Download reactions taking place within cells

Thermodynamics Study of conversion of energy between heat and other forms
Thermochemistry Relationship between chemical reactions and heat changes
Enthalpy H Measure of energy(heat content) Change Final value minus the initial value
Enthalpy change H kJmol –1 Heat energy transferred in a reaction at constant pressure H = (H)Products - (H)Reactants
Heat changes depend on conditions in which they are measured(open containers/atmospheric/constant pressure)
Standard conditions H θ 298K 100kPa Internationally agreed conditions under which a standard H should be measured
Endothermic reactions Heat absorbed, products have more enthalpy than reactants, decrease in temp of surroundings(photosynthesis)
Exothermic reactions Heat evolved, products have less enthalpy than reactants, increase in temp of surroundings
(oxidation H+(aq) + OH–(aq) H2O(l))
Change in physical state is accompanied by H
Allotrope A structurally different form of an element in the same physical state caused by the possibility of more than one arrangement of
atoms(graphite and diamond are allotropes of carbon)
G ‘free energy’ difference between the products and reactants
Standard enthalpy of combustion Hc θ(exothermic) Enthalpy change when 1 mole of a substance is completely burned in O2(excess
air), at 298K 100kPa
Combustion won’t take place under standard conditions but measurement of H must be made when the conditions at the start and end
of the reaction are standard
Energy value of fuel/food based on Hc since combustion and processes that fuel/food undergoes in the body gives rise to the same
product. 1 cal = 4.18J, calorific values of food important for people controlling their energy intake for dietary reasons
Standard state of an element The most stable form of the element under standard conditions H2(g), O2(g), Cu(s)
Standard enthalpy of formation Hf θ Enthalpy change when one mole of a compound is formed from its elements, at 298K 100kPa
Hf of any element in its standard state is zero
Enthalpy of neutralisation HN(exothermic) Enthalpy change when one mole of water is formed from a reaction of an acid with a base
Hc of graphite is the same as Hf of CO2 C(s, graphite) + O2(g)  CO2(g) H θ = 393kJmol –1
C(s, diamond) + O2(g)  CO2(g) H θ = 395kJmol –1Hc of diamond but not Hf of CO2 since carbon is not in its standard state
Graphite is the standard state for carbon(it’s more thermodynamically stable at 298K)
2Li(s) + ½O2(g)  Li2O(s)
Hf of Li2O is not the Hc of lithium since 2 moles of lithium are involved in the equation
Enthalpy of fusion of H2O
H for the process H2O(s)  H2O(l)
Enthalpy of vapourisation of H2O H for the process H2O(l)  H2O(g)
1°C ≡ 1K Heat transfer(J) = m(g) x c x T
Specific heat capacity (c)(Jg–1 °C–1) Amount of heat required to raise the temperature of 1g of substance by 1K
Temperature Measurement of KE of particles(independent of the amount)
Heat Measurement of total energy in a substance(dependent on the amount)
Calorimetry Measurement of heat transferred to known mass of another substance and measuring temp rise
Calorimeter Apparatus used to measure heat given off in a chemical reaction
Why T may not be accurate: • Loss of substance(ethanol) by evaporation
• Heat transferred to calorimeter/lost to surroundings instead of the water(avoided by calibrating apparatus appropriately)
• Incomplete combustion due to inadequate supply of O 2 leading to CO or C(indicated by deposit of soot on the bottom of calorimeter)
1st law of thermodynamics Energy can’t be created or destroyed, only converted from one form to another
Hess’s law H for a reaction is independent of the route it takes provided that the temperatures, pressures and physical states of the
reactants and products are the same
H of reverse reaction is the same but with opposite sign
Hess’s law allows for calculation of enthalpy changes for reactions which can’t be measured experimentally, if statements weren’t true
it’d be possible to create energy without any consumption of material
•
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•
•
•
•
•
150g 23°C water in a glass beaker and stirred • After certain time final temp noted
Spirit burner reweighed • m of spirit burner gives mass of ethanol burned (0.9g)
Heat gained by water in calorimeter = mxcxT = 150g x 4.2Jg–1 °C–1 x 20°C = 12600J
Heat produced by burning 0.9g of ethanol = 12.6kJ
Mr(C2H5OH) = 46gmol –1 Ethanol used = 0.9g/46gmol –1 = 0.01956mols
So heat produced by burning 1 mol of ethanol = 12.6 x 1/0.01956 = 644kJ
Hc = 640kJmol –1
• Note temp, 50cm3 1moldm–3 HCl in insulated polystyrene cup(beaker from expanded polystyrene)
• Add 50cm3 1.1moldm–3 NaOH(excess to ensure complete reaction)which is at same temp
• Stir continuously and note max temp reached
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Assuming no heat losses to surroundings and c of solution is 4.2Jg –1 °C –1
Heat absorbed by solution = m x c x T = 100g x 4.2Jg –1 °C –1 x 6.5°C = 2730J
Acid= 0.05mol, Heat given by 1 mol of acid = 2730J/0.05mol = 54.6kJmol –1 HN = 55kJmol –1
Heat loss to surroundings increases with slower reactions as
heat lost over a longer period
The greater the heat loss to the surroundings the greater the
correction of the temperature and the steeper the line
• 1st reagent placed in polystyrene cup, temperature noted at 1min
intervals for 4min stirring continuously
• 2nd reagent added, temperature noted more frequently until max
reached
• As solution starts to cool, temperature recording and stirring
continued for 5min
• Check that sign of H is correct
• If equation is multiplied, also multiply H value
• Hf of elements is 0
Conversion of graphite to requires high
temperatures and pressures as huge amount of
energy is needed to disrupt bonds before atoms can
be rearranged into the diamond structure(with
release of all but 2kJmol –1 of the energy put in)
(1)Calculate H for N2O4(g)  2NO2(g)
2NO2(g)  2NO(g) + O2(g) H θ = kJmol –1
½N2(g) + ½O2(g)  NO(g) H θ = kJmol –1
N2(g) + 2O2(g)  N2O4(g) H θ = 8kJmol –1
(1)
H = [sum of H θf products] [sum of H θf reactants]
H θc(NH3)? H θf NH3(g)= kJmol –1H θfH2O(g)= kJmol –1
4NH3(g) + 3O2(g)  2N2(g) + 6H2O(g)
{6x(242) + 2x0} xx} = 1268kJmol –1

