THIS IS A NEW SPECIFICATION

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1 TIS IS A NEW SPEIFIATIN ADVANED GE EMISTRY A Rings, Polymers and Analysis F324 * E / * andidates answer on the Question Paper R Supplied Materials: Data Sheet for hemistry A (inserted) ther Materials Required: Scientific calculator Monday 28 June 2010 Morning Duration: 1 hour * F * INSTRUTINS T ANDIDATES Write your name clearly in capital letters, your entre Number and andidate Number in the boxes above. Use black ink. Pencil may be used for graphs and diagrams only. Read each question carefully and make sure that you know what you have to do before starting your answer. Answer all the questions. Do not write in the bar codes. Write your answer to each question in the space provided. If additional space is required, you should use the lined pages at the end of the booklet. The question number(s) must be clearly shown. INFRMATIN FR ANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. Where you see this icon you will be awarded marks for the quality of written communication in your answer. This means for example you should: ensure that text is legible and that spelling, punctuation and grammar are accurate so that meaning is clear; organise information clearly and coherently, using specialist vocabulary when appropriate. You may use a scientific calculator. A copy of the Data Sheet for hemistry A is provided as an insert with this question paper. You are advised to show all the steps in any calculations. The total number of marks for this paper is 60. This document consists of 16 pages. Any blank pages are indicated. R 2010 [M/500/7836] D (N/GW) 15373/4 R is an exempt harity Turn over

2 2 BLANK PAGE PLEASE D NT WRITE N TIS PAGE R 2010

3 3 Answer all the questions. 1 Benzene is an important industrial chemical and is used in a wide range of manufacturing processes. ver time our understanding of the structure and bonding of benzene has changed and various models have been proposed. (a) In 1865, Kekulé proposed a model for the structure and bonding of benzene, but there is considerable evidence to suggest that Kekulé s model may not be correct. Scientists have proposed alternative models for the structure and bonding of benzene. Explain the evidence that led scientists to doubt the model proposed by Kekulé.... [3] R 2010 Turn over

4 4 (b) Sudan II is an azo dye which was used as a colourant in chilli powder. owever, scientists advised the Food Standards Agency that Sudan II was linked to an increased risk of cancer and it is now no longer used as a food colourant. The flowchart below shows how Sudan II could be prepared in the laboratory from 1,3-dimethylbenzene. (i) Draw the structures of the organic compounds A, B, and D in the boxes below. Display the functional group in compound. N 3, 2 S 4 1,3-dimethylbenzene compound A Sn, conc. l, heat NaN 2, l(aq) compound compound B alkaline solution of compound D N N compound D Sudan II [4] R 2010

5 5 (ii) ompound A is formed by reacting 1,3-dimethylbenzene with N 3 and 2 S 4. Explain, with the aid of curly arrows, the mechanism for the formation of compound A. Your answer should clearly show the role of 2 S 4 as a catalyst. [5] (iii) Deduce how many other structural isomers of compound A could have been formed from the mononitration of 1,3-dimethylbenzene.... [1] [Total: 13] R 2010 Turn over

6 6 2 A student was researching the development of polymers and discovered three polyesters, PET, PEN and PGA, that are used in the manufacture of plastic bottles. (a) The student discovered that the first polyester developed was Terylene which is also known as poly(ethylene terephthalate) or PET. PET can be made by reacting benzene-1,4-dicarboxylic acid with ethane-1,2-diol. (i) Draw the displayed formula of the repeat unit in PET. [2] (ii) The industrial manufacture of PET involves two main stages. The first stage, known as pre-polymerisation, forms compound F with molecular formula Draw the structure of compound F. (b) PEN is a new kind of polyester. PEN is rigid at high temperature whereas PET readily softens. The repeat unit of PEN is shown below. [1] PEN (i) What is the empirical formula of the repeat unit in PEN?... [1] R 2010

7 (ii) Draw the structures of two monomers that could be used to make PEN. 7 (c) Polyglycolic acid, PGA, is a polymer that is being developed as an inner coating for PET bottles. A short section of PGA is shown below. [2] PGA (i) ompared with other synthetic polymers, PGA can be easily hydrolysed. Draw the skeletal formula of the organic product formed from the complete hydrolysis of PGA by Na(aq). (ii) Explain why scientists now think that polymers such as PGA are better for the environment than hydrocarbon-based polymers. In your answer, you should use appropriate technical terms, spelt correctly.... [1] [2] R 2010 [Total: 9] Turn over

8 3 A student was given three compounds, an aldehyde, a ketone, and a carboxylic acid. 8 (a) The student carried out the same two chemical tests on each compound. This allowed her to distinguish between all three compounds. Describe two suitable tests that the student could have used. Show how the observations would allow her to distinguish between the compounds.... [4] (b) Explain how the student could use infrared spectroscopy to confirm which compound is a carboxylic acid.... [1] (c) The aldehyde has the molecular formula The 1 NMR spectrum of the aldehyde contains a doublet at δ = 0.9 ppm with a relative peak area of six compared with the aldehyde proton. Analyse this information to deduce the structure of the aldehyde. Explain your reasoning. R [3]

9 9 (d) The ketone also has the molecular formula There are three structural isomers of this formula that are ketones. (i) Two of these isomers are shown below. Draw the structural formula of the third structural isomer in the box below ketone 1 ketone 2 ketone 3 [1] (ii) The 13 NMR spectrum of the ketone given to the student is shown below. Use the spectrum to identify the ketone. Explain your reasoning. Identify the carbon responsible for the peak at δ = 210 ppm. absorption [3] R 2010 [Total: 12] Turn over

