A Level Chemistry (9701) — MCQs & Practice Questions

Proton: charge +1, mass 1. Neutron: charge 0, mass 1. Electron: charge −1, mass negligible (about 1/1840).

Atoms of the same element (same number of protons / same proton number) that have different numbers of neutrons, and so different mass numbers.

Protons = 17, electrons = 17 (neutral atom), neutrons = 35 − 17 = 18.

1. Both isotopes have the same number of electrons / the same electronic configuration (2,8,7)[1]
2. Chemical properties depend on the (outer) electrons, so they react identically[1]
3. Ar = (35 × 75 + 37 × 25) ÷ 100sets up weighted mean[1]
4. Ar = (2625 + 925)/100 = 35.5[1]
5. 7 electrons in the outer shell[1]
6. Gains 1 electron, so the chloride ion is Cl⁻ (charge −1)[1]

Same electron arrangement → identical chemistry; Ar = 35.5; 7 outer electrons, so Cl⁻.

Aᵣ = (10 × 20 + 11 × 80) ÷ 100 = (200 + 880) ÷ 100 = 1080 ÷ 100 = 10.8.

1. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²[1]
2. electron count totals 20[1]
3. Ca²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ (loses the two 4s electrons)[1]
4. The first two electrons are removed from the outer 4s sub-shell[1]
5. The third is removed from the 3p, a shell closer to the nucleus[1]
6. which is held much more strongly, so far more energy is needed — showing two electrons in the outer shell[1]

Ca = 1s²2s²2p⁶3s²3p⁶4s²; Ca²⁺ = 1s²2s²2p⁶3s²3p⁶; the jump after 2 electrons shows only two electrons in the outer shell.

Mr of CH₄ = 12 + 4 = 16. Moles = mass ÷ Mr = 8.0 ÷ 16 = 0.50 mol.

The sum of the relative atomic masses of all the atoms shown in the molecular formula of a substance.

4 Al + 3 O₂ → 2 Al₂O₃.

1. Mr(CaCO₃) = 40 + 12 + 48 = 100[1]
2. Moles CaCO₃ = 25 ÷ 100 = 0.25 mol[1]
3. Moles CaO = 0.25 (1:1 ratio); Mr(CaO) = 56[1]
4. Mass CaO = 0.25 × 56 = 14 g[1]
5. Moles CO₂ = 0.25 (1:1 ratio)[1]
6. Volume = moles × 24 = 0.25 × 24[1]
7. Volume CO₂ = 6.0 dm³[1]

Mr = 100; 14 g CaO; 6.0 dm³ CO₂.

V = 500 ÷ 1000 = 0.50 dm³. c = n ÷ V = 0.20 ÷ 0.50 = 0.40 mol/dm³.

1. Mr(Fe₂O₃) = 112 + 48 = 160[1]
2. Moles Fe₂O₃ = 320 ÷ 160 = 2 (tonne-moles)[1]
3. Moles Fe = 2 × 2 = 4; Ar(Fe) = 56[1]
4. Mass Fe = 4 × 56 = 224 tonnes[1]
5. Desired product mass in equation = 2 × 56 = 112; total product mass = 112 + 3 × 44 = 244[1]
6. Atom economy = (112 ÷ 244) × 100[1]
7. = 45.9% (≈ 46%)[1]

Mr = 160; 224 tonnes of iron; atom economy ≈ 46%.

Sodium loses one electron to chlorine; sodium becomes Na⁺ and chlorine becomes Cl⁻, and the oppositely charged ions attract by strong electrostatic forces.

It has a giant ionic lattice; the many strong electrostatic forces between oppositely charged ions need a large amount of energy to overcome.

Methane has weak intermolecular forces between molecules, which need little energy to overcome. Diamond is a giant covalent structure with strong covalent bonds throughout, needing very large energy to break.

1. Magnesium loses 2 electrons to oxygen, forming Mg²⁺ and O²⁻[1]
2. The oppositely charged ions attract by strong electrostatic forces[1]
3. When molten, the ions are free to move and carry charge[1]
4. When solid, the ions are fixed in the lattice and cannot move[1]
5. CO₂ is simple molecular with weak intermolecular forces[1]
6. Little energy is needed to separate the molecules, so it is a gas[1]
7. MgO is a giant ionic lattice with strong forces needing much energy, so it is a high-melting solid[1]

Mg→Mg²⁺ + O→O²⁻ (electrostatic attraction); MgO conducts molten (mobile ions) not solid (fixed ions); CO₂ gas (weak intermolecular forces) vs MgO solid (giant lattice, strong forces).

Down the group the atoms have more electrons, so the van der Waals (London) forces between them are stronger and need more energy to overcome, raising the boiling point.

1. Methane is tetrahedral[1]
2. with a bond angle of 109.5°[1]
3. because four bonding pairs repel to get as far apart as possible[1]
4. Water molecules form hydrogen bonds (O–H, H bonded to O)[1]
5. methane has only weak van der Waals forces[1]
6. hydrogen bonds are much stronger, so water needs far more energy to boil[1]

Methane: tetrahedral, 109.5°, four repelling bonding pairs; water boils higher because of strong hydrogen bonding vs methane's weak van der Waals forces.

Particles are closely packed in a regular, fixed arrangement and can only vibrate about fixed positions.

Gas particles move rapidly in all directions and collide with the walls; the force of these collisions per unit area is the pressure.

Heating gives particles more kinetic energy so they vibrate more; at the melting point the forces holding the lattice break and particles move past each other (liquid); at the boiling point particles gain enough energy to escape completely (gas).

