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Chemistry, Module Five

Chemistry, Module Five. xX … TheDitzyBlonde …Xx. What are electrochemical cells?. Made from two different metals dipped in salt solutions of their own ions and connected by a wire. The wire is the external circuit.

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Chemistry, Module Five

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  1. Chemistry, Module Five xX…TheDitzyBlonde…Xx

  2. What are electrochemical cells? • Made from two different metals dipped in salt solutions of their own ions and connected by a wire. • The wire is the external circuit. • Half-cells involving solutions of two aqueous ions of the same element use a platinum electrode. • Conversion of ions happens on the surface of the electrode. • Half-cells are connected by a salt bridge made of filter paper soaked in KNO3(aq). • K+ and NO3- ions flow through the salt bridge and balance out the charges in the two half-cells. • Redox reaction occurs in an electrochemical cell. • Electrons flow through the wire from the most reactive metal to the least reactive metal. • Voltmeter in the external circuit shows the voltage between the two half-cells. • This is called the cell potential or emf, Ecell. • Ecell = EᶱRHS - EᶱLHS • More negative standard electrode potential put on the left, so the cell potential is always positive. • Swapping the sides the half-cells are on changes the sign of the cell potential.

  3. What are electrode potentials? • Standard electrode potential, Eᶱ, of a half-cell is the voltage measured under standard conditions when the half-cell is connected to a standard hydrogen electrode. • Standard conditions are solutions at 1M concentration, temperature of 298K and pressure of 101kPa. • Ecell = EᶱRHS - EᶱLHS • Electrode potential can be positive or negative, depending on which way the electrons flow. • Standard hydrogen electrode is hydrogen gas bubbled through a 1M solution of H+ ions with a platinized platinum electrode submerged in, at 101kPa pressure. • Platinized platinum electrode means the surface of foil is finely powdered with platinum. • The platinized surface absorbs hydrogen gas, causing an equilibrium to be set up. • H2(g) ↔ H+(aq) + 2e- • Electrode potential of a standard hydrogen electrode is 0. • Reference cell because it is used to measure the electrode potentials of other half-cells. • Standard hydrogen electrode always put on the left.

  4. What is the electrochemical series? • The more reactive the metal, the more likely it is to lose electrons to form a positive ion. • More reactive metals have more negative standard electrode potentials. • The more reactive a non-metal, the more likely it is to gain electrons to form a negative ion. • More reactive non-metals have more positive standard electrode potentials. • Electrochemical series shows standard electrode potentials. • More positive electrode potentials mean the left-hand substance is more easily reduced and the right-hand substance is more stable. • More negative electrode potentials mean the right-hand substance is more easily oxidised and the left-hand substance is more stable.

  5. What is the anticlockwise rule? • Predicts whether a reaction is likely to happen. • Write half-equations out with the most negative standard electrode potential on top. • Draw on anticlockwise arrows to show the direction of the half-reaction. • Swap the top half-equation, and change the sign of the electrode potential. • Combine the half-equations and find the cell potential. • Electrode potential chart is another way of showing the anticlockwise rule; upside down y-axis showing electrode potentials with the more negative electrode potential on top.

  6. What problems are there with predicting electrode potential? • If the electrode potential is positive then the reaction is feasible under standard conditions. • Above 0.4V, the reaction goes to completion. • Between 0 and 0.4V then the reaction will be reversible. • Changing concentration or temperature of the solution can cause the electrode potential to change. • If you increase the concentration of the ions in the top half-equation, the equilibrium shifts left, making electron loss more difficult and the cell potential lower. • If you increase the concentration of the ions in the bottom half-equation, the equilibrium shifts right, making electron gain easier and the cell potential higher. • The kinetics may not be favourable: • Reactions occurs so slowly that it appears that nothing is happening. • Reaction has a high activation energy, which stops it occurring. • Bigger cell potential = bigger total entropy change.

  7. What are redox titrations? • Titrations using transition element ions. • Used to determine the amount of oxidising agent needed to exactly react with a quantity of reducing agent or vice versa. • Measure out a volume of reducing agent into a conical flask. • Add 20cm3 of dilute sulfuric acid to the conical flask. • Add a solution of oxidising agent to the conical flask while swirling the conical flask. • Record the volume of oxidation added when the end point is reached (rough titration). • Repeat the titration, adding the oxidising agent dropwise when you are within 2cm3 of the end point, until you get two or more readings within 0.1cm3 of each other. • Find average titre.

