Chemistry Notes HSC PDF

CHEMISTRY NOTES HSC Luo Jiajun Table of Contents CHEMISTRY NOTES HSC 1 Jason Yang 1 Module 5...

CHEMISTRY NOTES HSC Luo Jiajun Table of Contents CHEMISTRY NOTES HSC 1 Jason Yang 1 Module 5 4 IQ1: What happens when chemical reactions don’t go through to completion? 4 Reversible reactions 4 Cobalt (II) chloride hydration 4 Iron nitrate(III) and Potassium thiocyanate equilibrium 5 Burning magnesium and steel wool 5 un Dynamic equilibrium 5 Static equilibrium 6 Open and closed systems 6 Enthalpy and entropy 7 Combustion and photosynthesis 7 Collision theory and reaction rate 7 Le chatelier’s principle Concentration Temperature Pressure/volume aj IQ2: What factors affect equilibrium and how? 8 8 9 9 10 Ji Catalyst 11 Effect of activation energy and heat of reaction 11 Additional: How to answer LCP questions 11 IQ3: How can the position of equilibrium be described and what does the equilibrium constant represent? 12 o Equilibrium constant (Keq) 12 Iron thiocyanate equilibrium 13 IQ4: How does solubility relate to chemical equilibrium? 13 Lu Intermolecular forces and solubility 13 Dissolution equilibrium 14 Toxins in Cycad fruit 14 Precipitation reactions 15 Solubility of an ionic substance 15 Ionic product (Qsp ) and formation of precipitates 16 Common ion effect 16 Module 6 17 IQ1: What is an acid and what is a base? 17 Acids and bases 17 Properties 18 Indicators 18 Reactions 19 Everyday applications 19 Industrial applications 19 Early definitions 20 2  Arrhenius theory 20 Bronsted-Lowry theory of acids 20 Advantages and limitations of each theory 21 Measuring pH 21 Calculating pH 22 Factors that affect pH 22 Acidic, basic and neutral salts 23 Conjugate acid/base pairs 24 Amphiprotic salts 25 Module 7 26 IQ1: How do we systematically name organic chemical compounds 26 Organic compounds 26 Organic nomenclature 26 un Hydrocarbons 27 Structural isomerism 37 IQ2: How can hydrocarbons be classed based on their structure and reactivity? 39 Homologous series 40 Shape of molecules 40 Physical properties of alkanes 42 Properties of haloalkanes Handling and disposure aj Physical properties of alkenes and alkynes IQ3: What are the products of reactions of hydrocarbons and how do they react? 42 43 43 44 Ji Alkenes and Alkynes reactivity 44 Alkanes reactivity 46 IQ4: How can alcohols be produced and what are their properties? 47 Properties of alcohols 48 Combustion of alcohols 49 Reactions of alcohols 49 o Substitution 50 Fermentation 51 Oxidation of alcohols 51 Lu IQ5: What are the properties of organic acids and bases? 52 Aldehydes and ketones 52 Carboxylic acids 53 Amines 53 Amides 54 Properties of esters 54 Production of esters 54 Purification of esters 56 Organic acids 56 Organic bases 56 Reactions between organic acids and bases 57 Saponification 57 Structure of surfactants 57 Cleaning action of soaps 58 Module 8 59 IQ2: How is information about the reactivity and structure of organic compounds obtained? 59 3  Carbon-Carbon double bonds 60 Carboxylic acids 60 Hydroxyl groups 60 NMR Spectroscopy 61 Mass Spectrometry 63 Infrared Spectroscopy 64 un Module 5 aj IQ1: What happens when chemical reactions don’t go through to completion? Ji 5.1.1 conduct practical investigations to analyse the reversibility of chemical reactions, for example: – cobalt(II) chloride hydrated and dehydrated – iron(III) nitrate and potassium thiocyanate – burning magnesium – burning steel wool (ACSCH090) o Reversible reactions Lu Many chemical reactions are reversible o Reactants react to form products, products react to re-form reactants o Reverse reaction will have exact opposite thermodynamic properties of the forward reaction ▪ Exothermic🡪Endothermic Dynamic equilibrium is where both forwards and reverse reactions are occurring at the same rate o Can only be achieved in a closed system o Concentration of products and reactants are constant Cobalt (II) chloride hydration To analyse the reversibility of hydration of cobalt(II) chloride 4  o Dissolving anhydrous (containing no water) cobalt(II) chloride in water produces a pink solution of cobalt(II) chloride hexahydrate (CoCl2.6H2O) o This solution can be heated again, where the waters of crystallization are vaporized to reform blue anhydrous cobalt(II) chloride. ▪ An intermediate purple cobalt(II) chloride dihydrate (CoCl2.2H2O) is observed during heating Iron nitrate(III) and Potassium thiocyanate equilibrium Reaches a dynamic equilibrium where rates of forwards and reverse reaction are equal and concentration of products and reactants are constant un Iron(III) nitrate + Potassium thiocyanate 🡪 Potassium nitrate + iron (III) thiocyanate Net equation: aj Burning magnesium and steel wool Forward combustion reaction is initiated by heating magnesium and steel wool (Fe) under a Bunsen burner Ji o Combustion of magnesium produces a bright white light and white MgO solid o Combustion of steel wool produces reddish brown Fe2O3 solid Reversibility of this reaction is tested by placing the products of combustion in an ice o bath to cool, in an attempt to provide conditions to favour the reverse reaction o Products do not react again to reform reactants and thus these reactions are irreversible Lu 5.1.2 Model static and dynamic equilibrium and analyse the differences between open and closed systems. Dynamic equilibrium A reversible reaction can reach a state of dynamic equilibrium Rate of forward reaction equals rate of reverse reaction and is non-zero 5  o At first the rate of forward reaction is high and reverse reaction is low as there is a high concentration of reactants o As more product molecules are formed, the rate of the reverse reaction begins to increase until it is equal with the forward reaction (collision theory) Concentration of products and reactants are constant (not necessarily equal) un ∆𝐺 = 0 at chemical equilibrium Macroscopic o No observable changes Microscopic aj o In flux, constant conversion Energy profile diagram Ji o Lu Static equilibrium Rates of forward and reverse reaction are effectively zero Concentrations of products and reactants are constant o Same graph as above Can be reached in an irreversible reaction o Limiting reagent is used up, reaction goes to completion o Rates of forward and reverse reaction are effectively zero 6  Diamond-Graphite static equilibrium o C(diamond) ⇌ C(graphite) o Activation energy of both forwards and reverse reaction are immense o At room temperature negligible amounts of graphite are being converted into diamond Open and closed systems Systems are an integrated whole composed of matter and energy o Chemicals involved in a reaction form a system and the rest of the universe that exists outside are called the surroundings Open systems allow exchange of matter and energy between the system and its un surroundings o Test tube without lid Closed systems only allow exchange of energy between the system and its surroundings o Test tube with lid Additional: Isolated systems do not allow exchange of matter nor energy 5.1.3 aj o These are hypothetical and do not occur in nature analyse examples of non-equilibrium systems in terms of the effect of entropy and enthalpy, for Ji example:– combustion reactions – photosynthesis Enthalpy and entropy Enthalpy (∆𝐻) is a measure of internal heat of a system o 𝑜 ∆𝐻𝑓 is the change in enthalpy associated with the formation of 1 mol of a substance at standard conditions Lu 𝑜 𝑜 o ∆𝐻 = ∆𝐻𝑓(𝑃) − ∆𝐻𝑓(𝑅) Entropy (∆𝑆) is a measure of disorder or microstates of a system o ∆𝑆(𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑒) = ∆𝑆 (𝑠𝑦𝑠𝑡𝑒𝑚) + ∆𝑆 (𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠) > 0 Reactions will proceed to increase entropy and decrease enthalpy Combustion and photosynthesis Both non-reversible reactions and go to completion o Note: Respiration is not the reverse reaction of photsynthesis Enthalpy and entropy can be easily analysed for non-reversible reactions Combustion o Forwards entropy and enthalpy drive Photosynthesis 7  o Reverse enthalpy and entropy drive o Non-spontaneous and continuous energy supply from sunlight and other coupled reactions is needed 5.1.4 investigate the relationship between collision theory and reaction rate in order to analyse chemical equilibrium reactions Collision theory and reaction rate Collision theory states that chemical reactions occur when molecules collide with sufficient energy and correct orientation High concentration 🡪 High frequency