1. Diffusion is the movement of particles from a region of higher concentration to lower concentration[1]
2. ...until they are evenly spread / down a concentration gradient[1]
3. Bromine molecules are heavier / have greater mass than hydrogen molecules[1]
4. At the same temperature heavier particles move more slowly, so diffuse more slowly[1]
5. Raising temperature increases the rate of diffusion[1]
6. Because particles gain kinetic energy and move faster[1]

Diffusion = spreading down a concentration gradient; lighter particles diffuse faster; higher temperature → faster diffusion.

Before melting the temperature rises as particles gain kinetic energy. During melting it stays constant (a plateau) because the energy overcomes the forces holding the lattice. After melting it rises again as the liquid warms.

1. Particles move rapidly and randomly, colliding with the walls[1]
2. the force of these collisions per unit area is the pressure[1]
3. Heating gives particles more kinetic energy so they move faster[1]
4. collisions are harder and more frequent, so pressure rises[1]
5. Halving the volume means the same particles are in a smaller space[1]
6. so collisions with the walls are twice as frequent and the pressure doubles[1]

Pressure = wall collisions per area; heating raises it (faster, more frequent collisions); halving the volume doubles it (more frequent collisions).

A reaction that transfers energy to the surroundings (temperature of surroundings rises), e.g. combustion / neutralisation.

Bond breaking absorbs energy and bond making releases energy; in an exothermic reaction more energy is released making bonds than is absorbed breaking them.

Reactants at a higher energy than products; a 'hump' above the reactants marks the activation energy; ΔH is the (negative) drop from reactants to products.

1. Bonds broken = 2(H–H) + 1(O=O) = 2(436) + 498[1]
2. = 872 + 498 = 1370 kJ[1]
3. Bonds made = 4 O–H = 4 × 464[1]
4. = 1856 kJ[1]
5. ΔH = energy in − energy out = 1370 − 1856 = −486 kJ/mol[1]
6. Negative ΔH, so the reaction is exothermic[1]

Broken = 1370 kJ; made = 1856 kJ; ΔH = −486 kJ/mol → exothermic.

q = mcΔT = 200 × 4.2 × 25 = 21 000 J = 21 kJ.

1. q = mcΔT = 100 × 4.2 × 24[1]
2. = 10 080 J = 10.08 kJ[1]
3. moles methanol = 0.32 ÷ 32 = 0.010 mol[1]
4. ΔH = 10.08 ÷ 0.010 = 1008 kJ/mol (magnitude)[1]
5. ΔH ≈ −1010 kJ/mol (exothermic, so negative)[1]
6. Heat lost to the surroundings / incomplete combustion / evaporation[1]

q = 10.08 kJ; ΔH ≈ −1010 kJ/mol; less negative than the true value because heat escapes / combustion is incomplete.

Oxidation is an increase in oxidation number; reduction is a decrease in oxidation number.

The voltage of that half-cell measured against a standard hydrogen electrode under standard conditions (298 K, 1 mol dm⁻³, 1 atm).

EMF = E°(cathode) − E°(anode) = +0.34 − (−0.76) = +1.10 V. The zinc electrode is the negative terminal (it is oxidised).

1. Zn → Zn²⁺ + 2e⁻[1]
2. This is oxidation (loss of electrons)[1]
3. The salt bridge completes the circuit[1]
4. and allows ions to flow to balance the charge in each half-cell[1]
5. Electrons flow from zinc to copper[1]
6. because zinc is more easily oxidised / has the more negative electrode potential[1]

Zn → Zn²⁺ + 2e⁻ (oxidation); salt bridge completes circuit and balances charge; electrons flow zinc → copper (zinc more easily oxidised).

Silver has the more positive E°, so it is reduced (cathode) and copper is oxidised (anode). E°cell = 0.80 − 0.34 = +0.46 V.

1. The zinc electrode is negative[1]
2. Zn → Zn²⁺ + 2e⁻ (oxidation)[1]
3. E°cell = E°(cathode) − E°(anode) = 0.34 − (−0.76)[1]
4. = +1.10 V[1]
5. Electrons flow from zinc to copper in the external circuit[1]
6. because zinc has the more negative electrode potential, so it is oxidised and releases electrons[1]

Zinc electrode negative (Zn → Zn²⁺ + 2e⁻); EMF = +1.10 V; electrons flow zinc → copper (zinc more easily oxidised).

Kc = [NH₃]² ÷ ([N₂][H₂]³).

A large Kc means the equilibrium lies to the right — the mixture contains mostly products.

Increasing temperature shifts the equilibrium in the endothermic (backward) direction, so the yield of product falls and Kc decreases.

1. Higher pressure increases the yield of SO₃[1]
2. The equilibrium shifts to the side with fewer gas molecules[1]
3. 3 gas molecules on the left, 2 on the right, so it shifts right[1]
4. A catalyst does not change the yield / position of equilibrium[1]
5. It speeds up the forward and backward reactions equally (equilibrium reached faster)[1]
6. Increasing temperature decreases Kc[1]

Higher pressure → more SO₃ (fewer gas moles on right); catalyst → no change in yield (faster only); higher temperature → Kc decreases.

Kc = [SO₃]² ÷ ([SO₂]² [O₂]).

1. Higher pressure increases the yield of ammonia[1]
2. the system shifts to the side with fewer gas molecules[1]
3. 4 gas molecules on the left, 2 on the right, so it shifts right[1]
4. Increasing temperature decreases Kc[1]
5. the forward reaction is exothermic, so higher temperature favours the endothermic reverse[1]
6. less product forms, so Kc falls[1]

Higher pressure → more NH₃ (fewer gas moles on the right); higher temperature → lower Kc (favours the endothermic reverse).

The change in concentration of a reactant or product per unit time.

The power to which the concentration of that reactant is raised in the rate equation.

First order in A, second order in B, overall third order. Doubling [B] multiplies the rate by 2² = 4.

