Why did I lose the mark?

The answer that scoresMultiply the mass of each isotope by its relative abundance, add the products and divide by 100: (24 x 79.0) + (25 x 10.0) + (26 x 11.0) = 1896 + 250 + 286 = 2432, and 2432 / 100 = 24.3. The relative atomic mass has no units, because it is a mass compared with one twelfth of the mass of an atom of carbon-12.

The answer that scoresDown a group the first ionisation energy decreases. The outer electron is in a shell further from the nucleus and is shielded by more inner shells, so less energy is needed to remove it: the value falls from 520 kJ mol-1 for lithium to 496 for sodium and 419 for potassium. The number of electrons in the atom is irrelevant; what matters is how strongly the outer electron is held.

The answer that scoresChromium is [Ar] 3d5 4s1 and copper is [Ar] 3d10 4s1, because the half-filled and completely filled d sub-shells have a lower energy than the alternatives. Because 3d lies below 4s in energy once it is occupied, a transition element loses its 4s electrons first when it forms a positive ion: iron is [Ar] 3d6 4s2, Fe2+ is [Ar] 3d6 and Fe3+ is [Ar] 3d5.

The answer that scoresA 2p sub-shell contains three orbitals, labelled 2px, 2py and 2pz, and each holds a maximum of two electrons, so the sub-shell holds six electrons in total. An s sub-shell holds two electrons in its single orbital and a d sub-shell holds ten in its five orbitals. In electrons in boxes notation a 2p sub-shell is drawn as three boxes, filled singly with parallel spins before any box is given a second electron.

The answer that scoresThe relative formula mass of sodium chloride is 23.0 + 35.5 = 58.5. Relative molecular mass is reserved for substances that genuinely exist as molecules, such as H2O, CO2 and C2H5OH. Use relative formula mass for any ionic compound and for giant covalent structures.

The answer that scoresThe empirical formula is CH2O. If Mr is 60.0, the empirical formula mass is 12.0 + 2.0 + 16.0 = 30.0, so the multiplier is 60.0 / 30.0 = 2 and the molecular formula is C2H4O2, which is ethanoic acid. With Mr = 180 the multiplier would be 6 and the molecular formula would be C6H12O6, glucose.

The answer that scoresPercentage yield = (mass actually obtained / maximum mass from the equation) x 100 = 4.20 / 5.00 x 100 = 84.0%. The answer must always be less than 100%; a value above 100% means the product is wet or impure, or that the fraction has been inverted.

The answer that scoresConvert the volume first: 250 cm3 = 250 / 1000 = 0.250 dm3. Then c = 0.100 / 0.250 = 0.400 mol dm-3. The same conversion is needed in every titration: divide the cm3 from the pipette or burette by 1000 before using c = n / V.

The answer that scoresSodium chloride is a giant ionic lattice containing Na+ and Cl- ions, and the lattice is held together by the electrostatic attraction between oppositely charged ions. This attraction acts between every ion and all the oppositely charged ions around it, so a large amount of energy is needed to break the lattice. Sodium chloride therefore has a high melting point, about 800 degrees C. Ionic bonding, covalent bonding and metallic bonding are all much stronger than the van der Waals' forces that act between molecules.

The answer that scoresWater has two bonding pairs and two lone pairs around the oxygen, so the four pairs are arranged tetrahedrally. The two lone pairs repel more strongly than the bonding pairs, so the H-O-H angle is squeezed from the tetrahedral 109.5 degrees to 104.5 degrees. Methane, with four bonding pairs and no lone pairs, keeps the full 109.5 degrees, and ammonia, with one lone pair, sits in between at 107 degrees.

The answer that scoresEthene is made of three sp2 hybrid orbitals on each carbon. Two of them on each carbon form C-H sigma bonds, and one forms the C-C sigma bond by end-on overlap, giving 120 degree angles in a plane. The remaining unhybridised 2p orbital on each carbon, perpendicular to that plane, overlaps sideways with its partner to give one pi bond with electron density above and below the sigma bond. The double bond is therefore one sigma plus one pi, and the weaker pi bond is the one that breaks when bromine adds across the C=C to give 1,2-dibromoethane.

The answer that scoresIn ice each water molecule is hydrogen bonded to four others, using its two O-H groups to donate and the two lone pairs on its oxygen to accept, and this tetrahedral arrangement builds an open lattice containing hexagonal rings. As ice melts, some of those hydrogen bonds break and the structure collapses, so the molecules can pack closer together and the liquid is denser than the solid. That is why ice floats, and why the relatively high melting and boiling points, and the high surface tension of water, are all described as anomalous for such a small molecule.

The answer that scoresConvert everything into SI units first. A gas measured at 25 C is at 25 + 273 = 298 K, so with p in Pa, V in m3 and T in K, n = pV / (8.31 x 298). Mixed units are the most common reason for losing marks in pV = nRT questions, so write the unit beside each value as you substitute.

The answer that scoresIodine is a simple molecular solid in which I2 molecules are held in a lattice by weak id-id forces. Melting supplies enough energy to overcome those weak forces and separate the molecules, but not enough to break the strong covalent bonds within the molecules, which is why the purple vapour that forms is still made of I2 molecules. Similar molecules with stronger intermolecular forces, such as hydrogen-bonded ice, melt at much higher temperatures.

The answer that scoresIn graphite each carbon atom is bonded to three others in flat hexagonal layers, and the fourth outer electron from each atom is delocalised across the layer and free to move, so graphite conducts electricity along its layers. In diamond each carbon atom forms four tetrahedral covalent bonds, so all its outer electrons are localised in bonds and diamond does not conduct. In both substances melting requires the breaking of strong covalent bonds, which is why both have very high melting points.

The answer that scoresSolid sodium chloride does not conduct, because its ions are locked in the lattice and cannot move. When the crystal is melted, or dissolved in water, the lattice breaks down and the Na+ and Cl- ions become free to move, so both molten sodium chloride and aqueous sodium chloride conduct electricity. This difference between the solid and the molten or dissolved substance is exactly how you identify a giant ionic lattice.

The answer that scoresThe standard enthalpy change of formation of methane is the enthalpy change when one mole of methane is formed from its elements, carbon and hydrogen, in their standard states at 298 K and 101 kPa. Its value is about -75 kJ mol-1. The standard enthalpy change of combustion of methane is a different quantity entirely: it is -890 kJ mol-1, the energy transferred when one mole of methane is burned completely in excess oxygen.

The answer that scoresThe temperature rise measured is smaller than it should be, because energy is lost to the surroundings and absorbed by the cup and the thermometer. The measured value of -56.8 kJ mol-1 is therefore slightly less exothermic than the accepted value of about -57 kJ mol-1 for a strong acid with a strong base. The experiment can be improved by insulating the cup more effectively, fitting a lid, mixing the reagents quickly, and plotting temperature against time so that the graph can be extrapolated back to the moment of mixing to correct for the heat lost during the measurement.

The answer that scoresdelta Hr = energy absorbed breaking bonds - energy released making bonds = (436 + 242) - (2 x 431) = 678 - 862 = -184 kJ mol-1. The negative sign shows that more energy is released when the two H-Cl bonds form than is absorbed when the H-H and Cl-Cl bonds break, so the reaction is exothermic.

The answer that scoresThe difference is expected and is not an arithmetic error. Mean bond energies average the same bond over many compounds, in which the surroundings of the bond differ, so they do not reproduce a measured enthalpy change exactly. Bond energies also apply only to gaseous species, so the value they give refers to gaseous water, not to the liquid water formed in the standard combustion reaction. A calculation from exact bond energies, in a diatomic molecule such as H2 or HCl, does agree with the measured value.

