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A2 Unit 4 Kinetics, Equilibria and Organic Chemistry AQA A2-LEVEL Student Guide to Unit 4 Kinetics, Equilibria and Organic Chemistry See me in glorious All programs Shared Areas at: Chemistry Read Chemistry Mr Lund’s Classes A2 Chemistry Unit 4 Mr Lund 08 May 2017 1 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry How Science Works: A, B Equilibria dynamic (not static - do you know the difference!) equilibrium is achievable in a closed system (e.g. solutions in a test tube if there are no gaseous reactants/products). rates of the forward and reverse reactions at equilibrium are identical concentrations are unlikely to be 50 – 50 Le Chatelier’s Principle and Qualitative Aspects can you unambiguously write down what LCP states (see 148 of the AS textbook)? remember this is a predictive tool used to determine the effect on the position of equilibria when a change in concentration, temperature or pressure is made it is NOT an explanation of WHY it happens so avoid statements such as ‘because of LCP’, ‘LCP causes …’ and learn to state ‘LCP predicts that …..’ LCP is not suggesting that a system completely reverses a temperature change when establishing a new equilibrium as the new equilibrium will be that for the changed temperature it implies that the shift in the position (in terms of reactants and products) of equilibria is in the direction that seems to minimize the effect of that change a new position of equilibria in which the relative rates of the forward and backward reaction are once again in balance under the new set of conditions is eventually arrived at the position of equilibria is changed by: concentration (which can be easily understood using rates/collision theory) temperature (EA will be larger for the endothermic process so it will be relatively more favoured by a rise in temperature) pressure (only applicable where there is an imbalance between the number of moles of gaseous particles on either side of the equation) catalysts - do not affect position of equilibrium, just the time to achieve it Monitoring Equilibria remote sensing – this is non intrusive (e.g. level of absorbance of a given wavelength of light by a coloured solution) so will not effect the position of equilibrium titrimetric analysis – this will effect the position of equilibria (since the concentration of one of the reactants or products will be changed) so is only applicable to a system with a slow response to a change in conditions or where quenching (dilution or cooling) is used to slow down the rate of reaction and thus the effect of the investigative technique A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 20 and Chapter 9 in the AS Book 16 – 17 Q 1 - 3 on page 16 – 17 Q 1 on page 32 Chemical equilibrium Mr Lund 08 May 2017 2 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry This equation can vary and simply illustrates the principle behind its construction Equilibrium Law (i.e. equation) [C]c aA + cC + dD Kc = [A]a eqm eqm [B]b eqm eqm only applies to systems at equilibrium so don’t use initial concentrations Kc is calculated from concentrations at equilibrium (note that the concentration of a solid is constant and so solids do not appear in the equilibrium expression) equilibrium constant (Kc) is related to reaction stoichiometry (a, b, c etc) and is a constant at constant temperature the value of Kc is an indicator of the position of equilibrium (reverse reaction = inverse value) the value of Kc is not indicative of how fast the reaction proceeds you must be able to calculate the numerical value of Kc (possibly using data from an experiment you will carry out yourself) note that concentrations are used i.e. moles/volume not moles although quite often V will cancel down or cancel out completely (when the ∑powers is the same on the top as the bottom) – but show ALL working in exams determination of the units of Kc – you must show workings in the exam but ….. check using (moldm-3)∑top powers - ∑bottom powers Summary Questions Exam Style Questions bB [D]d Page 21 Page 28 1-3 1 determination of the concentrations of reactants/products present at equilibrium, given appropriate data including the numerical value of Kc this may involve the determination of an unknown value of x to solve an equation (but not a quadratic one on this syllabus) you may be required to realise that the equation can be simplified by taking the square root of both the top and the bottom terms and the value of Kc(see the example on page 22 – 23) you might also be asked to determine the amount of given reactant required in order to produce a given amount of product (see the example on page 23 – 24) Summary Questions Exam Style Questions Page 24 Page 28 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 1 2 20 – 24 18 – 20, 22 - 27 Q 4 – 6, 8 - 13 on pages 18 – 27 Q 3 on page 33 Equilibrium equation Mr Lund 08 May 2017 3 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Changing Conditions and Equilibria Pressure or Concentration Change the value of Kc does not change with variations in concentration or pressure to improve yield wrt a more expensive reactant a cheaper reactant could be added. removing the product as it is formed (assuming that the response of the system is fast enough) would also improve yield whist also allowing reactants to be recycled (see Haber) pressure only has an effect IF gaseous particles are involved and in addition the stoichiometric ratio of gaseous particles is unequal on either side of the equation in all cases a pressure increase will increase the rate of reaction involving gaseous reactants Kc and Temperature Change value of Kc increases as temperature increases for endothermic reactions i.e. the equilibrium shifts to the RHS i.e. more products value of Kc decreases as temperature increases for exothermic reactions i.e. the equilibrium shifts to the LHS i.e. less products Temperature increases decreases Exothermic reaction Kc decreases Kc increases Endothermic reaction Kc increases Kc decreases Kc and Catalysts value of Kc DOES NOT CHANGE when a catalyst is used therefore catalysts do not change the position of equilibria i.e. change the yield HOWEVER, the rate of the reaction will be faster (for both the forward and backward reaction) hence equilibrium will be achieved sooner adding a catalyst to a system already at equilibrium will not change its position How science works Summary Questions Exam Style Questions Page 26 Page 27 Page 28-29 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide Development of the Haber Process 1-3 3–5 25 - 27 18 – 27 Q 4 – 13 on pages 18 – 27 Q 3, 5 on page 33 Equilibrium equation Mr Lund 08 May 2017 4 How Science Works: E, F A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Kinetics review basic ideas of collision theory and activation energy from AS Rate Equations the rate of reaction is the rate of change of either the reactants or products in a chemical reaction the rate can be determined from concentration versus time graphs the mathematical expression for the graph on page 4 of reaction A + 2B →C: Rate = Don’t panic !! – this can be explained easily in words – ask your teacher there are different ways to express rate (i) (ii) (iii) d [C ] d [ A] 1 d [ B] dt dt 2 dt average rate instantaneous rate (this is what the equations above depict) which is the gradient of the tangent of the curve at selected values of concentration or time initial rate is the instantaneous rate right at the start of a reaction initial rate (t0) – the gradient is easiest to determine with confidence since it goes through zero Measuring Reaction Rates see AS Module 2 guide for general practical techniques for monitoring progress in a given reaction generally the main methods comprise two different strategies 1 following a single reaction – measuring colour change, change in conduction/pH of one product or reactant then plotting a concentration time graph and determining the rates at given concentrations 2 clock techniques – measuring the time to an observable event from known different initial conditions e.g. the ‘thiosulphate cross’ (the one you did at GCSE) or the ‘iodine clock’ this must be relatively early in the reaction so that the concentration of other reagents (other than the one being varied) can be deemed unchanged. initial rate How science works Summary Questions 1 t Page 5 Page 6 ‘Damn fast reactions indeed’ 1-5 Mr Lund 08 May 2017 5 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Rate Equations and Order of Reaction each reactant may or may not affect reaction rate (so it’s not quite as clear cut as it seemed at GCSE/AS) ALSO it is not necessarily directly proportional when it does – we shall see why later for a given reactant we can state Rate [reactant]n for two reactants A and B the general rate expression is: Rate [A]m[B]n moldm-3s-1 Note that rate always has these same units. m and n are the order of reaction for each reagent (values are limited to 0, 1 or 2 at A-Level) overall order = individual orders) unlike with equilibria the orders of reaction and thus the overall rate equation cannot be determined from reaction stoichiometry they can only by determined by experiments e.g. 2H2(g) + 2NO(g) 2H2O(g) + N2(g) doubling [NO] quadruples the rate i.e. double2 = 4 doubling [H2] doubles the rate Rate [H2][NO]2 i.e. [NO]2 The effect on rate of the change here… NOT as suggested by the ratio of reactants rate [H2]2[NO]2 k is the rate constant (units depend upon overall order) rate = k[H2][NO]2 ..is raised to this power i.e. its second order wrt to [NO] moldm-3s-1 units of k will vary depending upon the overall order of the reaction determination of the units of k for 0, 1, 2 and 3rd order reactions overall should be attempted as they are likely in an exam and here is another ‘cheat’ for checking your answer: units of k = (moldm-3)1 - overall order (s-1) note that if the order is 0 this term will not appear in the rate equation e.g. A2BC0 = A2B x 1 = A2B Mr Lund 08 May 2017 6 Maths tip: Any number raised to the power zero = 1 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Catalysts and Rate Equations catalysts will be involved in the rate expression (this might only be a modification of the value of k itself for example a solid in heterogeneous catalysis) H+ e.g. CH3COCH3(aq) + I2(aq) CH2ICOCH3(aq) + H+(aq) + I-(aq) by experimentation it was found that the reaction is: first order wrt [H+] and [CH3CO CH3] zero order wrt [I2] i.e. second order overall rate [CH3COCH3]1[ H+]1[I2]0 rate = k[CH3COCH3][ H+] moldm-3s-1 what are the units of k? Rate Determining Step most reactions occur in several steps (this is exemplified by organic reaction mechanisms) each step will take place at a different rate the slowest step will determine the overall rate of the reaction and is known as the rate determining step the order of the reaction regarding each reagent can provide information regarding its involvement in the rate determining step obviously a reactant with 0 order will not be involved in the rate determining step study of reaction kinetics can yield important information regarding the mechanism of a multi-step reaction in the reaction above iodine would not be involved in the rate determining (slow) step at this point you might ask your teacher to explain why there are variations in the mechanism for the hydrolysis of a haloalkanes as discussed on pages 15 – 16 (or perhaps research SN1 and SN2 yourself) Summary Questions Page 9 Page 16 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 1–4 1 4 – 9, 14 - 16 4–6 Q1, 2, 3 on page 5-6 Order of reaction, rate expression, SN2 Mr Lund 08 May 2017 7 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Concentration/Time and Rate/Concentration Graphs rate/concentration graphs can show the order with respect to a given reagent it is glossed over at A-Level that the other reagents will have to be present in xs so that their concentration can be deemed to be unchanged during the course of a reaction where a series of measurements of the concentration of a given chemical are measured. or – in a clock technique the event measured in relatively early in the reaction so better reflects the known initial concentration of each reagent zero order e.g. decomposition of ammonia (tungsten catalyst), rate is independent of concentration i.e. graphically it is a flat horizontal line 2NH3(g) W 3H2(g) + Do you have an idea as to why it might be zero order? N2(g) first order e.g. thermal decomposition of dinitrogen monoxide to nitrogen and oxygen (gold catalyst), rate is directly proportional to concentration and will be a straight line gradient = k (any points not on the straight line will be anomalies and require identification and explanation (e.g. temperature variations)) second order e.g. thermal