1. k increases with temperature[1]
2. Particles have more energy and collide more often[1]
3. and a greater fraction exceed the activation energy, so more collisions are successful[1]
4. The rate-determining step is the slowest step in a multi-step reaction[1]
5. It controls the overall rate[1]
6. s⁻¹[1]

k rises with temperature (more successful collisions); rate-determining step = the slowest step, which limits overall rate; first-order k has units s⁻¹.

Overall order = 2 + 1 = 3 (third order). Doubling [A] multiplies the rate by 2² = 4.

1. The slowest step in a multi-step reaction[1]
2. which controls (limits) the overall rate[1]
3. A catalyst provides an alternative pathway with lower activation energy[1]
4. so a greater fraction of collisions are successful, raising k[1]
5. Higher temperature gives molecules more energy[1]
6. so more exceed the activation energy — more successful collisions and a larger k[1]

Rate-determining step = slowest step controlling the rate; catalyst and higher temperature both raise k by increasing the fraction of successful (Ea-exceeding) collisions.

In order of increasing proton (atomic) number, arranged in periods (rows) and groups (columns) of similar properties.

Reactivity increases down Group I because the outer electron is further from the nucleus and more easily lost.

Group VII atoms all have 7 outer electrons, so they react similarly (gaining/sharing one electron). Noble gases have a full outer shell, so they have no tendency to gain, lose or share electrons and are unreactive.

1. Reactivity decreases down the group[1]
2. The outer shell is further from the nucleus / more shielded[1]
3. So it is harder to attract/gain an electron[1]
4. Chlorine is more reactive than bromine, so it displaces bromine[1]
5. The solution turns orange (bromine formed)[1]
6. Cl₂ + 2KBr → 2KCl + Br₂[1]

Reactivity falls down the group (outer shell further/shielded, harder to gain e⁻); chlorine displaces bromine: Cl₂ + 2KBr → 2KCl + Br₂.

Reactivity increases down the group because the outer electron is further from the nucleus and more shielded, so it is lost more easily.

1. Melting point rises from Na to a peak at Si, then falls sharply to Ar[1]
2. Silicon is a giant covalent (macromolecular) structure[1]
3. so a huge number of strong covalent bonds must be broken — very high melting point[1]
4. First ionisation energy generally increases across the period[1]
5. because nuclear charge increases while shielding stays similar[1]
6. so the outer electron is held more tightly and needs more energy to remove[1]

Melting point rises to a giant-covalent peak at Si then falls to molecular Ar; first ionisation energy rises (more nuclear charge, similar shielding).

Reactivity increases down the group because the outer electron is further from the nucleus and more easily lost.

It floats, melts into a ball, fizzes and moves around, giving off hydrogen: 2Na + 2H₂O → 2NaOH + H₂.

Thermal stability increases down the group. The cations get larger with less charge density, so they polarise/distort the carbonate ion less, making it harder to decompose.

1. Magnesium reacts very slowly with cold water (faster with steam)[1]
2. Calcium reacts readily with cold water, fizzing[1]
3. Ca + 2H₂O → Ca(OH)₂ + H₂[1]
4. balanced equation[1]
5. Reactivity increases down the group[1]
6. because the outer electrons are further from the nucleus and more easily lost[1]

Mg very slow with cold water, Ca reacts readily; Ca + 2H₂O → Ca(OH)₂ + H₂; reactivity rises down the group (electrons lost more easily).

Down the group the cations get larger and have lower charge density, so they polarise (distort) the nitrate ion less. The nitrate is therefore harder to break down, so a higher temperature is needed and thermal stability increases.

1. Reactivity increases down the group[1]
2. the outer electrons are further from the nucleus and more shielded[1]
3. so they are lost more easily, making the metal more reactive[1]
4. Hydroxides become more soluble down the group[1]
5. sulfates become less soluble down the group[1]
6. so barium sulfate is highly insoluble (used in the sulfate test)[1]

Reactivity rises down Group 2 (outer electrons lost more easily); hydroxides get more soluble, sulfates less soluble.

Reactivity decreases down the group because the outer shell is further from the nucleus and more shielded, so it is harder to attract an electron.

Damp red litmus paper turns blue in ammonia gas (ammonia is alkaline).

Chlorine is more reactive (a stronger oxidising agent) than bromine, so it removes electrons from bromide ions: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂.

1. Sulfur dioxide and nitrogen oxides are released when fossil fuels burn[1]
2. They dissolve in rainwater[1]
3. forming acidic solutions (sulfurous/sulfuric and nitric acids)[1]
4. SO₂ + H₂O → H₂SO₃[1]
5. Damages plants/forests, or acidifies lakes killing fish[1]
6. or corrodes buildings/limestone and metals (any valid effect)[1]

SO₂/NOₓ from burning fuels dissolve in rain forming acids; SO₂ + H₂O → H₂SO₃; acid rain damages forests, lakes, buildings.

With potassium bromide the solution turns orange as bromine is displaced; with potassium iodide it turns brown as iodine is displaced. Chlorine is more reactive than both. Equation: Cl₂ + 2KBr → 2KCl + Br₂.

1. Halogens react by gaining one electron[1]
2. down the group the outer shell is further from the nucleus and more shielded[1]
3. so an electron is attracted/gained less easily — reactivity falls[1]
4. The inert pair effect is the reluctance of the outer s-electron pair to bond in heavier p-block elements[1]
5. so they favour an oxidation state two lower than the group maximum[1]
6. lead is more stable as Pb²⁺ than Pb⁴⁺[1]

Halogen reactivity falls down the group (electron gained less easily); the inert pair effect makes heavier elements like lead favour the lower (+2) state.

Reactivity decreases down the group because the outer shell is further from the nucleus and more shielded, so it is harder to attract an electron.

