Shapes of molecules and ions Demonstrate an understanding of the use of electron-pair repulsion theory to interpret and predict the shapes of simple molecules and ions

pairs of electrons around the central atom repel each other to a position of maximum separation and minimum energy

lone pairs have a greater repulsive effect than bonding pairs o because they are not shared by two atoms so are attracted to a single nucleus and

so are closer to the central atom

lone pairs reduce the angle by ~2.5 Limitations of bonding model

dot-and-cross only show electrons shared, not the lengths of bonds or shape of molecule

most bonds are neither pure ionic or pure covalent, but an intermediate due to bond polarisation

Recall and explain the shapes of BeCl2, BCl3, CH4, NH3, NH4+, H2O, CO2, PCl5, SF6 and simple organic

molecules

BeCl2 o linear o 180

BCl3 o trigonal planar o 120

CH4 o tetrahedral o 109.5

NH3 o pyramidal o 3 bonding pairs and 1 lone pair repel each other to a position of maximum

separation o the lone pair is more repulsive so decreases the bond angle slightly to 107 o 107

NH4+ o tetrahedral o 109.5

H2O o bent o 2 bonding pairs, 2 lone pairs o 104.5

CO2 o linear o 180

PCl5 o trigonal bipyramid o 120 and 90

SF6 o octahedral o 90

Demonstrate an understanding of the terms bond length and bond angle and predict approximate bond angles in simple molecules and ions

Pairs of electrons Shape Bond angle

2 Linear 180 3 Trigonal planar 120 4 Tetrahedral 109.5 5 Trigonal bipyramid 90 and 120 6 Octahedral 90

Apply the electron-pair repulsion theory to predict the shapes of molecules and ions of the above

Discuss the different structures formed by carbon atoms, including graphite, diamond, fullerenes and carbon nanotubes and the applications of these, eg the potential to use nanotubes as vehicles to carry drugs into cells

Pyramidal

Diamond

each carbon atom forms four identical bonds to other carbon atoms

tetrahedral arrangement

strong covalent bonds o very high melting point o sublimes o hard; used in drill tips and saws o good thermal conductor as vibrations travel easily through the stiff lattice o doesn't conduct electricity because electrons are held in covalent bonds o insoluble in all solvents

Graphite

exists in layers

carbon atoms bonded to three other carbon atoms at 120 o very high melting point; sublimes o insoluble in all solvents

layers are weakly bonded to each other by London forces o sheets slide over each other; used as lubricant/pencils

4th outer electron from each carbon atom is delocalized o free to move so conducts electricity

poor thermal conductor/good electrical conductor

used in strong, lightweight sports equipment

Fullerenes

32 or more carbon atoms

each carbon bonded to 3 others in pentagons, giving a ball-shaped molecule

red and soluble in organic solvents

4th outer electron from each carbon atom is delocalized o however, they cannot move between molecules so fullerenes do not conduct

electricity Nanotubes

fullerenes in the form of tubes o sigma bonds very strong, used to reinforce graphite in tennis rackets and to make

strong, lightweight building materials o used in wires in circuits for computer chips

carbon hexagons

very small and stiff

can cage other molecules forming capsules; delivery of drugs to specific cells in the body

Intermediate bonding and bond polarity Explain the meaning of the term electronegativity as applied to atoms in a covalent bond

ability of an atom to attract a pair of electrons to itself in a covalent bond

highest for elements in top right hand corner o fluorine, followed by oxygen and then chlorine and nitrogen

Recall that ionic and covalent bonding are the extremes of a continuum of bonding type and explain this in terms of electronegativity differences leading to bond polarity in bonds and molecules, and to ionic bonding if the electronegativity is large enough

Between the extremes of pure ionic and pure covalent bonding there is a range of intermediate bonds. Polarization of ions leads to distorted ionic bonds. If the polarization is large, the electron density is distorted so much it will resemble a covalent bond.

if the difference in electronegativity is high, the electron density in the bond is distorted, causing a dipole

the bond is polarized Distinguish between polar bonds and polar molecules and be able to predict whether or not a given molecule is likely to be polar Non-polar if;

no polar bonds

symmetric molecule o dipoles cancel

Carry out experiments to determine the effect of an electrostatic force on jets of liquids and use the results to determine whether the molecules are polar or non-polar

rub rod with a piece of cloth to charge it

stream of polar liquid will be deflected away from a charged rod

Intermolecular forces Demonstrate an understanding of the nature of intermolecular forces resulting from interactions between permanent dipoles, instantaneous dipoles and induced dipoles (London forces) and from the formation of hydrogen bonds London forces

movement of electrons causes an unequal electron distribution in a particle, creating a temporary dipole

this causes an induced dipole in the opposite direction on a neighbouring particle, so the two dipoles attract

the net result is a weak attractive force

weakest of intermolecular force o strength increases with number of electrons o strength increases with increasing surface area

Permanent dipole-dipole forces

occurs in polar molecules

the delta+ is attracted to the delta- of another molecule

stronger than London forces Hydrogen bonds

stronger intermolecular force

exists when NOF (nitrogen/oxygen/fluorine) is directly bonded to hydrogen o these are the most electronegative elements

electrostatic attraction between the delta+ hydrogen and delta- NOF of another molecule

H is an exposed proton as NOF will pull all the electrons towards itself

Between lone pairs and hydrogen

180o

Relate the physical properties of materials to the types of intermolecular force present, eg.

