Chemistry A Level Flashcards

Question

How do you predict spontaneity by looking at the ΔH and ΔS?

Answer 1

- When ΔH is negative and ΔS is positive: Using the equation, ΔH is negative and TΔS is positive so -TΔS is negative. Regardless of temperature, ΔG is always negative so the reaction is spontaneous at all temperatures. - When ΔH is positive and ΔS is negative: Using the equation, ΔH is positive and TΔS is negative so -TΔS is positive. Both terms are positive regardless of temperature so ΔG is always positive and the reaction is never spontaneous. - When ΔH and ΔS are positive: Using the equation, ΔH is positive and TΔS is positive so -TΔS is negative. Increasing the temperature causes -TΔS to get more negative. At high temperatures, -TΔS will outweigh ΔH and ΔG will be less than 0 so the reaction will be spontaneous. The reaction won’t be spontaneous at low temperatures. - When ΔH and ΔS are negative: Using the equation, ΔH is negative and TΔS is negative so -TΔS is positive. At high temperatures, -TΔS becomes more positive so will outweigh ΔH meaning the reaction won’t be spontaneous. The reaction will be spontaneous at low temperatures.

Answer 2

- A reaction or process is spontaneous/feasible when ΔG is less than or equal to 0. - If ΔG is positive, the reaction is not spontaneous at the temperature used in the calculation.

Answer 3

ΔG = ΔH - TΔS ΔG - Gibbs free energy change (kJ mol-1) ΔH - enthalpy change (kJ mol-1) T - temperature (K) ΔS - entropy change (kJ K-1 mol-1)

Answer 4

- To find the minimum temperature at which a reaction is spontaneous when given ΔH and ΔS, rewrite the equation as: ΔH - TΔS < 0 and rearrange to find temperature.

Answer 5

Electrolysis is the decomposition of a compound into its element by an electric current

Answer 6

- Electrolyte is a compound that is decomposed (molten/ionic) - Electrodes are the rods which allow current to enter electrolyte (inert/reactive)

Answer 7

When the electrolyte is molten, cations (positive ions) move to the cathode and gain electrons to form atoms. Anions move towards the anode, lose electrons and become atoms

Answer 8

- At the cathode: positive ions reduced to atoms Pb2+(l) + 2e- → Pb(s) - At the anode: negative ions oxidized to atoms/molecules 2NO3-(l) → NO2(g) + O2(g) + 2e- SO42-(l) → SO2(g) + O2(g) + 2e-

Answer 9

Standard electrode potential (EӨ): the emf (electromotive force) of a half cell compared with a hydrogen half cell under standard conditions.

Answer 10

- When using inert electrodes to electrolyse a solution, if the metal ions are below hydrogen in the electrochemical series (more positive EӨ), the metal atoms will be produced at the cathode. - If the metal is above hydrogen in the electrochemical series, hydrogen gas will be produced at the cathode. - Metals from lead to zinc in the electrochemical series depend on the concentration, with the more concentrated ions becoming atoms. - A very dilute solution will form hydrogen and when the metal ion concentrations are similar, both may form atoms.

Answer 11

- When electrolysing a solution with non-inert electrodes, the same reaction takes place at the cathode. - At the anode, atoms in the anode lose electrons to form ions in the electrolyte. - The net charge is a transfer of an element from the anode to the cathode.

Answer 12

Aqueous Electrolyte: At the cathode By Eθ - Eθ > 0‎ then metal less reactive than H so metal formed - Eθ < 0‎ then metal more reactive than H so H2(g) formed By Concentration - Following results from above, if solution very dilute, this occurs greater 2H+(aq)+2e- → H2(g)

Answer 13

By Eθ - Eθ > 0.4‎ then non-metal less reactive than OH so non-metal formed - Eθ < 0.4‎ ‎ then non-metal more reactive than OH so H2O(l) formed By concentration 4OH−(aq) → 2H2O(l) + O2(g) + 2eZ-2

Answer 14

Standard conditions: - Temperature: 298 K - Pressure: 1 atm (101 kPa) - Concentration: 1.00 mol dm-3

Answer 15

Standard cell potential (EӨcell): the emf (electromotive force,) when two half cells are connected under standard conditions

Answer 16

Emf: electromotive force, the voltage when no current flows

Answer 17

The Faraday constant (F) is equal to the Avogadro constant (L) multiplied by the charge of an electron in coulombs (e): F = Le 1 faraday = 9.65 x 104 C mol-1 (coulombs per mole) - A coulomb is a quantity of electricity. Number of coulomb = current (amps) x time (seconds). - Avogadro’s constant is the number of particles in one mole of any substance. This is equal to 6.02 x 10^23

Answer 18

Number of coulomb (Q) = current (amps) (I) x time (seconds) (t) Q=It - Mass of a substance liberated is proportional to the quantity of electricity which passes through electrolyte

Answer 19

1. Multiplying current (amps) by time (seconds) to find the number of coulombs. 2. Convert the number of coulombs to faradays using the conversion 1 faraday = 96500 C mol-1 3. Divide the Faraday constant by the Avogadro constant to find the quantity of charge in coulombs

Answer 20

1. Multiply the current (amps) by the time (seconds) to find the number of coulombs. 2. Use the cathode reaction, - 1 mol of electrons forms 1 mol of silver 3. Calculate the number of moles by dividing the mass change by the atomic mass of the atom. 4. Find the number of coulombs in one mole of electrons by dividing the calculated number of coulombs from step 1 by the number of moles in step 3. 5. To calculate the Avogadro constant, divide the number of coulombs in one mole of electrons by the charge on an electron

Answer 21

EӨ cell = EӨ reduction (positive terminal) - EӨ oxidation (negative terminal).

Answer 22

Half cells with more negative electrode potentials are more likely to be oxidised and lose electrons. When two half cells are connected, the more negative half cell will always undergo oxidation

Answer 23

- A cell can be made by connecting two half cells with different electrode potentials. One half cell will release electrons while the other will gain them. - The half cell with the more positive EӨ (positive terminal) will gain electrons while the half cell with the more negative EӨ (negative terminal) will release electrons

Answer 24

- It is a reference cell, used to measure Eθ of any other half cell - Consists of H2(g) at 298K and 1atm bubbling around a Pt electrode in contact with an aq. solution of H+ ions at 1.00 mol dm-3 conc. (e.g. HCl(aq) or ½ H2SO4(aq) - Pt electrode is inert, allows conduction and is coated with finely divided Pt serving as a catalyst. Being porous, retains large amounts of H2(g)

Answer 25

- Used to complete the electrical circuit allowing the movement of ions between two half cells so that ionic balance is maintained; electrically neutral - A strip of filter paper soaked in KNO3(aq) - Salt used should not react with ions forming a ppt. e.g. KCl in a cell containing Ag2+ would form insoluble ppt.

Answer 26

Each half cell must contain one element in two oxidation states. To create a closed circuit and allow charge to flow, a salt bridge must be placed between the two solutions and a wire must connect the two electrodes. A voltmeter measures the potential difference of the cell.

Answer 27

A reaction is feasible, or spontaneous, if the cell potential is positive

Answer 28

- As the value of EӨ becomes more positive, the position of the equilibrium lies further to the right and the element is more likely to be reduced. - An oxidising agent is a species which oxidises another species while being reduced. - Fluorine is the most powerful oxidising agent in the group because it has the most positive EӨ value.

Answer 29

- Two platinum coated porous electrodes that allow gases to pass through - Electrolyte; either acidic or alkaline

Answer 30

- Negative Electrode (anode): 2H2(g) 4H+(aq) + 4e- - positive Electrode (cathode): O2(g) + 4H+(aq) + 4e- 2H2O(l) Overall reaction: O2 + 2H2(g) 2H2O(l) - H+ ions removed at positive electrode replaced by those produced at negative electrode so concentration of electrolyte remains constant

Answer 31

- Negative Electrode (anode): 2H2(g) + 4OH-(aq) 4H2O(l) + 4e- - Positive Electrode (cathode): O2(g) + 2H2O(l) + 4e- 4OH-(aq) - Overall reaction: O2 + 2H2(g) 2H2O(l) - OH- ions removed at -ve electrode replaced by those produced at +ve electrode so conc. of electrolyte remains constant

Answer 32

- No pollutants – by product is water - More energy per gram of fuel than other fuels - Lightweight - Operate with high efficiency

Answer 33

Half equations can be combined to form a redox equation. The number of electrons on both sides of the redox equation must be the same so that the electrons are cancelled out.