Damp red litmus paper turns blue in ammonia gas (ammonia is alkaline).

Chlorine is more reactive (a stronger oxidising agent) than bromine, so it removes electrons from bromide ions: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂.

1. Sulfur dioxide and nitrogen oxides are released when fossil fuels burn[1]
2. They dissolve in rainwater[1]
3. forming acidic solutions (sulfurous/sulfuric and nitric acids)[1]
4. SO₂ + H₂O → H₂SO₃[1]
5. Damages plants/forests, or acidifies lakes killing fish[1]
6. or corrodes buildings/limestone and metals (any valid effect)[1]

SO₂/NOₓ from burning fuels dissolve in rain forming acids; SO₂ + H₂O → H₂SO₃; acid rain damages forests, lakes, buildings.

With potassium bromide the solution turns orange as bromine is displaced; with potassium iodide it turns brown as iodine is displaced. Chlorine is more reactive than both. Equation: Cl₂ + 2KBr → 2KCl + Br₂.

1. Halogens react by gaining one electron[1]
2. down the group the outer shell is further from the nucleus and more shielded[1]
3. so an electron is attracted/gained less easily — reactivity falls[1]
4. The inert pair effect is the reluctance of the outer s-electron pair to bond in heavier p-block elements[1]
5. so they favour an oxidation state two lower than the group maximum[1]
6. lead is more stable as Pb²⁺ than Pb⁴⁺[1]

Halogen reactivity falls down the group (electron gained less easily); the inert pair effect makes heavier elements like lead favour the lower (+2) state.

Compounds with the same molecular formula but different structural arrangements of their atoms.

Propan-1-ol; the functional group is the hydroxyl group, –OH (an alcohol).

In homolytic fission each atom takes one electron, forming two radicals. In heterolytic fission one atom takes both electrons, forming a positive and a negative ion.

1. An electrophile is an electron-pair acceptor (electron-poor species)[1]
2. A nucleophile is an electron-pair donor (electron-rich species)[1]
3. Bromine + ethene is an addition reaction[1]
4. Haloalkane with hot alcoholic KOH is an elimination reaction[1]
5. A functional group is the atom or group of atoms[1]
6. that gives a molecule its characteristic chemical properties[1]

Electrophile accepts an electron pair, nucleophile donates one; Br₂ + ethene = addition, elimination gives the alkene; functional group = the reactive part defining the chemistry.

But-2-ene has a C=C double bond, and there is no rotation about it, so the methyl groups are fixed either on the same side (cis) or opposite sides (trans). Butane has only single bonds, which rotate freely, so no fixed arrangement — and no cis–trans isomers.

1. A nucleophile is an electron-pair donor (electron-rich)[1]
2. an electrophile is an electron-pair acceptor (electron-poor)[1]
3. (i) Bromine acts as an electrophile[1]
4. in an electrophilic addition to the C=C double bond[1]
5. (ii) Hydroxide acts as a nucleophile[1]
6. in a nucleophilic substitution of the haloalkane[1]

Nucleophile = electron-pair donor, electrophile = acceptor; Br₂/ethene = electrophilic addition; OH⁻/bromoethane = nucleophilic substitution.

1. Step 1: hydrate ethene with steam and an acid catalyst to give ethanol[1]
2. (~300 °C, ~60 atm, phosphoric acid)[1]
3. Step 2: oxidise the ethanol with acidified potassium dichromate[1]
4. under reflux, to give ethanoic acid[1]
5. React bromoethane with potassium cyanide in ethanol to give propanenitrile[1]
6. useful because it adds a carbon, lengthening the chain[1]

Ethene → ethanol (steam + acid catalyst) → ethanoic acid (oxidise with acidified dichromate, reflux); bromoethane + KCN → propanenitrile (adds a carbon).

A family of organic compounds with the same general formula and similar chemical properties, where each member differs from the next by CH₂.

Add bromine water: an alkene decolourises it (orange to colourless); an alkane does not (stays orange).

Many ethene monomers (with C=C double bonds) join together, with no other product, to form a long-chain polymer. Repeat unit: –(CH₂–CH₂)–.

1. C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂[1]
2. correct formulae / balanced[1]
3. Steam / water, ~300 °C, ~60 atm, phosphoric acid catalyst[1]
4. C₂H₄ + H₂O → C₂H₅OH[1]
5. balanced equation[1]
6. It is a continuous process / faster / gives purer ethanol (any one)[1]

Fermentation: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂; hydration: C₂H₄ + H₂O → C₂H₅OH (steam, ~300 °C, ~60 atm, H₃PO₄); hydration is continuous/faster/purer.

Add bromine water: ethene (an alkene) decolourises it from orange to colourless; ethane (an alkane) leaves it orange (no reaction).

1. Cracking breaks long-chain hydrocarbons into shorter ones[1]
2. using heat and a catalyst[1]
3. to make more useful small molecules (fuels) and alkenes for plastics[1]
4. Many ethene monomers (with C=C double bonds) join together[1]
5. the double bonds open and link into a long chain, with no other product[1]
6. The product is poly(ethene)[1]

Cracking breaks long hydrocarbons (heat + catalyst) into useful short molecules and alkenes; addition polymerisation joins many ethene monomers into poly(ethene).

Free-radical substitution. Initiation: Cl₂ → 2Cl• (UV). Propagation: Cl• + CH₄ → •CH₃ + HCl, then •CH₃ + Cl₂ → CH₃Cl + Cl•.

1. The mechanism is electrophilic addition[1]
2. the electron-rich C=C double bond attracts the electrophile (the δ+ H of HBr)[1]
3. The H can add to either carbon, giving either a secondary or a primary carbocation[1]
4. the secondary carbocation is more stable than the primary[1]
5. so the H adds to the terminal (CH₂) carbon (Markovnikov's rule)[1]
6. giving mainly 2-bromobutane[1]

Electrophilic addition; the C=C attracts the electrophile; via the more stable secondary carbocation the major product is 2-bromobutane.