The trends in boiling and melting temperatures of alkanes with increasing chain length o London forces are stronger for larger molecules as they have more electrons

The effect of branching in the carbon chain on the boiling and melting temperatures of alkanes

o Straight chains can pack o There is a greater surface area o London forces act over a large area and are therefore stronger

The relatively low volatility (higher boiling temperatures) of alcohols compared to alkanes with similar number of electrons

o Both have similar number of electrons so similar London forces o But alcohols can form additional hydrogen bonds between molecules

The trends in boiling temperatures of the hydrogen halides HF to HI o HCl, HI, HF o HCl has less electrons than HI so its London forces are weaker o Hydrogen Fluorine can form additional hydrogen bonds, so has the higher boiling

temperature Fluorine is very electronegative Hydrogen bonding is stronger So more energy needed to separate molecules

Carry out experiments to study the solubility of simple molecules in different solvents Interpret given information about solvents and solubility to explain the choice of solvents in given contexts, discussing the factors that determine the solubility including:

The solubility of ionic compounds in water in terms of the hydration of the ions o soluble in water as dipoles on H2O are attracted to oppositely charged ions o the energy required to overcome the strong electrostatic forces in the ionic lattice

and separate the ions is supplied by the energy released when polar water molecules are attracted to the ions

hydration enthalpy>ionic lattice enthalpy o energy released is called hydration energy o ions are said to be hydrated as they are surrounded by water molecules

The water solubility of simple alcohols in terms of hydrogen bonding o both have strong hydrogen bonding o the energy given out when new attractions form between the two liquids will be

sufficient to overcome the energy required to break the existing forces o however as carbon chain length increases, the solubility decreases (increasing

London forces) - pentan-1-ol is not soluble as the London forces between the long carbon chains are too strong to break

The insolubility of compounds that cannot form hydrogen bonds with water molecules eg polar molecules such as halogenalkanes

o only weak London force and dipole-dipole interactions can be formed between water and halogenalkane molecules

o energy released by new interactions is not enough to break strong hydrogen bonds between water molecules

The solubility in non-aqueous solvents of compounds which have similar intermolecular forces to those in the solvent

o have similar intermolecular forces o the energy given out when new attractions form between the two liquids will be

sufficient to overcome the energy required to break the existing forces

Redox Demonstrate an understanding of:

Oxidation number - the rules for assigning oxidation numbers o elements have an oxidation number of 0 o group 1 and 2 always have an oxidation number of +1 and +2 o hydrogen is always +1, except in hydrides, where it is -1 o oxygen is -2 except in peroxides where it forms a O-O single bond (-1) or is bonded

with Fluorine (which is always -1) o fluorine is always -1 o other halogens are usually -1, except when bonded to oxygen or a higher up halogen

Oxidation and reduction as electron transfer o Oxidation Is Loss Reduction Is Gain

Oxidation and reduction in terms of oxidation number changes o Oxidation:

the number goes up addition of oxygen loss of hydrogen (strong oxidising agents have high oxidation numbers)

o Reduction: the number goes down loss of oxygen gain of hydrogen

How oxidation number is a useful concept in terms of classification of reactions as redox and disproportionation

o redox is when oxidation and reduction both occur o disproportionation is where one of the reactants is both oxidised and reduced

Write ionic half-equations and use them to construct full ionic equations Writing half equations

balance elements which are not oxygen or hydrogen

balance oxygen by adding H2O

balance hydrogen by adding H+

balance charge by adding electrons To construct full ionic equations, write two half equations and then combine them by finding the lowest multiple of the number of electrons to cancel them.

Periodic table - groups 2 and 7 1. Properties down group 2 Explain the trend in first ionization energy down group 2 Decreases down the group

The radius of the atom increases

So the outermost electron is further from the nucleus

There are more quantum shells

Meaning there is more electron shielding

So the outermost electron has a reduced attraction to the protons in the nucleus

Less energy is required to remove it

Recall the reaction of the elements in group 2 with oxygen, chlorine and water Oxygen:

All react to form solid metal oxides

Reactivity increases down group

Mg (s) + 0.5O2(g)--> MgO (s)

Chlorine:

All react to from solid metal chlorides

Mg(s) + Cl2(g) --> MgCl2(s) Water:

Beryllium won't react unless dust due to insoluble oxide layer.

Reactivity increases down group

Ca(s) + 2H2O(l) --> Ca(OH)2(aq) + H2(g)

Magnesium reacts very slowly with cold water (Almost insoluble Mg(OH)2 forms barrier on magnesium preventing further reaction)

Magnesium reacts vigorously with steam o Mg(s) + H2O(g) --> MgO(s) + H2(g)

Recall the reactions of the oxides of group 2 elements with water and dilute acid Water:

BeO doesn't react

Form aqueous hydroxides

CaO(s) + H2O(l) --> Ca(OH)2(aq) Dilute Acid:

All react

MgO(s) + 2HCl(aq) --> MgCl2(aq) + H2O(l)

CaO(s) + 2HNO3(aq) -->Ca(NO3)2(aq) + H2O(l) Hydroxides react in a similar way

Sr(OH)2(aq) + 2HCl(aq) --> SrCl2(aq) + 2H2O

Recall the trends in solubility of the hydroxides and sulfates of group 2 elements Solubility of hydroxides increases down the group. (eg. CaOH is only slightly soluble) Solubility of sulphates decreases down the group. Recall the trends in thermal stability of the nitrates and the carbonates of the elements in groups 1 and 2 and explain these in terms of size and charge of cations involved Group 1:

Only lithium carbonate decomposes.

