IB Chemistry Revision Notes Based on 2016 syllabus Core & Option D Higher Level Yifan Chen © 2017 NFLS Tutoring Club Table of Contents Topic 1: Stoichiometric relationships........................................................................................................................................... 3 Topic 2: Atomic structure............................................................................................................................................................. 7 Topic 12: Atomic structure (HL)................................................................................................................................................. 11 Topic 3: Periodicity .................................................................................................................................................................... 13 Topic 13: The periodic table --- the transition metals (HL) ........................................................................................................ 18 Topic 4: Chemical bonding and structure.................................................................................................................................. 22 Topic 14: Chemical bonding and structure (HL) ....................................................................................................................... 30 Topic 5: Energetics/ thermochemistry ....................................................................................................................................... 34 Topic 15: Energetics/thermochemistry (HL) .............................................................................................................................. 37 Topic 6: Chemical kinetics......................................................................................................................................................... 40 Topic 16: Kinetics (HL) .............................................................................................................................................................. 41 Topic 7: Equilibrium ................................................................................................................................................................... 47 Topic 17: Equilibrium (HL) ......................................................................................................................................................... 49 Topic 8: Acids and bases .......................................................................................................................................................... 51 Topic 18: Acids and bases (HL) ................................................................................................................................................ 54 Topic 9: Redox processes ......................................................................................................................................................... 58 Topic 19: Redox processes (HL) ............................................................................................................................................... 62 Topic 10: Organic chemistry...................................................................................................................................................... 65 Topic 20: Organic chemistry (HL).............................................................................................................................................. 74 Topic 11: Measurement and data processing ........................................................................................................................... 83 Topic 21: Measurement and analysis(HL) ................................................................................................................................ 87 Option D: Medicinal chemistry................................................................................................................................................... 89 Topic 1: Stoichiometric relationships 1.1 Introduction to the particulate nature of matter and chemical change U1 Atoms of different elements combine in fixed ratios to form compounds, which have different properties from their components elements U2 Mixtures contain more than one element and/or compound that are not chemically bonded together and so retain their individual properties U3 Mixtures are either homogeneous or heterogeneous. A1 Deduction of chemical equations when reactants and products are specified. A2 Application of the state symbols (s), (l), (g) and (aq) in equations. A3 Explanation of observable changes in physical properties and temperature during changes of state. State of matter Solid Liquid Gas Fixed volume, fixed shape Fixed volume, no fixed shape No fixed shape, no fixed volume Cannot be compressed Cannot be compressed Can be compressed Attractive forces between particles hold the particles in a close-packed arrangement Forces between particles are weaker than in solids Forces between particles are taken as zero Particles vibrate in a fixed position Particles move around(vibrate, rotate, translate) Particles move around freely, much faster than a liquid The way the particles move depends on the temperature. Temperature (K) = temperature (C) + 273.15 Elements and compounds • • • • • Elements: contains atoms of only one type Compounds: atoms of elements combines in a fixed ratio Mixtures contain more than one element and/or compound that are not chemically bonded together and so retain their individual properties Homogeneous mixture: has uniform composition and uniform properties (metal alloys) Heterogeneous mixture: has non-uniform composition and non-uniform properties (paints) 1.2 The mole concept U1 The mole is a fixed number of particles and refers to the amount, n, of substance U2 Masses of atoms are compared on a scale relative to 12C and are expressed as relative atomic mass (Ar) and relative formula/molecular mass (Mr). U3 Molar mass (M) has the units g mol-1. U4 The empirical formula and molecular formula of a compound give the simplest ratio and the actual number of atoms present in a molecule respectively A1 Calculation of the molar masses of atoms, ions, molecules and formula units. A2 Solution of problems involving the relationships between the number of particles, the amount of substance in moles and the mass in grams. A3 Interconversion of the percentage composition by mass and the empirical formula. A4 Determination of the molecular formula of a compound from its empirical formula and molar mass. A5 Obtaining and using experimental data for deriving empirical formulas from reactions involving mass changes. Moles A mole (mol) of a substance (atoms / molecules / ions): • • • contains as many elementary entities (particles, atoms, molecules etc.) as there are atoms in 12 grams of pure carbon-12 is the relative atomic (or molecular or formula) mass expressed in grams 6.02 × 1023elementary entities (Avogadro’s constant) 3 Molar mass The molar mass (g mol-1) is the mass in grams of one mole of a substance. Number of particles = number of moles × Avogadro’s constant Relative abundance Isotopes are atoms of the same element that have the same number of protons in the nucleus but different numbers of neutrons. Relative abundance 1 × atomic mass 1 + Relative abundance 2 × atomic mass 2 = Relative atomic mass of this substance e.g. 75% [35Cl] × 35.0 + 25% [37Cl] × 37.0 = 35.5 Relative mass Definitions Relative atomic mass (Ar) – the relative atomic mass is the average mass of an atom, taking into account the relative abundances of all the naturally occurring isotopes of the element, relative to one atom of C-12. It is a relative term so it has no units. Relative molecular mass (Mr)–the relative molecular mass is the average mass of a molecule, calculated by adding the relative atomic masses of its constituent atoms. It is a relative term so it has no units. For formula units (the single units of ionic compounds), the term relative formula mass is used. The mass of a mole of a species is the relative mass expressed in grams. The number of moles is given by: Moles (mol) = Formulas Mass (g) Molar mass (g mol−1 ) The molecular formula is the chemical formula of a substance showing the actual number of atoms of each element in a molecule. The empirical formula is the formula in which the ratio is simplified into the smallest integers. Problem solving Finding the empirical formula from percent composition: Just simplify to the smallest integer ratio (round with 0.1 difference O1.9– O2) Finding the empirical formula from percent mass: Divide the percentages by the relative atomic masses and then simplify to the smallest integer ratio Finding the molecular formula given the empirical formula and relative molecular mass: 1. Use the empirical formula to find the empirical mass 2. Divide the molecular mass by the empirical mass, and round to an integer, n 3. Multiply the empirical formula by n Experimental Analysis • • Qualitative analysis: determine which elements are in the compound, verify the purity of the compound Quantitative analysis: relative mass of elements, work out exact composition 1.3 Reacting masses and volumes U1 Calculate theoretical yields from chemical equations. U2 The experimental yield can be different from the theoretical yield. U3 Avogadro’s law enables the mole ratio of reacting gases to be determined from volumes of the gases. U4 The molar volume of an ideal gas is a constant at specified temperature and pressure. 4 U5 The molar concentration of a solution is determined by the amount of solute and the volume of solution. U6 A standard solution is one of known concentration. A1 Solution of problems relating to reacting quantities, limiting and excess reactants, theoretical, experimental and percentage yields. A2 Calculation of reacting volumes of gases using Avogadro’s law. A3 Solution of problems and analysis of graphs involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. A4 Solution of problems relating to the ideal gas equation. A5 Explanation of the deviation of real gases from ideal behaviour at low temperature and high pressure A6 Obtaining and using experimental values to calculate the molar mass of a gas from the ideal gas equation. A7 Solution of problems involving molar concentration, amount of solute and volume of solution. A8 Use of the experimental method of titration to calculate the concentration of a solution by reference to a standard solution. Theoretical yield Theoretical yields assume that the limiting reagent is 100% used up. The actual yield is called the experimental yield. Percentage yield (%) = Percentage Purity Percentage purity (%) = experimental yield × 100% theoretical yield mass of the pure compound × 100% total mass of the mixture Limiting and excess reactants Limiting reagent as a means of controlling the amount of product obtained, will be completely consumed during the experiment. Limiting reagents determines amount of products produced Problem solving If A + B → C 1. 2. 3. 4. Calculate how much of reactant B (in moles) is needed to react with the amount of reactant A provided. If the amount needed is less than that provided, B is in excess and vice versa. A is therefore the limiting reagent. The difference between the amount of B needed and the amount of B provided is equivalent to the moles of excess B. The amount of product formed if A reacts completely is the theoretical yield. Avogadro’s law Avogadro’s law states that equal volumes of all gases under identical conditions of pressure and temperature contain the same number of molecules i.e. the molar volume of gas A is the same as that of gas B if the conditions (pressure and temperature) are the same. Under STP conditions (273K and 1.013 × 105 Pa or 1 atm) the volume per mole is 22.7 dm3. (STP) Ideal gases The ideal gas equation is: -1 -1 Where R is the gas constant which has a value of 8.31 J K mol ππππ = ππππππ The ideal gas equation implies Avogadro’s law. We can also use it to derive laws for specific cases by moving the variables to the left hand side and the constants to the right hand side. Law Combined gas law Result Formula ππππ = ππ ππ ππ1 ππ1 ππ2 ππ2 = ππ1 ππ2 5 Gay-Lussac’s law Boyle’s law Charles’s law ππ = ππ ππ ππππ = ππ ππ = ππ ππ ππ1 ππ2 = ππ1 ππ2 ππ1 ππ1 = ππ2 ππ2 ππ1 ππ2 = ππ1 ππ2 Definitions Solute – the solute is the substance dissolved in a solvent in forming a solution. Solvent – the solvent is the liquid that dissolves another substance or substances to form a solution. Solution – a solution is a homogeneous mixture of a liquid (the solvent) with another substance (the solute). There is usually some interaction between the solvent and solute molecules. Concentration – concentration is the amount of solute in a known volume of solution. It can be expressed either as g dm-3 or as mol dm-3. Concentration in mol dm-3 is often represented by square brackets around the substance. ππ = ππ ππ Molarity – is the concentration of the solution in terms of mol/L Unit: mol dm-3 Standard Conditions for temperature and pressure (RTP)(STAP) 298K, 1.01× 105Pa ---- 24.0 dm3/mol STP Condition 273K, 1.01× 105Pa ---- 22.7 dm3/mol 6 Topic 2: Atomic structure 2.1 The nuclear atom U1 Atoms contain a positively charged dense nucleus composed of protons and neutrons (nucleons). U2 Negatively charged electrons occupy the space outside the nucleus. U3 The mass spectrometer is used to determine the relative atomic mass of an element from its isotopic composition. A1 Use of the nuclear symbol notation to deduce the number of protons, neutrons and electrons in atoms and ions. A2 Calculations involving non-integer relative atomic masses and abundance of isotopes from given data, including mass spectra. Atomic structure Protons and neutrons (nucleons) are found in the nucleus. Electrons are found in energy levels or ‘shells’ surrounding the nucleus. particle relative mass relative charge proton 1 +1 neutron 1 0 electron 1 1836 -1 IB definitions Mass number (A) – the mass number of an atom is the total number of protons plus neutrons in its nucleus. Atomic number (Z) – the atomic number of an atom is the number of protons in its nucleus. It is also equal to the number of electrons it contains. The atomic number defines the element and its position in the Periodic Table. Isotopes – atoms of the same element (and so have the same atomic number, Z) but have different numbers of neutrons (and so have different mass number, A). Calculations • • • Number of protons = atomic number Number of electrons = atomic number – charge e.g. O-2→ 8 – (-2) = 10 electrons Number of neutrons = mass number – atomic number Properties of isotopes Isotopes show the same chemical properties as neutrons do not participate in chemical reactions. A larger relative atomic mass means a larger density and a slower movement of atoms for a given temperature. These differences affect the melting and boiling points and can be used in separation of isotopes Uses of radioisotopes Uses of radioisotopes include: nuclear power generation, the sterilisation of surgical instruments in hospitals, crime detection, finding cracks and stresses in metals and the preservation of food, specifically: • • • • Carbon-14 is used in carbon-dating. Cobalt-60 is used in radiotherapy.(treat cancer) Iodine-131 is used as a tracer in medicine for treating and diagnosing illnesses Uranium-235 is used for nuclear fission in power plants. 7 The mass spectrometer A mass spectrometer is used to determine relative atomic masses. (Work under a vacuum) The stages of operation are: 1. Vaporisation: a vaporised sample is injected into the instrument; this allows individual atoms to be analysed 2. Ionisation: atoms are bombarded with a stream of high energy electrons, knocking off valence electrons, generating positively charged species X(g) + e- ---- X+(g) + 2e- 3. Acceleration: the positive ions are attracted by the negatively charged plates; they are accelerated (focused) by the electric field 4. Deflection: the ions are deflected by an external magnetic field placed at right angles to their path; the amount of deflection is proportional to the charge/mass ratio. Lighter, highest charged atoms ----- deflected most Heavier, smallest charged atoms ----- deflected least 5. Detection: ions of a particular mass/charge ratio are detected (other ions will hit the wall and get neutralised) and a signal is sent to a recorder; the strength of the signal is a measure of the number of ions Calculating the Ar Using a mass spectrum, we can find out the relative atomic mass. The relative atomic mass is equal to the sum of the relative abundances multiplied by their respective masses divided by the sum of the relative abundances (which should be 100 if you’re dealing with percentages): π΄π΄ππ = ∑(relative abundance × mass) ∑ relative abundances Problems on this subject are usually quite easy; however there is one case which is not immediately obvious: working out the abundances of isotopes given that there are only two and that you know the Ar. The trick is that if one isotope has an abundance x, then the other must have an abundance of 100 – x, thus there is only one variable so the problem is capable of being solved. Problem solving Chlorine has two isotopes – 35Cl and 37Cl and a relative atomic mass of 35.5. What are the abundances of the two isotopes? Let x represent the abundance of 35Cl. 35.5 = π₯π₯ × 35 + (100 − π₯π₯) × 37 100 35.5 × 100 = 35π₯π₯ + 3700 − 37π₯π₯ 2π₯π₯ = 150 π₯π₯ = 75% So the abundance of 35Cl is 75% and the latter is 25% abundant. 2.2 Electron configuration U1 Emission spectra are produced when photons are emitted from atoms as excited electrons return to a lower energy level. U2 The line emission spectrum of hydrogen provides evidence for the existence of electrons in discrete energy levels, which converge at higher energies. U3 The main energy level or shell is given an integer number, n, and can hold a maximum number of electrons, 2n2 U4 A more detailed model of the atom describes the division of the main energy level into s, p, d and f sub-levels of successively higher energies. U5 Sub-levels contain a fixed number of orbitals, regions of space where there is a high probability of finding an electron. U6 Each orbital has a defined energy state for a given electronic configuration and chemical environment and can hold two electrons of opposite spin. 8 A1 Description of the relationship between colour, wavelength, frequency and energy across the electromagnetic spectrum. A2 Distinction between a continuous spectrum and a line spectrum. A3 Description of the emission spectrum of the hydrogen atom, including the relationships between the lines and energy transitions to the first, second andthird energy levels. A4 Recognition of the shape of an s atomic orbital and the px, py and pz atomic orbitals. A5 Application of the Aufbau principle, Hund’s rule and the Pauli exclusion principle to write electron configurations for atoms and ions up to Z = 36. Students should be able to identify the ultraviolet, visible and infrared and variation of wavelength and frequency across the spectrum: A continuous spectrum shows an unbroken sequence of frequencies, such as thespectrum of visible light.A line spectrum is an emission spectrum that has only certain frequencies of light. It is produced by excited atoms and ions as they fall back to a lower energy level Energy of eletromagnetic radiation is inversely proportional to wavelength and proportional to frequency of the radiation. Planck’s Equation: E = h × f (h – Planck Constant = 6.634 × 10-34Js) c f = (λ —wavelength) λ 700nm 400nm Low energy High energy The Bohr model According to the Bohr model, an electron moves into a higher energy level when it absorbs energy, from the ground state to an excited state. This absorption produces the absorption lines. When the electrons fall back to the ground state they emit energy, producing the emission lines. The fact that only a few distinct colours can be observed shows there are only certain orbits in which an electron can be placed. (Dexcitation of electron releases energy and captured on the emission spectrum) βπΈπΈelectron = πΈπΈphoton = βππ Energy of a particular energy level 1 1 1 πΈπΈ = −π π π»π» ( 2 ) ππ βππ Energy of dexicitation of an electronπ₯π₯π₯π₯ = π π π»π» οΏ½ππ 2 − ππ 2 οΏ½ = βπ£π£ = ππ The emission spectrum of hydrogen ππ ππ The observed emission spectrum show above results from the energy differences between energy levels that correspond to frequencies in the visible light region (the Balmer series). These are jumps from higher energy levels to the second energy levels. Jumps to the first energy level (the ground state) produce higher frequency emissions (ultraviolet) and jumps down to the third level produce lower frequency infrared radiation. The lines converge at higher energies because the energy levels inside the atoms become closer together. Only electrons in Balmer series can be capture within the visible light range (Electrons coming back to n=2 energy level) 9 Relationships between lines in emission spectrum of hydrogen As shown in the emission spectrum, lines towards violet colour are getting closer and closer. Also there are only certain distinct lines in the spectrum. So we can conclude that energy levels inside the atom are discrete and certain, and energy level are getting closer and closer to together as they become more energetic (e.g. n=4 and n=3 is closer than n=3 and n=2) 2n2Rule Maximum number of electrons in a given energy level = 2n2 (e.g. n=2 maxi. 8 electrons; n=3 maxi. 18 electrons) Four Quantum Numbers: #1 n – principal quantum number n: Values of the energy level (1,2,3,4,5,……∞) #2 l – angular moment quantum number l: shape of the 3D space (sub-shell) where the electrons may be found (0,1,2,3,4……n-1) Energy level (n) L value Sub-shell 1 0 s 2 0,1 s,p 3 0,1,2, s,p,d #3 m – magnetic quantum number m: orientation direction of 3D spaces (define the shape of the orbital) (-l,…0,…l) Energy level (n) 1 2 3 L value 0 0,1 0,1,2 M value 0 -1,0,1 -2,-1,0,1,2 Orbitals S Px,Py,Pz dxy,dyz,dzx,dx2-y2,dz2 s orbital – sphere shape #4 s – spin quantum number s: how electrons move inside an orbital (± 12 clockwise and anticlockwise) Filling up electrons Pauli’s Exclusion Principle No more than two electrons can occupy any one orbital and if two electrons are in the same orbital they must spin in opposite directions (no two electrons can have the same n,l,m,s value) Aufbau Principle Electrons are placed into orbitals of lowest energy first (n+l -- energy) 10 4 0,1,2,3, s,p,d,f Lower n+l value orbital will be filled first Same n+l value, lower n will be filled first Hund’s Third Rule Sub-shell of more than one orbital, no pair of electrons will happen till all orbitals have 1 electron each Removing electrons Removing electrons always is from outermost electrons. For example: Fe: 1s22s22p63s23p64d23d6(4d2 is the outermost electrons) Fe2+: 1s22s22p63s23p63d6(removing 4d2) There are two exceptions: (they will then have half filled (or full filled) 3d orbitals --- more stable) 1. Chromium: [Ar]3d54s1[1s22s22p63s23p64s13d5] 2. Copper: [Ar]3d104s1 [1s22s22p63s23p64s13d10] Topic 12: Atomic structure (HL) 12.1 Electrons in atoms U1 In an emission spectrum, the limit of convergence at higher frequency corresponds to the first ionization energy. U2 Trends in first ionization energy across periods account for the existence of main energy levels and sub-levels in atoms. U3 Successive ionization energy data for an element give information that shows relations to electron configurations. A1 A2 Solving problems using πΈπΈ = βπ£π£. (v-- frequency) A3 Deduction of the group of an element from its successive ionization energy data. A4 Explanation of the trends and discontinuities in first ionization energy across a period. Calculation of the value of the first ionization energy from spectral data which gives the wavelength or frequency of the convergence limit.. First Ionisation Energy Definition: Amount of energy required to remove one mole of electrons from one mole of atom in its gaseous state. X(g) ------ X+(g) + eShielding effect Inner electrons act as shield to block the charge form nucleus. So outermost electrons don’t receive all the forces from the nucleus. • Distances have more effect than charges, that why as we go down the group, first ionisation energy gets smaller Effective charges Effective charges show the effective forces on the electrons. Larger Zeff means stronger forces which the electrons experienced. Effective charges (Zeff) are depended on numbers of protons and amount of shielding effects. More protons and weaker shielding effect will lead to large Zeff 11 first ionisation energy for the first twenty elements first ionisation energy / kJ mol -1 2500 Ne He 2000 F N 1500 H 1000 C Be 500 0 O Ca Al Na 0 P Ar S B Li Si Mg Cl 5 K 10 15 20 atomic number • • • • In a group: from top to bottom, first ionisation energy decreases (more shielding effect, electrons are further away from the nucleus – Zeff decrease) In a period: generally increase in the period (more protons but same shielding effect – Zeff increases) HOWEVER, there is a decrease from group 2 to group 3 (e.g. Boron has smaller 1stIE value than beryllium); that is because that electrons are taken from the p orbitals, which is further from nucleus than s orbitals (remember, distance always have most effect) Group 5 elements (e.g. nitrogen and phosphorus) have much higher 1st IE value because they have half-filled p-shell, half filled shell is more stable. log (ionisation energy) succesive ionisation energies of aluminium the next eight are taken from the second energy level the first three electrons are removed from the third energy level 1 2 the last two are taken from the first energy level 3 4 5 6 7 8 9 10 11 12 13 ionisation number Use E=hvto calculate first ionisation energy First ionisation energy (in J) = hv First ionisation energy (in kJ mol-1) = hv NA (Avogardro’s constant) Shortest wavelength – highest frequency and energy – always dexictate from n = ∞to n =1; limit of convergence at higher