CHEM4. General Certificate of Education Advanced Level Examination June Unit 4 Kinetics, Equilibria and Organic Chemistry

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1 entre Number Surname andidate Number For Examiner s Use ther Names andidate Signature Examiner s Initials General ertificate of Education Advanced Level Examination June 202 Question 2 Mark hemistry Unit 4 Kinetics, Equilibria and rganic hemistry Wednesday 3 June am to 0.45 am For this paper you must have: the Periodic Table/Data Sheet provided as an insert (enclosed) a calculator. HEM TTAL Time allowed hour 45 minutes Instructions Use black ink or black ball-point pen. Fill in the boxes at the top of this page. Answer all questions. You must answer the questions in the spaces provided. Do not write outside the box around each page or on blank pages. All working must be shown. Do all rough work in this book. ross through any work you do not want to be marked. Information The marks for questions are shown in brackets. The maximum mark for this paper is 00. You are expected to use a calculator, where appropriate. The Periodic Table/Data Sheet is provided as an insert. Your answers to the questions in Section B should be written in continuous prose, where appropriate. You will be marked on your ability to: use good English organise information clearly use accurate scientific terminology. Advice You are advised to spend about 70 minutes on Section A and about 35 minutes on Section B. (JUN2HEM40) WMP/Jun2/HEM4 HEM4

2 2 Do not write outside the box Section A Answer all questions in the spaces provided. (a) A mixture of.50 mol of hydrogen and.20 mol of gaseous iodine was sealed in a container of volume V dm 3. The mixture was left to reach equilibrium as shown by the following equation. H 2 (g) + l 2 (g) 2Hl(g) At a given temperature, the equilibrium mixture contained 2.06 mol of hydrogen iodide. (a) (i) alculate the amounts, in moles, of hydrogen and of iodine in the equilibrium mixture. Moles of hydrogen... Moles of iodine... (2 marks) (a) (ii) Write an expression for the equilibrium constant (K c ) for this equilibrium. ( mark) (a) (iii) K c for this equilibrium has no units. State why the units cancel in the expression for K c ( mark) (a) (iv) A different mixture of hydrogen, iodine and hydrogen iodide was left to reach equilibrium at the same temperature in a container of the same volume. This second equilibrium mixture contained 0.38 mol of hydrogen, 0.9 mol of iodine and.94 mol of hydrogen iodide. alculate a value for K c for this equilibrium at this temperature. (2 marks) (Extra space)... (02) WMP/Jun2/HEM4

3 3 Do not write outside the box (b) This question concerns changes made to the four equilibria shown in parts (b) (i) to (b) (iv). In each case, use the information in the table to help you choose from the letters A to E the best description of what happens as a result of the change described. Write your answer in the box. Each letter may be used once, more than once or not at all. Position of equilibrium Value of equilibrium constant, K c A remains the same same B moves to the right same moves to the left same D moves to the right different E moves to the left different (b)(i) hange: increase the temperature of the equilibrium mixture at constant pressure. H 2 (g) + l 2 (g) 2Hl(g) H = +52 kj mol ( mark) (b)(ii) hange: increase the total pressure of the equilibrium mixture at constant temperature. 3H 2 (g) + N 2 (g) 2NH 3 (g) H = 92 kj mol (b)(iii) hange: add a catalyst to the equilibrium mixture at constant temperature. (g) + H 2 (g) 2 (g) + H 2 (g) H = 4 kj mol ( mark) (b)(iv) hange: add chlorine to the equilibrium mixture at constant temperature. Pl 5 (g) Pl 3 (g) + l 2 (g) H = +93 kj mol ( mark) ( mark) 0 Turn over (03) WMP/Jun2/HEM4

4 4 Do not write outside the box 2 Gases P and Q react as shown in the following equation. 2P(g) + 2Q(g) R(g) + S(g) The initial rate of the reaction was measured in a series of experiments at a constant temperature. The following rate equation was determined. rate = k[p] 2 [Q] 2 (a) omplete the table of data for the reaction between P and Q. Experiment Initial [P]/moldm 3 Initial [Q]/moldm 3 Initial rate / mol dm 3 s (3 marks) (Space for working)... 2 (b) Use the data from Experiment to calculate a value for the rate constant (k) at this temperature. Deduce the units of k. alculation... Units... (3 marks) 6 (04) WMP/Jun2/HEM4