All nitrates decompose, but Lithium does differently. o 4LiNO3(s) --> 2Li2O(s) + 4NO2(g) + O2(g) o KNO3(s) --> KNO2(s) + 0.5O2(g)

More stable down the group. Group 2:

2Ca(NO3)2 --> 2CaO + 4NO2 + O2

MgCO3 ---> MgO + CO2

Stability increases down group o Cation becomes larger o Therefore polarisation/distortion is reduced of the carbonate ion

Recall the characteristic flame colours formed by group 1 and 2 compounds and explain their origin in terms of electron transitions

Sodium: yellow (sunny sodium)

Potassium/Caesium: lilac

Lithium/Calcium/Strontium: red

Barium: green (BaG)

Magnesium: colourless

Electrons are given energy and are excited to higher energy levels. However, they fall back down to lower energy levels, causing energy to be released in the form of visible light (expect Mg) Describe and carry out the following:

Experiments to study the thermal decomposition of group 1 and 2 nitrates and carbonates o Boil the solid in a boiling tube o Melts on heating, releasing a brown gas and relighting a glowing splint (or turning

lime water cloudy)

Flame tests on compounds of group 1 and 2 o Clean nichrome wire in conc HCl to form soluble chloride salts which are colourless o Dip wire in compound and hold in roaring bunsen flame

Simple acid-base titrations using a range of indicators, acids and alkalis, to calculate solution concentrations in gdm-3 and moldm-3 eg measuring the residual alkali present after skinning fruit with potassium hydroxide

o Acid of known conc in burette (standard solution) o Alkali of known vol in conical flask (using a pipette) o Use methyl orange (red in acid, yellow in alkali, at end-point=orange) o For a weak acid use phenolphthalein (colourless in acid, pink in alkali, at end-

point=very pale pink)

Demonstrate an understanding of how to minimise the sources of measurement uncertainty in volumetric analysis and estimate the overall uncertainty in the calculated result Record in a table

first try is inaccurate; used as a guide; inaccurate

record reading to nearest 0.05cm3 - precision

to improve reliability, repeat until 2 concordant titres are within 0.2cm3 and then calculate mean

Systematic errors

caused by set-up or equipment Random errors

caused by reading from equipment

reduced by repeating experiment Calculating uncertainties;

- if reading to 0.05cm3 and you take two readings, each titre has an uncertainty of +-0.1cm3

2. Inorganic chemistry of group 7 (bromine, iodine and chlorine) Recall the characteristic physical properties of the elements limited to the appearance of solutions of the elements in water and hydrocarbon solvents Chlorine:

- Pale yellow-green gas - In water; pale yellow-green solution - In hydrocarbon; pale yellow-green solution

Bromine:

- Red-brown gas/liquid (very volatile) - Partially soluble in water - Very soluble in a hydrocarbon solvent - Red-brown solution in both cases

Iodine: - Grey-black solid - Sublimes on heating giving a purple gas - Slightly soluble in water giving a brown solution - Very soluble in hydrocarbon giving a pink solution

Describe and carry out the following chemical reactions of halogens:

Oxidation reactions with metal and non-metallic elements and ions such as iron(II) and iron(III) ions in solution

Cl2 is the most reactive and the best oxidation agent Metals/Ions

o Oxidise metals o 3Cl2(g) + 2Fe(s) --> 2FeCl3(s)

Bromine is a weaker oxidising agent so a mixture of iron(II) and iron(III) bromide is formed

Iodine is an even weaker oxidising agent so only iron(II) iodide forms o Fe2+(aq) + Cl2(aq) --> 2Fe3+(aq) + 2Cl-(aq)

All halogens will due this except iodine Non-metals

o Act as an oxidising agent The non-metal is oxidised The halogen is reduced

o Halogen can oxidise other halide ions if the halide is below it in the group Cl2(aq) + 2Br-(aq) --> Br2(aq) + 2Cl- (aq)

o Will not react directly with carbon/nitrogen/oxygen

Disproportionation reactions with cold and hot alkali, eg. hot potassium hydroxide with iodine to produce potassium iodate(V)

Cold/dilute - Produces halides and halates(I) - Cl2(aq) + 2NaOH(aq) --> NaCl(aq) + NaOCl(aq) + H2O(l)

Hot/concentrated

Further decomposition of halate(I) ions occurs to produce halide and halate(V) ions - 3I2(aq) + 6KOH(aq) --> KI(aq) + KIO3(aq) + 3H2O(l)

Carry out an iodine/thiosulfate titration, including calculation of the results and evaluation of the procedures involved, eg determination of the purity of potassium iodate(V) by liberation of iodine and titration with standard sodium thiosulfate solution Way of finding out the concentration of an oxidising agent. A known volume of the oxidising agent, eg potassium iodate(V) is reacted with excess iodide ions (potassium iodide)

– IO3-(aq) + 5I-

(aq) + 6H+(aq) 3I2(aq) + 3H2O(l)

Iodine formed is then titrated with sodium thiosulfate of known concentration. - I2(aq) + 2S2O3

2-(aq) --> 2I-(aq) + S4O62-(aq)

The colour change is yellow to colourless. The accuracy of the end-point can be enhanced by adding starch, the blue-black colour disappears at the end-point, showing all the iodine has just reacted.

