A-level CHEMISTRY (7405/2)

Transcription

1 A-level EMISTRY (7405/2) Paper 2: Organic and Physical hemistry Mark scheme Specimen paper

2 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students reactions to a particular paper. Assumptions about future mark schemes on the basis of one year s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Further copies of this mark scheme are available from aqa.org.uk

3 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN Question Marking guidance Mark AO omments 0. onsider experiments and 2: [B constant] [A] increases 3: rate increases by 3 2 therefore 2nd order with respect to A onsider experiments 2 and 3: [A] increases 2: rate should increase 2 2 but only increases 2 Therefore, halving [B] halves rate and so st order with respect to B AO3 a AO3 a Rate equation: rate = k[a] 2 [B] AO3 b 0.2 rate = k [] 2 [D] therefore k = rate / [] 2 [D] AO2h k = = 57.0 AO2h Allow consequential marking on incorrect transcription (.9 0 ) (3.5 0 ) mol 2 dm +6 AO2h s Any order 0.3 rate = 57.0 ( ) = (mol dm 3 s ) OR Their k ( ) AO2h 3 of 22

4 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN 0.4 Reaction occurs when molecules have E E a Raising T by 0 causes many more molecules to have this E Whereas doubling [E] only doubles the number with this E AOa AOa AOa 0.5 E a = RT(lnA lnk)/000 AOb Mark is for rearrangement of equation and factor of 000 used correctly to convert J into kj E a = (23.97 ( 5.03))/000 = 72.3 (kj mol ) AOb 4 of 22

5 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN Question Marking guidance Mark AO omments 02. Gradient drawn on graph AO3 a Line must touch the curve at 0.02 but must not cross the curve. 5 of 22

6 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN 02.2 Extended response Stage : Rate of reaction when concentration = mol dm 3 From the tangent hange in [butadiene] = and change in time = AO3 a Gradient = ( )/(7800 0) = Rate = (mol dm 3 s ) AO3 a Stage 2: omparison of rates and concentrations Initial rate/rate at = ( )/( ) = 2.23 AO3 a Marking points in stage 2 can be in either order Inital concentration/concentration at point where tangent drawn = 0.08/0.02 =.5 AO3 a Stage 3: Deduction of order If order is 2, rate should increase by factor of (.5) 2 = 2.25 this is approximately equal to 2.23 therefore order is 2nd with respect to butadiene AO3 b 6 of 22

7 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN Question Marking guidance Mark AO omments 03. 2,2,4-trimethylpentane AOa AO2b AO2b 03.4 Mainly alkenes formed AOb (monochloro isomers) AO2b AO2a l AO2a l 7 of 22

8 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN l = = 48.0 and l = = 50.0 AOb Both required M r of this 8 7 l ( ) + ( ) = AOb = 2.05 : 2.56 : Simplest ratio = 2.05 : 2.56 : = :.25 : AO2b Whole number ratio ( 4) = 4 : 5 : 4 AO2b MF = 8 0 l 8 AO2b 8 of 22

9 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN Question Marking guidance Mark AO omments methylbutan-2-ol AOa AO2g Allow ( 3 ) 2 O O 04.3 Elimination AOa AO2g Allow ( 3 ) 2 = AO2g Allow ( 3 ) 2 = of 22

10 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN 04.5 Position AOa 04.6 B A AO3 b AO2g Allow ( 3 ) 2 (O) 2 3 O AO2e Allow ( 3 ) 3 2 O 3 2 O 3 0 of 22

11 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN Question Marking guidance Mark AO omments 05. Secondary AOa 05.2 Nitrogen and oxygen are very electronegative Therefore, =O and N are polar Which results in the formation of a hydrogen bond between O and In which a lone pair of electrons on an oxygen atom is strongly attracted to the δ+ AOa AOa AOa AOa of 22

12 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN Question Marking guidance Mark AO omments N 2 AO2a O ( 2 ) 4 2 N N N OO O O glycine threonine lysine 06.2 AO2a 3 N OO AO2a Allow 3 N 3 O O (Br ) ( 3 ) 3 N 2 OO (Br ) amino-3-hydroxybutanoic acid AO2a 2 of 22

13 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN 06.5 N 3 AO2a ( 2 ) 4 3 N OO 3 of 22

14 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN Question Marking guidance Mark AO omments AOa 3 l l Addition AOa 07.2 AO2e O O 3 3 O O 3 3 O O OR l O 3 3 O l AO2e 07.3 Q is biodegradable Polar =O group or δ+ in Q (but not in P) Therefore, can be attacked by nucleophiles (leading to breakdown) AO2g AO2c AO2c 4 of 22