  8. What can you calculate from redox titration? • Concentration of reducing agent: • Write the balanced equation for the reaction. • Work out the number of mols of oxidising agent added to the conical flask. • Use the stoichiometry of the equation to find the number of mols of the reducing agent that has reacted. • Scale up if needed. • Work out concentration of the reducing agent. • Percentage purity: • Write the balanced equation for the reaction. • Work out the number of mols of the oxidising agent added to the conical flask. • Use the stoichiometry of the equation to find the number of mols of reducing agent that has reacted. • Scale up if needed. • Work out the mass of reducing agent. • Work out the percentage purity.

  9. What are iodine-sodium thiosulfate titrations? • Used to find the concentration of oxidising agents. • The more concentrated the oxidising agent is, the more ions will be oxidised. • Potassium iodate(V), KIO3, (oxidising agent) is added to excess acidic potassium iodide solution, oxidising some of the iodide ions to iodine. • IO3-(aq) + 5I-(aq) + 6H+(aq)  3I2(aq) + 3H2O(l) • Solution produced is titrated with sodium thiosulfate, Na2S2O3. • I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) • Colour change from blue/black to colourless. • Find average titre. • Work out the number of mols of thiosulfate that reacted. • Use the stoichiometry of the second equation to work out the number of mols of iodine that reacted. • Use the stoichiometry of the first equation to work out the number of mols of iodate(V) ions that reacted. • Scale up if needed. • Calculate the concentration of potassium iodate(V).

  10. How can iodine-sodium thiosulfate titrations be used to calculate percentage copper in a copper alloy? • Weighed amount of copper alloy dissolved in conc nitric acid and made up to 250cm3 with deionised water in a volumetric flask. • 25cm3 portion added to conical flask • Excess nitric acid is neutralised with sodium carbonate solution, until white ppate appears. • White ppate removed by adding ethanoic acid. • Excess potassium iodide added, which reacts with copper(II) ions. • 2Cu2+(aq) + 4I-(aq)  2CuI(s) + I2(aq) • CuI is seen as a white ppate. • Product mixture is titrated against sodium thiosulfate solution. • I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) • Find average titre. • Work out the number of mols of sodium thiosulfate that reacted. • Use the stoichiometry of the equations to find the number of mols of copper. • Scale up. • Work out the mass of copper. • Work out percentage of copper in the copper alloy.

  11. What errors can occur in iodine-sodium thiosulfate titrations? • Starch indicator must be added when most of the iodine has reacted, otherwise the blue colour will be very slow to disappear. • Starch indicator must be freshly made, otherwise it won’t behave as expected. • Ppate of CuI makes seeing the colour of the solution quite difficult. • Iodine produced in the reaction can evaporate from the solution, giving a false titration reading, leading to a percentage of copper lower than the actual value. • Avoid this by keeping the solution cool.

  12. What are fuel cells? • Produces electricity by reacting a fuel such as hydrogen with an oxidant such as oxygen. • At the anode (fuel), the platinum catalyst splits H2 into protons and electrons. • Polymer electrolyte membrane (PEM) only allows H+ ions across, and forces electrons to travel around the circuit to get to the cathode. • This created an electric current in the circuit, with voltage about 0.6V, which is used to power things. • At the cathode, O2 combines with H+ from the anode, and electrons from the circuit to make water, which is the only waste product. • Cell continues to produce a current as long as there is hydrogen and oxygen available. • Hydrogen has to be made to use the fuel cell: • Natural gas reacted with steam to produce hydrogen gas gives carbon dioxide as a waste product and needs fossil fuels to heat the process. • Electrolysis of water to produce hydrogen requires electricity, which is generated using fossil fuels, but could be generated sustainably using renewable resources.

  13. What fuels can be used in fuel cells other than hydrogen? • Hydrogen-rich fuels are fuels with a high percentage of hydrogen in their molecules, e.g. ethanol or methanol. • These are converted into hydrogen by a reformer. • Some fuel cells can use alcohols without have to reform them to produce hydrogen first. • Alcohol oxidised at the anode in the presence of water. • CH3OH + H2O  CO2 + 6e- + 6H+ • H+ ions pass through the electrolyte and are oxidised to water. • 6H+ + 6e- + (3/2)O2 3H2O • Advantages of using alcohols: • Higher hydrogen density than liquefied hydrogen. • Methanol and ethanol can be made on a large scale using renewable resources. • Alcohols are liquid at room temperature so don’t need refrigerated storage. • Methanol can be made from carbon dioxide, so can be used to reduce carbon dioxide levels in the atmosphere.