frequency corresponds to the first ionization energy (removing one electron) Example Question: Calculate the first ionisation energy in KJ mol-1, for hydrogen given that its shortest-wavelength line in the Lyman series is 91.16nm. Solution: IE = βπ£π£ = βππ = ππ 6.626×10 −34 ×2.998×10 8 91.16×10 −9 = 2.719 × 10−18 J Converted in to KJ mol-1IE = 2.719 × 10−18 J × 6.022 × 1023 = 1.312 × 106 Jmol−1 = πππππππππππππ¦π¦π¦π¦π¦π¦−ππ 12 Topic 3: Periodicity 3.1 The periodic table U1 The periodic table is arranged into four blocks associated with the four sublevels—s, p, d, and f. U2 The periodic table consists of groups (vertical columns) and periods (horizontal rows). U3 The period number (n) is the outer energy level that is occupied by electrons. U4 The number of the principal energy level and the number of the valence electrons in an atom can be deduced from its position on the periodic table U5 The periodic table shows the positions of metals, non-metals and metalloids. A1 Deduction of the electron configuration of an atom from the element’s position on the periodic table, and vice versa. The elements are arranged in the periodic table in increasing order of atomic mass from left to right. Going down one row increase the number of electron shells by one. Going one column to the right increases atomic mass by 1. Group - a group is a vertical column of elements in the Periodic Table. The atoms of the elements in the group all have the same outer shell structure but an increasing number of inner shells. Period – a period is a horizontal row of elements in the Periodic Table. Within a period, the atoms of the elements have the same number of shells but with an increasing number of electrons in the outer shell. Before calcium (Z = 20), • • group number = number of valence electrons period number = number of electron shells 3.2 Periodic trends U1 Vertical and horizontal trends in the periodic table exist for atomic radius, ionic radius, ionization energy, electron affinity and electronegativity. U2 Trends in metallic and non-metallic behaviour are due to the trends above. U3 Oxides change from basic through amphoteric to acidic across a period. A1 Prediction and explanation of the metallic and non-metallic behaviour of an element based on its position in the periodic table. A2 Discussion of the similarities and differences in the properties of elements in the same group, with reference to alkali metals (group 1) and halogens (group17). A3 Construction of equations to explain the pH changes for reactions of Na2O,MgO, P4O10, and the oxides of nitrogen and sulfur with water. IB definitions First ionisation energy – the first ionization energy is the minimum energy required to remove a mole of electrons from a mole of gaseous atoms to form a mole of univalent cations in the gaseous state. X(g) → X + (g) + e− Electronegativity – ability of an atom to attract electrons towards itself (man-made scale from 4.0 to 0) 13 Electron Affinity – energy released when one mole of electrons is added to one mole of gas atoms Atomic Radius The atomic radius is defined as the distance from the nucleus to the outermost electron or, in practice, half the distance between two bonded nuclei. • • In a period: atomic size decreasing, as larger Zeff (same amount of shielding but more proton numbers) In a group: from top to bottom, the atom size increases (adding more energy level, so the amount of shielding increases, so Zeff decreases) Ionic Radius • Positive ion’ radius is smaller than its atom: losing an electron means less outermost electrons, larger forces experienced by each outermost electrons (effective nuclear charges increase), so the radius gets smaller. • Negative ion’ radius is larger than its atom: gaining an electron means more outermost electrons, smaller forces experienced by each outermost electrons (effective nuclear charges decrease), so the radius gets larger Electron Affinity Value of electron affinity is negative because energy is released (-ve), not represents magnitude • • • • • • From period 2 to period 3, electron affinity increases: putting electrons in n=2 shell (maximum hold 8 electrons), electrons will face relatively larger repulsion, so it needs more energy to hold the electrons in place, that means less energy will be released. However, in the case of n=3 shell (maximum hold 18 electrons), there is more space for electrons to move, so it takes relatively less energy to hold the electrons in place, that means more energy will be released. (inter electron repulsion) Group 17 have largest electron affinity value: outer energy level will be fulfilled, so F- will be more stable than F, releasing a lot of energy Electron affinity is opposite from first ionisation energy In a period: Electron affinity generally increases (more protons but same shielding, larger Zeff, larger attraction to electrons) However, for example nitrogen has no EA value, because nitrogen itself is stable(half-filled orbitals); Beryllium also doesn’t have EA value, because adding an electron in will create a new 2p orbitals for beryllium, the distance from the nucleus and more electron shielding makes it very energy-consuming. Lithium EA value is larger than boron, that is because electron is added into 2s sub-shell, which is closer to the nucleus than 2p sub-shell. Remember, distance matters most; so the attraction form the nucleus to electron is far more larger In a group, electron affinity generally decreases as Zeff decreases except the case discussed above from period 2 to period 3 which has more space and less inter electron repulsion. Group 18 elements don’t have EA value – they are already stable Electronegativity Ability of an atom to attract electrons towards itself • • • • Electronegativity is all about effective nuclear charge, higher Zeff higher electronegativity In a period: electronegativity increases (more protons but same shielding, Zeff increases) In a group: from top to bottom, electronegativity decreases (electrons are future form the nucleus and more shielding effect, Zeff decrease) Fluorine has highest electronegativity value 4.0 --- least shielding, closer to nucleus 14 Trends across period 3 Trend Explanation Across the period the number of protons in the nucleus increases (so there is a greater attractive force) but there is the same amount of shielding effect. The effective nuclear charge therefore increases across the period, pulling the energy levels closer. This means that atomic nuclei decrease across the period. The same trend is observed with ionic radii except because the cations have 1 shell fewer they have a smaller radius. atomic and ionic radii P31 S2Cl- 2 atomic radius 3 4 Na+ 5 6 7 Mg2+ The graph shows the Si4+ ion (radius: 42), the Si4- ion also exists with a radius of 271. Al3+ Si4+ 1 2 3 4 5 group number 6 8 Ar first ionisation energies first ionisation energy 7 Cl There are, however, two decreases: P • Si Mg As mentioned previously, the effective nuclear charge increases across the period. Because of this, the first ionisation energies generally increase across the period. S • Na Al From Mg to Al – electrons in p orbitals are of higher energy and further away from the nucleus, thus easier to remove From P to S – an electron in a doubly occupied orbital is repelled by its partner and so is easier to remove than an electron in a half-filled orbital Mg: [Ne] 3s2 1 2 3 4 5 6 7 8 group number Al:[Ne] 3s2 3p1 P: [Ne] 3s2 3px13py1 3pz1 S: [Ne] 3s2 3px23py13pz1 electronegativity electronegativities 1 The electronegativity increases across the period as the size of the atoms decreases Cl S Mg Al Si P Na 2 3 4 5 6 7 8 group number 15 Group 1 – the alkali metals • • • • • they are very reactive metals they form ionic compounds with non-metals the elements further down the group are more reactive that the higher ones since the valence electron is further from nucleus and has more shielding they react with water to form hydrogen and metal hydroxide alkaline solution they react with halogens to form ionic salts Group 7 - halogens • • • • they are very reactive non-metals reactivity decreases down group they form ionic compounds with metals and covalent compounds with other non-metals higher halogens can displace lower ones from compounds Reactions with water The reactions specified by the syllabus are: Na2 O(s) + H2 O(l) → 2NaOH(aq) MgO(s) + H2 O(l) → 2Mg(OH)2 (aq) P4 O10 (s) + 6H2 O(l) → 4H3 PO4 (aq) SO3 (l) + H2 O(l) → H2 SO4 (aq) Acidity of oxides across period 3 Formula of oxide Na2O (s) Acid-base character MgO (s) basic • • • SiO2 (s) SO3 (l) SO2 (g) P4O10 (s) P4O6 (s) amphoteric Cl2O7 (l) Cl2O (g) acidic giant ionic Structure • Al2O3 (s) giant covalent molecular covalent metal oxides are solid because they have strong ionic bonds and therefore form an ionic lattice sulphur dioxide is a macromolecular covalent structure like diamond the remaining three are molecular covalent ionic compounds conduct best because of the mobile charges in molten or aqueous phase Period 3 chlorides Formula of oxide NaCl solid Physical state Oxidation number MgCl2 +1 +2 Electrical conductivity in molten state high Structure giant ionic • • • AlCl3 / Al2Cl6 SiCl4 PCl5 / PCl3 S2Cl2 Cl2 solid / gas liquid solid / liquid liquid gas +3 +4 +5 / +3 +1 0 poor none molecular covalent the more polar the substance is the better it conducts; ionic compounds conduct well the ionic compounds have the strongest forces of attraction and are therefore solid the non-metal chlorides have dipole-dipole or van der Waals’ forces (depending on whether the dipoles cancel) 16 • Cl2 is non-polar so the IMFs are weak, therefore it is a gas Reactions of period 3 chlorides with water Chlorine reacts in a reversible disproportionation (both reduced and oxidised) reaction, producing hydrochloric acid and chloric(I) acid: Cl2 (aq) + H2 O(l) β HCl(aq) + HOCl(aq) The solution is acidic. This is why chlorine turns damp litmus paper red. Ionic chlorides split into ions in solution, which get surrounded by water molecules (partially charged oxygen to cation and partially charged hydrogen to anion). They become hydrated. NaCl(s) → Na+(aq) + Cl−(aq) MgCl2 (s) → Mg 2+ (aq) + 2Cl−(aq) The pH of aqueous magnesium chloride is slightly less than 7 because the Mg2+ ion is polarising. Aluminium chloride dissociates into ions when added to water: AlCl3 (s) → Al3+(aq) + 3Cl−(aq) The aluminium ion has a high charge density due to it having a 3+ charge and a small ionic radius. The ion attracts water molecules which form dative(or coordinate) bonds with the ion to form an octahedral complex ion, [Al(H2O)6]3+ The hydrated ion is acidic as the Al3+ ion attracts the electrons of the OH bond of the surrounding water molecules, and releases the H+ ion to form an acidic solution. [Al(H2 O)6 ]3+ (aq) β [Al(H2 O)5 OH]2+ (aq) + H +(aq) This process can go on: [Al(H2 O)5 OH]2+(aq) β [Al(H2 O)4 (OH)2 ]+ (aq) + H +(aq) Silicon and phosphorus react to form hydrochloric acid: SiCl4 (l) + 2H2 O(l) → SiO2 (s) + 4HCl(aq) PCl3 (l) + 3H2 O(l) → H3 PO3 (aq) + 3HCl(aq) PCl5 (s) + 4H2 O(l) → H3 PO4 (aq) + 5HCl(aq) In summary: Formula of oxide NaCl MgCl2 Acid-base character neutral weakly acidic AlCl3 / Al2Cl6 SiCl4 PCl5 / PCl3 acidic 17 S2Cl2 Cl2 Topic 13: The periodic table --- the transition metals (HL) 13.1 First-row d-block elements U1 Transition elements have variable oxidation states, form complex ions with ligands, have coloured compounds, and display catalytic and magnetic properties. U2 Zn is not considered to be a transition element as it does not form ions within complete d-orbitals. U3 Transition elements show an oxidation state of +2 when the s-electrons are removed. A1 Explanation of the ability of transition metals to form variable oxidation states from successive ionization energies. A2 Explanation of the nature of the coordinate bond within a complex ion. A3 Deduction of the total charge given the formula of the ion and ligands present. A4 Explanation of the magnetic properties in transition metals in terms of unpaired electrons. Properties of transition elements • • • • • • • variable oxidation states formation of complex ions coloured complexes catalytic behaviour the normal metallic properties (good conductors, malleable, ductile) formation of alloys high melting point and boiling point Scandium and zinc A transition metal is defined as an element that possesses an incomplete, or partially filled, d sub-level in one or more of its oxidation states. Scandium is not a typical transition metal as its common ion Sc3+ has no d electrons. Zinc is not a transition metal as it contains a full d sub-level in all its oxidation states. Explanation of variable oxidation number of transition elements The fact that transition metals can have various oxidation numbers has to do with ionisation energies. The diagram below shows the difference between calcium and titanium’s ionisation energies. ionisation energy/ kJ mol-1 12000 10000 8000 6000 Calcium 4000 Titanium 2000 0 0 1 2 3 4 5 6 ionisation number The increase in ionisation energies for titanium is more gradual as the 3d and 4s orbitals are close in energy level. Titanium can exist in +2, +3 and +4 oxidation states, but not +5 because the jump in ionisation energies is too large. Students should know that all transition elements can show an oxidation number of +2. In addition, they should be familiar with the oxidation numbers of the following: Cr (+3, +6), Mn (+4, +7), Fe (+3) and Cu (+1). 18 Ligand IB definitions Ligand – an ion or molecule that donates a pair of electrons to a metal atom or ion in forming a coordination complex. Ligands are Lewis bases e.g. H2O, CN-, Cl- and NH3 Monodentate ligand: it can form one co-ordinate bond in one molecule, such as NH3, H2O Bidentate ligand: it can form two co-ordinate bonds in one molecule, such as ethylenediamine Polydentate ligand: it can form more than two co-ordinate bonds in one molecule, such as EDTA It has potential to donate 6 of its electron pair and become EDTA4- and a hexadentate ligand. Chelate effect Complexes with polydentate ligand (e.g. EDTA) are most stable and favorable to be formed. [Cu(H2O)6]2+ + EDTA4-→ [Cu(EDTA)]2- + 6H2O Because polydentate ligands have more binding sites, so it requires fewer moles of polydentate ligands to satisfy the “need” of the metal ions. As shown in the chemical equation, the reaction is moved from order to disorder (more moles of products than reactants), leading to a highly positive value of entropy. According to the equation π₯π₯πΊπΊ β = π₯π₯π»π»β − πππ₯π₯ππ β , highly positive entropy will give an highly negative value of Gibb’s free energy, so the reaction is spontaneous, meaning that the formation of [Cu(EDTA)]2- is favourable. Complexes Transition metal ions have relatively high charges and small sizes allowing them to attract ligands’ lone pairs of electrons. The number of dative bonds from ligands to the central ion is called the coordination number. Ligands can be exchanged. Complexes: co-ordinate bonds between a metal ions and negative ions or substances with a lone pair. Metal ion is Lewis acid – electron pair acceptor Ligand is Lewis base – electron pair donor Coordination Number Numbers of co-ordinate bonds around a metal ion in a complexes = coordination number Coordination number = 2: The shape of the complexes is linear 19 Coordination number = 4: The shape of the complexes is tetrahedral or square planar Coordination number = 6: The shape of the complexes is octahedral When water is used as a ligand, it always forms a coordination number of 6 – octahedral shape. 13.2 Coloured complexes U1 The d sub-level splits into two sets of orbitals of different energy in a complexion. U2 Complexes of d-block elements are coloured, as light is absorbed when an electron is excited between the d-orbitals. U3 The colour absorbed is complementary to the colour observed. A1 Explanation of the effect of the identity of the metal ion, the oxidation number of the metal and the identity of the ligand on the colour of transition metal ion complexes. A2 Explanation of the effect of different ligands on the splitting of the d-orbitals in transition metal complexes and colour observed using the spectrochemical series. Splitting d-orbitals When ligands approach the metal, in octahedral shape, they move in along the axis, so the orbital on the axis face more repulsion by electric field created by the incoming ligands. Therefore, they have more energy. Similarly, orbitals between the axis are influenced less by the electric field created by incoming ligands, Therefore, they have less energy. 20 The d-orbitals are split into two energy levels; the difference between two energy levels is denoted as Δππ Colour formation Colour of the light absorbed by the substance is complementary to the colour observed by us, showing on the colour wheel. For example, Fe2+ is green in colour, which means it must absorb red colour from the visible light. When visible light passes through the substance, one electron will be excited to the higher energy d-orbital. When it falls back, it gives out energy which falls in the visible light range due to the small gap between two d-orbital energy levels. This dexcitation will show colour. The difference between the d-orbital energy levels (Δππ ) will affect the energy released by dexication, thereby showing different colours. Factors affecting d-orbital splitting • The nature of the ligand: • Stronger electric field (CO), more splitting, more energy will be absorbed, longer wavelength colour will be shown. • Weaker electric field (I-), less splitting, less energy will be absorbed, shorter wavelength colour will be shown. • • • Electron configuration of metal ions (same elements) • Higher charge of the ions, more splitting, more energy will be absorbed, longer wavelength colour will be shown. For example, Fe2+ has less charge, less splitting, red colour (less energy) is absorbed, so show green colour (shorter wavelength); Fe3+ has more charge, more splitting, green colour (more energy) is absorbed, so show red colour (longer wavelength). Nature of metal ions • More d-orbital electrons, more splitting, more energy will be absorbed, longer wavelength colour will be shown. For example, Fe2+ has less d electrons, less splitting, red colour (less energy) is absorbed, so show green colour (shorter wavelength); Cu2+ has more d electrons, more splitting, orange colour (more energy) is absorbed. Geometry of complexes (tetrahedral/octahedral) Magnetic properties of metals The magnetic properties are about unpaired d-orbital electrons. • • • Diamagnetic: have all electrons paired up, so has no effects under magnetic field. Paramagnetic: have unpaired electrons; slightly attracted by a magnetic field and the material does not retain the magnetic properties when the external field is removed. Ferromagnetic: have unpaired electrons; exhibit a strong attraction to magnetic fields and are able to retain their magnetic properties after the external field has been removed. 21 Topic 4: Chemical bonding and structure 4.1 Ionic bonding and structure U1 Positive ions (cations) form by metals losing valence electrons. U2 Negative ions (anions) form by non-metals gaining electrons. U3 The number of electrons lost or gained is determined by the electron configuration of the atom. U4 The ionic bond is due to electrostatic attraction between oppositely charged ions. U5 Under normal conditions, ionic compounds are usually solids with lattice structures. A1 Deduction of the formula and name of an ionic compound from its component ions, including polyatomic ions. A2 Explanation of the physical properties of ionic compounds (volatility, electrical conductivity and solubility) in terms of their structure. Ionic bonding Ionic bonding occurs between elements that have a large difference in electronegativity (typically larger than 1.8). The atom with the low electronegativity (the metal) loses an electron and becomes a cation. The atom with the high electronegativity (the non-metal) gains the electron and becomes an anion. As a result there is an electrostatic attraction(not a real bond)between the two ions. In NaCl, for example, chlorine already has noble gas configuration (it is in the form of diatomic molecules), but the energy given out when the ionic lattice is formed is sufficient to break the bond in the chlorine molecule to give atoms. Each sodium atom gives an electron to a chlorine atom. In the lattice each cation is surrounded by 6 anions and vice versa. Ionic bonding only happens when a electron transferred from a metal to a non-metal Properties of ionic compounds • • • • • Dissolve in water (ions are free to move, so it conducts electricity) Molten ionic compound conducts electricity (ions are free to move) In solid state, ionic compound doesn't conduct electricity because ions are not free High melting point (in solid state ionic compound has giant lattice structure) Electronegativity difference ranges from 4 to 1.8 4.2 Covalent bonding U1 Describe the covalent bond as the electrostatic attraction between a pair of electrons and positively charged nuclei. U2 Describe how the covalent bond is formed as a result of electron sharing. U3 Deduce the Lewis (electron dot) structures of molecules and ions for up to four electron pairs on each atom. U4 State and explain the relationship between the number of bonds, bond length and bond strength. A1 Predict whether a compound of two elements would be covalent from the position of the elements in the periodic table or from their electronegativity values. Covalent bonding In a covalent bond, the electrons are shared and attracted electrostatically by both positive nuclei resulting in a directional bond between the two atoms to form a molecule. Generally the difference in electronegativity has to be less than 1.8 for covalent bonding. It’s the overlap of orbitals and the attraction between the nucleus and shared paired of electrons. 22 Lewis structures In Lewis structures, lone pairs of electrons can be depicted by two crosses, two dots or a line. A bond can be shown as a line (or double line for double bonds). Properties of covalent compounds Low melting point (smaller structure, except SiO2) Doesn’t form giant structure (except SiO2) Insulation of electricity (no free electrons) Form simple molecules • • • • Polar bonds When the atoms are different the more electronegative atom exerts a greater attraction for the electron pair, becoming more electron rich resulting in a polar bond. Bigger difference in electronegativities means a more polar bond. Polar bond ranges from 1.8 – 0.7 electronegativity difference. Ionic bonds 4 Non-polar covalent Polar covalent 1.8 0.7 0 Bond strength and bond length • Triple is the stronger than a double bond but short than a double bond. Shorter bonds are stronger. • Bond length is a measure of the distance between two nuclei. Bond strength is described in terms of bond enthalpy. Multiples bonds have a greater number of shared electrons and so have a stronger force of electrostatic attraction between the bonded nuclei. Thus there is a greater pulling power on the nuclei, bringing them closer together, resulting in bonds that are shorter and stronger than single bonds. 4.3 Covalent structures U1 Lewis (electron dot) structures show all the valence electrons in a covalently bonded species. U2 The “octet rule” refers to the tendency of atoms to gain a valence shell with a total of 8 electrons. U3 Some atoms, like Be and B, might form stable compounds with incomplete octets of electrons. U4 Resonance structures occur when there is more than one possible position for a double bond in a molecule. U5 Shapes of species are determined by the repulsion of electron pairs according to VSEPR theory. U6 Carbon and silicon form giant covalent/network covalent structures. A1 Deduction of Lewis (electron dot) structure of molecules and ions showing all valence electrons for up to four electron pairs on each atom. A2 The use of VSEPR theory to predict the electron domain geometry and the molecular geometry for species with two, three and four electron domains. A3 Prediction of bond angles from molecular geometry and presence of non-bonding pairs of electrons. A4 Prediction of molecular polarity from bond polarity and molecular geometry. A5 Deduction of resonance structures, examples include but are not limited toC6H6, CO32- and O3. A6 Explanation of the properties of giant covalent compounds in terms of their structures. Extended Octet The exceptions to the octet rule are: • small atoms like B and Be form stable molecules with fewer than eight electrons (an incomplete octet) 23 • atoms of elements in the third period and below may expand their octet by using d orbitals in their valence shell. This arrangement is possible because the d orbitals available in the valence shell of these atoms have energy values relatively close to those of the p orbitals. • In period 3, after phosphorus, we can see extended octet happening, due to an empty d-orbital where electrons can be excited to. Resonance Delocalisation is a characteristic of π bonds where there is more than one possible position for a double bond within a molecule (when there are resonance structures). Properties caused by delocalisation • • • intermediate bond lengths and strengths (between single and double) greater stability: delocalisation spreads electrons as far from each other as possible making the molecule more stable (electrons are not available for chemical reaction) electrical conductivity in metals and graphite Co-ordinate bond (Dative bond) • • • • Type of bonding involves giving out lone pairs of electrons One atom share its 2 electrons together to another atom, which electron contribution is 0 Arrows are used to shown the co-ordinate bonds (dative bonds) Example: carbon monoxide with triple bond (two covalent and one dative bond) VSEPR theory VSEPR (valence shell electron pair repulsion) theory states that pairs of electrons arrange themselves so that they are as far apart from each other as possible. Each molecule has bond pairs of lone pairs of electron, where they can be found is called electron dense region or charged centres. Charged centres are indentified only around centre atoms. Depending on how many charged centre, we can figure out the geometry of electron pair distribution; therefore, we can then find the actual shape of molecules. Bond pairs have more space to move due to overlap of orbitals so the repulsion between them is relatively low. However, lone pairs have less space because they are restricted to the certain orbital, so the repulsion between them is larger. Due to this, whenever lone pair presents, it takes more space so will push bond pairs closer. Therefore, for every one lone pair presents, 2.5°will be reduced from the main angle. Lone pair electrons don’t account for the shape of the molecules. 