The answer that scoresHydrogen has oxidation number +1 in most compounds, including water, methane and sulphuric acid. The exception is the metal hydrides, such as sodium hydride NaH and calcium hydride CaH2, where hydrogen is -1. Check with the rule that the sum of the oxidation numbers equals the charge: in NaH, sodium is +1 and the compound is neutral, so hydrogen must be -1.

The answer that scoresOxygen is -2 in most compounds but -1 in the peroxides, which include hydrogen peroxide H2O2, sodium peroxide Na2O2 and barium peroxide BaO2. Work it out from the charge: 2(+1) + 2x = 0, so x = -1. Oxygen is +2 in OF2, because fluorine is the more electronegative element and takes the shared electrons.

The answer that scoresDisproportionation is the oxidation and reduction of the same element in the same reaction. In Cl2 + 2NaOH -> NaCl + NaClO + H2O, chlorine is 0 in Cl2, -1 in the chloride and +1 in the chlorate(I) ion. A second example is 2Cu+ -> Cu2+ + Cu, where copper goes from +1 to +2 in one product and from +1 to 0 in the other.

The answer that scoresIn electrolysis, positive ions (cations) move to the cathode, which is the negative electrode, and are reduced there by gaining electrons. Negative ions (anions) move to the anode, which is the positive electrode, and are oxidised there by losing electrons. In the electrolysis of concentrated aqueous sodium chloride, hydrogen is formed at the cathode by reduction of water and chlorine is formed at the anode by oxidation of chloride ions.

The answer that scoresA catalyst lowers the activation energy of both the forward and the reverse reactions equally, so both rates increase by the same factor. The position of equilibrium is unchanged and the value of Kc is unchanged. The mixture simply reaches equilibrium sooner, which is why a catalyst lets an industrial plant work at a compromise temperature that gives a better equilibrium yield at an acceptable rate.

The answer that scoresIncreasing the pressure does shift the position of equilibrium to the right, because four moles of gas become two, but the equilibrium concentrations change in exactly the ratio that leaves Kc the same. Kc changes only when the temperature changes. For this exothermic forward reaction, raising the temperature would decrease the value of Kc, and lowering the temperature would increase it.

The answer that scoresHydrochloric acid is fully dissociated, so 0.100 mol dm-3 HCl gives 0.100 mol dm-3 of H+(aq) ions and a pH of about 1. Ethanoic acid sets up the equilibrium CH3COOH(aq) <-> H+(aq) + CH3COO-(aq), and in a 0.100 mol dm-3 solution only a small fraction of the molecules has donated a proton, so the concentration of H+(aq) ions is only about 1.3 x 10^-3 mol dm-3 and the pH is about 2.9.

The answer that scoresThe yield of ammonia increases as the temperature is lowered and as the pressure is raised, so 450 °C is a compromise between a reasonable yield and a workable rate, and the iron catalyst makes that rate acceptable. About 200 atm is the highest pressure that is economical: it shifts the equilibrium further towards ammonia, but the plant must be built to withstand it. The ammonia formed is liquefied out and the unreacted gases are recycled, so the overall conversion is high even though each pass is not.

The answer that scoresA catalyst provides an alternative reaction pathway of lower activation energy. At the same temperature a much larger fraction of the molecules now has energy greater than or equal to that lower activation energy, so the frequency of effective collisions increases and the reaction goes faster. The catalyst takes part in the reaction and is regenerated at the end, so it is chemically unchanged.

The answer that scoresTwo effects operate. The molecules move faster, so collisions are a little more frequent, but far more importantly the higher temperature shifts and flattens the Boltzmann distribution, so a much greater proportion of the molecules has energy greater than or equal to the activation energy. The area under the curve beyond EA increases greatly, so the frequency of effective collisions rises sharply and the rate increases far more than the collision frequency alone would explain.

The answer that scoresThe catalyst accelerates both directions equally, because it lowers the energy barrier of the forwards and the backwards reaction. The position of equilibrium and the value of the equilibrium constant are unchanged, so the final yield is the same. What changes is the time taken to reach equilibrium: the reaction gets there sooner, which is why industrial catalysts are used to make a workable rate at a moderate temperature.

The answer that scoresDraw a tangent to the curve at t = 30 s and find its gradient: rate = change in concentration / change in time, taken between two points on the tangent. For example, if the tangent passes through 0.750 mol dm-3 at 20 s and 0.450 mol dm-3 at 40 s, the rate at 30 s is 0.300 / 20 = 0.0150 mol dm-3 s-1. The unit is mol dm-3 s-1, and the answer should be given to the same number of significant figures as the data.

The answer that scoresAluminium oxide is amphoteric: it reacts with both an acid and a base. With hydrochloric acid, Al2O3 + 6HCl -> 2AlCl3 + 3H2O, giving a colourless solution of aluminium chloride. With sodium hydroxide solution, Al2O3 + 2NaOH + 3H2O -> 2NaAl(OH)4, giving a solution of sodium tetrahydroxoaluminate(III). Aluminium hydroxide behaves in the same way: Al(OH)3 + 3HCl -> AlCl3 + 3H2O and Al(OH)3 + NaOH -> NaAl(OH)4.

The answer that scoresSilicon is a giant covalent macromolecule: every silicon atom is covalently bonded to four neighbours in a continuous three-dimensional network. To melt it, the strong covalent bonds themselves must be broken, which needs a great deal of energy, so the melting point is very high, about 1414 C. The low melting points in Period 3 belong to P4, S8, Cl2 and Ar, where the molecules are separate and only weak London forces hold them together.

The answer that scoresAlCl3 hydrolyses vigorously to give an acidic solution of pH about 3, with a white precipitate of Al(OH)3 and fumes of HCl: AlCl3 + 3H2O -> Al(OH)3 + 3HCl. PCl5 hydrolyses violently to give a strongly acidic solution of pH about 1: PCl5 + 4H2O -> H3PO4 + 5HCl. The acidic character is caused by the partial positive charge on the aluminium and phosphorus atoms, which attracts a water molecule and displaces the chloride as HCl. Only NaCl, with the low charge density of Na+, dissolves to give a neutral solution of pH 7.

The answer that scoresLet x be the oxidation number of phosphorus. 4x + 10(-2) = 0, so 4x = 20 and x = +5. Phosphorus has five outer-shell electrons and uses all five in P4O10, so +5 is consistent with the position of phosphorus in Period 3, between silicon at +4 and sulfur at +6 in SO3.

The answer that scoresWith cold water: Mg(s) + 2H2O(l) -> Mg(OH)2(s) + H2(g), and the reaction is extremely slow because the sparingly soluble hydroxide coats the metal and keeps the water away from it. With steam: Mg(s) + H2O(g) -> MgO(s) + H2(g); the magnesium glows brightly and a white solid of magnesium oxide is left in the tube.

The answer that scoresThermal stability increases down the group. MgCO3(s) -> MgO(s) + CO2(g) happens on fairly gentle heating, at about 350 °C, whereas CaCO3(s) -> CaO(s) + CO2(g) needs about 825 °C, and the strontium and barium carbonates need higher temperatures still. The larger cation down the group has a lower charge density and polarises the carbonate ion less.