of successful collisions 🡪 High reaction rate Initially, there is a high concentration of reactants in a chemical system, and thus by un collision theory there is a high rate of reactants being converted to products o Concentration of reactants decrease whilst concentration of products increases Afterwards, rate of forwards reaction decreases whilst reverse reaction increases o Continues to a point where they are equal (dynamic equilibrium) aj Note: In questions that mention collision theory and reaction rate, always define and relate Ji IQ2: What factors affect equilibrium and how? 5.2.1 investigate the effects of temperature, concentration, volume and/or pressure on a system at equilibrium and explain how Le Chatelier’s principle can be used to predict such effects, for example: – heating cobalt(II) chloride hydrate – interaction between nitrogen dioxide and dinitrogen tetroxide – iron(III) thiocyanate and varying concentration of ions (ACSCH095) o 5.2.2 Explain the overall observations about equilibrium in terms of the collision theory Lu 5.2.3 examine how activation energy and heat of reaction affect the position of equilibrium. Le chatelier’s principle Le chatelier’s principle states that a system at dynamic equilibrium will shift to counteract changes until a new equilibrium is established At dynamic equilibrium, rate of forward reaction equals rate of reverse reaction and concentration of products and reactants are constant Shift refers to a reaction from a chemical system to counteract disturbances or changes o If the system shifts right, the rate of forwards reaction begins to exceed the rate of reverse reaction and concentration of products increases until a new equilibrium is established 8  o If the system shifts left, the rate of reverse reaction begins to exceed the forwards reaction, increasing the concentration of reactants until a new equilibrium is established Factors that cause changes to a system include concentration, temperature and pressure/volume A shift in equilibrium will never completely counteract the change introduced to the system o Concentrations of products and reactants will not return to original un aj o Consider the ammonia equilibrium where additional H2(g) is added Ji o [H2](New equilibrium) > [H2](Original equilibrium) o Rate of forwards and reverse reaction at the new equilibrium will also be higher than initially Concentration o Concentration of products/reactants is changed o System will shift to favor the reaction that removes excess reactants/products Lu − − + 𝐶𝑙2(𝑔) + 𝐻2𝑂(𝑙) ⇌𝐶𝑙(𝑎𝑞) + 𝐶𝑙𝑂(𝑎𝑞) + 2𝐻(𝑎𝑞) o If KCl is added to the system, concentration of product Cl- ions will increase and thus the system will shift left to favour the reverse reaction in order to remove excess Cl- o If H+ ions are removed, concentration of reactants will be in excess and equilibrium will shift right to favor production of H+ Collision theory o Increased concentration of product Cl- 🡪 Increased frequency of collision between product particles 🡪 Increased reverse reaction o As the reverse reaction is increased, the concentration of reactant particles will begin to increase and similarly, the forwards reaction rate will also increase until it is equal to the reverse reaction at a new equilibrium 9  Adding or remove pure solids or pure liquids from a system will generally have no effect on the equilibrium o S8(s) , Br2(l) However, adding or removing water will impact the equilibrium as it can change the concentration of aqueous products/reactants o Adding water will favour the reaction that produces more aqueous moles ▪ The side with more aqueous moles will have its concentration reduced proportionally more o Contrastingly, removing water will favour the reaction that produces less aqueous moles o If both sides have the same aqueous moles, then the equilibrium will be un unaffected Temperature The direction in which a system will shift when heat is added or removed depends on whether the reaction is endothermic or exothermic. aj Consider the synthesis of ammonia, forward reaction if exothermic, reverse is endothermic Ji o If the temperature is increased ▪ System shifts to favour the backwards endothermic reaction to remove heat from the system, lowering temperature o If the temperature is decreased o ▪ System shifts to favour the forwards exothermic reaction to produce heat, increasing temperature Collision theory Lu o In an endothermic equilibrium, activation energy of forward reaction > reverse reaction o By increasing the temperature, both forwards and reverse reaction rates will increase ▪ Greater effect on the forwards reaction as the activation energy is high and a greater proportion of particles will now have enough energy to overcome activation energy o Thus, an endothermic reaction is favoured when the temperature is increased o Similarly, by decreasing the temperature, the endothermic reaction with the higher activation energy is affected proportionally greater and thus the reverse exothermic reaction is favored. 10 Pressure/volume un Pressure and volume of system is changed o Increase volume = decreased pressure (Boyles law) o Increased moles of gaseous substances = increased pressure (avagadros law) o Only has effect on the concentration of gaseous products/reactants o Direction that equilibrium shifts depends on the molar ratio of gaseous aj reactants to products Ji Consider the synthesis of ammonia, molar ratio of reactants to products is 4:2 o If volume is decreased, pressure is increased and system will favour the forward reaction that produces less moles of gases to counteract the change by decreasing pressure o If volume is increased, pressure is decreased and system will favour the o reverse reaction that produces more moles of gases to counteract the change by increasing pressure Addition of an inert gas will have no effect on the equilibrium give that volume is Lu constant o Eventhough this raises the total pressure of the system, the partial pressures of the reactant/product gases will remain unchanged. Collision theory o Changing volume/pressure = Changing concentration o By decreasing the volume, gaseous reactant and product molecules are brought together, increasing concentration and thus reaction rate of both forward and reverse reactions ▪ However, the side with the higher molar ratio will be affected proportionally greater and thus the reaction rate will increase to a greater extent o Thus, increasing pressure favors the reaction with the reactants having the higher molar ratio. 11  o Similarly, by decreasing pressure, the concentration and reaction rate of the side with the higher molar ratio is decreased to a greater extent. ▪ Thus, increasing pressure favors the reaction with the reactants having the lower molar ratio o Addition of an inert gas has no effect on the partial pressures of reactant and product gases ▪ However, inert gas molecules will collide unsuccessfully with reacting molecules, obstructing potentially successful collisions ▪ This results in a decreased reaction rate for both forward and reverse reactions and thus it will take longer to reach equilibrium. un Catalyst Catalyst is neither consumed nor produced over the course of a reaction o Has no effect on equilibrium as it only increases reaction rate not production of reactant/products ▪ ▪ aj o Decreases the time taken for a system to reach equilibrium Forwards, reverse reaction rates increased Ji Reactants depleted quickly, products produced rapidly Effect of activation energy and heat of reaction Changes to the activation energy have no effect on the position of the equilibrium o o Forwards and reverse reaction will be affected similarly, no change to relative concentrations of products and reactants Lu Additional: How to answer LCP questions 2-3 mark LCP question 1. Define Le Chatelier’s principle 2. Identify change in system in terms of concentration, pressure and temperature 3. Explanation on which reaction the system will favour, and hence whether the equilibrium will shift left or right 4. Any observations you would expect (Only if question specifies) IQ3: How can the position of equilibrium be described and what does the equilibrium constant represent? 