24 Number of charge centres Geometrical arrangement of charge centres Number of bonding pairs Number of nonbonding pairs 2 linear 2 0 2 linear 1 1 3 Triangular planar 3 0 3 Triangular planar 2 1 4 tetrahedral 4 0 4 tetrahedral 3 Shape and angle Visual aid linear 180° Linear 180° Triangular planar 1 120° V-shaped 117.5° tetrahedral 109.5° Triangular pyramidal 107° 4 tetrahedral 2 2 v-shaped 104.5° Carbon allotropes, silicon and silicon dioxide Diamond sp3hybridization each C forms 4 single bonds tetrahedral structure hardest substance, high m.p. and b.p. Graphite C60fullerene sp2 hybridzation sp2 hybridization each C forms 3 single bonds each C forms 2 single bonds and 1 double parralel layers of hexagon one ‘ball’ contains 60 C atoms layer silde over each other – lubricant; high b.p. and m.p.; goode conductor of electricity conductor of electricity, low m.p. and b.p.; can be used as lubricant 25 Silicon Silicon dioxide Polarity The prerequisite of polar molecules is polar covalent bonds (electronagetivity difference 1.8-0.7) Polar bonds will result in a denser region of electron cloud closer to an atom, creating a dipole moments. ‘ If dipole moments cancel out, the molecule is non-polar, if not, this molecule is polar. 26 Number of charge centres Geometrical arrangement of charge centres Number of bonding pairs Number of nonbonding pairs 2 linear 2 0 2 linear 1 1 3 Triangular planar 3 0 3 Triangular planar 2 1 4 tetrahedral 4 0 4 tetrahedral 3 1 Shape and angle linear 180° Linear 180° Triangular planar 120° V-shaped 117.5° tetrahedral 109.5° Triangular pyramidal Polarity (one type of atom) Non-polar Non-polar Non-polar polar Non-polar Polar 107° 4 5 tetrahedral Triangular bipyramidal 2 2 5 0 V-shaped 104.5° Triangular bipyramidal Polar Non-polar 90°,120° 5 Triangular bipyramidal 4 1 5 Triangular bipyramidal 3 2 5 Triangular bipyramidal 2 3 6 Octahedral 6 0 6 Octahedral 5 1 6 Octahedral 4 2 6 Octahedral 3 3 6 Octahedral 2 4 27 See saw 117.5°,87.5° T-shape 90° Linear 180° Octahedral 90° Square pyramidal 87.5° Square planar 90° T-shape 90° Linear 90° Polar Polar Non-polar Non-polar Polar Non-polar Polar Non-polar 4.4 Intermolecular forces U1 Intermolecular forces include London (dispersion) forces, dipole-dipole forces and hydrogen bonding. U2 The relative strengths of these interactions are London (dispersion) forces <dipole-dipole forces < hydrogen bonds. A1 Deduction of the types of intermolecular force present in substances, based on their structure and chemical formula. A2 Explanation of the physical properties of covalent compounds (volatility, electrical conductivity and solubility) in terms of their structure and intermolecular forces. Dipole-dipole force (all polar molecules) Dipole-dipole occurs because polar molecules are attracted to each other by electrostatic forces. It is stronger that Van der Waals’ but weaker than H-bonding. Hydrogen bonding (only in polar molecules with –OH and –NH group) Hydrogen bonding occurs when hydrogen is bonded directly to a small highly electronegative element, such as fluorine, oxygen or nitrogen. As the electron pair is drawn from the hydrogen atom, it has a small relative charge and therefore attracts a lone pair from a neighbouring molecule. Hydrogen bonds are not real bonds, they are just attraction between molecules. London dispersion force (all molecules) Electrons can at any one moment be unevenly spread. This produces temporary induced dipoles. These can induce another dipole in neighbouring molecules. This force is called the London force. Number of electrons and size of molecules affects the magnitude of London force due to larger electron region. Strength of intermolecular forces 1. Hydrogen bonding 2. Dipole-dipole 3. London dispersion force IMFs & boiling points and melting points The stronger the intermolecular forces, the harder it is to separate the molecules i.e. the more energy it takes which corresponds to a higher temperature, so the stronger the IMFs, the higher the boiling point and melting point. Distillation can happen due to difference in intermolecular forces between molecules. IMFs & solubility Polar substances dissolve in polar solvent and non-polar substances dissolve in non-polar solvent. This is because the intermolecular forces of solute and solvent matches. The solution therefore will be homogenous. IMFs & vapour pressure Weaker intermolecular force will result in higher vapour pressure. This is because the liquid cannot hold particles together due to weak intermolecular forces, so particles leave the liquid, leading to more particles presented in form of gas, creating larger pressure. Also, higher vapour pressure leads to low boiling and melting point. 4.5 Metallic bonding U1 A metallic bond is the electrostatic attraction between a lattice of positive ions and delocalized electrons. U2 The strength of a metallic bond depends on the charge of the ions and the radius of the metal ion. U3 Alloys usually contain more than one metal and have enhanced properties. 28 A1 Explanation of electrical conductivity and malleability in metals. A2 Explanation of trends in melting points of metals. A3 Explanation of the properties of alloys in terms of non-directional bonding. Metallic bonding In the metal, the valence electrons detached from atoms creating a sea of electrons. A metallic bond is the attraction between positive metal ions and electron cloud. It can be described as “metal ions surrounded by a cloud of delocalised electrons (a sea of electrons)” Properties of metals • • • Metals are good conductors of heat and electricity because they contain mobile charges (free electrons) Metals are malleable and ductile because the layers of cations can slide over each other without breaking more bonds than are made. Metals have high melting point and boiling point due to strong forces between positive ions and electrons. Bonding Properties Metallic bonding Macromolecular covalent Simple covalent high melting point, good conductor high melting point, poor conductor (except graphite) Weak IMFs low melting/boiling points Strong IMFs higher melting/boiling points Some polar covalent molecules, however, in conditions where they can ionise will conduct electricity. For example, HCl dissolved in water (hydrochloric acid) is an electrical conductor. Strength of metallic bonding • • Radius of the ion (larger radius means electrostatic attraction is in larger distance, weaker bonding) Charge of the ion (higher the charge, strong the attraction) Trends in melting point & boiling point • • • In group 17, from F2 to I2, melting point gets larger. This is because atoms have larger size and more electrons, so the London dispersion forces become stronger. Stronger IMF leads to higher melting point. In group 1, from Li to Fr, melting point gets smaller. This is because atoms have larger size, so the electrostatic attraction is in the larger distances. IMF gets smaller, leading to decreasing melting point. In period 2, from N2 to F2, melting point decrease. This because the atom size gets smaller, leading to smaller London dispersion forces (however, electron number increases but with less effect); thus, it result in lower melting point. Alloys Mixture of metal, is a kind of metallic solution Transition metals always form alloys because they have closer atomic radius. In general, alloys have poor electricity and heat conductivity than pure metal. This is because their size is different, so the flow of electrons is blocked. 29 Topic 14: Chemical bonding and structure (HL) 14.1 Further aspects of covalent bonding and structure U1 Covalent bonds result from the overlap of atomic orbitals. A sigma bond (σ) is formed by the direct head-on/end-to-end overlap of atomic orbitals, resulting in electron density concentrated between the nuclei of the bonding atoms. A pi bond (π) is formed by the sideways overlap of atomic orbitals, resulting in electron density above and below the plane of the nuclei of the bonding atoms. U2 Formal charge (FC) can be used to decide which Lewis (electron dot) structure is preferred from several. The FC is the charge an atom would have if all atoms in the molecule had the same electronegativity. FC = (Number of valence electrons)-½(Number of bonding electrons)-(Number of non-bonding electrons). The Lewis (electron dot) structure with the atoms having FC values closest to zero is preferred. U3 Exceptions to the octet rule include some species having incomplete octets and expanded octets. U4 Delocalization involves electrons that are shared by/between all atoms in a molecule or ion as opposed to being localized between a pair of atoms. U5 Resonance involves using two or more Lewis (electron dot) structures to represent a particular molecule or ion. A resonance structure is one of two or more alternative Lewis (electron dot) structures for a molecule or ion that cannot be described fully with one Lewis (electron dot) structure alone. A1 Prediction whether sigma (σ) or pi (π) bonds are formed from the linear combination of atomic orbitals. A2 Deduction of the Lewis (electron dot) structures of molecules and ions showing all valence electrons for up to six electron pairs on each atom. A3 Application of FC to ascertain which Lewis (electron dot) structure is preferred from different Lewis (electron dot) structures. A4 Deduction using VSEPR theory of the electron domain geometry and molecular geometry with five and six electron domains and associated bond angles. A5 Explanation of the wavelength of light required to dissociate oxygen and ozone. A6 Description of the mechanism of the catalysis of ozone depletion when catalysed by CFCs and NOx. Sigma (σ ) bonds Sigma bonds result from axial (head-on) overlap of s, p and hybrid orbitals in different combinations. Pi (π) bonds Pi bonds result from sideways overlap of parallel orbitals and consist of two regions of electron density (two overlaps). Pi bonds are weaker than sigma bonds as their electron density is further from the positive charge of the nucleus. Double and triple bonds Double bonds consist of one sigma bond and one pi bond. Triple bonds consist of one sigma and two pi bonds. The first bond formed is always sigma bond. Sigma bond is shorter because head-on overlap happens closer to the nucleus. Pi bond is longer because sideway overlap happens further away from the nucleus. Formal Charge FC = (Number of valence electrons)-½(Number of bonding electrons)-(Number of non-bonding electrons). Least formal charge means the most stable atom, suggesting that this kind of structure is correct. If FC is the same, using electro negativity to deduce, more electronegative atom carries negative formal charge. As shown in the molecule [OCN]- above, left configuration has largest formal charge, so it cannot be the correct structure. Middle and the right one have the same formal charge; but oxygen in the right one, as more electronegative atom, has the negative formal charge. So the right one is the correct structure of [OCN]30 VSEPR Theory on 5 and 6 charged centre No. of charge centres 5 6 Geometry Triangular No. bonding pairs No. nonbonding pairs 5 0 4 1 3 Bond angles Shape 90° Triangular 120° bipyramidal 87.5° see saw 117.5° (distorted tetrahedral) 2 90° T-shaped 2 3 180° linear 6 0 90° octahedral 5 1 87.5° square pyramidal 4 2 90° square planar 3 3 90° T-shaped 2 4 180° Linear bipryramidal octahedral 31 Visual aid 14.2 Hybridisation U1 A hybrid orbital results from the mixing of different types of atomic orbitals on the same atom. A1 Explanation of the formation of sp3, sp2 and sp hybrid orbitals in methane, ethene and ethyne. A2 Identification and explanation of the relationships between Lewis (electron dot)structures, electron domains, molecular geometries and types of hybridization. Hybridisation Hybridisation is the process whereby unequal atomic orbitals within an atom mix to form new hybrid atomic orbitals which are the same as each other, but different from the original orbitals. Hybrid orbitals form stronger bonds by allowing greater overlap. Experimental evidence suggests that bonds in an molecule have equal length, which support the idea of hybridisation. sp3 hybridisation When carbon forms four single bonds (e.g. CH4), it undergoes sp3 hybridisation: sp2 hybridisation When carbon forms a double bond (e.g. C2H4), it undergoes sp2 hybridisation: sp hybridisation When carbon forms a triple bond (e.g. C2H2), it undergoes sp hybridisation: 32 Hybridisation and molecular shape The shape of a molecule can be used to determine the type of hybridisation that has occurred: • • • tetrahedral arrangement planar triangular arrangement linear arrangement Bond types and hybridisation ↔ ↔ ↔ sp3 hybridised sp2 hybridised sp hybridised Hybridisation only occurs when a sigma bond forms. Pi bonds only form when two unmixed orbitals sideway overlap. sp3 will means 4 sigma bonds formed; sp2 will means 3 sigma bonds formed; sp will means 2 sigma bonds formed. 33 Topic 5: Energetics/ thermochemistry 5.1 Measuring energy changes U1 Heat is a form of energy. U2 Temperature is a measure of the average kinetic energy of the particles. U3 Total energy is conserved in chemical reactions. U4 Chemical reactions that involve transfer of heat between the system and the surroundings are described as endothermic or exothermic. U5 The enthalpy change (ΔH) for chemical reactions is indicated in kJ mol-1. U6 ΔH values are usually expressed under standard conditions, given by ΔH°, including standard states. A1 Calculation of the heat change when the temperature of a pure substance is changed using ππ = mcΔππ. A2 A calorimetry experiment for an enthalpy of reaction should be covered and the results evaluated. IB definitions Exothermic reaction – An exothermic reaction is one that releases heat to the surroundings as a result of forming products with stronger bonds than the reactants. Exothermic reactions have a negative ΔH value. Endothermic reaction – An endothermic reaction is one that absorbs heat from the surroundings as a result of forming products with weaker bonds than the reactants. Endothermic reactions have a positive ΔH value. Standard enthalpy change of reaction –Standard enthalpy change is the heat transferred during a reaction carried out under standard conditions: Unit: heat per mole (KJmol-1) • • • pressure 100 kPa temperature 298 K all substances pure and in their standard state Combustion, neutralisation and simple displacement are always exothermic reactions. Temperature changes and enthalpy changes • • a decrease in temperature of the system means that energy is absorbed in the reaction which means it is endothermic an increase in temperature of the system means that the reaction gives out energy which means it is exothermic Energy profile diagrams Bonding in reaction • • • bond making release energy (more stable) bond breaking require energy (less stable) reaction energy release = bond making – bonding breaking = energy of products –energy of reactants Measurement of energy Q = m × c × Δt Specific heat capacity: amount of energy required to change the temperature of 1kg of substance by 1 degree 34 Enthalpy of formation Enthalpy change when 1 mole of a molecule is formed from its element in their standard state 1 H2 (g) + O2 (g) → H2 O(l) 2 **Only 1 mole of product should be made** ΔHf°(H2) = 0: For elements, its enthalpy of formation = 0 Enthalpy of combustion Enthalpy change is released when 1 mole of a substance burns completely with oxygen under standard condition **Only 1 mole of substance can be burnt** C2 H5 OH(s) + 7O2 (g) → 6CO2 (g) + 3H2 O(l) 5.2 Hess's Law U1 The enthalpy change for a reaction that is carried out in a series of steps is equal to the sum of the enthalpy changes for the individual steps. A1 Application of Hess’s Law to calculate enthalpy changes. A2 Calculation of Δπ»π» reactions using ΔHf° data A3 Determination of the enthalpy change of a reaction that is the sum of multiple reactions with known enthalpy changes. Enthalpy experiments Measuring enthalpy of combustion Measuring enthalpy of neutralisation Enthalpy calculations βπ»π» = ππ ππππβππ = ππ ππ Where n is the number of moles reacted (in the combustion experiment this can be calculated from the change in mass of the fuel burner) Determining temperature change from graphs You need to extrapolate backwards in order to compensate for the heat loss. 35 Hess’s Law Total enthalpy change = the sum of all intermediated steps Problem solving Using the equations below: C(s) + O2 (g) → CO2 (g) Mn(s) + O2 (g) → MnO2 (s) βπ»π» = −390 kJ mol−1 What is the enthalpy change for the following reaction? βπ»π» = −520 kJ mol−1 MnO2 (s) + C(s) → Mn(s) + CO2 (g) As you can see, if you reverse the second reaction (and change the sign of the enthalpy), it cancels to give the above reaction, therefore the enthalpy change is -390 + 520 = 130 kJ mol-1. 5.3 Bond enthalpies U1 Bond-forming releases energy and bond-breaking requires energy. U2 Average bond enthalpy is the energy needed to break one mol of a bond in a gaseous molecule averaged over similar compounds. A1 Calculation of the enthalpy changes from known bond enthalpy values and comparison of these to experimentally measured values. A2 Sketching and evaluation of potential energy profiles in determining whether reactants or products are more stable and if the reaction is exothermic or endothermic. A3 Discussion of the bond strength in ozone relative to oxygen in its importance to the atmosphere. IB definitions Bond enthalpy–enthalpy change is required when 1 mole of bonds are broken to form atoms in their gaseous state. For example, X2(g) → 2X(g) It is an average value because it takes account of the different energies in a bond between the same atoms in different molecules. Bond enthalpies and reaction enthalpies • • If the sum of the bond enthalpies in the reactants is greater than in the products then the reaction is exothermic. If it is less, the reaction is endothermic βH°reaction = ΣBE(reactants) − ΣBE(products) βH°reaction = ΣβHf° (products) − ΣβHf° (reactants) Bond-breaking is an endothermic process; bond-making is an exothermic process CCl4 (g) → C(g) + 4Cl(g) **All in gaseous state and only 1 mole of substance is breaking down** 36 Topic 15: Energetics/thermochemistry (HL) 15.1 Energy cycles U1 Representative equations (eg. M+(g)ο M+(aq)) can be used for enthalpy/energy of hydration, ionization, atomization, electron affinity, lattice, covalent bond and solution. U2 Enthalpy of solution, hydration enthalpy and lattice enthalpy are related in an energy cycle. A1 Construction of Born-Haber cycles for group 1 and 2 oxides and chlorides. A2 Construction of energy cycles from hydration, lattice and solution enthalpy. For example dissolution of solid NaOH or NH4Cl in water. A3 Calculation of enthalpy changes from Born-Haber or dissolution energy cycles. A4 Relate size and charge of ions to lattice and hydration enthalpies. A5 Perform lab experiments which could include single replacement reactions in aqueous solutions. IB definitions Standard enthalpy of solution–one mole of an ionic solid breaks into its aqueous ions in solution in standard state Standard enthalpy of hydration –one mole of gaseous ions becomes aqueous ions in standard state Lattice enthalpy –enthalpy change when one mole of an ionic compound breaks down into its ions in the gaseous state Enthalpy of atomisation –enthalpy change when one mole of substance change state from solid to gas Born- Haber Cycle Oxides Chlorides ° ° −βππππ° + βπππ’π’π’π’π’π’π’π’π’π’π’π’π’π’π’π’π’π’π’π’ + ππππππππ ππππππππππππππππ + ππππππππππππππππππππ ππππππππππππ + ππππππππππππππππ ππππππππππππππππ = βπππ₯π₯π₯π₯π₯π₯π₯π₯π₯π₯π₯π₯π₯π₯ Factors affect lattice enthalpy value: 37 Charge of the ions (greater charge, stronger interaction, greater lattice enthalpy value) Distance (smaller size, stronger interaction, greater enthalpy value) • • Enthalpy of solution and hydration: Enthalpy of solution is always positive (except the group 1 element) – endothermic reaction Enthalpy of hydration is always negative– exothermic reaction Charges of ions and size of ions may affect the enthalpy of solution value due to the interaction between ions and water • • • Energy cycle: Lattice enthalpy: solid to gaseous ions Enthalpy of hydration: gaseous ions to aqueous ions Enthalpy of solution: solid into aqueous ions ° ° ° = βHlattice + βHhydration βHsolution 15.2 Entropy and spontaneity U1 Entropy (S) refers to the distribution of available energy among the particles. The more ways the energy can be distributed the higher the entropy. U2 Gibbs free energy (G) relates the energy that can be obtained from a chemical reaction to the change in enthalpy (ΔH), change in entropy (ΔS), and absolute temperature (T). U3 Entropy of gas>liquid>solid under same conditions. A1 Prediction of whether a change will result in an increase or decrease in entropy by considering the states of the reactants and products. A2 Calculation of entropy changes (ΔS) from given standard entropy values (Sº). A3 Application of πΊπΊ° = Δπ»π»° − ππΔππ° in predicting spontaneity and calculation of various conditions of enthalpy and temperature that will affect this. Relation of ΔG to position of equilibrium. A4 Entropy: how energy is distributed in a system • • • • Entropy increases from order to disorder More ways to distribute energy, greater the entropy This tends to increase over time in a closed system. Gases mixing together increase the entropy. Unit of entropy: JK-1mol-1 Factors that increase the entropy of a system (ΔS> 0) • mixing different types of particles • a change in state where the distance between the particles increases (solid<liquid<gas) • the increased movement of particles • increasing the number of particles • the greatest increase is usually found where the number of particles in the gaseous state increases βππ β = βππ β (products) − βππ β (reactants) Gibbs’ free energy: energy available to do useful work The sign of ΔG must be negative for a process or reaction to be spontaneous. Spontaneous reaction: reaction happens itself without outside energy. π₯π₯πΊπΊ β = π₯π₯π»π» β − πππ₯π₯ππ β The fact that we use standard conditions implies that the temperature is 298K. 38 Calculating ΔG β βπΊπΊreaction = οΏ½ βπΊπΊππβ (products) − οΏ½ βπΊπΊππβ (reactants) Depending on the values of ΔH and ΔS, a reaction might become spontaneous at a certain temperature. ΔHβ ΔSβ T ΔG Spontaneity positive positive low not spontaneous positive positive high positive ≈ βπ»π»β positive negative low not spontaneous positive negative high positive ≈ βπ»π»β negative positive low spontaneous negative positive high negative ≈ βπ»π»β negative negative low spontaneous negative negative high negative ≈ βπ»π»β negative ≈ −ππβππ β spontaneous positive ≈ −ππβππ β not spontaneous negative ≈ −ππβππ β spontaneous positive ≈ −ππβππ β not spontaneous 39 Topic 6: Chemical kinetics 6.1 Collision theory and rates of reaction U1 Species react as a result of collisions of sufficient energy and proper orientation. U2 The rate of reaction is expressed as the change in concentration of a particular reactant/product per unit time. U3 Concentration changes in a reaction can be followed indirectly by monitoring changes in mass, volume and colour. U4 Activation energy (Ea) is the minimum energy that colliding molecules need in order to have successful collisions leading to a reaction. U5 By decreasing Ea, a catalyst increases the rate of a chemical reaction, without itself being permanently chemically changed. A1 Description of the kinetic theory in terms of the movement of particles whose average kinetic energy is proportional to temperature in Kelvin. A2 Analysis of graphical and numerical data from rate experiments. A3 Explanation of the effects of temperature, pressure/concentration and particle size on rate of reaction. A4 Construction of Maxwell–Boltzmann energy distribution curves to account for the probability of successful collisions and factors affecting these, including the effect of a catalyst. A5 Investigation of rates of reaction experimentally and evaluation of the results. A6 Sketching and explanation of energy profiles with and without catalysts. Collision theory According to collision theory, for a reaction to occur three criteria must be met: • the particles must collide • they must collide with the appropriate geometry or orientation so that the reactive parts of the particles come into contact • they must collide with sufficient energy (the activation energy) Quantity Effect of increasing quantity Explanation particle size decreases rate The greater the particle size, the smaller the exposed surface area. Reduced frequency of successful collision. So the rate of reaction decreases temperature increases rate Increase temperature leads to an increase in average kinetic energy, so particles move faster. More particles have energy above the activation energy. Therefore, there is an increase frequency of successful collision. concentration increases rate Increased concentration leads to an increase in frequency of successful collision. So the rate of the reaction increases without a change in activation energy. pressure increases rate In gas, increased pressure means decrease in volume, leading to a increase in concentration. So the frequency of successful collision increases. catalysts increases rate Catalyst will decrease the activation energy, providing an alternative route for the reaction. More particles will have energy beyond the activation energy. So the frequency of successful collision will increase. (surface area) Catalyst: • • • Catalyst is able to decrease the activation energy Heterogeneous catalyst: different state between catalyst and reactants Homogeneous catalyst: same state between catalyst and reactants The Maxwell-Boltzmann distribution It represents fixed numbers of moles of gas in a given volume. It tells us fraction of molecules with a given value of kinetic energy. Only if the molecules have energy beyond activation energy, can they perform successful collision. 