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂, using yeast at about 30 °C in the absence of air. An advantage is that it uses a renewable resource (sugar/plant crops) rather than crude oil.

1. Oxidise the ethanol with acidified potassium dichromate (warm/reflux)[1]
2. (or ferment ethanol in air) to give ethanoic acid[1]
3. The magnesium fizzes / dissolves, giving off hydrogen gas[1]
4. Mg + 2CH₃COOH → (CH₃COO)₂Mg + H₂[1]
5. Ethyl ethanoate is formed, plus water[1]
6. using a concentrated sulfuric acid catalyst[1]

Oxidise ethanol → ethanoic acid; with magnesium it fizzes giving H₂ (Mg + 2CH₃COOH → (CH₃COO)₂Mg + H₂); with ethanol it makes ethyl ethanoate + water (conc. H₂SO₄ catalyst).

The bond is polar; the halogen is more electronegative, leaving the carbon slightly positive (δ+), which attracts electron-pair-donating nucleophiles.

Ethanol is formed by nucleophilic substitution (OH⁻ replaces Br⁻).

The nitrogen atom has a lone pair of electrons which can accept a proton (H⁺), forming a positively charged ion; so amines are proton acceptors (bases).

1. Hot ethanolic (alcoholic) potassium hydroxide[1]
2. under reflux / heat[1]
3. This is elimination (loss of HBr forms the C=C)[1]
4. Excess ammonia[1]
5. (in a sealed tube / ethanolic solution)[1]
6. This is nucleophilic substitution[1]

→ ethene: hot alcoholic KOH, elimination; → ethylamine: excess ammonia, nucleophilic substitution.

The nitrogen lone pair can accept a proton (H⁺), so amines are bases. Ethylamine + HCl gives the salt ethylammonium chloride (C₂H₅NH₃⁺Cl⁻).

1. Warm aqueous sodium hydroxide (or water)[1]
2. gives ethanol[1]
3. Potassium cyanide (in ethanol) gives propanenitrile[1]
4. useful because it adds a carbon, lengthening the chain[1]
5. The C–I bond is weaker than C–Br[1]
6. so it breaks more easily, making iodoethane react faster[1]

OH⁻ → ethanol; CN⁻ → propanenitrile (adds a carbon); iodoethane is faster because C–I is weaker than C–Br.

1. In SN2 the nucleophile attacks the δ+ carbon as the halogen leaves[1]
2. in a single concerted step[1]
3. favoured by primary haloalkanes (the carbon is easy to approach)[1]
4. In SN1 the halogen leaves first to form a carbocation[1]
5. then the nucleophile attacks the carbocation[1]
6. tertiary haloalkanes form a stable tertiary carbocation, so this route is favoured[1]

SN2: one-step, nucleophile attacks as halide leaves, favoured by primary. SN1: two-step via a carbocation, favoured by tertiary (stable carbocation).

Secondary — the carbon bearing the –OH group is attached to two other carbon atoms.

First an aldehyde, then on further oxidation a carboxylic acid; the dichromate turns from orange to green.

When phenol loses H⁺, the negative charge is delocalised into the benzene ring, stabilising the ion, so H⁺ is released. Ethanol's anion is not stabilised in this way, so it does not release H⁺.

1. Propan-1-ol (primary) → propanal then propanoic acid[1]
2. Propan-2-ol (secondary) → propanone (a ketone)[1]
3. 2-methylpropan-2-ol (tertiary) → no reaction[1]
4. Those that react turn the dichromate from orange to green[1]
5. The tertiary alcohol has no hydrogen on the carbon bearing the –OH[1]
6. so there is nothing for the oxidising agent to remove[1]

1° → aldehyde/acid, 2° → ketone, 3° → no reaction (orange→green for the first two); tertiary resists because there is no H on the OH carbon.

Hydration is a fast, continuous process that gives pure ethanol directly, whereas fermentation is slow, works in batches and produces a dilute impure mixture that must be distilled.

1. When phenol loses H⁺, the negative charge is delocalised into the benzene ring[1]
2. this stabilises the phenoxide ion, so H⁺ is released — weakly acidic[1]
3. ethanol's anion is not stabilised in this way, so it does not release H⁺[1]
4. Add sodium carbonate solution to each[1]
5. the carboxylic acid fizzes, releasing CO₂[1]
6. phenol shows no reaction (too weak an acid)[1]

Phenol is acidic because the ring stabilises the phenoxide ion; ethanol's is not. Sodium carbonate fizzes with a carboxylic acid but not with phenol.

Both contain the carbonyl group C=O; in an aldehyde it is at the end of the chain (–CHO), in a ketone it is in the middle, bonded to two carbon groups.

Warm with Tollens' reagent: an aldehyde forms a silver mirror; a ketone gives no change.

Reduction (e.g. with NaBH₄) gives propan-1-ol; the carbonyl C=O is reduced to a C–OH group (a primary alcohol).

1. Propanal (aldehyde) gives a positive result[1]
2. the blue solution forms a brick-red precipitate[1]
3. Propanone (ketone) gives no change (stays blue)[1]
4. The characteristic reaction is nucleophilic addition[1]
5. The C=O bond is polar with a δ+ carbon[1]
6. so nucleophiles are attracted to and attack the carbonyl carbon[1]

Aldehyde → brick-red precipitate with Fehling's; ketone → stays blue; carbonyl undergoes nucleophilic addition at its δ+ carbon.

Reduction (e.g. with NaBH₄): propanal (an aldehyde) gives propan-1-ol (a primary alcohol); propanone (a ketone) gives propan-2-ol (a secondary alcohol).

