecules combine to form a larger molecule with the simultaneous elimination of a small molecule, usually water.
24
Q
Describe the reflux method.
A
Reflux is a technique that involves heating a reaction mixture in a vessel fitted with a cooling condenser so that the volatile reactants and products are returned to the reaction mixture without any loss.
25
Organic compound
A class of chemical compounds constituted primarily of hydrogen, oxygen and nitrogen covalently bonded onto a carbon chain backbone.
26
Explain why alkanes are known as saturated molecules but alkenes and alkynes are known as unsaturated.
- Alkanes are known as saturated molecules due to the presence of only carbon-carbon single bonds, meaning it is not possible to free any more bonds and attach another atom.
- Alkenes and alkynes are unsaturated due to the presence of carbon-carbon multi bonds, which are able to be turned into single bonds, freeing up bonds to attach another atom.
27
Structural isomers
Molecules that have the same molecular formula but different structural formulae. Their atoms are arranged differently resulting in completely different compounds. This also leads to different chemical and physical properties. (chain, position and functional group isomers are all types of structural isomers)
28
Chain isomers
Isomers which involve a rearrangement of the carbon atoms, causing changes in the length of the backbones by adding/removing/branching.
29
Position isomers
Molecules with the same carbon chain and same functional group, but with the functional group being in a different location.
30
Functional group isomers
Share the same molecular formula but have different functional groups. These isomers are in an entirely different homologous series resulting in significantly different chemical and physical properties.
31
Explain why boiling point increases with each member in a homologous series.
- With each member in a homologous series, there is added electrons due to molecular mass.
- Strength of dispersion forces increases with molecular mass, which adds electrons, thus with each added member the strength of intermolecular forces increases.
- This means more energy is required to overcome these forces to turn the molecules into a gaseous phase.
32
Explain why alkynes have a higher bp than alkanes which have a higher bp than alkenes.
- Boiling points are a measure of the thermal energy needed to overcome the IMF of a compound
- For all 3 homologous series, dispersion forces are the only type of IMF existent
- Thus, it would stand that alkanes have a higher bp than alkenes as it has a greater molecular mass, meaning more electrons, increasing the strength of the dispersion forces and thus the energy needed to overcome them
- However, alkynes have the highest bp's because of the linear geometry of the triple bond allows it to overcome its MM deficiency
- This allows their molecules to benefit from the packing phenomenon as they can align more closely with one another, thus increasing the strength of their dispersion forces.
33
Density
Density is a measure of the mass of a substance per unit volume. It increases with increasing molecular mass, which adds electrons and thus, increases dispersion force strength. This will, as a result, reduce the distance between molecules.
34
What are the environmental implications of obtaining and using hydrocarbons from the Earth? Give examples.
- Burning of hydrocarbons release CO2 in the atmosphere contributes to greenhouse effect.
- Incomplete burning of hydrocarbons results in the formation of CO, a fatal gas that can cause death within minutes to exposure.
- CFC, a widely used hydrocarbon in refrigerators and spray cans when exposed to the environment causes harm to the ozone layer => Greenhouse effect
- Aldehydes and other toxic materials released from burning plywood inhibit photosynthesis in plants, cause eye and lung irritations, and even possibly cause cancer.
- Benzene molecules have been found to deplete red blood cells, cause cancer in mammals and damage bone marrow.
- Obtaining petroleum and natural gas from the Earth requires drilling through rocks deep in the Earth’s crust whereby hydrocarbons from the drill machine’s lubricants can be dispersed into surrounding water pollutes the surrounding seawater or ocean. These hydrocarbons are toxic to aquatic organisms that reside in the sea.
35
What are the economic implications of obtaining and using hydrocarbons from the Earth? Give examples.
Positive
- The oil and gas industry employs millions of people all over the world. These are usually high-paying jobs that make a great percentage of them live above the average income, spend within the community and pay taxes to the governments.
- Crude oil → starting point for industries
- Producing countries have the potential to earn and save foreign exchange
Negative
- monopoly of market → Large corporations controlling extraction → expensive for countries without their own oil reserves
- Values of properties located in the proximity of oil and gas fields will decrease
- Obtaining these hydrocarbons could result in the death and reduction in the biodiversity of aquatic organisms, which are a major source of economic revenue for many countries, For e.g, aquatic organisms sold from the Philippines account for over $550 billion USD dollars annually
36
What are the sociocultural implications of obtaining and using hydrocarbons from the Earth? Give examples.
Sociocultural refers to how people live.