5.3.1 deduce the equilibrium expression (in terms of Keq) for homogeneous reactions occurring in solution 12 5.3.2 perform calculations to find the value of Keq and concentrations of substances within an equilibrium system, and use these values to make predictions on the direction in which a reaction may proceed 5.3.3 qualitatively analyse the effect of temperature on the value of Keq Equilibrium constant (Keq) Quantitative relationship between the concentrations of reactants and products at equilibrium Consider equation o 𝑎𝐴 + 𝑏𝐵⇌𝑐𝐶 + 𝑑𝐷 𝑐 𝑑 [𝐶] ×[𝐷] o 𝐾𝑒𝑞 = 𝑎 𝑏 [𝐴] ×[𝐵] un o Where A, B, C, D are gaseous or aqueous substances (cannot be pure solids or liquids Size of Keq denotes the position of equilibrium o Large K 🡪 Equilibrium lies to the right (Products) o Middling K (0.1 < K < 10) 🡪 Comparable concentrations of products and reactants aj o Small 🡪 Equilibrium lies to the left (Reactants) System not yet at equilibrium is given the reaction quotient (Q) Ji o Q≠K o If Q < K ▪ System will proceed right to increase concentration of products o If Q < K o ▪ System will process left to increase concentration of reactants Pressure, volume and concentration changes will not affect Keq Lu o Concentration of products and reactants are changed initially, however they will shift back to the same ratio and retain the initial Keq value Temperature changes will affect Keq o Initially, a temperature change does not induce a change in concentration, instead the system will shift to different concentrations and a different Keq o The way K changes with temperature depends on whether the reaction is exothermic or endothermic o Shifts left = decrease K o Shifts right = increase K 5.3.4 conduct an investigation to determine Keq of a chemical equilibrium system, for example: -Keq of the iron(III) thiocyanate equilibrium 13 Iron thiocyanate equilibrium Using colorimetry to determine Keq o Method of analysis which relates the absorption of light of a coloured solution to the concentration of its solute un o Greater absorption = Higher concentration o Concentration of FeSCN2+ will be determined by colorimetry which can then be used to calculate Keq Calibration curve o Concentration vs Absorbance o Graphed by testing absorbance from several FeSCN2+ standard solutions (0.1M, 0.2M, etc) aj o Line of best fit can be drawn and used to determine concentration of a solution which records a specific absorbance Ji 5.3.5 explore the use of Keq for different types of chemical reactions, including but not limited to: – dissociation of ionic solutions – dissociation of acids and bases IQ4: How does solubility relate to chemical equilibrium? o 5.4.1 describe and analyse the processes involved in the dissolution of ionic compounds in water Intermolecular forces and solubility Lu “Like dissolves like” o Polar solutes dissolve in polar solvents, non-polar solutes in non-polar solvents o Due to the different strengths of intermolecular forces ▪ A solute will only dissolve if the formation of IMF’s between solute and solvent are more favourable than existing IMF’s. Ionic compounds dissolve in water to form solvation spheres held together by ion-dipole forces o Whether the dissolution is exothermic or endothermic depends on the energy changes involved in breaking and forming bonds ▪ Hydrogens bonds broken (water) 14  ▪ Ionic bonds broken (salt) ▪ Ion dipole forces formed Dissolution equilibrium Dissolution of every substance is an equilibrium Dissolution of calcium chloride CaCl2(s) ⇌ Ca2+(aq) + 2Cl-(aq) o A static equilibrium is formed when a small sample of CaCl2 is added to water and it completely dissolves (unsaturated solution) un o A dynamic equilibrium is formed when sufficient CaCl2 is dissolved such that the solution is saturated ▪ Dissolution and precipitation is occurring and at the same rate The maximum amount of solute that can be dissolved can be increased or decreased by changing the temperature o Supersaturated solutions allow for more solute to be dissolved than at aj standard conditions (25oC) [ Keq for the calcium chloride equilibrium = 𝐶𝑎 × [𝐶𝑙 ] o Also known as the solubility product Ksp 2+ ] − 2 Ji 5.4.2 investigate the use of solubility equilibria by Aboriginal and Torres Strait Islander Peoples when removing toxicity from foods, for example: – toxins in cycad fruit Toxins in Cycad fruit o Cycad fruit/cycad seeds contains the toxin cycasin (C8H16N2O7) Cycasin is carcinogenic and a neurotoxin which can cause vomiting, diarrhea and damage to liver Lu Water soluble Detoxification 1. Strip outer flesh layer 2. Crushed to increase surface area (collision theory) 3. Submerged in boiling water to leach out toxins 4. Water is drained and seeds are leached again in boiling water When the cycad fruit is leached in water, cycasin dissolves and forms a dynamic equilibrium o C8H16N2O7(s) ⇌ C8H16N2O7(aq) ∆𝐻 > 0 o Since it is an endothermic equilibrium, increasing the temperature will shift the equilibrium to the right, increasing the amount of cycasin dissolved ▪ Therefore, boiling water is used 15  o Repeating the process with clean water allows the solution to be saturated again, increasing the total amount of cycasin leached out 5.4.3 conduct an investigation to determine solubility rules, and predict and analyse the composition of substances when two ionic solutions are mixed, for example: – potassium chloride and silver nitrate – potassium iodide and lead nitrate – sodium sulfate and barium nitrate Precipitation reactions Two ionic solutions are mixed, resulting in the formation of an insoluble solid (Precipitate) un Use NAGSAG for memorization of solubility rules aj Ji Potassium chloride and silver nitrate o KCl(aq) + AgNO3(aq) 🡪 KNO3(aq) + AgCl(s) o o White AgCl(s) precipitate Potassium iodide and lead nitrate Lu o 2KI(aq) + Pb(NO3)2(aq) 🡪 PbI2(s) + 2KNO3(aq) o Yellow PbI2(s) precipitate Sodium sulfate and barium nitrate o Na2SO4(aq) + Ba(NO3)(aq) 🡪 2NaNO3(aq) + BaSO4(s) o White BaSO4(s) precipitate 5.4.4 derive equilibrium expressions for saturated solutions in terms of Ksp and calculate the solubility of an ionic substance from its Ksp value Solubility of an ionic substance When a slightly soluble ionic substance is placed in water, some of it dissolves whilst the remaining exists as a solid o Dynamic equilibrium is formed Ksp value of an ionic compound can be used to determine its solubility (g/100ml) 16 5.4.5 predict the formation of a precipitate given the standard reference values for Ksp Ionic product (Qsp ) and formation of precipitates If Qsp < Ksp o Solution is unsaturated and ionic solid will fully dissolve If Qsp = Ksp o Solution is at equilibrium (saturated) If Qsp > Ksp o Solution may be supersaturated and some of the ionic solid will precipitate out un Common ion effect It is difficult to dissolve salts into solutions that already contain its ions dissolved Consider dissolution of PbBr2 into a solution containing Br- ions o PbBr2(s) ⇌ Pb2+(aq) + 2Br-(aq) o Existing Br- ions will increase concentration of Br- ions in the equation, this aj shifts the above equation to the left (LCP) making the dissolution of the salt unfavourable Ji o Lu 17  un Module 6 6.1.1 aj IQ1: What is an acid and what is a base? investigate the correct IUPAC nomenclature and properties of common inorganic acids and bases Ji Acids and bases Acids are compounds that ionise to form hydronium (H3O+) ions in aqueous solution o pH < 7 (acidic) Bases are compounds that ionise to form hydroxide (OH-) ions in aqueous solution o pH > 7 (alkaline/basic) o o Alkalis are compounds that produce a basic solution when dissolved (water soluble base) Lu Acids Bases Hydrochloric acid HCl Sodium hydroxide NaOH Nitric acid HNO3 Ammonia NH3 Ethanoic acid CH3COOH Potassium hydroxide KOH Sulfuric acid H2SO4 Calcium hydroxide Ca(OH)2 Carbonic acid H2CO3 Magnesium hydroxide Mg(OH)2 Phosphoric acid H3PO4 Sodium carbonate Na2CO3 Citric acid C6H8O7 Naming Inorganic acid o ‘Hydro’ + element – ‘ide’+ ‘ic’ + ‘acid’ ▪ HCl = Hydrochloric acid o Polyatomic ion – ‘ate’ + ‘ic’ + ‘acid’ 18  ▪ H2SO4 = sulfuric acid o Oxyanion – ‘ite’ + ‘ous’ + ‘acid’ ▪ H2SO3 (Sulfite anion) = sulfurous acid Properties Acids o Sour o Corrosive o Conductive in solution o Turns blue litmus red un Bases o Bitter o Soapy feel o Caustic o Conductive in solution 6.1.2 o aj Turns red litmus blue conduct an investigation to demonstrate the preparation and use of indicators as illustrators of the characteristics and properties of acids and bases and their reversible reactions Ji Indicators Substances that change colour based on the pH of the environment Cannot determine exact pH, instead gives a range o Can be used to determine whether a substance is acidic or basic o Indicator pH Acidic range Transition Basic range transition range range Lu Methyl orange 3.1-4.4 Red Orange Yellow Bromothymol blue 6.0-7.6 Yellow Green Blue Litmus 5.5.