H θc(NH3) = 1268/4 = -317 kJmol –1

(1)(a)Find Hf θ(CO)using a Hess’s law cycle, C(graphite) + ½O2(g)  CO(g)
Hcθ(C(graphite))= –394 kJmol–1
Hcθ(CO(g))= –283 kJmol–1
e.g. C + 1/2 O2
( H comb Carbon)
CO
(H form CO)
(H combCarbon monoxide)
CO 2
(1)
(a)
= 394 (283) = 111kJmol–1
(b)Suggest why it’s not possible to find Hf(CO) directly
(b)(some)CO2 is always produced in the reaction
Enthalpy of dissociation Enthalpy change when 1 mole of a gaseous substance is broken up into free gaseous atoms
(measure of strength of covalent bonds)
Bond breaking is endothermic, need energy to break bonds Bond forming is exothermic, energy released when bonds are formed
Average bond enthalpy CH4(g)C(g) + 4H(g) H θ = +1664kJmol –11664/4 = 416kJmol –1
(specific bond enthalpies for the 4CH bonds are different)
H = [sum of average bond enthalpies of reactants] [ sum of average bond enthalpies of products]
(1)Calculate Hf(H2O) H2(g) + ½O2(g)H2O(l) E(HH)(g) = 436kJmol –1;
E(O=O)(g) = 498kJmol –1; E(OH)(g) = 464kJmol–1
–1
(a)H = [E(HH) + ½E(O=O)]  [2E(OH)] = [436 + 249] 928 = kJmol 
H for H2O(g)  H2O(l); H = kJmol –1
243 + (44) = 287kJmol–1
(b)Suggest why the value obtained in (a)may not be accurate (b)Average values used in bond enthalpies
(2)When hydrogen reacts with chlorine in a container of fixed volume the pressure increases but the number of molecules has not
changed. What is the sign of H?
(2)H2(g) + Cl2(g)  2HCl(g) As the number of molecules doesn’t change during the reaction, the only reason for an increase in pressure
can be an increase in temperature. This means that the reaction is exothermic and H is negative
(3)Is Hc(H2) same as Hf (H2O)? (3)Yes, provided the conditions are the same H2(g) + ½O2(g)  H2O(l)
(4)Is Hc(CO) the same as Hf(CO2)?
(4)Enthalpy changes are not the same even though they both produce 1 mole of CO 2
θ
(5)Define Hf (urea) (5) • Enthalpy change for the formation of 1 mol of urea, from its elements, in their standard states/100kPa 298K
(6)(a) Find H for: 2C(graphite) + 2H2(g) + O2(g)  CH3COOH(l)
Hc(C(graphite))= –394kJmol–1 Hc(H2(g))= –286kJmol–1 Hc(CH3COOH(l))= –874kJmol–1
(a)H = [2(Hc(C)) + 2(Hc(H2))]  [Hc(CH3COOH)] = 486 kJ mol–1
(b)What is the H obtained in (a) called? (b) Hf
30
29
Temperature / ºC
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27
26
25
24
23
22
21
20
0
4
8
12
16
20
24
28
32
36
40
Volume of NaOH(aq) / cm –3
(7)(a) 20cm3 1moldm–3copper salt(aq) in a polystyrene cup. Burette filled with 2moldm–3 NaOH(aq)
2cm3 NaOH(aq) was run into the copper salt(aq) and temp measured immediately. As soon as possible a further 2cm3 of NaOH(aq)
was run in and temp measured again, process continued until 36cm3 of NaOH(aq) had been added
(i)Explain why the temp reaches a max and then falls slightly on addition of further NaOH(aq)
(i)Reaction is complete, addition of cooler NaOH causes temp to fall
(ii)Calculate the amount of copper ions that have reacted
(ii)201/1000 =0.02mol
(iii)Write the formula of the copper hydroxide that is produced
(iii)Cu(OH)2
(b)Identify a source of error in this experiment, and say what you would do to reduce its effect
(b) • Poor mixing, use(magnetic stirrer/swirl cup)between additions, to ensure even temp/reaction faster so less heat loss with time
• Solutions at different initial temperatures, allow them to stabilise at RT
• Measure temperature more often. Allows for more accurate  from graph
(8)A student calculates Hc (C2H5OH) = 853.54966 kJ mol–1 Give 3 criticisms of the value giving reasons
(8) • Shown as standard value. Experiment not conducted under ‘standard’ conditions
• Too many d.ps/significant figures. Accuracy of apparatus doesn’t warrant
• Not shown as negative. Exothermic reaction
Organic chemistry The study of the chemistry of carbon compounds
Homologous series A set of compounds with • similar chemical properties • same functional groups • a general formula where
successive members differ by CH2
Structural formula How the various atoms are bonded to one another within the molecule
- Factors affecting reactivity in organic compounds • single or double bonds • bond polarity • bond enthalpy
- Larger alkanes release a lot more energy per mole because they have more bonds to react
- Carbon can catenate(form stable covalent bonds with itself)
- Organic compounds thermodynamically unstable in the presence of oxygen CH4(l) + 2O2(g)  CO2(g) + 2H2O(g) (-)H
But activation energies of the reactions with oxygen are high so organic compounds are kinetically stable at temperatures on earth
If alkane and oxygen mixed first explosion will result on ignition
Carbon
2s
2p
Carbon
2s 2p
ground state
excited state
- If one electron is promoted from the 2s orbital to the 2p orbital, 4 unpaired electrons become available
- Energy to do this is more than compensated for by the energy released in the formation of 4 bonds instead of 2
- Electrons not all equal since they are in different types of orbital(s & p, methane has 3 bonds of different length from the other one, so
electrons are hybridised)
Isomer Same molecular formula, different structural formulae
Structural isomerism Occurs when 2 or more different structural formulae can be written for the same molecular formula
Chain isomers Different arrangements of carbon skeleton
Similar chemical properties, differ in physical properties(Mt)because of change
in shape of molecule
Positional isomers Same skeleton and functional group, side chains/functional
groups are in different positions on the carbon chain
Differ in physical properties
Functional group isomers Same atoms arranged into different functional groups
Differ in physical & chemical properties
Evidence that the 2 bonds in the C=C double bond are not the same:
• Bond energy of C=C (612kJmol –1) is greater than C-C (348kJmol –1) but not as twice as big hence pi bond is weaker than the sigma
bond
• Greater strength of C=C bond supported by shorter bond length(hence greater overlap)
• Existence of geometric isomers
Stereoisomerism Molecules have same molecular formula, same structural
Stereoisomerism found in any molecule of the type:
formula, but atoms have a different 3d arrangement(orientation in space).
Differ in physical properties
Geometric Isomerism Occurs when there’s restricted rotation about a bond(C=C double bond where each of the two C atoms carries
2 different atoms/groups)differ in physical properties(different positions of groups, chains affects shape, dipoles, intermolecular forces)
Single sigma bond Free rotation about this bond without any reduction in degree of overlap
Double bond Restricted rotation about C=C double bond because rotation would lead to a decrease in overlap of p orbitals that give the
pi bond. Requires energy to break the bond and doesn’t happen at RT. Heating geometric isomers may cause their interconversion.
Nucleophiles Species which seek out + centres. A molecule, atom or ion which can donate a lone pair of electrons to form a new dative
covalent bond
Electrophiles Species which seek out – centres. An electron deficient molecule, atom or ion, capable of accepting a lone pair of electrons
to form a new dative covalent bond
Aromatic compounds Always contain rings, don’t show the properties expected of compounds with double bonds(benzene)
Aliphatic compounds Don’t contain rings, contain double bonds and show the expected reactions eg alkenes
Alkanes CnH2n+2 Simplest homologous series, saturated hydrocarbons
(every C atom has 4 single bonds with other atoms and it’s impossible for carbon to make more than 4 bonds hence alkanes are saturated)
n
Formula Name
Bt °C Mt °C
Colourless gases
Methane used in homes for cooking, heating
1
CH4
Methane -162
-182
2
C2H6
Ethane
-89
-183
Propane/butane, mobile sources of heat/light
3
C3H8
Propane -42
-188
for camping
4
C4H10
Butane
-0.5
-138
5
C5H12
Pentane 36
-130
C5 – C15 colourless liquids, liquid alkanes used in petrol for cars
6
C6H14
Hexane
69
-95
C15 + white waxy solids
7
C7H16
Heptane 98
-91
- Main sources of alkanes from crude oil, natural gas
- Organic materials manufactured from alkanes, detergents, plastics, synthetic fibres
8
C8H18
Octane
126
-57
- Inorganic materials manufactured from alkanes, hydrogen from processed alkanes is
9
C9H20
Nonane
used to make ammonia
10 C10H22
Decane
Mt of alkanes depends on size & shape:
• Alkanes have covalent bonds within molecules and intermolecular van der waals forces
• A branched chain alkane has a lower Mt than straight chain isomer as branched chain alkanes can’t pack as closely together and have
smaller molecular surface areas so van der waals forces are reduced. Therefore Mt decreases as branching increases
- Fractional distillation of crude oil: Separation of mixture of alkanes and hydrocarbons into groups of compounds with similar Bts
Alkyl groups-CnH2n + 1 Not capable of independent existence but occurs within other molecules, CH3 methyl group, CH2CH3 ethyl group
Nomenclature A systematic way of naming chemical compounds
- Names based on longest continuous C chain
Alkyl group names come before name of longest C chain preceded, by a number to indicate C atom at which substitution occurred
Structural Isomerism In alkanes, only way in which different structures can be obtained is by rearranging the C chain
• Straight chain isomer C atoms joined together in a continuous chain(bending chain doesn’t change length of C chain)
• Branched chain isomer C atoms form ‘side chains’ which can’t be of greater length than main chain
3 structural isomer of C5H12
CH3CH2CH2CH2CH3
CH3
CH3
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2,2-dimethylpropane Bt10°C
Pentane Bt36°C
CH3CH2CHCH3
CH3CCH3
CH3C(CH3)3
CH3(CH2)3CH3
2-methylbutane Bt28°C
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CH3CH2CH(CH3)2
CH3
• Alkanes non-polar, not reacting with polar chemicals(doesn’t dissolve in water)
• Alkanes react with non-polar substances given enough energy
Burning fuels • Greenhouse effect(Increases greenhouse gases(CO2))which absorb IR radiation and stop Earth’s IR radiation getting
out) earth warms up slowly, climate changes, polar ice caps melt leading to flooding.
Control global warming by using cars less, replacing fossil fuels with other sources of energy (natural energy, wind, water)
• Power stations add sulphur oxides to the air(scrubbers reduce emissions)which are poisonous causing problems for people with asthma,
causes acid rain which kills trees, damages buildings, makes lakes acidic killing aquatic life
• Incomplete combustion(insufficient O2 supply)produces poisonous gas CO, vehicle engines make nitrogen oxides which add to acid
rain problem, fuel which comes out without burning (unburned hydrocarbons) escape into the air as pollutants
Renewable biofuels From plants, produce CO2 when burned but plants take in CO2 so if replaced, won’t add to CO2 in the atmosphere.
• Biodiesel made from rapeseed oil, used in vehicles • Ethanol made by fermenting sugar cane, mixed with petrol, fuel used in vehicles
Cyclic alkanes Bond angles for smaller rings are different from 109° of the sp3
CH2
CH2 – CH2
CH2
hybrid orbitals indicating poor overlap of the orbitals and considerable strain on
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the ring.
CH2 – CH2
CH2 – CH2
CH2 CH2
- They behave as normal alkanes concerning the C-H bond but are more reactive
Cyclopropane Cyclobutane
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towards reagents which break the C-C bond
CH2 CH2
\ /
CH2
Cyclohexane
Halogenation Halogen atom replaces 1 or more of the H atoms in an alkane by a substitution reaction(atom/group of atoms in a
molecule replaced by another atom/group of atoms)
- ALL halogens react with ALL alkanes, rate of reaction quicker with Cl than with Br than with I, I reaction results in equilibrium
- Rate of reaction decreases as the relative molecular mass of the alkane increases
Free radical substitution reaction substitution involving a free radical(species which have a single unpaired electron, highly
reactive)
Photochemical reaction Reaction initiated by light, Sunlight(UV)light energy essential for reactions to proceed at a reasonable rate
UV
C6H14 + Br2  C6H13Br + HBr(steamy acidic fumes)
Rapid decolourisation of bromine, if covered in a darkened test tube the colour remains longer
- All C-H bonds equivalent so no way of determining which H atom will be replaced
Initiation reactions Free radicals produced Photo dissociation Cl22Cl•
Homolytic bond fission (Covalent)bond splits equally to give 2 free radicals
X:Y  X• + Y•
Heterolytic bond fission Bond splits unequally and both electrons kept by one atom
Bond fission Bond breaking
Lysis Breakdown
Homo Same
Hetero Different
Heterolytic Bond cleavage where + & – ions produced
Homolytic Bond cleavage where 2 neutral species produced
Propagation reactions Free radicals used up and created in chain reaction
Cl• + CH4  CH3• + HCl
CH3• + Cl2  CH3Cl + Cl• etc until no more Cl2 / CH4 molecules
Termination reactions Free radicals mopped up
2 free radicals join together making a stable molecule, some products forming trace impurities in final sample
Lots of possible termination reactions
Cl• + Cl•  Cl2 or Cl• + CH3•  CH3Cl or CH3• + CH3•  CH3Cl + C2H6
More substitutions Dependent on amount of excess chlorine or methane
Excess chlorine Cl• free radicals start attacking chloromethane giving dichloromethane CH 2Cl2 trichloromethane CHCl3
tetrachloromethane CCl4
Excess methane Product will be mostly chloromethane
Alkenes CnH2n Unsaturated molecules because of C=C double bond
- Alkenes have lower Mt because less H atoms, so lower van der waals forces but more reactive because of C=C double bond
Structural Isomerism Occurs by moving the double bond to different positions in the C chain indicated by the number between the
prefix & -ene, number being the smallest possible counting from each end which takes precedence in numbering of the C atoms of the
longest C chain(isomer 3-methylbut-1-ene is not 2-methylbut-3-ene)
- 3 possibilities of C4H8
but-1-ene CH3CH2CH=CH2
(-cis or -trans)but-2-eneCH3CH=CHCH3
Addition reaction 2 molecules react together forming a single product
H H
H H
Electrophilic addition Addition reaction where, electrophile attacks a molecule at a region of high
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RT | |
electron density
–C=C– + A–B  –C–C–
- Electophilic addition reaction of alkenes at high electron density(pi bond, double bond is nucleophilic)
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pi bond breaks, bonds formed with reactant molecule
A B
Tests for unsaturation C=C bond
Pure bromine a safety hazard, avoided by dissolving it in an organic solvent(hexane)
RT, shake(as hydrocarbons are immiscible with water)
CH3CH=CHCH3 + Br2(aq)  CH3CHBrCHBrCH3 decolourised
But-2-ene
2,3-dibromobutane(dibromoalkane)
CH2=CH2 + H2O + [O]  HOCH2CH2OH
[O] oxygen from the oxidising agent
Ethene
ethane-1,2-diol
Oxidation of alkenes by purple alkaline KMnO4 (potassium manganate(VII)) (warning, other reducing agents give positive results)
• CH2=CH2 + 2MnO4– + 2OH–  HOCH2CH2OH + 2MnO42–
Manganate ions first reduced to green manganate ions, green solution
• 3CH2=CH2 + 2MnO4– + 4H2O  HOCH2CH2OH + 2MnO2 + 2OH–
(neutral or acidic conditions, no OH–/H+on LHS)
then to dark brown ppt manganese dioxide
Oxidation of alkenes by acidic purple KMnO4 turns colourless, manganate(VII) ions reduced to manganese(II) ions
5CH2=CH2 + 2H2O + 2MnO4– + 6H+  5HOCH2CH2OH+ 2Mn2+
Alkene with H2 Addition, reduction reaction
- Cheaper nickel catalyst used to convert unsaturated oils into saturated
fats for use in margarine
Finely divided nickel/platinum catalyst
H
H
150°C
H H
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/
High Pressure | |
C=C + H2(g) 
H–C–C–H
/
\
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H
H
H H
Ethene(Alkene)
Ethane(Alkane)
Alkene with HBr
RT Electrophilic addition
H2C=CH2 + HBr(not aq)  CH3CH2Br
Ethane
Bromoethane(Bromoalkane colourless liquid)
H2C=CHCH2CH3 + HBr 2 products