40 • • Increasing the temperature lowers the peak of the curve and moves it to the right, more particles have enough energy to react Adding a catalyst reduces the activation energy, more particles have enough energy to react Topic 16: Kinetics (HL) 16.1 Rate expression and reaction mechanism U1 Reactions may occur by more than one step and the slowest step determines the rate of reaction (rate determining step/RDS). U2 The molecularity of an elementary step is the number of reactant particles taking part in that step. U3 The order of a reaction can be either integer or fractional in nature. The order of a reaction can describe, with respect to a reactant, the number of particles taking part in the rate-determining step. U4 Rate equations can only be determined experimentally. U5 The value of the rate constant (k) is affected by temperature and its units are determined from the overall order of the reaction. U6 Catalysts alter a reaction mechanism, introducing a step with lower activation energy. A1 Deduction of the rate expression for an equation from experimental data and solving problems involving the rate expression. A2 Sketching, identifying, and analysing graphical representations for zero, first and second order reactions. A3 Evaluation of proposed reaction mechanisms to be consistent with kinetic and stoichiometric data. Rate expression π΄π΄ + π΅π΅ → products rate = ππ[π΄π΄]ππ [π΅π΅]ππ The rate constant (k) is the constant of proportionality in the rate expression. It has a specific value for a specific reaction at a specific temperature. The units vary depending on the expressions. The overall order of the reaction is m + n The order with respect to a reactant is the power it is raised to in the rate expression. Rate constant is only affected by temperature. Rate expression is only focused on reactants. 41 Units of rate constant Zero order First order Second order 2 Rate = k units of rate = mol dm−3 s −1 Rate = k[A] or Rate = k[A] Rate = k[A][B] mol dm−3 s −1 (mol dm−3 )2 mol dm−3 s −1 mol dm−3 = s−1 = mol−1 dm3 s−1 First, second and third order reactions Zero order reactants (Aο B) rate = − − dA =k dt dA =k dt dA = −k × dt Af t οΏ½ dA = − οΏ½ k dt A0 0 Af − A0 = −k × t • Af = A0 − kt In zero order reaction: Final concentration (Af) = Initial concentration (A0) – kt First order reaction: rate = k[A] = − dA dt 1 × dA = −k × dt A t 1 dA = − οΏ½ k dt A0 A 0 οΏ½ Af In(Af ) − In(A0 ) = −kt In οΏ½ Af οΏ½ = −kt A0 Af = Ao × e−kt 42 Third order Rate = k[A]3 or Rate = k[A][B]2 or Rate = k[A][B][C] mol dm−3 s −1 (mol dm−3 )3 = mol−2 dm6 s−1 • In first order reaction: Final concentration (Af) = Initial concentration (A0) ×e-kt Second order reaction: rate = k[A]2 = − dA dt 1 × dA = −k × dt A2 Af t 1 dA = − οΏ½ k dt 2 A0 A 0 οΏ½ − • In second order reaction: ππ ππ + = −ππππ ππππ π¨π¨ππ ππ ππ = + ππππ ππππ π¨π¨ππ ππ ππ = + π€π€π€π€ ππππππππππ ππππππππππππππππππππππππππ (ππππ ) π’π’π’π’π’π’π’π’π’π’π’π’π’π’ ππππππππππππππππππππππππππ (ππππ ) Combined graph 43 Half life: amount of time required for the final concentration to become half of the original amount Zero order reaction: 1 Af = A0 2 1 A = A0 − kt 2 0 kt = 1 A 2 0 ππ ππ = ππ First order reaction: ππππ ππππ 1 Af = A0 2 1 A = A0 × e−kt 2 0 1 −kt = In( ) 2 ππ. ππππππ π€π€ ππ ππ = ππ Second order reaction: 1 2 1 Af = A0 2 1 A0 = 1 + kt A0 2 = 1 + kt × A0 ππ ππ = ππ ππ π€π€ × ππππ The concentration vs. time graphs for first and second orders look quite similar however, first order reactions have a constant half-life, whereas with second order reactions each half-life is twice the preceding one. 44 Reaction mechanism Most reactions that occur at a measurable rate occur as a series of simple steps. The sequence of steps is known as the reaction mechanism. The individual steps (elementary steps) usually cannot be observed directly. Intermediates are substances that are made in one step and used up in another. The sum of the elementary steps should cancel to give the original equation. The term molecularity is used in reference to an elementary step to indicate the number of reactant species involved. If a reaction has three, four or more species combining, it is likely that it can be split into steps. The rate-determining step The rate equation can be known by knowing the reaction mechanism, specifically the slowest step, which acts as a limit on the rate and is therefore called the rate-determining step. The rate equation is also equal to the rate constant (k) multiplied by the concentrations of the substances raised to the power of their respective coefficients in the rate-determining step. The order of reaction is the exact numbers of molecules involves in the rate determining step. aA + bB → products (slowstep) rate = ππ [A]a [B]b 45 Problem solving A noteworthy example is the following Step 1: Step 2: Overall: NO(g) + NO(g) → N2 O2 (g) fast N2 O2 (g) + O2 (g) → 2NO2 (g) slow: the rate-determining step 2NO(g) + O2 (g) → 2NO2 (g) rate = ππ[N2 O2 ][O2 ] But therefore: NO(g) + NO(g) → N2 O2 (g) rate = ππ[NO]2 [O2 ] If a species is both on the reactant and product side of the overall equation, it is probably a catalyst. 16.2 Activation energy U1 The Arrhenius equation uses the temperature dependence of the rate constant to determine the activation energy. U2 A graph of 1/T against ln k is a linear plot with gradient – Ea / R and intercept, lnA. U3 The frequency factor (or pre-exponential factor) (A) takes into account the frequency of collisions with proper orientations. A1 Analysing graphical representation of the Arrhenius equation in its linear form A2 Using the Arrhenius equation A3 Describing the relationships between temperature and rate constant; frequency factor and complexity of molecules colliding. A4 Determining and evaluating values of activation energy and frequency factors from data. The rate of reaction increases with temperature (we know this already because of collision theory). A common relationship is that 10°C increase doubles the rate. Looking back at the Maxwell-Boltzmann distribution curve, an increase in temperature means that more particles have an energy greater or equal to the activation energy. This means that the value of the activation energy will determine temperature’s effect. When the Ea is larger a temperature rise will make a larger difference, and when the Ea is smaller the temperature change will have a less significant effect. The Arrhenius equation πΈπΈ ππ ππ = π΄π΄ππ −π π π π A = the Arrhenius constant– taking account of frequency of collision with correct geometry ln ππ = − πΈπΈππ + ln π΄π΄ π π π π When temperature increases, k increases exponentially, speed of reaction increases exponentially. The Arrhenius plot 46 Topic 7: Equilibrium 7.1 Equilibrium U1 A state of equilibrium is reached in a closed system when the rates of the forward and reverse reactions are equal. U2 The equilibrium law describes how the equilibrium constant (Kc) can be determined for a particular chemical reaction. U3 The magnitude of the equilibrium constant indicates the extent of a reaction at equilibrium and is temperature dependent. U4 The reaction quotient (Q) measures the relative amount of products and reactants present during a reaction at a particular point in time. Q is the equilibrium expression with non-equilibrium concentrations. The position of the equilibrium changes with changes in concentration, pressure, and temperature U5 A catalyst has no effect on the position of equilibrium or the equilibrium constant. A1 The characteristics of chemical and physical systems in a state of equilibrium. A2 Deduction of the equilibrium constant expression (Kc) from an equation for a homogeneous reaction. A3 Determination of the relationship between different equilibrium constants (Kc) for the same reaction at the same temperature. A4 Application of Le Châtelier’s principle to predict the qualitative effects of changes of temperature, pressure and concentration on the position of equilibrium and on the value of the equilibrium constant. Feature of equilibrium state Explanation Equilibrium is dynamic The reaction has stopped but both forward and backward reactions are still occurring at the same rate. 1 The speed of forward reaction is equal to the speed of backward reaction. 2 Equilibrium is achieved in a closed system A closed system prevents exchange of matter with the surroundings, so equilibrium is achieved where both reactants and products can react and recombine. 3 The concentrations of reactants and products remain constant at equilibrium They are being produced and destroyed at an equal rate. 4 At equilibrium there is no change in macroscopic properties Colour and density do not change as these depend on the concentrations. 5 Equilibrium can be reached from either direction The reaction can be started with all reactants, all products or a mixture. Equilibrium constant Given a reversible reaction, aA + bB β cC + dD The equilibrium expression is the concentration of the products raised to the coefficients in the balanced equation over those of the reactants: πΎπΎπΆπΆ = A rule of thumb is: [πΆπΆ]ππ [π·π·]ππ [π΄π΄]ππ [π΅π΅]ππ πΎπΎπΆπΆ > 1 → reaction is towards products πΎπΎπΆπΆ < 1 → reaction is towards reactants Special cases : 1. In aqueous reaction, if water takes part in, ignore water concentration in Kc expression. Cu(OH)2 (aq) + 2HCl(aq) β CuCl2 (aq) + H2 O(l) 2. [Cu Cl 2 ] Kc expression (ignore water): πΎπΎπΆπΆ = [πΆπΆπΆπΆ (ππππ) 2 2 ][π»π»π»π»π»π»] Solid doesn’t show up in Kc expression 47 Le Chatelier’s principle If any change is imposed on a system that is in equilibrium then the system tends to adjust to a new equilibrium counteracting the change. • • • • • • • In an exothermic reaction heat can be considered a product, so increasing the temperature will shift the equilibrium towards the reactants, and decreasing the temperature will shift it to the products In an endothermic reaction heat can be considered a reactant Increasing pressure shifts the equilibrium to the side with fewer moles of gas, so as to decrease the pressure Decreasing pressure shifts it to the side with more moles of gas Increasing concentration shifts it to the side with fewer moles of solute Decreasing concentration shifts it to the side with more A catalyst speeds up the reaction but does not affect the value of the equilibrium constant Only temperature affects the value of the equilibrium constant Catalyst has no effect on equilibrium position The Haber process The Haber process is the process used to produce ammonia, NH3 N2 (g) + 3H2 (g) β 2NH3 (g) Conditions: βπ»π» = −93kJ mol−1 Ideal conditions Actual conditions Reason (if ideal ≠ actual) high pressure 250 atm - low temperature 450°C faster rate of reaction catalyst iron - The Contact process The Contact process is the process used to produce sulphuric acid, H2SO4 2SO2 (g) + O2 (g) β 2SO3 (g) Conditions: βπ»π» = −197kJ mol−1 Ideal conditions Actual conditions Reason (if ideal ≠ actual) high pressure 2 atm 99% yield, higher pressure is unnecessary low temperature 450°C faster rate of reaction catalyst Vanadium(V) oxide - Reaction quotient Kc can only be calculated in equilibrium, while Qc can be calculated at any time in the reaction. [πΆπΆ]ππ [π·π·]ππ πππΆπΆ = [π΄π΄]ππ [π΅π΅]ππ for aA + bB β cC + dD πππΆπΆ >πΎπΎπΆπΆ → more products, less reactants, equilibrium will shift towards reactants till Qc = Kc πππΆπΆ <πΎπΎπΆπΆ →more reactants, less products, equilibrium will shift towards products till Qc = Kc πππΆπΆ =πΎπΎπΆπΆ →reaction is already in equilibrium 48 Topic 17: Equilibrium (HL) 17.1 The equilibrium law U1 Le Châtelier’s principle for changes in concentration can be explained by the equilibrium law. U2 The position of equilibrium corresponds to a maximum value of entropy and a minimum in the value of the Gibbs free energy. U3 A1 The Gibbs free energy change of a reaction and the equilibrium constant can both be used to measure the position of an equilibrium reaction and are related by the equationβG = −RT Ink A2 Relationship between ΔG and the equilibrium constant. A3 Calculations using the equation.βG = −RT Ink Solution of homogeneous equilibrium problems using the expression for Kc. ICE Method (initial change equilibrium) 1. Write the balanced equation 2. Under the equation write in the values of the concentrations of each component using three rows: initial, change and equilibrium • Initial represents the concentration originally placed in the flask; it is assumed to be 0 if not stated otherwise • Change represents the amount that reacts to reach equilibrium. The changes that occur must be in the same ratio as the coefficients in the balanced equation, so if we know one of these values we can deduce the others. • Equilibrium is the concentration present in the equilibrium mixture = initial + change 3. Write the expression for KC from the balanced equation. Substitute the values for equilibrium concentration and calculate KC. Problem solving Calculating the equilibrium concentrations given KC and initial conditions The equilibrium constant KC for the reaction SO3 (g) + NO(g) β NO2 (g) + SO2 (g) was found to be 6.78 at a specified temperature. If the initial concentrations of NO and SO3 were both 0.03 mol dm-3, what would be the equilibrium concentration of each component? Solution SO3(g) + NO(g) Initial 0.03 0.03 Change -x Equilibrium 0.03 - x Solving a quadratic gives the value of x. πΎπΎπΆπΆ = β NO2(g) + SO2(g) 0.00 0.00 -x x x 0.03 - x x x [NO2 ][SO2 ] π₯π₯ 2 = = 6.78 [SO3 ][NO] (0.03 − π₯π₯)2 Calculating equilibrium concentrations when KC is very small This situation allows us to make the following approximation: [reactant]initial ≈ [reactant]equilibrium The thermal decomposition of water has a very small value of KC. At 1000°C, KC = 7.3 × 10-18 for the reaction 2H2 O(g) β 2H2 (g) + O2 (g) A reaction is set up at this temperature with an initial H2O concentration of 0.10 mol dm-3. Calculate the H2 concentration at equilibrium. 49 Solution 2H2O(g) Initial 0.10 Change Equilibrium = 2H2(g) + O2(g) 0.00 0.00 -2x +2x +x 0.10 - 2x 2x x We will use the aforementioned approximation to give: The rest is simple arithmetic. πΎπΎπΆπΆ = [H2 ]2 [O2 ] 2π₯π₯ 2 = = 7.3 × 10−18 [H2 O] 0.102 Kc Changes Effect on equilibrium expression Effect on KC reversing the reaction inverts the expression KC-1 doubling the reaction coefficients squares expression KC2 halving the reaction coefficients square roots expression KC1/2 adding together two reactions multiplies two expressions KC ×KC’ Gibb’s free energy &equilibrium constant Remainder: when ΔG = -ve, the reaction is spontaneous; when ΔG = +ve, the reaction is not spontaneous ° πΎπΎπΆπΆ > 1 → reaction is towards products, so the forward reaction is spontaneous: βGforward reaction < 0 ° πΎπΎπΆπΆ < 1 → reaction is towards reactants, so the forward reaction is NOT spontaneous: βGforward reaction > 0 ° πΎπΎπΆπΆ >1→βGforward reaction = 0 Equation: βG = −RT Ink c In Kc 1 T InK c = × −βG° R Gradient = − βG° R 1 T 50 Topic 8: Acids and bases 8.1 Theories of acids and bases U1 A Brønsted–Lowry acid is a proton/H+ donor and a Brønsted–Lowry base is a proton/H+ acceptor. U2 Amphiprotic species can act as both Brønsted–Lowry acids and bases. U3 A pair of species differing by a single proton is called a conjugate acid-base pair A1 Deduction of the Brønsted–Lowry acid and base in a chemical reaction. A2 Deduction of the conjugate acid or conjugate base in a chemical reaction. IB definitions Brønsted–Lowry acid – a proton (H+) donor Brønsted–Lowry base – a proton (H+) acceptor Conjugate acid-base pairs HA + B β A− + BH + Conjugate Brønsted–Lowry acid-base pairs differ by just one proton. Bronsted acid will give a proton to become a conjugate base in a chemical reaction. Bronsted base will receive a proton to become a conjugate acid in a chemical reaction. Amphiprotic Molecules that behave both like a Bronsted acid and a Bronsted bas are amphiprotic. e.g. Water is an example of amphiprotic substance. 8.2 Properties of acids and bases U1 Most acids have observable characteristic chemical reactions with reactive metals, metal oxides, metal hydroxides, hydrogen carbonates and carbonates. U2 Salt and water are produced in exothermic neutralization reactions. A1 Balancing chemical equations for the reaction of acids. A2 Identification of the acid and base needed to make different salts. A3 Candidates should have experience of acid-base titrations with different indicators. Properties of acids Properties of bases pH below 7 react with hydroxides to form a salt and water react with metal oxides to form a salt and water react with ammonia to form a salt react with metals to form a salt and hydrogen gas react with carbonates forming a salt, water and carbon dioxide turns litmus pink • • • • • • • • • • • pH above 7 alkalis are bases that dissolve in water turns litmus blue is prepared by metal oxide and water Monoprotic acid: one H+ can be given away per molecule (HCl, CH3COOH) Diprotic acid: two H+ can be given away per molecule (H2SO4) Triprotic acid: three H+ can be given away per molecule (H3PO4) Neutralisation: • • Salt and water are produced in neutralisation Neutralisation reaction is always exothermic, the enthalpy of neutralisation is always negative. Amphoteric substances can react with acids and bases such as Al2O3 51 8.3 The pH scale U1 pH = -log[H+(aq)] and [H+] = 10-pH. U2 A change of one pH unit represents a 10-fold change in the hydrogen ion concentration [H+]. U3 pH values distinguish between acidic, neutral and alkaline solutions. U4 The ionic product constant, Kw = [H+][OH−] = 10-14 at 298 K. Acids have a pH below 7. Neutral substances have a pH of exactly 7. Bases have a pH greater than 7. The greater the pH the more alkaline a substance and the lower the pH the more acidic it is. pH = − log[H +] This means that decreasing one pH unit represents a ten-fold increase in hydrogen ion concentration and increasing one unit represents a tenfold increase. A two unit change represents a one hundred-fold change and so on. Ionic product of water H2 O + H2 O β H3 O+ + OH − K c = [H3 O+ ][OH −] In this case, Kc is equal to Kw, which is the ionic product of water, measuring the natural disassociation degree of water. Kw = 10-14 So –log Kw = 14, which evolves and becomes the pH scale. 8.4 Strong and weak acids and bases U1 Strong and weak acids and bases differ in the extent of ionization. U2 Strong acids and bases of equal concentrations have higher conductivities than weak acids and bases. U3 A strong acid is a good proton donor and has a weak conjugate base. U4 A strong base is a good proton acceptor and has a weak conjugate acid. A1 Distinction between strong and weak acids and bases in terms of the rates of their reactions with metals, metal oxides, metal hydroxides, metal hydrogencarbonates and metal carbonates and their electrical conductivities for solutions of equal concentrations. Strong and weak acids and bases A strong acid or base dissociates completely into its ions in aqueous solution (the dissociation reaction goes to completion) A weak acid or base only dissociates partially (the dissociation reaction is an equilibrium that lies to the left) Strong acids and bases conduct electricity better in solution and react more vigorously than weak acids and bases. Strong Bronsted acid/base gives weak conjugate base/acid Weak Bronsted acid/base gives strong conjugate base/acid Examples Strong acid Strong base hydrochloric acid, HCl sodium hydroxide, NaOH nitric acid, HNO3 potassium hydroxide, KOH sulphuric acid, H2SO4 barium hydroxide, Ba(OH)2 Weak acid Weak base ethanoic acid, CH3COOH ammonia, NH3 carbonic acid, H2CO3 Aminoethane, C2H5NH2 52 Distinguish between strong and weak acid Using universal pH paper or pH sensor: the one with lower pH is the stronger acid and the one with higher pH is the stronger base. (same concentration Acid-base titration: faster rate of titration leads to stronger acid/base, lower rate of titration leads to a weaker acid/base. However, the volume required is the same between the weak and strong acid. The weak acid will also disassociate completely since the equilibrium will shift to the product side due to decreasing amount of H+ Conductivity of solutions: stronger the acid/base, it will have high conductivity due to the presence of more ions. Rate of reaction will metal oxides, metals and carbonates. 8.5 Acid deposition U1 Rain is naturally acidic because of dissolved CO2 and has a pH of 5.6. Acid deposition has a pH below 5.6. U2 Acid deposition is formed when nitrogen or sulfur oxides dissolve in water to form HNO3, HNO2, H2SO4 and H2SO3. U3 Sources of the oxides of sulfur and nitrogen and the effects of acid deposition should be covered. A1 Balancing the equations that describe the combustion of sulfur and nitrogen to their oxides and the subsequent formation of H2SO3, H2SO4, HNO2 and HNO3. A2 Distinction between the pre-combustion and post-combustion methods of reducing sulfur oxides emissions. A3 Deduction of acid deposition equations for acid deposition with reactive metal sand carbonates. Rain is always acidic because carbon dioxide from the atmosphere dissolves in rain water and gives natural pH of 5.6 When the rain water has a pH below 5.6, it is considered as acid rain. Sources of the acid rain include: HNO3, HNO2, H2SO4 and H2SO3 H2 O + SO2 → H2 SO3 H2 O + SO3 → H2 SO4 H2 O + NO2 → HNO3 + HNO2 HNO2 + O2 → HNO3 53 Topic 18: Acids and bases (HL) 18.1 Lewis acids and bases U1 A Lewis acid is a lone pair acceptor and a Lewis base is a lone pair donor. U2 When a Lewis base reacts with a Lewis acid a coordinate bond is formed. U3 A nucleophile is a Lewis base and an electrophile is a Lewis acid. A1 Application of Lewis’ acid–base theory to inorganic and organic chemistry to identify the role of the reacting species. Lewis theory IB definitions Lewis acid – electron pair acceptor Lewis base – electron donor Lewis base: molecules usually have a lone pair of electron to give. e.g. ammonia and water Lewis acid: molecules usually don’t have a complete octet e.g. BF3, AlCl3, BeCl2 Coordinate bonds will be formed between a Lewis acid and a Lewis base. Lewis acid (incomplete octet): they are electrophiles (electron loving, accepting electron pairs) Lewis base (lone pair of electrons): they are nucleophiles (electron hating, donating electrons) 18.2 Calculations involving acids and bases U1 The expression for the dissociation constant of a weak acid (Ka) and a weak base (Kb). U2 For a conjugate acid base pair, Ka× Kb= Kw. U3 The relationship between Ka and pKa is (pKa= -log Ka), and between Kb and pKb is (pKb = -log Kb). A1 Solution of problems involving [H+ (aq)], [OH–(aq)], pH, pOH, Ka, pKa, Kb and pKb. A2 Discussion of the relative strengths of acids and bases using values of Ka, pKa, Kb and pKb. Ionic product constant of water H2 O(l) β H +(aq) + OH −(aq) πΎπΎππ = At room temperature, KW has a value of 1.00 × 10-14 [H+ ][OH − ] [H2 O] πΎπΎπ€π€ = πΎπΎππ [H2 O] = [H +][OH −] pH and pOH pH = − log[H +] pOH = − log[OH − ] 54 pH + pOH = pπΎπΎππ (= 14 at 25°C) Weak acids A weak acid dissociates partially in water: HA(aq) + H2 O(l) β H3 O+ + A−(aq) From this we can work out an equilibrium expression: πΎπΎππ = [H3 O+ ][A−] → [H2 O][HA] Weak bases Weak bases also dissociate partially. This can be expressed as: πΎπΎππ = πΎπΎππ [H2 O] = [H3 O+ ][A−] [HA] B(aq) + H2 O(l) β BH + (aq) + OH − (aq) This gives the expression: πΎπΎππ = [BH +][OH − ] [B][H2 O] → πΎπΎππ = πΎπΎππ [H2 O] = [BH + ][OH − ] [B] Since weak acids and bases only dissociate slightly they have low Ka or Kb respectively (remember that if the equilibrium lies to the left, Kc<< 1). Since strong acids and bases dissociate fully, they will have larger Ka or Kb respectively (remember if equilibrium lies to right, Kc>> 1). 