decomposition of ethanal to methane and carbon monoxide will produce a graph that curves upwards (rate against concentration2 is a straight line) Initial Rates Method initial reaction rates are determined by plotting the tangent to the time/concentration graph for different initial reagent concentrations at the start of the reaction (t = 0) when reaction concentrations are accurately known (and at a fixed temperature/catalyst) the gradient of this line is the initial rate orders of reaction can be determined from initial rates data by inspection the value of k can also be determined from this data the best way to grasp this idea is to try examples Summary Questions Exam Style Questions Page 13 1 Page 17 - 19 1, 3, 5 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 10 - 12 7 – 11 Q4 on page 8 Q1 on page 14 rate Mr Lund 08 May 2017 8 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Variation of k with Temperature Maxwell-Boltzmann Distribution Curve distribution of energies amongst particles at different temperatures give rise to the MaxwellBoltzmann distribution curves based on the Arrhenius expression k Ae A R T = = = EA RT Arrhenius constant (determined by collision frequency and orientation factor) gas constant (8.31 JK-1mol-1) absolute temperature (K) the main significance of this equation (which you don’t need to know – unless you are trying to understand the subject) is that a small rise in temperature has an exponential (i.e. big) effect on rate Extra info for those who do Maths – how can we get values for EA and A ln(A) – ln(k) = 1 E - A R T + ln(A) m + c y = Plot ln(k) against 1/T gives gradient = - Exam Style Questions Page 17 - 18 Page 158 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide EA RT ln(k) = x EA R and intercept = ln(A) 2,4 1 10 - 12 12 – 13 Q2-4 on page 14 - 15 Q1 on page 150 Arrhenius expression Mr Lund 08 May 2017 9 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Acid-Base Equilibria Arrhenius definition of an acid - hydrogen ions and oxonium ions (H3O+(aq)) soluble base = alkali alkaline solutions have relatively high [OH-(aq)] Brønsted-Lowry Theory of Acids and Bases Arrhenius limited to aqueous solutions but acid-base concept is broader e.g.: NH3(g) + HCl(g) NH4Cl(s) acid-base equilibria involve proton transfer acids are proton donors, bases are proton acceptors water is amphoteric it behaves as an acid with NH3 and as a base HCl conjugate acid-base pairs undergo proton exchange competition occurs for protons between bases on either side of the equilibria relative strength of the base determines the equilibria bias a relatively strong base has a relatively weak conjugate acid and visa-versa note that in the protonation of nitric acid in the nitration of benzene – nitric acid acts as a base You must make the effort to learn the definition of a BrønstedLowry acid and base Ionic Product of Water (Kw) water undergoes slight auto-ionisation (dissociation of the water molecule) ionic product of water – again the equation defines the concept – and don’t forget units Kw = variation of Kw Summary Questions [H+(aq)][OH-(aq)] 1 x 10-14 mol2dm-6 at 298K its value increases as temperature increases - LCP - (hence [H+(aq)] increases and therefore the pH of a neutral solution decreases!) Page 31 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide = 1-3 30 - 31 34 – 35, 36, 42 - 43 Q1 on page 35 Q6 on page 37 Brønsted, ionic product Mr Lund 08 May 2017 10 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry pH Calculations define pH using an equation NOT in words as it is more certain to get full marks. pH = -log10[H+(aq)] learn to use your calculator ! pH of monoprotic (release a single proton into aqueous solution) acids e.g. HCl (note: pH’s lower than 1, including –ve values, are possible) you should be able to calculate pH after a strong acid of known volume and concentration (or pH) is diluted by a known volume of water pH of diprotic acids calculating [H+(aq)] from pH values [H+(aq)] = 2 x [H2SO4(aq)] NOTE: [H+(aq)] = 10- pH learn to use your calculator !!!!!! calculating the pH of alkalis using Kw to calculate [H+(aq)] note for Ca(OH)2 that [OH-(aq)] = 2 x [Ca(OH)2(aq)] but why might it give a weaker acid than sodium hydroxide? [OH-(aq)] = Kw [H (aq) ] How science works Summary Questions Page 33 Page 31 Measuring pH Questions 1 - 5 Exam Style Questions Page 50 Questions 3 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 32 – 35 38 – 40, 42 - 45 Q7 - 10 on pages 39 - 40 Q13, 14 on pages 37 – 38 Q 3, 4 on page 55 Q 7 on page 56 Strong acids pH Mr Lund 08 May 2017 11 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Weak Acids and Ka weak acids are only partially ionised HA(aq) + Kc = H3O+(aq) H2O(l) + A-(aq) [H 3 O (aq)][A - (aq)] [HA(aq)][H 2 O (l) ] [H2O] ~ constant [A-(aq)] = [H3O+(aq)] = [H+(aq)] given that it is a weak acid we can assume the degree of dissociation is minimal hence: Ka = [acid]equilibrium ~ [acid]initial Kc[H2O] = [H (aq)][A - (aq)] [HA(aq)] Ka = [H (aq)] 2 [HA(aq)] pKa = -log10Ka Ka = 10- pKa relatively higher Ka / relatively lower pKa = stronger acid (given the same concentration) value is independent of concentration and therefore more useful you should be able to calculate the pH of a weak acid of known concentration using Ka and calculate Ka from the pH of a weak acid of known concentration Summary Questions Exam Style Questions Page 38 Page 51 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 1-3 6 32 – 35 35 – 36, 40 – 42 Q11 - 12 on pages 41 - 42 Q 1 on page 55 Weak acid Mr Lund 08 May 2017 12 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Titration Curves as an alkali is added to an acid the pH increases and visa-versa you should also be able to calculate the pH at any point in the addition of sodium hydroxide to a monoprotic acid (and visa-versa) including weak acids typical errors to avoid are not converting to moles and getting the stoichiometric ratio wrong most likely you will forget to use the total volume of the solution created and thus get the concentration wrong and therefore the pH another likely error with a weak acid where xs alkali has not been added is assuming that the remaining acid is fully dissociated i.e. forgetting to use Ka to determine [H+(aq)] equivalence point is where two solutions have reacted in stoichiometrically the correct molar ratio – this will be the vertical point on the titration curve where the pH changes markedly pH’s at equivalence point and appropriate curves for: WASB SASB SAWB WAWB http://www.avogadro.co.uk/chemeqm/acidbase/titration/phcurves.htm pay particular attention to the position of the initial pH for strong and weak acids and look carefully at how it changes at the start also carefully note the position of the equivalence point – the mid point of the vertical section finally ensure that a sensible final pH is shown to reflect the use of a strong or a weak base. you should be able to calculate concentrations of an unknown acid or alkali from the results of a titration – pretty much as was the case for AS level – but there will be more likely hood of diprotic acids cropping up (e.g. sulphuric acid) Summary Questions Page 41 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 1-2 39 – 41 44 – 48 Q15 on page 45 Q 16 on page 48 Q 2, 8 on pages 55 - 56 Titration curves Mr Lund 08 May 2017 13 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Ka of Weak Acids from Titration Curves at the half equivalence point (half neutralisation point as stated in your text book) given that: [HA] = Ka = pKa = [A] [H+(aq)] pH the half equivalence point can be determined practically by determining the pH at half the equivalence point (half the volume) from a plotted titration curve note that it is not half the pH value itself at the equivalence point that is used! alternatively a titration can be repeated with half the volume of the already determined equivalence point and the pH then measured using a pH meter End Point of an Indicator indicators can only be used for a titration curves with a vertical section of >2 pH units suitable indicators for acid-base titration’s will have an end point and range that lie within that vertical section i.e. will thus exhibit a sharply defined colour change pH meters can be used on coloured solutions indicators are weak acids H+(aq) + In-(aq) HIn(aq) [H(aq) ][In-(aq) ] [HIn(aq) ] Ka (or Kin) = at end point [In-(aq)] = [HIn(aq)] Summary Questions Exam Style Questions Kin = [H+(aq)] pKa = pKin Page 41 Page 50 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide = 1-2 4 39 - 41 49 – 51 Q17 - 18 on pages 50 – 51 Q 5 on page 56 Indicators Mr Lund 08 May 2017 14 pH A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Buffer Solutions the effect of pH changes e.g. lemon juice and the proteins in milk buffers are designed to maintain pH stability you must understand buffer solution in terms of the response of an equilibrium system to the addition of hydrogen or hydroxide ions Acidic Buffer Added H+(aq) is removed by A-(aq) provided by the salt as the equilibrium shifts to the left HA ⇋ H+ + A- reservoir of HA provided by acid Added OH-(aq) is removed by reacting with H+(aq) provided by the acid as the equilibrium shifts to the right NaA An acidic buffer (for pH <7) consists of a weak acid and its soluble salt Na+ + A- reservoir of A- provided by the sodium salt of the acid Basic Buffer Added H+(aq) is removed by NH3 as the equilibrium shifts to the right NH3 + H+ ⇋ NH4+ reservoir of NH3 provided by weak base Added OH-(aq) is removed by reacting with NH4+(aq) provided by the salt and the equilibrium shifts to the left NH4Cl An alkaline buffer (for pH >7) consists of a weak alkali and its soluble salt NH4+ + Cl- reservoir of NH4+ provided by the ammonium salt Mr Lund 08 May 2017 15 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Important Buffer Solutions Carbonic acid-Bicarbonate Buffer in the Blood is by far the most important process for maintaining the acid-base balance (in our bodies there are also phosphate and protein buffers). Added H+(aq) is removed as the equilibrium shifts to the right Added OH-(aq) is removed by reacting with H+(aq) shitting the equilibrium to the left a pH change of over ± 0.5 can be fatal buffer solutions are also used in: the food industry (‘acidity regulators’) fabric dyeing – where ‘AZO’ dyes are used hair care products – which are kept slightly acidic ‘pH 5.5’ as alkaline conditions make hair look rough as microscopic scales on the surface of the hair are made to stand up. calibrating pH meters biochemical research (enzymes are denatured by pH extremes) and there are numerous other areas Preparation of Buffer Solutions a buffer solution can be made in two different ways: 1 2 adding a suitable soluble salt to an acid partially neutralising a weak acid with a strong base up to the required pH when [HA] = [A-] the buffer solution is equally able to deal with the addition of acid and base by equal sized reservoirs for this reason a weak acid with a pKa relatively close to desired pH is selected for more effective buffering (see the calculations in the next section) Mr Lund 08 May 2017 16 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Calculating the pH of a Buffer Solution HA ⇋ H+ + A- but [H (aq)][A - (aq)] [HA(aq)] [HA] ~ [ACID]i and [A-] = [salt] Ka = Ka = [H (aq)][SALT ] [ACID)] [H+(aq)]= [ACID] [SALT] it is assumed that the volume of the weak acid solution is unchanged by the addition of a small quantity of its solid salt since all particles are present in the same total volume of solvent we can make life easier by appreciating that the acid to salt: moles ratio Ka Will the dilution of a buffer solution change the pH? = concentration ratio the best buffer will be that obtained at half the equivalence point: and so [ACID] = [SALT] [H+(aq)] = Ka = pKa pH equimolar amounts of acid and salt produces a buffer solution with a pH of the same numerical value as pKa you should be able to determine the required combination of acid and salt to produce a buffer solution of a given pH you should also be able to demonstrate by calculation that adding acid or alkali to a buffered solution changes the pH by less than for an un-buffered solution Summary Questions Exam Style Questions Page 48 Page 49 Page 51 Page 160 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 1 1, 2 5 5 45 - 48 52 – 54 Q17 - 20 on pages 52 – 54 Q6, 9 on page 56 Buffer Mr Lund 08 May 2017 17 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry How Science Works: H Organic Nomenclature displayed formula – every bond and every atom should be shown in examinations!!! structural formula e.g. CH3CH2CH2OH skeletal formula may be useful for prospective medical etc students it is important that you understand the difference between 2-D displayed formula and the actual 3-D