BrCH2CH2CH2CH3
1-bromobutane

CH3CHBrCH2CH3
2-bromobutane
Major product because more stable with more attached alkyl groups
Alkenes with H2SO4(catalyst) Electrophilic addition, hydrolysis
cold conc H2SO4
H2C=CH2  CH3CH2OH
Alkene with H2O(g)
Reaction yield low 5%, recycle unreacted gas to get yield 95%
300°C 60atm H3PO4(s) Phosphoric(V) acid catalyst
H2C=CH2(g) + H2O(g)
CH3CH2OH(g)
LDPE(Low Density Polythene) Polymer chains very branched, not packing
Poly(ethene)‘polythene’
Addition polymerisation, electrophilic addition reaction closely, amorphous(non crystalline)
• Strong • Flexible • Deformed by heat
(H2C=CH2)n

(–CH2–CH2–)n
Monomer ethene(repeating unit, essentially an alkane) - Used in bags, bottles, packaging, electrical insulation
- LDPE made by ethene at 2000atm 500K
Ziegler Natta catalysts are mixtures of titanium
HDPE(High Density Polythene) Polymer chains branched little, chains line up
compounds like titanium(III)chloride TiCl3 or
packing closely, crystalline structure
titanium(IV)chloride TiCl4 and compounds of
• Strong • Rigid • Not deformed by heat • Easy to mould
aluminium like aluminium triethyl Al(C2H5)3
- Used in water pipes, petrol tanks, containers, hospital equipment needing
sterilising
- HDPE made by ethene at 25atm 330K Ziegler Natta catalyst
- Polymerisation of propene produces stereoregular polymers, regular geometrical
Poly(propene)
arrangement of methyl groups in a spiral which stiffens the structure allowing the long
H CH3 Ziegler Natta
H CH3
molecules to line up close to one another, increasing
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( C=C )n