Answer 34

1. Half equations 2. Decide the positive and negative terminals: The more positive standard electrode potential so this is the positive terminal and the forward reduction reaction occurs. It is more negative and so undergoes oxidation. 3. Balance the electrons 3. Combine equations 4. Cancel species that appear on both sides

Answer 35

The Nernst equation is used to calculate the electrode potential of a half cell when the concentration of aqueous ions changes.

Answer 36

E = Eθ + (0.059/z) log ([oxidised form] / [reduced form])

Answer 37

- If concentration less than 1 mol dm-3,log[oxidized form] is negative and E is less than Eθ - If conc. more than 1 mol dm-3, log[oxidized form] is positive and E is more than Eθ

Answer 38

When acids and alkalis are added to water, they dissociate/ionize and a dynamic equilibrium is set up

Answer 39

An acid and a base that are only different by one proton are called a conjugate acid-base pair

Answer 40

- HA(aq) + H2O(l) ‎→‎ H+(aq) + A-(aq) - Equation position to the right - Dissociates completely ‎therefore [Acid] = [Ions]

Answer 41

- HA(aq) + H2O(l) ‎⇌‎ H+(aq) + A-(aq) - Equation position to the left - Dissociates partially ‎therefore [Acid] ‎≠‎ [Ions]

Answer 42

pH = -log10[H+] [H+] = 10-pH

Answer 43

Ka = [H+][A−] / [HA] - For the equation HA ⇋ H+ + A - Ka is the acid dissociation constant. The lower the Ka value, the further left the equilibrium lies. - For weak acids, the Ka is very small so Ka values can be converted into pKa values - As pKa increases, the strength of the acid decreases.

Answer 44

- pKa = -log10(Ka) - As pKa increases, the strength of the acid decreases.

Answer 45

- Gives strength of acid depending on H+ - Use –log of values as they are very small giving pH - Lower pH value, stronger the acid

Answer 46

−log[H+]=pH ⇔ [H+]=10−pH

Answer 47

1. Calculate the concentration of hydroxide ions. 2. Calculate the concentration of H+ ions using Kw 3. Convert [H+] into pH, pH = -log10[H+].

Answer 48

1. For the equation HA(aq) ⇋ H+(aq) + A-(aq): - Assume that a very small number of hydrogen ions come from the dissociation of water, so [H+] = [A-] - Assume that the concentration of acid at equilibrium is the same as the original concentration because it is a weak acid so little dissociation has occurred. 2. Rewrite Ka, Ka = [H+]^2 / [HA] 3. Rearrange Ka before entering the values to calculate [H+] 4. Convert [H+] into pH

Answer 49

Kw = [H+][OH-] - Kw is the ionic product of water. The value of Kw varies with temperature. - At room temperature, this is equal to 1.00 x 10-14 mol2 dm-6

Answer 50

Kc = [H+][OH−] / [H2O] - Water molecules can function as both acids and bases; one molecule (base) accepts H+ from second one (acid) H2O(l) ⇌ H+(aq) + OH-(aq)

Answer 51

−log throughout equation therefore −log1×10−14=14∴ 14 = pH + pOH

Answer 52

A solution in which pH doesn’t change significantly when small amounts of acids or alkali added

Answer 53

A buffer solution can be: - Acid: Weak acid + its salt e.g. CH3COH and CH3COONa - Alkali: Weak alkali + its salt e.g. NH3 and NH4Cl

Answer 54

An acidic buffer has a pH less than 7 and contains a large amount of weak acid and its conjugate base (from a salt) and relatively few H+ ions in equilibrium

Answer 55

Blood pH must remain between 7.35 and 7.45

Answer 56

An alkaline buffer has a pH greater than 7 and contains a weak base and its salt.

Answer 57

-Respiration in cells produces carbon dioxide and carbon dioxide combines with water in blood CO2(aq) + H2O(aq) ⇌ H+(aq) + HCO3-(aq) - The enzyme carbonic anhydrase is present to supply HCO3- ions and increase its concentration similar to acidic buffers

Answer 58

If pH increases (hydrogen ion concentration decreases), the position of equilibrium will shift to the right to increase the concentration of hydrogen ions. If pH falls (hydrogen ion concentration increases), the position of equilibrium will shift to the left to remove the added H+ ions.

Answer 59

CO2(aq) + H2O(aq) ⇌ H+(aq) + HCO3-(aq) - Equation shifts to the left - Reduces H+ - pH of blood unchanged

Answer 60

CO2(aq) + H2O(aq) ⇌ H+(aq) + HCO3-(aq) - Equation shifts to the right - Increases H+ pH of blood unchanged

Answer 61

pH=pKa +log([Salt] / [Acid])

Answer 62

For weak acids, two assumptions were made: 1) [H+] = [A-] - this is no longer true as the buffer contains a much larger concentration of A- . The concentration of A- is assumed to be the same as the concentration of the salt as the number of A- ions coming from HA is tiny (as it is a weak acid). 2) [HA] at equilibrium = [HA] at the start - this is assumed for the buffer because the added A- ions from the salt push the equilibrium to the left meaning there is little dissociation of the acid.

Answer 63

Its a product of concentration of each ion in a saturated solution of sparingly soluble salt at 298K raised to the power of moles in dissociation equation

Answer 64

- The concentration of ions is independent of amount of solid - Ksp values change only with temperature - Concentration of ions = solubility of salt

Answer 65

The solubility product for a compound will always be the same under the same conditions provided that the solution is saturated. If two compounds are mixed together, a precipitate will only form if the ionic concentrations give a value greater than the solubility product (otherwise, the solution won’t be saturated).

Answer 66

Precipitation is when two aq. solutions of ionic substance are mixed, a ppt. will form provided the Ksp of the insoluble solid is less than ionic product of ions present

Answer 67

The lowering of solubility of an ionic compound by addition of a common ion to the solution

Answer 68

- Titration is the method used to find conc. of a solution by gradually adding to it a second solution of known conc. until neutralization point is reached - Progress of neutralization reaction monitored using a pH meter and shown on a graph

Answer 69

Equivalence point: mid-point of near-vertical portion of the graph where no. of mols H+ = no. of mols OH- End point: volume of alkali/acid added when indicator just changes colour

Answer 70

- Sharp colour change at end point - End point must be ±2 pH of equivalence point - Indicator should give distinct colour change

Answer 71

Indicator - Phenolphthalein Methyl orange Colour at low pH - Colourless Red pH Range - 8.2 – 10.0 3.2 – 4.4 Colour at high pH - Pink Orange

Answer 72

- General solubility rule: like dissolves like - Covalent substances dissolve/mix in other covalent substances - Ionic substances do not dissolve/mix in other covalent substance (exception is water)

Answer 73

- Although covalent, highly polar nature makes it excellent solvent for ionic substances - Dissolves some covalent substances because they either react or form H-bonds with water molecules

Answer 74

- The Partition law is a solutes solubility will be differed in two different liquids that are immiscible. The solute will be more soluble in one of the two. - Measured in [conc.]: g cm-3 or mol dm-3

Answer 75

- Partition coefficient (Kpc): equilibrium constant that relates the concentration of solute partition between two immiscible solves at a particular temperature. - Kpc = Conc. of Solute in Solvent 1 / Conc. of Solute in Solvent 2 - Place concentration of the first solvent mentioned on top of the Kpc expression

Answer 76

- Two immiscible liquids will form two separate layers in a separating funnel with the less dense layer on top. - If substance X is dissolved in both of these liquids, it may be more soluble in one layer than another. - A dynamic equilibrium will be set up at the boundary between the two liquids: X(in more dense liquid) ⇋ X(in less dense liquid) There are no units for the partition coefficient because any units cancel out.

Answer 77

Rate equation: for the reaction A + B → C, the rate equation is rate = k[A]m[B]n (m is the order of the reaction with respect to A, n is the order of reaction with respect to B)

Answer 78

Rate constant: the constant (k) linking rate of reaction and the concentration of reactants. This value changes with temperature and use of a catalyst but remains the same if only concentrations of reactants are changed.