The answer that scoresWith dilute hydrochloric acid the vigour increases from magnesium to barium, because the ionisation energies fall down the group and all the chlorides are soluble. With dilute sulfuric acid the reaction becomes slower down the group: magnesium reacts steadily because magnesium sulfate is soluble, calcium slows down and stops as sparingly soluble calcium sulfate coats the metal, and barium reacts hardly at all because barium sulfate is essentially insoluble.

The answer that scoresAcidify the unknown solution with dilute hydrochloric acid, then add barium chloride solution. A white precipitate that does not dissolve in the dilute hydrochloric acid confirms sulfate ions: Ba2+(aq) + SO4^2-(aq) -> BaSO4(s). If the precipitate fizzes or dissolves in the acid, the solution contained carbonate or sulfite ions instead.

The answer that scoresChlorine is a gas and iodine a solid because the instantaneous dipole-induced dipole forces between the molecules are much stronger in iodine. An iodine molecule has more electrons and a larger, more polarisable electron cloud than a chlorine molecule, so the temporary dipoles it can set up are larger and the molecules are held together more strongly. More energy is therefore needed to separate the molecules, so iodine has the higher boiling point and the lower volatility. The X-X covalent bond is not broken when a halogen boils.

The answer that scoresCl2(g) + 2NaOH(aq) -> NaCl(aq) + NaClO(aq) + H2O(l). The oxidation number of chlorine is 0 in Cl2, -1 in NaCl and +1 in NaClO, so one portion of the chlorine has been reduced and another portion oxidised at the same time. This is a disproportionation. With hot concentrated sodium hydroxide the oxidation goes further and the product is the chlorate(V): 3Cl2(g) + 6NaOH(aq) -> 5NaCl(aq) + NaClO3(aq) + 3H2O(l), chlorine going from 0 to -1 and from 0 to +5.

The answer that scoresSilver chloride is a white precipitate and dissolves in dilute aqueous ammonia. Silver bromide is cream and dissolves only in concentrated aqueous ammonia. Silver iodide is yellow and does not dissolve even in concentrated aqueous ammonia. A yellow precipitate that is insoluble in concentrated ammonia therefore confirms iodide ions, by Ag+(aq) + I-(aq) -> AgI(s). Always quote the colour and the ammonia result together.

The answer that scoresSodium bromide with concentrated sulfuric acid first gives steamy fumes of hydrogen bromide, NaBr(s) + H2SO4(l) -> NaHSO4(s) + HBr(g), and the hydrogen bromide is then oxidised to bromine while sulfur is reduced from +6 to +4: 2HBr(g) + H2SO4(l) -> Br2(g) + SO2(g) + 2H2O(l). The observations are orange-brown bromine vapour and choking sulfur dioxide. Yellow sulfur and hydrogen sulfide appear only with iodide, because iodide is the stronger reducing agent and can take sulfur to 0 and then to -2: 6HI(g) + H2SO4(l) -> 3I2(g) + S(s) + 4H2O(l) and 8HI(g) + H2SO4(l) -> 4I2(g) + H2S(g) + 4H2O(l).

The answer that scoresIn the N2 molecule the two atoms are held by a triple bond with a bond enthalpy of about 945 kJ mol-1, so an enormous amount of energy is needed to break it and any reaction has a very high activation energy. The molecule is also non-polar, because two identical atoms share the electrons equally, so there is no partially charged end for a polar reagent such as H+ or OH- to attack. Nitrogen reacts only at very high temperature, as in lightning or an engine, or with a catalyst, as in the Haber process with iron.

The answer that scoresThe ammonium ion, NH4+, is tetrahedral with bond angles of about 109.5 °. It is formed in an acid-base reaction when the lone pair on the nitrogen atom of ammonia accepts a proton from an acid. The new N-H bond is a dative (coordinate) covalent bond, with both electrons donated by the nitrogen atom, and once it has formed all four N-H bonds are identical in length and strength.

The answer that scoresNitrogen oxides are formed inside the cylinder, where the high temperature makes N2 and O2 react: N2 + O2 -> 2NO. The catalytic converter then removes them from the cooled exhaust gas by reducing them to nitrogen, 2NO + 2CO -> N2 + 2CO2, while at the same time oxidising carbon monoxide to carbon dioxide, 2CO + O2 -> 2CO2, and unburned hydrocarbons to carbon dioxide and water. The catalyst is platinum, palladium and rhodium, it must be hot to work, and it is poisoned by lead and by sulfur in the fuel.

The answer that scoresNitrogen dioxide takes part in the oxidation of sulfur dioxide: NO2 + SO2 -> SO3 + NO, and the NO is oxidised back to NO2 by the air, 2NO + O2 -> 2NO2, so that the overall change is 2SO2 + O2 -> 2SO3 with NO2 unchanged at the end. The sulfur trioxide is the acid anhydride, and it is this that forms sulfuric acid with water: SO3 + H2O -> H2SO4. Separately, NO2 makes acid rain directly by dissolving in water to give nitric acid, 2NO2 + H2O -> HNO3 + HNO2.

The answer that scoresChain isomers have the same functional group but a different carbon skeleton, so but-1-ene, CH2=CHCH2CH3, and 2-methylpropene, (CH3)2C=CH2, are chain isomers, not functional group isomers. Functional group isomers have different functional groups, like propanal and propanone, CH3CH2CHO and CH3COCH3, which are both C3H6O.

The answer that scoresPropene has no geometrical isomers. Restricted rotation about the C=C means the groups cannot swap ends, but one carbon of the double bond carries two hydrogen atoms, so there is no second arrangement to form. But-2-ene, CH3CH=CHCH3, has a methyl group and a hydrogen atom on each carbon of the double bond, so it does exist as a cis and a trans isomer.

The answer that scoresIn the dark there is no reaction at all. Ultraviolet light supplies the energy for the Cl-Cl bond to break homolytically, forming two chlorine radicals; the radicals then take a hydrogen atom from methane and give chloromethane in a chain of propagation steps, which end when two radicals combine in a termination step.

The answer that scoresThe correct arrows run the other way round. The pi bond of the C=C is the electron-rich species, so the first curly arrow starts at the pi bond and points to the hydrogen atom of HBr. A second curly arrow starts in the middle of the H-Br bond and points to the bromine, which takes both electrons and leaves as a bromide ion, leaving a carbocation that the bromide ion then attacks.

The answer that scoresCracking is used to obtain more useful alkanes and alkenes of lower Mr from heavier crude oil fractions. Long-chain alkanes from the heavy fractions are heated with Al2O3 or passed over a hot zeolite catalyst, and they break into shorter-chain alkanes such as octane, which are the petrol components, and into alkenes such as ethene and propene, which are the raw materials for polymers and for ethanol. It is carried out because distilling crude oil gives too much of the heavy, low-demand fractions and too little of the light, high-demand ones, and because crude oil contains almost no alkenes of its own.

The answer that scoresUltraviolet light splits the halogen molecule by homolytic fission: Cl2 -> 2Cl, giving two chlorine atoms, each a free radical with one unpaired electron. In the propagation step a chlorine radical removes a hydrogen atom from ethane, C2H6 + Cl -> C2H5 + HCl, and the ethyl radical then reacts with a chlorine molecule, C2H5 + Cl2 -> C2H5Cl + Cl, which gives chloroethane and regenerates a chlorine radical. The Cl-Cl bond is much weaker than a C-H bond, and that is exactly why the light breaks only the halogen molecule.

