Paper 4 Corrections Flashcards
9701/42/M/J/23:
Explain why phenol is brominated much more easily than benzene is brominated. (3)
- lone pair on O atom is delocalised into the ring
- greater pi-electron density around the ring
- polarises electrophiles more easily
9701/43/O/N/23:
CH3CH2COOH, CH3CCl2COOH, & H2SO4 are all acidic.
Suggest the trend in the relative acid strength of these three compounds.
Explain your answer.
- (From strongest to weakest acid) H2SO4 > CH3CCl2COOH > CH3CH2COOH
Any 4:
- H2SO4 is fully dissociated / strong acid
- CH3CH2COOH / CH3CCl2COOH are partly dissociated / weak acids
- Cl is electron-withdrawing / electronegative
- alkyl group in CH3CH2COOH is electron-donating
- stabilises / destabilises anion OR weakens / strengthens O-H bond (linked correctly)
- correct ref. to donation of H+
Define partition coefficient, Kpc.
Ratio of the concentrations (of a solute between) 2 solvents (at equilibrium).
Explain why transition elements behave as catalysts. (2)
- more than 1 (stable) oxidation state
- empty d-orbitals can form dative bonds with ligands
9701/43/O/N/23 Q5div:
Suggest why a solution of Cu2+ is coloured but solid CuI is white.
- Cu2+ is d^9 / orbitals not full AND Cu+ is d^10 / orbitals full
- d-orbital electron promotion not possible
9701/42/O/N/23 Q4f:
A solution of [Cu(EDTA)]2- ions is pale blue while a solution of [Cu(NH3)4(H2O)2]2+ ions is deep blue.
Explain this difference in colour.
- different energy gap between d-orbitals
- absorption of different wavelengths
9701/42/O/N/23 Q6aii:
Define transition element complex.
- Metal atom or ion bonded to one or more ligands.
9701/42/O/N/23 Q6aiii:
Explain why transition elements form complexes.
- Has vacant d-orbitals which are energetically accessible.
9701/42/O/N/23 Q8bii:
Name the substance responsible for the peak at ppm = 0.0.
TMS / tetramethylsilane
9701/42/O/N/23 Q8cii:
Suggest suitable substances or types of substances, that could be used as the mobile & stationary phases.
- mobile phase: an unreactive gas
- stationary phase: a non-polar liquid
9701/42/O/N/23 Q8ciii:
Describe how the percentage composition of the mixture can be determined from the gas/liquid chromatogram.
Area of peak divided by total area of all peaks x 100%
9701/42/O/N/23 Q9a:
State the reactants & conditions for 2 different types of reactions that both produce diethylamine, CH3CH2NHCH2CH3.
- Reaction 1: CH3CH2Cl + CH3CH2NH2, heat under pressure in alcoholic solution
- Reaction 2: CH3CONHC2H5, CH3CH2NH2, LiAlH4
Chlorobenzene & phenol both show a lack of reactivity towards reactants that cause the breaking of the C-X bond (X=Cl or OH). Explain why. (9701/43/o/n/18 Q7a)
- C-X bond is stronger (in chlorobenzene / phenol)
- lone pair on Cl / OH
- overlap / delocalise with pi-electron cloud AND electrons (of the Cl / O)
9701/42/F/M/23 Q1aiv:
Explain why the lattice energy of ZnO is more exothermic than that of ZnS.
- O2- has smaller ionic radius that S2-
- greater attraction between Zn2+ & O2-
9701/42/F/M/23 Q2d:
Describe the mode of action of a heterogeneous catalyst. (3)
- reactants adsorb (to surface of catalyst)
- bonds (in reactant) weaken
- (reaction occurs & the) products are desorbed
9701/42/F/M/23 Q3a:
Define transition element.
A d-block element that forms 1 or more stable ions with incomplete filled d-orbitals.
9701/42/F/M/23 Q3bii:
Explain why transition elements have variable oxidation states.
The d & s orbitals are similar in energy (level).
9701/42/F/M/23 Q4b (adapted):
What determines the relative basicities of ammonia, ethylamine & phenylamine? (1)
Ability of base to accept a proton
9701/42/F/M/23 Q4dii:
State the reagents used to convert phenylamine to benzenediazonium chloride.