5 5 Do not write outside the box 3 This question is about several Brønsted Lowry acids and bases. 3 (a) Define the term Brønsted Lowry acid. ( mark) 3 (b) Three equilibria are shown below. For each reaction, indicate whether the substance immediately above the box is acting as a Brønsted Lowry acid (A) or a Brønsted Lowry base (B) by writing A or B in each of the six boxes. 3 (b)(i) H 3 H + H 2 H 3 + H (b)(ii) H 3 NH 2 + H 2 H 3 NH H ( mark) 3 (b)(iii) HN 3 + H 2 S 4 H 2 N HS 4 ( mark) ( mark) 3 (c) A 25.0cm 3 sample of mol dm 3 hydrochloric acid was placed in a beaker. Distilled water was added until the ph of the solution was.25 alculate the total volume of the solution formed. State the units. (3 marks) (Extra space)... Question 3 continues on the next page Turn over (05) WMP/Jun2/HEM4

6 6 Do not write outside the box 3 (d) At 298 K, the value of the acid dissociation constant (K a ) for the weak acid HX in aqueous solution is mol dm 3. 3 (d)(i) alculate the value of pk a for HX at this temperature. Give your answer to 2 decimal places. ( mark) 3 (d)(ii) Write an expression for the acid dissociation constant (K a ) for the weak acid HX. ( mark) 3 (d)(iii) alculate the ph of a 0.74 mol dm 3 solution of HX at this temperature. Give your answer to 2 decimal places. (3 marks) (Extra space)... (06) WMP/Jun2/HEM4

7 7 Do not write outside the box 3 (e) An acidic buffer solution is formed when 0.0 cm 3 of 0.25 mol dm 3 aqueous sodium hydroxide are added to 5.0 cm 3 of 0.74 mol dm 3 aqueous HX. The value of K a for the weak acid HX is mol dm 3. alculate the ph of this buffer solution at 298 K. Give your answer to 2 decimal places. (6 marks) (Extra space)... 8 Turn over (07) WMP/Jun2/HEM4

8 8 Do not write outside the box 4 Acyl chlorides and acid anhydrides are important compounds in organic synthesis. 4 (a) utline a mechanism for the reaction of H 3 H 2 l with H 3 H and name the organic product formed. Mechanism Name of organic product... (5 marks) 4 (b) A polyester was produced by reacting a diol with a diacyl chloride. The repeating unit of the polymer is shown below. H 2 H 2 H 2 H 2 H 2 H 2 H 2 4 (b)(i) Name the diol used. ( mark) 4 (b)(ii) Draw the displayed formula of the diacyl chloride used. ( mark) (08) WMP/Jun2/HEM4

9 9 Do not write outside the box 4 (b)(iii) A shirt was made from this polyester. A student wearing the shirt accidentally splashed aqueous sodium hydroxide on a sleeve. Holes later appeared in the sleeve where the sodium hydroxide had been. Name the type of reaction that occurred between the polyester and the aqueous sodium hydroxide. Explain why the aqueous sodium hydroxide reacted with the polyester. Type of reaction... Explanation... (3 marks) 4 (c) (i) omplete the following equation for the preparation of aspirin using ethanoic anhydride by writing the structural formula of the missing product. H H + H 3 H H 3 + H 3 aspirin... ( mark) 4 (c) (ii) Suggest a name for the mechanism for the reaction in part (c) (i). ( mark) 4 (c) (iii) Give two industrial advantages, other than cost, of using ethanoic anhydride rather than ethanoyl chloride in the production of aspirin. Advantage... Advantage 2... (2 marks) Question 4 continues on the next page Turn over (09) WMP/Jun2/HEM4

10 0 Do not write outside the box 4 (d) omplete the following equation for the reaction of one molecule of benzene-,2-dicarboxylic anhydride (phthalic anhydride) with one molecule of methanol by drawing the structural formula of the single product. + H 3 H ( mark) 4 (e) The indicator phenolphthalein is synthesised by reacting phthalic anhydride with phenol as shown in the following equation. H + 2 H conc H 2 S 4 heat * H + H 2 phenol phenolphthalein 4 (e) (i) Name the functional group ringed in the structure of phenolphthalein. ( mark) 4 (e) (ii) Deduce the number of peaks in the 3 n.m.r. spectrum of phenolphthalein. ( mark) 4 (e) (iii) ne of the carbon atoms in the structure of phenolphthalein shown above is labelled with an asterisk (*). Use Table 3 on the Data Sheet to suggest a range of δ values for the peak due to this carbon atom in the 3 n.m.r. spectrum of phenolphthalein. ( mark) (0) WMP/Jun2/HEM4

