9701 CHEMISTRY. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers.
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1 CAMBRIDGE INTERNATINAL EXAMINATINS GCE Advanced Level MARK SCHEME for the May/June 2014 series 9701 CHEMISTRY 9701/41 Paper 4 (Structured Questions), maximum raw mark 100 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2014 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some rdinary Level components.
2 Page 2 Mark Scheme Syllabus Paper Section A 1 (a) (i) m. pt. is high(er) / large(r) / greater (for iron) density is high(er) / large(r) / greater (for iron) (ii) (higher m. pt. due to) strong attraction between cations and electrons or more delocalised electrons (higher density due to) greater A r and smaller radius (b) (i) components to be added: voltmeter or V salt bridge [must be labelled] (ii) M1: A and B copper (metal) or Cu and iron (metal) or Fe M2: either C or D as 1 mol dm 3 / 1 M M3 C and D Cu 2+ or CuS 4 or Cu 2 or Cu (N 3 ) 2 etc. and Fe 2+ or FeS 4 etc. (iii) E o cell = = 0.78 (V) (iv) if C is Fe 2+ ; (as [C] increases), the E of the Fe 2+ / Fe increases / becomes more positive / less negative so the overall cell potential / E cell would decrease / become less positive / more negative or if C is Cu 2+ ; (as [C] increases), the E of the Cu 2+ /Cu increases / becomes more positive / less negative so the overall cell potential / E cell would increase / become more positive / less negative (c) (i) (colour change is) colourless to pink/pale purple or (end point is the first) permanent (pale) pink/pale purple colour (ii) {n(mn 4 ) = /1000 = mol} n(fe 2+ ) = 5 n(mn 4 ) = mol mass of Fe = 55.8 x = g (M2 55.8) ecf M r = mass / moles = 0.500/ = ecf [Total: 16] 2 (a) (i) A complex is a compound / molecule / species / ion formed by a central metal atom / ion surrounded by / bonded to one or more ligands / groups/ molecules / anions A ligand is a species that contains a lone pair of electrons that forms a dative bond to a metal atom / ion / or a lone pair donor to metal atom / ion
3 Page 3 Mark Scheme Syllabus Paper (ii) H H 2 H 2 Cu H 2 H 2 H 2 and Cu correct 3D structures: + octahedral and tetrahedral (iii) H 3 N Pt NH 3 or H 3 N Pt NH 3 Pt H 3 N or NH 3 Pt NH 3 NH 3 both structures geometric or cis-trans (b) (i) Cu(II) is [Ar] 3d 9 Cu(I) is [Ar] 3d 10 (ii) Cu(II): d orbitals / subshell are split (in ligand field) and electron moves from lower to upper orbital or an electron is promoted / excited in doing so it absorbs a photon / light [2] Cu(I): no gap in upper orbital / all orbitals are full (c) (i) H o = = (+) 212 kj mol 1 ecf [2] (ii) H o = = (+)146 kj mol 1 allow ecf from (c)(i) high T / temperature since H is positive / endothermic [Total: 16] 3 (a) heat in dilute H(aq) (or H 2 S 4 (aq)) (b) (i) four isomers
4 Page 4 Mark Scheme Syllabus Paper (ii) must be skeletal trans-cis H cis-trans H H cis-cis (iii) C 2 H + C 2 or H 2 C-C 2 H (c) (i) K w = [H + ][H ] (ii) In 0.15 mol dm 3 NaH, [H - ] = 0.15 mol dm 3 [H + ] = K w / [H ], so [H + ] = / 0.15 = mol dm 3 ph = -log 10 [H + ] = (13.2) ecf from [H + ] (iii) piperidine is a poorer proton acceptor or piperidine is partially ionised (iv) piperidine should be a stronger base/more basic than ammonia because of the electron-donating (alkyl/ch 2 ) groups (d) (i) n(h) at start = /1000 = mol n(h) at finish = = / mol (ii) this is in 30 cm 3 of solution, so [H] at finish = /0.030 = mol dm 3 ph = log 10 ( ) = 1.78 ecf from (d)(i) (iii) ph / vol curve: start at ph 11.9 vertical portion at V = 15 cm 3 levels off at ph 1.8 (iv) indicator is B [Total: 16] 4 (a) three from phenol (secondary) alcohol (primary) amine arene / aryl / benzene 3
5 Page 5 Mark Scheme Syllabus Paper (b) (i) H Compound Z is H CH CN H step 1: HCN + NaCN or HCN + base step 2: H 2 + Ni or LiAlH 4 or Na + ethanol (ii) bromine decolourises or goes from orange to colourless or white ppt. formed Br H CH e.g. H Br (2 2 or 3 x bromines Br ring) in ring (c) (i) (i) (ii) Na Na H H H H NH 2 (or ionic) NH 3 NH 3 (or ionic) (iii) CH 3 C CCH 3 NHCCH 3 CH 3 C 2 M1: amide M2: alcoholic ester M3: both phenolic esters [5] max [4] (d) amide ester [Total: 14]
6 Page 6 Mark Scheme Syllabus Paper 5 (a) (i) H or hydroxyl groups (allow alcohol groups) (ii) alkenes or C=C (double) bonds or carbon double bonds (iii) CH 3 CH(H) or CH 3 C- groups (b) V is CH 3 CH(H)CH=CH 2 W is CH 3 CH=CHCH 2 H (c) compound V shows optical isomerism (ecf for 'geometric(al)' if candidate's V is capable of cis-trans) CH 3 CH 3 H 2 C C CH H H H H C CH CH 2 (d) H H H or CH 3 CH(H)CH(H)CH 2 H [Total: 8]
7 Page 7 Mark Scheme Syllabus Paper 6 (a) feature formation of α-helix formation of disulfide bonds formation of ionic bonds level of bonding secondary tertiary tertiary linking amino acids primary [3] (b) block letter J K L name Deoxyribose Cytosine Phosphate M Thymine 4 (c) (i) H/hydrogen (bonds between bases) (ii) Bonds are weak and so require relatively little energy to break / are easily broken (d) (sugar, J) (base, M) DNA deoxyribose thymine / T RNA ribose uracil / U [Total: 10] (a) Expression: n = or equivalent n = 3.1 hence G has three carbon atoms (b) (i) (δ 1.1) RCH 3 or RCH 2 R or methyl or CH 3 (δ 2.2) (R)CH 2 C(R) or CH 3 C(R) (δ 11.8) (R)CH or (R)CNH(R) 3
8 Page 8 Mark Scheme Syllabus Paper (ii) The ( H) peak at δ 11.8 (disappears) because of ()H-D exchange or equation showing this (e.g. R-H + D 2 R-D + HD) (iii) CH 3 CH 2 C 2 H (c) (i) H H C H 3 C CH 3 or or or H 3 C or N C H 3 C H (ii) If methyl ethanoate: δ δ r if 1, 3-dioxolane: δ δ r if 1, 2-dioxolane: δ δ r if dihydroxycyclopropane: δ δ [Total: 11] 8 (a) (i) Amide or ester or peptide (ii) Hydrolysis (iii) Drug B (iv) two ester and one amide groups circled [2] (b) (i) At point Q because the hydrocarbon tails region is hydrophobic/non-polar/ form van der Waals only or can dissolve in the fat-soluble area (ii) They all contain polar or hydrogen-bonding (groups) (c) (i) range to m (ii) (higher frequency radiation could) cause tissue/cell damage or mutation or harmful to cells [Total: 9]
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