NaNO2 + dilute HCl
9701/42/F/M/23 Q4fiv:
When CH3NH2 (aq) is added to Cd2+ (aq), a mixture of [Cd(CH3NH2)4]2+ (aq) & [Cd(OH)4]2- forms.
Suggest how [Cd(OH)4]2- is formed. (1)
CH3NH2 is basic so reacts with water to produce OH- that reacts with Cd2+.
9701/42/F/M/23 Q5a:
Describe & explain the shape of benzene, include:
- the bond angle between carbon atoms
- the hybridisation of the carbon atoms
- how orbital overlap forms sigma & pi bonds between the carbon atoms. (5)
- bond angle = 120° AND shape is trigonal planar
- (carbons are) sp2 hybridised
- contains delocalised electrons in the pi bonds / system
- (sp2 orbitals) overlap head-on to form sigma bonds
- a p-orbital (from each carbon atom) overlaps sideways (with each other above & below the ring) forming pi bonds
9701/42/F/M/23 Q5ei:
Define enantiomers.
- rotates the plane of polarised light in the opposite direction
- molecules that are non-superimposable mirror images
9701/42/F/M/23 Q5eii:
Suggest one disadvantage of producing 2 enantiomers in this synthesis.
- need to separate the optical isomers to form the pure active isomer
- OR reduced / different biological activity of ‘other’ enantiomer
- OR lower yield of biologically active molecule / desired molecule
9701/42/F/M/23 Q5eiii:
Suggest a method of adapting the synthesis to produce a single enantiomer.
- Chiral catalyst
- OR use of an enzyme
9701/43/O/N/22 Q1di:
Define enthalpy change of atomisation.
Enthalpy change when 1 mole of gaseous atoms formed from the element (in its standard state at 298K).
9701/43/O/N/22 Q1dii:
Define first electron affinity.
Enthalpy change when 1 mole of gaseous atoms gains 1 mole of electrons.
9701/43/O/N/22 Q1ei:
Explain what is meant by entropy, S.
Number of possible arrangements of particles & energy in a system.
9701/43/O/N/22 Q2aiii:
Iodide ions react with peroxydisulfate ions.
2I- + S2O82- –> I2 + 2SO42-
This reaction is slow, but it is catalysed by Fe2+ ions.
Suggest why the alternative route in the presence of Fe2+ ions has a lower activation energy than the route in the absence of a catalyst.
Reactants are both anions AND so they repel each other.
9701/43/O/N/22 Q3aiii:
Equal volumes of 1.0 mol dm-3 Sn2+ (aq) & 1.0 mol dm-3 Cl- (aq) are mixed.
Use relevant E° values to explain whether a reaction occurs between these 2 ions.
- no (reaction) AND both E° values (Sn2+/Sn) -0.14 V & Cl2/Cl- +1.36 V
- E°cell is negative OR E° of Sn4+/Sn2+ is more negative than E° Cl2/Cl-
9701/43/O/N/22 Q5bi:
(Transition Metals) Explain why 2 solutions are coloured, & why the colours are different. (4)
- d-orbitals or different energy / d-d splitting occurs
- electron(s) promoted / excited
- wavelength of visible light absorbed AND complementary colour seen
- different energy gap OR different frequency of light is absorbed
9701/43/O/N/22 Q5c:
Iron(II) carbonate, FeCO3, & nickel(II) carbonate, NiCO3, both decompose when heated. FeCO3 decomposes at a lower temperature than NiCO3.
Suggest a possible reason for this difference. Explain your answer. (2)
- Fe2+ has a smaller radius OR Fe2+ has a greater charge density
- polarises CO3^2- more
9701/43/O/N/22 Q7di:
Explain what is meant by isoelectric point. (1)
- pH where the species is electrically neutral
- pH where the species is a zwitterion is a dominant form
9701/43/O/N/22 Q8ci:
Chlorobenzene & chloroethane have different reactivities in nucleophilic substitution reactions.
Identify a suitable reagent to illustrate this difference in reactivity.
The reagent chosen should give visibly different results with chlorobenzene & chloroethane. (1)
9701/43/O/N/22 Q8cii:
Write equations to describe any reactions that occur. (1)
(i) AgNO3
(ii) C2H5Cl + H2O –> C2H5OH + HCl / C2H5Cl + NaOH –> C2H5OH + NaCl
AND Ag+ + Cl- –> AgCl
AND NO equation shown for C6H5Cl
