9701 CHEMISTRY. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers.

Transcription

1 CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME f the May/June 2013 series 9701 CHEMISTRY 9701/42 Paper 4 (A2 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 befe 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 Rept f Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes f the May/June 2013 series f most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

2 Page 2 Mark Scheme Syllabus Paper 1 (a) (i) RBr + R+ Br (ii) nucleophilic substitution [2] (b) (i) plotting of all points (plotted to within ½ small square) good line of best fit (ii) t ½ = 118 min 79 min (± 5 min) construction lines f two half-lives and mention that half-life is constant calculate the ratio of two rates at two different concentrations (iii) either ratio of initial rates (slopes) ratio of t ½ ratio of times f [RBr] to fall to the same level: all should be = 1.5 therefe reaction is first der w.r.t. [ ] (iv) rate = k [RBr] [ ] initial rate = 0.01 / 185 = (mol dm 3 min 1 ) k = 5.4 x 10-5 / ( ) = (mol 1 dm 3 min 1 ) [8 max 7]

3 Page 3 Mark Scheme Syllabus Paper (c) energy/kj mol i1 (2NO + Br 2 ) NOBr extent of reaction four marking points: one activation "hump" 2NOBr (not just NOBr) H labelled crectly (arrow down, double headed, just a line) E a labelled crectly (arrow up, double headed, just a line) all four points [2] three two points [2] [Total: 11]

4 Page 4 Mark Scheme Syllabus Paper 2 (a) (i) hydrogen gas at 1 atm Pt H + /HCl at 1 mol dm -3 and 298K H 2 (g) going in (i.e. not being produced) platinum electrode in contact with solution, with H 2 bubbling over it H + HCl H 2 SO 4 solution at 1 mol dm 3 ( 0.5 M if H 2 SO 4 ) and T=298 K, p=1 atm (ii) E o = 1.33 (-0.41) = 1.74 V Cr 2 O H + + 6Cr 2+ 8Cr H 2 O (iii) Colour would change from ange to green [8] (b) there are two ways of calculating the ratio: pk a = log 10 (K a ) = log 10 (1.79 x 10 5 ) = (4.75) [H + ] = = 3.16 x 10-6 log 10 ([B] / [A]) = ph pk a = (0.75) [salt] / [acid] = K a / [H + ] [B] / [A] = = 5.66 = 1.79 x 10 5 / 3.16 x 10 6 = 5.66 ( [A] / [B] = 0.177) (crect ratio = marks) since B + A = 100, (100 A) / A = 5.66 vol of acid = 15 cm 3 vol of salt = 85 cm 3 [4]

5 Page 5 Mark Scheme Syllabus Paper (c) (i) CH 3 CO 2 Na + HCl CH 3 CO 2 H + NaCl (ii) CH 3 CO 2 H + Na CH 3 CO 2 Na + H 2 O [2] (d) e.g. hydrolysis of esters RCO 2 R' (+ H 2 O) RCO 2 H + R' its reverse hydrolysis of amides: RCONH 2 (+ H 3 O + ) RCO 2 H + NH 4 + hydrolysis of nitriles: RCN (+ H 3 O + + H 2 O) RCO 2 H + NH 4 + nitration of benzene ( any arene): C 6 H 6 + HNO 3 C 6 H 5 NO 2 (+ H 2 O) dehydration of alcohols, e.g. : CH 3 CH()CH 3 CH 3 CH=CH 2 + H 2 O ( the reverse) halogenation of ketones, e.g. : CH 3 COCH 3 + X 2 CH 3 COCH 2 X (+ HX) [Total: 17] 3 (a) (i) HNO 3 + H 2 SO 4 conc (both acids) and 30 o C < T < 60 o C warm (ii) dilute HNO 3 HNO 3 (aq) and room temp. (allow T 30 o C) (b) (allow intermediate from methylbenzene) H NO 2 (c) Sn/tin ( SnCl 2, Fe) + HCl (NOT H 2 SO 4 H +, Zn, LiAlH 4.)