molecular shape as this will prove particularly relevant later functional group – one or more reactive sites on a hydrocarbon skeleton homologous series – same general formula + same functional group (hence same chemistry as the increase in chain length has little effect on their chemical reactivity) nomenclature is based on four criteria: root suffix prefix locant longest unbranched hydrocarbon chain (including main functional group) determines principal functional group (a pecking order exists) other changes to root molecule (e.g. side chains, functional group) position of branch or substituent (e.g. double bond) on the main chain look for the longest chain NOT the longest ‘straight’ chain of carbons start numbering at the end of the chain that results in the lowest numbers in the name or for the primary functional group position alphabetical order is used where more than one type of functional group or branch is required in the prefix mono, di, tri and tetra indicate multiple functional groups or branches of a given type (don’t change alphabetical order) commas and dashes are important - so learn how to do this correctly !!!! An “e” is dropped if the next letter is a vowel: “propan-2-ol, propane-1,2-diol” An “a” is added if inclusion of di, tri, etc., would put two consonants consecutively: “buta1,3-diene”, not “but-1,3-diene” “propanenitrile, not propannitrile or propanitrile.) you will need to aware of nomenclature examples of the following (including cyclic variations): alkanes alkenes haloalkanes alcohols aldehydes and ketones carboxylic acids esters acyl chlorides amines amides amino acids benzene and its derivatives Summary Questions Exam Style Questions Page 56 Page 64 A2 Chemistry (Nelson Thornes) AQA Chemguide 1-3 1 52 - 56 nomenclature Mr Lund 08 May 2017 18 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry How Science Works: H Isomerism isomers have the same molecular formula but different chemical and/or physical properties remember that there are a number of ways in which isomerism exists that you have already met: UNIT 1 isomers structural isomers position chain stereoisomers UNIT 2 functional group UNIT 4 geometrical optical Structural Isomerism structural isomers have the same molecular formula but different structural formula one form of structural isomerism is called chain isomerism – unbranched chain and branched chain – i.e. different hydrocarbon skeleton e.g. How many versions of C4H8 can you find that represent structural isomers? functional groups that are present at different positions are called positional isomers (there will be different numbers in the name) e.g. propan-2-ol and propan-1-ol functional group isomerism exists where the molecular formula is the same but different functional groups (and therefore chemical properties exist) e.g. propane-l-ol and methoxy ethane, but-2-ene and cyclobutane Mr Lund 08 May 2017 19 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Stereoisomerism Stereoisomers have the same molecular and structural formula but differ in the spatial arrangement of their atoms. How Science Works: A Geometrical isomerism this is consequential of the non-rotation of a double bond (unlike in alkanes) – which is what you will state as the fundamental requirement in the exam this lack of free rotation is consequential of the π bond present in the alkene a single carbon-carbon bond cannot give rise to this type of isomerism as (unless a chain is present) there is unrestricted rotation Geometrical isomers have the same molecular formula, same structural formula but a different spatial arrangement of the atoms due to the non rotation of the carbon-carbon double bond whilst the minimum requirement is the presence of restricted rotation consideration should also be given to the substituent’s on each carbon the two molecules on the left above are identical even though there is a carbon-carbon double bond as simply flipping vertically makes them super imposable across a carbon-carbon double bond each carbon in turn must have different substituent’s it doesn't matter whether the two groups are the same e.g. in the example on the right no amount of flipping or rotating makes them super imposable. For geometrical isomerism to be possible both carbon atoms on the double bond must have different atoms/groups attached to themselves, however, the carbon atoms can still both be identical in that respect. How Science Works: H E and Z are used to distinguish between the two isomers (its quite EZy to do will a little practice) E isomers have the main grouping diagonally across the double bond Z isomers have the main grouping on the same side of the double bond cis and trans were previously used and in most cases E corresponds to trans and Z to cis BUT NOT ALWAYS How science works How Science Works: L Find out about the food industry and trans fats and the hydrogenation of vegetable oils. Mr Lund 08 May 2017 20 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry the trick is to identify what the main groupings are: where the two atoms directly bonded to the carbons of the double bond with the largest atomic numbers (highest priority) are diagonally opposite then it is deemed an E isomer where the two atoms directly bonded to the carbons of the double bond with the largest atomic numbers (highest priority) on the same side then it is deemed an Z isomer if on one of the carbons the atoms directly bonded are identical then to establish the highest priority grouping a tie break situation arises in which you look at the next highest priority atom attached to each of them e.g -CH2Br beats CH2Cl and so on take care here when using older text books as some molecules deemed ‘trans’ in the old system would actually be ‘Z’ in the new system i.e. across the double bond in one system does not directly yield across the double bond in the other e.g. 3-bromobut-2-ene Example: but-2-ene Step 1: split the alkene Step 2: assign the relative priorities. The two attached atoms are C and H, so since the atomic numbers C > H then the -CH3 group is higher priority. Step 3: look at the relative positions of the higher priority groups : same side = Z, hence (Z)-but-2-ene. The two attached atoms are C and H, so since the atomic numbers C > H then the -CH3 group is higher priority. Therefore the two high priority groups are on the opposite side, then this is (E)-but-2-ene. E-Z transformations are possible given an energy source e.g. photochemistry and eyesight: nerve impulse to the brain CH3 CH3 CH3 O light to eye CH3 CH3 CH3 CH3 CH3 CH3 CH3 How many functional groups can you see here? How can you test to provide evidence for each one? O Mr Lund 08 May 2017 21 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Optical Isomerism Optical isomerism exists where there is an asymmetric (i.e. chiral) carbon with four different groups attached. Can you give the systematic name for this molecule? Optical isomers are non-super imposable molecules (enantiomers) which are mirror images of one another. don’t just put ‘mirror images’ as they can sometimes be superimposable where a plane of symmetry exists if two groups are the same i.e. there is a plane of symmetry then a simple rotation yields the same spatial arrangement so they are not enantiomers many chemicals synthesised in the lab produce equal amounts of both enantiomers (see lactic acid later) which is called a racemic mix this is because the reagents are not stereo specific (rather like a left handed screwdriver) however, living organisms tend to manufacture one enantiomer in preference to the other as determined by the reactive sites of the optically active enzyme used to construct it enzymes are stereo specific reagents the way that these enantiomeric molecules interact with biological systems can be different, for example: carvone limonene one enantiomer tastes of spearmint the other caraway one smells of lemons the other oranges (see if you can spot the chiral centre in these enantiomers of limonene) Mr Lund 08 May 2017 22 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Enantiomers rotate plane polarised light in opposite directions (clockwise and anticlockwise). this physical property is the means by which they can be distinguished one enantiomer rotates the polarised light clockwise (to the right) and is the (+) enantiomer; the other rotates the polarised light anticlockwise (to the left) and is called the (–) enantiomer. a racemic mix consists of a 50:50 mix of both isomers and this will therefore NOT rotate plane polarised light as the two enantiomers cancel one another out How Science Works: L How science works Summary Questions Exam Style Questions Page 62 Page 59 Page 64 A2 Chemistry (Nelson Thornes) AQA Chemguide The thalidomide tragedy 1-4 2, 4, 5 57 - 59 isomerism Mr Lund 08 May 2017 23 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Aldehydes and Ketones Physical properties lower Mr aldehydes and ketones are miscible with water due to polar C=O bonds ability to hydrogen bond with water molecules higher Mr molecules have greater VdW with one another due to increasing size of hydrocarbon tail hence miscibility is reduced for energetic reasons Preparation primary alcohol aldehyde heat with K2Cr2O7(aq)/H2SO4(aq), distil off aldehyde (boiling point lower than alcohol) secondary alcohol ketone reflux under heat with K2Cr2O7(aq)/H2SO4(aq) orange dichromate(VI) ions (Cr2O72-aq)) are reduced to green chromium(III) ions (Cr3+(aq)) Distinguishing Between Aldehydes and Ketones ACIDIFIED POTASSIUM DICHROMATE SOLUTION tests are based on the fact that aldehydes can be easily oxidised to a carboxylic acid while ketones cannot be. Aldehyde + [O] reflux with K2Cr2O7(aq)/H2SO4(aq) Carboxylic Acid orange dichromate(VI) ions (Cr2O72-(aq)) are reduced to green chromium(III) ions (Cr3+(aq)) Care: The above test is only applicable if it is clear that the unknown sample is not a primary or secondary alcohol. Mr Lund 08 May 2017 24 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry NOTE: PRIMARY AND SECONDARY ALCOHOLS CANNOT BE OXIDISED BY FEHLINGS’ OR TOLLENS’ HENCE DO NOT GIVE POSITIVE RESULTS WITH THE TESTS BELOW. writing balanced redox equations under alkaline conditions is a little more involved than acidic conditions but you will find an excellent strategy on CHEMGUIDE – see the link below FEHLINGS’ TEST warm with Fehlings’ solution (an alkaline solution of a complexed copper ion) blue complexed Cu2+(aq) is reduced to brick red Cu2O(s) (this is the basis of the test for reducing sugars) RCHO(aq) + 2Cu2+(aq) + 4OH-(aq) RCOOH(aq) + Cu2O(s) + 2H2O(l) TOLLENS’ REAGENT complex ion [Ag(NH3)2]+(aq) is reduced to Ag(s), hence the silver mirror effect warm with Tollens’ reagent (aqueous silver nitrate in xs ammonia) RCHO(aq) + 2[Ag(NH3)2]+(aq) + 2OH-(l) RCOOH(aq) + 2Ag(s) + 4NH3(aq) + H2O strictly speaking you will get the carboxylate anion RCOO-(aq) under alkaline conditions rather than the carboxylic acid itself additionally it’s worth knowing that methanoic acid (which has a hydrogen present HCOOH) can be oxidised to carbon dioxide via carbonic acid H2CO3 which then easily breaks down into CO2 and H2O (see if you can work out the equations) you should be aware of changes in the IR spectra during the oxidation reactions of compounds containing one or more oxygen atoms (and also the reduction reactions of said molecules) Summary Questions Page 67 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 1-4 66 – 67, 69 65 – 67 Q 1 on page 66 Ionic alkaline, Tollens’, Fehlings’ Mr Lund 08 May 2017 25 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Nucleophilic Addition Reduction of Aldehydes and Ketones NA polar nature of the carbonyl group - you would be wise to revise electro negativity and the nature of nucleophiles aldehyde + 2[H] primary alcohol ketone + 2[H] secondary alcohol sodium tetrahydridoborate(III) (NaBH4) in water (+ ethanol as a universal solvent for higher (longer chain) members) BH4- provides the hydride ion, H- which acts as a nucleophile mechanism is required (nucleophilic attack by hydride ion) intermediate ion then gains H+ from the water present in the aqueous solvent – just show H+ in the mechanism itself NA Reaction with Hydrogen Cyanide ethanal + HCN 2-hydroxypropanenitrile HCN is made ‘in situ’ using acidified sodium or potassium cyanide HCN is a toxic gas so there are health and safety issues in its use cyanide ions are toxic so this will not be done in the lab note: an extra carbon is introduced into the chain so this is an important synthesis step mechanism for reaction = nucleophilic addition trigonal planar carbonyl group can be attacked from either side hence products from aldehydes other than methanal exhibit optical isomerism i.e. there is an asymmetric (i.e. chiral) carbon with four different groups attached hence two nonsuper