  14. What are the advantages and disadvantages of fuel cell cars compared to normal cars? • Advantages: • Produce less pollution when they are used. • Only waste product of hydrogen fuel cell is water. • Engines running on hydrogen-rich fuels produce less carbon dioxide than petrol or diesel engines. • More efficient at converting fuel to power. • Disadvantages: • Expensive. • Made using toxic chemicals, which are difficult to dispose of. • Limited life-span.

  15. What are fuel cells used for? • Powering cars. • Space Shuttle uses 3 fuel cell power plants, each made up of many fuel cells, which produce up to 12kW of power. • Waste product of water can be used as drinking water for astronauts. • In breathalysers, as the amount of ethanol in someone’s breath is directly related to the amount of ethanol in the blood. • Used to be measured using the reaction of ethanol with potassium dichromate(VII), and measuring the colour change using a photocell system. • Breathalysers in police stations use IR spectrometry, which is accurate but not easily portable. • Ethanol fuel cells can be used, as the amount of alcohol present is proportional to the current produced when the person’s breath is fed to the anode of the cell. • Less susceptible to producing false readings to other substances in the breath. • Easily portable. • Accurate.

  16. What are the transition metals? • D-block is in the middle of the periodic table. • D-block elements are those that are found in the d-block of the periodic table. • Transition metals are elements that can form one or more stable ions with a partially filled d-subshell. • Scandium is not a transition metal because it only forms one ions, Sc3+, which has an empty d-subshell. • Zinc is not a transition metal because it only forms one ions, Zn2+, which has a full d-subshell. • Electron configuration is given as [Ar]3dx4sy. • All orbitals are occupied singly before any orbitals have two electrons. • 4s subshell fills before the 3d subshell except chromium and copper, which have only one electron in the 4s subshell as this gives them more stability. • Special chemical properties: • Form complex ions. • Form coloured ions in solution. • Are good catalysts. • Can exist in variable oxidation states.

  17. How are ions of transition metals formed? • Electrons are removed, first from the 4s subshell, then from the 3d subshell. • Ionisation energy is the energy required to remove an electron from an element. • First ionisation energies are similar for d-block elements, showing they have similar structures, and lose the first electron from the same shell. • Second ionisation energies increase steadily along the period, with exceptions of Cr and Cu, which have a slightly higher IE than expected because the 2nd electron is removed from the 3d-subshell, which is much closer to the nucleus than the 4s-subshell. • Third ionisation energies increase steadily, with a step down in ionisation energy at iron because form that point onwards the third electron is being removed from a paired 3d orbital, which is easier to remove due to repulsion between the electrons.

  18. What are complex ions? • Complex ions are metal ions surrounded by dative covalently bonded ligands. • Dative covalent bonds are covalent bonds where both electrons in the shared pair come from the same atom. • Ligands are atoms, ions or molecules that donate a pair of electrons to a central metal ion. • Coordination number is the number of dative covalent bonds that are formed with the central metal ion. • Small ligands such as H2O, NH3 and CN- result in a coordination number of 6. • Octahedral shape. • Larger ligands such as Cl- result in a coordination number of 4. • Tetrahedral shape if all Cl- ligands. • Square planar shape if a mix of small and larger ligands. • Some complexes only have 2 dative covalent bonds so are linear. • (CuCl2)- and [Ag(NH3)2]+ • Ligand substitution is when one ligand is replaced with a different ligand. • Causes a colour change. • Oxidation state of metal ion = total oxidation state – oxidation states of ligands.