11 Do not write outside the box 4 (f) Phenolphthalein can be used as an indicator in some acid alkali titrations. The ph range for phenolphthalein is (f) (i) For each acid alkali combination in the table below, put a tick ( ) in the box if phenolphthalein could be used as an indicator. Acid sulfuric acid hydrochloric acid ethanoic acid nitric acid Alkali sodium hydroxide ammonia potassium hydroxide methylamine Tick box ( ) 4 (f) (ii) In a titration, nitric acid is added from a burette to a solution of sodium hydroxide containing a few drops of phenolphthalein indicator. Give the colour change at the end-point. (2 marks) ( mark) 2 Turn over for the next question Turn over () WMP/Jun2/HEM4

12 2 Do not write outside the box 5 A possible synthesis of the amino acid X is shown below. H 3 H 2 H Step HN H H 3 H 2 H N Step 2 Br H 3 H 2 H N Step 3 NH 2 H 3 H 2 H Step 4 NH 2 H 3 H 2 H H N X 5 (a) Name and outline a mechanism for Step. Name of mechanism... Mechanism 5 (b) Give the IUPA name of the product of Step 2. (5 marks) ( mark) (2) WMP/Jun2/HEM4

13 3 Do not write outside the box 5 (c) For Step 3, give the reagent, give a necessary condition and name the mechanism. Reagent... ondition... Name of mechanism... (3 marks) 5 (d) At room temperature, the amino acid X exists as a solid. 5 (d)(i) Draw the structure of the species present in the solid amino acid. 5 (d)(ii) With reference to your answer to part (d) (i), explain why the melting point of the amino acid X is higher than the melting point of H 3 H 2 H(H)H ( mark) (2 marks) (Extra space)... Question 5 continues on the next page Turn over (3) WMP/Jun2/HEM4

14 4 Do not write outside the box 5 (e) There are many structural isomers of X, H 3 H 2 H(NH 2 )H 5 (e) (i) Draw a structural isomer of X that is an ethyl ester. ( mark) 5 (e) (ii) Draw a structural isomer of X that is an amide and also a tertiary alcohol. 5 (e) (iii) Draw a structural isomer of X that has an unbranched carbon chain and can be polymerised to form a polyamide. ( mark) 5 (f) Draw the structure of the tertiary amine formed when X reacts with bromomethane. ( mark) ( mark) 6 (4) WMP/Jun2/HEM4

15 5 Turn over for the next question D NT WRITE N THIS PAGE ANSWER IN THE SPAES PRVIDED Turn over (5) WMP/Jun2/HEM4

16 6 Do not write outside the box Section B Answer all questions in the spaces provided. 6 Benzene reacts with ethanoyl chloride in a substitution reaction to form 6 H 5 H 3 This reaction is catalysed by aluminium chloride. 6 (a) Write equations to show the role of aluminium chloride as a catalyst in this reaction. utline a mechanism for the reaction of benzene. Name the product, 6 H 5 H 3 (6 marks) (6) WMP/Jun2/HEM4

17 7 Do not write outside the box 6 (b) The product of the substitution reaction ( 6 H 5 H 3 ) was analysed by mass spectrometry. The most abundant fragment ion gave a peak in the mass spectrum with m/z =05 Draw the structure of this fragment ion. ( mark) 6 (c) When methylbenzene reacts with ethanoyl chloride and aluminium chloride, a similar substitution reaction occurs but the reaction is faster than the reaction of benzene. Suggest why the reaction of methylbenzene is faster. (2 marks) 9 Turn over for the next question Turn over (7) WMP/Jun2/HEM4

18 8 Do not write outside the box 7 (a) A chemist discovered four unlabelled bottles of liquid, each of which contained a different pure organic compound. The compounds were known to be propan--ol, propanal, propanoic acid and -chloropropane. Describe four different test-tube reactions, one for each compound, that could be used to identify the four organic compounds. Your answer should include the name of the organic compound, the reagent(s) used and the expected observation for each test. (8 marks) (Extra space)... (8) WMP/Jun2/HEM4