( –C–C– )n • Hardness • Wear-resistance, strength of the material • Softening temp
- Longer chains means greater van der waals forces which get tangled so less flexibility
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- Used to make ropes, sacking, carpets, fishing nets
H H
H H
Poly(chloroethene) PVC
H Cl
H Cl
- Amorphous, large chlorines stick out randomly from the chains so they don’t pack closely
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- PVC unusually hard & rigid because C–Cl bonds are polarised(Cl more EN than C)dipole-dipole
( C=C )n  ( –C–C– )n
interactions exist between chains
To make PVC more flexible, plasticisers are added
| |
| |
To make PVC more hard, mineral fillers are added
H H
H H
- Used for water pipes(rigidity, resistance to wear), rainwear, coating on electric cables(at high temps can Chloroethene
PVC
melt causing short circuits and give toxic chlorine containing compounds)
Poly(fluoroethene)PTFE or Teflon - Solid is • Hard, strong • Slippery • High Mt
F F
F F
- PTFE inert because • C–F bond is strong • Resistant to hydrolysis
| |
| |
• Big F atoms protect C chain from chemical attack
( C=C )n  ( –C–C– )n
- PTFE has high electron density due to F atoms and close packing,
| |
| |
Vdw forces stronger than HDPE
F F
F F
- Used in surface coating for non-stick ovenware, low friction bearings, seals, pipes, skis,
Tetrafluroethene Poly(tetrafluroethene)
stain–proofing of fabrics
- Polymers cause environmental problems when disposed, when burned, give off toxic fumes, not biodegradable so would accumulate in
rubbish tips never to disappear
- Polymers can be recycled(but expensive to sort out plastics)break waste polymers into smaller molecules by cracking, use the small
molecules as raw materials for making new polymers or other chemicals
Ideal fuel should 1 Be abundant Methane, butane, octane, coal, finite resources. Ethanol replaceable by fermentation of sugars of
vegetable origin, however if earth’s energy needs were to be met by ethanol it is debatable whether enough vegetable matter could be
produced.
Hydrogen obtained by electrolysis of water, burning it converts it back to its source, regenerated by electrolysis where oxygen used to
burn it is recovered. But fuel required to produce energy, and more than all the energy is used to recover the fuel
2 Be easy & safe to store & move (high EACT for combustion reaction) Transport of liquid fuel is dangerous but gases more so.
- Liquid fuels are easy to store/transport whereas gases will need to be stored under pressure in a special container as a liquid/in a large
container as a gas
Hydrogen is difficult, a highly flammable gas, can’t be liquefied under pressure at RT, the more stored the higher the pressure, the
stronger and heavier the container must be, so storage & transport of hydrogen is dangerous due to risk of explosion
Liquid oxygen & liquid nitrogen are transported on roads daily, light weight insulated containers safely ‘leak away’ excess pressure
3 Be non toxic Hydrogen & methane can be breathed in, in small amounts without harm
Butane, octane have narcotic and hallucinogenic properties in small amounts, and poisonous in larger quantities
Ethanol poisonous in small amounts, taken continuously causes long term damage to organs of the body
4 Have a high calorific value (high energy density) Hydrocarbons of increasing molecular mass are superior giving lots of energy due to
large amounts of bonds, uncompressed (hydrogen)gases are inferior
Hc(C8H18) = –5510kJmol–1 density 0.703gcm–3 Calculate the calorific value per gram and per cm3
1 mole C8H18 = 114g
1cm3 of octane has a mass of 0.703g
114g of octane yields 5510kJ on combustion
–1
1g yields 5510kJ/114g = 48.3kJg = calorific value
1cm3 yields 0.703g x 48.3kJg–1 = 34.0kJcm–3 = calorific value
5 Give rise to harmless combustion products (little pollution)
Hydrocarbons can give CO(poisonous) and carbon(soot) if there’s insufficient air or fuel is not correctly mixed
High temp of combustion can form toxic nitrogen oxides from the air used to burn the fuel: N2(g) + O2(g)  2NO(g) followed in
atmosphere by 2NO(g) + O2(g)  2NO2(g) contributing to formation of acid rain, as nitric acid and by oxidising sulphur dioxide to
sulphuric acid
H
H
H H H H
(1)(i)Write the full structural formula of buta-1,3-diene
H
C
C
C
C
H
H C C C C H
(ii)Give structural formula of product of reaction of buta-1,3-diene with alkaline
solution of potassium manganate(VII)
H
H
OH OH OH OH
(i)
(ii)
(2)(a)Explain the difference in reactivity of ethane and ethene with bromine in terms bonding
(a)Ethane single bonds/sigma only • C-H must be broken • Ethene also has sigma and pi bonds • where electrons are more
accessible/pi bond is weaker(and breaks)
(3)(a)State with reason the empirical formula of polypropene (a)CH2 as polymer made by addition reaction/no loss of small molecules
(4)(a)Write equation for complete combustion of hydrogen
(a)2H2 + O2  2H2O
(b)Why does methane not react with air unless a flame or spark is applied to the mixture?
(b) • Activation energy needed • before reaction can proceed at reasonable rate
(5)What does methane do in the industrial production of ammonia? (5) • Source of hydrogen • Source of heat (energy)to run process
(6)(i)Draw the structural formula of a compound which is an isomer of but-2-ene but which does not show geometric isomerism
(ii)Explain why the isomer drawn in (i) does not show geometric isomerism
H
H
C 2 H5
C
C
H
C
C
H
C
CH3
H
C
C
CH 3
CH 2
CH 2
CH 2
CH 2
H
H
CH3
H
H
(6)(i) H
(ii)One end of C=C bond has 2 identical atoms/groups attached OR if cyclobutane, no C=C
(7)Draw full structural formulae for
(i)The organic product of the reaction of ethene C2H4 with
(aq)potassium manganate(VII) and H2SO4
(ii)3,4-dimethy1hex-2-ene
H
H
H
C
C
OH OH
H
H
H
C
H
H
H
C
C
C
H
H
H
C
C
C
H
H
H
H
H
C
H
H
H
(7)(i)
(ii)
(8)(a)State the relationship between 2,2,4-trimethylpentane and octane
(a)structural isomers
(b)Octane has to be vaporised before burning in an engine. Determine fuel to air ratio by volume for complete combustion of octane(g)
(b)2C8H18 + 25O2  16CO2 + 18H2O
air is 20%O2 therefore
2:125
(c)Lead tetraethyl used to be added to petrol to boost its Relative Octane Number but this has now been replaced by compounds such as
benzene or MTBE(not so effective as the lead compound and so need to be added in larger quantities causing solubility problems)
(i)Why has the addition of lead tetraethyl to petrol been stopped in the UK? (i)lead is poisonous/ruins catalyst in catalytic converters
(ii)How might the difficulty in keeping MTBE in solution in the petrol cause a problem in the running of the car?
(ii)RON may not be maintained OR fuel can cause knocking/pre-ignition
(iii)Apart from solubility, state one problem associated with the use of benzene as an additive to petrol
(iii)benzene is carcinogenic
(9)(a)Draw a diagram to show the shape of the chloromethane molecule and explain why it has this shape
(b)Explain why the Bt of chloromethane is higher than that of methane
(c)Explain why the Bt of methanol is higher than that of chloromethane
H
(a)4 pairs of electrons around C arranged to minimise repulsion
(b)chloromethane has a (permanent)dipole, methane only has van der Waals forces
H
Cl
• attraction(forces) between dipoles stronger than Vdw in CH4 Increase in number of electrons in molecule, causes
(a) H
increase in VDW forces of attraction between molecules
(c)H bonding in methanol between molecules stronger than dipole-dipole/VDW forces
(10) Ethane and chlorine react when exposed to light
C
H
CH 3
C H
H
H
+
Cl
CH 3
C
+
H
Cl
CH 3
C
H
+ Cl
Cl
CH 3
C
Cl
+ Cl
H
H
H
H
step 1
step 2
(a)Explain the movement of the C-H bond electron pair in step 1
(a)1 electron goes to the C atom (to form a radical) the other goes to form a bond with the Cl atom
Haloalkanes CnH2n + 1X Compounds formed when a member of the halogen group is substituted into an alkane
Number indicating halogen position on C chain- halogen name- alkane name
H
H H H
|
| | |
CH3Cl
Chloromethane
Carbon–Halogen bond is polar
Cl–C–Cl
H–C–C–C–H
\ δ+ δ–
CH3CH2Br
Bromoethane
|
| | |
CH3CH2CH2Br 1-bromopropane – C  X (halogen more EN than carbon)
Cl
H I H
/ displacement of bonding electron pair
CH3CHBrCH3
2-bromopropane
trichloromethane
2-iodopropane
CH2Br
CH3
F Br
|
|
| |
CH3–CH–CH3
CH3–C–CH2–Br
F–C–C–H
1-bromo-2-methylpropane
|
| |
CH3
F Cl
1-bromo-2,2-dimethylpropane 1,3-dimethylcyclopentane 2-bromo-2-chloro-1,1,1-trifluoroethane
Functional group An atom/group of atoms/structural feature in a molecule which has chemical properties not shown by an
alkane(determines reactions)
- All reactions of haloalkanes are reactions of the halogen atom(functional group) Atoms other than C and H are more reactive than
hydrocarbon chain which can only react like alkanes. Carbon–halogen bond easier to break than C–H bonds
Structural isomerism • Moving the halogen atom to different positions on the C chain • Branching of the C chain in larger molecules
Hydrocarbon chain to which a functional group is attached can exist in one of 3 possible forms
R1 R2 R3 are alkyl groups which maybe the same or different but must contain at least one C atom
- Same types of reaction for all 3 types of haloalkane but difference in rate of reaction
- No compound with fewer than 4 C atoms per molecule can form a tertiary C compound
H No fewer than 2H atoms attached
H One H atom attached
R3 No H atom on functional group C atom
|
to functional group C atom
|
to functional group C atom
|
R1–C–X
R1–C–X
R1–C–X
|
–CH2X primary 1°
|
–CHX secondary 2°
|
–CX tertiary 3°
H
R2
R2
Rates of reaction of haloalkane depends on:
• Nature of the halogen Iodides

Bromides

Chlorides
React quickest because
React slowest because
- Largest atoms
- Smallest atoms
- C–I bond is longer easier to break
- Larger bond energy
- Lower bond energy
• Type of haloalkane
1° compound