1. Add 2,4-DNPH (Brady's reagent) to each[1]
2. both give an orange precipitate, confirming a carbonyl group[1]
3. Warm each with Tollens' reagent (ammoniacal silver nitrate)[1]
4. the aldehyde forms a silver mirror[1]
5. the ketone gives no change[1]
6. (alternatively Fehling's: aldehyde gives a brick-red precipitate, ketone stays blue)[1]

2,4-DNPH gives an orange precipitate with both (carbonyl present); Tollens' gives a silver mirror only with the aldehyde.

The carboxyl group, –COOH; carboxylic acids are weak acids (only partly ionised in water).

Sodium ethanoate, water and carbon dioxide are formed; effervescence (fizzing) is seen.

When a carboxylic acid loses H⁺, the negative charge on the carboxylate ion is delocalised over both oxygen atoms, stabilising it; an alcohol's anion is not stabilised in this way.

1. The reaction is esterification (condensation)[1]
2. The product is ethyl ethanoate (an ester)[1]
3. CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O[1]
4. balanced with water as the other product[1]
5. The sulfuric acid is a catalyst (and dehydrating agent)[1]
6. Esters are used as flavourings / perfumes / solvents (any one)[1]

Esterification → ethyl ethanoate; CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O; H₂SO₄ is the catalyst; esters used as flavourings/perfumes/solvents.

Propyl ethanoate, formed with a concentrated sulfuric acid catalyst (which also removes water).

1. CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O[1]
2. conditions: warm with a concentrated sulfuric acid catalyst[1]
3. the reaction is reversible[1]
4. (i) with dilute acid: ethanoic acid + ethanol[1]
5. (ii) with sodium hydroxide: sodium ethanoate[1]
6. plus ethanol[1]

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (conc. H₂SO₄); acid hydrolysis → acid + alcohol; alkaline hydrolysis → sodium ethanoate + ethanol.

The bond is polar; the halogen is more electronegative, leaving the carbon slightly positive (δ+), which attracts electron-pair-donating nucleophiles.

Ethanol is formed by nucleophilic substitution (OH⁻ replaces Br⁻).

The nitrogen atom has a lone pair of electrons which can accept a proton (H⁺), forming a positively charged ion; so amines are proton acceptors (bases).

1. Hot ethanolic (alcoholic) potassium hydroxide[1]
2. under reflux / heat[1]
3. This is elimination (loss of HBr forms the C=C)[1]
4. Excess ammonia[1]
5. (in a sealed tube / ethanolic solution)[1]
6. This is nucleophilic substitution[1]

→ ethene: hot alcoholic KOH, elimination; → ethylamine: excess ammonia, nucleophilic substitution.

The nitrogen lone pair can accept a proton (H⁺), so amines are bases. Ethylamine + HCl gives the salt ethylammonium chloride (C₂H₅NH₃⁺Cl⁻).

1. Warm aqueous sodium hydroxide (or water)[1]
2. gives ethanol[1]
3. Potassium cyanide (in ethanol) gives propanenitrile[1]
4. useful because it adds a carbon, lengthening the chain[1]
5. The C–I bond is weaker than C–Br[1]
6. so it breaks more easily, making iodoethane react faster[1]

OH⁻ → ethanol; CN⁻ → propanenitrile (adds a carbon); iodoethane is faster because C–I is weaker than C–Br.

1. In SN2 the nucleophile attacks the δ+ carbon as the halogen leaves[1]
2. in a single concerted step[1]
3. favoured by primary haloalkanes (the carbon is easy to approach)[1]
4. In SN1 the halogen leaves first to form a carbocation[1]
5. then the nucleophile attacks the carbocation[1]
6. tertiary haloalkanes form a stable tertiary carbocation, so this route is favoured[1]

SN2: one-step, nucleophile attacks as halide leaves, favoured by primary. SN1: two-step via a carbocation, favoured by tertiary (stable carbocation).

Carbohydrates (from simple sugars), proteins (from amino acids) and fats/lipids (from fatty acids and glycerol).

Add biuret reagent (sodium hydroxide then copper(II) sulfate solution); a positive result is a colour change from blue to purple/violet.

Starch is broken down (hydrolysed) into simple sugars such as glucose. Enzymes are biological catalysts that speed up this reaction at body temperature, which would otherwise be far too slow.

1. It speeds up a reaction[1]
2. without being used up / by lowering the activation energy[1]
3. As temperature rises up to the optimum, activity increases[1]
4. because particles collide more frequently and with more energy[1]
5. Above the optimum, activity falls sharply[1]
6. because the enzyme is denatured — its shape (active site) changes so substrate no longer fits[1]

Catalyst = speeds a reaction without being used up; activity rises to an optimum then falls as the enzyme denatures (active site changes shape).

Condensation joins two monomers together and releases a small molecule (usually water), building a polymer. Hydrolysis is the reverse: it adds water to split a polymer back into its monomers, as in digestion.

1. Starch: add iodine solution[1]
2. a blue-black colour is a positive result[1]
3. Protein: add biuret solution[1]
4. a purple/violet colour is a positive result[1]
5. A cloudy emulsion shows a lipid (fat) is present[1]
6. dissolve the sample in ethanol then add water — a cloudy white emulsion forms[1]

Starch → iodine blue-black; protein → biuret purple; lipid → cloudy emulsion (ethanol then water).

A zwitterion is an amino acid that has transferred a proton internally from its –COOH to its –NH₂, giving –NH₃⁺ and –COO⁻ ends and no overall charge. It can accept H⁺ (onto –COO⁻) in acid and donate H⁺ (from –NH₃⁺) in alkali, so it resists pH change — acting as a buffer.