- (however, if starch is added too soon, iodine makes an insoluble compound with the starch and won't react as expected with the thiosulfate)

Evaluation of the results: - difficult to determine accurately the volume of the liquid in a burette if the meniscus lies

between two graduation marks - the material used to prepare a standard solution may not be 100% pure - a burette is calibrated to be used at 20oC; a higher temperature could result in a

difference of the actual volume of liquid when filled to a calibration mark - difficult to make an exact judgement on the exact end-point of a titration

Finding the purity of a sample of potassium iodate(V)

Weighed sample of potassium iodate(V) is dissolved in the minimum amount on concentrated nitric acid and the solution is made to a known volume (eg 250)

25cm3 sample of this solution is pipetted into a conical flask and acidified with sulfuric acid. An excess of potassium iodide solution is added. The iodate(V) ions oxidise iodide ion to iodine in acidic solution.

The iodine is titrated with sodium thiosulfate. Describe and carry out the following reactions:

Potassium halides with concentrated sulfuric acid - All halides react with conc sulfuric acid to produce a halogen halide.

KF or KCl with H2SO4: - KX(s) + H2SO4(l) KHSO4(s) + HX(g) - See misty fumes of HX gas as it comes into contact with the air. - HF and HCl aren’t strong enough to reduce sulfuric acid. - NOT A REDOX REACTION; the oxidation states remain the same.

KBr with H2SO4: - KBr(s) + H2SO4(l) KHSO4(s) + HBr(g) - See misty fumes of HBr gas. - HBr reduces H2SO4 in a redox reaction. - 2HBr(aq) + H2SO4(l) Br2(g) + SO2(g) + 2H2O(l) - See orange fumes of Br2 gas. - SO2 turns sodium dichromate paper yellow to green.

KI with H2SO4: - KI(s) + H2SO4(l) KHSO4(s) + HI(g) - HI reduces H2SO4 in a redox reaction. - 2HI(g) + H2SO4(l) I2(s) + SO2(g) + 2H2O(l) - HI reduces SO2 to H2S in a redox reaction. - 6HI(g) + SO2(g) H2S(g) + 3I2(s) + 2H2O(l) - H2S gas is toxic and smells of rotten eggs. - H2S turns lead ethanoate paper white to black

Potassium halides with halogens

Displaced from solution by more reactive halogens.

Shake mixture with an organic solvent to see reaction more clearly: - Halogen dissolves in organic solvent. (Iodine is purple in hexane so can distinguish it

from bromine which is orange) - Halide ions dissolve in water. - The organic solvent and water settle as two layers in a test tube. Organic is on top.

Chlorine displaces bromide and iodide. - Cl2(aq) + 2Br-

(aq) 2Cl-(aq) + Br2(aq)

Orange aqueous solution of Br2 - Cl2(aq) + 2I-

(aq) 2Cl-(aq) + I2(aq)

Brown aqueous solution of I2

Bromine displaces only iodide. - Br2(aq) + 2I-

(aq) 2Br-(aq) + I2(aq)

Brown aqueous solution of I2

Iodine cannot displace halide ions.

Potassium halides with silver nitrate solution Precipitate of the silver halide is formed.

– Ag+(aq) + X-

(aq) AgX(s) – F- forms no precipitate. – Cl- forms a white precipitate. – Br- forms a cream precipitate. – I- forms a yellow precipitate.

Silver halides with sunlight and their solubilities in aqueous ammonia solution AgCl :

- white ppt (darkens in sunlight - 2AgCl 2Ag + Cl2) - dissolves in dilute ammonia giving colourless solution

AgBr: - cream ppt (darkens in sunlight - 2AgBr 2Ag + Br2) - dissolves in conc ammonia giving colourless solution

AgI: - yellow ppt (will not darken) - insoluble in ammonia

Hydrogen halides with ammonia and with water (to produce acids) With water:

- HX(g) H+(aq) + Cl-

(aq) all fume in moist air very soluble in water dissolve in water to give strong acids

With ammonia: - HX(aq) + NH3(aq) --> NH4X(aq)

React with ammonia (use glass rod dipped in conc NH4OH) gas to give white fumes of ammonium hydride

Make predictions about fluorine and astatine and their compounds based on the trends in the physical and chemical properties of halogens

Fluorine should be a gas as hardly any electrons (low London forces). Astatine a solid.

Low boiling point for fluorine, high for astatine.

Fluorine will be most oxidising as elements get less oxidising down the group

Compounds of both should be soluble in water, forming acids.

Bond enthalpies for carbon-halogen bonds decrease down the group. Fluorocarbon carbons should be the most stable, whilst astatocarbons should be the least.

Extra facts on Fluorine:

Weak F-F bond as it is so short non-bonding electrons repel one another weakening the bond.