Positive
- Work increases income for individuals → increased quality of life
- Transport (fuel) → ease of movement allows for increased income
- Food storage (food safe by plastics)
Negative
- Respiratory issues from burning fuels → increased air pollution (smog)
- Ingestion of microplastics through food chain → results in illness through the exposure to chemicals of plastic and/or chemicals absorbed by plastics in waterways
- Accidents/injuries involving workers due to faulty equipment when drilling for hydrocarbons, Workers also exposed to drilled rocks that are covered in toxic hydrocarbon lubricants
- Potential adverse impacts to groundwater and contamination of water reservoirs → health
- Decrease in seafood supply due to hydrocarbon pollution → increase in price → less affordable for general population
- Exposure to certain hydrocarbons can increase risk of cancers and affect eyes, kidneys and liver
37
Explain why the boiling points of alcohols decrease when examining higher-order alcohols.
- Bp is a measure of the energy required to overcome the IMFs of a particular substance to turn it into a gaseous state
- As alcohols increase in magnitude, there are more alkyl groups surrounding the polar hydroxyl group
- This crowding prevents the hydroxyl group from closing in towards the hydroxyl group of another alcohol molecule, which it must do to form a strong H-bond, the main IMF of alcohols
- Hence due to weaker H-bonding, higher-order alcohols have a lower bp
38
Why is the bp of alcohols much higher than those of alkanes with comparable molar mass?
- Alcohols have a higher bp compared to alkanes due to stronger IMFs
- They have a highly polar hydroxyl group, which allows for the formation of strong H-bonds between molecules, forming a network of H-bonds which requires significant energy to be overcome
- Alkanes do not have significantly polar bonds and as such can only form weak dispersion forces, which do not require as much energy to overcome.
39
Why is it for longer chain alcohols and alkanes, the difference between bp's becomes minimal?
- For longer carbon chains, the strength of dispersion forces becomes significant due to the large number of electrons
- This means that dispersion forces actually exceed H-bonds in terms of strength and become the main IMF of both alcohols and alkanes
- Since both alcohols and alkanes are now relying on the same type of IMF, the difference between bps becomes minimal
40
Explain how the polar and non-polar section of alcohols determine their physical characteristics. (6 marks)
Polar section allows strong H-bonds to be formed:
- with each other and other polar molecules => high bp and good solubility for short carbon chains
- between molecules which requires significant amount of energy to be input for them to be broken => raised bp
- with other polar molecules such as water => adhesive forces formed are strong enough to overcome the cohesive forces => solubility
Non-polar section allows:
- Them to form dispersion forces with each other and other non-polar organic molecules => poor solubility in water and good solubility in organic solvents (for longer chains)
- longer chain => physical hinderance preventing the hydroxyl group from H-bonding with water = weak adhesive forces + cohesive forces ˄ due to stronger dispersion forces => no dissolution
- in organic solvents, longer chain = ˄ molecular mass = ˄ electrons = ˄ dispersion forces =>adhesive forces are relatively strong and able to overcome the cohesive forces => solubility occurs
41
Assess the validity of an experiment you performed to test the enthalpy of combustion of a range of alcohols.
- experiment is valid because variables are controlled and we are testing the aim
- repetition makes its valid
42
Outline any improvements that could have been made to the experiment you performed to test the enthalpy of combustion of a range of alcohols.
- heat lost to the environment --> use a heat shield to reduce this error
- we had incomplete combustion --> whereas theoretical is based on complete combustion -->, therefore, we need to supply more oxygen to increase combustion
- reduce the distance between the flame and the can
43
Dehydration of an alcohol.
Conc H2SO4
| Alcohol ---> alkene + H2O
44
Substitution reactions with alcohols.
Heat
Alcohol + HX ---> haloalkane + H2O
UV
45
Oxidation with primary alcohols
H+/ Cr2O7 2- H+/ Cr2O7 2-
Primary alcohol --> aldehyde --> carboxylic acid
Heat Heat
46
Oxidation with secondary alcohols
H+/ Cr2O7 2-
Alcohol ---> Ketone
Heat
47
How can we distinguish between 2-methylbutan-2-ol and ethanol?
- Oxidation reaction
- If using acidified dichromate, the ethanol solution will very rapidly turn from orange to green, as the primary alcohol oxidised to become carboxylic acid
- However, the 2-methylbutan-2-ol solution will not undergo a colour change as tertiary alcohols cannot undergo an oxidation reaction
primary - rapid colour change
secondary - moderately fast
tertiary - not at all
48
Combustion of propanol
C3H7OH(l) + 9/2O2(g) ---> 3CO2(g) + 4H2O(g)
49
Substitution reactions of halogenated organic compounds to form alcohol.
XOH
Haloalkane (e.g bromoethane) ---> alcohol + XBr
Heat
50
Production of alcohol by fermentation
Reaction conditions:
- Dissolved in water (aq)
- Anaerobic environment
- Zymase (from yeast)
- Temp of 30-40 °C (37°C)
glucose ---> carbon dioxide + ethanol
C6H12O6(aq) ---> 2CO2(g) + 2C2H5OH(aq)
fermentation is exothermic
51
Outline and justify the conditions of fermentation.