Compounds with the same molecular formula but different structural arrangements of their atoms.

Propan-1-ol; the functional group is the hydroxyl group, –OH (an alcohol).

In homolytic fission each atom takes one electron, forming two radicals. In heterolytic fission one atom takes both electrons, forming a positive and a negative ion.

1. An electrophile is an electron-pair acceptor (electron-poor species)[1]
2. A nucleophile is an electron-pair donor (electron-rich species)[1]
3. Bromine + ethene is an addition reaction[1]
4. Haloalkane with hot alcoholic KOH is an elimination reaction[1]
5. A functional group is the atom or group of atoms[1]
6. that gives a molecule its characteristic chemical properties[1]

Electrophile accepts an electron pair, nucleophile donates one; Br₂ + ethene = addition, elimination gives the alkene; functional group = the reactive part defining the chemistry.

But-2-ene has a C=C double bond, and there is no rotation about it, so the methyl groups are fixed either on the same side (cis) or opposite sides (trans). Butane has only single bonds, which rotate freely, so no fixed arrangement — and no cis–trans isomers.

1. A nucleophile is an electron-pair donor (electron-rich)[1]
2. an electrophile is an electron-pair acceptor (electron-poor)[1]
3. (i) Bromine acts as an electrophile[1]
4. in an electrophilic addition to the C=C double bond[1]
5. (ii) Hydroxide acts as a nucleophile[1]
6. in a nucleophilic substitution of the haloalkane[1]

Nucleophile = electron-pair donor, electrophile = acceptor; Br₂/ethene = electrophilic addition; OH⁻/bromoethane = nucleophilic substitution.

1. Step 1: hydrate ethene with steam and an acid catalyst to give ethanol[1]
2. (~300 °C, ~60 atm, phosphoric acid)[1]
3. Step 2: oxidise the ethanol with acidified potassium dichromate[1]
4. under reflux, to give ethanoic acid[1]
5. React bromoethane with potassium cyanide in ethanol to give propanenitrile[1]
6. useful because it adds a carbon, lengthening the chain[1]

Ethene → ethanol (steam + acid catalyst) → ethanoic acid (oxidise with acidified dichromate, reflux); bromoethane + KCN → propanenitrile (adds a carbon).

The mass/charge value of the molecular ion peak (highest m/z, ignoring isotope peaks) gives the relative molecular mass of the compound.

It identifies bonds/functional groups present; for example a strong absorption near 1700 cm⁻¹ indicates a C=O group (or a broad band ~3300 cm⁻¹ indicates O–H).

Number of peaks = number of different hydrogen environments; integration (area) = the ratio of hydrogens in each environment; splitting = the number of hydrogens on neighbouring carbons (n+1 lines for n neighbours).

1. Relative molecular mass = 46[1]
2. The broad ~3300 cm⁻¹ band suggests an O–H group (alcohol)[1]
3. Three peaks means three different hydrogen environments[1]
4. Ethanol, CH₃CH₂OH, has Mr 46[1]
5. It has three H environments (CH₃, CH₂, OH)[1]
6. so X is ethanol[1]

Mr = 46; O–H group present; three H environments (CH₃, CH₂, OH) → X is ethanol, CH₃CH₂OH.

Rf = distance moved by spot ÷ distance moved by solvent = 3.6 ÷ 4.8 = 0.75.

1. Relative molecular mass = 60[1]
2. 1710 cm⁻¹ suggests C=O and the broad 3000 cm⁻¹ suggests the O–H of a carboxylic acid[1]
3. Two peaks means two hydrogen environments[1]
4. A carboxylic acid of Mr 60 is ethanoic acid, CH₃COOH[1]
5. it has two H environments (CH₃ and COOH)[1]
6. and contains both C=O and O–H, matching the IR — so X is ethanoic acid[1]

Mr = 60; C=O + O–H (carboxylic acid); two H environments → X is ethanoic acid, CH₃COOH.

The heat energy change for a reaction measured at standard conditions (298 K, 1 atm), with substances in their standard states.

The total enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same. It lets us calculate enthalpy changes that cannot be measured directly.

q = mcΔT, where q = heat energy, m = mass of solution, c = specific heat capacity, ΔT = temperature change.

1. q = mcΔT = 100 × 4.2 × 30[1]
2. = 12600 J = 12.6 kJ[1]
3. Moles ethanol = 0.50 ÷ 46 = 0.0109 mol[1]
4. ΔH = energy ÷ moles = 12.6 ÷ 0.0109[1]
5. = 1156 kJ/mol (magnitude)[1]
6. ΔH is negative (exothermic): ≈ −1160 kJ/mol[1]
7. Heat is lost to the surroundings / incomplete combustion[1]

q = 12.6 kJ; 0.0109 mol; ΔH ≈ −1160 kJ/mol; lower magnitude due to heat loss / incomplete combustion.

Bonds broken = 436 + 242 = 678 kJ. Bonds made = 2 × 431 = 862 kJ. ΔH = 678 − 862 = −184 kJ/mol (exothermic).

1. The energy released when gaseous ions[1]
2. combine to form one mole of solid ionic lattice[1]
3. MgO contains Mg²⁺ and O²⁻ ions with 2+ and 2− charges[1]
4. NaCl contains only 1+ and 1− ions[1]
5. the ions in MgO are also smaller[1]
6. higher charges and smaller radii give stronger electrostatic attraction, so a much larger (more exothermic) lattice energy[1]

Lattice energy = energy released forming a mole of lattice from gaseous ions; MgO's higher charges and smaller ions give much stronger attraction than NaCl.

Oxidation is loss of electrons; reduction is gain of electrons (OIL RIG).

The breakdown (decomposition) of an ionic compound, when molten or in aqueous solution, by the passage of an electric current.