  19. How do ligands form bonds? • Using lone pairs of electrons: • Monodentate ligands have one lone pair of electrons. • E.g. H2O, NH3, Cl-, CN- • Bidentate ligands have two lone pairs of electrons. • E.g. ethane-1,2-diamine and ethanedioate. • Polydentate ligands have more than two lone pairs of electrons. • E.g. EDTA+ has six lone pairs of electrons and haemoglobin has four lone pairs of electrons. • Ligands split the 3d subshell into two energy levels, creating a visible colour. • Usually 3d orbitals of a transition metal ion all have the same energy. • Repulsion between electrons in the ligand and the 3d electrons of the metal increases the energy of the 3d orbitals of the transition metal ion in the complex. • Some orbitals are increased in energy level more than others, splitting the 3d orbital into two different energy levels. • Electrons occupy lower orbitals but can jump to higher energy level orbitals by absorbing energy from visible light equal to the energy gap. • The remaining frequencies of light are transmitted, causing the complement colour to be seen. • The central metal ion, the ligands and the coordination number affect the size of the energy gap, and thus the colour seen. • If there are no 3d electrons, there are no electrons to jump energy levels so no energy is absorbed and no colour is seen. • If the 3d subshell is full then there is no room in the upper orbitals for electrons to jump to, so no energy is absorbed and no colour is seen.

  20. What is ligand exchange? • Swapping of a ligand in a complex ion with a different ligand. • Causes a colour change. • If the ligands are similar size then the coordination number doesn’t change. • If the ligands are different sizes then the coordination number changes and so does the shape of the complex. • Ligand exchange can occur partially. • E.g. [Cu(H2O)6]2+(aq) + 4NH3(aq) [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l) • Shape changes from octahedral to elongated octahedral. • E.g. [Cr(H2O)6]3+(aq) + SO42-(aq)  [Cr(H2O)5(SO4)]+(aq) + H2O(l) • Shape changes from octahedral to distorted octahedral. • Can be reversed, unless the new complex is more stable than the original complex. • If the new bonds between the ligand and the transition metal ion are stronger than the original bonds. • Bidentate ligands for more stable complexes than monodentate ligands. • Polydentate ligands form more stable complexes than bidentate and monodentate ligands. • This is because the number of molecules increases so the entropy is likely to increase.

  21. What reactions do ligands undergo? • Addition of sodium hydroxide produces insoluble metal hydroxides by disproportionation. • Metal aqua 3+ ions: • [M(H2O)6]3+(aq) + H2O(l) (equil)[M(H2O)5(OH)]2+(aq) + H3O+(aq) • Adding OH- ions shifts the equilibrium to the right because H3O+ ions are removed. • [M(H2O)5(OH)]2+(aq) + H2O(l) (equil) [M(H2O)4(OH)2]+(aq) + H3O+(aq) • Adding OH- ions shifts the equilibrium right to replace the H3O+ ions that are removed. • [M(H2O)4(OH)2]+(aq) + H2O(l) (equil) M(H2O)3(OH)3(s) + H3O+(aq) • M(H2O)3(OH)3(s) is insoluble in water because it is uncharged, therefore is seen as a white ppate. • Metal aqua 2+ ions: • Same process occurs as with 3+ ions. • [M(H2O)6]2+(aq) + H2O(l) (equil) [M(H2O)5(OH)]+(aq) + H3O+(aq) • [M(H2O)5(OH)]+(aq) + H2O(l) (equil) M(H2O)4(OH)2(s) + H3O+(aq) • Ammonia dissolves in water: • NH3 + H2O (equil) NH4+ + OH- • Adding a small amount of ammonia solution gives the same result as adding sodium hydroxide because OH- ions are formed in the equilibrium. • Adding excess ammonia solution can cause ligand exchange to occur, producing a soluble metal ion complex.

  22. What colour changes occur when complexes react?

  23. What oxidation states can copper exist as? • +1 (Cu+): • Stable in solid copper(I) compounds. • Can be coloured, such as Cu2O, which is red. • Unstable in aqueous solution, disproportionating to Cu2+(aq) and Cu(s) • 2Cu+(aq)  Cu2+(aq) + Cu(s) • Can form some stable ligands such as [CuCl2]- forms if there are excess Cl- ions to stabilise the copper(I) and prevent it disproportionating. • Complexes cannot be coloured because the 3d subshell of copper(I) ions is full. • +2 (Cu2+): • Stable in aqueous solution. • Reduced to Cu metal by more electrophilic metals via a displacement reaction. • E.g. Cu2+ + Zn  Cu + Zn2+

  24. What oxidation states can chromium exist as? • Chromium is used to make stainless steel and added to steel to make it harder. • +2 (Cr2+): • Least stable oxidation state. • +3 (Cr3+): • Most stable oxidation state. • Violet colour solution when surrounded by 6 water ligands. • Green colour solution when some of the water ligands are substituted for impurities in water such as Cl- ligands. • +6 (CrO42-): • Chromate(VI) ions. • Good oxidising agent as it is easily reduced to Cr3+. • +6 (Cr2O72-): • Dichromate(VI) ions. • Good oxidising agent as it is easily reduced to Cr3+ using a reducing agent such as zinc and dilute acid.