Small size means ionic compounds have high lattice enthalpy

Highly exothermic formation of hydrated fluoride ions (though high lattice energies means many fluorides are not water soluble)

Strong covalent bonds formed with non-metallic elements as the bond is short.

Kinetics Recall the factors that influence the rate of chemical reaction, including concentration, temperature, pressure, surface area, and catalysts Explain the changes in rate based on a qualitative understanding of collision theory Collision theory: To successfully react, particles must collide with an energy greater than the activation energy with the correct orientation, to break existing bonds and react. Concentration:

increasing the concentration of solutions increases the collision frequency in a liquid system

the distance between colliding species is reduced, so there is less distance to travel before encountering another molecule

more successful collisions per second Pressure:

increasing the pressure increases the collision frequency in a gaseous system

the distance between colliding species is reduced, so there is less distance to travel before encountering another molecule

Temperature:

collision frequency increases with temperature

the molecules have more kinetic energy and move faster, meaning they travel the necessary distance more quickly

more particles will have energy greater than the activation energy

more successful collisions per second Surface area:

increasing the surface area increases rate

allows more molecules to come into contact with one another and react

makes successful collisions more likely Catalysts:

provide alternative route of lower activation energy hence a greater proportion of molecules can react at a given temperature

increase rate but not yield

Use, in a qualitative way, the Maxwell-Boltzmann model of the distribution of molecular energies to relate changes of concentration and temperature to the alteration in the rate of a reaction

Particles in liquids and gases don't all have the same amount of kinetic energy. Shows how kinetic energy of particles is distributed Notice that:

no molecules have zero energy and the graphs start at the origin

each curve rises to a peak and then falls away to approach the x-axis without ever meeting it

the distribution is skewed

the area under the curve is the same for both temperatures because there are the same no of particles

Demonstrate an understanding of the concept of activation energy and its qualitative relationship to the effect of temperature changes on the rate of reaction Activation energy: minimum collision energy needed for particles to react. Particles that collide with energy greater than the activation energy will react if their orientation is correct.

increasing temperature increases the proportions of particles that collide with energies greater than the activation energy, and react.

Demonstrate an understanding of the role of catalysts in providing alternative reaction routes of lower activation energy and draw the reaction profile of a catalysed reaction including the energy level of the intermediate formed with the catalyst A catalyst provides an alternative route for a reaction, which has a lower activation energy.

the proportion of successful collisions per second at a given temperature increases One of the reactants must combine with the catalyst as an initial step. The complex formed then reacts with another reactant to give the products and regenerate the catalyst.

The reaction profile therefore for a catalysed reaction has at least two humps, with an intermediate reactant-catalyst complex.

Carry out simple experiments to demonstrate the factors that influence the rate of chemical reactions, eg the decomposition of hydrogen peroxide

Reactant-catalyst

complex

E

n

t

h

a

l

p

y

Chemical Equilibria Demonstrate an understanding that chemical equilibria are dynamic Equilibria:

reactions that do not go to completion

reactions that are reversible

dynamic o only in closed systems o is undergoing no net change, the concentration of each substance is constant o is one in which the forward reaction is happening at the same rate as the reverse

reaction

Le Chatelier's Principle:

when a change is imposed on a chemical equilibrium, the reaction responds in such a way as to oppose the change. As a result, the position of equilibrium changes.

Deduce the qualitative effects of changes of temperature, pressure and concentration on the position of equilibrium, eg extraction of methane from methane hydrate Change in temperature:

If the temperature of an exothermic forward reaction is increased, the reverse reaction will be favoured in order to reduce the change.

Change in pressure:

If the pressure of a gaseous equilibrium is increased, the position of equilibrium will move to the side with fewer moles (and therefore, fewer molecules)

Change in concentration:

Lowering the concentration of a reactant will make the reaction reverse, attempting to make more of the reactant.

Removing the product formed shifts the equilibrium to the right as the reactions makes more product to restore the equilibrium concentration.

Conditions have to compromise eg. Methanol from hydrogen and carbon monoxide - For an exothermic reaction, lower temperatures give a better yield

However, a lower temperature reduces the rate of reaction, so a compromise of 250 degrees is used

- Fewer mols of gas on the RHS, so a high pressure favours a high yield and also increases the rate of reaction.

However, equipment needed for high pressures is expensive so a compromise of 50-100atm is used.

- A catalyst makes the reaction reach equilibrium more quickly, it does not affect the yield.

Interpret the results of simple experiments to demonstrate the effect of a change of temperature, pressure and concentration on a system at equilibrium, eg

N2O4 <-> 2NO2 (H=+58) yellow brown

o When a closed tube is put into hot water, the equilibrium is shifted to favour the production of NO2, the colour darkens.

o When the pressure is increased, the colour fades.

Organic Chemistry

Alcohols Give examples of, and recognise, molecules that contain the alcohol functional group Demonstrate an understanding of the nomenclature and corresponding structural, displayed and skeletal formulae of alcohols, and classify them as primary, secondary or tertiary

Primary alcohols only have one carbon atom attached to the carbon atom forming the -OH bond.