19 9 Do not write outside the box 7 (b) A fifth bottle was discovered labelled propan-2-ol. The chemist showed, using infrared spectroscopy, that the propan-2-ol was contaminated with propanone. The chemist separated the two compounds using column chromatography. The column contained silica gel, a polar stationary phase. The contaminated propan-2-ol was dissolved in hexane and poured into the column. Pure hexane was added slowly to the top of the column. Samples of the eluent (the solution leaving the bottom of the column) were collected. Suggest the chemical process that would cause a sample of propan-2-ol to become contaminated with propanone. State how the infrared spectrum showed the presence of propanone. Suggest why propanone was present in samples of the eluent collected first (those with shorter retention times), whereas samples containing propan-2-ol were collected later. (4 marks) (Extra space)... 2 Turn over for the next question Turn over (9) WMP/Jun2/HEM4

20 20 Do not write outside the box 8 When the molecular formula of a compound is known, spectroscopic and other analytical techniques can be used to distinguish between possible structural isomers. Draw one possible structure for each of the compounds described in parts (a) to (d). 8 (a) ompounds F and G have the molecular formula 6 H 4 N 2 4 and both are dinitrobenzenes. F has two peaks in its 3 n.m.r. spectrum. G has three peaks in its 3 n.m.r. spectrum. F G (Space for working) (2 marks) (20) WMP/Jun2/HEM4

21 2 Do not write outside the box 8 (b) ompounds H and J have the molecular formula 6 H 2 Both have only one peak in their H n.m.r. spectra. H reacts with aqueous bromine but J does not. H J (Space for working) (2 marks) Question 8 continues on the next page Turn over (2) WMP/Jun2/HEM4

22 22 Do not write outside the box 8 (c) Kand L are cyclic compounds with the molecular formula 6 H 0 Both have four peaks in their 3 n.m.r. spectra. K is a ketone and L is an aldehyde. K L (Space for working) (2 marks) (22) WMP/Jun2/HEM4

23 23 Do not write outside the box 8 (d) ompounds M and N have the molecular formula 6 H 5 N M is a tertiary amine with only two peaks in its H n.m.r. spectrum. N is a secondary amine with only three peaks in its H n.m.r. spectrum. M N (Space for working) (2 marks) 8 END F QUESTINS (23) WMP/Jun2/HEM4

24 24 There are no questions printed on this page D NT WRITE N THIS PAGE ANSWER IN THE SPAES PRVIDED opyright 202 AQA and its licensors. All rights reserved. (24) WMP/Jun2/HEM4

25 Version.2 General ertificate of Education (A-level) June 202 hemistry HEM4 (Specification 2420) Unit 4: Kinetics, Equilibria and rganic hemistry Final Mark Scheme

26 Mark schemes are prepared by the Principal Examiner and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme Report includes any on amendments the Examination made at the standardisation events which all examiners participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the candidates responses to questions and that every examiner understands and applies it in the same correct way. As preparation for standardisation each examiner analyses a number of candidates scripts: alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, examiners encounter unusual answers which have not been raised they are required to refer these to the Principal Examiner. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of candidates 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 opyright 202 AQA and its licensors. All rights reserved. opyright AQA retains the copyright on all its publications. However, registered centres for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to centres to photocopy any material that is acknowledged to a third party even for internal use within the centre. Set and published by the Assessment and Qualifications Alliance. The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales (company number ) and a registered charity (registered charity number ). Registered address: AQA, Devas Street, Manchester M5 6EX.

27 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 Question Marking Guidance Mark Additional Guidance (a)(i) mol H 2 = 0.47 mol I 2 = 0.7 If answers reversed, ie mol H 2 = 0.7 mol I 2 = 0.47 then allow one mark (for second answer). (a)(ii) 2 [HI] Penalise expression containing V [H ][I ] But mark on in (a)(iv) 2 2 Penalise missing square brackets in this part (and not elsewhere in paper) but mark on in (a)(iv) (a)(iii) equal number of moles (on each side of equation) R equal moles (top and bottom of Kc expression) (a)(iv) 2 [.94] [0.38][0.9] Ignore V If Kc wrong in (a)(ii) (wrong powers or upside down etc) no marks here = 52(.) (b)(i) D (b)(ii) B (b)(iii) A (b)(iv) 3