imposable molecules (enantiomers) exist which are mirror images of one another. a racemic mix is produced – i.e. one that is 50:50 of both isomers and this will therefore NOT rotate plane polarised light in this case as the two enantiomers cancel one another out symmetrical ketones do not yield enantiomers as no chiral centre is present hydroxynitriles can be converted to a carboxylic acid by undergoing acid hydrolysis this involves refluxing with a dilute acid (below I have shown the organic as named for clarity - although you MUST give its formula - as the other particles will always be the same irrespective of the number of carbons in the chain) the optical properties are preserved in the carboxylic acid 2-hydroxypropanenitrile + 2H2O + H+ 2-hydroxypropanoic acid (lactic acid) Mr Lund 08 May 2017 26 + NH4+(aq) A2 Unit 4 Kinetics, Equilibria and Organic Chemistry lactic acid synthesised this way will exist as racemates BUT that produced biologically will not be since enzymes are stereo specific yielding only one optical isomer hence this DOES rotate plane polarised light. Summary Questions Exam Style Questions Page 63 Page 70 Page 83 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 1–3 1-4 2 60 – 63, 68, 70 66 – 70, 57 – 62 Q 2 – 3 on pages 68 Q 4 -6 on page 70 Q 1, 3, 5 on page 82 Q 4 on page 149 Aldehydes, Isomerism Mr Lund 08 May 2017 27 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Carboxylic Acids and Esters Nomenclature be aware of different ways to write the acid functional groups RCOOH, RCO2H etc benzenecarboxylic acid C6H5CO2H general structural formula of esters: RCO2R` e.g. CH3CO2CH2 CH2CH3 you should be able to work out the acid/alcohol used to make an ester and visa versa esters and carboxylic acids are functional group isomers (easily distinguished by IR or nmr – see later - or a simple chemical test using sodium carbonate and testing for evolved CO2) Summary Questions Page 73 1–4 Physical properties lower members of the carboxylic acids and esters are miscible with water due to hydrogen bonding with water higher members are less miscible with water as the extent of VdW with themselves becomes prevalent (they are more soluble in sodium hydroxide solution – do you know why?) most carboxylic acids are crystalline solids (hydrogen bonding) whilst esters are typically oils and fats (no hydrogen bonding) compared to similar sized hydrocarbons melting points can be used to identify and determine the purity (to some extent) of organic solids A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 71 - 73 71 – 75 Q 7 – 10 on pages 71 - 74 Carboxylic acids, Esters Synthesis of carboxylic acids primary alcohol or aldehyde carboxylic acid reflux under heat with xs K2Cr2O7(aq) and H2SO4(aq) orange dichromate(VI) ions (Cr2O72-aq)) are reduced to green chromium(III) ions (Cr3+(aq)) A word about REAGENTS: When the examiner asks for a reagent then it is the name on the bottle NOT the active particle introduced e.g. H+(aq) is not a reagent but H2SO4(aq) is. Mr Lund 08 May 2017 28 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Reactions of Carboxylic Acids and Esters carboxylic acids are weak acids – understanding required in terms of equilibria chlorine-substituted ethanoic acids are more acidic due to negative inductive (–I) effect of chlorine atoms due to their relatively high electronegativity ethanoic acid ACID + + sodium hydroxide BASE sodium ethanoate SALT ethanoic acid + sodium carbonate sodium ethanoate + water + carbon dioxide + + water WATER the latter is a useful test for the presence of –COOH for which you will OBSERVE effervescence and that the gas evolved turns limewater (Ca(OH)2(aq)) cloudy and thus INFER that CO2(g) was produced suggesting a carboxylic acid. Ca(OH)2(aq) + CO2(g) CaCO3(s) + H2O(l) acid catalysed esterification with an alcohol is relatively slow and gives a poor yield ethanol + ethanoic acid ethyl ethanoate + water acid ester + alcohol Summary Questions Page 77 reflux under heat with cH2SO4 catalyst + water 1–4 Hydrolysis of Esters ethyl ethanoate + water ethanoic acid + ethanol sulphuric acid catalyst initiated by nucleophilic attack by the water molecule on the Cδ+ of the carbonyl group acid catalysed hydrolysis not complete due to an equilibrium being established alkali (hot NaOH) catalysed (saponification) hydrolysis is quicker and goes to completion sodium salt of the carboxylic acid is produced since the acid produced reacts with the sodium hydroxide which will drive the equilibrium RHS as acid is removed from the system adding xs sulphuric acid protonates the alkanoate anion carboxylic acid A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 74 - 76 71 – 74 Q 11 on page 74 Carboxylic acids, esters hydrolysis Mr Lund 08 May 2017 29 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Uses of Esters solvents – e.g. ethyl ethanoate for nail varnish they evaporate relatively easily since there is no hydrogen bonding esters have pleasant smells (unlike carboxylic acids which typically have unpleasant smells – rancid fats – ask to smell some butanoic acid – you’ll get the idea) so are used in perfumes and in the food industry as flavourings: butyl butanoate ethyl pentanoate 3-methylbutyl ethanoate The cost of synthesising esters is often far less expensive than extracting them from natural sources 2-methoxyphenol is a waste product from the paper industry and can be used to make methyl vanillin (4-hydroxy-3-methoxybenzaldehyde) an artificial vanilla – by the end of module 4 you might be able to draw its structure and suggest a possible synthesis strategy (although the industrial process is more complex) plasticizers – added to plastics (e.g. PVC) to make them softer and more flexible as they weaken the IMF between polymer strands allowing them to slide over each other more readily (loss over time makes the plastic brittle) some phthalate based plasticizers have a possible association with birth defects although this is still a subject of some disagreement How Science Works: I Fats and Oils animal fats + vegetable oils are triesters of propane-1,2,3-triol (glycerol) and fatty acids (long chained carboxylic acids) they are triglycerides - three alcohol groups esterified by up to three different carboxylic acids (take care with the hydrolysis equation regarding the actual products e.g. under alkaline conditions the anion of each acid group will be formed and each of these could be different) oils have a higher degree of unsaturation cf fats (research E-Z isomerism and trans fats) a greater number of double bonds reduces the relative flexibility (due to restricted rotation) of the molecule which in turn reduces the overlap efficiency of intermolecular forces if a triglyceride undergoes alkaline hydrolysis (NaOH(aq))(saponification) then the salt of a fatty acid is produced e.g. sodium octadecanoate which is also known as sodium stearate and better known as a soap being ionic, soaps are soluble the carboxylate anion RCOO- released is miscible with water (due to the hydrophilic carboxylate group) and miscible with grease (due to its hydrophobic tail) hence can solvate grease into water allowing its removal soap is precipitated out of solution by adding xs common salt (salting out) – which can be understood from an equilibria point of view Mr Lund 08 May 2017 30 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Glycerol (glycerine) the co-product of soap making, propane-1,2,3-triol (glycerol) has numerous uses: it is a useful solvent e.g. in medicines, food colouring its extensive hydrogen bonding makes it good at retaining water thus preventing drying out (e.g. facial creams) and can be used to make nitro-glycerine (which mixed with finely divided silicon(iv) oxide = dynamite) How Science Works: K Biodiesel a renewable fuel made from oils obtained from vegetable matter (e.g. rape seed) which generally consist of a combination of any three of five common carbon chains linked by a glycerol structure it can also be made from animal fat and waste oil small industrial plants tend to be batch processes but larger plants can use continuous flow methods which are more economical methyl esters are produced by reacting these oils with methanol and a strong alkali at around 60oC this is called base-catalysed transesterification the methyl groups replace the glycerol structure on each of the fatty acids Note: there may be three different methyl esters produced but the general formula of each is: CH3OOCCxHy where x and y depend on chain length and degree of unsaturation (this can be written the other way around CxHyCOOCH3) the ester produced does not readily mix with the propane-1,2,3-triol co product so can be separated using a separating tank or a centrifuge any remaining glycerol can be extracted using water (hydrogen bonding) there may be some soap bi-product so further processing will be necessary to achieve a level of purity acceptable for a biofuel rape seed (the yellow stuff you see in fields) produces rape methyl ester (RME) which is very similar to the diesel obtained from crude oil this can be used directly or as a small % of filling station diesel Exam Style Questions Page 83 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 4 76 - 77 75 Fats and oils, soap, glycerol Mr Lund 08 May 2017 31 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Acylation using Acyl Chlorides or Acid Anhydrides nomenclature – named after the parent carboxylic acid hence deemed acid derivatives (as are esters and amides) O R C X Acyl chlorides: X = Cl -oic acid replaced with: -oyl chloride e.g. ethanoyl chloride = an acyl chloride Acid anhydrides X = OR e.g. ethanoic anhydride both are readily attacked by nucleophiles (book error on 79) due to very polar carbonyl group the polarity of which is increased by the electron withdrawing effect of X thus this group of compounds are more useful than carboxylic acids in synthesis due to their high reactivity due to the enhanced δ+ of the carbonyl carbon and since -X is a good leaving group cf –OH used to join an acyl group O R C :Nu to the oxygen of water, alcohol or phenol or the nitrogen of ammonia or an amine (ethanoylation is specifically when R = CH3) H+ is eliminated in each case in the final step Hydrolysis NAE ethanoyl chloride + water ethanoic acid + hydrogen chloride very exothermic reaction, steamy fumes of hydrogen chloride are produced even when exposed to air (due to the reaction between HCl and water vapour), hence anhydrous conditions essential with acyl chlorides and must be stated in exams reaction is faster than with haloalkanes due to additional polarising effect of C=O hydrogen chloride fumes can be tested for using: conc. ammonia a drop at the end of a glass rod will create a white smoke of ammonium chloride silver nitrate solution a drop at the end of a glass rod will go cloudy as white silver chloride is precipitated mechanism is nucleophilic addition-elimination (condensation) NOTE: H is not abstracted by Cl- (think why HCl is a strong acid!) Ethanoic anhydride + water ethanoic acid Mr Lund 08 May 2017 32 + ethanoic acid A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Reaction with Alcohols and Phenols ethanoyl chloride ethanol ethyl ethanoate + HCl yield of ester better than with carboxylic acid since reaction goes to completion NOTE: + NAE acyl chlorides form esters with the phenol group unlike carboxylic acids. The lone pair of the O in phenol is less readily available since these electrons are delocalised into the ring system hence reducing the electron density and the thus the effectiveness of phenol as a nucleophile (higher activation energy) mechanism for reaction with an alcohol - ester formation is very similar to that with water treat the alcohol as RO-H cf water as HO-H so RO- is added to the carbon rather than HONOTE: H is not abstracted by Cl- ethanoic anhydride + ethanol ethyl ethanoate + ethanoic acid Reactions with Ammonia and Amines NAE ethanoyl chloride + ammonia ethanamide (a primary amide) + hydrogen chloride violent reaction with aqueous ammonia at room temperature ethanamide is the only product as further substitution does not occur due to strong electron withdrawing effect of C=O which makes the lone pair of the nitrogen less readily available than with amines (cf haloalkanes + ammonia) NAE ethanoyl chloride + ethylamine (primary amine) N-ethylethanamide + hydrogen chloride (secondary amide) ethanoyl chloride + phenylamine N-phenylethanamide + hydrogen chloride (i) (ii) the product (an acyl derivative) is a white crystalline solid with a sharp melting point - can be recrystallised and used in the identification of the original amine suggest reagents and mechanism for the synthesis of paracetamol N-(4hydroxyphenyl)ethanamide ethanoic anhydride + Summary Questions Exam Style Questions Page 82 Page 83 