18.3 pH curves U1 The characteristics of the pH curves produced by the different combinations of strong and weak acids and bases. U2 An acid–base indicator is a weak acid or a weak base where the components of the conjugate acid–base pair have different colours. U3 The relationship between the pH range of an acid–base indicator, which is a weak acid, and its pKa value. U4 The buffer region on the pH curve represents the region where small additions of acid or base result in little or no change in pH. U5 The composition and action of a buffer solution. A1 The general shapes of graphs of pH against volume for titrations involving strong and weak acids and bases with an explanation of their important features. A2 Selection of an appropriate indicator for a titration, given the equivalence point of the titration and the end point of the indicator. A3 While the nature of the acid–base buffer always remains the same, buffer solutions can be prepared by either mixing a weak acid/base with a solution of a salt containing its conjugate, or by partial neutralization of a weak acid/base with a strong acid/base. A4 Prediction of the relative pH of aqueous salt solutions formed by the different combinations of strong and weak acid and base. A buffer solution is resistant to changes in pH on the addition of small amounts of strong acid or alkali. They are a mixture of two solutions such that it contains the two species of a conjugate acid-base pair. Acidic buffers maintain the pH below 7 and basic buffers maintain it above 7. Composition of the buffer solution • • • • • Acidic buffer is made by a weak acid and the salt of this weak acid (same concentration and volume) e.g. CH3COOH and CH3COONa Basic buffer is made by a weak base and the salt of this weak base (same concentration and volume) e.g. NH4OH and NH4NO3 Because the weak acid/base disassociates partially in water, it will create equilibrium with the reactant side being the acid/base and product side being the ion of acid/base. When adding H+/OH-, large concentration of both sides (products/reactants) will resist the shift of the equilibrium, thereby stabilising the pH value. For example, when mixing the same concentration and volume of CH3COOH and CH3COONa together, we can the equilibrium like this: CH3COOHβ CH3COO- + H+ The concentration of both CH3COOH and CH3COO- is both high, it resists the shift of equilibrium. Calculating the pH (or pOH) of a buffer Step 1: Write both dissociation equations: HA β H + + A− Step 2: Make the following assumptions: MA → M + + A− 55 [HA]initial = [HA]equilibrium [MA]initial = [A− ]equilibrium Step 3: Write the equilibrium expression and rearrange depending on what you want to find: Step 4: Implement the assumptions: [H + ][A− ] [HA] → [π»π» + ] = πΎπΎππ [HA] [A− ] [H + ][MA]initial [HA]initial ππππ [H + ] = πΎπΎππ [HA]initial [MA]initial πΎπΎππ = πΎπΎππ = The same procedure can be done with pOH and Kb. The Henderson-Hasselbalch equations Acid-base titration pH = pπΎπΎππ + log [salt] [salt] & pOH = pπΎπΎππ + log [acid] [base] Strong acid + Strong base e.g. HCl + NaOH • • • • Weak acid + Strong base e.g. CH3COOH + NaOH The leftmost and rightmost points represent the pH of the acid and base respectively the big vertical jump is the point of inflection the equivalence point is where the solutions neutralise pH = pKa or pOH = pKb at the half-equivalence point, where a buffer is created Strong acid + Weak base e.g. HCl + NH3 • Weak acid + Weak base e.g. CH3COOH + NH3 when a weak acid is added to a weak base, it is difficult to determine the equivalence point. Instead other techniques are used. 56 Indicators Indicators are weak acids or weak bases in which the undissociated and dissociated forms have different colours. HIn(aq) β H + (aq) + In− (aq) colour A colour B Le Chatelier’s principle Increasing the H+ concentration moves the equilibrium to the left Increasing the OH- concentration moves the equilibrium to the right Colour change Indicators change colour when the pH is equal to their pKa, at the midpoint in the equilibrium, so that [HIn] = [In-] πΎπΎππ = [H+ ][In−] [H +][In−] = [HIn] [HIn] → pπΎπΎππ = pH This is known as the change point or the end point of the indicator. At this point, the addition of a very small volume of acid or base will shift the equilibrium as described above, and so cause the indicator to change colour. Choosing the appropriate indicator An indicator will be effective in signalling the equivalence point of a titration when its end point coincides with the pH at the equivalence point. Reactants in titration pH range at equivalence strong acid + strong base 3 – 11 weak acid + strong base 7 – 11 strong acid + weak base 3–7 weak acid + weak base no significant change in pH at equivalence 57 Topic 9: Redox processes 9.1 Oxidation and reduction U1 Oxidation and reduction can be considered in terms of oxygen gain/hydrogen loss, electron transfer or change in oxidation number. U2 An oxidizing agent is reduced and a reducing agent is oxidized. U3 Variable oxidation numbers exist for transition metals and for most main-group non-metals. U4 The activity series ranks metals according to the ease with which they undergo oxidation. U5 The Winkler Method can be used to measure biochemical oxygen demand (BOD), used as a measure of the degree of pollution in a water sample. A1 Deduction of the oxidation states of an atom in an ion or a compound. A2 Deduction of the name of a transition metal compound from a given formula, applying oxidation numbers represented by Roman numerals. A3 Identification of the species oxidized and reduced and the oxidizing and reducing agents, in redox reactions. A4 Deduction of redox reactions using half-equations in acidic or neutral solutions. A5 Deduction of the feasibility of a redox reaction from the activity series or reaction data. A6 Solution of a range of redox titration problems. A7 Application of the Winkler Method to calculate BOD. IB definitions Oxidation – the loss of an electron, the gain of oxygen, the loss of hydrogen or the increase in oxidation number Reduction – the gain of an electron, the loss of oxygen, the gain of hydrogen or the decrease in oxidation number OIL RIG: Oxidation is gain, Reduction is loss Oxidation number rules 1. Elements have an oxidation number of 0 2. For group one and group two ions: oxidation number = charges (+1/+2) 3. Hydrogen always has an oxidation of +1 4. Oxygen always has an oxidation number of -2 except in H2O2 5. Group 17 ions has an oxidation number of -1 when forming a compound with metal ions. 6. Addition of all individual oxidation number = total charges on this molecule Balancing redox equations Redox reaction is the reaction in which both reduction and oxidation happens simultaneously. Redox reactions will take place in H+ or OH- medium IB definitions Oxidising agent – a substance that readily oxidizes other substances. Oxidizing agents are thus reduced Reducing agent – a substance that readily reduces other substances. Reducing agents are thus oxidized. More reactive metals are stronger reducing agents than less reactive metals. A more reactive metal is able to reduce the ions of a less reactive metal. More reactive non-metals are stronger oxidising agents than less reactive non-metals. A more reactive non-metal is able to oxidise the ions of a less reactive non-metal. 58 Writing half-equations (in H+ medium) 1. Separating reduction and oxidation side of the reaction 2. Balance the atoms other than H and O 3. Balance each half-equation for O by adding H2O as needed. 4. Balance each half-equation for H by adding H+ as needed. 5. Balance each half-equation for charge by adding electrons to the sides with the more positive charge. Writing the redox equation 6. Equalize the number of electrons in the two-half equations by multiplying each appropriately. Add the two half-equations together, cancelling out anything that is the same on both sides, which includes electrons. Problem solving Balance the following reactions Solution 1. I − + HSO4 − → I2 + SO2 Separate the oxidation and reduction half of the equation: Oxidation: I− → I2 Reduction: HSO4 − → SO2 2. Balance all atoms without O and H 2I − → I2 HSO4 − → SO2 3. Balance Oxygen by adding H2O HSO4 − → SO2 + 2H2 O 4. Balance Hydrogen by adding H+ 3H+ + HSO4 − → SO2 + 2H2 O 5. Balance the charge by adding electrons 3H+ + HSO4 − + 2e− → SO2 + 2H2 O 6. 2I − → I2 + 2e− Combine two half equations together by multiplying the equations to cancel the electrons 3H+ + HSO4 − + 2I− → I2 + SO2 + 2H2 O Winkler method 1. 2. Fix oxygen: 2Mn2+ + ππππ + 4OH− → 2MnO(OH)2 Fix all oxygen in the water to MnO(OH)2 the solution colour changes from colourless to brown Convert to iodine: MnO(OH)2 + 2I− + 4H+ → ππππ + Mn2+ + 3H2 O The solution colour changes from brown to gold (fully dissolved); add starch into the solution as indication, the solution turns dark blue 59 3. Titration: I2 + 2S2 O3 2− → 2I− + S4 O6 2− The solution then turns from blue to colourless In summary: ππππππππ O2 in the sample → 2mol MnO(OH)2 → 2mol I2 → 2mol S4 O6 2− → ππππππππ S2 O3 2− added Gases in polluted water sample Elements Aerobic bacteria Anaerobic bacteria C CO2 CH4 H H2O CH4, NH3, H2S, PH3 N NO3- NH3 S SO42- H2S P PO43- PH3 Biochemical oxygen demand (BOD) BOD is the amount of dissolved oxygen required to decompose the organic matter in water biologically over a period of 5 days. 1. 2. 3. 4. Air bubbles through the water sample and measure the dissolved oxygen (DO1) using Winkler method Container is sealed with bacteria in the dark for 5 days Measure the dissolved oxygen (DO2) in the sample using Winkler method Calculate BOD = DO1 – DO2 Unit: ppm = mg/L 9.2 Electrochemical cells U1 Voltaic cells convert energy from spontaneous, exothermic chemical processes to electrical energy. U2 Oxidation occurs at the anode (negative electrode) and reduction occurs at the cathode (positive electrode) in a voltaic cell. U3 Electrolytic cells convert electrical energy to chemical energy, by bringing about non-spontaneous processes. U4 Oxidation occurs at the anode (positive electrode) and reduction occurs at the cathode (negative electrode) in an electrolytic cell. A1 Construction and annotation of both types of electrochemical cells. A2 Explanation of how a redox reaction is used to produce electricity in a voltaic cell and how current is conducted in an electrolytic cell. A3 Distinction between electron and ion flow in both electrochemical cells. A4 Performance of laboratory experiments involving a typical voltaic cell using two metal/metal-ion half-cells. A5 Deduction of the products of the electrolysis of a molten salt. Galvanic cell If zinc is placed in a copper (II) sulphate solution, electrons are transferred spontaneously from the zinc to copper and energy is released as heat. Instead of being lost as heat, the energy can be available as electrical energy by separating the two half-reactions into half-cells and allowing the electrons to flow through a circuit. This is known as an electrochemical, galvanic or voltaic cell. Half cells generate electrode potentials: when a metal is placed in a solution containing its ions, an equilibrium forms between the oxidised and normal forms. As atoms become ions they leave their electrons behind them creating a charge separation or electrode potential, the size of which is determined by the position of the equilibrium. In general, the more reactive a metal, the more negative its electrode potential in its half-cell. Half-cells connected by a wire are called electrodes. Oxidation always occurs at the anode; reduction always occurs at the cathode. In the voltaic cell, the anode has a negative charge and the cathode has a positive charge. Electrons flow from the anode to cathode. A salt bridge is a glass tube that contains an aqueous solution of ions that enables negative charge to be carried in the opposite direction to that of the electrons (from cathode to anode). This ion movement neutralises any build up of charge and maintains the potential difference. 60 Functions of the salt bridge 1. 2. 3. 4. Complete the circuit Physically separates the solutions Reduce liquid junction potential (creates when two solution with different conductivity meetes) Produce ions to neutralise the solution Notation The notation used for describing electrochemical cells is: Electrolytic cell An electrolytic cell uses an external source of voltage to bring about a redox reaction that would otherwise be non-spontaneous. The reactant is known as the electrolyte – a molten liquid or aqueous solution of an ionic compound. As the electric current passes through it, redox reactions occur at the electrodes, removing the charges of the ions or ‘discharging’ them. Oxidation occurs at the positive electrode (anode) – anions lose electrons - and reduction occurs at the negative electrode (cathode) – positive ions gain electrons. The current is not passed through the electrolyte by electrons but by the ions as they are mobile. When the electrolyte is a molten solution, for example NaCl, anions will appear in anode (Na solid) and cations will appear in cathode (Cl2 gas) 61 Topic 19: Redox processes (HL) 19.1 Electrochemical cells U1 A voltaic cell generates an electromotive force (EMF) resulting in the movement of electrons from the anode (negative electrode) to the cathode (positive electrode) via the external circuit. The EMF is termed the cell potential (Eº). U2 The standard hydrogen electrode (SHE) consists of an inert platinum electrode in contact with 1 mol dm-3 hydrogen ion and hydrogen gas at 100 kPa and 298 K. The standard electrode potential (Eº) is the potential (voltage) of the reduction half-equation under standard conditions measured relative to the SHE. Solute concentration is 1 mol dm-3 or 100 kPa for gases. Eº of the SHE is 0 V. U3 When aqueous solutions are electrolysed, water can be oxidized to oxygen at the anode and reduced to hydrogen at the cathode. U4 Gº = - nFEº. When Eº is positive, Gº is negative indicative of a spontaneous process. When Eº is negative, Gº is positive indicative of a non-spontaneous process. When Eº is 0, then Gº is 0. U5 Current, duration of electrolysis and charge on the ion affect the amount of product formed at the electrodes during electrolysis. U6 Electroplating involves the electrolytic coating of an object with a metallic thin layer. A1 Calculation of cell potentials using standard electrode potentials. A2 Prediction of whether a reaction is spontaneous or not using Eo values. A3 Determination of standard free-energy changes (ΔGo) using standard electrode potentials. A4 Explanation of the products formed during the electrolysis of aqueous solutions. A5 Perform lab experiments that could include single replacement reactions in aqueous solutions. A6 Determination of the relative amounts of products formed during electrolytic processes. A7 Explanation of the process of electroplating. Standard hydrogen electrode (SHE) • • • • • • • • • 1M H+ ions H2 gas bubbles through the platinum electrode at 100kPa At 298K the electrode is coated in very fine platinum the large surface area makes the reaction happen readily platinum is also fairly inert and can also catalyse the dissociation of H2 Act as a cathode (reduction): 2H+ + 2e− → H2 Act as an anode (oxidation): H2 → 2H + + 2e− SHE always connect to the negative terminal of the voltmeter Standard electrode potential When the standard hydrogen electrode is connected to another standard half-cell by an external circuit with a high resistance voltmeter and a salt bridge, the emf generated is known as the standard electrode potential of the half-cell. πΈπΈ β cell = πΈπΈ β half −cell where reduction occurs − πΈπΈ β half −cell where oxidation occurs = πΈπΈ β ππππππ βππππππ − πΈπΈ β ππππππππππ The values inserted into this equation must be reduction potentials. The stoichiometry of the equation does not matter. Gº = - nFEº. When Eº is positive, Gº is negative indicative of a spontaneous process. When Eº is negative, Gº is positive indicative of a nonspontaneous process. When Eº is 0, then Gº is 0. 62 If πΈπΈ β cell is positive, the reaction is spontaneous as written; if πΈπΈ β cell is negative, the reaction is non-spontaneous and in fact the reverse reaction is spontaneous. Electrolytic cell Predicting the products of electrolysis The products of electrolysis are influenced by various factors: 1. the relative Eβ values of the ions At the cathode the cation with the higher Eβ is value preferentially reduced. At the anode the anion with the lower Eβ is value is preferentially oxidised. 2. the relative concentrations of the ions in the electrolyte At the anode, in concentrated NaCl, Cl- will be oxidised to Cl2 gas. In dilute NaCl, water will be oxidised to produce O2 3. the nature of the electrode Electrolysis of copper (II) sulphate solution: when carbon (neutral) electrodes are used, electrolysis occurs as you would expect. With copper electrodes, at the cathode Cu2+ ions are discharged (becomes Cu). At the anode the copper electrode is oxidised (has more negative Eβ ), supplying electrons. So the net movement of Cu2+ ions is from the anode to the cathode. The anode shrinks and the cathode swells. Carbon electrodes Copper electrodes Determining the relative amounts of products The relative amounts of products depend on the stoichiometry of the equation e.g. 2H2 O(l) → 2H2 (g) + O2 (g) In this electrolysis of water, twice the number of moles (and because of Avogadro’s law, twice the volume too) of hydrogen will be produced than of oxygen. 63 Electroplating Electroplating is the process of using electrolysis to deposit a layer of a metal on top of another metal or other conductive substance. An electrolytic cell used for electroplating has the following features: • • • an electrolyte containing the metal ions to be deposited the object to be plated as the cathode sometimes the anode is made of the same metal which is to be coated because it may be oxidised to replenish the supply the ions in the electrolyte Electroplating can be used for various purposes: • • • decorative purposes corrosion control e.g. zinc is deposited on iron or steel in a process called galvanisation. This is called sacrificial protection. improvement of function e.g. electroplating of chromium on steel improves the wear on steel parts such as hand tools or crankshafts 64 Topic 10: Organic chemistry 10.1 Fundamentals of organic chemistry U1 A homologous series is a series of compounds of the same family, with the same general formula, which differ from each other by a common structural unit. U2 Structural formulas can be represented in full and condensed format. U3 Structural isomers are compounds with the same molecular formula but different arrangements of atoms. U4 Functional groups are the reactive parts of molecules. U5 Saturated compounds contain single bonds only and unsaturated compounds contain double or triple bonds. U6 Benzene is an aromatic, unsaturated hydrocarbon. A1 Explanation of the trends in boiling points of members of a homologous series. A2 Distinction between empirical, molecular and structural formulas. A3 Identification of different classes: alkanes, alkenes, alkynes, halogenoalkanes, alcohols, ethers, aldehydes, ketones, esters, carboxylic acids, amines, amides, nitriles and arenes. A4 Identification of typical functional groups in molecules eg phenyl, hydroxyl, carbonyl, carboxyl, carboxamide, aldehyde, ester, ether, amine, nitrile, alkyl, alkenyl and alkynyl. A5 Construction of 3-D models (real or virtual) of organic molecules. A6 Application of IUPAC rules in the nomenclature of straight-chain and branchedchain isomers. A7 Identification of primary, secondary and tertiary carbon atoms in halogenoalkanes and alcohols and primary, secondary and tertiary nitrogen atoms in amines. A8 Discussion of the structure of benzene using physical and chemical evidence. Homologous series The main features of a homologous series are: 1. Successive members of a homologous series differ by one CH2 group 2. All members of a homologous series can be expressed by the same general formula 3. Successive members show a gradation in physical properties 4. The different members have similar chemical properties The boiling points increase as the chain gets longer as the Van der Waals forces get stronger. Formulas The empirical formula shows the simplest ratio of atoms in a compound. The molecular formula shows the actual number of atoms in a compound. The structural formula is a representation of a molecule showing how the atoms are bonded together. Full structural formula H O β£ β₯ H− C −C−O−H β£ β₯ H O Condensed structural formula CH3COOH 65 Skeletal formula Structural isomers Structural isomers are compounds with the same molecular formula but with different arrangements of atoms, for example the isomers of hexane are: hexane 2-methylpentane 2,2-dimethylbutane • • 3-methylpentane 2,3-dimethylbutane Less branched isomer will have more surface area. So they are more likely to be polarised (induced dipole moment), leading to a higher boiling point. More branched isomer will have less surface area, leading to a lower boiling point. Nomenclature for organic compounds: the IUPAC system Rule 1: Identify the longest straight chain of carbon atoms Rule 2: Identify the functional group, which gives the suffix (ending). Rule 3: Identify the side chains or substituent groups, which give the prefix. Rule 4: Assemble the name of the compound, starting with halogens, then side chains, name of the longest chain and function groups IUPAC The name is in the form: Prefix – Parent – Suffix Homologous series Functional group Prefix Suffix alkane - alkyl- -ane* alkenyl- -ene* alkene alcohol −OH hydroxy- -ol* amine −NH2 *amino- -amine oxo- -anal* aldehyde 66 Ketone (carbonyl group) -oyl- -one* carboxylic acid carboxy- -oic acid* amide carboxamido- -amide* ester alkoxycarbonyl- -oate* nitrile cyano- -anenitrile Halogen *fluoro-, chloro, bromo-, iodo- C−X Phenyl group (arenes) *phenyl- Alkyne Alkynyl- -yne* ether -xyl -ether Primary, secondary and tertiary carbon atoms A primary carbon atom is attached to the functional group and also to at least two hydrogen atoms. A secondary carbon atom is attached to the functional group and also to one hydrogen atoms and two alkyl groups. A tertiary carbon atom is attached to the functional group and is also bonded to three alkyl groups and so has no hydrogen atoms. Primary Secondary Tertiary Volatility We can summarise the effect on volatility of the different functional groups as follows. most volatile least volatile alkane > halogenoalkane > aldehyde > ketone > alcohol > carboxylic acid van der Waals’ → dipole-dipole → hydrogen bonding increasing strength of molecular attraction → increasing boiling point 67 Solubility Molecules with functional groups that enable hydrogen bonds to form with water include the alcohols, the carboxylic acids and the amines. So the smaller members of these series are readily soluble in water. Aldehydes, ketones, amides and esters are less soluble, while halogenoalkanes, alkanes and alkenes are insoluble. 10.2 Functional group chemistry U1 Alkanes have low reactivity and undergo free-radical substitution reactions. U2 Alkenes are more reactive than alkanes and undergo addition reactions. Bromine water can be used to distinguish between alkenes and alkanes. U3 Alcohols undergo nucleophilic substitution reactions with acids (also called esterification or condensation) and some undergo oxidation reactions. U4 Halogenoalkanes are more reactive than alkanes. They can undergo (nucleophilic) substitution reactions. A nucleophile is an electron-rich species containing a lone pair that it donates to an electron-deficient carbon. U5 Addition polymers consist of a wide range of monomers and form the basis of the plastics industry. U6 Benzene does not readily undergo addition reactions but does undergo electrophilic substitution reactions. A1 Writing equations for the complete and incomplete combustion of hydrocarbons. A2 Explanation of the reaction of methane and ethane with halogens in terms of a free-radical substitution mechanism involving photochemical homolytic fission. A3 Writing equations for the reactions of alkenes with hydrogen and halogens and of symmetrical alkenes with hydrogen halides and water. A4 Outline of the addition polymerization of alkenes. A5 Relationship between the structure of the monomer to the polymer and repeating unit. A6 Writing equations for the complete combustion of alcohols. A7 Writing equations for the oxidation reactions of primary and secondary alcohols (using acidified potassium dichromate(VI) or potassium manganate(VII) as oxidizing agents). Explanation of distillation and reflux in the isolation of the aldehyde and carboxylic acid products. A8 Writing the equation for the condensation reaction of an alcohol with a carboxylic acid, in the presence of a catalyst (eg concentrated sulfuric acid) to form an ester. A9 Writing the equation for the substitution reactions of halogenoalkanes with aqueous sodium hydroxide. Low Reactivity of Alkanes Alkanes contain only C-C and C-H bonds. These are both strong bonds (bond enthalpies of 348 and 412 kJ mol-1 respectively) so alkanes are quite stable. They will only react when there is a strong source of energy. Alkanes are also non-polar, so are not susceptible to attack by most common reactants. Combustion of Alkanes Alkanes undergo complete combustion to give CO2 and H2O: C3 H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2 O When the oxygen supply is limited, alkanes undergo incomplete combustion, to produce carbon monoxide and water or carbon and water if the supply is even more limited: 2C3 H8 (g) + 7O2 (g) → 6CO(g) + 8H2 O(g) Free-radical substitution of Alkanes C3 H8 (g) + 2O2 (g) → 3C(s) + 4H2 O(g) Involves homolytic split – electron split evenly when bonds break, which requires UV light Alkanes undergo free-radical substitution (halogenations) in the presence of UV light e.g. CH4 (g) + Cl2 (g) → CH3 Cl(g) + HCl(g) The UV light is needed to break the covalent bond in the chlorine molecule. This creates two free radicals i.e. two chlorine atoms each with an unpaired electron. This is the start of a reaction described by a reaction mechanism. Use half arrow to show the movement of ONE eletron Use a dot to show the single electron on the atom 68 1. Initiation The homolytic fission of the chlorine molecule: 2. Propagation For example: UV light Cl2 οΏ½β―β―β―β―οΏ½ Cl β +Cl β Cl β +CH4 → CH3 ⋅ +HCl CH3 β +Cl2 → CH3 Cl + Cl β In this case, radical Cl β acts like a catalyst, being regenerated The reaction can continue and more chlorine atom will substitute hydrogen atoms. CH3 Cl + Cl β → CH2 Cl β +HCl CH2 Cl β +Cl2 → CH2 Cl2 + Cl β These reactions are called propagation because they both use and produce free radicals. 3. Termination Cl β +Cl β → Cl2 CH3 β +Cl β → CH3 Cl CH3 β +CH3 β → C2 H6 This is when all the radicals are used up in the reaction. Reaction stops. Graphic Display the movement of electrons Electrophilic Addition reactions of alkenes The π bond is weaker than the σ and so it breaks. Electrophile (electron lover) will accept a pair of electrons from the break of double bond With hydrogen Hydrogenation occurs in the presence of a nickel/Platinum catalyst at 150°C e.g. H H H H H H Ni catalyst β£ β£ β£ β£ β£ β£ 150°C H − C − C = C − H + H2 οΏ½β―β―β―β―β―β―οΏ½ H − C − C − C − H β£ β£ β£ β£ β£ β£ H H H H H H propene + hydrogen→propane 69 With halogens Dihalogeno compounds are produced in the reactions between halogens and alkenes. These reactions happen quickly at room temperature accompanied by the loss of colour of the halogen. There are called halo-ination (e.g. bromination) H H H H H H β£ β£ β£ β£ β£ β£ fffffffff H − C − C = C − H + Br2 οΏ½β―β―β―οΏ½ H − C − C − C − H β£ β£ β£ β£ β£ β£ H H H H Br Br propene + bromine → 1,2-dibromopropane With hydrogen halides The hydrogen halide (HX) with the weaker bond reacts faster. This means the reaction rate increases down the group. This takes place rapidly at room temperature e.g. H H H H H H β£ β£ β£ β£ β£ β£ fffffffff H − C = C − C − H + HCl οΏ½β―β―β―οΏ½ H − C − C − C − H β£ β£ β£ β£ β£ β£ H H H H Cl H The halide bonds to the carbon with fewer hydrogens bonded to it. With water Hydration is the reaction that converts an alkene into an alcohol. Conditions: heat with steam and catalyst of concentrated sulphuric acid.