Answer 79

- Larger the rate constant, faster the reaction - Depends on temperature - Unit depends on overall order of reaction

Answer 80

Order of reaction: the power to which the concentration of a reactant is raised in the rate equation (this could be 0, 1 or 2)

Answer 81

Half-life: the amount of time taken for the concentration of the reactants to halve

Answer 82

- If a reaction is 0 order with respect to reactant P, the rate is unaffected by changing the concentration of P - The overall order of a reaction is equal to the sum of all the orders of the reactants

Answer 83

- If a reaction is first order with respect to reactant Q, doubling the concentration of Q with double the rate (rate ∝ [Q] or rate = k[Q]) - The overall order of a reaction is equal to the sum of all the orders of the reactants

Answer 84

- If a reaction is second order with respect to X, doubling the concentration of X will quadruple the rate (rate ∝ [X]2 or rate = k[X]2) - The overall order of a reaction is equal to the sum of all the orders of the reactants

Answer 85

- Only used with first order reactions - The half life (t½) of a first order reaction is constant. This means the time taken for the concentration of reactants to half is always the same. - The relationship between the rate constant (k) k= 0.693 / t½

Answer 86

- Increasing the temperature, increases the value of k - k= R/[A][B] - When temperature is increased, the k.e. of reacting molecules increases resulting in more successful collisions - Reactants change faster to products therefore conc. of reactants decreases. - Using fraction above, numerator increases and denominator gets smaller therefore kk increases

Answer 87

A reaction mechanism describes the steps involved in making and breaking bonds during a reaction.

Answer 88

Rate-determining step: the stage with the slowest rate in a multi-step reaction.

Answer 89

- The rate of reaction is dependent on the slowest step that needs the highest activation energy. - Rate equation includes only reactants that are present in the rate-determining step. - The orders with respect to the reactants are the moles of the reactants in the rate determining step

Answer 90

- The slowest step of a reaction ( the rate- determining step) must include reactants which have an impact on the rate of reaction when their concentrations are changed. - Zero order reactants have no influence on the rate of a reaction so they cannot be included in the rate-determining step.

Answer 91

- When constructing a reaction mechanism, the powers in the rate equation indicate the number of molecules of each substance involved in the slowest step - Any intermediates generated in the slowest step must be reactants in another step as they are not present in the balanced overall equation.

Answer 92

Sampling: method that involves taking small sample of a reaction mixture at various times and then carrying out chemical analysis on sample.

Answer 93

Colorimeter measures amount of light absorbed as it passes through solution; recorded as absorbance.

Answer 94

- Measure change in volume of gas using a gas syringe - Take down readings at regular intervals - Plot graph and calculate rate of reaction

Answer 95

Continuous is a method that involves monitoring a physical property over a period of time

Answer 96

- Before experiment, create calibration curve by finding absorbance of different concentration and plot a graph of concentration against absorbance - During experiment, measure absorbance from meter at regular intervals, and use calibration curve to convert values into concentrations

Answer 97

A homogeneous catalyst is in the same phase as the reactants while a heterogeneous catalyst is in a different phase to the reactants

Answer 98

- Forms intermediate with reactant - Intermediate breaks down giving product

Answer 99

A catalyst is a substance which speeds up the rate of a reaction without being chemically changed at the end

Answer 100

Rate of reaction dependant on conc. of catalyst

Answer 101

- Catalyst provides surface on which reaction occurs - Reactant particles adsorbed on surface on collision with catalyst - Molecular rearrangement occurs - bonds in reactants break and new bonds in product formed - Product molecules desorbed from surface

Answer 102

Rate of reaction dependant on surface area of catalyst

Answer 103

Adsorption: weak bonds formed between reactant molecules & surface of catalyst (diffusion on to surface)

Answer 104

Desorption: releasing of product molecules from surface of catalyst (diffusion away from surface)

Answer 105

- The area under the curve is equal to the total number of particles present. - The Boltzmann distribution is for gases - No particles have no energy. - There is no maximum energy. - Only particles with energy above the activation energy have sufficient energy to react when they collide.

Answer 106

- The reactions of Group 2 metals, metal oxides, metal hydroxides and metal carbonates generally indicate that reactivity increases as you go down Group 2. - Reactivity increases down the group because ionisation energy decreases due to the increasing atomic radius and the shielding effect of electrons. This means further down the group, the electrons become easier to remove so reactivity increases.

Answer 107

Some exceptions to this trend can be seen when sulfates and hydroxides are produced, however this is due to the insolubility of some sulfates and hydroxides inhibiting reactions.

Answer 108

Both carbonates and nitrates become more thermally stable as you go down Group 2.

Answer 109

The stability of the compounds is influenced by charge density of the cation and how polarised the anion becomes.

Answer 110

- Ions formed from elements at the top of group 2 are smaller than those at the bottom. - A smaller +2 ion has a greater charge density because the same charge is packed into a small volume. - This means smaller group 2 ions have a greater polarising effect on neighbouring negative ions.

Answer 111

- The more polarised the anion is, the less heat is required for thermal decomposition to occur. - This means that thermal stability increases down the group because ions further down the group are worse at polarising the anion so more heat energy is required to break bonds for thermal decomposition.

Answer 112

- The trend in the solubility of sulfates is opposite to the trend in the hydroxides - Compounds with very low solubilities are often said to be sparingly soluble. - Most sulfates are soluble in warm water except barium sulfate which is insoluble.

Answer 113

Group 2 element - X Hydroxide - X(OH)2 Magnesium Least soluble Calcium Beryllium Barium Most soluble

Answer 114

Group 2 element - X Sulfate - XSO4 Magnesium Most soluble Calcium Beryllium Barium Least soluble

Answer 115

As you go down Group 2, the lattice enthalpy required to break up the compound decreases because the size of the positive ions increases

Answer 116

-As the cations increase in size down the group, the enthalpy change of hydration (the amount of energy released as the ions bond to water molecules) also decreases.

Answer 117

Larger cations means there is more space between ions in the compound so there are weaker forces of attraction between the ions.

Answer 118

For hydroxide ions (relatively small ions), the lattice enthalpy falls faster than the enthalpy change of hydration of the cations. This means the enthalpy change of solution will become more negative down the group (more exothermic)

Answer 119

- For sulfate ions (relatively large ions), the lattice enthalpy falls slower than the hydration enthalpy of the cations. This means the enthalpy of solution will become more positive down the group (more endothermic). - The more exothermic the enthalpy of solution is, the more soluble a compound is. Therefore sulfates become less soluble down the group and hydroxides become more soluble down the group.

Answer 120

Transition elements are found in the d-block of the periodic table. Transition elements form one or more stable ions with incomplete d orbitals.

Answer 121

- The 4s orbital is filled before the 3d so the 4s electrons are removed before the 3d electrons. - Also remember that the electron configurations of copper and chromium are different than expected as the 4s orbital only has one electron. - This is so that the 3d subshell is filled for copper and half filled for chromium, as these are more stable electron configurations.

Answer 122

Scandium and Zinc are not transition metals because: - Sc3+ has no e-s in the d-orbital - Zn2+ contains a full d-orbital

Answer 123

- dxy, dxz, dyz have d-orbitals in between axis - dx2−y2 and dz2 have d-orbitals along axis - dz2 is formed by the merging of dx2−z2 and dy2−z2

Answer 124

- Transition metals have higher melting points that metals found in the s-block, like calcium. - This is because the transition metals have the extra 3d electrons so metallic bonding is stronger, requiring more energy to overcome

Answer 125

- As you go across the periodic table, nuclear charge increases. - This draws the electrons inwards slightly, decreasing the atomic radius. As a result, transition metals have higher densities than metals in the s-block because the atoms are smaller so atoms fit into a given volume compared with s-block metal atoms. - Density is also greater because atomic mass increases across the period.