The answer that scoresThe pi electrons of the C=C attack the delta-positive hydrogen of HBr, forming a C-H bond and releasing a bromide ion. The hydrogen adds to the terminal carbon, so the carbocation is CH3-CH(+)-CH3, a secondary cation, which is more stable than the primary cation that the alternative route would give because the two electron-releasing alkyl groups spread and reduce the positive charge. The bromide ion then attacks the positive carbon, so the major product is 2-bromopropane. This is Markovnikov's rule: the hydrogen goes to the carbon that already carries more hydrogen atoms.

The answer that scoresCold dilute acidified KMnO4 reacts with ethene to give the diol, ethane-1,2-diol, HOCH2CH2OH, and the purple colour of the manganate(VII) ion disappears. Hot concentrated acidified KMnO4 ruptures the C=C, and a =CH2 group becomes CO2, a =CHR group becomes the carboxylic acid RCOOH and a =CR2 group becomes a ketone; ethene therefore gives carbon dioxide and water, propene gives ethanoic acid and carbon dioxide, and hex-2-ene gives ethanoic acid and butanoic acid.

The answer that scoresWith NaOH(aq) and heat under reflux the product is ethanol: C2H5Br + NaOH -> C2H5OH + NaBr, a substitution. With NaOH in ethanol and heat under reflux the product is ethene: C2H5Br + NaOH -> C2H4 + NaBr + H2O, an elimination. Always state the solvent, because it decides whether the nucleophile attacks the carbon or a hydrogen atom on the neighbouring carbon.

The answer that scoresThe C-X bond enthalpy decreases from C-F to C-I, so the activation energy for breaking the bond decreases and the rate of hydrolysis increases in the order RI > RBr > RCl > RF. The silver nitrate test shows it directly: 1-iodohexane gives a yellow precipitate of silver iodide in the cold, while 1-chlorohexane needs several minutes of heating before white silver chloride appears.

The answer that scoresIn SN1 the C-Br bond breaks heterolytically first, in the slow rate-determining step, with the pair of electrons moving onto the bromine to give a bromide ion and a carbocation. The hydroxide ion then attacks the carbocation in a fast second step. The rate depends only on the halogenoalkane, rate = k[RX], which is why SN1 is favoured by tertiary halogenoalkanes whose carbocations are stabilised by the inductive effect of their alkyl groups.

The answer that scoresBromoethane with KCN in ethanol, heated under reflux, gives propanenitrile: C2H5Br + KCN -> C2H5CN + KBr. The chain has grown by one carbon, and to reach an amine you would either reduce the nitrile, or use the separate reaction of bromoethane with ammonia in ethanol heated under pressure: C2H5Br + 2NH3 -> C2H5NH2 + NH4Br.

The answer that scoresHeating under reflux with an excess of acidified potassium dichromate(VI) oxidises propan-1-ol all the way to propanoic acid: propan-1-ol -> propanal -> propanoic acid, with [O] needed at each stage. Propanal is the product only if the reaction mixture is distilled and the aldehyde, which boils at about 49 degrees C, is collected as it forms and removed from the oxidising agent.

The answer that scoresA tertiary alcohol has no hydrogen atom on the carbon carrying the -OH group, so it is not oxidised by acidified potassium dichromate(VI) or by acidified potassium manganate(VII). The solution stays orange and no ketone is formed. This is the reaction that identifies an unknown tertiary alcohol, because a primary or a secondary alcohol in the same tube would turn the solution from orange to green.

The answer that scoresThe pale yellow precipitate is tri-iodomethane, CHI3, formed because ethanol contains the CH3CH(OH)- group; the ion left in solution is the methanoate ion, HCO2-, since R = H for ethanol. Naming the precipitate correctly, and giving the ion that goes with it, is what scores the marks.

The answer that scoresAcidified potassium dichromate(VI) is the oxidising agent and is used in excess when a full conversion to the carboxylic acid is wanted. The orange-to-green colour change is the oxidising agent being reduced, and it is also the observation that shows the alcohol has been oxidised. A true catalyst, such as the H3PO4 in the hydration of ethene, is not used up and is not reduced.

The answer that scoresTo stop at the aldehyde the mixture must be heated with acidified potassium dichromate(VI) and the propanal distilled off as it forms, because propanal is more volatile than propan-1-ol and leaves the flask before it can be oxidised further. Reflux is used when the carboxylic acid is wanted.

The answer that scoresThe orange precipitate shows only that a carbonyl group is present. To find out whether the compound is an aldehyde or a ketone, warm separate samples with Fehling's solution and with Tollens' reagent. An aldehyde gives a brick-red precipitate of Cu2O with Fehling's solution and a silver mirror with Tollens' reagent; a ketone gives no change with either.

The answer that scoresThe nucleophile is the cyanide ion, CN-. A few drops of KCN are added to provide it, because HCN is a weak acid. The lone pair on the carbon of CN- attacks the delta-positive carbonyl carbon, the pi bond of the C=O breaks and both electrons move onto the oxygen to give an alkoxide ion, and the alkoxide is then protonated by a molecule of HCN to give the hydroxynitrile. The cyanide ion is regenerated, so KCN acts as a catalyst.

The answer that scoresThe hydride ion from NaBH4 adds to the carbonyl carbon and the oxygen is protonated, giving propan-2-ol, CH3CH(OH)CH3, which is a secondary alcohol. A primary alcohol such as propan-1-ol can only be obtained by reducing an aldehyde, in this case propanal.

The answer that scoresUnder reflux with excess acidified dichromate(VI) the oxidation runs through both steps, primary alcohol -> aldehyde -> carboxylic acid, so propan-1-ol gives propanoic acid. To stop at the aldehyde you heat gently and distil the propanal off as it is formed, because an aldehyde boils well below the alcohol it came from.

The answer that scoresEthyl ethanoate with dilute sodium hydroxide and heat gives sodium ethanoate and ethanol, and the hydrolysis goes to completion because the alkali removes the carboxylic acid from the equilibrium as its salt. To obtain ethanoic acid itself the mixture must be acidified afterwards. Acid hydrolysis, by contrast, is reversible and gives ethanoic acid and ethanol in an equilibrium mixture.

The answer that scoresA carboxylic acid is a stronger acid than an alcohol. Ethanoic acid donates a proton to water to give the ethanoate ion, in which the negative charge is delocalised over both oxygen atoms; ethanol gives the ethoxide ion, in which the charge sits on a single oxygen atom, so that ion is far less stable and the equilibrium lies much further to the left. That is why ethanoic acid fizzes with sodium carbonate and ethanol does not react with it.

The answer that scoresEthanoic acid is reduced to ethanol, a primary alcohol, by lithium tetrahydridoaluminate, LiAlH4, in dry ether: CH3COOH + 4[H] -> CH3CH2OH + H2O. Dry ether is essential because LiAlH4 reacts violently with water, so the apparatus must be dry and no flame may be used near the ether. NaBH4 would be the correct answer only for an aldehyde or a ketone.

The answer that scores1-Bromopropane, CH3CH2CH2Br, has three carbon atoms, so the product is butanenitrile, CH3CH2CH2CN, which has four. The same counting applies after hydrolysis: butanenitrile gives butanoic acid, CH3CH2CH2COOH, not propanoic acid.

The answer that scoresAlkaline hydrolysis gives CH3CH2CN + NaOH + H2O -> CH3CH2COONa + NH3. The mixture must then be acidified, CH3CH2COONa + HCl -> CH3CH2COOH + NaCl, before propanoic acid can be distilled out. With dilute acid the carboxylic acid is formed directly, together with ammonium ions.