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide ammonia ethanamide + 1–4 1, 3 78 - 81 76 – 83 Q 12 - 15 on pages 77 – 81 Q 1 – 8 pages 82 - 83 Acyl Mr Lund 08 May 2017 33 ethanoic acid A2 Unit 4 Kinetics, Equilibria and Organic Chemistry How Science Works: I, J The Synthesis of Aspirin (aspirin is 2-ethanoyloxybenzoic acid – don’t panic you wont be asked for this on the exam) the benefits of willow bark, which contains salicylic acid (2-hydroxybenzoic acid) a similar compound to aspirin, have been known for millennia e.g. Hippocrates (~460 B.C - 377 B.C.), African Hottentots and North American Indians it acts as an analgesic (pain killer) and has an anti-pyretic effect (body temperature) salicylic acid was first isolated around 1829 the next step was to find a way to synthesise it rather than rely on extraction from a natural source as this can be problematic: the source might be rare, or seasonal, or have a low concentration, or have harmful contaminants in 1860 it was synthesised from phenol (a by product of the production of town gas from coal) using the Kolbe process (NaOH and high pressure CO2) but the problem was that it was tough on stomachs so alternatives with a similar structure (hence retaining the benefits) were searched for aspirin itself had been synthetically produced in 1853 by a French chemist named Charles Frederic Gerhardt but he didn’t realise its potential and took it no further in 1898, a German chemist named Felix Hoffmann rediscovered Gerhardt's formula he gave it to his father who was suffering from the pain of arthritis and with good results (he had tried other formulations before that!!) so convinced the German pharmaceutical company Bayer to patent it in 1900 (the patent was ignored by the allies during WW1 and thereafter – along with the patent they held for heroin!) its sales increased dramatically during the Spanish Flu epidemic of 1918 aspirin can be synthesised from salicylic acid using ethanoyl chloride or ethanoic anhydride + 2-hydroxybenzoic acid (salicylic acid) aspirin + Ethanoic anhydride + 2-hydroxybenzoic acid (salicylic acid) aspirin + ethanoic acid ethanoyl chloride HCl NAE both are more readily attacked by nucleophiles than the corresponding acid acid anhydrides offer certain advantages over acyl chlorides, despite being less reactive, in that they are: cheaper, less corrosive (no HCl liberated), less readily hydrolysed How science works Exam Style Questions Page 82 Page 83 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide Aspirin 1 82 81 Q 14 on page 81 Aspirin Mr Lund 08 May 2017 34 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Aromatic Chemistry involves compounds containing a benzene ring (aka arenes look out for C6H5-) empirical formula CH, Mr =78, molecular formula C6H6 Structure and Stability of Benzene hydrogenation with 3 moles of H2 suggests the equivalence of 3 carbon-carbon double bonds How Science How science works Page 85 Kekule’s dream Works: A Kekule did propose a cyclic structure – but it could not account for some major aspects of the chemistry of Benzene: 1. no electrophilic addition reactions (e.g. with Br2(aq) in the dark) unlike alkenes 2. you don’t get two isomeric (1,2) disubstituted compounds 3. X-ray diffraction studies found intermediate – between double and single – and equal C-C bond length i.e. a symmetrical structure 4. enthalpy of hydrogenation (208 kjmol-1) is less than 3 x cyclohexene (360 kjmol-1) Why is it drawn like this? How Science Works: F delocalised electrons increase relative stability (less electron - electron repulsion) Summary Questions Exam Style Questions Page 86 Page 93 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 1–3 6 84 - 86 84 - 87 Q 1 – 2 on page 87 Bonding benzene Mr Lund 08 May 2017 35 initially a resonance hybrid structure was suggested the true structure and shape of benzene can be explained in terms of the delocalisation of the electrons of 6 x 2p orbital’s (whilst its not on the syllabus, knowledge of orbital hybridisation would be helpful – have a look on Chemguide or in an older A-level book) A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Physical properties non-polar colourless liquid and does not mix with water (no hydrogen bonding) boiling point similar to 6 carbon aliphatic hydrocarbons but melting point is higher a planar structure allows better packing therefore more effective VdW. Nomenclature C6H5- phenyl group and simple monosubstituted aromatic compounds (arenes) normally named as derivatives of benzene so ‘benzene’ often forms the root of the name mono-substituted arenes are generally of formula C6H5X e.g. benzaldehyde C6H5CHO and yield a peak of 77 on mass spectra due to the fragment C6H5+ (see later) methylbenzene (toluene) chlorobenzene benzenecarboxylic acid (benzoic acid) nitrobenzene ethylbenzene benzaldehyde (chloromethyl)benzene some names also use ‘phenyl’ or variations of it when the benzene is regarded as a side chain phenol (instead of hydroxybenzene) phenylamine (aniline) (instead of aminobenzene) 4-hydroxyphenyl ethanoate phenylethene (instead of ethenylbenzene) aromatic compounds with more than one substituent 1 2 lowest numbering possible alphabetical order for substituent’s (ignore di, tri etc) 2-hydroxybenzoic acid 2,4,6-trinitrotoluene 2,4,6-trinitrophenol (picric acid) benzene-1,4-dicarboxylic acid (Terephthalic acid) take heart – naming aromatic compounds is complex but you will only have to deal with simple examples as it is far more important that you understand the chemistry! Summary Questions Page 88 A2 Chemistry (Nelson Thornes) AQA Chemguide 1–4 87 - 88 Naming aromatic Mr Lund 08 May 2017 36 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Reactions of aromatic compounds Electrophilic Substitution high electron density in the ring attracts electrophiles Br2(aq) is not decolourised (in the dark) – electrophile must be powerful (+ve not just +) why benzene resists attack by poor electrophiles – stability of benzene compared to alkenes electrophilic substitution rather than addition (c.f. alkenes) since this retains the relatively stable benzene ring structure hence is energetically more favourable Nitration of Benzene and Methylbenzene ES benzene nitrobenzene (a yellow oil) cH2SO4/cHNO3 refluxed at 50oC (nitrating mixture) mechanism of formation of NO2+ the nitronium (old name = nitryl) cation (sulphuric acid acts as a homogeneous catalyst) ES methylbenzene 2(and 4)-nitromethylbenzene cH2SO4/cHNO3 refluxed at 50oC the methyl group is 2, 4, and 6 directing and activates the ring towards electrophilic substitution (hence faster rate) since it donates electron density into the ring thus making it relatively less stable and more susceptible to attack by electrophiles. ES further substitution requires more vigorous conditions (higher acid conc. and temperature) as the nitro group deactivates the ring towards electrophilic substitution by withdrawing electron density from the ring (i.e. increasing the extent of delocalisation thus further stabilising the ring) – TNT is an explosive nitrobenzene (a yellow oil) phenylamine cHCl/Sn or H2/Ni Don’t use H2 for cHCL/Sn C6H5NO2 + 6[H] C6H5NH2 + 2H2O the above reaction is much simplified – (whilst this equation is acceptable do you see why the initial product would not be phenylamine? – a possible A* question perhaps?) phenylamine is important for the production of AZO dyes since the end of 19th century (these replaced the old technique of mordant dying – why not read about this – it is interesting) How Science Works Exam Style Questions Page 90 Page 92 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide TNT 1, 3 89 - 90 88 - 89 Q 3, 4 on page 89 Electrophilic substitution, nitration Mr Lund 08 May 2017 37 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Friedel-Crafts Acylation role of the AlCl3 - halogen carrier – why is it a lewis acid anhydrous conditions - AlCl3 readily hydrolysed by water before it does its job electrophilic substitution mechanism proceeds via an acylium ion intermediate you should be able to write a full balanced equation for the above benzene + ethanoyl chloride Summary Questions Exam Style Questions phenylethanone Page 91 Pages 92 – 93 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide warm with AlCl3 catalyst anhydrous conditions 1-4 2-5 91 91, 184 Q 7, 8 on page 91 Q 1 – 8 on pages 94 - 95 Acylation Mr Lund 08 May 2017 38 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Amines nomenclature of primary, secondary and tertiary amines (by the way amine has ONE m in it!!!) note how this differs from alcohols and haloalkanes amino can sometimes be used (see amino acids later on) Physical properties boiling points are elevated by the ability to hydrogen bond but are lower than similar sized alcohols due to the relative electronegativity of O and N compared (ASK if you don’t understand the significance of this) lower members are gases liquid amines smell like rotting (fishy) flesh – adding acid removes this smell – WHY? smaller primary amines are water soluble (hydrogen bonding) producing alkaline solutions solubility decreases with chain length (as with alcohols) due to increased mutual VdW phenylamine is not very soluble in water as the VdW between the rings is significant compared to hydrogen bonding between the amine group and water produce alkaline solutions in water when they dissolve Summary Questions Page 95 1-4 Basic properties lone pair on N can be a nucleophile, base or ligand depending on the context how good it is depends on the availability of that lone pair i.e. what the N is bonded to Brønsted-Lowry bases – proton acceptors, Lewis base = lone pair donor ethylamine is more basic than ammonia due to +I inductive effect of the alkyl group (its also a better nucleophile than ammonia - see alkyl halides) this explains the relatively more basic nature of small secondary amines but this does not hold true for tertiary amines where reduced solubility is a factor phenylamine is less basic than ammonia since the lone pair on the nitrogen is less available due to delocalisation in the ring structure – diagram (phenylmethyl)amine is as basic as primary amines as the N is NOT bonded directly to the ring (note brackets in name – why ?) acid amides are relatively poor bases/nucleophiles/ligands as the strongly electronegative oxygen causes the lone pair on the nitrogen to be more withdrawn Mr Lund 08 May 2017 39 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry pKa value of conjugate acid increases (i.e. is poorer) with increased basicity of the conjugate amine amines react with acids to form salts in a similar manner to ammonia ethylamine + hydrochloric acid ethylammonium chloride + water solvation of insoluble phenylamine achieved by the addition of HCl to form a soluble salt phenylammonium chloride (reversed by adding NaOH) phenylamine + hydrochloric acid Summary Questions Exam Style Questions phenylammonium chloride + water Page 97 Pages 108 – 9 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 1-3 1 94 - 97 96 – 97, 99 - 100 Q 1, 2 on pages 96 – 97 Q 5, 6 on pages 99 - 100 Q 4 on page 104 Amine name, amine base Mr Lund 08 May 2017 40 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Preparation From haloalkanes NS bromoethane ethylamine alcoholic solution of NH3 under pressure (lots of by-products so not a good method) mechanism – NH3 acts as nucleophile excess ammonia is used to improve the yield of the primary amine if xs bromoethane is used, since the ethylamine produced is also a nucleophile (stronger than ammonia due to the +I inductive effect of the alkyl group) it can react with the xs bromoethane to give diethylamine further substitution can then occur to produce: triethylamine and tetraethylammonium bromide (a quaternary ammonium salt cf ammonium ions) note that acyl chlorides only yield primary amide (mechanism reminder) – lone pair withdrawn by strong d+ on C due to the polarity of C=O caused by the electronegativity of the O From nitriles Ethanenitrile RCN + ethylamine 4[H] reduction by H2/Ni RCH2NH2 (note: at AS you were also told that acid hydrolysis of nitrile yields a carboxylic acid) LiAlH4 but not NaBH4 (not a powerful enough reducing agent) can also be used (don’t put H2 in balanced equation in this case!!) Aromatic Amines benzene nitrobenzene (a yellow oil) * nitrobenzene (a yellow oil) phenylamine cH2SO4/cHNO3 refluxed at 50oC (nitrating mixture) reduction cHCl/Sn (or H2/Ni) phenylammonium chloride* is initially produced, and addition of NaOH yields phenylamine (via deprotonation) which is then obtained by steam distillation strictly speaking its actually phenylammonium hexachlorostannate(IV) Mr Lund 08 May 2017 41 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Uses of Amines quaternary ammonium salts are used as cationic