  25. What reactions do different oxidation states of chromium do? • Dichromate(VI) ions can be reduced to Cr3+ ions using a reducing agent such as zinc and dilute acid. • Cr2O72-(aq) + 14H+(aq) + 3Zn(s) 3Zn2+(aq) + 2Cr3+(aq) + 7H2O(l) • Colour change from orange to green. • Cr3+ ions can be reduced to Cr2+ ions in an inert atmosphere using a zinc. • 2Cr3+(aq) + Zn(s) Zn2+(aq) + 2Cr2+(aq) • Colour change from green to blue. • Cr2+ is so unstable that it oxidises straight back to Cr3+ ions in air. • Cr3+ can be oxidised to chromate(VI) ions by hydrogen peroxide in alkaline solution. • 2Cr3+(aq) + 10OH-(aq) + 3H2O2(aq) 2CrO42-(aq) + 8H2O(l) • Colour change from green to yellow. • Dichromate(VI) ions can be converted to chromate(VI) ions using a strong alkali. • Cr2O72- + 2OH- 2CrO42- + H2O • Chromate(VI) ions can be converted to dichromate(VI) ions using an acid. • 2CrO42- + 2H+ Cr2O72- + H2O

  26. What is chromium(II) ethanoate? • Cr2(CH3COO)4(H2O)2 • Made using chromium(II) chloride solution. • Chromium(II) chloride is reduced with zinc in acid solution to produce Cr2+ ions. • 2Cr3+ + Zn  2Cr2+ + Zn2+ • Colour change from green to blue. • Sodium ethanoate is mixed with the solution to produce a red ppate of chromium(II) ethanoate. • 2Cr2+ + 4CH3COO- + 2H2O  [Cr2(CH3COO)4(H2O)2] • The complex is easily oxidised so it has to be made in an inert atmosphere such as nitrogen. • HCl added to a flask containing chromium(II) chloride solution and zinc mesh to reduce Cr3+ ions to Cr2+ ions. • Hydrogen produced can escape through a rubber tube into a beaker of water. • When the solution turns clear blue, pinch the rubber tube shut so hydrogen can no longer escape. • Pressure builds up in the flask, forcing Cr2+ ions through the open glass tube into a flask of sodium ethanoate. • Red ppate of chromium(II) ethanoate forms. • Ppate filtered off and washed using water, then ethanol, then ether.

  27. Why is chromium hydroxide amphoteric? • Can react with both acids and alkalis. • [Cr(H2O)3(OH)3](aq) + 3H+(aq)  [Cr(H2O)6]3+(aq) • [Cr(H2O)3(OH)3](aq) + 3OH-(aq)  [Cr(OH)6]3-(aq) + 3H2O(l) • Zn is the only other 3d element that can do this. • Not ligand exchange, but ligand is chemically modified.

  28. How do transition metals catalyse reactions? • Good catalysts because they can use their s and d orbitals to form bonds with reactants and can change oxidation state easily. • Can act as homogeneous or heterogeneous catalysts. • Platinum and rhodium are used in catalytic converters in cars to convert nitric oxide and carbon monoxide to nitrogen and carbon dioxide. • Catalyst used is a mesh or fine powder to increase surface area. • 2NO(g) + 2CO(g)  N2(g) + 2CO2(g) • Reactant molecules bond to the surface of the solid catalyst (adsorption). • Bonds between the reactant atoms are weakened and break, forming radicals. • Radicals join to make the new molecules. • The Contact process produces large quantities of sulfuric acid using a vanadium catalyst. • 2SO2(g) + O2(g)  2SO3(g) • Vanadium(V) oxide reacts with sulfur dioxide and is reduced to vanadium(IV). • V2O5(s) + SO2(g) V2O4(s) + SO3(g) • Vanadium(IV) oxide is oxidised back to vanadium(V) by oxygen. • 2V2O4(s) + O2(g) 2V2O5(s)