-8.0 Red Purple Blue Phenolphthalein 8.3-10 Colourless Pale pink Pink/magenta 19 6.1.3 predict the products of acid reactions and write balanced equations to represent: ICT – acids and bases – acids and carbonates – acids and metals Reactions Acid + Base 🡪 Salt + Water o Neutralisation reaction o Exothermic o Salt may be a precipitate un Acid + Carbonate 🡪 Salt + Water + Carbon dioxide o Type of neutralization reaction Acid + Metal 🡪 Salt + hydrogen gas 6.1.4 investigate applications of neutralisation reactions in everyday life and industrial processes Everyday applications Antacid tablets aj Ji o Stomach produces hydrochloric acid to aid chemical digestion ▪ pH around 1.5-3.5 o Acid reflux ▪ Acid from stomach regurgitates into oesophagus, causing pain o o Antacid tablets neutralise HCl in stomach to reduce irritation from acid reflux ▪ Commonly made from Mg(OH)2 and Al(OH)3 Lu ▪ Mg(OH)2(s) + 2HCl(aq) 🡪 MgCl2(aq) + 2H2O(l) Toothpaste o Acids from foods/drinks cause tooth decay ▪ Bacteria convert sugars in teeth into acids o Weakly alkaline to neutralise any acids ▪ Contain bases such as CaCO3, Al2O3 or MgCO3 Baking soda o NaHCO3(s) reacts with acids in baking powder and in the dough to produce CO2(g), allowing the dough to rise Body spills o Neutralisation with acid or base is not desired as it is exothermic and can cause burns 20  o Put under running water to dilute the acid/base Industrial applications Acid and base spills o Acids are corrosive and bases and caustic ▪ Can harm environment and humans if spilt o Acids and bases can be used to neutralise spills o Common neutralising agent is NaHCO3(s) as it has many desirable properties ▪ Cheap, easy to store un ▪ Weakly acidic or basic less dangerous to use ▪ Solid powders Contain spill without spreading it ▪ ▪ aj Amphiprotic Can neutralise both acids and bases Useful for unknown spills Produces bubbles (CO2(g)) to indicate process or end of reaction Ji Treatment of wastewater o Industrial wastewater may be acidic or basic, potentially causing harm if discharged into the environment o HF(aq) is commonly found in wastewater in the semiconductor industry o ▪ NaOH and Ca(OH)2 are used to neutralise it o Basic industrial wastewater can be treated with citric acid (C6H8O7) or acetic Lu acid (CH3COOH) 6.1.5 conduct a practical investigation to measure the enthalpy of neutralisation 6.1.6 explore the changes in definitions and models of an acid and a base over time to explain the limitations of each model, including but not limited to: – Arrhenius’theory – Brønsted–Lowry theory. Early definitions Antoine Lavoisier o Acids were substances that contained oxygen o Bases were able to neutralize acids Humphry davy o Acids contained replaceable hydrogens 21  ▪ Replaced by metals in acid metal reactions ▪ Ca(s) + 2HCl(aq) 🡪 CaCl2(aq) + H2(g) Arrhenius theory Acids ionized to form H+ ions in aqueous solution o HA(aq) 🡪 H+(aq) + A-(aq) Bases ionized to form OH- ions in aqueous solution o XOH(aq) 🡪 X+(aq) + OH-(aq) Bases and acids neutralized each other in solution un o H+(aq) + OH-(aq) 🡪 H2O(l) Bronsted-Lowry theory of acids aj Acids are substances that tend to donate protons (H+) o HA(aq) + H2O(l) 🡪 H3O+(aq) + A-(aq) Ji A base is simply a substance that tends to accept protons o NH3(aq) + H2O(l) 🡪 NH4+(aq) + OH-(aq) An acid can only be defined in relation to a base o Any substance that donates a proton is the acid and the species that accepts it is the base o Amphiprotic substances can act as an acid or a base depending on the environment o Able to donate or accept protons (H+) Lu Advantages and limitations of each theory Theory Advantages Limitations Lavoisier -First scientific theory -Some oxygen containing substances regarding the definition of an were basic, e.g., CaO, Na2O acid -Some acidic substances lacked -Realised that bases were oxygen, e.g, HCl, HF able to neutralize acids Humphry Davy -Worked for many common -Some acidic substances did not acids contain hydrogen, e.g., Acidic oxides CO2, NO2 Arrhenius -Explained the mechanism for -Could only account acidity and acid-base neutralization basicity for acids and bases in aqueous solution 22  -Worked for many acids and -Does not recognize the solvent in bases determining the strength and -Could explain the difference weakness of an acid in potency between strong -Could not explain the basic nature of and weak acids non-hydroxide bases -Could not explain why some salts were acidic, e.g., NH4Cl -Only accounted for neutralization reactions in aqueous ionized form (H+, OH-) Bronsted-Lowry -Explained behaviour of acids -Cannot explain amphoteric and bases in non-aqueous substances solvent -Cannot explain Lewis acids and bases un -Considers role of solvent in (do not accept or donate electrons) determining strength of acid -Explains the mechanism of amphiprotic substances -Explains existence of non-hydroxide bases 6.2.1 aj conduct a practical investigation to measure the pH of a range of acids and bases Ji Measuring pH pH probes o Measures [H3O+] to calculate pH o Can determine exact pH level o Very sensitive equipment o ▪ Kept in buffer solution to prevent degredation ▪ Lu Calibrated solutions of known pH before testing ▪ Cleaned with demineralised water before testing to prevent accumulation of impurities Indicators o Provide pH range of solution o HInd(aq) + H2O(l) ⇌ Ind-(aq) + H3O+(aq) + − 6.2.2 calculate pH, pOH, hydrogen ion concentration ([H ]) and hydroxide ion concentration ([OH ]) for a range of solutions Calculating pH pH is dependant on the concentration of hydronium ions in solution; [H3O+] + o 𝑝𝐻 = − 𝑙𝑜𝑔10[𝐻3𝑂 ] 23  o Number of decimal places in final pH = Number of significant figures in [H3O+] o Generally ranges from 0-14 o Higher [H3O+] = Lower pH = more acidic pH scale is logarithmic o A decrease of 1 pH = x10 increase in [H3O+] pOH = -log10[OH-] o pH +pOH = 14, at 25oC Water self ionises to form an equilibrium mixture of OH- and H3O+ ions o 2H2O(l) ⇌ H3O+(aq) + OH-(aq) ∆𝐻 > 0 ▪ The Keq for this reaction is termed the self-ionisation constant of water (Kw) un ▪ Kw = [OH- ] x [H3O+] = 1.0 x 10-14 at 25oC aj Ji o Since the self ionisation reaction is endothermic, increasing the temperature will increase Kw ▪ Neutral pH < 7 for solutions warmer than 25oC ▪ Neutral pH > 7 for solutions cooler than 25oC o Factors that affect pH Lu Concentration of ions Strength of acid o Strong acids have 100% degree of first ionisation ▪ Completely ionises in aqueous solution into hydronium ions ▪ [H3O+] final = [HA]initial ▪ HCl(aq) + H2O(l) 🡪 Cl-(aq) + H3O+(aq) Unidirectional arrow to show that reaction goes to completion o Weak acids have < 100% degree of ionisation ▪ HF (aq) + H2O(l) 🡪 H3O+(aq) + F-(aq) ▪ Bidirectional arrow indicates that this reaction does not go to completion 24  o Equimolar solutions of a strong acid will have lower pH than a weak acid o Common strong and weak acids Strong acid Weak acid HCl HF HI CH3COOH HBr H2CO3 HNO3 H3PO4 H2SO4 C6H8O7 un Proticity (strong acids) o Number of protons an acid can ionise with o Monoprotic: HCl, HNO3 , CH3COOH o Diprotic: H2CO3, H2SO4 o Triprotic: H3PO4, C6H8O7 ▪ aj o Strong acids with a higher proticity will tend to have a lower pH ▪ Equimolar solutions of H2SO4 will have lower pH than HCl Both 1st ionisation is 100%, however H2SO4 2nd ionisation will also Ji contribute to [H3O+] ▪ Note: di- or tri- protic strong acids will be weak for their 2nd or 3rd ionisations Degree of ionisation (weak acids) o o Weak acids with higher degrees of ionisation will generally have lower pH o Proticity doesn’t matter here as weak acid’s 2nd or 3rd ionisations will contribute neglible amounts of [H3O+] Lu Acidic, basic and neutral salts Acid + base 🡪 salt + water o Endpoint of a neutralisation reaction would not always have a neutral pH as the salt produced can be acidic, basic neutral o Properties of the salt will be determined by what types of acids and bases are reacted together (e.g., strong acid with weak base) Strong acid and strong base o Conjugates of strong acids/bases will be extremely weak ▪ Thus, neutral salts will be produced 25  ▪ Conjugates may be a spectator ion and does not accept or donate electrons (e.g., Na+) Strong acid and weak base o Weak conjugate acid and extremely weak conjugate base ▪ Thus, acidic