(H+) *There is no need to remember the intermediate product Test to distinguish alkanes and alkenes We can distinguish between alkanes and alkenes in terms of their chemical properties. Alkenes will react with bromine water, decolourising it, whereas alkanes will only do this in the presence of UV light. Since alkenes have a higher carbon to hydrogen ratio, they contain much more unburned carbon than alkanes when they burn, giving a smokier, dirtier flame. Polymerisation Alkenes (the monomers) can join together to produce long chains called polymers by addition under high pressure. The hydration of ethene is of industrial significance because ethanol is a very important solvent and so is manufactured on a large scale. The hydrogenation reaction is used in the margarine industry to saturate oil compounds, making the liquid solid. Polymers can be made into plastic for bags, water pipes, ropes and so on. Combustion of Alcohol When alcohols undergo complete combustion (when oxygen is plentiful), the products are water and carbon dioxide: 2CH3 OH(l) + 3O2 (g) → 2CO2 (g) + 4H2 O(g) The longer the chain the greater the βH When oxygen is less plentiful, the combustion will produce water and carbon or carbon monoxide. 70 2CH3 OH(l) + O2 (g) → 2C(s) + 4H2 O(g) CH3 OH(l) + O2 (g) → CO(g) + 2H2 O(g) Oxidation of Alcohols Combustion involves the complete oxidation of the alcohol molecules, but it is also possible for them to react with oxidising agents which selectively oxidise the carbon atom attached to the OH group, keeping the carbon skeleton intact, so that useful compounds can be made. In the lab, acidified potassium dichromate(VI)/potassium manganese(VII) is used as the oxidising agent. Owing to the Cr(VI), this solution is bright orange. When the mixture is heated, the colour changes to green as the Cr(VI) is reduced to Cr(III). The oxidising agent is usually represented by [O]. Primary alcohols Primary alcohols are oxidised in a two-step reaction, first forming the aldehyde and then the carboxylic acid: H H H O H O β£ β£ β£ β₯ +[O], heat β£ β₯ +[O], heat H − C − C − OH οΏ½β―β―β―β―β―β―οΏ½ H − C − C − H οΏ½β―β―β―β―β―β―οΏ½ H − C − C − OH reflux β£ β£ β£ β₯ β£ β₯ H H H O H O ethanol → ethanal → ethanoic acid If you want to obtain the aldehyde, it is possible to remove it by distillation (since the aldehyde is the only of the three that does not have Hbonding – low boiling point). To obtain the carboxylic acid, on the other hand, a reflux condenser is used. Secondary alcohols Secondary alcohols are oxidised to the ketone: H H H H H H β£ β£ β£ β£ β£ +[O], heat β£ H − C − C − C − H οΏ½β―β―β―β―β―β―οΏ½ H − C − C − C − H reflux β£ β₯ β£ β£ β£ β£ H OH H H O H Tertiary alcohols propan-2-ol → propanone Tertiary alcohols are not readily oxidised under similar conditions, as this would involve breaking the carbon skeleton of the molecule, which requires significantly more energy: no colour change. 71 Esterification Esterification is the combination of carboxylic acid and primary alcohol to produce an ester and water under heat and concentrated sulfuric acid. Naming: Alcohol (prefix-yl) + carboxylic acid (prefix-oate) Esterifcation giving out water; thus, it is also called condensation. And its reverse reaction is called hydrolysis, with NaOH as an catalyst Nucleophilic Substitution of halogenoalkanes In a halogenoalkane, the C-X bond is polar because the halogen is more electronegative. As a result the halogen gains a partial negative charge (δ -), whereas the carbon gains a partial positive charge (δ+ ) and is said to be electron deficient. Nucleophiles are reactants that are themselves electron rich (they have a lone pair of electrons or they are anions) such as H2O, OH-, NH3 and CN-. They are attracted to positive charged molecules such as the carbon. A stronger nucleophile can replace the weaker one. The overall reaction with NaOH is: H β£ H − C − Cl β£ H CH3 Cl + OH − → CH3 OH + Cl− Using – NH3 can make amine Using –OH- can make alcohol Using –CN- can make nitrile *By adding reducing agent [H], nitrile can be reduced (LiAlH4) to amine with an one unit increase in carbon chain length Electrophilic substitution of benzene It needs a Lewis acid as catalyst (accept pair of electrons) such as AlCl3, FeCl3, AlBr3 72 As more and more bromine is added, there could more substitution occurs, but only on 1,3,5 carbon on the benzene ring. The reason behind it is unnecessary to know for SL. Reaction Summary Table Reaction Name Reactant 1 Reactant 2 Catalyst Reaction Condition Product 1 Product 2 Comment Halogenation Alkane Halogen N/A UV light Halogenoalkane hydrogen halides Free-radical substitution Hydrogenation Alkene Hydrogen Nickle/Platium High Temp 150°C Alkane N/A Saturate oil compounds Bromination Alkene Bromine N/A Room Temp Dibromoalkane N/A Distinguish between alkene and alkane Hydration Alkene Water Conc. Sulfuric Acid (H+) Heat Alcohol N/A To make alcohol Polymerisation Alkene Many more alkene N/A High Pressure Polymer N/A To make plastic Oxidation of Alcohols Primary Alcohol Oxidising Agent Acidity condition Heat Distillation Aldehyde Oxidation of Alcohols Primary Alcohol Oxidising Agent Acidity condition Heat Reflux Carboxylic Acid Oxidation of Alcohols Secondary Alcohol Oxidising Agent Acidity condition Heat Reflux Ketone Reduced state of oxidising agent Oxidised Once Esterification Primary Alcohol Carboxylic Acid Conc. Sulfuric Acid (H+) Heat Ester Water A.K.A condensation Hydrolysis Ester Water Conc. NaOH (OH-) Heat Primary Alcohol Carboxylic Acid Reverse of Esterification Nucleophilic Substitution Halogenoalka -nes Nucleophile N/A Heat Alkane with new nucleophile Halogen ions Order follows a series Reduction of nitrile Nitrile Reducing Agent N/A Heat Primary Amide Oxidised state of reducing agent one unit increase in carbon chain length Electrophilic substitution of benzene Benzene Halogen Lewis Acid Room Temp Halogenobenzene hydrogen halides Electrophilic substitution 73 Reduced state of oxidising agent Reduced state of oxidising agent Oxidised Once Oxidised Twice Topic 20: Organic chemistry (HL) 20.1 Types of organic reactions U1 SN1 represents a nucleophilic unimolecular substitution reaction and SN2 represents a nucleophilic bimolecular substitution reaction. SN1 involves a carbocation intermediate. SN2 involves a concerted reaction with a transition state. U2 For tertiary halogenoalkanes the predominant mechanism is SN1 and for primary halogenoalkanes it is SN2. Both mechanisms occur for secondary halogenoalkanes. U3 The rate determining step (slow step) in an SN1 reaction depends only on the concentration of the halogenoalkane, rate = k[halogenoalkane]. For SN2, rate = k[halogenoalkane][nucleophile]. SN2 is stereospecific with an inversion of configuration at the carbon. U4 SN2 reactions are best conducted using aprotic, non-polar solvents and SN1 reactions are best conducted using protic, polar solvents. U5 An electrophile is an electron-deficient species that can accept electron pairs from a nucleophile. Electrophiles are Lewis acids. U6 Markovnikov’s rule can be applied to predict the major product in electrophilic addition reactions of unsymmetrical alkenes with hydrogen halides and interhalogens. The formation of the major product can be explained in terms of the relative stability of possible carbocations in the reaction mechanism. U7 Benzene is the simplest aromatic hydrocarbon compound (or arene) and has a delocalized structure of π bonds around its ring. Each carbon to carbon bond has a bond order of 1.5. Benzene is susceptible to attack by electrophiles. U8 Carboxylic acids can be reduced to primary alcohols (via the aldehyde). Ketones can be reduced to secondary alcohols. Typical reducing agents are lithium aluminium hydride (used to reduce carboxylic acids) and sodium borohydride. A1 Explanation of why hydroxide is a better nucleophile than water. A2 Deduction of the mechanism of the nucleophilic substitution reactions of halogenoalkanes with aqueous sodium hydroxide in terms of SN1 and SN2 mechanisms. Explanation of how the rate depends on the identity of the halogen (ie the leaving group), whether the halogenoalkane is primary, secondary or tertiary and the choice of solvent. A3 Outline of the difference between protic and aprotic solvents. A4 Deduction of the mechanism of the electrophilic addition reactions of alkenes with halogens/interhalogens and hydrogen halides. A5 Deduction of the mechanism of the nitration (electrophilic substitution) reaction of benzene (using a mixture of concentrated nitric acid and sulphuric acid). A6 Writing reduction reactions of carbonyl containing compounds: aldehydes and ketones to primary and secondary alcohols and carboxylic acids to aldehydes, using suitable reducing agents. A7 Conversion of nitrobenzene to phenylamine via a two-stage reaction. Nucleophilic Substitution of halogenoalkanes In a halogenoalkane, the C-X bond is polar because the halogen is more electronegative. As a result the halogen gains a partial negative charge (δ -), whereas the carbon gains a partial positive charge (δ+ ) and is said to be electron deficient. Nucleophiles are reactants that are themselves electron rich (they have a lone pair of electrons or they are anions) such as H2O, OH-, NH3 and CN-. SN stands for nucleophilic substitution. They are attracted to positive charged molecules such as the carbon. A stronger nucleophile can replace the weaker one. Primary halogenoalkanes: SN2 mechanism Primary halogenoalkanes have at least two hydrogen atoms attached to the carbon with the C-X bond e.g. The overall reaction with NaOH is: H β£ H − C − Cl β£ H CH3 Cl + OH − → CH3 OH + Cl− As the hydrogen atoms are so small, the carbon atom is relatively open to attack by the nucleophile. An unstable transition state is formed in which the carbon is weakly bonded simultaneously to both the halogen and the nucleophile. The C-X bond breaks heterolytically, which means a pair of electron is transferred to a single atom, releasing X- and forming the product. 74 This is known as a bimolecular reaction because the rate is dependent on both the concentration of the halogenoalkane and of the hydroxide ion. It is called SN2: substitution nucleophilic bimolecular. Rate = k[halogenoalkane][nucleophile] The product is stereo-specific since nucleophile can only attack at the opposite direction of the halogen. It is an inversion of configuration at the carbon SN2 reaction is likely to happen in aprotic solvent such as propan-2-one and diethyl ether. Tertiary halogenoalkane: SN1 mechanism Tertiary halogenoalkanes have three alkyl groups attached to the carbon with the C-X bond e.g. The overall reaction that occurs with NaOH is: CH3 β£ CH3 − C − Cl β£ CH3 CH3 C(CH3 )ClCH3 + OH − → CH3 C(CH3 )OHCH3 + Cl− The presence of the alkyl groups around the carbon of the C-X bond causes steric hindrance, meaning that these bulky groups make it difficult for an incoming group to attack this carbon atom. The first reaction involves the C-X bond breaking heterolytically. The carbon is left with a positive charge. It is called a carbocation. This is then attacked by the nucleophile in the second step. A factor which favours this mechanism in tertiary halogenoalkanes is that the carbocation is stabilised by the presence of the three alkyl groups, as each of these has an electron-donating or positive inductive effect, shown by the blue arrows in the diagram above. This stabilising effect helps the carbocation to persist for long enough for the second step to occur. This is a unimolecular reaction. Slow step is the rate determining step. Rate = k[halogenoalkane] Since the halogen comes off first, nucleophile can attack the carbocation both from above and below the plane, thus creating a racemic mix with 50% of each enantimers.(inversion and retention) SN1 reaction prefers protic solvent such as water, methanol and ethanol 75 Secondary halogenoalkanes Secondary halogenoalkanes undergo both SN1 and SN2 mechanisms depending on the solvent and strength of the nucleophile. Water vs. the hydroxide ion Water can be used as the nucleophile in a hydrolysis reaction instead of NaOH(aq), but H2O is a weaker nucleophile than the OH- ion as it does not have a negative charge, just a δ -. The reaction will be slower, except for the SN1 mechanism (unimolecular). The effect of the halogen on the rate 1. The strength of the C-X bond – the C-X bond decreases in strength from fluorine to iodine. The size of halogen increases, so less orbital will overlap with carbon. The bonds will become weaker down the group. The weaker the bond, the faster the reaction. The data suggest that the strength of the bond is the dominant factor, thus: iodoalkanes > bromoalkanes > chloroalkanes > fluoroalkanes The effect of the mechanism Since SN1 reaction give the fastest rate. Reactions with tertiary alkanes happen fastest (SN1) then secondary (SN1 and SN2) then primary (SN2). SN1 happens in tertiary halogenoalkane because the positive charge can be well stabilised around 3 carbons; thus the molecule is relatively stable to proceed the remaining part of the reaction. In primary halogenoalkane, this positive charge cannot be balanced, so it has to give to the nucleophile as halogen group coming off. Markovnikov’s rule The rules states “when an unsymmetrical alkene reacts with a hydrogen halide to give an alkyl halide, the hydrogen adds to the carbon of the alkene that has the greater number of hydrogen substituents, and the halogen to the carbon of the alkene with the fewer number of hydrogen substituents” • When a hydrogen halide undergoes eletrophilic addition with an alkene, the carbocation intermediate will form. • As the double bond of the alkene breaks, it will leave a positive charge on one of two carbons around the double bonds. • If there is more carbons around the positive charge, this charge can be stabilised around those carbon; thus, this intermediate is more stable and more likely to be produced • If there is only a few carbons around the positive charge, this charge cannot be stabilised, leaving the entire molecule unstable and less like to be produced. Stabilised by 2 carbons Stabilised by 1 carbons Major product • Miner product This type of electrophilic reaction will create a miner product (2%) and a major product (98%) from different intermediate. Nitration of benzene • • Benzene is the simplest aromatic hydrocarbon compound (or arene) and has a delocalized structure of π bonds around its ring. Each carbon to carbon bond has a bond order of 1.5. Benzene is susceptible to attack by electrophiles. Benzene reacts with mixture of concentrated nitric acid and sulphuric acid(catalyst) to give nitro benzene 76 • • • • • • Presence of H2SO4 kicks off OH- group on nitric acid to forms water, leaving a NO2+ electrophile Electrophile attacks one double bond on benzene ring, giving a intermediate with one positive charge resonating on the ring HSO4- attacks hydrogen. Nucleophilic gives electron to the ring to balance the charge. H2SO4 is regenerated as a catalyst As more and more HNO3 is added, there could more substitution occurs, but only on 1,3,5 carbon on the benzene ring. There is partial negative charge (δ-) on the carbon attached to NO2, which induces a partial positive charge (δ+) on the next carbon along the ring. One partial negative carbon followed by one partial negative carbon. Attack partially negative carbon δ+ δ- δ- δ+ δ+ δ- Conversion of nitrobenzene to phenylamine Stage 1: conversion of nitrobenzene into phenylammonium ions Nitrobenzene is reduced to phenylammonium ions using a mixture of tin and concentrated hydrochloric acid. The mixture is heated under reflux in a boiling water bath for about half an hour. Under the acidic conditions, rather than getting phenylamine directly, you instead get phenylammonium ions formed. The lone pair on the nitrogen in the phenylamine picks up a hydrogen ion from the acid. 77 The nitrobenzene has been reduced by gaining electrons in the presence of the acid. The electrons come from the tin, which forms both tin(II) and tin(IV) ions. Stage 2: conversion of the phenylammonium ions into phenylamine All you need to do is to remove the hydrogen ion from the -NH3+group. Sodium hydroxide solution is added to the product of the first stage of the reaction The phenylamine is formed together with a complicated mixture of tin compounds from reactions between the sodium hydroxide solution and the complex tin ions formed during the first stage. The phenylamine is finally separated from this mixture. The separation is long, tedious and potentially dangerous - involving steam distillation, solvent extraction and a final distillation. Overall Reduction of alcohol Carboxylic acids can be reduced to primary alcohols (via the aldehyde). Ketones can be reduced to secondary alcohols. Typical reducing agents are lithium aluminium hydride(LiAlH4) (used to reduce carboxylic acids) and sodium borohydride(NaBH4) 78 20.3 Stereoisomerism U1 Stereoisomers are subdivided into two classes—conformational isomers, which interconvert by rotation about a σ bond and configurational isomers that interconvert only by breaking and reforming a bond. Configurational isomers are further subdivided into cis-trans and E/Z isomers and optical isomers. U2 Cis-trans isomers can occur in alkenes or cycloalkanes (or heteroanalogues) and differ in the positions of atoms (or groups) relative to a reference plane. According to IUPAC, E/Z isomers refer to alkenes of the form R1R2C=CR3R4 (R1 ≠ R2, R3 ≠ R4) where neither R1 nor R2 need be different from R3 or R4. U3 A chiral carbon is a carbon joined to four different atoms or groups. U4 An optically active compound can rotate the plane of polarized light as it passes through a solution of the compound. Optical isomers are enantiomers. Enantiomers are non-superimposeable mirror images of each other. Diastereomers are not mirror images of each other. U5 A racemic mixture (or racemate) is a mixture of two enantiomers in equal amounts and is optically inactive. A1 Construction of 3-D models (real or virtual) of a wide range of stereoisomers. A2 Explanation of stereoisomerism in non-cyclic alkenes and C3 and C4 cycloalkanes. A3 Comparison between the physical and chemical properties of enantiomers. A4 Description and explanation of optical isomers in simple organic molecules. A5 Distinction between optical isomers using a polarimeter. Stereoisomerism Stereoisomers differ from each other in the spatial arrangement of their atoms. There are two types of stereoisomerism: configurational (rotation will result in breaking a bond – double bond) and conformational (rotation about aσ bond) Configurational isomers When there is some constraint in a molecule that restricts the free rotation of bonded groups, they become fixed in space relative to each other. This restriction can be caused by a double bond or a cyclic structure. The double bond is made of a σ and a π bond. Rotating would break the π bond. For cyclic molecules, the substituted groups do not have to be on adjacent carbon atoms – only their position relative to the plane matters. Cis/Trans Isomer Cis means same function groups (hydrogen) around the double bond is on the same side Trans means same function groups (hydrogen) around the double bond is on the different side 79 Physical properties The physical properties depend on various factors, including: • the polarity of the molecules • the shape or symmetry of the molecules cis-1,2-dichloroethene net dipole boiling point 60°C melting point -80°C trans-1,2-dichloroethene non-polar molecule boiling point 48°C melting point -50°C The boiling point of the cis isomer is higher because the molecule has dipole-dipole forces of attraction as well as van der Waals’. The melting point is generally more influenced by the symmetry of the molecules as this affects the packing in the solid state, therefore the trans isomers has the higher melting point. *Trans fat has higher melting point and is unhealty cis-butenedioic acid melting point 139°C intramolecular h-bonding trans-butenedioic acid melting point 287°C intermolecular h-bonding Chemical properties The chemical properties of geometric isomers are usually very similar. An exception is butenedioic acid. When the trans isomer is heated it sublimes, but does not react chemically. E/Z Isomer Z: highest molecular mass group (higher priority group) is on the same side E: highest molecular mass group is on the different side 80 Z E Optical isomers - conformational isomers A carbon atom bonded to four different atoms or groups is known as asymmetric or chiral. Optical isomers are non-superimposable mirror images of a molecule. The optical isomers are called enantiomers. A mixture containing equal amounts of the two enantiomers is known a racemic mixture (racemate). Optical isomers have identical chemical and physical properties with two exceptions: • optical activity • reactivity with other chiral molecules Optical activity Ordinary light consists of electromagnetic waves that oscillate in an infinite number if planes at right angles to the direction of travel. When passed through a polariser, only the light waves oscillating in a single plane pass through. This is plane-polarised light. The amount and direction of rotation can be measured with a polarimeter. The light that comes out of the solution passes through a second polariser called an analyser, which is rotated until the maximum amount of light passes through. In order to compare different solutions, the concentrations, the wavelength of light used and the sample path-length must all be kept the same. 81 Optical isomers rotate plane-polarised light in equal amounts but in opposite directions. A racemic mixture is optically inactive. Polarimetry can be used to test the purity of chiral compounds. Reactivity with other chiral molecules When a racemic mixture is reacted with a single enantiomer of another chiral compound, the two components of the mixture, the (+) (rotates clockwise) and (-) (anticlockwise), react to produce different products, having different chemical and physical properties which allow them to be separated, by a process called resolution Stereoisomerism in alkanes - conformational isomers • In ethane, when we look towards the carbon-carbon bond, there are two ways that hydrogen can be aliened to the carbon. On with 60 degrees apart, called staggered; and another form which hydrogen overlaps, called eclipsed • • • • Staggered form the protons have space between them so this form is energetically favourable. Eclipsed form is energetically unfavourable since there is stereo hindrance among protons. Different spatial arrangements due to rotations of groups of atoms about a bond can cause stereo-isomers in non-cylic alkanes. This can also happen in cyclohexane – boated shaped and chair shaped cyclohexane. Stereoisomerism in cycloalkanes - configurational isomers • Cycloalkanes cannot rotate around the bond as rotation would break the ring structure • Thus, cycloalkanes will have Cis/Trans isomerism 82 Topic 11: Measurement and data processing 11.1 Uncertainty and error in measurement U1 Qualitative data includes all non-numerical information obtained fromobservations not from measurement. U2 Quantitative data are obtained from measurements, and are always associatedwith random errors/uncertainties, determined by the apparatus, and by humanlimitations such as reaction times. U3 Propagation of random errors in data processing shows the impact of theuncertainties on the final result. U4 Experimental design and procedure usually lead to systematic errors inmeasurement, which cause a deviation in a particular direction. U5 Repeat trials and measurements will reduce random errors but not systematicerrors. A1 Distinction between random errors and systematic errors. A2 Record uncertainties in all measurements as a range (+) to an appropriateprecision. A3 Discussion of ways to reduce uncertainties in an experiment. A4 Propagation of uncertainties in processed data, including the use of percentageuncertainties. A5 Discussion of systematic errors in all experimental work, their impact on theresults and how they can be reduced. A6 Estimation of whether a particular source of error is likely to have a major orminor effect on the final result. A7 Calculation of percentage error when the experimental result can be comparedwith a theoretical or accepted result. A8 Distinction between accuracy and precision in evaluating results. Accuracy and Precision Accuracy is an indication of how close a measurement is to the accepted, literate value (a measure of correctness). Precision is an indication of the agreement among a number of measurements made in the same way (a measure of exactness). Random and systematic errors A random error is produced by unknown and unpredictable variations in the experimental situation, such as temperature fluctuations and estimations when reading instruments. They affect the precision of the results and can be reduced but not eliminated by repeating trials. They appear as error bars on a graph. A systematicerror is an error associated with a particular instrument or experimental technique that causes the measured value to be off by the same amount each time. They affect the accuracy of the results and can be mitigated by fixing source of error. They appear as a non-zero yintercept. Significant figures and Decimal Places The result of addition (or subtraction) is quoted to the same number of decimal places as the input value with the fewest. The result of multiplication (or division) is quoted to the same number of significant figures. Calculating uncertainties Given a measurement 5 ± 1, • • the absoluteuncertainty is ± 1 the percentage uncertainty is 20% When adding or subtracting, the uncertainties are added together. π΄π΄ + π΅π΅ = πΆπΆ → βπΆπΆ = βπ΄π΄ + βπ΅π΅ When multiplying or dividing, percentage uncertainties are added and the result is multiplied by the value to get the absolute uncertainty. 