surfactants (long carbon chains help) in fabric conditioners and hair conditioning products the positive charges present will repel and add body to the hair/fabric whilst the ‘tail’ section associates with the fabric if you are particularly interested in laundry see: http://www.scienceinthebox.com/en_UK/glossary/s urfactants_en.html http://www.chemistryquestion.com/English/Questions/ChemistryInDailyLife/27c_nonionic_s urfactant.html aryl amines are used in synthetic dyes aryl amines are used to make certain drugs e.g. paracetamol amines are used to make polymers e.g. polyurethane’s (cavity wall insulation) and polyamides (nylons) How Science Works Summary Questions Exam Style Questions Pages 100 - 1 Page 101 Pages 108 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide Sulfa drugs 1, 2 2-4 99 - 101 97 – 98, 101 - 105 Q 7 on page 103 Q 13, 5-8 on pages 104 - 105 Amine preparation, amine uses Mr Lund 08 May 2017 42 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Amino acids there are 20 important naturally occurring amino acids (amine is on the C next to the acid group –CO2H) R can vary R=H glycine R = CH3 alanine R = CO(OH)CH2 aspartic acid etc you should recognise that there is a CHIRAL centre hence amino acids exhibit optical isomerism (name the exception) some amino acids have been identified in space (those of you who are interested in science might read the next link) http://www.newscientist.com/article/dn7895-space-radiation-may-select-amino-acids-for-life.html the amino acid proline is a 2o amine all the others are primary amines CH 2 NH CH 2 C COOH CH2 H Extra Info for potential medical/biochemistry/pharmacy students: amino acids in proteins are all L-isomers this does not necessarily mean plane polarised light is rotated the same way if L(+) then D(-) for a given amino acid and L(-)/D(+) don’t exist in our natural system CORN law – I have provided some info about this on my site – it is not on the exam – but it may be helpful at your interview for medical school given the biochemical significance Summary Questions Page 103 Page 107 1, 2 1 Mr Lund 08 May 2017 43 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Zwitterions the amino group and the carboxyl group of each amino acid are both ionisable the acidic carboxyl group (with a pKa of about 3 – it’s a weak acid) is deprotonated the basic amino group (with a pKa of around 9 – it’s a weak base) is protonated amino acids thus exist as zwitterions with both a positive and negative charge present Zwitterions are amphoteric i.e. they exhibit both acidic and basic properties in solution because of the two functional groups they thus form salts with both acids and bases (note all similar groups ionised as appropriate) amino acids can thus act as buffers (hence regulate pH) Adding H+ pH decreasing Adding OHpH increasing charge on the zwitterion ion depends on pH, -ve at high pH and +ve at low pH at a given pH, the isoelectric point of the amino acid there will be no overall charge as the + and – cancel this pH varies from amino acid to amino acid and provides a means of separating them using a technique called electrophoresis. Extra Info for potential medical/biochemistry/pharmacy students: you might be interested to find out how electrophoresis works by looking at this link http://www.saburchill.com/IBbiology/chapters01/003.html Amino acids exist as zwitterions in the solid state and thus have strongly ionic character this explains their high solubility in polar solvents e.g. water it also explains the high MPt (white crystalline solid when pure) MPt too high to be accounted for by hydrogen bonding alone - supporting existence of Zwitterions and ionic nature of amino acids in the solid state Exam Style Questions Pages 108 5, 8, 11 Mr Lund 08 May 2017 44 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Condensation Polymerisation and hydrolysis peptide (amide) links are formed between 2 amino acids by condensation reactions (where a small molecule such as water is eliminated) to form a dipeptide Polypeptide (~50 amino acids) are formed by condensation polymerisation and catalysed by enzymes sequence of amino acids in a protein is called its primary protein structure hydrogen bonding between C=O and N-H controls shape - secondary protein structure - -helix (coiled) and -pleated sheet (folded) tertiary structure involves further bending and twisting (a good analogy is a knotted, multicoloured telephone coil) the stretching of wool is dependent on hydrogen bonding the length of which can be reversibly increased up to a limit (very hot water can break these ruining the fluffyness) hydrolysis of proteins is achieved by refluxing with acid, base or enzyme catalyst (cf hydrolysis of an amide) in effect reversing the process shown in the diagram above the liberated amino acids can then be separated by paper chromatography (developed by treating with ninhydrin which colours amino acids violet some enzymes are selective and only partially hydrolyse certain proteins enabling amino acid sequences to be identified enzymes are themselves proteins and are very specific in what they catalyse (substrate) due to their shape their shape is dependent on hydrogen bonding hence their activity is sensitive to elevated temperatures where they are denatured How Science Works Summary Questions Exam Style Questions Pages 107 Page 107 Pages 109 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide Robots in the lab 1, 2 6, 8 - 10 102 - 107 111 – 114, 57 – 62 Q 8 - 12 on pages 111 – 113 Q 2 – 4, 5 - 6 on pages 115 - 116 Zwitterions, peptide Mr Lund 08 May 2017 45 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Polymerisation Addition Polymerisation addition across the carbon-carbon double bond of a single unsaturated monomer occurs by a free radical mechanism (details not required) started by an initiator (e.g. a peroxide) which is incorporated at the start/end of the polymer chain you should be able to work out monomer from polymer and visa-versa (tip use the >C=C< form of the monomer i.e. rewrite it) if asked for a single repeating unit don’t show brackets otherwise use –[C-C]-n format ethene poly(ethene) propene poly(propene) heat, pressure, catalyst in all cases chloroethane poly(chloroethene) (PVC) phenylethene poly(phenylethene) (polystyrene) tetrafluoroethene poly(tetrafluoroethene) (PTFE or Teflon) methyl 2-methylpropenoate Summary Questions Exam Style Questions Page 114 Pages 122 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide ‘perspex’ 1-4 4, 5 112 - 114 106 -108 Q 1-3 on page 106 Q 7, 8 on page 116 Addition polymerisation Mr Lund 08 May 2017 46 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Condensation Polymerisation Polyesters two different types of monomers, one a diol, the other a dicarboxylic acid this polymer is connected by ester linkages you should be able to draw repeating units and determine the monomers used to make a given polymer – colour coding diagrams helps here condition usually involve heat + catalyst look for –CO2- in structural formula balanced equation (don’t forget 2n H2O !!!!) ethane-1,2-diol (ethylene glycol) + benzene-1,4-dicarboxylic acid (terephthalic acid) PET the chain repeat unit of PET is: commonly just called ‘polyester’ PET, poly(ethylene terephthalate) was initially used as a fibre (e.g. Terylene and Dacron) it is now used extensively in plastic containers e.g. for fizzy drink bottles – it does not smash on impact you can read about the invention of polyester at: http://inventors.about.com/library/inventors/blpolyester.htm those of you doing textiles might also like to visit the ‘polyester story’ at http://schwartz.eng.auburn.edu/polyester/polyester.home.html poly(2-hydroxypropanoic acid) – used in surgery as its broken down by enzymes/body fluids over a number of days (why is this good?) how would you synthesise it from ethene? How Science Works: I You will find it useful to look at the reaction pathways template that I have put on line you can find out more about plastics from renewable raw materials and biologically degradable plastics at this site (its also a good example of (or idea for) an EPQ project!): http://www.rsc.org/education/teachers/learnnet/green/docs/plastics.doc Mr Lund 08 May 2017 47 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Polyamides two different types of monomers, one a diamine, the other a dicarboxylic acid this polymer is connected by amide linkages an diacyl chloride can be used instead of the acid but what would be the pros and cons of this nylon-6,6 is so called as both monomers have 6 carbons nylon is a polyamide (just like proteins) 1,6-diaminohexane + hexanedioic acid nylon-6,6 How Science Works: I you might be interested in the history of nylon – it’s an ideal stocking filler for Xmas see: http://www.cha4mot.com/p_jc_dph.html and http://inventors.about.com/od/nstartinventions/a/nylon.htm polyamides have extensive hydrogen bonding between parallel strands (cf protein structure) when nylon is spun into fibres, amide groups on adjacent chains form hydrogen bonds making nylon yarn strong. kevlar is an example of an aromatic polyamide and is made from the monomers benzene-1,4dicarboxylic acid and benzene-1,4-diamine Medieval chainmail similarity? the molecule is flat because of the aromatic groups the uses of Kevlar are related to its strength (its several times stronger than steel) this is related to the packing together of sheets of molecules held together by hydrogen bonds formed between N – H groups and C = O groups on adjacent molecules here is a scuba diving site that has an excellent overview of natural and synthetic polymers – have a look it is very good: http://njscuba.net/artifacts/matl_polymers.html Mr Lund 08 May 2017 48 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Environmental Issues How Science Works: J polyalkenes are saturated, have no polar bonds and have strong C-C and C-H bonds hence they are relatively unreactive e.g. with acids, alkali or oxidants this in turn makes them difficult to dispose of non biodegradability means they last a long time but you can use landfill – not ideal though combustion yields toxic and greenhouse gases e.g CO and carbon particulates, NO2 and HCN from polyurethane in older upholstery or HCl and dioxins released through combustion of halogenated plastics such as PVC and of course there will always be CO2 produced recycling is expensive as the plastics must be identified and separated from other waste energy production is an option but as this involves combustion there will be CO2 produced using as chemical feedstock after cracking use biodegradable/photodegradable polymers instead is an option polyesters and polyamides both have polar bonds hence they are potentially biodegradable they are broken down by (catalytic) hydrolysis in acid/alkaline solution or with enzymes although this can take a long time recycling these materials e.g. terylene will save on natural resources and energy in their initial production (as well as reducing the need for landfill), however, as they are biodegradable their structural integrity will diminish after repetitive usage as with polyalkanes the cost of collection, separation and transportation (require energy and are labour intensive) must be taken into account How Science Works Summary Questions Exam Style Questions Pages 120 - 1 Page 119 Page 121 Pages 121 - 123 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide Hermann Staudinger + Q 1,2 1, 2 1-4 1, 2, 3, 6, 7 115 - 120 108 -110 Q 4 - 8 on page 109 - 110 Q 1, 5 on pages 115 - 116 Condensation polymerisation Mr Lund 08 May 2017 49 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Aliphatic synthesis You should be able to write full chemical equations, and identify the type for all the reactions listed: Oxidation, Reduction, Addition, Elimination, Addition-Elimination (Condensation), Substitution, Hydration, Dehydration, Hydrogenation, Dehydrogenation, Hydrolysis Mechanisms: Nucleophilic Addition NA, Nucleophilic Substitution NS, Electrophilic Addition EA, Nucleophilic Addition-Elimination NAE, Free Radical Substitution FRS or Elimination E. Alkanes haloalkane, alkene methane + Cl2 chloromethane FRS alkanes can be used to produce alkenes by thermal cracking Alkenes haloalkane, alkane, alcohol, alkoxyalkane ethene + HBr bromoethane EA ethene + Br2 1,2-dibromoethane EA ethene + H2 ethane o Ni catalyst, ~200 C (catalytic hydrogenation) + Method 1 Method 2 cH3PO4 on a silica support with high temperature and pressure, first react with cold cH2SO4, then warm with water Alkyl Halides H2 O ethene ethanol EA amine, alcohol, nitrile, alkene bromoethane + NH3 ethylamine + HBr heat with xs ammonia under pressure to minimise further substitution NS bromoethane + OH-(aq) ethanol reflux with dilute NaOH dissolved in water NS + Br- bromoethane + CN- propanenitrile + boil under reflux with alcoholic NaCN or KCN (NOT HCN) propanenitrile + 4[H] propylamine reduction by H2/Ni Br- bromoethane + OH- reflux with NaOH dissolved in ethanol + ethene Mr Lund 08 May 2017 50 + H2 O NS Br- E A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Alcohols carbonyl (aldehyde and ketone), carboxylic acid, ester, haloalkane, alkene ethanol + [O] ethanal + H2 O mild conditions - K2Cr2O7(aq)/H2SO4(aq) distil off aldehyde as formed to prevent further oxidation to a carboxylic acid ethanol + 2[O] reflux with K2Cr2O7(aq)/H2SO4(aq) under heat ethanoic acid + H2 O propan-2-ol + [O] reflux with K2Cr2O7(aq)/H2SO4(aq) under heat propanone + H2 O ethanol + ethanoic acid ethyl ethanoate + reflux with glacial ethanoic acid and concentrated sulphuric acid catalyst H2 O ethanol ethene heat with excess cH2SO4 at 170oC + Aldehydes and Ketones alcohol, carboxylic acid, hydroxynitrile H2 O ethanal + 2[H] ethanol sodium tetrahydridoborate(III) (NaBH4) in aqueous ethanol propanone + 2[H] NA propan-2-ol NA ethanal + HCN 2-hydroxypropanenitrile in alkaline solution to increase [CN ] as HCN is a weak acid 2-hydroxypropanenitrile + 2H2O + H+ Carboxylic Acids NA 2-hydroxypropanoic acid (lactic acid) + NH4+(aq) water WATER a salt, ester ethanoic acid ACID + + sodium hydroxide BASE sodium ethanoate SALT + + ethanoic acid + sodium carbonate sodium ethanoate ACID + METAL CARBONATE SALT + water + carbon dioxide + WATER + CO2 ethanol ACID + + ethanoic acid ALCOHOL ethyl ethanoate ESTER + + Mr Lund 08 May 2017 51 water WATER A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Esters carboxylic acids (or its salt – look for OH-(aq)) and alcohol ethyl ethanoate + OH-(aq) ethanoate ion + ethanol ethanoate ion + H+(aq) ethanoic acid (protonation of the anion of a weak acid by a stronger acid – think about the equilibria here) Acyl Chlorides carboxylic acid, ester, amide ethonoyl chloride + H2 O ethanoic acid + hydrogen chloride NAE ethanoyl chloride + ethanol ethyl ethanoate + HCl ethanoyl chloride + 2-hydroxybenzoic acid (salicylic acid) ethanoyl chloride + NH3 aspirin ethanamide + NAE HCl NAE + HCl NAE ethanoyl chloride + ethylamine (primary amine) N-ethylethanamide + (secondary amide) HCl NAE ethanoyl chloride + phenylamine N-phenylethanamide + HCl NAE Acid Anhydrides carboxylic acid, ester, amide ethanoic anhydride + H2 O 2 ethanoic acid ethonoic anhydride + ethanol ethyl ethanoate + ethanoic acid ethonoic anhydride + 2-hydroxybenzoic acid (salicylic acid) warm with conc. sulphuric acid catalyst aspirin ethonoic anhydride + ammonia ethanamide ethonoic anhydride + 4-aminophenol (primary amine) paracetamol + (secondary amide) Mr Lund 08 May 2017 52 + ethanoic acid + ethanoic acid ethanoic acid A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Aromatic Synthesis Electrophilic Addition EA, Electrophilic Sustitution ES, Nucleophilic Addition-Elimination NAE Substitution on the ring benzene + HNO3 nitrobenzene cH2SO4/cHNO3 refluxed at 50oC (nitrating mixture) + H2 O ES methylbenzene + HNO3 cH2SO4/cHNO3 refluxed at 50oC 2(and 4)-nitromethylbenzene + H2O ES benzene + ethanoyl chloride AlCl3 catalyst anhydrous conditions phenylethanone + HCl ES phenylamine + 2H2O phenylammonium chloride + H2 O N-phenylethanamide + HCl NAE HCl NAE Modification of substituents nitrobenzene cHCl/Sn or H2/Ni + 6[H] phenylamine + HCl phenylamine + ethanoyl chloride Polymerisation phenylethene free radical peroxide initiator, high pressure phenylamine + ethanoyl chloride ethane-1,2-diol (ethylene glycol) + 1,6-diaminohexane + poly(phenylethene) N-phenylethanamide + benzene-1,4-dicarboxylic acid (terephthalic acid) hexanedioic acid Mr Lund 08 May 2017 53 nylon-6,6 Terylene (a polyester) A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Amine Nitrile (hydroxynitrile) Haloalkane Alkane Alkene Carbonyl Alcohol Amide (polyamide) Carboxylic Acid Acyl chloride (Acid anhydride) Ester (polyester) Reagents/conditions Mechanism required also2017 Mr Lund 08 May 54 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Testing for functional groups Bromine Water (Br2(aq)) Test for alkenes (unsaturated hydrocarbons) Add bromine water dropwise to ~1 cm3 of the unknown substance. OBSERVATION INFERENCE EXPLANATION orange colourless alkene electrophilic addition. product is colourless Acidified Potassium Dichromate Solution Test for 10 and 20 alcohols (CARE it is also positive with an aliphatic aldehyde) To ~ 1 cm3 of the substance under test add ~ 1 cm3 of acidified (1 mol dm-3 sulphuric acid) potassium dichromate solution then heat gently in a water bath. Acidified purple potassium 23+ manganate(VII) will be Redox reaction in which orange Cr2O7 is reduced to green Cr decolourised in a similar test 2+ 3+ Cr2O7 (aq) + 14H (aq) + 6e 2Cr (aq) + 7H2O(l) OBSERVATION INFERENCE orange green 1o or 2o alcohol EXPLANATION 1o alcohol aldehyde acid 2o alcohol ketone aldehyde aldehyde carboxylic acid Methanoic acid can also be oxidized to carbon dioxide NOT 3o alcohol or ketone Tollens’ Reagent (Silver Mirror Test) Test for aldehydes NOT ketone Don’t confuse with the test for alkyl halides To ~1 cm3 of your sample add ~ 1 cm3 of Tollen’s reagent* and warm in a hot water bath. (*to ~2 cm3 of ~ 0.1 mol dm-3 silver nitrate solution add ~2.0 mol dm-3 sodium hydroxide solution 1 drop at a time until a brown precipitate just forms. Add ~2.0 mol dm-3 ammonia solution to it dropwise until the precipitate just dissolves - (ammoniacal silver nitrate)). OBSERVATION INFERENCE silver mirror formed on test tube aldehyde EXPLANATION + Ag(NH3)2 (aq) + e- Ag(s) + 2NH3(aq) aldehyde acid NOTE THAT ALCOHOLS DO NOT GIVE A POSITIVE RESULT WITH THIS TEST Mr Lund 08 May 2017 55 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Fehlings’ Test Test for aldehydes NOT ketone NOT benzaldehyde To ~1 cm3 of your sample add ~ 1 cm3 of Fehling’s solution* (BLUE) and warm in a water bath. (*made by mixing equal volumes of: Fehling’s A (dissolve 17g of CuSO4.5H2O in 250 cm3 of water) and; Fehling’s B (dissolve 86 g of Rochelle salt (potassium sodium 2,3-dihydroxybutanedioate (tartrate) and 30 g of NaOH in 250 cm3 of water with gentle warming). OBSERVATION INFERENCE EXPLANATION orange-red precipitate aldehyde reduction of copper(II) copper (I) Cu2O is precipitated aldehyde acid NOTE THAT ALCOHOLS DO NOT GIVE A POSITIVE RESULT WITH THIS TEST Test for carboxylic acids Sodium Carbonate Solution To a small quantity of the unknown substance in a boiling tube add ~1 cm3 of sodium carbonate solution. Test any gas evolved using a drop of lime water at the end of a glass rod. OBSERVATION INFERENCE EXPLANATION effervescence gas evolved turns lime water cloudy carboxylic acid acid + metal carbonate salt + water + CO2 Acidified Silver Nitrate Solution Test for alkyl halides Don’t confuse with Tollen’s! To test for an alkyl halide, the halogen atom must first be released as a halide ion by hydrolysis. Dissolve ~ 1 cm3 of the alkyl halide in ~1 cm3 of ethanol, then add ~ 1 cm3 of sodium hydroxide solution and warm in a water bath. Add ~ 1 cm3 of silver nitrate solution that has been acidified with dilute nitric acid (this removes excess OH-(aq) which would otherwise precipitate out Ag2O(s) thus masking the test results). OBSERVATION white precipitate soluble in ammonia pale yellow precipitate soluble in xs ammonia yellow precipitate insoluble in ammonia INFERENCE EXPLANATION alkyl chloride AgCl(s) AgCl(s) + 2NH3(aq) Ag(NH3)2+(aq) + 2Cl-(aq) AgCl(s) AgBr(s) + 2NH3(aq) Ag(NH3)2+(aq) + 2Br-(aq) AgI(s) alkyl bromide alkyl iodide Relative rates of hydrolysis of alkyl halides are: iodo > bromo > chloro Mr Lund 08 May 2017 56 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Mass Spectrometry this provides us with: Mr, Relative abundance and fragmentation patterns do you know how a mass spectrometer works: vaporisation ionisation acceleration deflection detection you must be able to interpret simple mass spectra to determine the elements present from their relative isotopic mass and also their relative abundance the value of the relative isotopic mass can be read directly from the m/z value of a given peak on the spectra from a mass spectrometer relative abundance is reflected in the heights of the peaks when a molecule is subject to ionisation a molecular ion (parent ion), which is a radical cation, is formed (note BOTH + must be shown on the molecular ion) M(g) e- This molecular ion can then undergo fragmentation which produces a radical (not detected) and a positive ion (detected) which are shown with a + only M(g)+ M(g)+ + X(g)+ + Y(g) can be either way around – so both give m/z peaks – but not necessarily in equal proportions higher peaks correspond to more stable ions and the highest is called the base peak and this determines the 100% benchmark on the spectra m/z Interpretation value 15 CH3+ Notes Not much use without other data + Look for a 2H quartet and 3H triplet on 1H nmr 29 CH3CH2 31 CH3O+ Look for a 3H singlet on 1H nmr 43 HIGH peak = CH3CO+ CH3CH2CH2+ 57 CH3CH2CO+ 43 and 29 could be butanone 43 and 15 could be propanone 43 with no 15 ethanal or acyl chloride possible Check IR for a carbonyl group Check IR for a carbonyl group 77 C6H5+ 105 C6H5CO+ 77 + 29 consider ethylbenzene 77 + 43 propylbenzene or phenylethanone Benzaldehyde or Benzoic acid Mr Lund 08 May 2017 57 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry fragmentation occurs most easily at the weakest bond Carbonyl compounds have a tendency to break adjacent to the C=O yielding the relatively stable, hence high abundance, acylium ion RCO+ - look for m/z 43 the stability of a fragment determines its abundance on the spectra e.g. primary, secondary and tertiary carbocations – the positive inductive effect (+I) explains their relative stability http://www.chem.arizona.edu/massspec/example_html/examples.html Have a look at these examples isomers can be determined from fragmentation patterns: o this is particularly useful in resolving aldehydes and ketones which will both give similar >C=O wave numbers on IR o isomeric acids and esters should be obvious from IR (look for OH peak) data but again will give different fragmentation patterns o isomeric esters will give similar IR and nmr and may need fragmentation patterns to resolve – these are harder to do M+ + 1 peak (satellite peak) due to carbon-13 height = 1% of M+ per carbon in the compound e.g. 5% suggests 5 carbons 7% (or higher) is likely to involve a substituted benzene derivative note: the number carbons can be used, in conjunction with Mr to make an estimate of the possible number of oxygen’s for example chloro-alkanes will produce two molecular ion peaks when mono substituted in a ratio commensurate with halogen abundance. e.g. chloromethane gives M(g)+ peaks at 50 and 52 in a 3:1 ratio Di-substituted chloro-alkanes will yield three molecular ion peaks of relative intensity for dichloro 9:6:1 – those who do Maths Statistics will be able to explain that: e.g. dichloromethane gives M(g)+ peaks at 85, 87 and 89 in a 9:6:1 ratio high resolution mass spectra can distinguish between compounds with similar M r e.g. ~123 for C6H5NO2 and C7H7O2 when dealing with mass spec in conjunction with either/both IR and nmr it is sensible to use mass spec initially only to establish the Mr from which a reasonable estimate of the number of carbons can be made this is particularly useful with IR data to indicate whether there are any oxygen atoms in the structure thus giving a maximum for the number of carbon atoms Summary Questions Exam Style Questions Page 136 Page 159 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 1, 2 4 132 - 136 117 – 123 Q2 on page 138 Q1 – 6 on pages 118 - 123 Mass spectrometry Mr Lund 08 May 2017 58 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry IR Spectroscopy used to identify certain functional groups in particular –OH and C=O in carbonyl compounds, acids, alcohols and esters beam splitting occurs in the IR spectrometer through a reference and the organic sample and comparison determines the relative % transmission creating troughs in the spectra molecules absorb IR energy at values corresponding to a natural vibration frequency associated with asymmetric stretching and bending of the covalent bonds present which is dependent on the bond energy and mass involved (the ability of CO2 to do this is why there is a relationship between atmospheric CO2 and global warming) the spectra produced uses a scale of % transmission vs. wave number (cm-1) above 1500 cm-1 two important absorption values are: 1680 – 1750 cm-1 C=O strong and sharp with slight variations in position depending on the type of compound (see table 1 page 139) 3230 – 3550 cm-1 O-H this is broad (due to hydrogen bonding) cf the narrow C– H of 2850 – 3300 cm-1 note that the O-H for alcohols lies further to the