  29. How can catalyst development be used to make chemistry greener? • Allow reaction to occur at a lower temperature or pressure. • Saves energy and reduces cost. • Can improve atom economy. • Reduces amount of raw materials that need to be used. • Reduces amount of waste products. • E.g. Making ethanoic acid from methanol and carbon monoxide: • Originally made by oxidising butane but atom economy was very low. • Methanol first used to make ethanoic acid in the 1960’s using a catalyst of cobalt/iodine. • Methanol is a cheaper, more environmentally friendly raw material than butane. • Higher atom economy than using butane. • Rhodium/iodine catalyst developed in 1966, which requires lower temperature and pressure than cobalt/iodine catalyst. • Iridium/iodine catalyst developed in 1980’s made reaction faster and more efficient.

  30. What are transition metals used for? • Construction because they are strong and unreactive. • Jewellery because they are malleable and inert, and often look shiny. • Paint pigments/coloured glass. • Chemotherapy: • Cisplatin is a complex of platinum(II) with two chloride ions and two ammonia molecules. • Square planar shape. • Chloride ions next to each other. • Prevents cancer cells from dividing by crosslinking the DNA, causing the cells to die. • Also prevents normal cells dividing including in the blood, so can supress the immune system, increasing the risk of infection. • Can damage the kidneys. • Polychromic sunglasses: • Sunglasses embedded with a silver halide which decomposes in UV radiation to form silver atoms. • Silver atoms darken the lenses. • Process reverses when there is no UV, reforming the silver halide, lightening the lenses.

  31. What are aromatic compounds? • Derived from benzene. • Also known as arenes. • Named as substituted benzene rings or as compounds with a phenyl group. • Benzene = C6H6 • Kekule’s structure and delocalised structure are the two ways of representing benzene. • Kekule’s structure consists of carbon atoms in a ring with alternating single and double bonds. • X-ray diffraction studies show all carbon-carbon bonds in benzene are the same length, between the length of C-C and C=C bonds. • IR studies have shown no carbon-carbon bonds in benzene are C-C or C=C bonds because they absorb energy at different frequencies. • Delocalised model is the now accepted model. • P-orbitals of all 6 carbon atoms overlap to create pi bonds. • Two ring shaped electron clouds are formed; one above the plane and one below the plane. • Electrons in the ring are delocalised, and represented by a circle inside the carbon ring. • All bonds are the same so are the same length.

  32. How does delocalisation of electrons give benzene stability? • Cyclohexane has one double bond. • Hydrogenation of cyclohexane to hexane gives an enthalpy change of -120kJmol-1. • If benzene had Kekule’s structure, you would expect the enthalpy of hydrogenation to be -360kJmol-1. • Actual enthalpy of hydrogenation of benzene is -208kJmol-1. • Less exothermic than the expected enthalpy of hydrogenation. • More energy is needed to break the bonds of benzene than would be required to break the bonds if benzene had Kekule’s structure. • This is because the delocalised system gives benzene more stability.

  33. How does benzene react with oxygen? • Benzene is a hydrocarbon. • Burns in oxygen to produce carbon dioxide and water. • 2C6H6 + 15O2 12CO2 + 6H2O • Burns in air to give a smoky flame because there is too little oxygen so the benzene does not burn completely. • Carbon atoms stay as carbon and form soot particles in hot gas.

  34. How do arenes react with nitronium ions? • Electrophilic substitution. • Benzene warmed with conc nitric acid and sulfuric acids to produce nitrobenzene. • Nitrobenzene forms as a layer floating on top of the acid. • Sulfuric acid acts as a catalyst to make the nitronium ion, NO2+. • HNO3 + H2SO4 H2NO3+ + HSO4- • H2NO3+  NO2+ + H2O • H+ + HSO4-  H2SO4 • The nitronium ions acts as an electrophile. • Temperature must be kept below 55 degrees for mononitration, as above this, extra substitutions occur.

  35. How do arenes react with sulfur trioxide molecules? • Electrophilic substitution. • Benzene warmed with fuming sulfuric acid at 40 degrees for about 30 mins. • Fuming sulfuric acid is sulfur trioxide, SO3, dissolved in sulfuric acid. • Electrophile is sulfur trioxide, SO3.