Answer 126

- Transition metals can exist in variable oxidation states. - Variable oxidation states form because the electrons in the 4s and 3d orbitals have very similar energies. - This means that a relatively similar amount of energy is required to gain or lose different numbers of electrons. - Most common oxidation state +2 when 2e-s from 4s lost

Answer 127

- Transition elements show highest oxidation states when they combine with O or F (most electro-ve) - When transition elements form compounds with high oxidation states above +4, they form large oxyanions and are covalent (acidic oxides) - When transition elements in lower oxidation states they form ionic compounds (basic oxides)

Answer 128

A complex is an ion or molecule formed by a central metal atom/ion surrounded by one or more ligands

Answer 129

- Central transition metal ion (positive) that can accept e-s - Ligand (negative) is a species that contains a lone pair of e-s that forms a dative bond to a central metal atom/ion

Answer 130

Coordination number is a number of coordinate or dative bonds to the central metal atom/ion

Answer 131

- Ion are small in size so they have a strong electric field around them which attract e--rich ligands - They have empty 4s and 4p orbitals that are hybridised and can accept e- - Different metal ions show different coordination number with same ligands

Answer 132

Ligand: a species that has a lone pair of electrons that forms a dative covalent bond with a central metal atom or ion

Answer 133

Monodentate ligands: forms only one coordinate bond with central metal ion (donates one pair of e-s)

Answer 134

When ligands swap places with one another, this is called ligand exchange or ligand substitution. The coordination number and shape of the ion stays the same if the ligands are a similar size and charge.

Answer 135

Bidentate ligands: forms 2 coordinate bonds with central metal ion (donates 2 pairs of e-s per molecule)

Answer 136

- If neutral and -ve ligands present, first name -ve ligand - If ligands all neutral -ve or contain more than one of each, name in alphabetical order

Answer 137

- For monodentate ligands, use prefixes mono-, di-, tri-, tetra-, hexa- - For bi and polydentate ligands, use prefixes mono-, bis-, tris-, tetrakis-, pentakis-

Answer 138

- If complex ion +ve/ neutral – use normal name - If complex ion –ve, use name ending –ate; special: - Iron = Ferrate - Lead = Plumbate - Copper = Cuprate - Oxidation no. of metal ion in Roman numeral and ‘ion’

Answer 139

- Metal ion written first followed by ligand - Place charge on formula in square brackets - Total charge on complex = sum of charge on metal ions and charge on ligand - If ligands all neutral, charge on complex = charge on metal ion

Answer 140

Transition metal complexes can be described as linear, octahedral, tetrahedral or square planar, depending on the ligands and the metal ions in the complex

Answer 141

A linear complex is most commonly formed by a silver metal ion with two coordinate bonds. The bond angles are 180

Answer 142

An octahedral complex is formed when a metal ion has 6 coordinate bonds. This happens with water and ammonia because they are small so 6 ligands are able to surround the central metal ion. Two octahedral complexes are shown below, one with a single type of ligand and the second with a mixture of ligands. The bond angles are 90. .

Answer 143

If there are four coordinate bonds in a complex the shape is usually tetrahedral. The bond angles are 109.5.

Answer 144

Sometimes a square planar shape when there are four coordinate bonds instead of a tetrahedral shape. The bond angle is 90. The compound shown is cisplatin, an anti cancer drug.

Answer 145

Square planar complexes with two pairs of ligands show cis-trans isomerism. A trans isomer is when the two paired ligands are opposite each other in the complex. A cis isomer is when the pair of ligands are next to each other.

Answer 146

Optical isomerism is where an ion can exist in two forms which are non-superimposable mirror images. This occurs in octahedral complexes with three bidentate ligands.

Answer 147

- DNA contains bases, one of which is guanine. A nitrogen atom in guanine forms a coordinate bond with the platinum ion in cisplatin, replacing a chloride ion ligand. - A second nitrogen from a different guanine base bonds to the platinum ion, replacing the other chloride ion ligand. - The cisplatin has bonded to the DNA strands, creating a kink in the strands and preventing correct cell replication.

Answer 148

- Cisplatin can also bind to DNA in healthy cells. This is a problem for cells which replicate frequently, like hair cells and blood cells. This is why the drugs can cause hair loss and why it can suppress the immune system. - The drug can also cause kidney damage. - The issues with the drug are minimised by giving patients small dosages and by targeting the tumour directly.

Answer 149

Cisplatin is a complex of platinum(II) containing two ammonia ligands and two chloride ion ligands. It has a square planar shape. Cisplatin is used as an anticancer drug.

Answer 150

Metal aqua ions react with hydroxide ions to form precipitates. Hydroxide ions are monodentate ligands Not ligand exchange; hydroxide ions remove hydrogen ions from water ligand

Answer 151

- Small amount of ammonia - Initially, ammonia acts as a base and hydrogen ions are pulled off the hexa aqua ion as above. - Excess ammonia - Ammonia replaces water as a ligand - Only 4 of the 6 water molecules replaced

Answer 152

Chloride ions act as monodentate ligands when they react with metal aqua ions. They are charged and larger so only four Cl- ions are able to fit around the metal ion. This means the coordination number decreases from 6 to 4.

Answer 153

Stability constant, Kstab: the equilibrium constant for the formation of the complex ion in a solvent from its constituent ions or molecules Water not included in Kstab expression because concentration almost constant

Answer 154

- Exchange of ligands can be explained in terms of completing equilibria of forward & backward reaction - Equilibrium position lies towards more stable complex - Adding excess weak ligand can shift equilibrium backward and form weaker complex

Answer 155

- Stability constants are used to compare stability of any two ligands - The greater the Kstab value, the more stable the complex - Kstab values may be given on log10 scale as values large - Stability constants for complexes with bi/polydentate ligands very high

Answer 156

- At least one d-orbital must be occupied by an e- - At least one d-orbital must not be fully occupied

Answer 157

Hence, some compounds of d-block metals colourless: - All Sc compounds colourless because 3d0 - All Zn compound colourless because 3d10

Answer 158

Factors that determine colour of complex (which affects the splitting of d-orbitals): - Magnitude of ΔE - Strength of ligands - Oxidation state of metal ion - Geometry of complex

Answer 159

Large ΔE = light absorbed from blue = complex red-ish Small ΔE = light absorbed from red = complex blue-ish

Answer 160

1. Identify the longest carbon chain that contains the functional group. 2. Identify the functional group on the chain. This gives you the suffix or prefix of the compound. 3. Count along the carbon chain so that the functional group has the lowest number. 4. If there are any side chains, add these as prefixes (e.g. methyl-) to the beginning of the name. Do the same if there are other (less important) functional groups. Put these at the start of the name in alphabetical order. 5. If there are two or more identical functional groups or side chains use the prefixes di-, tri- and tetra- before that section of the name.

Answer 161

Functional group - a group of atoms responsible for the characteristic reactions of an organic compound.

Answer 162

Electrophile - a molecule or substance that acts as an electron pair acceptor.

Answer 163

Addition - a reaction where two or more molecules react to form a larger molecule.

Answer 164

Elimination - a reaction in which two substituents are removed from a molecule in a mechanism with one or two steps.

Answer 165

A chiral centre is an atom with four different groups bonded to it. This arrangement creates compounds which are non-superimposable mirror images of each other. If a compound has a chiral centre, it will display optical isomerism. It is possible for compounds to have more than one chiral centre.

Answer 166

Chiral centres can be spotted in organic compounds by identifying a carbon atom which is bonded to four different groups.

Answer 167

- A chiral drug extracted from a natural source often contains a single optical isomer - More effective - Fewer side effects - Reduces cost long term as money isn’t wasted producing the optical isomer that is ineffective.

Answer 168

- Optical isomers of a molecule have different arrangements of bonds so they have different shapes. -This means that only one of the enantiomers may fit into the active site (binding site) of an enzyme. As a result, only one of the enantiomers will cause the desirable effect

Answer 169

- Optical isomerism is another branch of stereoisomerism. It occurs when there is a chiral centre. - Optical isomers have the same molecular and structural formula but they are mirror images of one another

Answer 170

The two mirror images are called enantiomers, they have identical physical and chemical properties except one

Answer 171

Optical isomers enantiomers can only be identified by their ability to rotate plane polarised light in equal and opposite directions.