15 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN Question Marking guidance Mark AO omments deoxyribose AOa 08.2 Base A Top N forms hydrogen bonds to lone pair on O of guanine AO3 b AO2a If Base B stated, allow mark only for response including hydrogen bonding The lone pair of electrons on N bonds to N of guanine AO2a A lone pair of electrons on O bonds to lower N of guanine AO2a Allow all 4 marks for a correct diagram showing the hydrogen bonding Students could also answer this question using labels on the diagram 08.3 Allow either of the nitrogen atoms with a lone pair NOT involved in bonding to cytosine AO2a 08.4 Use in very small amounts / target the application to the tumour AO2e 5 of 22

16 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN Question Marking guidance Mark AO omments 09. (nucleophilic) addition-elimination AOa Not electrophilic addition-elimination Allow 6 5 or benzene ring M2 M3 Allow attack by :N O O O 3 3 l 3 l 6 5 N N 6 5 N AO2a M RN 2 M4 for 3 arrows and lp M2 not allowed independent of M, but allow M for correct attack on + M3 for correct structure with charges but lone pair on O is part of M4 M4 (for three arrows and lone pair) can be shown in more than one structure 6 of 22

17 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN 09.2 The minimum quantity of hot water was used: To ensure the hot solution would be saturated / crystals would form on cooling AOb The flask was left to cool before crystals were filtered off: Yield lower if warm / solubility higher if warm AOb The crystals were compressed in the funnel: Air passes through the sample not just round it AOb Allow better drying but not water squeezed out A little cold water was poured through the crystals: To wash away soluble impurities AOb 09.3 Water AO3 b Do not allow unreacted reagents Press the sample of crystals between filter papers AO3 2b Allow give the sample time to dry in air 09.4 M r product = 35.0 AO2h Expected mass = = 7.33 g 93.0 AO2h Percentage yield = = = 65.8(%) 7.33 AOb Answer must be given to this precision 7 of 22

18 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN 09.5 AO2c NO 3 NO 3 + NO OR NO NO 3 + NO (NO 3 )NO Electrophilic substitution AOa 09.7 ydrolysis AO3 a 09.8 Sn/l AOb Ignore acid concentration; allow Fe/l 8 of 22

19 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN Question Marking guidance Mark AO omments 0 IR Extended response M NMR Absorption at 3360 cm shows O alcohol present AO3 a Deduction of correct structure without explanation scores maximum of 4 marks as this does not show a clear, coherent line of reasoning. M2 There are 4 peaks which indicates 4 different environments of hydrogen M3 The integration ratio =.6 : 0.4 :.2 : 2.4 AO3 a AO3 a Maximum of 6 marks if no structure given OR if coherent logic not displayed in the explanations of how two of O, 3 and 2 3 are identified. The simplest whole number ratio is 4 : : 3 : 6 M4 The singlet (integ ) must be caused by in O alcohol AO3 a M5 The singlet (integ 3) must be due to a 3 group with no adjacent AO3 b M6 Quartet + triplet suggest 2 3 group AO3 b M7 Integration 4 and integration 6 indicates two equivalent 2 3 groups AO3 b 2 3 M8 3 O AO3 b of 22

20 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN Question Marking guidance Mark AO omments. 3 2 O 3 + 2[] 3 2 (O) 3 AOb.2 This question is marked using levels of response. Refer to the Mark Scheme Instructions for Examiners for guidance on how to mark this question. Level 3 All stages are covered and the explanation of each stage is generally correct and virtually complete. 5 6 marks Level marks Level 2 marks Level 0 0 marks Answer is communicated coherently and shows a logical progression from stage to stage 2 then stage 3. All stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies OR two stages are covered and the explanations are generally correct and virtually complete. Answer is mainly coherent and shows progression from stage to stage 3. Two stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies, OR only one stage is covered but the explanation is generally correct and virtually complete. Answer includes isolated statements but these are not presented in a logical order or show confused reasoning. Insufficient correct chemistry to gain a mark. 6 AOa 5 AO2a Indicative hemistry content Stage : Formation of product Nucleophilic attack Planar carbonyl group attacks from either side (stated or drawn) Stage 2: Nature of product Product of step shown This exists in two chiral forms (stated or drawn) Equal amounts of each enantiomer/racemic mixture formed Stage 3: Optical activity Optical isomers/enantiomers rotate the plane of polarised light equally in opposite directions With a racemic/equal mixture the effects cancel 20 of 22

21 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN Question Marking guidance Mark AO omments 2. Br OR l OR 2 SO 4 AOb Allow I or Y 2.2 Electrophilic addition AOa M 2 3 M2 Br M Br M4 Br 4 AO2a Allow consequential marking on acid in 2. and allow use of Y 2.3 The major product exists as a pair of enantiomers The third isomer is -bromobutane (minor product) Because it is obtained via primary carbocation AO2a AO2a AO2a 2 of 22

22 MARK SEME A-LEVEL EMISTRY 7405/2 - SPEIMEN 22 of 22