left than for acids revealing the narrower C-H absorption which might be partially or totally obscured with an acid A data sheet will be provided in the exam – similar to table 1 on page 138 used with mass spec they will help establish a maximum number of carbon atoms in the compound by indicating the presence of at least one oxygen. significant spectral changes involving the above upon redox or esterification can be revealing regarding the functional groups present the fingerprint region 400 – 1500 cm-1 is unique to each organic molecule determination of the sample uses computerised comparisons with a database impurities will yield additional peaks in this region Summary Questions Page 138 Page 141 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 1, 2 1-3 137 - 141 124 – 127 Q7 – 10 on pages 124 – 127 Q 1 page 137 Q 5 on page 149 Infra red spectroscopy Mr Lund 08 May 2017 59 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Nuclear Magnetic Resonance Spectroscopy nmr can be used on small samples to provide details of structure cf wet chemical tests - limited primarily to identifying functional groups nmr can be applied where there is an odd nucleon number e.g. 1H and 13C these nuclei has a small magnetic field associated with spin an external magnetic field results in either alignment with ( - lower energy) or against ( higher energy) resonance between these two states occurs when radio waves of appropriate energy are applied this results in energy being absorbed which is detected by the instrumentation Carbon-13 (13C) nmr organic chemicals will contain 1% carbon-13 their resonance field strength will vary depending on the number of surrounding electrons the field strengths of other carbon environments only differ slightly so the scale is measured in ppm electrons shield the nucleus thereby reducing the effective magnetic field and requiring an applied radio frequency of a lower energy to cause resonance when electrons are withdrawn from a nucleus, the nucleus is deshielded and feels a stronger magnetic field requiring more energy (higher frequency) to cause resonance as the energy gap is greater NOTE: If the radio frequency is kept constant and the magnetic field varied instead to achieve resonance then a relatively weaker magnetic field will be required for more deshielded carbon atoms the nmr spectra is based on the scale of chemical shifts relative to a standard reference peak – tetramethylsilane TMS has 4 carbon atoms in the same environment which produces a single strong peak Si is relatively less electronegative than C so the electrons in the C-Si bonds are closer to the carbons than in other organic compounds resulting in a peak on the spectrum at the extreme right-hand side Mr Lund 08 May 2017 60 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry for TMS = 0 by definition thus, nmr provides information about a carbon’s electronic environment. carbons attached to electron withdrawing atoms/groups such as oxygen tend to resonate at higher frequencies yielding peaks downfield from TMS i.e. a higher (chemical shift) carbonyl carbons aromatic carbons alcohols, esters, ethers unsaturated aliphatic saturated aliphatic Note: 3 carbons but 2 peaks 41 170 160-220 110-160 50 - 90 90-150 5-40 C=O C-O C=C C-C 170 ppm ppm ppm ppm ppm the major differences that you will notice in 13C nmr in comparison to 1H-nmr spectra include: 1. no integration of carbon spectra 2. wide range (0-220 ppm) of resonances for common carbon atoms (typical range for protons 0-12 ppm) CDCl3. (i.e. hydrogen has been replaced by its isotope, deuterium) is commonly used as a solvent (also see proton-NMR) and the line for this carbon is removed from the final spectrum Summary Questions Page 145 A2 Chemistry (Nelson Thornes) AQA Chemguide 1, 2 143 - 145 Carbon 13 nmr Mr Lund 08 May 2017 61 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Low Resolution 1H Spectra 1 H is far more abundant in organic compounds than 13C hence the spectra are much easier to obtain their resonance field strength will vary depending on the number of surrounding electrons electrons shield the nucleus thereby reducing the effective magnetic field and requiring an applied radio frequency of a lower energy to cause resonance when electrons are withdrawn from a nucleus, the nucleus is deshielded and feels a stronger magnetic field requiring more energy (higher frequency) to cause resonance NOTE: If the radio frequency is kept constant and the magnetic field varied instead to achieve resonance then a relatively weaker magnetic field will be required for more deshielded carbon atoms the nmr spectra is based on the scale of chemical shifts relative to a standard reference peak – tetramethylsilane TMS is used as o it provides a single strong signal from 12 equivalent H atoms o it is chemically inert o the peak will be up field of most other peaks given the relatively low electronegativity of silicon compared to carbon for TMS = 0 by definition hydrogen’s on the same atom in a molecule will experience an identical electromagnetic environment and will thus yield a single peak at a given value on a low resolution spectra greater electron density results in relatively more shielding hence the requirement for a higher external field to cause resonance electronegative atoms e.g. oxygen (or unsaturated regions e.g. a benzene ring) cause deshielding hence a lower external energy requirement number of peaks = number of non equivalent groups of H atoms chemical shift identifies adjacent environment e.g. nearby oxygen integrated spectra indicate the relative proportion of equivalent H atoms i.e. those in an identical environment Mr Lund 08 May 2017 62 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Below are common situations: 1 hydrogen strongly suggests an alcohol, acid or aldehyde 9 equivalent H’s suggests –C(CH3)3 6 equivalent H’s suggests –C(OH)(CH3)2 – NOT –CH(CH3)2 as this would be outside the limitations for the AQA syllabus (see later) 3 equivalent H’s suggests –CH3 e.g. –COC H3 or -CH(OH)CH3 etc 2 equivalent H’s suggests –CH2- a proton free solvent is used e.g. CCl4, D2O CDCl3 (2H gives no peak) Summary Questions How Science Works Page 147 Pages 151-2 A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide 1 The birth of nmr 146 - 148 128 – 132 Q11 – 13 on pages 130 – 132 nmr spectroscopy Mr Lund 08 May 2017 63 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Interpretation of High Resolution Spectra non equivalent hydrogen’s on adjacent carbons will themselves behave as little magnets and their effect can be seen on high resolution nmr each creates a small magnetic field aligned with or against the external field – spin coupling this creates small differences in the local field which modifies the radio frequency energy required for resonance hydrogen’s on the same atoms are identical and will not modify one another NOTE: hydrogen’s bonded to an oxygen are not split and do not cause others to split the more adjacent hydrogen’s there are, the more possible permutations exist peaks are thus split into sub sets by neighbouring non equivalent H atoms n+1 rule – if there are n H atoms in total on adjacent C atoms then the peak is split into n+1 signals of relative intensities derived from Pascal’s triangle: these two together are strongly indicative of an ethyl group NOTE: A2 Chemistry (Nelson Thornes) AQA A2 Chemistry (Heinemann) AQA Chemguide Your syllabus is limited to samples yielding singlet’s, doublets, triplets and quartets Summary Questions Page 152 1, 2 Exam Style Questions Pages 155-7 Pages 158 – 161 1–6 2, 3, 5, 6 148 - 152 128 – 136 Q11 – 15 on pages 130 – 136 Q 1 – 4 on pages 137 -138 Q 6,11 on pages 151 - 153 Proton nmr Mr Lund 08 May 2017 64 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry INTEGRATED SPECTROSCOPY you may be required to use data from more than one spectra for identification on my CHEMISTRY - READ site I have placed a suggested flow chart it would be useful for you to think about and develop your own strategy IR will enable you to determine if there is/are one or two oxygen’s present coupled with Mass Spectrometry this will allow you to establish a maximum number of carbon atoms in the structure (which may also be derived from the M+ + 1 peak (satellite peak) due to carbon-13) and may also indicate the number of hydrogen’s hence give clues to the degree of saturation in the compound fragmentation patterns in Mass Spectrometry are best left until needed or as a confirmatory check at this point you might want to write down the pieces of the jigsaw that are present and then use nmr to work out what is connected to what be alert to the possibility of carbons/hydrogen’s in identical environments e.g. in propanone or in branched alkyl components Sample spectra: http://home.clara.net/rod.beavon/spectra.htm Here are some useful tables: Proton nmr MS data m/z value Interpretation Notes 15 CH3+ Not much use without other data 29 CH3CH2+ 31 CH3O+ Look for a 2H quartet and 3H triplet on 1H nmr Look for a 3H singlet on 1H nmr 43 HIGH peak = CH3CO+ CH3CH2CH2+ 57 CH3CH2CO+ 77 C6H5+ 105 C6H5CO+ Infra-red absorption data Bond Wavenumber 43 and 29 could be butanone 43 and 15 could be propanone 43 with no 15 ethanal or acyl chloride possible Check IR for a carbonyl group Check IR for a carbonyl group 77 + 29 consider ethylbenzene 77 + 43 propylbenzene or phenylethanone Benzaldehyde or Benzoic acid Mr Lund 08 May 2017 65 C—H C—C C=C C=O C—O O—H O—H 2850–3300 750–1100 1620–1680 1680–1750 1000–1300 (alcohols) 3230– 3550 (acids) 2500–3000 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry CHROMATOGRAHY essentially the physical separation of the constituents of a mixture a mobile phase (solution or gases called the eluent) flows through a stationary phase (a solid or liquid supported on a solid) and in so doing the components of a mixture are separated as they travel at different rates. the more soluble a solid, the further (faster) it moves in a given time the stationary phase will also determine the rate of progress of the components as a function of its affinity for each component hence the trick is to use a suitable combination of solvent and stationary phase to achieve the separation of similar components these can then be analysed separately using other techniques where required Column Chromatography the solid phase is a powder e.g. silica or alumina packed into a tube a solvent (the eluent) is added at the top the components of a mixture are separated as they travel at different rates down the column so can be collected separately for further analysis the separation depends on the balance between the solubility in the mobile phase and the retention in the stationary phase as in TLC (felt tip pens and paper is the simplest form), there is an equilibrium established between the solute adsorbed on the silica gel or alumina and the eluting solvent flowing down through the column. this method is effective at separating mixtures of amino acids it can effectively separate fairly large amounts of substance different eluents can be used to extract different components here is a practical guide to how it is done: http://www.wfu.edu/academics/chemistry/courses/CC/index.htm Here is a video clip of some people in white coats doing Chemistry: Video clip Mr Lund 08 May 2017 66 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Gas-liquid Chromatography GC this is particularly useful in the separation of volatile liquids a sample is injected into the device in which there is a long (typically 100 m) thin (0.5 mm) tube (coiled for compactness) containing the stationary phase (an inert powder coated with oil) a carrier gas (unreactive e.g. N2 or He) is used as the eluent to provide the mobile phase the carrier gas transports the sample along the tube separating the mixture in the process separation depends on rate of movement with the gas and retention by the coil the time taken for each component to complete its journey through he coil is called it’s retention time data will also be collected regarding the relative amount of each component the output samples can be analysed by MS, IR or nmr under computer control this method is effective even with small sample sizes/trace amounts due to its high sensitivity (e.g. food additives, drug usage detection) Here is a ‘How Science Works’ link that considers the controversy surrounding the interpretation of GC/MS data regarding whether life has/has not been detected on Mars – there are many like this http://www.msss.com/http/ps/life/life.html GC/MS equipment on Mars lander Summary Questions Page 154 A2 Chemistry (Nelson Thornes) AQA Chemguide 1-3 153 - 154 Chromatography Mr Lund 08 May 2017 67 A2 Unit 4 Kinetics, Equilibria and Organic Chemistry Mr Lund 08 May 2017 68 AS Chemistry Unit 2 69 AS Chemistry Unit 2 70