The answer that scoresThe reaction is nucleophilic substitution. A lone pair on the nitrogen atom of the ammonia attacks the delta-positive carbon atom of the C-Br bond, and the C-Br bond breaks heterolytically so that the pair of electrons goes to the bromine, which leaves as Br-. The product is the protonated amine, CH3CH2NH3+, which loses a proton to give ethylamine; a second molecule of ammonia removes the HBr.

The answer that scoresThe cyanide ion is the nucleophile: a lone pair on its carbon atom attacks the delta-positive carbon of the C=O group, the pi bond of the C=O moves onto the oxygen, and the negative oxygen then takes a proton. A small amount of KCN supplies CN- ions so the reaction goes at a usable rate, and because the CN- is regenerated at the end of the mechanism only a catalytic amount is needed. The product is 2-hydroxy-2-methylpropanenitrile, CH3C(OH)(CH3)CN.

The answer that scoresThe monomer is chloroethene, CH2=CHCl, also called vinyl chloride. It is unsaturated, so the C=C opens and the units join to give the repeat unit -CH2-CHCl-. Chloroethane has no double bond, so it cannot be the monomer and the repeat unit could not be formed from it.

The answer that scoresThe repeat unit is (-CH2-CH2-)n, with a single bond between the two carbons and the two joining bonds drawn through the brackets. The polymer is saturated, which is why poly(ethene) does not decolourise bromine water while ethene does.

The answer that scoresBurning PVC gives carbon dioxide and water, but also hydrogen chloride, because the polymer contains chlorine. The hydrogen chloride is acidic and corrosive and dissolves in rain to give acid deposition, and incomplete combustion also produces toxic carbon monoxide and soot. Incinerators therefore burn PVC in a controlled furnace and pass the flue gases through an alkaline scrubber to neutralise the hydrogen chloride: Ca(OH)2 + 2HCl -> CaCl2 + 2H2O.

The answer that scoresThe theoretical mass of polymer equals the mass of monomer, so 10.0 g of ethene gives up to 10.0 g of poly(ethene). The empirical formula is unchanged, CH2, and any shortfall in a real plant is the unreacted monomer that was recovered rather than a product that was lost.

The answer that scoresHeat propan-1-ol with acidified potassium dichromate(VI) and distil the product out as it forms. Propanal boils at a lower temperature than propan-1-ol, so it leaves the mixture before the dichromate can oxidise it further to propanoic acid. Reserve reflux for the day you want the carboxylic acid.

The answer that scoresUse LiAlH4 in dry ether, which reduces a carboxylic acid to the primary alcohol. It also reduces aldehydes, ketones, esters and nitriles, so it cannot be used when another group has to be left untouched.

The answer that scoresBromoethane and KCN in aqueous ethanol, heated under reflux, give propanenitrile, CH3CH2CN. Hydrolysis with dilute hydrochloric acid under reflux then gives propanoic acid, CH3CH2COOH. Count the carbons across the arrow: C2 to C3 is the whole point of the step.

The answer that scoresHeating 1-bromobutane with NaOH(aq) gives butan-1-ol as the main product, contaminated with but-1-ene formed by elimination. If the alkene is what you want, change the solvent: KOH in ethanol makes elimination the main reaction and the alkene the main product.

The answer that scoresIdentify the molecular ion as the peak at the highest m/e value in the spectrum, ignoring the small [M + 1] and [M + 2] isotope peaks just above it. In ethanol that peak is at m/e 46, so Mr is 46, which matches C2H5OH ((2 x 12) + (6 x 1) + 16 = 46). The peak at m/e 31 is the base peak and is the fragment ion CH2OH+, formed when the molecular ion loses a methyl radical, and 31 + 15 = 46 confirms it.

The answer that scoresOne chlorine atom gives an [M + 2] peak about one third the height of the M+ peak, because chlorine is roughly 75% chlorine-35 and 25% chlorine-37. One bromine atom gives an [M + 2] peak about the same height as the M+ peak, because bromine is almost exactly 50% bromine-79 and 50% bromine-81. A peak at about half the height of M+ therefore indicates one bromine atom, and if the compound also contained two chlorine atoms the spectrum would show an M+, M+2, M+4 pattern in the ratio 9 : 6 : 1.

The answer that scoresThe broad absorption between 2500 and 3300 cm-1 is the O-H of the carboxylic acid, and the sharp strong absorption at about 1710 cm-1 is the C=O. It is the combination that matters: a broad O-H with no C=O would be an alcohol, whereas a broad O-H together with a C=O is a carboxylic acid. An O-H group never gives a sharp, narrow absorption, and a carbonyl group never gives a broad one.

The answer that scoresMultiply each isotope mass by its fractional abundance and add: Ar = (63 x 69.2 + 65 x 30.8) / 100 = (4359.6 + 2002) / 100 = 63.6, to one decimal place. The answer lies closer to 63 than to 65 because copper-63 is the more abundant isotope, and it is not a whole number precisely because it is a weighted mean.

The answer that scoresLattice energy is the enthalpy change when one mole of a solid ionic lattice is formed from its gaseous ions, so for sodium chloride it is -788 kJ mol-1; energy is released as the oppositely charged ions come together. Separating the lattice takes in the same amount of energy, +788 kJ mol-1, and that is the endothermic step you draw in an enthalpy of solution cycle, written as -delta H_latt.

The answer that scoresUse the first and second ionisation energies separately and add them: +738 + +1451 = +2189 kJ mol-1 for Mg(g) -> Mg2+(g) + 2e-. The same principle applies to a 2- anion such as oxide, where the first electron affinity is negative (about -142 kJ mol-1) but the second is positive (about +798 kJ mol-1), so the two must be added with their own signs.

The answer that scoresA low temperature does make the Haber reaction more feasible, because delta H is negative and delta S is negative so T delta S is a positive term that is subtracted and grows with temperature. But feasibility is not rate, so the plant uses a compromise temperature of about 700 K, an iron catalyst to speed up the reaction, a high pressure to favour the side with fewer gas molecules, and continuous removal of the ammonia to shift the position of equilibrium.

The answer that scoresConvert first: delta S = +160.4 J K-1 mol-1 = +0.1604 kJ K-1 mol-1, so T delta S = 298 x 0.1604 = 47.8 kJ mol-1 and delta G = 178 - 47.8 = +130.2 kJ mol-1. The positive value is exactly what the sign rules predict for a reaction with positive delta H and positive delta S at a low temperature, and the decomposition only becomes feasible above T = 178 / 0.1604 = 1110 K.

The answer that scoresZinc is oxidised and copper(II) ions are reduced, so E(cell standard) = E(Cu2+/Cu) - E(Zn2+/Zn) = +0.34 - (-0.76) = +1.10 V. The value is positive, so the cell works, zinc is the negative electrode and electrons flow from zinc to copper through the external circuit. Subtracting the other way round simply describes the reverse reaction, and its negative value correctly says that the reverse reaction is not feasible.

The answer that scoresFor Cu2+ + 2e- <-> Cu the Nernst equation is E = 0.34 + (0.059 / 2) log[Cu2+], so lowering [Cu2+] from 1.00 to 0.0100 mol dm-3 gives E = 0.34 + 0.0295 x (-2) = +0.28 V. Increasing the concentration of the oxidised species makes E more positive, and increasing the concentration of the reduced species makes E less positive.