28 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 Question Marking Guidance Mark Additional Guidance 2(a) Exp Exp Exp R R R Min 2sf If three wrong answers, check their value of k in 2(b). They can score all 3 if they have used their (incorrect) value of k. see below. Exp 2 rate = k ( ) Exp 3 [Q] = 0.02/k Exp 4 [P] = 0.093/ k 2(b) k = ( ) 2 5 ( ) Mark is for insertion of numbers into a correctly rearranged rate equ, k = etc If upside down, score only units mark from their k AE (-) for copying numbers wrongly or swapping two numbers = 4.4(4) (allow 40/9) mol 2 dm +6 s Any order If k calculation wrong, allow units conseq to their k expression 4

29 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 Question Marking Guidance Mark Additional Guidance 3(a) Proton donor or H + donor Allow donator 3(b)(i) B B Both need to be correct to score the mark 3(b)(ii) A A Both need to be correct to score the mark 3(b)(iii) B A Both need to be correct to score the mark 3(c) M [H + ] = 0.25 R M2 mol Hl = ( ) (= ) M3 vol ( = ) = dm 3 or 37.8 cm allow dm 3 or cm 3 Mark for Working Units and answer tied Lose M3 if total given as ( ) = 62.8 cm 3 Ignore vol added = 2.8cm 3 after correct answer 3(d)(i) 4.52 Must be 2dp 3(d)(ii) K a = + - [H ] [X ] [HX] ignore = [H ] but this may score M [HX] in d(iii) Must have all brackets but allow ( ) Allow HA etc N mark for 0 -pka 3(d)(iii) M K a = [H + ] 2 or with numbers [HX] Allow [H + ] = (Ka [HA]) for M M2 [H + ] = ( ( ) = ( ) ) Mark for answer M3 ph = 2.64 = (allow more than 2dp but not fewer) Allow for correct ph from their wrong [H + ] If square root forgotten, ph = 5.28 scores 2 for M and M3 5

30 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 3(e) M mol H = ( ) 0.25 = Mark for answer M2 orig mol HX = ( ) 0.74 = Mark for answer M3 mol HX in buffer = orig mol HX mol H = = ([HX] = / = ) Mark for answer Allow conseq on their (M2 M) If no subtraction, max 3 for M, M2 & M4 (ph = 4.20) If [H + ] = [X - ] & used, max 3 for M, M2 & M3 (ph = 2.89) M4 mol X in buffer = mol H = ([X - ] = / = 0.05) May be scored in M5 expression M5 [H + ] ( = Ka x [HX] [X - ] ) If use K a = [H + ] 2 no further marks [HX] = x R x If either value of HX or X - used wrongly or expression upside down, no further marks (= ) M6 ph = 4.48 or 4.49 (allow more than 2dp but not fewer) Do not allow M6 for correct calculation of ph using their [H + ] - this only applies in 3d(iii) - apart from earlier AE 6

31 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 Question Marking Guidance Mark Additional Guidance 4(a) M2 H 3 H 2 l M H 3 H methyl propanoate H 3 H 2 H 3 H l M3 for structure M4 for 3 arrows and lone pair (N mark for name of mechanism) 4 M2 not allowed independent of M, but allow M for correct attack on + + rather than δ+ on = loses M2 If l lost with = breaking, max for M M3 for correct structure with charges but lp on is part of M4 only allow M4 after correct/very close M3 ignore l removing H + 4(b)(i) pentane-,5-diol Second e and numbers needed Allow,5-pentanediol but this is not IUPA name 4(b)(ii) l H H H H l Must show ALL bonds 4(b)(iii) All three marks are independent M (base or alkaline) Hydrolysis M2 δ+ in polyester M3 reacts with H - or hydroxide ion (allow close spelling) Allow (nucleophilic) addition-elimination or saponification Not reacts with NaH 4(c)(i) H 3 Allow H 3 H or H 3 2 H H 7

32 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 4(c)(ii) (nucleophilic) addition-elimination R (nucleophilic) addition followed by elimination Both addition and elimination needed and in that order Do not allow electrophilic addition-elimination / esterification Ignore acylation 4(c)(iii) any two from: ethanoic anhydride is less corrosive less vulnerable to hydrolysis less dangerous to use, less violent/exothermic/vigorous reaction R more controllable rxn does not produce toxic/corrosive/harmful fumes (of Hl) R does not produce Hl less volatile 2 max NT ST List principle beyond two answers 4(d) H 3 Allow H 3 2 H 3 or H H 2 H 4(e)(i) ester Do not allow ether Ignore functional group/linkage/bond 4(e)(ii) 2 or twelve (peaks) 8