salts will be produced Weak acid and strong base o Weak conjugate base and extremely weak conjugate acid ▪ Thus, a basic salt will be formed ▪ Conjugates may be a spectator ion and does not accept or donate un electrons (e.g., Na+) Note: Weak acid and weak base cannot be determined as the reaction is complex 6.2.3 conduct an investigation to demonstrate the use of pH to indicate the differences between the strength of acids and bases 6.2.4 write ionic equations to represent the dissociation of acids and bases in water, conjugate acid/base pairs in solution and amphiprotic nature of some salts, for example: aj – sodium hydrogen carbonate – potassium dihydrogen phosphate Conjugate acid/base pairs Ji Acid that donates its proton becomes a conjugate base o Can now accept protons in reverse reaction Base that accepts a proton becomes a conjugate acid o Can now donate proton in reverse reaction o HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq) Lu Strength of conjugate species o Consider degree of ionisation of acid/base and determine position of equilibrium o Conjugate base of weak acid is a weak base o Conjugate acid of a weak base is a weak acid o Conjugate base of a strong acid is an extremely weak base o Conjugate acid of a strong base is an extremely weak acid Amphiprotic salts Salt NaHCO3 contains the amphiprotic anion HCO3- 26  Salt KH2PO4 contains the amphiprotic anion H2PO4- un aj Ji o Module 7 Lu IQ1: How do we systematically name organic chemical compounds 7.1.1 investigate the nomenclature of organic chemicals, up to C8, using IUPAC conventions, including simple methyl and ethyl branched chains, including: (ACSCH127) – alkanes – alkenes – alkynes – alcohols (primary, secondary and tertiary) – aldehydes and ketones – carboxylic acids – amines and amides 27  – halogenated organic compounds Organic compounds Study of carbon compounds Carbon atoms covalently bonded o Single, double or triple o Chains, branched chains, rings o Functional groups can be placed on carbon atoms ▪ Dictate reactivity and chemical properties of compound o Compounds with the same functional group are called homologous series Representation un o Structural formula o Semi-structural/condensed formula ▪ ▪ aj CH3(CH)2(CH2)2CH3 Single bonds are not shown however double and triple bonds are Ji o Skeletal structures Organic nomenclature o Locant + Multiplier + prefix(es) + stem + -ane, -ene, -yne + locant + suffix Stem Lu o Stem prefix is given to number of carbons in a molecule’s main chain (Longest chain) Locant o Position of principal functional group (highest priority) Multiplier o Indicates how many times a functional group occurs (di, tri, tetra..) ▪ E.g, 2,3,3-tri-methylheptane Hydrocarbons Consist solely of Carbon and Hydrogen atoms Saturated compounds o All covalent bonds between carbon atoms are single 28  o Maximum number of hydrogens in chain Unsaturated o Double or triple bonds between carbon atoms o Not maximum number of hydrogens possible Alkanes, alkenes and alkynes o Homologous series ▪ Structurally related as each member differs by CH2 ▪ Determine longest continuous carbon chain (main chain) and give prefix accordingly o Alkanes un ▪ Only single carbon bonds (saturated) ▪ CnH2n+2, where n = 1, 2, 3… ▪ Ends in -ane o Alkenes ▪ ▪ aj One or more double carbon bonds (unsaturated) CnH2n, where n = 2,3,4… Ji ▪ Note: When there is more than one double or triple bond present, an ‘a’ is added between the name and locant, and the number of double/triple bonds is specified E.g., buta-1,3-diene o Alkynes o ▪ One or more triple carbon bonds (unsaturated) ▪ Lu CnH2n-2, where n = 2,3,4… Alkyl groups o A carbon group branching off the main chain o Methyl: -CH3, Ethyl: -CH2 CH3, Propyl: -CH2CH2CH3 o Naming ▪ Substituent written before the main chain, e.g., 2-methylbutane ▪ If there are multiple alkyl groups, list in alphabetical order, e.g., 2-ethyl-3-methyloctane ▪ If there are two or more of the same alkyl group, use prefixes to denote (di, tri, tetra , e.g., 2,2-dimethylpentane Halogenated o Halogens substitute hydrogen atom 29  o Naming ▪ Bromo, chloro, fluoro, etc as prefix ▪ If more than one type of halogen, priority is based on alphabetical order First point of difference rule o Principle functional group is given lowest locant number o At the first point where there is a difference in locant number of functional groups, numbering scheme is given to name with the lower locant un Carboxylic acid aj o General formula CnH2n+1COOH o Carboxyl functional group (-COOH) o Suffix added is -oic acid Ji ▪ E.g., Propanoic acid o Carbon atom in carboxyl functional group is counted as number 1 in main chain o Terminal as the -COOH group can only be at the ends of the main chain o Lu Alcohols o General formula CnH2n+1OH o Alkanols are a subset of alcohols that only contain C, H and O atoms o Hydroxyl functional group (-OH) ▪ Hydroxy is prefix when there is functional group of higher priority o Suffix used in -ol o Primary alcohol: C-OH groups attached to one other C atom o Secondary alcohol: C-OH groups attached to two other C atoms o Tertiary alcohol: C-OH groups attached to three other C atoms 30  o Compounds with 2 or more of the same dominant principle functional group do not omit the -e (Ethane-1,2-diol is correct rather than Ethan-1,2-diol) Aldehydes o General formula R-CHO o Have carbonyl functional group (-CHO) un o Terminal as the -CHO group can only be at the end of the main chain o Suffix is -al ▪ Prefix is aldo- aj Ji Ketones o General formula: R-CO-R’ o Similar to Aldehydes however there is no hydrogen attached to carbonyl o functional group (C=O) o Suffix is -one ▪ Prefix is keto-, oxo- Lu o Always specify location even for propan-2-one (propanone is incorrect even though it is the only possible isomer) Esters o Esterification reaction (Type of condensation reaction) ▪ Produces water ▪ Carboxylic acid + Alcohol 🡪 Ester molecule + Water molecule 31  o Ester functional group (-OCO-) o -OH group is contributed by carboxylic acid and a H atom is contributed by the alcohol o Naming ▪ Alkyl group attached to the double bonded oxygen is the main group and is given the suffix -oate ▪ Alkyl group attached to the oxygen with a single bond is named as an alkyl chain prefix un Amines aj o Derivatives of ammonia (NH3) where hydrogen atoms have been replaced with carbon chains o Suffix used is -amine Ji o Lu ▪ Prefix is amino ▪ Primary amine: One alkyl chain attached Determine position of NH2 group on carbon chain and give lowest locant 32  ▪ Secondary amine: Two alkyl chains attached Determine longest alkyl group attached being the main chain Other alkyl group provide N- prefix ▪ Tertiary amine: Three alkyl chains attached un Determine longest alkyl group and give other alkyl groups N- prefix in alphabetical order aj Ji Amides o Nitrogen atom attached to carbonyl group (C=O) o o Terminal as carbonyl group and nitrogen form 3 bonds o Suffix used is -amide Lu o Like amines, if a hydrogen is substituted for a carbon chain, these substituents are given N- as the locant 33  Priority list o Name based on principal functional group o Compounds with multiple functional groups must take the naming of the functional group with the highest priority o Highest priority functional group is given lowest locant un Justification in exams aj o Longest carbon chain o Alphabetical order arrangement of functional groups o Priority of most dominant functional group Ji o Number from side closest to principle functional group 4.1.2 explore and distinguish the different types of structural isomers, including saturated and unsaturated hydrocarbons, including: (ACSCH035) o – chain isomers – position isomers Lu – functional group isomers Structural isomerism Molecules that contain the same number and type of atoms but are arranged differently o Same molecular formula but different structural formula Can have different chemical and physical properties Chain isomers o Different arrangements of a molecule’s carbon skeleton because of branching o Can contain different alkyl groups branched to the main carbon chain o Similar chemical properties due to presence of same functional group o Different physical properties due to degree of branching 34  ▪ Greater degree of branching 🡪 Less surface area/inefficient un packing🡪Spherical shape🡪Weaker intermolecular (dispersion) forces 🡪 lower BP aj Chain isomers of hexane Ji o Lu Positional isomers o Same functional group o Different positions of functional groups along carbon chain ▪ Different positions of double or triple bonds on the carbon chain also form positional isomers 35  o Similar chemical properties (same functional group) however have different physical properties un o Isomers can be chain and positional at the same time ▪ aj Above: 2-methylpropan-1-ol and 2-methylpropan-2-ol are chain and positional isomers of butanol Functional group isomers Ji o Different functional groups o Different chemical and physical properties o Lu IQ2: How can hydrocarbons be classed based on their structure and reactivity? 