Answer 172

If there is an equal mixture of each enantiomer then this is called a racemic mixture. This mix has no effect on plane polarised light as the equal and opposite rotations cancel out to zero

Answer 173

- Clear, colourless, non-polar liquid at r.t.p - It is volatile and has a low b.p. - Relatively unreactive due to strong bonding - Burns with smoky flame

Answer 174

- Benzene is a planar molecule with bond angles of 120. - All the carbon-to-carbon bonds are identical so it has a regular hexagonal shape. - There are C-H and C-C sigma bonds as well as a delocalised pi system above and below the carbon ring. - One electron from each carbon is delocalised into the pi (π) system which contains 6 electrons

Answer 175

- E+ attacks (pi) π e- ring, accepting a pair of e- from benzene - An intermediate +ve specie is formed in which +ve charge is delocalized around the ring (more stable than a normal carbocation) - Proton eliminated from +ve intermediate to restore stability and \piπ system of delocalized e-s reformed

Answer 176

- An H-atom on the ring can be substituted by other atoms/groups - All carbon atoms identical and benzene ring symmetrical hence any H-atom can be substituted - An electrophile (E+) can attack the ring of (pi) π e-s but benzene highly stable so E+ must be very strong. - A +ve ion can be generated by heterolytic fission of a covalent bond hence catalyst needed to generate E+

Answer 177

C6H6(l) + X2(g/l) →{catalyst}→ C6H5X(l) + HX(g) - Reagent: dry Cl2 gas / pure Br2 liquid - Conditions: Temperature/pressure: r.t.p. - Catalyst: anhydrous AlCl3(s) / FeBr3(s) - Generating electrophile: catalysts are e- deficient halogen carriers, form dative bond drawing e- from halogen molecule producing X+ - AlCl3 + Cl2 →[AlCl4]- + Cl+ - Regenerating catalyst: H+ + [AICl4]− → HCl + AlCl3

Answer 178

C6H6(l) + H2SO4/HNO3 → C6H5NO2(l) + H2SO4(g) + H2O - Reagent: nitrating mixture; conc. HNO3 / conc. H2SO4 - Condition: Temp.: 45-55C under reflux - Generating electrophile: 2H2SO4 + HNO3 → 2HSO4- + NO2+ + H3O+ - Regenerating catalyst: 2HSO4- + H3O+ + H+→ 2H2SO4 + H2O

Answer 179

C6H6(l) + RCl/RCOCl → {AlCl_3} → C6H5R/C6H5RCO + HCl(g) - Reagent: alkyl halide (RCl) / acyl chloride (RCOCl) - Condition: Temp.: heat under reflux - Catalyst: anhydrous AlCl3 - Generating electrophile: AlCl3 + RCl/RCOCl [AlCl4]- + R+/RCO+ - Regenerating catalyst: H+ + [AlCl4]- HCl + AlCl3

Answer 180

C6H6(l) + 3H2(g) →{Nickel}→ C6H12 - Type of reaction: electrophilic addition - Conditions: Temp.: 150ºC - Catalyst: nickel

Answer 181

C6H5CH3(l) + Cl2(g) → C6H5CH2Cl(l) + HCl - Type of reaction: free radical substitution - Reagent: Cl2(g) - Condition: u.v. light - With excess Cl2 a mixture of di, tri chloromethyl benzene can be obtained

Answer 182

C6H5CH3(l) + 3[O] → C6H5COOH(l) + 2H2O - Type of reaction: oxidation - Reagent: KMnO4(aq) or K2Cr2O7(aq) - Condition: heat under reflux (2 to 3 hours) - Alkyl gp. always oxidized regardless of chain length - If more than one alkyl gp., each one oxidized

Answer 183

- Alkyls, NH2, OH - Donates e- to the ring - positive inductive effect - electro-sub faster - Increase e- density at position 2 and 4

Answer 184

- Cl, NO2, COOH - Withdraws e- from ring - negative inductive effect - electro-sub slower - Sub. occurs at pos. 3 & 5

Answer 185

Friedel-Crafts reactions involve electrophilic substitution of aromatic rings, there are two types - Alkylation and Acylation. Alkylation involves the substitution of alkyl groups such as CH3, C2H5 and C3H7, acylation involves the substitution of acyl groups such as CH3C=O in both cases a catalyst is needed this is because the attacking species isn’t a strong enough electrophile

Answer 186

- Chlorine is a ring deactivating group - Chlorine is more electronegative than carbon, and draws electrons in the ring towards itself. - Hence electron density around ring decreases, less attractive for electrophiles and so reaction is slower

Answer 187

- the halogen atom in halogenoarene is much less reactive than that in a halogenoalkane - Lone pair of e‑s of Cl overlap slightly with p orbitals of benzene ring giving C–Cl a double bond character - Makes C–X bond much stronger therefore nucleophilic substitution difficult - High density of e-s on benzene ring tends to repel the approaching –ve charged nucleophile

Answer 188

C6H5Cl(l) + NaOH(aq) → C6H5OH(l) + NaCl(aq) - Type of reaction: nucleophilic substitution - Reagent: sodium hydroxide - Condition: Temp.: 300-330C, Press: 150-300 atm - Rate of Hydrolysis: (slowest to fastest) R-CH2F → R-CH2CI → R-CH2Br → R-CH2I

Answer 189

Arenes are aromatic hydrocarbons

Answer 190

- Aromatic hydrocarbons, are able to undergo electrophilic substitution. - Aromatic hydrocarbons delocalise electron systems and makes it a very stable molecule. Aromatic hydrocarbons is highly attractive to electrophiles because of the high electron density in the ring. - Electrophiles can be a positive ion or a positive end of a polar molecule. When the reaction takes place, the delocalised electron system remains in the compound and one of the hydrogen atoms swaps places with the electrophile.

Answer 191

The reaction of benzene with chlorine requires an aluminium chloride catalyst. C6H6 + Cl2 → C6H5Cl + HCl The mechanism for the electrophilic substitution reaction: Stage 1: The aluminium chloride catalyst generates the electrophile from chlorine. Cl2 + AlCl3 → AlCl4- + Cl+ Stage 2: The electrophile reacts with the benzene molecule. Stage 3: The hydrogen ion reacts with the AlCl4-, reforming the AlCl3 catalyst. AlCl4- + H+ → AlCl3 + HCl

Answer 192

The reaction of benzene to form nitrobenzene requires a mixture of concentrated nitric acid and concentrated sulfuric acid to generate the electrophile. Stage 1: The electrophile is generated from concentrated nitric acid and concentrated sulfuric acid. H2SO4+ HNO3 → H2O + NO2+ + HSO4- Stage 2: The NO2+ electrophile reacts with the benzene molecule in electrophilic substitution. Stage 3: The hydrogen ion reacts with the HSO4-, reforming the H2SO4 catalyst. H+ + HSO4- → H2SO4

Answer 193

Acylation of benzene involves the substitution of an acyl group. This reaction takes place when benzene is reacted with ethanoyl chloride in the presence of an aluminium chloride catalyst. Stage 1: The electrophile is generated from ethanoyl chloride and aluminium chloride. CH3COCl + AlCl3 → AlCl4- + CH3CO+ Stage 2: The CH3CO+ electrophile reacts with the benzene molecule. CH3CO+ + C6H6 → C6H5COCH3 + H+ Stage 3: The aluminium chloride catalyst is regenerated. H+ + AlCl4- → AlCl3 + HCl

Answer 194

Alkylation of benzene involves the substitution of an alkyl group. The reaction occurs in the same way as the acylation reaction takes place above. It requires a halogenoalkane as a reactant as they contain a polar bond between the carbon and halogen. Stage 1: CH3Cl + AlCl3 → AlCl4- + CH3+ Stage 2: CH3+ + C6H6 → C6H5CH3 + H+ Stage 3: H+ + AlCl4- → AlCl3 + HCl

Answer 195

- Hydrogenation takes place when hydrogen atoms are added all around the benzene ring, removing the delocalised electron system. - A cycloalkane is formed as the product. - The reaction is carried out with hydrogen, in the presence of a nickel catalyst. A temperature around 150C is required.

Answer 196

When a halogen reacts with methylbenzene, two possible reactions can take place depending on the conditions. The halogen can bond to the benzene ring or the methyl group. - Substitution in the benzene ring: This reaction happens in the presence of aluminium chloride and in the absence of UV light. - Substitution in the methyl group: This reaction happens in the presence of UV light, without a catalyst. The reaction can continue to undergo substitution so that all the hydrogen atoms in the methyl group are replaced by chlorine atoms, forming C6H5CCl

Answer 197

- Chlorobenzene is much less reactive than chloroalkane. This is because the C-Cl bond in chlorobenzene is much stronger than in an halogenoalkane. - The aromatic C-Cl bond is stronger due to one of the lone pairs on the chlorine atom interacting with the delocalised electron system, strengthening the bond. - For reactions, like nucleophilic substitution, the C-Cl bond would require breaking. Since the chlorobenzene bond requires more energy to break, chlorobenzene is less reactive.