33 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 4(e)(iii) Allow a number or range within these limits Penalize extra ranges given Ignore units 4(f)(i) sulfuric acid sodium hydroxide 2 4 correct scores 2 hydrochloric acid ammonia X or blank ethanoic acid potassium hydroxide nitric acid methylamine X or blank 3 correct scores 2 or correct scores 0 4(f)(ii) Pink to colourless Allow red R purple R magenta instead of pink Do not allow clear instead of colourless 9

34 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 Question Marking Guidance Mark Additional Guidance 5(a) nucleophilic addition M2 H 3 H 2 H M N M4 for lp and arrow to H + H+ H 3 H 2 H N M3 for structure H 3 H 2 H N H 4 allow :N M2 not allowed independent of M, but allow M for correct attack on + + rather than δ+ on = loses M2 M3 is for correct structure including minus sign but lone pair is part of M4 Allow 2 H 5 M and M4 for lp and curly arrow 5(b) 2-bromobutanenitrile Allow 2-bromobutane--nitrile 5(c) M ammonia or NH 3 M2 excess (ammonia) M3 nucleophilic substitution excess tied to NH 3 and may score in M unless contradicted Ignore temp or pressure Ignore concentrated or sealed container, Acid loses conditions mark Allow close spelling 5(d)(i) H 3 H 2 NH 3 H Allow 2 H 5 Allow 2 Allow + NH 3 Don t penalize position of + on NH 3 5(d)(ii) M electrostatic forces between ions in X QL Marks independent M2 (stronger than) hydrogen bonding between H 3 H 2 H(H)H Allow ionic bonding. E mention of molecules of X or inter molecular forces between X loses both marks 0

35 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 5(e)(i) H NH 2 H H 2 H 3 Isomer of 4 H 9 N 2 Allow NH 2 R H 3 H N H 2 H 3 5(e)(ii) H 3 H 3 H NH 2 Isomer of 4 H 9 N 2 allow NH 2 Allow H 3 H 3 N H H H 5(e)(iii) H 2 N H 2 H 2 H 2 H or H 2 N (H 2 ) 3 H Isomer of 4 H 9 N 2 allow NH 2 R Do not allow 3 H 6 - NH 2 H 3 HH 2 H Beware do not credit X itself 5(f) H 3 H 2 N(H 3 ) 2 H H Answer has 6 carbons so NT isomer of X Allow 2 H 5 Must have bond from to N not to methyl group

36 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 Question Marking Guidance Mark Additional Guidance 6(a) H 3 l + All 3 H 3 + All 4 All 4 + H + All 3 + Hl Allow RHS as H 3 δ+ δ l All 3 Allow + on or in equation but + must be on in mechanism below Ignore curly arrows in equation even if wrong. R M M H 3 H 3 + M3 H H 3 M2 M3 H H 3 3 M arrow from within hexagon to or to + on + must be on of R in mechanism + in intermediate not too close to gap in horseshoe must be centred approximately around M3 arrow into hexagon unless Kekule allow M3 arrow independent of M2 structure ignore base removing H for M3 N mark for name of mechanism M2 Phenylethanone ignore in name, penalise other numbers Note: this is the sixth marking point in 6a 6(b) or + must be on But allow [ 6 H 5 ] + 2

37 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 6(c) M about electrons methyl group has (positive) inductive effect R increases electron density on benzene ring R pushes electrons R is electron releasing Ignore reference to delocalisation M2 about attraction electrophile attracted more or benzene ring better nucleophile Allow intermediate ion stabilised M2 only awarded after correct or close M 3

38 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 Question Marking Guidance Mark Additional Guidance 7(a) If 2 stage test for one compound, award no marks for that compound, eg no mark for RH or RX to alkene then Br 2 test. If reagent is wrong or missing, no mark for that test; if wrong but close/incomplete, lose reagent mark but can award for correct observation. In each test, penalise each example of wrong chemistry, eg Agl 2 propan--ol M acidified potassium dichromate sodium Named acid + conc H 2 S 4 named acyl chloride Pl 5 M2 (orange) turns green effervescence Sweet smell Sweet smell /misty fumes Misty fumes propanal M3 M4 add Tollens or Fehlings / Benedicts Tollens: silver mirror or Fehlings/ Benedicts: red ppt acidified potassium dichromate (orange) turns green Bradys or 2,4- dnph Yellow or orange ppt if dichromate used for alcohol cannot be used for aldehyde propanoic acid M5 Named carbonate/ hydrogencarbonate water and UI (paper) Named alcohol + conc H 2 S 4 sodium or magnesium M6 effervescence orange/red Sweet smell effervescence Pl 5 Misty fumes if sodium used for alcohol cannot be used for acid if Pl 5 used for alcohol cannot be used for acid -chloro propane M7 NaH then acidified AgN 3 AgN 3 If acidification missed after NaH, no mark here but allow mark for observation M8 white ppt white ppt 4