7.2.1 construct models, identify the functional group, and write structural and molecular formulae for homologous series of organic chemical compounds, up to C8 (ACSCH035) ICT o - alkanes o - alkenes o - alkynes 36 7.2.2 conduct an investigation to compare the properties of organic chemical compounds within a homologous series, and explain these differences in terms of bonding (ACSCH035) Homologous series A group of molecules with similar structures formulae and both chemical and physical properties o Alkanes, alkenes, alkynes form their own homologous series Boiling point o Measure of the energy required to overcome intermolecular forces between molecules with phase change of liquid 🡪 gas o Stronger IMF’s 🡪 more energy to overcome 🡪 Higher boiling point un Solubility o Ability for molecules to form strong adhesive forces with the solvent and break cohesive solute-solute and solvent-solvent forces o Similar solute-solvent intermolecular forces favour dissolution (like dissolves like) o Primary amide o Carboxylic acid aj General trend of decreasing BP, MP Ji o Alcohol o Ketone o Aldehyde o Amine o Haloalkane o o Alkyne o Alkane o Alkene Lu 7.2.3 analyse the shape of molecules formed between carbon atoms when a single, double or triple bond is formed between them Shape of molecules Carbon atoms can bond to form o Single bonds o Double bonds o Triple bonds Different types of bonds can affect the geometric arrangement of the molecule 37  Ground state of carbon is 1s22s22p2 When energy is applied, electrons move to higher energy orbitals referred to as the excited state un o 1s22s12p3 Sp3 hybridization aj o S and p orbitals overlap and produce sp hybrid orbitals Ji o 4 sp hybrid orbitals form in sp3 hybridization, 3 from the original p orbitals and the other from the s orbital o All 4 orbitals are equivalent and gives carbon a valency of 4 when bonding o Gives carbon bonds a tetrahedral structure Sp hybridization 2 o o Only 3 orbitals (an s and two p’s) hybridize, the remaining p orbital is unchanged o Forms a planar structure Lu o Sp hybrid orbitals in carbon overlap to form C-C single bonds (sigma bond) o The remaining p orbitals also overlap to form the C=C double bond (Pi bond) o Pi bonds present in double and triple bonds are what restrict the free rotation of the bond 38 7.2.4 explain the properties within and between the homologous series of alkanes with reference to the intermolecular and intramolecular bonding present Physical properties of alkanes o “Like dissolves like” o Polar substances are more likely to dissolve in polar solvents o Non-polar substances are more likely to dissolve in non-polar solvents o Solubility o Alkanes are non-polar and thus are only soluble in non-polar solvents o In alkanes, electrons in covalent bonds are equally shared resulting in no net dipole moment, thus it is only able to form dispersion forces with other un non-polar solvents o Not soluble in water o During the dissolution of alkanes, more energy is required to break strong hydrogen bonds between water molecules than the energy released from the formation of dispersion forces between the alkane and water molecules. This o Boiling point aj results in a non-favourable endothermic chemical process o Generally low as only weak dispersion forces can form between alkane molecules Ji o More carbon atoms 🡪 Higher BP o Alkanes with a higher degree of branching will have lower BP as they have less surface area for interaction and greater distance between molecules o Melting point o o Similarly low o More carbon atoms 🡪 Higher MP o Alkanes with even numbers of carbon atoms have higher melting points than Lu alkanes with odd numbers of carbon atoms, this is due to increased symmetry in even alkanes allowing it to pack more efficiently o Branching can however have the opposite effect on alkane’s melting points, more branching decreases the surface area of alkanes and allows it to pack together in a more compact and efficient way increasing the strength of the dispersion forces Physical properties of alkenes and alkynes Boiling point o Relatively low due to non-polar nature, only able to form dispersion forces o Double and triple bonds provide more area for interaction, increasing the strength of dispersion forces 39  o Alkynes have higher BP than alkanes however alkenes have lower BP than alkanes due to its decreased molecular mass Properties of haloalkanes Able to form dipole-dipole forces o Halogens with higher electronegativity will be able to form stronger dipole-dipole forces Greater molecular mass Generally, have higher BP and MP Molecular mass and electronegativity of halogen both play a role in determining the BP and MP of a haloalkane un 7.2.5 describe the procedures required to safely handle and dispose of organic substances Handling and disposure incorrectly aj Many organic substances are potentially hazardous when handled or disposed of Always consult the Materials Safety Data Sheet (MSDS) before using any compound Hazards Ji o Volatile ▪ Forms a vapour at room temperature o Flammable ▪ Low flashpoint, can ignite at room temperature o o Corrosive or caustic ▪ Organic acids and bases Lu Exposure methods o Inhalation through lungs o Absorption through skin o Ingestion Safe handling o Wearing PPE (personal protective equipment) ▪ Safety glasses, gloves, lab coat o Fume cupboard ▪ When handling volatile substances, so harmful gases don’t escape o Have emergency eye showers nearby Safe disposal 40  o Not to be flushed down sink as they can contaminate waterways potentially harming the environment and human health o Waste is segregated into categories to be treated and disposed of differently ▪ Acidic and basic waste Neutralise with NaHCO3 and dispose in waste container ▪ Heavy metal Place in heavy metal waste container to be treated by chemical waste disposal companies ▪ Halogenated Place in halogenated waste container un ▪ Non-halogenated Place in waste container IQ3: What are the products of reactions of hydrocarbons and how do they react? 7.3.1 aj investigate, write equations and construct models to represent the reactions of unsaturated hydrocarbons when added to a range of chemicals, including but not limited to: Ji – hydrogen (H2) – halogens (X2) – hydrogen halides (HX) – water (H2O) (ACSCH136) o Alkenes and Alkynes reactivity Lu Generally, more reactive than alkanes due to higher electron density available in pi bonds Addition reactions o Small molecule adds across a double or triple bond of an unsaturated hydrocarbon molecule o Double or triple bond becomes a single bond or triple bond becomes a double bond o Include hydrogenation, hydrohalogenation, Halogenation, Hydration 41  o Catalysts may be required and need to be stated below or above the arrow Hydrogenation o Hydrogen gas reacts with an unsaturated hydrocarbon in the presence of a catalyst o Alkenes can be fully hydrogenated using a Ni, Pt or Pd/C catalyst o Alkynes can be fully hydrogenated using a Pt or Pd/C catalyst or partially hydrogenated into an alkene using Lindlar’s catalyst (pd/c poisoned with Pb) un o Reaction mechanism aj Ji Halogenation o Halogen reacts with an unsaturated hydrocarbon o Does not require a catalyst o Halogen adds across the double or triple bond o Lu Markovnikov’s rule o When H-X adds across the double bond of an alkene, the H atom is more likely to become attached to the C atom with the larger number of H atoms already attached o Due to the rule of stability of carbocations 42  Hydrohalogenation o Addition reaction where a hydrogen halide (HCl, HBr, HI) reacts with an unsaturated hydrocarbon o Does not require catalyst or UV light un Hydration o Water is added to an unsaturated hydrocarbon to produce an alcohol o Only will react when heated in the presence of a suitable acid catalyst such as dilute sulfuric acid aj Ji 7.3.2 Investigate, write equations and