2° compound

3° compound
Reacts quickest
Reacts slowest
- Haloalkanes are polarised molecules, Mt higher than alkanes of the same length, dipole-dipole interactions exist between chains
Elimination reaction Elements of a simple molecule(H2O)are removed from the organic molecule and not replaced by any other
atom/group of atoms
Nucleophilic substitution δ+ C atom can be attacked by a nucleophile
OH–, CN–, NH3 nucleophiles which react with haloalkanes
:OH– provides a pair of electrons for C
C–Br bond breaks
–
heterolytically, both electrons from the bond taken by Br then OH– bonds to C
Test for haloalkanes
nucleophilic substitution reaction
• Warm haloalkane with NaOH(aq)
CnH2n + 1X + OH–(nucleophile) CnH2n + 1OH + X–(halide ion)
• Silver nitrate test, Acidify with (dil)nitric acid to remove excess OH– which could react with Ag+
• Add silver nitrate(aq) Ag+(aq) + X–(aq)  AgX(s)
Result: - AgCl, White ppt, dissolves in (dil)NH3 to give a colourless solution
- AgBr, Cream ppt, partially dissolves in (dil)NH3 but dissolves in (conc)NH3 to give a colourless solution
- AgI, Yellow ppt, insoluble in NH3 solution of any concentration
C–Cl bonds in compounds used as herbicides, weedkillers for crops, insecticides, DDT, C–Cl bond is sufficiently inert to give compound
longish life but survival in environment causes problems as DDT is toxic to insect eating birds which accumulates toxic compound in
body fat
Haloalkane reaction with NaOH/KOH
2-bromopropane (haloalkane)
CH3CHBrCH3 + OH–
Heat under reflux, hydrolysis(halogen atom displaced by OH group)
KOH/NaOH(aq)
KOH/NaOH(ethanol)
Aqueous conditions favour nucleophilic substitution,
Anhydrous conditions favour elimination,
OH– acts as a nucleophile
OH– acts as a base
–
Propan-2-ol(Alcohol)CH3CHOHCH3 + Br
Propene(Alkene) CH3CH=CH2 + Br– + H2O
1-chloropropane CH3CH2CH2Cl(l) + NaOH(aq)  propan-1-ol CH3CH2CH2OH(aq) + NaCl(aq)
CH3CH2CH2Cl + OH –  CH3CH2CH2OH + Cl –
CH3CHBrCH3 + KOH(ethanol)  CH3CH=CH2 + KBr + H2O
Depending on which H atom is removed it’s possible to form 2 different alkenes in the same reaction
H Br
Br H
|
|
Heat under reflux
| |
Heat under reflux
CH2CHCH2CH3 + KOH(ethanol)  CH2=CHCH2CH3 or
CH3CHCHCH3 + KOH(ethanol)  CH3CH=CHCH3
2-bromobutane
but-1-ene
2-bromobutane
but-2-ene
Haloalkane reaction with NH3(ethanol) Nucleophilic substitution reaction :OH– :NH3 nucleophiles, :OH– stonger base than :NH3
heat in sealed container
heated in sealed container
(ethanol)H3N: + RI  (primary amine)RNH2 + HI
CH3CH2Br + NH3(ethanol)  CH3CH2NH2 + HBr
ethylamine
For every ammonia molecule reacting in this way one maybe prevented from reacting by the acid produced
H3N: + HI  NH4+ I – Ammonia ion no longer a nucleophile, thus minimum of 2 moles of ammonia to every mole of haloalkane
2H3N + RI  RNH2 + NH4I More ammonia must be used because amine product is a nucleophile
RN:H2 + RI  (secondary amine) R2N:H + HI
- Anhydrous conditions, ammonia dissolved in alcohol not water as water is another competing nucleophile
NH3 + H2O
NH4+ + OH–
- Huge excess of cheap ammonia used, reaction normally done in a closed vessel under pressure because of
• volatility of ammonia • need to heat
- Boiling under reflux not suitable for reactions of haloalkanes with ammonia because, ammonia is a volatile material boiling well
below RT and escaping from top of condenser as Bt of ammonia is well below temp of water in cooling jacket of condenser
Haloalkane reaction with KCN(ethanolic)
Heat under reflux, Nucleophilic substitution
CH3CH2I
+
CN – 
CH3CH2CN
CH3CHBrCH3 + KCN(ethanol)  CH3CHCNCH3 + KBr
Iodoethane(Haloalkane)
Nucleophile
Propanenitrile(Cyanides/Nitriles)
Heat acid, Hydrolysis
(Nitrile)CH3CH2CN + 2H2O

(Carboxylic acid)CH3CH2COOH + NH3
(1)(a)Why is the rate of reaction slower with bromobutane than with iodoethane
(a)C–Br bond stronger than C–I • Ea for C–Br is higher than C–I
(b)CH3CHBrCH2CH3 + KOH  CH3CHOHCH2CH3 + KBr Experiment repeated except 2-iodobutane replaced 2-bromobutane
Explain effects that this change would have on rate of reaction (b)Rate increased C-I bond weaker(than C-Br bond)/lower bond energy
(2)Product C2H4Br2 is a bromoalkane. Suggest the structural formulae of each of the products of the reaction of C 2H4Br2 with the
reagents given below and identify the type of reaction involved
(i)NaOH(aq)Product? Type of reaction?
(ii)NaOH(ethanol) (heated under reflux) Product? Type of reaction?
H
H
H
C
C
OH OH
H or H
H
H
C
C
Br OH
CH 3
H
H
H
(1)
C
C
H
or
H
C
C
H
C
Br
C
H
(1)
CH 3 CH 3
CH3
H
H
A
(i) nucleophilic substitution
(ii) elimination
(iii)Suggest, giving the reagents and conditions, how compound A could be converted in 2 steps into compound B
(iii)Step 1 is production of halogen intermediate(HCl)at RT
Step 2 alcohol or hydrogen sulphate followed by alcoholic KCN • heat under reflux
Aldehydes & Ketones CnH2nO R1 & R2 are alkyl groups which maybe the same or different
Aldehydes
Aldehydes
Alkanal
Ketones
Ketones
HCHO
Methanal C atom in functional group
CH3COCH3
CH3CHO
Ethanal
CH3CH2COCH3
CH3CH2CHO
Propanal
CH3CH2CH2COCH3
CH3CH2CH2CHO Butanal
CH3CH2COCH2CH3
2-methylpropanal
1
2
R1 can be a H atom (CH3)2CHCHO
R & R must
contain at least
one C atom
C
C
H
CN
H
B
Alkanone
Propanone
Butanone
Pentan-2-one
Pentan-3-one
Test for carbonyl(C=O)group(aldehydes & ketones)
Add excess of(2, 4-dinitrophenylhydrazine)Brady’s reagent solution and ppt of orange/yellow
Test for aldehydes(-CHO) group(not ketones) Aldehydes are good reducing agents unlike ketones –CHO + [O]  -COOH
• Fehling’s/Benedict’s solution both reduced from blue Cu2+ complexes  brick-red Cu2O (copper(I) oxide)
• Tollen’s reagent [Ag(NH3)2]+ when warmed is reduced to silver (silver coats inside of apparatus)
(Making Tollen’s reagent: few drops of dil NaOH to silver nitrate solution, then adding dil ammonia solution until brown ppt dissolves)
• Heated under reflux, potassium dichromate(VI) solution changes from orange  green
Carboxylic Acids CnH2n + 1COOH
Alkanoic Acid
n = 0 HCOOH
Methanoic acid
n = 1 CH3COOH
Ethanoic acid
CH3CH2COOH
Propanoic acid
CH3CH2CH2COOH
Butanoic acid
Alcohols CnH2n + 1OH Number inserted before –ol indicates position of –OH(functional group)on C chain
CH3OH
Methanol
Fuel(unleaded petrol), plastics, dyes
CH3CH2OH
Ethanol
Alcoholic drinks, methylated drinks, fuel, as a solvent in perfumes/cosmetics
CH3CH2CH2OH Propan-1-ol
CH3CHOHCH3 Propan-2-ol
1° alcohol C atom which carries the –OH
2° alcohol C atom which carries the –OH
3° alcohol
C atom which carries
group attached to only 1 alkyl group
group attached to only 2 alkyl groups
the –OH group attached to 3 alkyl groups
Combustion, oxidation reaction
CH3CH2OH(l) + 3O2(g)  2CO2(g) + 3H2O(g)
Test for –OH compound Sample of alcohol in a clean dry test tube(steamy acidic fumes with any –OH compound, water contains
–OH group producing fumes of HCl which invalidates test)
RT
RT
ROH + PCl5(s)  RCl + POCl3 + HCl
CH3CH2OH + PCl5(s)  CH3CH2Cl + POCl3 + HCl
Phosphorus pentachloride
Ethanol
Chloroethane
Heat under reflux
2P + 3I2  2PI3(l)
PI 3(l) + 3ROH(l)  H3PO3(l) + 3RI(l) Iodoalkane
Moist red phosphorus
Phosphorus trihalide
(distilled off)
Nucleophilic substitution reaction
ROH + HX
RX + H2O
Hal– nucleophile, attracted towards polarised C atom in Cδ+–Oδ– and –OH snaps off
- HI acid not a laboratory reagent since it’s rapidly oxidised in air
- HBr made in situ(in the place where reaction is occurring) alcohol mixed with Na/KBr + concH2SO4
Reflux
KBr + (conc)H2SO4  KHSO4 + HBr
HBr(g) + CH3CH2OH(l)  CH3CH2Br(l) + H2O(l)
3° alcohols react reasonably rapidly with conc HCl but 1°,2° alcohols react slow
OH
Cl
|
Shake at RT
|
CH3– C – CH3 + conc HCl