39 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 7(b) M oxidation (of alcohol by oxygen in air) M2 absorption at (due to =) M3 comparison of polarity of molecules or correct imf statement: propanone is less polar R propan-2-ol is more polar R propanone has dipole-dipole forces R propan-2-ol has hydrogen bonding M4 - about attraction to stationary phase or solubility in moving phase Propan-2-ol has greater affinity for stationary phase or vice versa R propanone is more soluble in solvent/moving phase or vice versa l Must refer to the spectrum 5

40 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 Question Marking Guidance Mark Additional Guidance 8(a) F G N 2 2 N N 2 N 2, Penalize 2 N once Penalise missing circle once Don t penalise attempt at bonding in N 2 8(b) H J H 3 H 3 H 3 H 3, If both H and J correct but reversed, award one mark A carbon in saturated ring structures should be shown as H H R but not R R H 2 6

41 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 Question Marking Guidance Mark Additional Guidance 8(c) K L H, R R R H 3 H 3 H 3 H 3 H 3 H H 3 8(d) M N R H 3 H 2 N H 2H 3 H 2 H 3 H 3 H 3 N H 3 H 3, Allow 2 H 5 but NT allow 4 H 9 or 3 H 7 H 3 N H 3 H H H (H 3 ) 3 7

42 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 General principles applied to marking HEM4 papers by MI+ (June 202) It is important to note that the guidance given here is generic and specific variations may be made at individual standardising meetings in the context of particular questions and papers. Basic principles Examiners should note that throughout the mark scheme, items that are underlined are required information to gain credit. ccasionally an answer involves incorrect chemistry and the mark scheme records E = 0, which means a chemical error has occurred and no credit is given for that section of the clip or for the whole clip. A. The List principle and the use of ignore in the mark scheme If a question requires one answer and a candidate gives two answers, no mark is scored if one answer is correct and one answer is incorrect. There is no penalty if both answers are correct. N.B. ertain answers are designated in the mark scheme as those which the examiner should Ignore. These answers are not counted as part of the list and should be ignored and will not be penalised. B. Incorrect case for element symbol The use of an incorrect case for the symbol of an element should be penalised once only within a clip. For example, penalise the use of h for hydrogen, L for chlorine or br for bromine.. Spelling In general The names of chemical compounds and functional groups must be spelled correctly to gain credit. Phonetic spelling may be acceptable for some chemical terminology. N.B. Some terms may be required to be spelled correctly or an idea needs to be articulated with clarity, as part of the Quality of Language (QoL) marking. These will be identified in the mark scheme and marks are awarded only if the QoL criterion is satisfied. 8

43 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 D. Equations In general Equations must be balanced. When an equation is worth two marks, one of the marks in the mark scheme will be allocated to one or more of the reactants or products. This is independent of the equation balancing. State symbols are generally ignored, unless specifically required in the mark scheme. E. Reagents The command word Identify, allows the candidate to choose to use either the name or the formula of a reagent in their answer. In some circumstances, the list principle may apply when both the name and the formula are used. Specific details will be given in mark schemes. The guiding principle is that a reagent is a chemical which can be taken out of a bottle or container. Failure to identify complete reagents will be penalised, but follow-on marks (e.g. for a subsequent equation or observation) can be scored from an incorrect attempt (possibly an incomplete reagent) at the correct reagent. Specific details will be given in mark schemes. For example, no credit would be given for the cyanide ion or N when the reagent should be potassium cyanide or KN; the hydroxide ion or H when the reagent should be sodium hydroxide or NaH; the Ag(NH 3 ) 2 + ion when the reagent should be Tollens reagent (or ammoniacal silver nitrate). In this example, no credit is given for the ion, but credit could be given for a correct observation following on from the use of the ion. Specific details will be given in mark schemes. In the event that a candidate provides, for example, both KN and cyanide ion, it would be usual to ignore the reference to the cyanide ion (because this is not contradictory) and credit the KN. Specific details will be given in mark schemes. F. xidation states In general, the sign for an oxidation state will be assumed to be positive unless specifically shown to be negative. 9