- Reaction temp (30-40°C) => Zymase is optimised at this range
- Anaerobic conditions => oxygen would have resulted in different products
52
Outline the properties of aldehydes and Ketones.
- solubility => shorter => higher, that decreases as they lengthen (more CH2 groups --> more symmetry => less polar due to dispersion forces => decreased solubility)
- bp & mp => high, due to presence of =O, increases with increased chain length due to the presence of more dispersion forces
- density - low compared to water
- Oxidise to form carboxylic acids
53
esterification
- Alcohol and carboxylic acid reacted together to form an ester and water
- Requires H2SO4 catalyst
- Boiling chips are also used to prevent 'bumping'
54
Describe a practical investigation you conducted to measure and reliably compare the enthalpy of combustion of a range of alcohols.
1. Add 200mL water to the copper can. record the initial temp.
2. Clamp the can above the spirit burner so that the tip of the flame will just touch the can when lit.
3. Add fuel to the spirit burner and weigh. Record the mass and place the burner under the can.
4. Light the wick and gently stir the water with a stirring rod.
5. When the temp has risen by about 10°C, extinguish the flame by replacing the cap.
6. Accurately record the maximum temp of the water.
7. Reweigh the burner and record its final mass
8. Repeat with a different fuel. (Ethanol, pentanol, butanol)
55
Outline the procedure you used when conducting an investigation to compare an alkene with its corresponding alkane.
Equal quantities (2 mL) of cyclohexane and cyclohexene were placed in separate test tubes. The same number of drops of Br water was added to each. They were shaken in the same way, in absence of UV light (away from direct sunlight or fluorescent lights) and observed. An alkene (being unsaturated) will react more readily than an alkane (being saturated) with the Br water, so the sample that immediately decolourised the bromine water was recorded as hexene.
56
Outline the risks associated with the practical investigation you conducted to measure and reliably compare the enthalpy of combustion of a range of alcohols.
Fuels were toxic if ingested/inhaled and highly flammable.
57
Identify potential hazards in your investigation to compare an alkene with its corresponding alkane, and outline how you addressed this hazard
Cyclohexene is:
- Highly flammable liquid + vapour - may form unstable peroxides over an extended period of time
- May be fatal if swallowed and enters airways
- Toxic in contact with skin
PPE
- Gloves and safety goggles
58
Account for the ability of soaps to clean dirty surfaces
- Dissolution
Soaps are salts of fatty acids with a hydrophilic carboxylate ion head and a hydrophobic hydrocarbon tail. When dissolved in water they form micelles which gradually find their way to a grease molecule.
- Adsorption
The hydrophobic tail of soaps is able to dissolve into grease molecules because they are both non-polar. This means they are attracted to each other via dispersion forces
Meanwhile, the polar hydrophilic head is still interacting with passing water molecules via ion-dipole forces and hence the soap molecule can be pulled off with the grease still attached
- Emulsification
with agitation, the grease molecules can be broken into smaller, spherical droplets which are centred within the soap micelle structure. The resulting emulsion solution can simply be washed away with water, leaving behind a clean surface.
59
Structure of a soap molecule
Draw and check
60
Polyethylene
draw structure and check
Addition polymer
Properties
PE exists in either a low density (LDPE) or high density form (HDPE).
LDPE is made from branched chains resulting in
- flexible
- light weight
- not dense or hard properties.
HDPE is made from unbranched chains, resulting in
- rigid
- hard
- high density properties
Uses
LDPE: Garbage bags, water bottles
HDPE: cutting boards, garbage bins, buckets
61
Polyvinyl chloride
draw structure and check - Addition polymer
Properties
- Very hard
- Rigid
- Brittle (This is due to the chlorine side group. Although it doesn’t seem bulky, the fact that it’s quite electronegative means that it will repulse any other chlorines on the same chain, thus keeping the chain rigidly linear)
- Flame Retardant
- Chemical Resistance (due to the chlorine side group)
Uses
Piping, gutters, waste water pipes, credit cards
62
Polystyrene
draw structure and check - Addition polymer
| Bulky side group causes rigidity
63
Polytetrafluoroethylene
draw structure and check - Addition polymer
To deduce the structure just think it’s tetrafluoroethylene - 4 fluorines on an ethene.
Electronegative effects of fluorine lock chain in linear structure, just like with the chlorines in PVC
64
Nylon/Polyamide
draw structure and check
Condensation polymer - The monomers are joined by an amide linkage, to form an amide, we need to react together a carboxylic acid and amine functional group.
Polyamides could be formed from a single monomer which has both acid or amine groups, or from two monomers which are difunctional for one of those groups each.
65
Polyesters
draw structure and check
Condensation polymer - Polymers where the monomers are joined together with an ester link.