construct models to represent the reactions of saturated hydrocarbons when substituted with halogens o Alkanes reactivity Lu Combustion o Burn in air to form carbon dioxide and water Substitution o An atom in an alkane molecule is replaced by another or a group of atoms o Alkanes undergo substitution reactions with halogens such as chlorine or bromine o UV light is used to produce reactive free radicals 43  o Process (Green = chlorine, black = carbon, white = hydrogen) un aj IQ4: How can alcohols be produced and what are their properties? Ji 7.4.1 investigate the structural formulae, properties and functional group including: – primary – secondary – tertiary alcohols 7.4.2 explain the properties within and between the homologous series of alcohols with reference to the o intermolecular and intramolecular bonding present Lu Properties of alcohols Polar hydroxyl head and non-polar alkyl chain Hydroxyl functional group (-OH) o Highly electronegative O atom bonded to H atom o Polar covalent bond 44  ▪ Hydrogen bonding ▪ Dipole-dipole bonding C-O bond can also form dipole-dipole forces Boiling point o Higher boiling points than their corresponding alkanes ▪ Can form hydrogen bonding and dipole dipole forces ▪ Always has higher MM than corresponding alkanes due to the OH group o As the chain length increases, the difference in BP between alcohols and un alkanes decrease ▪ Greater proportion of the alcohol molecule is non-polar 🡪 dispersion forces increase such that hydrogen bonding has minimal effect in contrast o BP of Primary alcohols > Secondary > Tertiary ▪ ▪ aj Primary alcohols have the most accessible -OH group Crowding of alkyl groups around -OH group in secondary and tertiary alcohols hinders the ability for it to form hydrogen bonding Ji Solubility o Small (first 4) alcohol molecules are readily soluble in water ▪ Due to polar head, alcohols can form hydrogen bonding with water, favouring dissolution o o As chain length increases, alcohols become increasingly insoluble in water ▪ Longer alkyl chain 🡪 Interacts with other alcohol molecules via strong Lu dispersion forces 🡪 Requires more energy to break 🡪 Unfavourable for dissolution ▪ Longer alkyl chain 🡪 greater proportion of molecule becomes non-polar🡪 can only form dispersion forces with water molecules 🡪 opposes tendency to dissolve o Solubility of Tertiary alcohols > Secondary > Primary ▪ Primary alcohols have strongest H-bonding and dispersion forces (least branching) between molecules 🡪 requires most energy to break 🡪 Unfavourable for dissolution 45 7.4.3 conduct a practical investigation to measure and reliably compare the enthalpy of combustion for a range of alcohols Combustion of alcohols Molar heat of combustion o Energy released when 1 mole of an alcohol undergoes combustion Heat of combustion is a positive value, enthalpy of combustion is a negative value Assumptions ▪ All heat is transferred to water and not radiated or conducted to surroundings ▪ Complete combustion occurs un ▪ During dissolution, the heat capacity of the solution is same as water (University level) Problems affecting enthalpy of combustion ▪ Not stirring water such that water at base is warmer ▪ Thermometer is resting at base instead of being in the centre ▪ Heat radiated to surroundings aj ▪ Flame is too close to apparatus resulting in lack of O2 leading to incomplete combustion ▪ Sooty base leading to insufficient heat transfer ▪ Impurities in the spirit burner Ji 7.4.4 write equations, state conditions and predict products to represent the reactions of alcohols, including but not limited to (ACSCH128, ACSCH136): – combustion o – dehydration – substitution with HX Lu – oxidation Reactions of alcohols Combustion 1. Alcohol + O2(g) 🡪 CO2(g) + H2O(l) Dehydration 1. Alcohol 🡪 Alkene + H2O 2. Reaction conditions ▪ Concentrated H2SO4 or Al2O3 catalyst ▪ Heat 46  3. Dehydration of tertiary alcohols can occur at room temperature 4. Zaitsev’s rule ▪ Dehydration of an alcohol can form two products (major and minor) ▪ Product with the least hydrogen atoms on double bonded carbons will un aj Ji be the major product (more stable) Substitution with HX 1. Alcohol + HX 🡪 Haloalkane + H2O o ▪ HX is a hydrogen halide (e.g, HF, HCl, HBr) 2. Reaction conditions Lu ▪ ZnCl2 for HCl substitution ▪ Heat 3. Alcohol reactivity order: Tertiary > Secondary > Primary > Methanol ▪ Tertiary alcohols do not require heat 4. Hydrogen halide reactivity order: HI > HBr > HCl 7.4.5 investigate the production of alcohols, including: 47  – substitution reactions of halogenated organic compounds – fermentation Substitution Substitution with metal hydroxides 1. Reaction conditions: Heat 2. Strong bases such NaOH and KOH 3. Haloalkane + Metal Hydroxide 🡪 Alcohol + salt ▪ Tertiary haloalkanes can also undergo substitution with water ▪ Tertiary haloalkane + H2O 🡪 Tertiary alcohol + HX un Fermentation Biochemical reaction where sugars are converted into alcohols under the action of yeast Reaction conditions aj Ji 1. Alcohol tolerant yeast ( zymase enzyme) 2. 37oC (body temperature) 3. Anaerobic environment (No oxygen) a. If oxygen is present, it will react to form CO2 and H2O 4. Low pH (3.7-4.6) o a. Prevent pathogenic growth Fermentation can only produce a solution with ethanol concentration of 10-20% (v/v) Lu o Many by products ( aldehydes, ketones, carboxylic acids) are produced o Fractional distillation can be used to reach high concentrations of ethanol ▪ However, water cannot be completely removed as it forms strong hydrogen bonding with ethanol 7.4.6 investigate the products of the oxidation of primary and secondary alcohols Oxidation of alcohols Primary, secondary alcohols can be oxidised under the presence of strong oxidising agents and reflux o Acidified potassium dichromate solution ( H+/ Cr2O72- ) ▪ Orange 🡪 Green 48  o Acidified potassium permanganate solution ( H+/ MnO4-) ▪ Purple 🡪 Clear Primary alcohols 🡪 Aldehyde 🡪 Carboxylic acid o Aldehyde under a strong oxidising agent and reflux can oxidise into a carboxylic acid un o Reaction can be stopped at aldehyde by distilling it immediately after it is formed since it has a lower BP than alcohols Secondary alcohols 🡪 Ketone aj Ji Tertiary alcohols will not oxidise o Lack of hydrogen atoms on the carbon atom to which the hydroxyl group is attached Testing for presence of aldehyde groups using other oxidising agents o Aldehyde oxidised to carboxylic acid o o Fehling’s solution ▪ Clear blue solution (Cu2+) reduced to opaque red precipitate of Cu+ Lu ions o Tollens solution ▪ Ag+ reduced to Ag and forms a silver mirror 7.4.7 compare and contrast fuels from organic sources to biofuels, including ethanol IQ5: What are the properties of organic acids and bases? 7.5.1 investigate the structural formulae, properties and functional group including: – primary, secondary and tertiary alcohols – aldehydes and ketones (ACSCH127) 49  – amines and amides – carboxylic acids 7.5.2 explain the properties within and between the homologous series of carboxylic acids amines and amides with reference to the intermolecular and intramolecular bonding present Aldehydes and ketones Polar carbonyl (C=O) group Higher boiling points than corresponding alkanes but lower than carboxylic acids and alcohols o Able to form dipole-dipole forces o un aj Small molecules are soluble in water o C=O bond can hydrogen bond with water, favouring dissolution ▪ However, cannot hydrogen bond with itself Ji o o Solubility decreases with chain length Lu Carboxylic acids Carboxyl group (-COOH) o Hydrogen bonding (-OH) o Dipole dipole forces (C=O) MP and BP higher than alcohols o Greater molecular weight 🡪 Greater dispersion forces o Forms a dimer ▪ C=O group hydrogen bonds with OH group 50  ▪ Dimers are very stable 🡪 Require more energy to overcome Shorter chained acids are highly soluble in water o Able to form hydrogen bonds with water, favouring dissolution un o Solubility decreases with chain length as a greater proportion of the molecule becomes non polar Amines Can form hydrogen bonding due to polar N-H bonds aj o Primary amines: Two N-H bonds capable o Secondary amines: One N-H bond capable o Tertiary: Cannot hydrogen bond Higher boiling points than alkanes but lower than alcohols Ji o O atom is more electronegative than N atom and thus OH bond is more polar than NH bond o Except tertiary amines as they can only form dispersion forces Solubility decreases with chain length o Amides Lu Contain polar N-H and