The answer that scoresIn dilute sodium chloride the chloride concentration is far below 1.00 mol dm-3, so the oxygen couple is the one reached and oxygen is given off from 2H2O -> O2 + 4H+ + 4e-. In concentrated brine the high chloride concentration makes the chlorine couple easier to reverse, so chlorine is the anode product from 2Cl- -> Cl2 + 2e-. Hydrogen is produced at the cathode in both cases, because water is reduced rather than sodium ions, E(Na+/Na) = -2.71 V.

The answer that scoresMoles of electrons = It / F = (2.00 x 1800) / 96500 = 0.0373 mol. From Cu2+ + 2e- -> Cu, n = 2, so moles of copper = 0.0373 / 2 = 0.0187 mol and the mass deposited is 0.0187 x 63.5 = 1.19 g. Always take n from the half-equation of the substance you are weighing or measuring.

The answer that scoresUse Ka rather than the concentration of the acid: [H+] = sqrt(Ka x c) = sqrt(1.8 x 10^-5 x 0.100) = sqrt(1.8 x 10^-6) = 1.34 x 10^-3 mol dm-3, so pH = -log10(1.34 x 10^-3) = 2.87. Only about 1.3 per cent of the acid is dissociated, which is what makes the approximation valid.

The answer that scores[OH-] = 2 x 0.0100 = 0.0200 mol dm-3, so [H+] = Kw / [OH-] = 1.00 x 10^-14 / 0.0200 = 5.00 x 10^-13 mol dm-3 and pH = -log10(5.00 x 10^-13) = 12.30. Always look at the formula of the alkali before converting.

The answer that scoresKsp = [Ag+][Cl-] = 1.8 x 10^-10 mol2 dm-6. In 0.100 mol dm-3 sodium chloride solution, [Cl-] is fixed at 0.100 mol dm-3 by the sodium chloride, so [Ag+] = 1.8 x 10^-10 / 0.100 = 1.8 x 10^-9 mol dm-3, compared with 1.34 x 10^-5 mol dm-3 in pure water. The solubility is about seven thousand times lower, because the extra chloride ions push the equilibrium AgCl(s) <-> Ag+(aq) + Cl-(aq) to the left and precipitate silver chloride.

The answer that scoresWrite Kpc = [solute in ethoxyethane] / [solute in water] and let the equilibrium concentrations be 4c and c. With equal volumes, 4c x 0.100 + c x 0.100 = 2.00 x 10^-3 mol, so c = 0.00400 mol dm-3 and 4c = 0.0160 mol dm-3. The amounts are 1.60 x 10^-3 mol in the ethoxyethane and 0.400 x 10^-3 mol in the water.

The answer that scoresThe rate equation must be deduced from experimental data or from the mechanism, and it can be anything the data show. If doubling [B] makes no difference to the rate, the reaction is zero order in B and B does not appear in the rate equation at all. Only the species involved in the rate-determining step (and in any faster steps before it) appear.

The answer that scoresTake the gradient of the concentration-time curve at t = 0, which is the tangent drawn at the start, or use a clock method that times a small, fixed change at the beginning of the reaction. Change one concentration at a time, keep every other concentration the same, hold the temperature constant and repeat the runs, so that each measured rate belongs to known starting concentrations.

The answer that scoresCompare the ratio of the two rates with the ratio of the two concentrations. A concentration ratio of 2 with a rate ratio of 2 is first order, a rate ratio of 4 is second order, and a rate ratio of 1 is zero order. If the concentration is multiplied by 3, the rate multiplies by 3 for first order and by 9 for second order.

The answer that scoresFor a first-order reaction the half-life is constant, and k = 0.693 / t1/2. Starting from 0.800 mol dm-3, the concentration falls to 0.400, then 0.200, then 0.100 mol dm-3 in equal intervals of time. If the half-life measured from the graph grows as the reaction proceeds, the reaction is second order, and the constant half-life test has failed.

The answer that scoresThermal stability increases down the group because polarising power falls as the cation radius increases. Mg2+ is small and highly polarising, so it distorts the carbonate ion strongly, weakens a C-O bond and lowers the decomposition temperature. Ba2+ is large and only weakly polarising, so the carbonate ion is distorted less, its C-O bonds are left stronger and a much higher temperature is needed before carbon dioxide is released.

The answer that scoresMagnesium hydroxide is only sparingly soluble because dissolving it is endothermic, so it is not favoured by the enthalpy change; its low solubility is the reason its saturated solution contains so few hydroxide ions. Going down the group the lattice enthalpy falls faster than the enthalpy of hydration of the cation, so the enthalpy change of solution becomes less endothermic and finally exothermic, and the hydroxides become more soluble.

The answer that scoresThe enthalpy of hydration of the cation falls faster than the lattice enthalpy for the sulfates, because the large sulfate ion means the lattice is not very sensitive to the size of the cation. The enthalpy change of solution therefore becomes less exothermic and eventually endothermic down the group, dissolving becomes less favourable, and solubility falls - which is why barium sulfate is effectively insoluble while magnesium sulfate is very soluble.

The answer that scoresAcidify the unknown solution with dilute hydrochloric acid first - this removes carbonate and sulfite ions by converting them to carbon dioxide and sulphur dioxide - then add barium chloride or barium nitrate solution. A white precipitate that does not dissolve in the dilute acid is barium sulfate, so the solution contains sulfate ions; barium carbonate and barium sulfite would have dissolved in the acid and given no precipitate.

The answer that scoresA transition element is a d-block element which forms one or more stable ions with incomplete d orbitals. Zinc is a d-block element but not a transition element, because its only common ion, Zn2+, has a full 3d sub-shell with the configuration [Ar]3d10. Scandium is the other exception: it forms Sc3+, which is [Ar]3d0, with an empty d sub-shell.

The answer that scoresThe ligands split the degenerate d orbitals into two non-degenerate sets separated by an energy gap dE. An electron absorbs a photon of frequency matching dE and is promoted, so the solution absorbs part of the visible spectrum. The colour observed is the complementary colour, the light that is not absorbed: a solution that absorbs orange light looks blue, which is why [Cu(H2O)6]2+ solutions, which absorb mainly in the red, look pale blue.

The answer that scoresWith a few drops of aqueous ammonia a pale blue precipitate of copper(II) hydroxide forms: Cu2+(aq) + 2OH-(aq) -> Cu(OH)2(s). Adding excess ammonia dissolves the precipitate and gives a deep blue solution containing the complex [Cu(NH3)4(H2O)2]2+: [Cu(H2O)6]2+ + 4NH3 <-> [Cu(NH3)4(H2O)2]2+ + 4H2O. Only four of the six water ligands are replaced, so the coordination number stays at 6.

The answer that scoresOn the free-ion potentials alone the reaction would not be feasible, because 0.15 - 0.54 = -0.39 V. It happens anyway because copper(I) iodide is precipitated, and removing Cu+ from solution pulls the equilibrium over and makes the effective potential of the copper couple much more positive. The reaction is 2Cu2+ + 4I- -> 2CuI + I2, and the iodine formed is titrated with sodium thiosulphate: 2S2O3^2- + I2 -> S4O6^2- + 2I-.

The answer that scoresBenzene and bromine react by electrophilic substitution, and only in the presence of a halogen carrier such as iron(III) bromide, which polarises the bromine molecule. The delocalised ring attacks the polarised bromine, giving a delocalised carbocation intermediate, and a hydrogen ion is then lost to restore the aromatic ring: C6H6 + Br2 -> C6H5Br + HBr. The product is bromobenzene, not a dibromocyclohexane, because losing H+ keeps the delocalisation energy that an addition would destroy.