44 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 G. Marking calculations In general A correct answer alone will score full marks unless the necessity to show working is specifically required in the question. An arithmetic error may result in a one mark penalty if further working is correct. A chemical error will usually result in a two mark penalty. H. rganic reaction mechanisms urly arrows should originate either from a lone pair of electrons or from a bond. The following representations should not gain credit and will be penalised each time within a clip... H 3. H 3 Br H 3 Br Br For example, the following would score zero marks _ : H H.. _ H H 3 H H Br When the curly arrow is showing the formation of a bond to an atom, the arrow can go directly to the relevant atom, alongside the relevant atom or more than half-way towards the relevant atom. In free-radical substitution The absence of a radical dot should be penalised once only within a clip. The use of double-headed arrows or the incorrect use of half-headed arrows in free-radical mechanisms should be penalised once only within a clip In mass spectrometry fragmentation equations, the absence of a radical dot on the molecular ion and on the free-radical fragment would be considered to be two independent errors and both would be penalised if they occurred within the same clip. 20

45 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 I. rganic structures In general Displayed formulae must show all of the bonds and all of the atoms in the molecule, but need not show correct bond angles. Bonds should be drawn correctly between the relevant atoms. This principle applies in all cases where the attached functional group contains a carbon atom, e.g nitrile, carboxylic acid, aldehyde and acid chloride. The carbon-carbon bond should be clearly shown. Wrongly bonded atoms will be penalised on every occasion. (see the examples below) The same principle should also be applied to the structure of alcohols. For example, if candidates show the alcohol functional group as H, they should be penalised on every occasion. Latitude should be given to the representation of bonds in alkyl groups, given that H 3 is considered to be interchangeable with H 3 even though the latter would be preferred. Similar latitude should be given to the representation of amines where NH 2 will be allowed, although H 2 N would be preferred. Poor presentation of vertical H 3 bonds or vertical NH 2 bonds should not be penalised. For other functional groups, such as H and N, the limit of tolerance is the half-way position between the vertical bond and the relevant atoms in the attached group. By way of illustration, the following would apply. H 3 H 3 H 3 H 2 H H allowed allowed not allowed not allowed not allowed NH 2 N 2 NH 2 NH 2 allowed allowed allowed allowed not allowed NH 2 2

46 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 N N H H H not allowed not allowed not allowed not allowed not allowed H l H H l l not allowed not allowed not allowed not allowed not allowed not allowed In most cases, the use of sticks to represent H bonds in a structure should not be penalised. The exceptions will include structures in mechanisms when the H bond is essential (e.g. elimination reactions in haloalkanes) and when a displayed formula is required. Some examples are given here of structures for specific compounds that should not gain credit H 3 H for ethanal H 3 H 2 H for ethanol HH 2 H 3 for ethanol 2 H 6 for ethanol H 2 H 2 for ethene H 2.H 2 for ethene H 2 :H 2 for ethane N.B. Exceptions may be made in the context of balancing equations Each of the following should gain credit as alternatives to correct representations of the structures. H 2 = H 2 for ethene, H 2 =H 2 H 3 HHH 3 for propan-2-ol, H 3 H(H)H 3 22

47 Mark Scheme General ertificate of Education (A-level) hemistry Unit 4: Kinetics, Equilibria and rganic hemistry June 202 J. rganic names As a general principle, non-iupa names or incorrect spelling or incomplete names should not gain credit. Some illustrations are given here. but-2-ol 2-hydroxybutane butane-2-ol 2-butanol ethan-,2-diol 2-methpropan-2-ol 2-methylbutan-3-ol 3-methylpentan 3-mythylpentane 3-methypentane propanitrile aminethane 2-methyl-3-bromobutane 3-bromo-2-methylbutane 3-methyl-2-bromobutane 2-methylbut-3-ene difluorodichloromethane should be butan-2-ol should be butan-2-ol should be butan-2-ol should be butan-2-ol should be ethane-,2-diol should be 2-methylpropan-2-ol should be 3-methylbutan-2-ol should be 3-methylpentane should be 3-methylpentane should be 3-methylpentane should be propanenitrile should be ethylamine (although aminoethane can gain credit) should be 2-bromo-3-methylbutane should be 2-bromo-3-methylbutane should be 2-bromo-3-methylbutane should be 3-methylbut--ene should be dichlorodifluoromethane 23