  36. How can halogen carriers be used to make good electrophiles? • Electrophiles have a strong positive charge. • Halogen carriers can be used as a catalyst to make strong electrophiles. • E.g. aluminium halides, iron halides and iron. • Halogen carriers accept lone pairs of electrons from the halogen in the electrophile. • This causes polarisation of the R-X bond to increase, forming a carbocation. • RCl: - - - > AlCl3 R+AlCl4-

  37. What is a Friedel-Crafts alkylation reaction? • Halogenoalkane reacts with benzene via electrophilic substitution to produce alkylbenzenes. • Refluxed in dry ether because aluminium chloride is sensitive to hydrolysis. • Halogen carrier of aluminium chloride is used as a catalyst. • RCl + AlCl3 R+ + AlCl4- • Alkylbenzenes are more reactive than benzene so more than one substitution occurs (polyalkylation).

  38. What is a Friedel-Crafts acylation reaction? • Acyl chlorides react with benzene by electrophilic substitution to produce phenylketones. • Heated under reflux with dry ether. • Acyl chlorides are polar so can easily lose chlorine atoms to form carbocations. • Halogen carrier such as aluminium chloride is used as a catalyst. • RCOCl + AlCl3 RCO+ + AlCl4- • Phenylketones are less reactive than benzene so only one substitution occurs.

  39. How do arenes react with halogens? • Cycloalkene shaken with bromine water. • Bromine adds to the cycloalkene to give dibromoalkane. • Bromine water is decolourised (can be used to test for alkenes). • Bromine isn’t a strong enough electrophile to react with benzene unless a halogen carrier catalyst is also used. • Halogen carrier polarises the halogen, creating a strong electrophile that can attack the benzene ring. • Br2 + FeBr3 Br+ + FeBr4- • Bromobenzene is formed.

  40. How does benzene react with hydrogen? • Finely divided nickel catalyst used. • Temperature of 150 degrees and pressure of 10 atm required. • Cyclohexane is produced. • C6H6 + 3H2 C6H12

  41. What reactions do methylbenzene and methoxybenzene undergo? • CH3 and OCH3 groups donate electrons to the delocalised ring, increasing the rings electron density, making it more reactive. • React faster than benzene. • Less toxic than benzene so they are safer to use. • Products will be slightly different but the mechanism is the same as when benzene is used.

  42. What is phenol? • Benzene ring with an –OH group substituted for an H-atom. • Phenol reacts by electrophilic substitution when shaken with bromine water, decolourising bromine water and forming 2,4,6-tribromophenol, which is an insoluble white ppate in water that smells of antiseptic. • Phenol reacts with bromine water because one of the electron pairs in the p-orbital of oxygen overlaps with the delocalised ring of benzene, increasing the electron density of the ring. • Phenol reacts with dilute nitric acid to produce two isomers of nitrophenol, and water. • Easier to nitrate than benzene because it has a higher electron density in the ring.

  43. What are amines? • Ammonia with one or more hydrogen replaced with an organic group. • Aliphatic amines are chained/branched. • Aromatic amines have a benzene ring. • Small amines have a slightly fishy smell, larger amines smell very fishy. • Solubility: • Small amines are soluble in water because amines can form hydrogen bonds to water molecules. • Bigger amines are less soluble in water because there are greater London forces between the amine molecules. • Amines form alkaline solutions when dissolved by forming alkyl ammonium ions and hydroxide ions. • React with acids to form salts: • Can act as bases because they have a lone pair on the nitrogen atom so can form a dative covalent bond with H+ ions (accept protons). • CH3CH2NH2 + HCl CH3CH2NH3+Cl-

  44. How do amines form complex ions? • Copper(II) sulfate solution forms [Cu(H2O)6]2+ complexes with water. • Adding a small amount of methylamine solution produces a pale blue ppate of [Cu(H2O)4(OH)2] by acting as a Bronsted-Lowry base and taking two H+ ions from the complex. • Adding excess methylamine solution causes the ppate to dissolve, producing a deep blue solution of [Cu(CH3NH2)4(H2O)2]2+ complex ions by ligand exchange.