6 Page 6 Mark Scheme Syllabus Paper (d) (i) NH 3 + Br NH 2 Br Br N N -ONa -O - NOT -NaO A B + (ii) NaNO 2 + HCl H 2 SO 4 H + HNO 2 T 10 o C [4 max 3] (e) (i) amide (ii) M r = = 149 %N = (14 x 100)/149 = 9.4% (iii) NHCOC 2 H 5 [Total: 11] 4. (a) (i) Many electrons of similar energy in a valence-shell bital successive ionisation energies rise steadily (no big jumps) ability to fm bonds with ligands can stabilise very low very high oxidation states 4s + 3d bitals/shells/energy levels have similar / same energies (ii) VO 2 + : +5 CrF 6 2 : +4 MnO 4 2 : +6 [3 1] [4]

7 Page 7 Mark Scheme Syllabus Paper (b) (colour due to) absption of light/photons/frequencies/wavelengths colour seen is complement of colour absbed. d-bitals/d-subshell split (by ligand field) (when photon is absbed), electron is promoted moves (from lower) to higher (d )bital energy difference/gap E splitting cresponds to photon/frequency/wavelength in visible region in s-block elements the energy gap is too large (to be able to absb visible light) [any four 4 1] [4] (c) (i) 2MnO 4 + 2H 2 O + 5SO 2 2Mn SO H + (ii) solution will go from purple to colourless (d) (pale) blue solution gives a (pale) blue ppt. which re-dissolves, fms a solution, which is dark/deep blue purple [Total: 14]

8 Page 8 Mark Scheme Syllabus Paper 5 (a) (i) O O HO O HO * * * * two three centres crectly identified four centres crectly identified [2] (ii) C 16 H 18 O 9 (iii) 3 moles of H 2 (iv) in cold: 3 moles of Na on heating: 4 moles of Na [6] (b) (i) hydrolysis (ii) alkene C=C (iii) with Na 2 CO 3 (aq): carboxylic acid with Br 2 (aq): phenol (iv) HO 2 C F ( can be at the 3, 4, 5 positions, but not the 2 6 positions) Br () CO 2 H Br() Br Br G (ring subst. allow 2 3 Br in ring) (addition to C=C: allow one of the aliphatic Br to be, but not both) (v) geometrical cis-trans E-Z

9 Page 9 Mark Scheme Syllabus Paper (vi) HO 2 C skeletal structural [9 max 8] (c) M r (E) = 180, so 0.1 g = 1/1800 (5.56 x 10-4 ) mol 3 mol Na react with 1 mol of E, so n(na) = 3/1800 = 1/600 mol = mol volume of 0.1M Na = 1000/(600 x 0.1) = 16.7 cm 3 [Total: 17]

10 Page 10 Mark Scheme Syllabus Paper 6 (a) substance protein synthesis fmation of DNA cysteine cytosine glutamine guanine (b) (i) Hydrogen bonding Between bases between A,T, C and G (all four needed) (ii) Bonds are (relatively) weak easily broken This enables strands to separate DNA to unzip/unwind/unravel. [4] (c) changes / mutations in DNA by the addition / insertion /deletion / substitution / replacement of a base adds / deletes / replaces an amino acid changes the amino acid sequence this causes a loss of function changes the shape / tertiary structure of the protein any three points [Total: 10]

11 Page 11 Mark Scheme Syllabus Paper 7. (a) (i) 43.3 = x n n = 100 x 3.35 = 7.03 = 7 (calculation must be shown) 43.3 x 1.1 (ii) The M and M+2 peaks are in the ratio 3 : 1 hence the halogen is chline/cl (iii) L contains 7 hydrogen atoms there are 3 types/environments of proton/h (iv) The multiplet with 4 hydrogens peaks at δ 7.3 suggests a benzene ring The singlet with 2 hydrogens peak at δ 4.7 suggests a CH 2 group The singlet with 1 hydrogen peak at δ 2.3 suggests an group reaction with Na suggests an group must be an alcohol, not a phenol (due to its δ value) Since L also contains 7 carbon atoms and chline, this accounts f 126 of the 142 mass, the remaining atom must be oxygen Thus L is Cl CH 2 (allow the 2-, 3-4- isomer) [6] [9 max 7] (b) (i) we expect propene to have a CH 3 peak a peak at m/e 15 cyclopropane would have fewer peaks (ii) cyclopropane would have 1 peak (igne splitting) propene would have 2 ( 3, 4) peaks (igne splitting) propene would have peaks in the δ (alkene) region no splitting of cyclopropane peak (any two points) [2] [Total: 10]

12 Page 12 Mark Scheme Syllabus Paper 8 (a) (i) CH 2 = CH CO 2 H CH 2 = CH CO 2 R CH 2 = CH COCl [2] (ii) addition (polymerisation) (iii) C(CH 2 ) 4 (iv) water [5] (b) (water is bonded to the polymer by) hydrogen bonding hydrogen bonds are weak easily broken [2] (c) (i) cross-linking causes no reduction in the number of groups cross-linking molecules also have groups (ii) property e.g. becomes harder / me rigid / less flexible / stronger / higher melting point. because the chains are me strongly / tightly held [Total: 10]