CH3– C – CH3 + H2O
|
|
CH3
CH3
2-methyl-propan-2-ol
2-chloro-2-methylpropane
3° alcohol
3° haloalkane
-OH with Na produces alkoxides
2CH3CH2OH(l) + 2Na(s)  2CH3CH2O– Na+ + H2(g)
The longer the alcohol chain, the less reactive it is with Na
Ethanol
Ionic sodium ethoxide(colourless solution)
Polar –OH groups on alcohols helps them to form H bonds
H bonds form between –OH and H2O
Larger the alcohols, the less miscibility in water, larger alcohols are mostly non-polar C chains, so less attraction for polar H2O
because of H bonding alcohols have high Bt compared to non-polar compounds(alkanes of similar size, weaker Vdw forces)
Isomeric alkenes with more than one adjacent CH group
Dehydration
(alcohol should have an H atom on α–C atom(C atom next to
H OH H H
that which carries the –OH))
| | | |
Elimination reaction
Butan-2-ol H–C–C–C–C–H
| | | |
H H H H
170°C excess(conc)H2SO4/70°C H3PO4 phosphoric(V)acid
Method 1 Heating alcohol with excess conc H2SO4
(dehydrating agent in elimination reaction)
Alkene produced collected over water
Method 2 Hot alcohol vapour passed over a hot catalyst of
aluminium oxide(vapour phase dehydration, air must be
excluded)Al2O3 catalyst(large surface area for reaction,
crushing into blocks/powder)
Al2O3
Ethanol CH3CH2OH  Ethene CH2=CH2 + H2O(g)
Catalyst conc H2SO4 is a strong oxidising agent and oxidises some of the
alcohol to CO2 and is reduced itself to SO2 both gases needing removal
from the alkene, therefore gases passed through NaOH(aq). Alkene is
collected over water
[O] = (one atom of) oxygen has been added from an oxidising agent
• Reactants aren’t mixed in bulk at start because of vigour of the reaction.
Ethanol and conc H2SO4 slowly added from a separating funnel to [O]
• 1° alcohol converted to an aldehyde if excess alcohol is used(so there’s insufficient oxidising agent to carry out 2nd stage) and aldehyde
formed is distilled off immediately before further oxidation(aldehyde boils at lower temp than alcohol). Fractional distillation for
increasing purity of ethanal and obtaining a good specimen
3CH3CH2OH + Cr2O72– + 8H+  3CH3CHO + 2Cr3+ + 7H2O
ethanol
ethanal
• 1° alcohol converted directly to carboxylic acid by heating under reflux with excess[O]ensuring reaction does not stop at aldehyde stage
3CH3CH2OH + 2Cr2O72– + 16H+  3CH3COOH + 4Cr3+ + 11H2O
simplified
CH3CH2OH + 2[O]  3CH3COOH + H2O
ethanol
ethanoic acid
Yield obtained x 100
% Yield = Theoretical Yield
Organic preparations are designed to consume a material completely because
• Material difficult to remove unchanged from the product • Material is expensive
Calculate the theoretical yield of bromoethane from 12g of potassium bromide CH3CH2OH + KBr + H2SO4  CH3CH2Br +
KHSO4 + H2O
Mr(KBr) = 119gmol –1
Amount of KBr = 12g/119gmol –1 = 0.101mol = Amount of CH3CH2Br
Mr(CH3CH2Br) = 109gmol –1
Theoretical yield = 109gmol –1 x 0.101mol = 11g
If 5g of bromoethane is obtained then %Yield = 5 x 100/11 = 45%
(1)Alcohol Z heated with (conc)sulphuric acid, gas Y produced which reacts with bromine solution decolourising it.
Draw functional group in Y
(1)C=C
(a)(i)Give the reagent and the conditions needed for step 1
(a)(i)HBr(g)
step 1
C 3 H6
C 3 H 7 Br (major product) (ii)State the type of reaction in, step 1, the conversion of S to P
Propene
S
(ii)electrophilic addition, nucleophilic substitution/hydrolysis
step 2
(b)(i)Give the reagent and the conditions needed for step 2
KMnO 4 in
aqueous NaOH
(b)(i)170°C (conc)H2SO4/70°C H3PO4/aluminium oxide heat
alkali
(ii)Give the reagents and the conditions needed for step 3
C3 H8 O
(ii)H2SO4 acidified potassium dichromate(VI) heat under reflux
Q
P
(c)Give the structural formula for Q (c)CH3CH(OH)CH2OH
(d)Draw the full structural formula showing all the bonds for the isomer of
step 3
butan-2-ol that is a tertiary alcohol
oxidation
CH3 COCH 3
H
H
H
C
H
CH 3
H
or
H
(d)Methylpropan–2–ol
C
C
C
H
OH H
H
CH 3
C
CH 3
OH
Four of the structural isomers of C4H10O are alcohols. One of these isomers is butan-2-ol
(a)Draw the structural formulae of 2 other alcohols with molecular formula C 4H10O and name them
H
H
H
H
C
C
C
C
(a)
H H
Butan-1-ol
H
H
H
OH
H
H
H
H C H
H
H
H C H
H
H
C
C
C
H
OH H
H
H
(2)-methylpropan-1-ol,
C
C
C
H
H
H
OH
(2)-methylpropan-2-ol
(b)butan-2-ol heated with H2SO4 acid and potassium dichromate(VI), name the product and name type of reaction
(b)butan-2-one, oxidation/redox
K 2 Cr 2O 7 in
•None of the compounds in the scheme shows cis-trans isomerism
dilute sulphuric
•D reacts with KCN to form 2-methylpropanonitrile
C 3 H 8 O acid
C3 H6 O
•An isomer of A will form C by the same route but will not produce B by reaction with potassium
A
B
dichromate(VI) acidified with (dil) H2SO4 Instead it makes E, C3H6O2
Identify using a name or structural formulae:A,B,C,D,E
conc sulphuric acid
C
A Propan-2-ol
B Propanone
C Propene
D 2-bromopropane
E Propanoic acid
D
HBr
(a)Propan-1-ol to 2-bromopropane in 2 stages?
(a)CH3CH2CH2OH  CH3CH=CH2  CH3CHBrCH3
One of the isomers of C4H10O is the alcohol 2-methylpropan-2-ol (b)Draw the structural formula of the final organic product of the
which has the structural formula
reaction when each of the three alcohols in (a)(i) is heated under
CH 3
reflux with a solution of potassium dichromate(VI) in (dil)H 2SO4
CH 3
C
CH 3
H
OH
(a)Draw structural formulae of 3 other structural isomers of
C4H10O which are also alcohols
H
H
H
H
H
C
C
C
C
H
H
H
H
OH
H
H
H
H
H
C
C
C
C
H
H
OH H
H
C
C
C
H
H
H
H
H
H
H
C
C
C
C
H
H
H
H
H
H
H
H
C
C
C
C
H
H
OH H
OH
H
H
H
H
H
H
OH
C
C
C
C
H
H
H
H
H
C
C
C
C
H
H
O
H
H
CH 3
C
C
C
O
H
H
OH
O
(1)
H
H
(1)
(b)
H
H
H
H C H
(a)
H
H
OH
H
H
CH3 H
C
C
C
H
H
H
OH
H
(1)
Propanone Bt 56°C prepared by oxidising propan-2-ol Bt 82°C (H2SO4 potassium dichromate(VI)) Boil mixture for 15minutes
(a)What safety precaution in addition to wearing eye protection must be taken when adding the concH2SO4?
(a)Gloves/add slowly/cool while adding NOT lab coats/be careful not to spill on hands/do in fume cupboard
(b)Outline how you would obtain propanone from this aqueous mixture after it had been boiled for 15 minutes
(b)(Fractionally)distil, collect fraction that distils at 55-57°C
• Catalysts provide an alternative route for a reaction which has
a lower Eact than normal route, thus more molecules having
enough energy to overcome the Eact
- Catalyst has no effect on H of the reaction or amounts of
products
- Catalysts used in petroleum industry(longer chain alkanes are
cracked into shorter more useful molecules using zedite catalyst)
S2O82–(aq) + 2I–(aq)  2SO42–(aq) + I2(aq)
Catalysed reaction:
Step1 S2O82–(aq) + 2Fe2+(aq)  2SO42–(aq) + 2Fe3+(aq)
Step2 2Fe3+(aq) + 2I–(aq)  2Fe2+(aq) + I2(aq)
Catalyst
Factors which affect rate of chemical reaction
• Concentration of reactants in solution • Temperature • Pressure of any gases present • Surface area of any solid reactants
• Light(certain reactions)
• Catalysts substances which increases rate of chemical reaction without being chemically changed themselves
2 theories of kinetics, collision theory and transition state theory
Collision theory Before 2 particles react they must collide, only a small fraction of the total collisions results in a reaction because:
Molecules must have correct Direction of approach (Steric factor )2 molecules traveling in approx same direction with high KE may
not produce a high energy collision
Orientation If bulk of molecule protects functional group from attack in a high energy collision then a reaction may not occur
Activation energy Certain minimum amount of KE that must be provided to reactants in a chemical reaction so they can start a
reaction (reach an activated state from which products can form)
Reaction rate increased if:
• Concentration, pressure increased More particles in a given volume, collisions occur more frequently
• Surface area increased More collisions between solid surface and other reactant
• Temperature increased Proportion of molecules having E act increased, KE’s of molecules are increased along with frequency of
collisions, overall increase in number of effective collisions per second
• Shaking a container helps to increase the rate of heterogenous reactions(solid settling out from a liquid) Shaking brings the
reactants together increasing frequency of collision
Maxwell Boltzmann distributions of molecular energy in a
Maxwell Boltzmann distribution Way in which energies are
sample of gas at T1 & T2
distributed, molecules in a gas/liquid don’t all have the same KE
since they don’t all have the same speed, only a few molecules
will have very low/high energies most being around the most
common value represented by the peak
- As temperature increases • Curve broadens out
• Peak value decreases moving towards a higher energy value
• Area under curve represents total number of molecules in
sample and is therefore constant
• Curve starts at (0,0) because no molecules have 0 energy
• Area under the curve beyond Eact represents number of
molecules having energies greater than or equal to E act which
increases with temperature and rate of reaction
Transition state theory As 2 molecules approach each other, repulsion between their electron clouds will push them apart unless they
have sufficient KE to overcome the repulsion.
If they do get sufficiently close to each other, a rearrangement of electron clouds will occur so that some bonds are broken and new
bonds form. While this is happening a highly unstable species is formed in which some bonds are partially broken/formed(activated
complex) and KE of collision is converted into PE
Thermodynamic stability A mixture is thermodynamically unstable Either as the reaction is very exothermic Or energy level of
products below energy level of reactants
- A substance(mixture) mostly converted into something else at equilibrium is thermodynamically unstable(exothermic reaction, products
more stable than reactants) Free energy needs to considered in determining whether or not a reaction is spontaneous(forward reaction)
Kinetic stability Reaction is kinetically stable when Eact is high (When reactants fail to undergo thermodynamically feasible reactions
they are kinetically stable)
(1)(a)If a mixture is thermodynamically stable, can it be kinetically stable?
(1)(a)No, kinetic stability describes a condition in which a thermodynamically unstable substance/mixture fails to react. If a
substance/mixture is thermodynamically stable it isn’t going to react(to give the products with which its stability is being compared)
(b)Explain the terms thermodynamic and kinetic stability with reference to the combustion of graphite
(b)• Reactants are at a higher energy level(than products) • So CO2 products are thermodynamically stable with respect to reactants
• C(+O2)activation energy high enough to prevent appreciable reaction at RT so mixture/C(+O2) is kinetically stable
(c)CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) Hc(CH4(g))= –890 kJ mol–1 Methane doesn’t burn unless lit
Use this information to explain the difference between thermodynamic and kinetic stability
(c) • CH4(and O2)/reactants thermodynamically unstable with respect to products
• Since reactants are at a higher energy level(than products) • Reactants/methane, oxygen are kinetically stable due to high Ea
(d)Why would a reaction expected to take place, proceed slowly as to appear never to happen?
(d)• Reaction has high Eact • Reactants kinetically stable
(e)CH4 (l) + 2O2 (g)  CO2 (g) + 2H2O (g) H θ = -890kJmol –1 Why would you expect the reaction to take place?
(e)Products more thermodynamically stable than reactants
(c)A mixture of nitrogen and hydrogen is kinetically stable at 25°C but kinetically unstable at 400°C. Explain why
(c)• At 25C few collisions have E  Ea/sufficient energy to react • At higher temp(average)energy of molecules increases
• At 400C more molecules/collisions have E  Ea/sufficient energy to react
Reversible reaction
A reaction that doesn’t go to completion and occurs in both the forward and reverse direction
Rate of reaction from left to right starts at a certain level but decreases as the
concentrations of the reactants decrease. Reverse reaction will not start until some
products have formed. Point reached when forward and reverse reactions will be
occurring at the same rate, where concentrations of substances will remain constant but
reactions haven’t stopped
Dynamic equilibrium
Rate of forward reaction and reverse
reactions equal and there are no concentration changes
• Dynamic equilibrium will only be achieved in a closed system(rates of
condensation & evaporation are equal, if gas escaped then reverse
reaction couldn’t occur)
• Same equilibrium achieved provided temperature is constant
Position of equilibrium Extent of a reaction when equilibrium is established. If reaction uses more than 50% of reactants before
reaching equil then POE lies to RHS
Factors which might affect POE: Changing concentration on POE
nA + mB + xC
pP + yQ
Increasing concentration of a reactant and POE moves to RHS. Increasing concentration of A means more A will react with B & C
giving more P & Q until equilibrium is restored.
Argument applies safely to concentration NOT amount, A + B
C + D Adding more A(all gases at atmospheric pressure)increases
total vol, conc of A increases and reduces concentrations of all other species
Changing pressure on POE • Only affects gaseous equilibria • And only if there is a change in total number of molecules
Increasing pressure on an equilibrium mixture pushes POE towards side with smaller number of molecules
Increasing pressure moves POE to RHS in this reaction
N2(g) + 3H2(g)
2NH3(g)
4 molecules
2 molecules
Changing temperature on POE Affects reactions which involve a H If temperature raised POE moves in the endothermic direction
In an endothermic reaction hydrogen formed favours high temperatures H2O(g) + CO(g)
H2(g) + CO2(g) H = + 41kJmol –1
Rate of attainment of equilibrium • Increase in concentration, temperature, pressure(if gases), increases rate of attainment of equilibrium
• Rates of both forward and reverse reactions increase(lower activation routes for both)so catalyst will increase rate at which equilibrium
is established but POE remains unaltered
Catalytic converter in the exhaust system of a car engine 2NO(g) + 2CO(g)
N2(g) + 2CO2(g)
(i)Which way will the POE shift if temp were lowered (i) • To RHS • Exothermic direction
(ii)Gases from engine aren’t cooled before entering converter, why? (ii)• Rate of reaction would be too slow if cooled • Yield too small
(a)Effect of catalyst on rate of reaction? (a)• Alternative routes • Lower Eact • Increase in rate because more successful collisions
(b)Why is the catalyst in the form of a gauze/mesh? (b)Increased surface area
(c)Explain why: When a tiny electric spark is produced in a mixture of methane and oxygen at 10°C the heat transferred is NOT
sufficient to raise the temperature of the mixture by even 1°C yet the reaction occurs with explosive violence
(c)When a tiny electric spark passes through the mixture molecules in immediate vicinity obtain enough energy to react. The resulting
highly exothermic reaction raises the temperature of the surrounding molecules which also react and a wave of reaction passes outward
from the spark until reaction is complete
(d)Suggest why neutralisation of HCl by NaOH solution is virtually instantaneous
(d)Enormous speed of neutralisation of acids by alkalis as Eact is low, since no bonds need to be broken and reaction occurs by collision
of oppositely charged ions which are drawn together by electrostatic attraction H +(aq) + OH– (aq)  H2O(l)
(e)Why do concentrations of substances in an equilibrium mixture remain constant? (e)Rates of forward and back reactions the same
(f)Industrial catalysts(platinum) are finely divided, and fine wire or mesh is preferred to powder, why?
(f)Fine powders, unless “stuck” on a support, either block the flow of gases or tend to be blown away and may contaminate the product
(g)Although speed is important in industry there is often quite a low upper limit to the temperature of industrial reactions, why?
(g)Many industrial reactions are exothermic equilibria. Raising the temperature increases the reaction rate but makes the POE less
favourable. A compromise has to be found between these conflicting effects(other methods of increasing the rate maybe used, such as
increasing the pressure/concentration, using catalysts)
(h)Sketch Maxwell-Boltzmann distribution representing energies of the molecules in the Ostwald reaction system at a given temp
Fraction
of
molecules
E a cat
E a uncat
Energy
(h)
2g of Mg ribbon reacts with 100cm³ of H 2SO4 acid 1moldm–3
(a)Explain why the hydrogen is produced at a faster rate at the beginning of the experiment than it is at the end of the experiment
• Higher conc of acid • Greater surface area of magnesium • More collisions per second and therefore a faster rate
Haber Bosh process N2(g) + 3H2(g)
2NH3(g) H = 92kJmol –1 200atm 400°C Iron catalyst
Raw materials: CH4(g) + H2O(g)  CO(g) + 3H2(g)
or
C(s) + H2O(g)  CO(g) + H2(g)
natural gas steam Ni catalyst hydrogen obtained
Coal
CO would poison the catalyst, CO removed through air injection, CO oxidised to CO2 and removed
Economic factors in increasing rate of reaction & yield: • Containers operating at high pressures are expensive to build and maintain,
high costs exceed value of extra product. Reduction in temp slows down rate at which equilibrium is attained but catalysts usually last
longer at lower temps
- Catalysts increase rate of reaction & rate of attainment of equil but doesn’t affect POE, they’re susceptible to poisoning by impurities so
incoming gases are purified
- Equilibrium mixture formed is passed into refrigeration plant because it allows NH 3 to be liquefied and removed, unreacted gases can
be recycled if product can be removed, thus allowing a continuous flow process (NH3(g)removed, N2(g) + 3H2(g) recycled)
- Compromise of conditions pushing POE to RHS balancing against keeping rate of reaction at a reasonable level
- Periodically gases in Haber plant are ‘purged’(cleared out)and fresh gas put in. Air contains 1% argon which builds up if it isn’t
removed(with traces of other noble gases)
Uses:NH3 used in detergents production of nitric acid and fertilisers NH4NO3
Ostwald process Nitric acid(HNO3)manufacture
5atm 1000K Platinum rhodium catalyst
- Combustion of CH4 gives CO2 + H2O, oxides (-ve)Hf thermodynamically stable with respect to their elements
- NH3 gives N2 + H2O, all common oxides of nitrogen have (+ve)Hf less thermodynamically stable with respect to their elements
- Ammonia not a flammable gas in air but will burn/oxidise in oxygen 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) H = –1636kJmol –1
Excess of hot air ensures complete oxidation of NH3 