Describe the following chemistry of alcohols:

Combustion o Primary, secondary and tertiary o Burn with a clean/blue flame o C2H5OH + 3O2 --> 2CO2 + 3H2O

Reaction with sodium o Primary, secondary and tertiary o 2ROH + 2Na --> 2RO-Na+ + H2 o Sodium disappears, effervescence, white solid product forms (sodium ethoxide) o Longer the hydrocarbon chain, the less reactive the alcohol is with sodium.

Substitution reactions to for halogenoalkanes, including reaction with PCl5 and its use as a qualitative test for the presence of the -OH group

o Primary, secondary and tertiary o PCl5(s) + C4H9OH(l) --> C4H8Cl(l) + HCl (g) + POCl3(l) o Misty fumes seen o Test for -OH bond

Oxidation using potassium dichromate(VI) in dilute sulfuric acid on primary alcohols to produce aldehydes and carboxylic acids and on secondary alcohols to produce ketones Aldehydes

Dilute sulfuric acid/potassium dichromate VI

Distill immediately

Primary alcohols

3CH3CH2CH2CH2OH(l) + Cr2O72-(aq) + 8H+(aq) --> 3CH3CH2CH2CHO(l)+ 2Cr3+(aq) +

7H2O(l)

Form a red precipitate when boiled with Benedict's (Cu2+ reduced to Cu+ - Cu2O) Carboxylic Acids

Conc sulfuric acid/potassium dichromate VI

Heated under reflux before distillation

Primary alcohols

3CH3CH2CH2CH2OH(l) + 2Cr2O72-(aq) + 16H+(aq) --> 3CH3CH2CH2COOH(l)+ 4Cr3+(aq) +

11H2O(l)

Turn blue litmus paper red Ketones

Conc sulfuric acid/potassium dichromate VI

Heated under reflux

Secondary alcohols

Give a yellow/orange precipitate with Brady's reagent.

No change when added to Benedict's Tertiary alcohols are not oxidised under normal conditions

Demonstrate an understanding of, and practise, the preparation of an organic liquid (reflux and distillation), eg oxidation of alcohols

(hot vapours will condense in condenser into a liquid, reactants/products do not escape) Purifying halogenoalkanes:

Shake in a separating funnel - discard the aqueous layer

Wash with sodium hydrogencarbonate solution to react with any residual acid - separate again, keeping the organic layer.

Repeat washing until no more CO2 given off

Wash with water to remove remaining inorganic impurities (excess sodium hydrogencarbonate etc)

Dry by shaking with anhydrous calcium chloride until clear

Filter off calcium chloride through glass wool

Redistil, collecting the liquid at the boiling temperature of the expected chloroalkane (removes organic impurities)

Dilute sulfuric acid/potassium dichromate VI in a round bottomed flask

condenser

round bottomed flask

After reflux, distil the carboxylic acid from the

mixture.

Why must sulfuric acid be added slowly? Water + sulfuric acid is very exothermic. If mixed rapidly, so much heat will be produced that the mixture will boil and spray out hot acidic solution. What are anti-bumping granules? Silicon dioxide. Prevent formation of large gas bubbles that cause 'bumping' Why is receiver surrounded by ice-water? Boiling temp. of ethanal is 21C, ensures there is no loss by evaporation. Why is the ethanol/dichromate added slowly to hot acid? Rapid addition would lead to an excess of oxidising agent and some of the ethanal formed would be oxidised to ethanoic acid. Why isn't the ethanol oxidised by the dichromate(VI) solution? The oxidation requires H+ ions, and the mixture has not yet been acidified.

Halogenoalkanes Demonstrate an understanding of the nomenclature and corresponding structural, displayed and skeletal formulae for halogenoalkanes, including the distinction between primary, secondary and tertiary structures

Contain C-X

Primary halogenoalkanes have one carbon attached to the carbon which is bonded to the halogen

Interpret given data and observations comparing the reactions and reactivity of primary, secondary and tertiary compounds Tertiary reacts first

More attached methyl groups weaken the carbon-halogen bond Iodo-R reacts first

Carbon-halogen bond must break

Size of halogen atom increases down the group

Carbon-halogen bond gets longer and weaker o Electrons further from the nucleus

Activation energy decreases

Rate of hydrolysis increases

Carry out the preparation of an halogenalkane from an alcohol and explain why a metal halide and concentrated sulfuric acid should not be used when making a bromoalkane or an iodoalkane ROH + PCl5-> HCl + POCl3 + RCl

Phosphorus pentachloride

Shake, the chloroalkane forms a separate layer ROH + NaBr + 50%H2SO4 -> RBr + NaHSO4 + H2O

Has to be dilute or Br- will be oxidised to Br2, reducing the yield

Heated under reflux, impure halogenoalkane is distilled off, collected over ice cold water, purified

Why is sulfuric acid added slowly? Why is cooling and shaking needed?

Sulfuric acid with water is very exothermic

Hot sulfuric acid causes oxidation of X- to X2 - reducing the yield of halogenoalkane

Why is sand bath used for heating?

Spreads heating uniformly over the base of the flask Why is the mixture heated?

Activation energy is high What impurities are present in the distillate?

Water, unchanged alcohol and sulfuric acid. What does shaking with water achieve?

Removes sulfuric acid and some alcohol How do you decide which layer to keep?

eg. 1-bromobutane denser than water Why is concentrated HCl added?

Protonates the alcohol, making it ionic and more soluble in water

Why is the mixture shaken with sodium carbonate solution?