Cathode (−): Pb²⁺ + 2e⁻ → Pb (lead metal). Anode (+): 2Br⁻ → Br₂ + 2e⁻ (bromine gas).

1. Melting frees the ions[1]
2. so they can move to the electrodes and carry the current[1]
3. Al³⁺ + 3e⁻ → Al[1]
4. balanced with electrons[1]
5. Oxygen is produced at the anode[1]
6. The hot oxygen reacts with the carbon anodes forming carbon dioxide[1]
7. so the anodes burn away and must be replaced[1]

Molten = mobile ions to carry current; cathode Al³⁺ + 3e⁻ → Al; anode makes O₂, which burns the carbon anodes to CO₂ so they wear away.

Oxygen is −2, so 4 × (−2) = −8. The whole ion has a charge of −1, so Mn + (−8) = −1, giving Mn = +7.

1. The anode (positive electrode) is the impure copper[1]
2. Cu → Cu²⁺ + 2e⁻ (copper dissolves)[1]
3. Pure copper deposits on the cathode[1]
4. Cu²⁺ + 2e⁻ → Cu[1]
5. Impurities do not dissolve / are not deposited[1]
6. they collect below the anode as anode sludge[1]

Anode = impure copper (Cu → Cu²⁺ + 2e⁻); cathode gains pure copper (Cu²⁺ + 2e⁻ → Cu); impurities drop as anode sludge.

Acids produce hydrogen ions, H⁺, in aqueous solution; their pH is below 7.

A strong acid is fully ionised in water; a weak acid is only partly ionised. It is about degree of ionisation, not concentration.

Hydrochloric acid + sodium hydroxide → sodium chloride + water. HCl + NaOH → NaCl + H₂O.

1. CuO + H₂SO₄ → CuSO₄ + H₂O (correct formulae)[1]
2. Equation balanced[1]
3. Warm the acid and add excess copper(II) oxide, then filter off the unreacted solid[1]
4. Heat the filtrate to evaporate some water / to the point of crystallisation[1]
5. Leave to cool so crystals form[1]
6. Filter and dry the crystals (e.g. between filter paper)[1]
7. Excess ensures all the acid reacts, and the excess solid can simply be filtered off[1]

CuO + H₂SO₄ → CuSO₄ + H₂O; react with excess base, filter, evaporate to crystallisation, cool, filter, dry; excess guarantees all acid used and is easily removed.

pH = −log₁₀(1 × 10⁻³) = 3. pH 3 is below 7, so the solution is acidic.

1. CuO + H₂SO₄ → CuSO₄ + H₂O (correct formulae)[1]
2. balanced[1]
3. Warm the acid and add copper(II) oxide in excess until no more reacts[1]
4. Filter to remove the excess copper(II) oxide[1]
5. Heat the filtrate to evaporate water to the point of crystallisation, then leave to cool[1]
6. Filter off the crystals and dry them (e.g. between filter paper)[1]

CuO + H₂SO₄ → CuSO₄ + H₂O; react with excess base, filter, evaporate to crystallisation, cool, filter, dry.

The change in concentration of a reactant or product per unit time.

The power to which the concentration of that reactant is raised in the rate equation.

First order in A, second order in B, overall third order. Doubling [B] multiplies the rate by 2² = 4.

1. k increases with temperature[1]
2. Particles have more energy and collide more often[1]
3. and a greater fraction exceed the activation energy, so more collisions are successful[1]
4. The rate-determining step is the slowest step in a multi-step reaction[1]
5. It controls the overall rate[1]
6. s⁻¹[1]

k rises with temperature (more successful collisions); rate-determining step = the slowest step, which limits overall rate; first-order k has units s⁻¹.

Overall order = 2 + 1 = 3 (third order). Doubling [A] multiplies the rate by 2² = 4.

1. The slowest step in a multi-step reaction[1]
2. which controls (limits) the overall rate[1]
3. A catalyst provides an alternative pathway with lower activation energy[1]
4. so a greater fraction of collisions are successful, raising k[1]
5. Higher temperature gives molecules more energy[1]
6. so more exceed the activation energy — more successful collisions and a larger k[1]

Rate-determining step = slowest step controlling the rate; catalyst and higher temperature both raise k by increasing the fraction of successful (Ea-exceeding) collisions.

Reactivity increases down the group because the outer electron is further from the nucleus and more easily lost.

It floats, melts into a ball, fizzes and moves around, giving off hydrogen: 2Na + 2H₂O → 2NaOH + H₂.

Thermal stability increases down the group. The cations get larger with less charge density, so they polarise/distort the carbonate ion less, making it harder to decompose.

1. Magnesium reacts very slowly with cold water (faster with steam)[1]
2. Calcium reacts readily with cold water, fizzing[1]
3. Ca + 2H₂O → Ca(OH)₂ + H₂[1]
4. balanced equation[1]
5. Reactivity increases down the group[1]
6. because the outer electrons are further from the nucleus and more easily lost[1]

Mg very slow with cold water, Ca reacts readily; Ca + 2H₂O → Ca(OH)₂ + H₂; reactivity rises down the group (electrons lost more easily).

Down the group the cations get larger and have lower charge density, so they polarise (distort) the nitrate ion less. The nitrate is therefore harder to break down, so a higher temperature is needed and thermal stability increases.

1. Reactivity increases down the group[1]
2. the outer electrons are further from the nucleus and more shielded[1]
3. so they are lost more easily, making the metal more reactive[1]
4. Hydroxides become more soluble down the group[1]
5. sulfates become less soluble down the group[1]
6. so barium sulfate is highly insoluble (used in the sulfate test)[1]

Reactivity rises down Group 2 (outer electrons lost more easily); hydroxides get more soluble, sulfates less soluble.