Answer 198

- Phenol is an aromatic alcohol; a white crystalline solid that is both corrosive and anaesthetic - OH group is ring activating so phenol more reactive than benzene and substituents attached at positions 2,4 and 6

Answer 199

- Benzoyl chloride is very unreactive because the -COCl group is directly attached to a stable benzene ring. To produce the ester phenyl benzoate, phenol is first converted into an ionic compound by dissolving it in sodium hydroxide. - This produces the phenoxide ion: The phenoxide ion is more reactive than the phenol so this ion can react with benzoyl chloride to produce phenyl benzoate

Answer 200

- Benzoyl chloride is very unreactive because the -COCl group is directly attached to a stable benzene ring. To produce the ester phenyl benzoate, phenol is first converted into an ionic compound by dissolving it in sodium hydroxide. - This produces the phenoxide ion: The phenoxide ion is more reactive than the phenol so this ion can react with benzoyl chloride to produce phenyl benzoate

Answer 201

C6H5OH(l) + NaOH(aq) →{r.t.p}→ C6H5O-Na+(aq) H2O - Type of reaction: Neutralization - Reagent: sodium hydroxide - Condition: r.t.p. Regenerating phenol: - When strong acid (HCl) added to solution of sodium phenoxide, phenol regenerated C6H5O-Na+(aq) + HCl(aq) → C6H5OH(l) + NaCl - Initially dense white emulsion formed then slowly oily liquid layer of phenol separates

Answer 202

C6H5OH(l) + Na(s) →{r.t.p}→ C6H5O-Na+(aq) + ½H2(g) - Type of reaction: metal and acid, redox - Condition: r.t.p. - Reacts with Na liberating H2 gas - phenol does not react with metal carbonate to liberate CO2 hence shows its fairly weaker than other carboxylic acids - Phenol reacts with metals in the same way that most acids. The reaction occurs slower than comparable acid-metal reactions because phenol is a weak acid.

Answer 203

-Type of reaction: electrophilic substitution - Reagent: aq. Bromine - Condition: r.t.p. (no catalyst) - Rapid reaction, forms a white ppt. - No catalyst needed since Br2 molecule easily polarized by increase e- density in ring

Answer 204

- Type of reaction: electrophilic substitution - Reagent: dilute nitric acid - Condition: r.t.p. - If concentrated nitric acid used, then trinitrophenol produced

Answer 205

- Higher the Ka value - More easily H+ is donated - More stable is the conjugate base

Answer 206

- phenol is an aromatic alcohol - the OH group is attached directly to the benzene ring - it is an almost colourless crystalline solid of formula C6H5OH

Answer 207

production of plastics - antiseptics - disinfectants - resins for paints

Answer 208

C2H5OH + H2O ⇌ C2H5O- + H3O+ - Ka: 10-18 mol dm-3 - Ability to donate H+ ions: C2H5 is an e- donating gp. which increases charge density on O of OH ‎- More attraction between O–H so loss of H+ difficult Stability of conjugate base: - In ethoxide ion, C2H5 gp. increases –ve charge - Makes ethoxide ion more basic than OH- - Ability to accept H+ increases, moving equi. to left

Answer 209

H2O + H2O ⇌ OH- + H3O+ - Ka: 10-16 mol dm-3 - No e- donating or withdrawing gp. present hence ability to donate and accept H+ ion is the same

Answer 210

C6H5OH + H2O ⇌ C6H5O- + H3O+ - Ka: 10-10 mol dm-3 - Ability to donate H+ ions: OH is a ring activating gp. & the lone pair of e-s on O becomes part of delocalized e- system - Decreases e- density on O of OH and attraction between O and H decreases so H+ lost more easily Stability of conjugate base: - In phenoxide ion, -ve charge on O delocalized around ring and reduces tendency to attract H+ - Conjugate base stable

Answer 211

Test 1: Reagent: Iron(III) Chloride (FeCl3(aq)) – yellowish brown Observation: violet colour obtained Test 2: Reagent: bromine water (Br2(aq)) – orange Observation: white ppt. obtained

Answer 212

- Alkyl groups are e- donating groups which have a positive inductive effect increasing strength of O–H bond, and making it more difficult to donate H+ - Alkyl groups increase negative charge on ion formed when dissociated, making it easier to accept the H+ and reform the molecule

Answer 213

- Chlorine atoms show a –ve inductive effect withdrawing –ve charge from the molecule, reducing the strength of the O–H, and hence making H+ be lost more easily - Conjugate base is more stabilized due to the extended delocalization of the negative charge on the –COO- and so less likely to bond with a H+ ion - If an e- donating gp. present molecule less acidic - If an e- withdrawing gp. present molecule more acidic

Answer 214

- Carboxylic acids are weak acids due to the structure of the carboxylate ion (RCOO-) which forms in solution. - In the carboxylate ion, a delocalised pi system develops over the -COO- group because the lone pair on the negatively charged oxygen overlaps with the p orbitals of the carbon atom. - This delocalised system distributes the negative charge through the -COO- group, making the carboxylate ion more stable.

Answer 215

- Carboxylic acids are weak acids and dissociate incompletely in aqueous solutions R–COOH + H2O ⇌ R–COO- + H3O+ - The O–H bond in carboxylic acid is weakened by the carbonyl group C==O - The carboxylate ion is stabilized by the delocalization of e-s around the –COO- group, spreading and reducing the density of -ve charge making it less likely to bond with H+

Answer 216

- Acyl chlorides can be formed from carboxylic acids using sulfur dichloride oxide (SOCl2). - In this reaction, sulfur dioxide, hydrogen chloride and an acyl chloride are produced. The products are relatively easy to separate since SO2 and HCl are both gases. CH3COOH + SOCl2 → CH3COCl + SO2 + HCl - Acyl chloride can also be produced from carboxylic acids by using phosphorus(V) chloride or phosphorus(III) chloride

Answer 217

HCOOH + [O] → CO2 + H2O - With Tollens’ Reagent: - Ag+ gets reduced to silver – silver mirror formed - With Fehling’s Solution: - Cu2+ gets reduced to Cu+ – copper(I) oxide (red) Note: can use stronger [O] agents; Mn or Cr

Answer 218

HOOC–COOH + [O] → CO2 + H2O - With warm Acidified Manganate(VII) - KMnO4/H+: Purple/pink to colourless - can also use Acidified Dichromate(VI) Changes from orange to green

Answer 219

- When esters react with water an acid catalyst is required. This reaction is reversible so excess water must be used to ensure the position of equilibrium is shifted as far towards the products as possible. - This reaction can occur when the ester is mixed with dilute acid. CH3CH2COOCH3 + H2O ⇌ CH3CH2COOH + CH3OH

Answer 220

- An ester can be hydrolysed by heating it under reflux with a dilute alkali (such as sodium hydroxide). These reactions are not reversible so the products are easier to separate. CH3CH2COOCH3 + NaOH → CH3CH2COONa + CH3OH - To convert the carboxylate salt product into a carboxylic acid, excess strong dilute acid can be added to the salt (after the alcohol has been removed from the products by distillation).

Answer 221

- Acyl chlorides, also known as acid chlorides are acid derivatives. They have a similar structure to carboxylic acids, with the -OH group being replaced by a chlorine atom. Acyl chlorides have the suffix -oyl chloride, e.g. propanoyl chloride.

Answer 222

- Esters are produced when alcohols and acyl chlorides react together. - When an alcohol reacts with an acyl chloride, a vigorous reaction takes place and steamy fumes of hydrochloric acid are produced.

Answer 223

- Phenols are aromatic compounds with an alcohol group bonded to the benzene ring. - The reaction between phenols and acyl chlorides produces an aromatic ester

Answer 224

- A primary amine contains an -NH2 group bonded to an alkyl group. Methylamine has the formula CH3NH2. - When a primary amine reacts with an acyl chloride, an N-substituted amide is produced. In the reaction below, N-methylpropanamide is formed (‘methyl’ comes from the alkyl group on the primary amine). - The hydrogen chloride produced reacts with the excess primary amine so the overall equation for the reaction can be written as: CH3CH2COCl + 2CH3NH2 → CH3CH2CONH2 + CH3NH3Cl

Answer 225

- Acyl chlorides react vigorously with cold water to produce a carboxylic acid and hydrogen chloride gas. E.g. CH3COCl + H2O → CH3COOH + HCl. - Alkyl chlorides have almost no reaction with water. - Aryl chlorides have no reaction with water.