The answer that scoresEnantiomers differ only in a chiral environment: they rotate the plane of plane-polarised light by equal amounts in opposite directions, and they behave differently towards enzymes and receptors. A racemic mixture is optically inactive because the rotations cancel. To separate the enantiomers you must use a chiral method, such as chromatography on a chiral stationary phase, or reaction with a single pure enantiomer of another compound to form salts that are no longer mirror images and so have different physical properties.

The answer that scoresDraw the arrow from the pi bond of the C=C to the hydrogen atom of H-Br, and a second arrow from the H-Br bond to the bromine. That gives the secondary carbocation, CH3CH+CH3, and a bromide ion, and the bromide then attacks the positive carbon to give 2-bromopropane. The secondary carbocation is preferred because the two alkyl groups release electron density inductively and stabilise the positive charge, and this is Markovnikov's rule.

The answer that scoresChlorobenzene does not hydrolyse with hot aqueous sodium hydroxide, so phenol is not made that way. Industrially phenol is made from benzene by the cumene process: Friedel-Crafts alkylation of benzene with propene gives cumene, oxidation with air gives cumene hydroperoxide, and acid-catalysed cleavage gives phenol and propanone. The mechanism you learn for halogenoalkanes, nucleophilic substitution at a saturated sp3 carbon, does not apply to a halogenoarene.

The answer that scoresBenzene has a delocalised pi system rather than three isolated double bonds, so it does not decolourise bromine water in the dark. To brominate the ring, bromine is used with a halogen carrier such as anhydrous AlBr3, and the reaction is electrophilic substitution giving bromobenzene and hydrogen bromide. An alkene such as cyclohexene does decolourise bromine water, and it does so by electrophilic addition.

The answer that scoresFor (chloromethyl)benzene, C6H5CH2Cl, use chlorine with methylbenzene and ultraviolet light, and leave the aluminium chloride out, so that the substitution is free-radical at the side-chain carbon. For chlorination of the ring, use chlorine with anhydrous AlCl3 in the dark, away from ultraviolet light: the methyl group is a 2,4-director, so the products are 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene, with hydrogen chloride as the by-product.

The answer that scores-NO2 is a 3-director and it deactivates the ring. It removes electron density especially from the 2-, 4- and 6-positions, so the arenium ion formed by attack at the 3-position avoids putting the positive charge on the carbon that carries the nitro group and is therefore the most stable intermediate. Nitration of nitrobenzene gives mainly 1,3-dinitrobenzene, and it is slower than the nitration of benzene.

The answer that scoresThe pi electrons of the ring attack the NO2+ ion, forming a new carbon-nitrogen bond and giving an arenium ion in which the positive charge is delocalised over the five remaining carbon atoms. The electrons of the C-H bond on that same carbon then move back into the ring and a proton, H+, is lost, which restores the delocalised ring and gives nitrobenzene. The H+ reacts with HSO4- to regenerate the sulphuric acid, so H2SO4 is a catalyst.

The answer that scoresChlorobenzene is unchanged by NaOH(aq) under reflux, and it gives no precipitate with AgNO3(aq) in ethanol, because no chloride ions are released. To hydrolyse it at all the conditions must be severe: aqueous sodium hydroxide at about 300 C under a high pressure in a sealed reactor, which gives phenol and sodium chloride. Chloroethane, by contrast, is hydrolysed at reflux to give ethanol and sodium chloride, because its C-Cl bond is a plain polar single bond that a nucleophile can break.

The answer that scoresChlorobenzene gives no precipitate with AgNO3(aq) in ethanol, because the C-Cl bond is not broken under those conditions. Chloroethane gives a white precipitate of silver chloride within a few minutes of warming, because its C-Cl bond breaks and the chloride ion released is precipitated as AgCl. The difference is the delocalisation of the chlorine lone pair into the ring in chlorobenzene, which strengthens the C-Cl bond and stops the halide leaving.

The answer that scoresChlorine on its own will not substitute a benzene ring. Anhydrous AlCl3 acts as a halogen carrier: it accepts a chloride ion from the chlorine molecule, forming AlCl4- and leaving a strongly electrophilic species with Cl+ character: Cl2 + AlCl3 -> Cl+ + AlCl4-. The pi electrons of the ring attack that electrophile to give the arenium ion, a proton is lost, and the H+ released reacts with AlCl4- to give HCl and regenerate the AlCl3, so the catalyst is used over and over again.

The answer that scoresThe reaction is substitution, not addition: chlorine replaces one hydrogen atom of the ring, and that hydrogen leaves as H+. The ring is restored to full aromaticity at the end of the reaction, so no stabilisation energy is lost. The product from benzene is chlorobenzene with hydrogen chloride as the by-product. Addition would only be possible with hydrogen under severe conditions, with a platinum or nickel catalyst.

The answer that scoresPhenol is the stronger acid. When it loses a proton the negative charge on the phenoxide ion is delocalised into the aromatic ring and spread onto the 2-, 4- and 6-positions, which stabilises the ion and pushes the dissociation further to the right. The ethoxide ion has no ring to delocalise the charge into and its electron-donating ethyl group makes the ion less stable, so ethanol is the weaker acid. Phenol has pKa about 10 and ethanol about 16, so phenol will dissolve in NaOH(aq) and ethanol will not.

The answer that scoresPhenol is brominated by Br2(aq) at room temperature with no catalyst and no need for dry conditions. The orange colour of the bromine water disappears and a white precipitate of 2,4,6-tribromophenol forms, with three hydrogens replaced. Benzene needs liquid bromine with a halogen carrier catalyst such as FeBr3 or AlCl3 and anhydrous conditions, and gives only bromobenzene, because its ring is not activated.

The answer that scoresWith NaOH(aq): C6H5OH + NaOH -> C6H5ONa + H2O, sodium phenoxide and water, with no gas evolved. With sodium metal: 2C6H5OH + 2Na -> 2C6H5ONa + H2, sodium phenoxide and hydrogen, and here the sodium does fizz as it reacts. The reaction with sodium metal is not a way of showing that phenol is acidic towards hydroxide: it works for ethanol as well.

The answer that scoresThe nitrous acid, made from NaNO2 and dilute HCl, is added to phenylamine while the mixture is kept below 10 degrees C in an ice bath, which forms benzenediazonium chloride. Only when that salt has formed is the mixture warmed, and the warming is done with water, which replaces the -N2+ group with -OH: benzenediazonium chloride + water -> phenol + nitrogen + HCl.

The answer that scoresChlorine is electronegative, so it withdraws electron density inductively (-I effect) along the sigma bonds of the chain. The withdrawal disperses and stabilises the negative charge in the chloroethanoate ion, so the conjugate base is more stable, the O-H bond is weakened and ionisation is easier: the pKa falls from about 4.8 for ethanoic acid to about 2.9 for chloroethanoic acid, which means the acid is stronger.

The answer that scoresEthanoyl chloride reacts with ethanol rapidly at room temperature with no catalyst: CH3COCl + C2H5OH -> CH3COOC2H5 + HCl. The products are ethyl ethanoate and hydrogen chloride, and the HCl fumes are the evidence that the reaction has taken place. The slow, reversible esterification of ethanoic acid with ethanol, which does need concentrated sulfuric acid as a catalyst, is a different reaction.