Platinum rhodium catalyst
- Too low a temp and slower reaction may not go to completion, presence of NH 3 in subsequent stages of nitric acid production disastrous
Too high a temp and some NH3 oxidised to nitrogen + water, an economic loss of NH3
- Cooling of gases(cold air)as it leaves the catalyst before resulting gases are absorbed in cold water
2NO(g) + O2(g)
2g)H = –94kJmol –1 pushing equilibrium to NO2 since reaction is exothermic
- Passed into water with excess air
4NO2(g) + O2(g) + 2H2O(l) 4HNO3(aq)
NH3(aq) + HNO3(aq)  NH4NO3(aq) (water soluble)powerful oxidising agent, dangerously explosive
ammonium nitrate used in agriculture/fertiliser
(1)Ammonia boils at -330°C but the liquefaction plant uses cold water to remove NH 3 from the gas stream?
(1)NH3 not dissolved in water, gas mixture at 200-300atm where Bt of NH3 is wabove RT thus easily liquefied at temp of cold water
(2)Industrially this reaction doesn’t usually achieve equilibrium 2NO(g) + O 2(g)
2NO2(g) why?
(2)Products removed from reaction system, not in the system long enough
Contact process Sulphuric acid manufacture
4atm 440°C vanadium(V)oxide catalyst (V2O5)
S(s) + O2(g)  SO2(g)
or
2ZnS(s) + 3O2(g)  2ZnO(s) + 2SO2(g)