Answer 226

- Acyl chlorides react vigorously with hydroxide ions to produce salts and water: CH3COCl + NaOH → CH3CONa + NaCl + H2O - Alkyl chlorides react with hydroxide ions to produce an alcohol and a salt: CH3Cl + NaOH → CH3OH + NaCl ● Aryl chlorides generally have no reaction with hydroxide ions

Answer 227

- Acyl chlorides contain the group -COCl. - Alkyl chlorides are chloroalkanes containing C-Cl. - Aryl chlorides are aromatic compounds with a chlorine atom bonded to a benzene ring - The reactions above show that the relative ease of hydrolysis is acyl chlorides > alkyl chlorides > aryl chlorides. Acyl chlorides are hydrolysed the easiest while aryl chlorides are the least likely to undergo hydrolysis.

Answer 228

R–X + NH3 → R–NH2 + HX(g) - Type of reaction: nucleophilic substitution - Reagent: hot conc. NH3 in ethanol - Condition: heat under reflux

Answer 229

- Heat the halogenoalkane in a sealed tube with concentrated ammonia in an ethanol solvent (reflux cannot be used as ammonia is too volatile). To ensure that a primary amine is formed, rather than an ammonium salt, an excess of ammonia must be used.

Answer 230

R–C ‎≡‎ N + 4\[H\] ‎→‎ R–CH2NH2 - Type of reaction: reduction of a nitrile - Reagent: hydrogen gas - Reducing Agents (catalyst): LiAlH4 in dry ether Nickel - Conditions: high temp. and pressure R–CO–NH2 + 4\[H\] → R–CH2NH2 + H2O - Type of reaction: reduction of an amide - Reducing Agent (catalyst): LiAlH4 in dry ether - Conditions: high temp. and pressure Conditions: r.t.p.

Answer 231

- Type of reaction: reduction of nitrobenzene - Reagent: conc. HCl with Tin - Condition: heat

Answer 232

- Bases are proton acceptors (electron donors) - N-atoms in amines have a lone pair of e-s - N donates lone pair and accepts H+ forming dative bond

Answer 233

- Type of reaction: electrophilic substitution - Reagent: aromatic amines or phenols - Azo compounds are complex compounds involving a minimum of two aromatic rings joined by N=N coupling - Benzene diazonium ion carries a +ve charge and readily reacts with cold alkaline solutions of aromatic amines and phenols to give brightly coloured azo-compounds - Bright orange dye formed - By using alterative aryl compounds to phenol, a range of brightly coloured dye can be formed.

Answer 234

- Type of reaction: electrophilic substitution - Reagent: aq. Bromine - Conditions: r.t.p. (no catalyst) - Forms a white ppt

Answer 235

- (–N2+ is called diazonium ion) - Type of reaction: diazotization reaction - Reagent: nitrous acid (NaNO2 with excess HCl(aq))(HNO2 weak, unstable acid so produced by reaction NaNO2 with excess HCl) - Condition: low temp. (5∘C) - Diazonium salts of aryl amine stabilized by delocalization of –N2+ ion’s e-s over benzene ring - However, diazonium ion highly unstable and can decompose above 10C to

Answer 236

- The presence of the electron withdrawing oxygen atom means that the lone pair on the amide’s nitrogen atom is not available to be donated to e.g. H+ ions - Hence, amides are neutral

Answer 237

Optical activity: all amino acids (except glycine) have a chiral carbon therefore they are optically active

Answer 238

- Amides contain the -CONH2 group. They are neutral despite the fact that they contain an -NH2 group. - This is because the lone pair on the nitrogen is delocalised into the pi bond between oxygen and carbon meaning the nitrogen atom is unable to attract a hydrogen ion, so is a much weaker base than amines. - The delocalisation also makes the molecule more stable so disrupting the delocalisation to accept a hydrogen ion would require a lot of energy.

Answer 239

- When acyl chlorides react with ammonia, an amide and hydrogen chloride gas is produced. Amides are organic compounds with the group -CONH2. - The suffix for naming an amide is -amide so for the reaction below, the product formed is propanamide. - The hydrogen chloride produced reacts with the excess ammonia to produce ammonium chloride so the overall equation for the reaction can be written as: CH3CH2COCl + 2NH3 → CH3CH2CONH2 + NH4Cl

Answer 240

- A primary amine is an organic compound with an -NH2 group bonded to an alkyl group, like methylamine, CH3NH2. - The product formed from the reaction between a primary amine and acyl chloride is called an N-substituted amide. - The hydrogen chloride produced reacts with the excess primary amine so the overall equation for the reaction can be written as: CH3CH2COCl + 2CH3NH2 → CH3CH2CONH2 + CH3NH3C

Answer 241

- Carboxyl group donates a proton to amino group - Amino group accepts proton and zwitterion formed - Amino acids solids at r.t.p. due to ionic bonds that exist between zwitterions - Presence of zwitterions means that amino acids are soluble in water

Answer 242

Zwitterion: ion that contains regions of +ve & –ve charge

Answer 243

1. A piece of moistened filter paper is placed on a microscope slide. 2. Crocodile clips are attached to each end of the paper and connected to a battery. 3. A drop of amino acid solution is added to the middle of the paper. 4. The apparatus is left for a period of time to enable separation to occur. 5. Ninhydrin is sprayed onto the paper to make the colourless amino acid solution visible. The paper is dried and warmed gently, causing the amino acids to be seen as coloured spots.

Answer 244

Hydrolysis: involves breaking of peptide links by reaction with water catalysed by an acid or alkali catalyst, giving back the amino acids, temp. nearly 90C

Answer 245

- Amide link formed by nucleophilic attack of –NH2 group of one amino acid on –COOH group of another - Reaction is a condensation reaction as H2O eliminated - Reaction can continue to occur as product still has –NH2 and –COOH group present - Dipeptide → Tripeptide → Polypeptide (protein) - Proteins are polymers of amino acids; many polypeptide chains held together by intermolecular forces

Answer 246

Electrophoresis is used to separate, identify and purify amino acids obtained when protein hydrolysed

Answer 247

Technique based on separating ions placed in an electric field. When sample placed between two electrodes: - +ve charge ions move towards –ve charged electrode - –ve charge ions move towards +ve charged electrode - Sample placed on absorbent filter paper (or gel) - Buffer solution carries ions along (back or forth)

Answer 248

- An amino acid contains both an acidic group (-COOH) and a basic group (-NH2). - The carboxylic acid donates a proton to the amine group to form a zwitterion (an ion containing a positive and negative charge). - A zwitterion has no overall electrical charge but it contains separate parts that are charged. - If an amino acid has a carboxylic acid or amine group in the R group, only one acidic and one basic group will be involved in the formation of the zwitterion. - In water, proton transfer will occur to form a charged ion.

Answer 249

- The -NH2 group in phenylamine activates the benzene ring meaning that it is more reactive. - This is because the lone pair on nitrogen is delocalised into the pi system, increasing the electron density meaning electrophiles are more attracted. - The -NH2 has a 2,4-directing effect meaning that groups will generally be substituted at the 2-position and the 4-position (with the carbon that is bound to the amine group being 1-position).

Answer 250

Condensation polymerisation is the joining together of monomers to form a polymer with the release of a small molecule such as water or HCl.

Answer 251

- Only C and H present = only VDW forces present - Presence of Cl or F = dipoles = permanent dipole interaction - Longer chains with have fewer side chains have stronger VDW forces

Answer 252

A polyester contains monomers linked with ester bonds. For this link to form, either of the following is needed: - A monomer containing 2 carboxylic acid groups and a monomer containing 2 alcohol groups. - A monomer containing both an alcohol and a carboxylic acid group

Answer 253

- The ester linkage is formed during a condensation reaction when H is lost from the OH of an alcohol and OH is lost from the COOH of a carboxylic acid. - The H+ and OH- combine to form water. - A monomer containing two -COCl (acyl chloride) groups may be used instead of the carboxylic acid. The only difference is that HCl forms instead of water

Answer 254

A polyamide contains monomers held together by amide bonds. For this link to form, either of the following is needed: - A monomer containing 2 carboxylic acid groups and a monomer containing 2 amine groups. - A monomer containing both a carboxylic acid and an amine group.