C=O groups o Hydrogen bonding and dipole dipole forces o N-H hydrogen bonds to C=O bond of another molecule Primary amides have the strongest intermolecular forces of all classes of organic compounds o Hydrogen bonding network 51  Solubility decreases as chain length increases 7.5.3 investigate the production, in a school laboratory, of simple esters Properties of esters Polar C-O and C=O bonds o Dipole dipole interactions Generally have lower boiling point than both the carboxylic acid and alcohols from which they are made Lack of hydrogen bonding, long non-polar alkyl chain🡪 poor solubility in water un Production of esters Esterification is a condensation reaction with the elimination of a water molecule o Forms an equilibrium (does not go to completion) o Slow process at room temperature and does not go to completion aj Alcohol + Carboxylic acid ⇌ Ester + Water Reaction conditions ( to ensure reaction proceeds at a reasonable rate) o 1-3ml Concentrated sulfuric acid catalyst (18molL-1) ▪ Ji Also acts as a dehydrating agent by removing water and shifting the equilibrium to favour production of the ester o o Heating via reflux Lu Highly fragrant o Pentyl ethanoate: Bananas o Ethyl butanoate: Pineapple o Methyl butanoate: apple Heating o Water bath or heating mantle is used to heat reaction mixture instead of a Bunsen burner ▪ Alcohol is flammable and may combust near a naked flame 52  o Reflux is used to condense reactants and products back into the reaction mixture ▪ Prevents loss of valuable reagents and products, to maximise yield ▪ Allows reaction to be performed at much higher temperatures, increasing rate of reaction ▪ Improves safety of procedure as flammable of harmful substances can’t escape o Boiling chips ▪ Prevent reaction mixture from superheating, by providing nucleation un points for bubbles to form and dissipate, promoting a smooth boiling aj Ji o Note Lu o Stopper is not put at top as build-up of pressure increases risk of explosion o Cold water is pumped in from the bottom so it can fill entire tube before flowing out Purification of esters Ester can be separated from impurities in reaction mixture by: 1. Mixture washed with water in a separating funnel 2. After settling, aqueous layer (bottom) flows out to remove water soluble substances (short chain alcohol/carboxylic acid) 3. Sodium carbonate solution is added to neutralise any acid present 4. Flask is swirled and carbon dioxide gas is let out 5. Remove aqueous layer again 53  6. Anhydrous magnesium sulfate (drying agent) is added to remove water from the mixture 7. Distillation may be performed to further purify the ester 7.5.4 investigate the differences between an organic acid and organic base Organic acids Organic substances that exhibit acidic properties Carboxylic acids o Weak acids, hydrolyse to a small extent in water un ▪ RCOOH + H2O ⇌ RCOO- + H3O+ o Undergo typical reactions of acids (acid-metal, neutralisation) o pH < 7 Organic bases Amines are basic aj Organic substances that exhibit basic properties o Weak bases, hydrolyse to small extent in water Ji o Undergoes typical reactions of bases o pH > 7 o Electronegative Nitrogen atom attracts positively charged H+ (protons) ▪ Can accept a proton as highly electronegative N atom draws electron density o o Tertiary amines are more basic as they attract higher electron density Lu Reactions between organic acids and bases Small amines react similarly to ammonia Neutralisation 54  o Carboxylic acid + amine 🡪 ammonium salt 𝐻𝑒𝑎𝑡, 𝑑𝑒ℎ𝑦𝑑𝑟𝑎𝑡𝑖𝑜𝑛→ amide + water ▪ If amine is replaced with ammonia, a primary amide is produced ▪ If amine is replaced with a secondary amine, a tertiary amide is produced un 7.5.5 investigate the structure and action of soaps and detergents Saponification Reaction between fatty ester (long ester chain) and an alkali to form fatty acid salts (soap) and glycerol aj o Heat is required for reaction to occur A triester is always used, commonly fats or oils o Fats: saturated molecule, solid at rtp o Oils: Unsaturated molecule, liquid at rtp Ji o Lu Structure of surfactants Soaps and detergents are surfactants o Soap is produced from natural sources o Detergents are artificial products Surfactants lower the surface tension of water by interfering with the hydrogen bonding between water molecules o Surface tension is a measure of the tension of the surface of a liquid caused by attractive forces 55  o Polar liquids (e.g, water) have strong cohesive and have high surface tension Surfactants are composed of two parts o Hydrophilic polar/charged head ▪ Interacts with water molecules via ion dipole or hydrogen bonding o Hydrophobic non-polar tail un ▪ Does not interact with water Cleaning action of soaps Soap structure aj Ji o Lu Soaps are usually present as sodium or potassium salts 56  Cleaning action of soap and water on oil or grease 1. Surfactant molecule will act on surface of water and break down surface tension 2. Some surfactant molecules will dissolve in water as it weakens surface tension 3. Non-polar tails will dissolve in grease and form a monolayer 4. Surface is agitated (vigorous scrubbing), grease is lifted off the surface 5. Micelles form around grease or oil in a emulsion Micelles o Spherical arrangement of surfactant molecules with hydrophilic heads facing out and hydrophobic tails trailing to the center un Emulsification aj o Hydrophilic head of surfactant molecules form ion-dipole forces with water, Ji lifting the micelle up o Micelles repel each other and disperse throughout the solution ▪ Like charges from polar head repel each other o Prevents oil or grease to set ground on surface again o o Water can then be drained away, taking the oil with it Lu Soaps and detergents Soaps are commonly used for hygiene o Mild, non irritating o Harder to rinse out of fabrics o Produced from natural sources Detergents act as cleaning agents o Skin irritant o Easier to rinse out of fabrics o Synthetic Soaps have an anionic carboxylate head Detergents can have anionic, cationic, or non-ionic head 57 Hard and soft water Hard water is impure and contains Ca2+ and Mg2+ ions un Soft water is water that is treated to contain only Na+ ions Soap will produce a lather of bubbles in soft water o In hard water, the calcium and magnesium ions precipitate with the anionic carboxylate head, to form solid scum o Anionic detergents similarly struggle to clean in hard water as they form soluble complexes aj Ji o Module 8 Lu IQ2: How is information about the reactivity and structure of organic compounds obtained? 8.2.1 conduct qualitative investigations to test for the presence in organic molecules of the following functional groups: – carbon–carbon double bonds – hydroxyl groups – carboxylic acids (ACSCH130) 58 Carbon-Carbon double bonds Testing for presence of C=C double bonds (unsaturation) Bromine water test o Solution of Br2 dissolved in water o Has an orange-red colour which becomes clear when an alkene is added o Alkene undergoes addition reaction with bromine to form a haloalkane un ▪ C=C double bond is broken allowing Br atoms to bond to molecule Carboxylic acids Carboxylic acids are weak organic acids containing the carboxyl (-COOH) functional group aj Able to undergo neutralisation with NaHCO3(s) o Produces CO2 gas which can be confirmed through the limewater test or bubbling Ji Hydroxyl groups Presence of an alcohol can be confirmed by reacting it with a small piece of sodium metal o o Presence of hydrogen gas can be confirmed by bubbling of solution or through pop test Lu Oxidation tests can also be performed to determine whether the alcohol is primary, secondary or tertiary o Alcohols can be oxidised by solutions of acidified potassium dichromate or acidified potassium permanganate ▪ Primary alcohol 🡪 aldehyde 🡪 carboxylic acid ▪ Secondary alcohol 🡪 ketone ▪ Tertiary alcohols don’t oxidise o Acidified potassium dichromate oxidation ▪ Orange 🡪 Green o Acidified potassium permanganate oxidation ▪ Deep purple 🡪 colourless 59 8.2.2 investigate the processes used to analyse the structure of simple organic compounds addressed in the course, including but not limited to: – proton and carbon-13 NMR – mass spectrometry – infrared spectroscopy (ACSCH130) NMR Spectroscopy Nuclear magnetic resonance (NMR) o Spectroscopy technique which can be used to determine the structure of complex molecules. un It takes advantage of the magnetic properties, particularly nuclear spin of certain atomic nuclei