  45. How are amines acylated to N-substituted amides? • Butylamine + ethanoyl chloride  N-butylethanamide • Butylamine + HCl  Butylammonium chloride • CH2COCl + 2C4H9NH2  CH3CONHC4H9 + [C4H9NH]+Cl- • Ethanoyl chloride must be added to a concentrated aqueous solution of amine. • Chloroethane + butylamine  ethylbutylamine + HCl • Lone pair of the nitrogen attacks the slightly positive carbon atom on the halogenoalkane, displacing the chlorine and making a salt. • Chlorine removes a hydrogen from the salt to produce the amine. • Ethanoyl chloride + phenylamine  N-phenylethanamide+ HCl • Phenylamine + HCl  phenylammonium chloride • Product stained brown with unreacted phenylamine. • Paracetamol is made by reacting ethanoyl chloride with p-aminophenol. • HCl is made as a waste product.

  46. What are amides? • Derivatives of carboxylic acids. • Functional group of –CONH2. • N-substituted amides are amides with one hydrogen replaced with an alkyl group. • Carbonyl group (C=O) pulls electrons away from the rest of the group. • Made by acylation of ammonia or amines. • Acyl chloride + ammonia  amide + HCl • Acyl chloride + amine  N-substituted amide + HCl • HCl can go on to react with ammonia/amine to form an ammonium salt.

  47. What are amines? • Made by reducing a nitro compound such as nitrobenzene. • Heated with tin and concHCl under reflux to make an ammonium salt. • Sodium hydroxide solution added to remove H+ ions from the salt, forming an aromatic amine. • Nitrobenzene + 6[H]  phenylamine + 2H2O. • Used to make azo dyes: • Azo dyes contain the group –N=N- • Made by reacting aromatic amines with phenols. • Azo group links two aromatic groups. • Two aromatic groups give stability to the molecule. • Azo group becomes part of the delocalized electron system. • Colours are a result of light absorption by the delocalised electron system.

  48. How are azo dyes made? • Diazonium salt made by reacting nitrous acid with phenylamine: • Nitrous acid made in situ from sodium nitrite and HCl because it is unstable. • NaNO2 + HCl HNO2 + NaCl • Nitrous acid reacts with phenylamine and HCl to form benzenediazonium chloride. • Temperature must be about 5 degrees to prevent phenol being made instead. • Diazonium salts are unstable and decompose quickly so the azo dye must be made straight away. • Azo dye made from diazonium salts by a coupling reaction with phenol. • Phenol is dissolved in sodium hydroxide to make sodium phenol. • Sodium phenol is stood in ice and chilled benzenediazonium chloride is added. • Azo dye ppates out of the solution. • Lone pairs on the oxygen of phenol act as a coupling agent, increasing the electron density of the benzene ring, giving the diazonium ion electrophile an electron rich area to attack.

  49. What are amino acids? • Contains amino group (NH2) and carboxyl group (COOH). • Amphoteric because the amino group is basic and the carboxyl group is acidic. • Alpha amino acids have the amino group and the carboxyl group attached to the same carbon atom. • General formula RCH(NH2)COOH. • Monomers used to make proteins. • Can exist as zwitterions. • Dipolar ions with both a positive and a negative charge in different parts of the molecule. • Can only form zwitterions near the amino acid’s isoelectric point; the pH where the average overall charge of the amino acid is zero. • In conditions more acidic than the isoelectric point, the NH2 group is likely to be protonated to NH3+. • At the isoelectric point, the carboxyl group and the amino group are ionised. • In conditions more basic than the isoelectric point, the COOH group is likely to lose protons, becoming COO-.

  50. What is paper chromatography? • Used to identify unknown amino acids. • Pencil line drawn near the bottom of a piece of chromatography paper. • Spot of amino acid mixture put on the line. • Bottom of the paper is dipped into solvent. • As the solvent spreads up the paper, the amino acids move at different rates so separate. • When the solvent reaches the top of the paper, the paper is removed form the solvent and a line is made to mark the distance the solvent travelled (solvent front). • Amino acids are colourless so they are located using ninhydrin solution. • Ninhydrin reacts with amino acids to produce ammonia, aldehydes, carbon dioxide and hydrindantin. • Hydrindantin reacts with ammonia and ninhydrin to give a purple pigment called Ruhemann’s purple. • Amino acid spots turn purple. • R1 value = distance travelled by spot/distance travelled by solvent. • Table of R1 values can be used to identify the amino acids in the mixture.

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