Answer 255

- The amide linkage is formed during a condensation reaction when H is lost from the NH2 of an amine and OH is lost from the COOH of a carboxylic acid. - Water forms when the H+ and OH- combine. - As with ester bond formation, a monomer containing two -COCl (acyl chloride) groups can be used instead of the carboxylic acid, forming HCl instead of water

Answer 256

A repeat unit is a structure that occurs in a molecule many times. Sometimes the repeat unit is made up of one monomer, sometimes it contains a pair of monomers.

Answer 257

- Proteins are polymers made from amino acids. - Amino acids contain an amine group (-NH2) and a carboxylic acid group (-COOH). - The general structure of an alpha amino acid is shown below (where R represents any group): - A dipeptide contains 2 amino acids joined together with an amide bond. A polypeptide contains many amino acids bonded together in a chain by amide bonds.

Answer 258

DNA are polynucleotides; made by condensation polymerization of nucleotides.

Answer 259

- Can make copies of itself so that genetic information can be pass on from generation to generation - Contains sequence of bases that form genetic code used to synthesize proteins

Answer 260

- A sugar called deoxyribose - A phosphate group - A nitrogen-containing base

Answer 261

Process of copying DNA during cell division

Answer 262

- H-bonds and VDWs between base pairs in part of a DNA are broken and this part of the double helix unwinds - In the nucleus, there are nucleotides to which two extra phosphates have been added (nucleotide triphosphate) - The bases of the nucleotide triphosphates pair up with complimentary bases on the old strand and H-bonds & VDW forces form between original and new strands. - Enzymes catalyse the polymerization reaction - Each new strand contains a sequence of bases that is complimentary to the original strand

Answer 263

- When a new strand built on template of old strand, the incoming nucleotide is selected for its ability to pair with base in the old strand - H-bonding plays an important role in the recognition as different number of H-bonds involved in each pair

Answer 264

- A polymer that contains silicon bonded to oxygen - They set by reacting with moisture in air; water hydrolyses silicon-oxygen parts, forming cross-linkages - Effectively bonds polymer chains to each other with strong covalent bonds of siloxane cross-linkage

Answer 265

- Example of thermosets: polymers that form extensive networks of covalent cross-links - Very strong and cannot be melted and remoulded - Formed by reacting a monomer with expoyethane (triangular molecule CH2CH2O) and a diamine are mixed

Answer 266

- Uses addition reaction to stick objects together - The monomer is CH2==C(CN)COOCH3, methyl cyanoacrylate, and addition takes place across C==C - Polymerisation initiated by presence of moisture

Answer 267

Polymers generally insulators but ethyne can conduct electricity - Contains alternate single & double carbon-carbon bonds - Can conduct electricity because p-orbitals on neighbouring carbon atoms overlap resulting in long bands of delocalized \piπ e-s free to move along the chain - Doping: other substances, such as iodine, added to improve electrical conductivity of polymer

Answer 268

- Do not corrode - Much less dense - Shaped more easily

Answer 269

- Non-biodegradable - On combustion, give out harmful vapours

Answer 270

- Can contain small amounts of starch: bacteria and fungi in moist soil can break it down into smaller compounds, increase surface area and easier to decompose - Can be made from monomers derived from plants e.g. PLA: soil microorganism can digest it easily

Answer 271

- Polyamides broken down to carboxylic acids & amines - Polyesters broken down to carboxylic acids & alcohols

Answer 272

- Carbonyl group (–C==O) can absorb energy from ultraviolet waves - Causes bonds in region of carbonyl group to weaken and break

Answer 273

- Plastic usually buried under \therefore∴ cannot get sunlight - If recycled, products are weaker

Answer 274

- Poly(alkenes) are chemically inert (very unreactive) meaning they are difficult to biodegrade. - Some polymers, such as LDPE, can be degraded using light. Light (typically UV) causes the polymer chains to break so the material crumbles. - Polyesters are biodegradable and can be hydrolysed using dilute acids or alkalis (slower hydrolysis with acids)

Answer 275

- Synthetic polyesters: terylene - Natural polyesters: fats

Answer 276

- Monomers are alkenes; polymer formed is polyalkene - Alkene joins to itself, no molecule lost - Molecular mass of polymer is multiple of monomer

Answer 277

Addition polymerisation - Monomers contain C=C double bonds. - Main chain of the polymer only contains C-C single bonds. - Main chain contains nitrogen or oxygen atoms as well as carbon atoms. - The polymer is the only product of the reaction. Condensation polymerisation - Monomers contain -OH and -COOH or -COCl for polyesters. Monomers contain -NH2 and -COOH or -COCl for polyamides. - Main chain contains nitrogen or oxygen atoms as well as carbon atoms. - The polymer and a small molecule like water or HCl are formed during the reaction

Answer 278

The acid hydrolysis of proteins is very similar to the acid hydrolysis of polyamides. There are two ways to complete this reaction: - Heat protein at 110°C with 6 mol dm-3 hydrochloric acid for 8 hours (slow method). - Place protein in a sealed tube containing 6 mol dm-3 hydrochloric acid and an atmosphere of nitrogen. Place in a microwave for 5-30 minutes (depending on the protein), using temperatures up to 200°C. Faster method for small samples of protein during analysis.

Answer 279

- if temperature is increased - TΔS value becomes more +ve - ΔG +ve therefore less likely to be spontaneous

Answer 280

- if temperature is increased - TΔS value become more -ve - ΔG -ve therefore more likely to be spontaneous

Answer 281

If ‎∆H > T∆S Enthalpy-driven reaction: flow of thermal energy provides most of the free energy in the reaction

Answer 282

If ‎∆H < T∆S Entropy-driven reaction: increased disorder provides most of the free energy in the reaction

Answer 283

ΔGθ reaction = ΔG2θ −ΔG1θ Form Hess’s cycle and calculate Free Energy as you would enthalpies (using moles etc.)

Answer 284

- At position of equilibrium, the total entropy change of forward reaction equals the total entropy change of backward reaction - Under standard reactions, the overall entropy change is zero

Answer 285

ΔG=−TΔStotal - Gibbs free energy (G): amount of energy in a system that is available to do useful work - Standard molar Gibbs free energy of formation (ΔGfθ ): the free energy change that accompanies the formation of one mole of a compound from its elements in their standard state

Answer 286

Firstly, calculate the entropy change for a reaction ΔSθ system =ΔSθ products −ΔSθ reactants Next, calculate entropy change in the surroundings TΔSθ surroundings = −ΔHreactionθ / T - T in Kelvin; std. temp. = 298K - If ΔHθ reaction in kJ mol-1, change to J by × 1000 - Negative sign part of equation, always put it Use values calculated to find total entropy change ΔSθ total = ΔSθ system −ΔSθ surroundings If ΔSθ total positive, reaction is feasible

Answer 287

- Energy released to surroundings - Increases ways of arranging the energy of molecules in surrounding - Entropy increases - Increased probability of chemical change occurring spontaneously

Answer 288

- Energy absorbed from surroundings - Decreases ways of arranging the energy - Entropy decreases - Increased probability of chemical change occurring spontaneously

Answer 289

- As a substance changes state from (s) → (l) → (g); its entropy increases as molecules become more disordered - When an ionic salt is dissolved in water (s) → (aq); the entropy increases as as the crystal breaks up and the ions find their way between the water molecules - Increasing the temperature of a substance (without changing state) increases entropy e.g. when heating a gas, the number of different possibilities for arranging the energy over the molecules increases. - If a reaction increases the number of gas molecules, the entropy increases as gases have higher entropies than solids or liquids - Simpler substances with fewer atoms have lower entropy values than more complex substances - For similar substances, harder substances have lower entropy values

Answer 290

- Entropy: measure of the ‘disorder’ of a system - A system becomes more stable when its energy is spread out in a more disordered state

Answer 291

- Spontaneous change: a change that tends to happen naturally; once started, the change will carry on - Spontaneous changes may not be instantaneous or happen rapidly; most are slow and need an input of energy to start

Chemistry A Level Flashcards

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