Just like normal esters, polyesters will be made from carboxylic acid and alcohol functional groups. This could mean either:
- 1 monomer has both (so different functional groups on each end)
- 2 monomers have 1 each (one monomer has the same functional group on either end, and the other monomer has the other functional group)
66
Define a polymer
A long chain molecule made up of repeating units called monomers, joined by covalent bonds.
67
Define addition polymers
Polymers made from reacting unsaturated monomers together in an addition reaction, without the elimination of any small molecules.
Addition or chain polymerisation can be divided into 3 steps – initiation, propagation, and termination.
Initiation
The polymerisation reaction is initiated through the addition of a free radical. The free radical acts to open the C=C double bond by joining to one side of the monomer. This allows the monomers to react with other open monomers on their other side. A possible initiator is free radicals formed from hydrogen peroxide through the addition of heat.
Propagation
Following initiation, the process continues with the successive addition of monomer units to the chains. This is known as propagation.
Termination
The third and final stage in the polymerisation process is known as termination. Termination of the reaction process can occur through the addition of a terminating free radical or when two chains combine.
68
Define a condensation polymer
- A condensation polymer is a polymer formed by difunctional monomers that are not necessarily identical, undergoing a condensation reaction with the elimination of a small molecule, usually water
69
Explain the differences between an addition polymer and a condensation polymer
Addition
- Typically one monomer
- Addition reaction
- No elimination of a small molecule
- Unsaturated monomers
Condensation
- Typically two different monomers
- Condensation reaction
- Elimination of small molecule
- Difunctional monomers
70
Outline the need for sustainable biofuels and how they differ from fossil fuels in terms of renewability.
- Fossil fuels consumed at unsustainable rates, thus biofuels needed
- To be sustainable, a biofuel must be replenished at the same rate it is consumed, and the wastes involved in making it must not harm the env
- Fossil fuels are non-renewable as they were produced over millions of yrs and their limited reserves are being rapidly used up and cannot be replaced
71
Describe the procedure you used to perform an esterification.
1. Place 10mL carboxylic acid and 10mL alcohol in a round bottomed flask with boiling chips.
2. Add 3mL conc sulfuric acid to the mixture.
3. Attach a reflux condenser and heat mixture under reflux for 1 hour, in a heating mantle.
4. Allow mixture to cool and transfer content to a separating funnel with a large vol of distilled water.
5. Shake the mixture then allow it to separate into two layers. Discard the aqueous layer.
6. Add a small amount of saturated sodium carbonate solution to the organic layer until bubbling ceases. Add water and shake the mixture. Allow the mixture to separate into 2 layers. Discard the aqueous layer.
7. Add a drying agent to the organic layer. Filter off the drying agent.
72
Draw set up of esterification, and justify parts of the setup.
Check picture
73
Outline the processes involved in the industrial production of ethanol from sugar cane.
- A high sugar content plant material such as sugarcane is mashed up with water and years is added
- Air is excluded so to make an anaerobic env and the temp is kept at 37 degrees
- When the reaction produces a reaction mixture of 15% ethanol conc, the reaction mixture is removed and distilled further to reach a higher concentration of ethanol.
74
Burning of petroleum.
C8H18 + 25/2 O2 --> 8CO2 + 9H2O
75
Describe the properties and action of a soapy detergent
A soapy detergent is made from fats and oils reacted with either sodium hydroxide or potassium hydroxide. It has a non-polar tail and polar head, meaning it works similarly to a soap. It forms a precipitate in hard water, that settles out as a 'scum'. Some of its properties include; it is readily biodegradable, can be hard to dissolve in water, does not lather well in hard water, deteriorates with age and is non-toxic.
76
Structure of a soapy detergent
draw and check.
77
Structure of a soapless detergent
draw and check.
78
Describe the action and properties of a soapless detergent
A soapless detergent is made from a hydrocarbon chain from petroleum reacted with sulfuric acid. It has a non-polar tail and polar head, meaning it works similarly to a soap. Some of its properties include that it is biodegradable when made with an unbranched structure, readily dissolves in water, does not deteriorate with age, may be toxic, stays dissolved in hard water and lathers in hard water.
79
Anionic detergents
```
Usually contain either a sulfate (SO4) head or a sulfonate (SO3) head
widely used due to cost and performance
-laundry detergents
-dishwashing liquids
-oven cleaners
```
80
Cationic detergents
Usually contain a modified ammonium ion as active site which produces a germicidal action.
More expensive than anionic detergents.
- cleaning plastics
- hair shampoos
- nappy washes
- fabric softeners and conditioners
81
Non-ionic detergents
Contains polar parts, eg, OH groups, to provide water solubility.
No ionic groups so no reaction in hard water.
Low lathering prevents foam build up in dishwashers.
- car shampoos
- dishwasher detergents
- cosmetics
Chemistry
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