11.2 Graphical techniques π΄π΄ × π΅π΅ = πΆπΆ → βπΆπΆ = πΆπΆ × οΏ½ βA βπ΅π΅ + οΏ½ π΄π΄ π΅π΅ **This section is too simple and straight forward for students. There is no need for composing revision notes. ** 83 11.3 Spectroscopic identification of organic compounds U1 The degree of unsaturation or index of hydrogen deficiency (IHD) can be used to determine from a molecular formula the number of rings or multiple bonds in a molecule. U2 Mass spectrometry (MS), proton nuclear magnetic resonance spectroscopy (1H NMR) and infrared spectroscopy (IR) are techniques that can be used to help identify compounds and to determine their structure. A1 Determination of the IHD from a molecular formula. A2 Deduction of information about the structural features of a compound from percentage composition data, MS, 1H NMR or IR. IHD (index of hydrogen deficiency) Use the equation to calculate IHD IHD gives you the number of how many H2 gas (2H atom) can be added, and how many rings and double bonds presented in this molecule. Mass spectrometry A mass spectrometer is used to determine relative atomic masses. (Work under a vacuum) The stages of operation are: 1. Vaporisation: a vaporised sample is injected into the instrument; this allows individual atoms to be analysed 2. Ionisation: atoms are bombarded with a stream of high energy electrons, knocking off valence electrons, generating positively charged species X(g) + e- ---- X+(g) + 2e- 3. Acceleration: the positive ions are attracted by the negatively charged plates; they are accelerated (focused) by the electric field 4. Deflection: the ions are deflected by an external magnetic field placed at right angles to their path; the amount of deflection is proportional to the charge/mass ratio. Lighter, highest charged atoms ----- deflected most Heavier, smallest charged atoms ----- deflected least 5. Detection: ions of a particular mass/charge ratio are detected (other ions will hit the wall and get neutralised) and a signal is sent to a recorder; the strength of the signal is a measure of the number of ions 84 Mass spectrometry can break a bond anywhere on a molecule. • The furthest peak is the molecular mass of the molecule, which is denoted as M+ (since it is a positive ion) • Other peak is denoted as M+- x (where x is the difference in Mr of the fragments and M+; known fragments also have a 1+ charge, label it) MR can identify: 1. Mr of a compound 2. Series of possible fragments 3. Can be compared to a data base of known compounds in order to determine the identity of the compound 4. Use other techniques together with MR to find the substance • Sample MR Infra red spectrometry • IR light interacts with the vibration in function groups in organic molecules. • Change in polarity of the molecule affects electromagnetic wave of the infra red light • Same function group always give similar infra red spectra • % transmittance on Y axis • Wavenumber = 1/wavelength on X axis • Different peak with different shape in different wavenumber suggests a different function group. Sample IR spectrometry 85 1 H NMR Spectrometry • Use hydrogen environment to distinguish function group. • Each bonding structure gives different hydrogen environment • If identical hydrogen environments bind to different function group, it also those two separate hydrogen environment. • Label each hydrogen environment (A,B,C….) and count how many hydrogen atoms are in each type of environment. • The numbers of hydrogen environments are equals to the peaks on NMR diagram. • Integration number(area under the peak) (trace) is at the bottom of the graph. The ratio between those integration numbers gives the number of hydrogen per peak. • Using chemical shift data (in ppm) can identify the hydrogen environment from data booklet. Sample graph Identifying hydrogen environment • A cluster of carbon atom and hydrogen atoms around it is a hydrogen environment • The same cluster structure (e.g. CH3) but with different adjacent structure is considered as different hydrogen environment. Hydrogen Environment A • • Hydrogen Environment C Hydrogen Environment B Hydrogen Environment C Hydrogen environment B and C has the same structure but they attach to different structures (A/D). So they are considered as different hydrogen environments. Therefore, the molecule will have four peaks on NMR. 86 Topic 21: Measurement and analysis(HL) 21.1 Spectroscopic identification of organic compounds U1 Structural identification of compounds involves several different analytical techniques including IR, 1H NMR and MS. U2 In a high resolution 1H NMR spectrum, single peaks present in low resolution can split into further clusters of peaks. U3 The structural technique of single crystal X-ray crystallography can be used to identify the bond lengths and bond angles of crystalline compounds. A1 Explanation of the use of tetramethylsilane (TMS) as the reference standard. A2 Deduction of the structure of a compound given information from a range of analytical characterization techniques (X-ray crystallography, IR, 1H NMR and MS). TMS (Tertra methyl silane) • Used to mark the zero in 1H NMR spectrum • There are several reasons to use TMS to calibrate: • • • • All protons are in the same hydrogen environment, so it will give a strong single peak to signal the zero. It is non-toxic and unreactive. So it will not interfere with the sample. It is volatile (low boiling point) so it is very easy to be removed. It will absorb upfield of protons in NMR, away from most protons in test sample. High resolution 1H NMR Spectrum • Peak splitting occurs in high resolution 1H NMR spectrum • The number of peaks tells us about the number of protons attached to the adjacent carbon atoms • For n number of protons on the adjacent carbon, there will be (n+1) peaks • The intensity of peak (height) follows the Pascal Triangle. 87 Peak splitting Number of protons on adjacent carbon atom (n) Number of peaks Type of splitting 0 1 singlet 1 2 doublet 2 3 triplet 3 4 quartet X-ray crystallography • X-ray crystallography can be used to: • • • • • • Determine bond length Determine bond angle Determine the crystalline structure Use for confirmation of structure of compound It requires 100% pure crystal of the sample in rotated X-ray beam. It usually takes 24-72 hours to complete and the data is analysed by the computer. 88 Sample peak splitting diagram Option D: Medicinal chemistry D1 Pharmaceutical products and drug action U1 In animal studies, the therapeutic index is the lethal dose of a drug for 50% of the population (LD50) divided by the minimum effective dose for 50% of the population (ED50). U2 In humans, the therapeutic index is the toxic dose of a drug for 50% of the population (TD50) divided by the minimum effective dose for 50% of the population (ED50). U3 The therapeutic window is the range of dosages between the minimum amounts of the drug that produce the desired effect and a medically unacceptable adverse effect. U4 Dosage, tolerance, addiction and side effects are considerations of drug administration. U5 Bioavailability is the fraction of the administered dosage that reaches the target part of the human body. U6 The main steps in the development of synthetic drugs include identifying the need and structure, synthesis, yield and extraction. U7 Drug–receptor interactions are based on the structure of the drug and the site of activity A1 Discussion of experimental foundations for therapeutic index and therapeutic window through both animal and human studies. A2 Discussion of drug administration methods. A3 Comparison of how functional groups, polarity and medicinal administration can affect bioavailability. Drugs and medicines • A medicine or drug is any chemical that does one or more of the following to the human body, for better or worse: • • • alters the physiological state, including consciousness, activity level, coordination or cellular chemistry alters incoming sensory sensations alters mood or emotions Physiological effects • Therapeutic effect: the desirable outcome of a pharmaceutical drug on the body • A drug achieves its therapeutic effect as its molecular structure allows it to bind (through ionic bonding, hydrogen bonding or Van Der Waal’s forces) with a receptor e.g. a protein molecule such an enzyme or a cellular receptor such as a cell membrane. • The binding prevents or inhibits the biological activity (e.g. enzyme activity) that allows the disease to develop. • The receptor part in the molecule or cellular structure is referred to as the site of activity. The placebo effect • Placebo: biologically inert substance which achieves therapeutic effect due to the body’s natural healing process • Placebo effect: the actual therapeutic effect achieved by a placebo • A double-blind study is one in which neither the participants nor the experimenters know who is receiving a particular treatment. • Using double-blinded study, researchers can find whether particular drug is effective or not. Research and development of drugs • • Identification of a ‘need’: e.g. a disease, could be new disease. Identification of a molecular target (in the disease causing organism, pathogen) that has a receptor e.g. an enzyme or a cell structure that has a biological activity in the disease e.g ebola) and that needs to be inhibited. 89 • • • Identification of ‘lead’ molecule i.e. molecules with a molecular structure that can bind onto the receptor site. Such lead molecules could be found in plants or microorganisms but often will need to be modified to improve their chemical fit or their bioavailability. Alternatively a lead molecule is designed using the drug-receptor interactions approach; designing on computers a molecule with a structure that fits chemically into the receptor site and can bonds with it. Using the lead molecule, many different molecules or derivatives are synthesized and their therapeutic effect tested. During this process the intended molecules are extracted. Preclinical trials: testing of medicine in laboratory, • • • Clinical trials • • • • ‘in vitro’: the lead molecule is tested on animal/human cells and tissues which have been removed from the body and are kept in an artificial environment. ‘in vivo’: testing in live animals (usually 3 different species) to establish ED50 or LD50 which is the amount which kills 50 % of the population. Testing of its effectiveness, its therapeutic window, tolerance and its side effects using the placebo effect. This is a ‘blind trial’ (double-blinded study) in which half of the people/patients involved are given the drug whilst the other half are given a similar substance that is not the drug (called ‘placebo’ but none of the patients (or even their administering doctors) know which half they are in. Structural modifications likely to be made to, for instance, improve effectiveness or reduce side-effects. Submission of reports on the drug and its trials to international or national regulatory bodies. Monitoring of the drug after it has been launched; molecule might need further structural changes. Administering drugs • oral – taken by mouth, it is very convenient • inhalation – vapour breathed in; smoking; treatment of respiratory conditions • skin patches; absorbed directly from the skin into the blood e.g. nicotine patches • suppositories – inserted into the rectum, it is easy to perform • eye or ear drops – treatment of infections in those areas • parenteral (by injection) • • • intravenous: into a vein of the blood stream – used for immediate impacts as it is the fastest method; drug is immediately pumped around the body by the blood e.g. anaesthetics. intramuscular i.e. into the muscles, e.g. many vaccines, antibiotics, usually used when a large dose needs to be administered and it needs to act locally subcutaneous: in the layer of the skin directly below the cutis (dermis and epidermis) e.g. dental injections, morphine, insulin. Slow absorption needed; also slow effect. 90 Consideration of drug administration • Tolerance: Tolerance refers to the body’s reduced response to a drug i.e. its therapeutic effect is less than what it is intended, usually as a result of taking the drug over a long period of time. As a result more of the drug needs to be taken to achieve the same initial physiological effect with the danger of exceeding the lethal dose. • Side effects: Side effects are physiological effects which are not intended and vary greatly from one drug to another, and with the same drug in different people. Some side effects are beneficial e.g. aspirin, taken for pain relief, helps protect against heart disease. Other drugs e.g. thalidomide have negative side effects. • Addiction: Physical dependence. A condition that occurs when a person needs the drug just to live normally and shows withdrawal symptoms when not taking the drug. • Dosage: the amount of drug used for each dose and how frequently the drug should be taken. The target of a dosage is to achieve constant, safe and effective levels of the medicine in blood Therapeutic window and therapeutic index • Effecitve Dose (ED50): the minimum dose of the drug required to produce the desired therapeutic effect in 50% of the subjects • Lethal Dose (LD50): the minimum dose of the drug that causes death in 50% of the subjects • Toxic Dose (TD50): the minimum dose of the drug required to cause toxicity in 50% of the subjects (TD used because unethical to kill humans to determine LD ) • The effectiveness and safety of a drug can be expressed using its therapeutic index (TI) • The higher the TI, the safer the drug 50 50 • Therapeutic Window: the range of doses where the drug provides the desired therapeutic effect without causing unacceptable adverse effects in most patients. It is between ED50 and TD50 Bioavailability of a drug • Drug bioavailability: the fraction of administered dose that reaches target part of body • By definition, intravenous injection has 100% drug bioavailability while other methods of administration decrease in bioavailability • Bioavailability of pharmaceutical drugs depends on their solubility, polarity, and the presence of certain functional groups. More soluble with polar groups leads to higher bioavailability. 91 D2 Aspirin and penicillin U1 Mild analgesics function by intercepting the pain stimulus at the source, often by interfering with the production of substances that cause pain, swelling or fever. U2 Aspirin is prepared from salicylic acid. U3 Aspirin can be used as an anticoagulant, in prevention of the recurrence of heart attacks and strokes and as a prophylactic. U4 Penicillins are antibiotics produced by fungi. U5 A beta-lactam ring is a part of the core structure of penicillins. U6 Some antibiotics work by preventing cross-linking of the bacterial cell walls. U7 Modifying the side-chain results in penicillins that are more resistant to the penicillinase enzyme. A1 Description of the use of salicylic acid and its derivatives as mild analgesics. A2 Explanation of the synthesis of aspirin from salicylic acid, including yield, purity by recrystallization and characterization using IR and melting point. A3 Discussion of the synergistic effects of aspirin with alcohol. A4 Discussion of how the aspirin can be chemically modified into a salt to increase its aqueous solubility and how this facilitates its bioavailability. A5 Discussion of the effects of chemically modifying the side-chain of penicillins. A6 Discussion of the importance of patient compliance and the effects of the overprescription of penicillin. A7 Explanation of the importance of the beta-lactam ring on the action of penicillin. Mild analgesics Mild analgesics, including aspirin, paracetamol, ibuprofen, function by intercepting the pain stimulus at the source, often by blocking the transmission of pain from source to brain. Prostaglandins are chemicals that send pain impulses to the brain and cause swelling or fever. Aspirin works by suppressing the production of pain-causing prostaglandins. Salicylic acid Salicylic acid is an analgesic that comes from the bark of the willow tree. To reduce its side-effects, OH group was replaced with an ester group and the compound creates is aspirin. Paracetamol is also salicylic derivatives. Salicylic acid aspirin paracetamol Aspirin vs. Paracetamol Aspirin Paracetamol Analgesic (painkiller) yes yes Antipyretic (reduces fever) yes yes Reduces inflammation yes no Side-effects stomach wall irritant – stomach ulcer; allergies does not irritate stomach wall Severe side-effects (over-dosage) Reye’s syndrome in children serious kidney, liver and brain damage Effects of aspirin • Relieve pain • Anticoagulant – stop the blood clotting • prevention of the recurrence of heart attacks and strokes • Prophylactic (medicines taken for preventative measures): prevent colon cancer and cardiovascular disease 92 Synthesis of aspirin • OH group was replaced with an ester group in salicylic acid to make aspirin • Esterification reaction using ethanoic anhydride is taken to make aspirin • Method • • Purification of the aspirin crystals: recrystallization • • The 2-hydroxybenzoic acid and ethanoic anhydride are warmed gently with concentrated sulphuric or phosphoric acid as a catalyst. The mixture is diluted with water and allowed to cool down so that aspirin crystals form as aspirin has a low solubility in water. The aspirin crystals are removed using filtration To increase the yield, the impure crystals are removed using filtration and then dissolved in hot ethanol to make a saturated solution. This solution is cooled slowly and the aspirin crystalizes again (or recrystallizes) out first (lower solubility of aspirin in ethanol than the impurities) and is removed using filtration Characterization of aspirin • • The identity of the product from the above synthesis can be determined using IR spectroscopy whilst the purity (and therefore the yield) is determined using melting point determination. The aspirin IR spectrum is different from salicylic acid due to the presence of ester group and absence of OH group Synergetic effect of aspirin with ethanol • Ethanol produces a synergic effect with a number of drugs including aspirin, this means that the effect of the drug is enhanced in the presence of alcohol which can be dangerous e.g. aspirin and ethanol together can increase risk of stomach bleeding. Bioavailability of aspirin • Since aspirin is almost insoluble, its bioavailability is limited • The solubility and thus bioavailability of drugs can be increase by converting them into ionic salts • In aspirin, the carboxylic group can be neutralized with sodium hydroxide, producing soluble sodium salt of aspirin Antibiotics • Antibacterials are drugs that kill or inhibit the growth of bacteria that cause infectious diseases. An example of antibacterials are penicillins. • Penicillins are a group of compounds that are produced by fungi and kill harmful micro-organisms; they are therefore called antibiotics. • Alexander Fleming noticed that a fungus or mould known as Penicillium notatum produced a chemical which inhibited bacterial growth. • Howard Florey and Ernst Chain isolated penicillin as the antibacterial agent. 93 How penecillins work Its core structure is a four-membered ring consisting of one nitrogen and three carbon atoms and known as beta-lactam ring, which is responsible for the antibacterial properties. This beta-lactam ring is highly reactive and irreversibly binds to the enzyme transpeptidase in bacteria. It prevents the development of cross-links in bacterial cell walls This weakens their cell wall causing water to flow into the bacteria until the water pressure bursts the bacteria open Human and other animal cells don’t have cell walls therefore are not affected by penicillin Modifying side chains Certain bacteria mutated and developed varying degrees of antibiotic resistance due to increased production of enzyme called penicillinase. This enzyme could deactivate penicillin and makes the beta lactam ring break open. Over time, this species of bacteria became the dominant species. • Modern or semi-synthetic penicillins, such as ampicillin, are penicillin molecules that have been modified by replacing the side-chain with other atoms or groups of atoms. • The replacement of side chain will make penicillinase useless since it can no longer identify the modified version of pencillin. • This continued production of penicillin triggered multidrug resistance (MDR) in bacteria. MDR bacterial infections require a combination of many different antibiotics and strict patient compliance to solve the issue Antibiotic resistance • • • Patient compliance: Patient compliance refers to patients not completing the full course of penicillin and this results in prolonging the disease as not all bacteria are killed. By allowing the bacteria to live longer there can be more mutations eventually producing bacteria with resistance. Overuse of penicillins by humans: Many doctors are too quick to prescribe penicillin. Patients should be encouraged to fight an infection using their own immune system as overprescription weakens it Use of penicillins in animal feedstock as growth promoters: Some penicillins are also effective in animals but they are more often administered without the animals having any disease; they are administered as a prophylactic to prevent the animals from developing any disease that could affect their growth. In most cases these penicillins are passed by the animals into the environment and eventually ending up in the food chain. 94 D3 Opiates U1 The ability of a drug to cross the blood–brain barrier depends on its chemical structure and solubility in water and lipids. U2 Opiates are natural narcotic analgesics that are derived from the opium poppy U3 Morphine and codeine are used as strong analgesics. Strong analgesics work by temporarily bonding to receptor sites in the brain, preventing the transmission of pain impulses without depressing the central nervous system. U4 Medical use and addictive properties of opiate compounds are related to the presence of opioid receptors in the brain. A1 Explanation of the synthesis of codeine and diamorphine from morphine. A2 Description and explanation of the use of strong analgesics. A3 Comparison of the structures of morphine, codeine and diamorphine (heroin). A4 Discussion of the advantages and disadvantages of using morphine and its derivatives as strong analgesics. A5 Discussion of side effects and addiction to opiate compounds. A6 Explanation of the increased potency of diamorphine compared to morphine based on their chemical structure and solubility. Opiates • Opiates, such as morphine, diamorphine (heroin) and codeine, are natural strong analgesics as they reduce severe pain by temporarily bonding to opioid receptors in the brain or other parts of the central nervous system. • It will prevent transmission of pain impulses without depressing the central nervous system. • It is originally derived from the opium poppy plants • All those three drugs need to be metabolized into morphine before binding to the opiod recepter • It is the most powerful conscious analgesic • Opiates are also called narcotics as they act on the brain, are potent analgesics, cause changes in mood and behaviour and can result in addiction. • Blood-brain barrier: a series of lipophilic cell membranes that coat the blood vessels in the brain and prevent polar molecules from entering the central nervous system (CNS) • To be able to bond with the opioid receptors the opiates need to cross the blood-brain barrier and how well they do this depends on their solubility in water (blood) and lipids (brain) and on their chemical structure.Strong analgesics Codeine Morphine Diamorphine (heroin) Functional groups benzene ring ether alkene alcohol (hydrogen-bonding) tertiary amine benzene ring ether alkene alcohol (hydrogen-bonding) tertiary amine benzene ring ether alkene ester – ethanoate tertiary amine Solubility Soluble - cross the blood brain barrier slowly Most soluble - cross the blood brain barrier slowest 10x stronger than codeine Least soluble – cross the blood brain barrier fastest Structure 95 Synthesis of codeine In the synthesis of codeine from morphine one of the hydroxyl groups on the morphine is converted into a methyl ether group through a process called methylation. This also makes codeine less polar but more lipid soluble although it results in less binding with the opioid receptors – weaker analgesic. Synthesis of diamorphine (heroin) Diamorphine’s structure is only slightly different from morphine. Both the hydroxyl groups in the morphine molecule have been converted into ester groups. This is achieved by reacting the morphine with ethanoic acid; as a result an esterification occurs during which also water is produced. This action makes diamorphine more soluble in lipids and therefore more potent as a pain killer but also more addictive. Pros and Cons of using strong analgesic Advantages • • • • • • Disadvantages • Strong analgesics and therefore can relieve extreme pain Fast acting as can be administered intravenously Wider therapeutic window/wider safety margin Relieves anxiety Induces relaxation/feeling of well-being. High bioavailability. • • Addictive/habit-forming or physical dependence which leads to withdrawal symptoms As heroin is often taken by injections many addicts get infections such as HIV and hepatitis as a result of sharing needles. Tolerance can become an issue with this type of drug as more of the drug needs to be taken to achieve the same effect; in order to achieve the desired effect heroin users may take doses which exceed the lethal dose. Side effects and addiction: • Constipation, loss of sex drive, poor appetite, induce sleep, addictive, constriction of pupil in the eye, depression, kidney and liver disorder • Withdrawal symptoms: Perspiration, diarrhoea, cramps, acute feelings of distress, fever Increase bioavailability of heroin The tertiary amine in the heroin structure is basic. So heroin can reactive with hydrochloric acid to become soluble (increase bioavailability). + HCl 96 D4 pH regulation of the stomach U1 Non-specific reactions, such