Burn in excess air
sulphide ore
heat(roast)
2SO2(g) + O2(g)
2SO3(g) H = –200kJmol –1

oxidation V2O5
SO3 reaction with water is an uncontrollable reaction creating a fog of sulphuric acid so H2SO4(l) + SO3(g)  H2S2O7(l) Oleum
H2O(l) + H2S2O7(l)  2H2SO4(l) H = –130kJmol –1
- Provided temperature is not too high little gained by high pressures with the difficulty and cost
- Excess air and sufficient pressure to push gases round the plant ensures efficient oxidation
- Competition between low rates/temp with favourable POE and high rates/temp with less favourable POE
- SO2 ‘scrubbed’ with water before returning to the atmosphere
2NH3(aq) + H2SO4(aq)  (NH4)2SO4(aq) (ammonium salt)fertiliser NH3(aq) + H+(aq)  NH4+(aq)
Ca3(PO4)2(s) + 2H2SO4(aq)  Ca(H2PO4)2(s) + 2CaSO4(s)
Natural calcium phosphate rock
Calcium dihydrogen phosphate
Uses:H2SO4 used in dyes, soaps, paint pigments, explosives, fertilisers, detergents, pharmaceuticals
Extraction of aluminium
Amphoteric oxide Can react like an acid or base forming water and salt
Aluminium found raw in aluminosilicates(and igneous rocks though there is no economic way of extracting it from clay)
1
Aluminium extracted from hydrated aluminium oxides(bauxite) Al2O3.H2O and Al2O3.3H2O
High negative Gf of its oxide Gf makes electrolysis the method of extraction
Raw material purified before extraction process(impurities in bauxite are iron(III)oxide and silica)
Al2O3 reacts because it is amphoteric and dissolves forming a solution
Impurities don’t react or dissolve because Fe2O3 is basic and silica is too unreactive
2
High energy costs from bauxite being crushed & heated under pressure with conc NaOH(aq)
Amphoteric aluminium oxide dissolves to give sodium aluminate solution + sodium silicate + iron(III)oxide
Silica which is acidic, dissolves
Iron(III)oxide doesn’t react with alkali because it’s a basic oxide and is left behind in a ‘red oxide mud’(can be removed by
filtration) used to manufacture protective paint for ironwork
3
Blow CO2
Sodium silicate remains in solution
Aluminium hydroxide is ppted and changes to insoluble hydrated aluminium oxide which is filtered off, washed and roasted to
give pure aluminium oxide used in electrolytic extraction of the metal
Electrolytic extraction - Aluminium oxide Mt 2040°C is high so unsuitable as an electrolyte(economically because of heat energy costs)
- Aluminium dissolved in bath of molten cryolite(NaAlF6)making electrolyte Mt1000°C by electric current 100000amps and 5V(DC) cell
(only approx 2V to decompose oxide rest to overcome electrical resistance)
- Carbon(graphite) lining cathode, when DC current is passed through, aluminium forms at cathode and sinks to bottom of the tank
- Carbon(graphite) anode electrodes are pure since aluminium is rarely purified after production, carbon reacts with oxygen and erodes,
2O2–(solution)  O2(g) + 4e–
C(s) + O2(g)  CO2(g)
- Anodes continuously fabricated by baking carbon paste(using cell heat)preventing need for stopping process when new anodes required
- Anodes moved continuously as they erode maintaining constant dist(5cm) from the aluminium and to allow for change of depth of
aluminium between removal of batches of metal
- Great demand in electricity of aluminium plant means cheap and plentiful electricity is desirable(hydroelectricity)
- High cost of aluminium makes recycling economically effective, conserving earths resources and energy, but quality makes it
unsuitable for critical uses(manufacture of aircraft alloy)
Uses: -Vehicle construction, low density, mechanical weakness of pure metal requires it to be alloyed to increase its strength.
- Manufacture of electrical cable for overhead power lines, good electrical conductivity, low density, cheaper than copper, supported
by a steel core to prevent it stretching and breaking
- High reflectivity, malleability means its used in reflectors of car headlamps
(1)Write 2 ionic half-equations to show corrosion of Fe by O2 and water
(1)Fe(s)Fe2+(aq) + 2e– or
½ O2 + H2O + 2e–  2OH–
(2)Explain why Fe corrodes faster than aluminium even though aluminium has a more negative standard electrode potential
(2)Al covered with a protective layer of oxide iron is not covered by a protective layer/layer is porous
(3)Explain why the electrolyte must be molten
(3)So that the ions are mobile
Downs Process Electrolysis of electrolyte brine NaCl(aq)
- Cell divided by a diaphragm which allows free passage of ions(under influence of electric field) but prevents mixing of products
because chlorine reacts with NaOH
Overall reaction occurring in membrane cell
2NaCl + 2H2O  2NaOH + H2 + Cl2
- Before brine is electrolysed it is treated with NaOH(or carbonate CO 3) to remove traces of iron and calcium salts by precipitation and
filtration(undesirable cations interfere with ion-exchange action of the membrane)
- Brine flows into the anode compartment where the chlorine is liberated, piped off and used 2Cl –(aq) – 2e–  Cl2(g)
Anodes made of graphite or titanium which resist attack by liberated chlorine
- Under influence of applied voltage, Na ions move through the membrane into the cathode compartment where water is provided.
Cathodes made of steel resisting attack by NaOH produced
- Hydrogen liberated and piped away for storage by reduction of water 2H2O(l) + 2e–  2OH–(aq) + H2(g)
- Na ions and OH ions constitute NaOH solution, they do not form it, they are it
Uses: Manufacture NaOH, H2 and Cl2
NaOH purification of bauxite, manufacture of soap, bleaches
Cl2 manufacture of PVC, insecticides, herbicides, bleaches
H2 manufacture of HCl/methanol/margarine Or rocket fuel/hydrogenation of oils
Cl2(g) + 2OH –(aq)
ClO–(aq) + Cl–(aq) + H2O(l)
NaOH
Sodium chlorate(I) household bleach
NaOH(aq) is cooled before the reaction since chlorate(I) (active ion in bleach) undergoes disproportionation if heated:
3ClO–(aq)  2Cl–(aq) + ClO3–(aq)
(1)Give one piece of everyday evidence that reaction 2NaCl + 2H2O  2NaOH + H2 + Cl2 can’t occur without electrolysis
(1)Salt water does not evolve Cl2
(2)Suggest why a catalyst maybe in the form of a gauze or mesh
(2)Increased surface area/more active sites
(3)(i)Explain how the process results in the formation of NaOH in the cathode compartment
(i)H+ removed or OH– formed • Na+ ions migrate to cathode/through the diaphragm
(ii)Forms hydroxides/ppts/compounds which block the diaphragm/the cell
(ii)Suggest why the brine must be purified to remove Ca and Mg ions
(iii)In modern processes, diaphragm replaced by membrane which is ion-selective. It will allow Na ions to pass through but will not
allow Cl ions through. Suggest advantage of such a process (iii)Purer product/faster production/no NaCl present in product
(4)(i)In production of NaOH why an electrolytic process rather than one where metal is first produced and then reacted with water?
(ii)Sodium has to be made electrolytically anyway or reaction violent/too exothermic or 2 stages is more expensive to melt NaCl
(ii)In production of NaOH why a titanium anode, rather than a steel one?
(ii)Steel is more reactive than titanium/titanium less reactive steel reacts with the chlorine