Removes HCl Why must the pressure be released?

CO2 increases pressure, stopper would be pushed out, losing product

What is the function of the calcium chloride?

Drying agent What is the significance of the temperatures quoted?

Range is narrow enough to ensure that the halogenoalkane is the only thing being collected.

Impurities include; HBr, Br2, SO2, ROH PI3 + 3ROH -> 3RI + H3PO3

2P + 3I2 -> 2PI3 (made in situ by refluxing alcohol with 'red phosphorus' and the halogen)

Describe the typical behaviour of halogenalkanes. This will be limited to treatment with:

Aqueous alkali, eg KOH (aq) o OH- acts as a nucleophile o Nucleophilic substitution o RX + OH- ROH + X-

Alcoholic potassium hydroxide o Reflux with conc alcoholic alkali o OH- acts as a base o Alkene + H2O + X- o Elimination

Water containing dissolved silver nitrate o RX + H2O ROH + H+ + X- o Hydrolysis releases halide ions, which immediately precipitate with the silver ions

Alcoholic ammonia o Nucleophilic substitution o Form amines o RX + 2NH3 -> RNH2 + NH4X

happens in two steps RX + NH3 -> RNH3

+X- RNH3

+X- + NH3 -> RNH2 + NH4X o Heated under pressure in ethanol solvent (prevent side reactions with OH-)

Carry out the reactions described above Discuss the uses of halogensoalkanes, eg as fire retardants and modern refrigerants Do not burn easily and are generally unreactive

- Fire retardants and refrigerants - Absorb heat from fire/prevent oxygen from reaching the fire/absorb free radicals in

combustion propagation/Strength of C-F bond makes molecules inert CFCs used to be used, but now HFCs are used as CFCs deplete the ozone layer.

Mechanisms Classify reaction as addition, elimination, substitution, oxidation, reduction, hydrolysis or polymerization Addition:

- two of more substances react to form one product Elimination:

- one reactant breaks down to produce a new substance and a small molecule, is eliminated Substitution:

- one atom or a group of atoms is replaced by another atom or a group of atoms Oxidation:

- Addition of oxygen/removal of hydrogen Reduction:

- Removal of oxygen/addition of hydrogen Hydrolysis:

- Molecule is broken down by the addition of water Polymerization:

- Many identical small molecules join to form a large chain

Demonstrate an understanding of the concept of a reaction mechanism and that bond breaking can be homolytic or heterolytic and that the resulting species are either free radicals, electrophiles or nucleophiles Homolytic bond breaking: Occurs when each atom involved in the bond keeps one of the shared electron pair in the bond that breaks Creates free radicals Cl2 → 2Cl∙ Heterolytic bond breaking: Occurs when both of the electrons in a bond are kept by one of the atoms, forming a positive and negative ion Creates nucleophiles (negative ion)/electrophiles (positive ion) HCl → H+ + :Cl- Give definition of the terms free radical, electrophile and nucleophile Free radical:

- atom/molecule with an unpaired electron Electrophile:

- atom/molecule/ion that accepts a pair of electrons to form a covalent bond Nucleophile:

- atom/molecule/ion that donates a pair of electrons to form a new covalent bond Demonstrate an understanding of why it is helpful to classify reagents Demonstrate an understanding of the link between bond polarity and the type of reaction mechanism a compound will undergo

- Polar bonds always break heterolytically - Non-polar bonds usually break homolytically - A nucleophile can attack the delta positive in a polar bond - An electrophile can attack an electron-rich part of a molecule (eg. pi bond)

Demonstrate an understanding of how oxygen, O2, and ozone, O3, absorb UV radiation and explain the part played by emission of oxides of nitrogen, from aircraft, in the depletion of the ozone layer, including the free radical mechanism for the reaction and the fact that oxides act as catalysts Ozone is produced from oxygen in the ozone layer. Oxygen molecules break up to form oxygen free radicals, due to absorbing UV from the sun. The oxygen free radicals formed then recombine with oxygen to form ozone. O=O ---UV---> O∙ + O∙ O∙ + O2 --> O3

Nitrogen oxides also damage the ozone layer by producing NO∙ free radicals - NO∙ + O3 → NO2 + O2 - NO2 + O∙ → NO∙ + O2

Describe the mechanisms of the substitution reactions of halogenoalkanes and recall those above

SN1 - tertiary/secondary SN2 - primary

Aqueous alkali

Alcoholic alkali

Asymmetric halogenoalkanes gives mixture of products. Hydrogen must be taken from carbon adjacent to C-X bond

Or this one: (no SN1/SN2)

Alcoholic ammonia

Water

slow

fast

Mass spectra and IR Interpret fragment ion peaks in the mass spectra of simple organic compounds, eg the difference between propanal and propanone Peak furthest to right is the parent ion and corresponds to the molecular mass. All ions produced are positive Parent ion is unstable, so breaks into smaller fragment ions of different masses The ion peaks indicate likely functional groups Use infrared spectra, or data from infrared spectra, to deduce functional groups present in organic compounds and predict infrared absorptions, given wavenumber data, due to familiar functional groups. This will be limited to:

C-H stretching absorptions in alkanes, alkenes and aldehydes o ~3000

O-H stretching absorption in alcohols and carboxylic acids o ~3200-3800 o Carboxylic acid will be VERY broad, stretching back as far as 2500.