as the use of antacids, are those that work to reduce the excess stomach acid. U2 Active metabolites are the active forms of a drug after it has been processed by the body. A1 Explanation of how excess acidity in the stomach can be reduced by the use of different bases. A2 Construction and balancing of equations for neutralization reactions and the stoichiometric application of these equations. A3 Solving buffer problems using the Henderson–Hasselbalch equation. A4 Explanation of how compounds such as ranitidine (Zantac) can be used to inhibit stomach acid production. A5 Explanation of how compounds like omeprazole (Prilosec) and esomeprazole (Nexium) can be used to suppress acid secretion in the stomach. Indigestion • A pH of less than 1.0 can also cause ulceration or damage (breaking down of tissue) to the lining of the stomach walls. • Symptoms of acid indigestion and heartburn can be relieved by either increasing the pH of the stomach by: • • • Reducing the effect of the excess acid after it has been released in the stomach by using antacids to neutralize some of the excess acid. Antacids have an immediate effect but only last for a short-term Preventing the production of the excess acid by using H2 –receptor antagonists or proton pump inhibitors. Both these antagonists and proton pump inhibitors take a longer time to provide relief but have a longer term effect. They can also be used to treat ulcers. Antacids Antacids are usually weakly basic compounds, often metal oxides or hydroxides, carbonates or hydrogencarbonates, which react with the acid to produce a salt and water e.g. Al(OH)3 (s) + 3HCl(aq) → AlCl3 (aq) + 3H2 O(l) Mg(OH)2 (s) + 2HCl(aq) → MgCl2 (aq) + 2H2 O(l) Some antacids use carbonates, which produce carbon dioxide during neutralization, which can cause stomach bloating and flatulence (too much gas in the stomach). To avert this, antifoaming agents are added such as dimethicone. Some antacids also contain alginates which float to the top of the stomach, forming a ‘raft’ which acts as a barrier preventing reflux into the oesophagus (heart burn). Buffer in the stomach • As antacids are usually weak bases that are added to a strong acid in the stomach a buffer is created. Calculating the pH (or pOH) of a buffer pH = pπΎπΎππ + log [salt] [salt] & pOH = pπΎπΎππ + log [acid] [base] Use Henderson-Hasselbalch equation to calculate the pH of the buffer. H2 receptor antagonists • The acidity of gastric juice can be controlled at cellular level by targeting biochemical mechanisms of acid production • The secretion of acid in stomach is triggered by histamine binding to H2 receptors in parietal cells of gastric lining • Ranitidine (Zantac) blocks H2 receptors and reduce secretion of acid. It also provides short term relief of symptoms of indigestion and require frequent administration • The drug competes with the histamine to interact with the H2 receptors. • Using this method have slow effects but effective in the long term. Proton pump inhibitors • Omeprazole (Prilosec) and esomeprazole (Nexium) reduce production of stomach acid by inhibiting enzyme known as gastric proton pump in parietal cells, which is directly responsible for secreting H+ ions into gastric juice. In contrast to ranitidine, these provide longer relief of up of three days • Omeprazole and esomeprazole have same molecular formula (C17H19N3O3S) and differ only in their stereoisomeric structure • These structures are chiral and exist as two enantiomers 97 • • Omeprazole is a racemic mixture of both enantiomers while esomeprazole is a single enantiomer Both enantiomers have low polarity and thus easily cross cell membrane, and undergo chemical transformations in the acidic environment to produce active metabolites which then bind to proton pump enzymes. This increases efficiency of drug and thus requires less frequency of administration Active metabolites An active metabolite is an active form of a drug that has been administered in an inactive form (called prodrug) because of a number of reasons. The inactive form is then metabolized by the body into its active form (bioactivated) that has a greater therapeutic effect than the inactive form. Possible reasons for using an inactive form include: • Has greater bioavailability because it is more soluble or is absorbed faster. • Easier to administer. • More selective in its interaction with healthy cells. • Has fewer side effects linked to the administration. • Can be stored longer. • Can withstand different storage conditions. Mechanism of H2 receptor antagonists/ Proton pump inhibitors 98 D5 Antiviral medications U1 Viruses lack a cell structure and so are more difficult to target with drugs than bacteria. U2 Antiviral drugs may work by altering the cell’s genetic material so that the virus cannot use it to multiply. Alternatively, they may prevent the viruses from multiplying by blocking enzyme activity within the host cell. A1 Explanation of the different ways in which antiviral medications work. A2 Description of how viruses differ from bacteria A3 Explanation of how oseltamivir (Tamiflu) and zanamivir (Relenza) work as a preventative agent against flu viruses. A4 Comparison of the structures of oseltamivir and zanamivir. A5 Discussion of the difficulties associated with solving the AIDS problem. Viruses vs. bacteria Bacteria Virus bacteria are self-reproducing i.e. by cell division – do not need a host viruses are not self-reproducing as they need a host cell to multiply; viruses insert DNA into host cells – after reproduction the host cell dies bacteria are able to grow, feed and excrete viruses lack any metabolic functions so they do not grow, feed or excrete bacteria contain organelles such as cytoplasm, cell wall and a nucleus which all perform specific functions viruses consist only of genetic material and protective coating, no cell wall, no nucleus and no cytoplasm bacteria are (many times) larger than viruses viruses are smaller than bacteria bacteria mutate/multiply slower than viruse viruses mutate/multiply (much) faster than bacteria As viruses lack the same cell structures as bacteria, antibacterials are ineffective; in addition viruses also live inside host cells they are also more difficult to target by drugs. The action of antiviral drugs • Blocking the virus from entering the host cell by causing changes in the cell membrane of the host cell. • Stop uncoating of virus and injection of viral DNA into cell • Prevent biosynthesis of viral components (zidovudine) • Stop the release of viruses from the cell (Oseltamivir, Zanamivir) Oseltamivir (Tamiflu) and Zanamivir (Relenza) • The flu virus has an enzyme called neuraminidase that binds with the active site in a substrate molecule called sialic acid that is part of the cell membrane. • This provides a pathway with a lower activation energy for a reaction that allows new viral particles (after multiplication) to leave the host cell and infect the rest of the body. • Two of these neuraminidase inhibitors are oseltamivir (Tamiflu) and zanamivir (Relenza) as they each have a similar structure to the sialic acid in the cell membrane and can therefore also bond with the active site on the neuraminidase of the virus preventing it from bonding with the sialic acid in the cell membrane. • Oseltamivir is inactive in original form due to ester group. In water, this is hydrolyzed into a carboxyl group producing active metabolite with enhanced antiviral activity • Both oseltamivir and zanamivir are very similar in structure. They both contain a six membered ring with three chiral carbons. However they have different side chains containing different functional groups ultimately affecting the properties of these drugs. oseltamivir zanamivir ether ether primary amine primary amine amide amide ester carboxylic acid hydroxyl (3) 99 Some bacterial resistance More bacterial resistance Oral Inhalation HIV/ADIS • HIV invades white blood cells or CD4+ T cells and causes the disease AIDS that causes the failure of the immune system but this also allows other life-threatening diseases such as pneumonia and cancer to affect the person carrying the HIV. • The HIV uses its genetic information in the form of RNA to instruct the while blood cell’s DNA to produce new viral particles. • HIV is an example of a retrovirus which uses reverse transcriptase to produce DNA strands from their RNA genomes. Difficulties of treating HIV • • • • • Extremely fast at replication Targets white blood cells which are in fact responsible for preventing attacks on the body HIV is able to incorporate itself into host DNA where it can remain dormant for years High price of antiretroviral drugs, cost to state, access to drugs. Increase drug use, prostitution, lack of condoms during sex Zidovudine • Since reverse transcriptase is used only by viruses, its inhibition does not affect normal cells • The antiviral drug called zidovudine uses this technique to combat AIDS and prevent HIV transmission • However, zidovudine does not eliminate HIV completely, allowing virus to become resistant over time • Thus, zidovudine is used in combination with other inhibitors to effectively target viruses D6 Environmental impact of some medications U1 High-level waste (HLW) is waste that gives off large amounts of ionizing radiation for a long time. U2 Low-level waste (LLW) is waste that gives off small amounts of ionizing radiation for a short time. U3 Antibiotic resistance occurs when micro-organisms become resistant to antibacterials. A1 Describe the environmental impact of medical nuclear waste disposal. A2 Discussion of environmental issues related to left-over solvents. A3 Explanation of the dangers of antibiotic waste, from improper drug disposal and animal waste, and the development of antibiotic resistance. A4 Discussion of the basics of green chemistry (sustainable chemistry) processes A5 Explanation of how green chemistry was used to develop the precursor for Tamiflu (oseltamivir). Green Chemistry • Green chemistry or sustainable chemistry is an approach to carrying out chemical processes such as the manufacture, administration and disposal of pharmaceuticals that involves some of the following: • Considering the atom economy in the manufacturing process – making the processes as effective as possible by maximizing raw materials. 100 • • • • • • Reducing the amount of waste or avoiding waste all together. Reduce amount of hazardous waste e.g. reduce use of radioisotopes for instance in diagnostic medicine by using alternatives such as dyes. Safe disposal of waste. Using solvents safely in the manufacture or extraction. Considering implications to human health of the synthesis and extraction processes. Reducing the impact of the pharmaceutical industry on the environment. Solvent Waste • The synthesis and extraction of drugs often involves the use of solvents as they can provide a medium in which the synthesis occurs or they are used to extract the product. (e.g. using solubility) that need to be disposed off at the end of the synthesis or extraction. • Chlorinated solvents ( CHCl , CCl , CH Cl ) present the hazard of causing ozone depletion and contribute to formation of “photochemical smog” • Issues with the use of solvents concern: 3 • • • • 4 2 2 Health issues of the solvent itself to the workers e.g. is the solvent carcinogenic or toxic. Safety issue with the synthesis or extraction process in which the solvent is used e.g. solvent can be explosive or could form toxic by-products as a result of the process. Environmental impact of solvent use and disposal e.g. emissions in the air and water of chlorinated solvents. Some are toxic to animals and plants. Greenhouse effect. Flammable. Possible solutions to green use of solvents: • • • Modify the synthesis or extraction so less solvent is used. Use an alternative safer solvent or one with zero environmental impact. Reuse and recycle solvents. Nuclear Waste • Nuclear chemistry is used often in the diagnosis and treatment of diseases and this approach often produces hazardous radioactive waste that needs to be disposed off. • Many medicinal procedures involve use of radionuclides which are unstable isotopes of certain elements that undergo spontaneous radioactive decay • Ionizing radiation is the hazardous radiation emitted by radioactive materials. • The radiation can cause changes in DNA, which will lead to cancer development. • It can also cause reproductive problems in human. High-level waste • • • • • • Low-level waste Emit large amounts of ionizing radiation/high activity. High environmental impact Long half-life • Bury underground in concrete chambers for thousands of years Vitrification: fuse with glass Incorporate with minerals • • • • • Small amounts of ionizing radiation emitted/low activity. Limited environmental impact Short half-life Treat as normal waste Store in the steel container for several weeks Except 60Co – bury underground due to long half-life Shikimic acid • Shikimic acid is the precursor (= a chemical from which a more active chemical is produced) in the production of oseltamivir(Tamiflu) and is found in low concentrations in a variety of plants. • For many years shikmic acid was extracted from a natural source but with very low yield. • Green scientists then produce shikimic acid from E. coli effectively solving shortages of oseltamivir in future 101 D7 Taxol—a chiral auxiliary case study (HL) U1 Taxol is a drug that is commonly used to treat several different forms of cancer. U2 Taxol naturally occurs in yew trees but is now commonly synthetically produced. U3 A chiral auxiliary is an optically active substance that is temporarily incorporated into an organic synthesis so that it can be carried out asymmetrically with the selective formation of a single enantiomer. A1 Explanation of how taxol (paclitaxel) is obtained and used as a chemotherapeutic agent. A2 Description of the use of chiral auxiliaries to form the desired enantiomer. A3 Explanation of the use of a polarimeter to identify enantiomers. Taxol • Paclitaxel (Taxol) is a drug used to treat several different forms of cancer • It is obtained from the bark of pacific yew tree • However, the yield of taxol from natural yew tree is so low 0.004%. • Also, it is insoluble, so hard for intravenous injection (into skin) • Has 11 chiral carbons so extremely difficult to replicate Semi-synthetic production • The semi-synthetic method of production taxol used 10-deacetylbaccatin (precursor of taxol) • 10-deacetylbaccatin production has the yield of 0.2%, which is still 50x higher than extraction from yew tree. • 10-deacetylbaccatin can be converted to taxol by several synthetic steps including condensation reactions • For intravenous, a mixture of the drug with chemically modified castor oil and ethanol was diluted with normal saline solution immediately before injection Chiral auxiliaries • Chiral auxiliary is an optically active substance that is temporarily incorporated into an organic synthesis so that it can be carried out asymmetrically with the selective formation of a single enantiomer • A chiral auxiliary is used to convert a non-chiral molecule into just the desired enantiomer, thus avoiding the need to separate enantiomers from a racemic mixture. This is a chiral molecule which binds to the reactant, physically blocking one reaction site through steric hindrance, so ensuring that the next step in the reaction can only take place from one side. This effectively forces the reaction to proceed with a specified stereochemistry. Once the specific enantiomer of the new product has been set, the auxiliary can be taken off and recycled. • • • Identity and purity of chiral compounds can be determined using a polarimeter Polarimeter is rotated until the maximum amount of light passes through in order to identify specific enantiomer. The racemix is optically inactive. 102 D8 Nuclear medicine (HL) U1 Alpha, beta, gamma, proton, neutron and positron emissions are all used for medical treatment. U2 Magnetic resonance imaging (MRI) is an application of NMR technology U3 Radiotherapy can be internal and/or external. U4 Targeted Alpha Therapy (TAT) and Boron Neutron Capture Therapy (BNCT) are two methods which are used in cancer treatment. A1 Discussion of common side effects from radiotherapy. A2 Explanation of why technetium-99m is the most common radioisotope used in nuclear medicine based on its half-life, emission type and chemistry. A3 Explanation of why lutetium-177 and yttrium-90 are common isotopes used for radiotherapy based on the type of radiation emitted. A4 Balancing nuclear equations involving alpha and beta particles. A5 Calculation of the percentage and amount of radioactive material decayed and remaining after a certain period of time using the nuclear half-life equation. A6 Explanation of TAT and how it might be used to treat diseases that have spread throughout the body. Radiation: Alpha Emitter: ππππππ ππ ππ+ ππππππ ππππππππ → ππππππππ + ππππ ππ− Emitter ππ+ Emitter (positron) ππ − ππππ ππππ ππππππππ → ππππππππ + −ππππ ππππ ππ + ππππ ππππ → ππππ + ππππ Half-life • The amount of time taken for half of radioactive substance to decay • Denoted as t 1 2 • • ππ – is the decay constant λ= In2 t1 2 Larger decay constant – faster the rate of decaying, meaning the compound has shorter half-life. t • • Nt = N0 (0.5)k Nt means the amount of radioactive substance left at time t. N0 means initial concentration. K is the rate constant. The equation can also be change to Nt = N0 e−λt 103 Diagnostic techniques in nuclear medicine • Diagnostic techniques usually involve first attaching a radionuclide(unstable nucleus), known as a tracer, to a biologically active molecule, making a drug called a radiopharmaceutical. • Tracer targets specific part of the body. e.g. iodine for the thyroid gland and glucose for the brain • The tracers used in radiopharmaceuticals must emit gamma rays with enough energy to escape from the body and must have a half-life just long enough for the scan to be complete before its decay. ππππ • The radiopharmaceutical most widely used in diagnosis is technetium-99m, ππππ ππππ • • • Its half-life is 6 hours, which means that activity in the body stays high for long enough for metabolic processes to be examined by scanning, but also decays quickly enough to minimize the exposure to the patient. Its decay involves the release of gamma rays and low-energy electrons. Without high energy beta emission, the radiation dose is low. Low-energy gamma rays escape the body and are accurately detected by the gamma camera. Technetium is chemically versatile, so acts as a tracer by bonding to a range of biologically active substances. These are chosen according to the organ to be studied. Positron emission tomography (PET) • This is a type of scanner that gives three-dimensional images of tracer concentration in the body. • Positron emitter 11C is coated with sugar, which is injected into the patient body. • Since cancer cells grow uncontrollably, they take in much sugar. Radioactive sugar will accumulate in the cancer cells • The image was taken and the image of the organ is formed. Magnetic resonance imaging (MRI) • MRI is an application of nuclear magnetic resonance (NMR) spectroscopy • Hydrogen atoms, 1H, have a magnetic moment due to their odd number of protons. In the presence of a powerful magnet, radio waves are used to generate an electronic signal that can be decoded by a computer • MRI does not use ionizing radiation • MRI scans give detailed images of almost any part of the body and are widely used in cancer detection External Radionuclide therapy (Gamma Ray Therapy) • An external source of radiation is directed at the site of cancer in the body from a radioactive source, usually cobalt-60. This undergoes beta decay producing the stable product nickel-60. • • 60 60 0 27 Co → 28 Ni + −1β + γ ray A beam of gamma radiation is focused on a tumor target The gamma rays destroy cells within the small focused area. Other areas will not be hurt since they are exposed to very small amount of radiation. Target Alpha Therapy (TAT) • Alpha emitter is coated with protein. • Those labeled protein enters the cancer cells • The alpha emitter then is activated – killing the cells around the emitter. • Alpha ray will ionized in a short distance, so it will only kill the cells around it. • TAT can also map the distribution of cancer in the body. So it can also serve as a diagnosis tool. • Boron neutron capture therapy (BNCT) • The entire molecule contains non-radioactive isotope boron-10 hidden in amino acid • The amino acid is absorbed into the cancer cells. • Then the molecule is hit with a beam of neutrons, turning boron-10 into boron-11, which is an alpha emitter. 104 Boron-11 will decay into lithium-7, which is toxic to the cells. • The alpha ray can also kill surrounding tissues. The general formula for BNCT • 10 11 1 7 4 5B + 0n → 5B → 3Li + 2α Side effects of radiotherapy • fatigue – rest and regular hydration are important during treatment • nausea – more common when the treatment is in the area of the digestive system • hair-loss – this occurs within the treatment area and is usually temporary • sterility – more likely if treatment is close to ovaries or testes • skin reaction – skin may become red, sore, or itchy in local area of radiation. yttrium-90 and lutetium-177 • They are pure beta emitter • Yttrium-90 is a common radiation source or cancer brachytherapy and palliative treatment of arthritis. • Brachytherapy: the radioactive source is planted near the tumour site • Lutetium-1 77 produces low-energy beta particles with reduced tissue penetration, which is very useful in the targeted therapy of small tumours. • lutetium-1 77 emits just enough gamma rays for visualizing tumours and monitoring the progress of their treatment D9 Drug detection and analysis (HL) U1 Organic structures can be analysed and identified through the use of infrared spectroscopy, mass spectroscopy and proton NMR. U2 The presence of alcohol in a sample of breath can be detected through the use of either a redox reaction or a fuel cell type of breathalyser. A1 Interpretation of a variety of analytical spectra to determine an organic structure including infrared spectroscopy, mass spectroscopy and proton NMR. A2 Description of the process of extraction and purification of an organic product. Consider the use of fractional distillation, Raoult’s law, the properties on which extractions are based and explaining the relationship between organic structure and solubility. A3 Description of the process of steroid detection in sport utilizing chromatography and mass spectroscopy. A4 Explanation of how alcohol can be detected with the use of a breathalyser. Extraction and purification of products • The partition of a solute between two immiscible liquids can be described as a heterogeneous equilibrium between different states of the same compound. • When iodine is partitioned between water and organic solvent I2 (aq) β I2 (org) • • • • • • The equilibrium constant of this reaction is partition coefficient Pc Pc = [I2 (org)] [I2 (aq)] Fractional distillation is another common method of isolation and purification of organic compounds Raoult’s law: p(A) = p∗ (A) β x(A) p(A) is the vapour pressure of A over the mixture (partial pressure) at a given temperature p*(A) is the vapour pressure over a pure sample of A at the same temperature x(A) is the mole fraction of A, which is the ratio of the amount of A to the sum of the amounts of all components in the mixture In the boiling mixture of several substances, the more volatile the molecule is, the easier it will vapourised. 105 • • The mixture of liquid will re-condense inside the distillation column. The more volatile molecule will have more mole fraction in the condensation; thus, it will gain higher partial pressure. Overall, the volatile molecule will gradually move up the distillation column. Steroid detection in sport • Athletes may use steroids to enhance their performance • Anabolic steroids are predominantly non-polar compounds, so they can be extracted from biological materials with organic solvents and concentrated for further studies. • Each steroid produces a characteristic mass spectrum which can be compared form the library • Athletes metabolites can be detected by a combination of gas chromatography (GC) or high performance liquid chromatography (HPLC) • gas chromatography separates the chemical mixture into pure chemicals • mass spectroscopy identifies and quantifies the components. Breathalyzer • The simplest breathalyzer consists of a glass tube filled with acidified crystals of potassium dichromate(VI) . • Alcohol can be oxidized into ethanol or ethanoic acid, which will change the colour of potassium dichromate from orange to green. • Another type of breathalyzer uses a fuel cell in which ethanol is oxidized at the anode by atmospheric oxygen on the surface of platinum electrodes. C2 H5 OH + H2 O → CH3 COOH + 4H + + 4e− O2 + 4H + + 4e− → 2H2 O • The electric current produced by the fuel cell is proportional to the concentration of ethanol in the breath Gas chromatography • The basic principle is that the components have different affinities for two phases, a stationary phase and a mobile phase • stationary phase – a microscopic layer of a non-volatile liquid, usually a polymer, which is coated on the walls of an inert solid support • mobile phase – an inert carrier gas, such as helium • Separation of the components of a mixture is determined by the different rates at which they move through the instrument • Depends on boiling points and solubility, molecules partition themselves between two phases. • Molecule spend more time in gas phase will move faster inside the tube. • Each component of the mixture will be eluted at a specific interval of time, known as its retention time. • In some cases, the eluted sample is then analysed by mass spectrum.\ • Gas Chromatography is used to separate different molecules inside a mix sample. • It can also use retention time to roughly identify a product. • Together with IR and MS, GS will give a more accurate reading of alcohol level. 106
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