The answer that scoresMethanoic acid, HCOOH, is oxidised to carbon dioxide and water by Tollens' reagent (silver mirror), Fehling's solution (red-brown precipitate of copper(I) oxide), acidified KMnO4 (purple decolourised) and acidified K2Cr2O7 (orange to green): HCOOH + [O] -> CO2 + H2O. Ethanedioic acid, HOOCCOOH, is oxidised by warm acidified KMnO4 to carbon dioxide and water. Ethanoic acid, CH3COOH, has a methyl group on the carboxyl carbon and is not oxidised by any of these reagents.

The answer that scoresEthanoyl chloride is hydrolysed rapidly by cold water because its carbon is strongly delta-positive and chloride is a good leaving group, so the addition-elimination route is easy: CH3COCl + H2O -> CH3COOH + HCl. Chlorobenzene is not hydrolysed by water or dilute alkali at all, because the chlorine lone pair is delocalised into the pi system of the ring and the C-Cl bond is strengthened. Hydrolysing chlorobenzene needs very harsh conditions, such as concentrated sodium hydroxide solution at high temperature and pressure. The order of ease of hydrolysis is acyl chloride > alkyl chloride (chloroethane, needing hot aqueous alkali) > halogenoarene.

The answer that scoresPhenylamine is a much weaker base than ammonia. In phenylamine the lone pair on the nitrogen is delocalised into the pi system of the benzene ring, so it cannot be used to accept a proton, and Kb for phenylamine is only about 4 x 10^-10 mol dm-3. Ethylamine is stronger than ammonia because its alkyl group pushes electron density towards the nitrogen by a positive inductive effect, and the order of base strength is ethylamine > ammonia > phenylamine, roughly 4 x 10^-4 > 2 x 10^-5 > 4 x 10^-10 mol dm-3.

The answer that scoresIn an amine the lone pair sits on the nitrogen and is available, so ethylamine is a weak base with Kb about 4 x 10^-4 mol dm-3. In an amide the lone pair is delocalised over the O-C-N system, so ethanamide is essentially neutral and gives no alkaline solution. The same reasoning explains why the peptide bond that holds amino acids together is not basic: it is an amide bond.

The answer that scoresBromoethane with an excess of concentrated ammonia in ethanol in a sealed tube gives ethylamine as the main product: CH3CH2Br + 2NH3 -> CH3CH2NH2 + NH4Br. The excess ammonia removes the hydrogen bromide formed and keeps the concentration of the amine low, so further substitution is limited. A secondary amine such as diethylamine is made by heating a halogenoalkane with an excess of the primary amine in ethanol in a sealed tube: CH3CH2Br + CH3CH2NH2 -> (CH3CH2)2NH + HBr.

The answer that scoresThe charge on an amino acid is decided by comparing the buffer pH with its isoelectric point. At pH 6, glutamic acid (isoelectric point about 3.2) is above its isoelectric point, so it carries a net negative charge and moves towards the anode; glycine (about 6.0) is at its isoelectric point, so it is mainly the zwitterion and does not move; and lysine (about 9.7) is below its isoelectric point, so it carries a net positive charge and moves towards the cathode.

The answer that scoresEthene has a C=C double bond and no second functional group, so poly(ethene) is made by addition polymerisation and every atom of the monomer ends up in the polymer. Hexane-1,6-diamine and hexanedioic acid each carry two functional groups, an amine and a carboxylic acid, so nylon-6,6 is made by condensation polymerisation, and one molecule of water is lost at each of the two links formed per repeat unit.

The answer that scoresHexane-1,6-diamine and hexanedioic acid condense with the loss of two molecules of water per repeat unit, so the repeat unit is -NH-(CH2)6-NH-CO-(CH2)4-CO-. Each -NH2 group has become an -NH- in an amide link and each -COOH group has become a -CO- in an amide link, and the two free bonds at the ends are where the neighbouring repeat units are joined.

The answer that scoresPoly(ethene) is not biodegradable: its backbone has no polar link that hydrolysis can open and no group that an esterase or a protease can attack, so microorganisms cannot break the chains. A polyester such as PET is biodegradable because it contains ester links, which are hydrolysed by acid, by alkali and by esterase enzymes, and a polyamide such as nylon-6,6 is biodegradable because its amide links are hydrolysed in the same way.

The answer that scoresUltraviolet light is absorbed by the carbonyl groups built into the chain and the chain is broken at that point, so the plastic becomes brittle and fragments. That is only the first step: the fragments must then be hydrolysed and metabolised by microorganisms before the material is actually removed. A polymer is only fully degradable if the bonds in it can also be attacked by water or by enzymes.

The answer that scoresHeat methylbenzene with hot alkaline potassium manganate(VII) to oxidise the side chain completely, then add dilute acid: the product is benzoic acid, C6H5COOH. Benzaldehyde and benzyl alcohol are the partially oxidised compounds you would get if the oxidation stopped early, which these conditions do not allow.

The answer that scoresHeat the nitrobenzene with tin and concentrated hydrochloric acid, cool the mixture, then add sodium hydroxide solution until it is alkaline. That liberates the free amine, phenylamine, and dissolves the tin salts. The phenylamine is then separated in a separating funnel or steam-distilled from the alkaline mixture, dried over a drying agent such as potassium hydroxide pellets, and distilled.

The answer that scoresMake phenylamine from benzene in two steps: nitration with concentrated nitric acid and concentrated sulphuric acid at 25-60 C gives nitrobenzene, and heating nitrobenzene with tin and concentrated hydrochloric acid, followed by sodium hydroxide solution, reduces the nitro group to an amino group. Chlorobenzene is not a starting material for phenylamine, just as it is not a starting material for phenol.

The answer that scoresThe nitration of phenol with dilute nitric acid at room temperature gives a mixture of 2-nitrophenol and 4-nitrophenol, which can be separated. Benzene has no activating group, so its ring is less reactive and it needs concentrated nitric acid with concentrated sulphuric acid at 25-60 C; under those conditions it gives the single product nitrobenzene, because all six positions of benzene are equivalent.

The answer that scoresThe polar compound is held strongly by adsorption on the polar stationary phase and is only poorly soluble in the non-polar mobile phase, so it spends most of its time on the stationary phase and moves a short distance, giving the smaller Rf value. The non-polar compound dissolves readily in the mobile phase and is carried along quickly, so it has the larger Rf value. Explaining an Rf always needs both phases and the polarity of the compound.

The answer that scoresRf = distance moved by the spot from the baseline / distance moved by the solvent front from the same baseline. Both distances are measured from the baseline, and since the solvent front always travels at least as far as any spot, Rf is always between 0 and 1. A spot at the solvent front has Rf 1 and a spot that has not moved at all has Rf 0.

The answer that scoresThe relative molecular mass is read from the molecular ion, the peak at the highest m/e value apart from the small [M + 1] and [M + 2] isotope peaks that sit above it. In this question the molecular ion is at m/e 88, so Mr is 88, which fits C4H8O2. The peak at m/e 43 is a fragment: it is the CH3CO+ ion formed when the molecular ion loses a neutral ethoxy radical of mass 45, and 43 + 45 = 88 confirms that assignment.

The answer that scoresPropanone, CH3COCH3, gives two peaks: one for the carbonyl carbon and one for the two equivalent methyl groups. The number of peaks in a carbon-13 NMR spectrum is the number of different carbon environments, and equivalent carbon atoms give a single peak, so symmetry must always be considered before the peaks are counted.