N-H stretching absorption in amines o ~3300-3500 o Broad cause hydrogen bonding

C=O stretching absorption in aldehydes and ketones o ~1700 o aldehydes slightly slower, ketones slightly higher

C-X stretching absorption in halogenoalkanes

as an analytical tool to show the change in functional groups during the oxidation of an alcohol to a carbonyl

Demonstrate an understanding that;

only molecules which change their polarity as they vibrate can absorb infrared radiation Demonstrate an understanding that;

H2O, CO2, CH4 and NO molecules absorb IR radiation and are greenhouse gases

Only molecules that can absorb IR are greenhouse gases

Oxygen/nitrogen do not contain polar bonds, so do not absorb IR and are not greenhouse gases

Green chemistry Demonstrate an understanding that the processes in the chemical industry are being reinvented to make them more sustainable ('greener') by:

Changing to renewable resources o Planet's resources are finite

Finding alternatives to very hazardous chemicals o Avoiding toxic intermediates/solvents

Discovering catalysts for reactions with higher atom economies o Makes reaction faster o Requires a lower temperature so less energy o Trying to develop highly specific catalysts so that a process only gives the material

required and suppresses side reactions o Less energy needed to then separate products

Making more efficient use of energy

o Microwave heating is more efficient and controllable than conventional heating, but cannot be used large-scale

o Running at low temperatures & pressure o High yield

Reducing waste and preventing pollution of the environment o Less to treat/dispose of o Recycling materials and creating biodegradable products

Discuss the relative effects of different greenhouse gases as absorbers of IR and hence on global warming Infrared radiation from the sun, which has short wavelength, mostly passes through the atmosphere and is absorbed by the Earth's surface. This heats the Earth, which re-emits longer wavelength IR. Any greenhouse gases in the atmosphere reflect this longer wavelength IR, warming the atmosphere. The relative greenhouse effect of a gas varies because molecules with different bonds absorb IR differently. The global warming potential of a gas combines its ability to absorb IR with its lifetime in the atmosphere. The concentration of a gas in the atmosphere also affects its potential to cause warming.

Carbon dioxide has a low global warming potential, but levels are rising

CFCs have a much higher global warming potential, but overall concentrations are very low. Discuss the difference between anthropogenic and natural climate change over hundreds of thousands of years Anthropogenic climate change:

Results from human activities such as burning fossil fuels that increase levels of CO2, CH4, N2O over short timescales

Natural climate change:

Due to natural processes such as the dissolving of CO2 in sea water, formation of carbonates in rocks, that remove CO2 from the atmosphere over much longer timescales.

Other natural processes such as volcanic eruptions or changes in solar activity can also cause climate change

Demonstrate understanding of the terms 'carbon neutrality' and 'carbon footprint' A carbon neutral process has no net contribution to the amount of carbon dioxide in the atmosphere. Carbon neutral fuel;

one for which the release of carbon dioxide in its manufacture and burning equals the absorption of carbon dioxide from the atmosphere as the raw material is grown/fuel is formed

only biofuels can be considered as carbon neutral within a human lifetime Carbon neutral process;

no net carbon dioxide emission to the atmosphere. Emissions are balanced by actions that remove an equivalent amount of carbon dioxide.

Carbon footprint; measure of the amount of carbon dioxide emitted through the use of fossil fuels. Apply the concept of carbon neutrality to different fuels such as petrol, bio-ethanol and hydrogen Fossil fuels:

Fossil fuels have locked up large amounts of carbon dioxide that used to be in the atmosphere

Burning of any fossil fuel returns this carbon dioxide to the atmosphere

However, they are not carbon-neutral, as they increase the amount of atmospheric carbon dioxide

Bio-ethanol:

Seems to be a carbon-neutral fuel.

Carbon dioxide is absorbed by the plant as it grows and is released with the fuel is burnt

However, large-scale; o fermentation/distillation require energy o lowers ability of land to absorb CO2 since crops use less of the gas for photosynthesis

than the carbon-rich rainforest that they replaced Hydrogen:

Only produces water when burned

However, involves carbon emissions as it is manufactured from water and methane.

This process requires energy for heating, both for the purification of the methane and for its endothermic reaction with steam.

Discuss and explain, including the mechanisms for the reactions, the science community's reasons for recommending that CFCs are no longer used due to their damaging effect on the ozone layer Ozone prevents harmful UV reaching the surface, but is broken down by reactions with CFCs and nitrogen oxides produced by car/jet engines. CFCs are dissociated by UV radiation to form highly reactive chlorine free radicals.

- CCl2F2 → ∙CClF2 + Cl∙ - Cl∙ + O3 → ∙ClO + O2 - The cycle is continued as Cl free radicals are regenerated and unchanged (catalytical radical):

∙ClO + O3 → 2O2 + Cl∙ Net effect is to produce two molecules of ozone to three molecules of oxygen. A chain reaction is set up, so just one molecule can do a lot of damage. Mark scheme answer:

Can release Cl free radicals

Cl free radicals reacts with ozone

Ozone layer depletes

Leading to greater levels of UV exposure

Greater risk of skin cancer C-F bond is strong so does not release F radicals.