10 10 4 Two esters, 3 ( 2 ) 2 ( 2 ) 3 3 and 3 ( 2 ) 2 2 3, contribute to the odour of pineapple. A food scientist analysed a sample of pineapple essence by separating the two esters using gas chromatography, G, and measuring their retention times. (a) (i) State what is meant by retention time.... [1] (ii) Explain the possible limitations of G in separating the two esters.... [1] (iii) Give the systematic name for the ester 3 ( 2 ) 2 ( 2 ) [1] (b) The unsaturated ester, ethyl deca-2,4-dienoate contributes to the flavour of pears. (i) Draw the structure of this ester. [2] (ii) When pears ripen, ethyl deca-2,4-dienoate is formed following the breakdown of triglycerides. Draw the general structure of a triglyceride with any functional groups fully displayed. You can use R to represent the carbon chains. R 2010 [1]

11 11 (c) The food scientist decided to synthesise the ester shown below, for possible use as a flavouring The only organic compound available to the food scientist was phenylethanal ( ). Explain how the food scientist was able to synthesise this ester using only phenylethanal and standard laboratory reagents. In your answer, you should use appropriate technical terms, spelt correctly.... [7] [Total: 13] R 2010 Turn over

12 5 ydroxyamines are organic compounds containing hydroxyl and amino functional groups. 12 (a) Salbutamol is a hydoxyamine used in the treatment of asthma and bronchitis. Salbutamol is an example of a chiral drug. (i) Draw a circle around the chiral carbon in the structure of salbutamol shown below. N salbutamol [1] (ii) Suggest possible problems of making a chiral drug such as salbutamol and describe two ways that the pharmaceutical industry might overcome these problems.... [4] R 2010

13 13 (b) Monoethanolamine, MEA, 2 N 2 2, is a hydroxyamine that is used in aqueous solution as a gas scrubber to remove acidic gases from emissions in incinerators. MEA is prepared industrially by reacting ammonia with epoxyethane. 2 2 epoxyethane (i) Write an equation for the industrial preparation of MEA. [1] (ii) During the manufacture of MEA, a compound with molecular formula 4 11 N 2 is also formed. Draw the structure of the compound with molecular formula 4 11 N 2. (c) The combustion of some polymers produces emissions containing toxic acidic gases such as l and 2 S. MEA can remove l and 2 S from the emissions. Give the formula of the organic salts formed when MEA removes: [1] (i) l, [1] (ii) 2 S. [1] R 2010 TURN VER FR QUESTIN 5 PARTS (d) AND (e) Turn over

14 (d) MEA, 2 N 2 2, can be oxidised to form an α-amino acid. 14 (i) Explain what is meant by an α-amino acid.... [1] (ii) Write an equation for the oxidation of MEA to form an α-amino acid. Use [] to represent the oxidising agent.... [1] (e) Isomers F and G are hydroxyamines each with the molecular formula 4 11 N. Isomer F can be dehydrated to form the cyclic compound N Isomer G has two chiral centres. Identify and draw the structural isomers F and G. isomer F isomer G [2] [Total: 13] END F QUESTIN PAPER R 2010

15 15 ADDITINAL PAGE If additional space is required, you should use the lined pages below. The question number(s) must be clearly shown. R 2010

16 16 ADDITINAL PAGE opyright Information R is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. R has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced in the R opyright Acknowledgements Booklet. This is produced for each series of examinations, is given to all schools that receive assessment material and is freely available to download from our public website ( after the live examination series. If R has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, R will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the opyright Team, First Floor, 9 ills Road, ambridge B2 1GE. R is part of the ambridge Assessment Group; ambridge Assessment is the brand name of University of ambridge Local Examinations Syndicate (ULES), which is itself a department of the University of ambridge. R 2010

17 GE hemistry A Advanced GE F324 Mark Scheme for June 2010 xford ambridge and RSA Examinations

18 R (xford ambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of pupils of all ages and abilities. R qualifications include AS/A Levels, Diplomas, GSEs, R Nationals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. R is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by Examiners. It does not indicate the details of the discussions which took place at an Examiners meeting before marking commenced. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the Report on the Examination. R will not enter into any discussion or correspondence in connection with this mark scheme. R 2010 Any enquiries about publications should be addressed to: R Publications P Box 5050 Annesley NTTINGAM NG15 0DL Telephone: Facsimile: publications@ocr.org.uk

19 F324 Mark Scheme June 2010 Allow Kekulé structures throughout Question Expected Answers Marks Additional Guidance 1 a Bond length intermediate between/different from (short) = and (long) hydrogenation less exothermic than expected (when compared to hydrogenation for cyclohexene) nly reacts with Br 2 at high temp or in presence of a halogen carrier / resistant to electrophilic attack Please annotate, use ticks to show where marks are awarded b i compound A N 2 if N 2 in wrong position penalise here and EF for rest of b(i) and b(ii) 3 ALLW all carbon carbon bonds the same length ALLW hydrogenation less (negative) than expected ALLW hydrogenation different from that expected D NT ALLW halogenation/hydration ALLW doesn t decolourise/react with/polarise Br 2 ALLW doesn t undergo addition reactions (with Br 2 ) ALLW any 4-nitro-1,3-dimethylbenzene drawn in any orientation ALLW N drawn in any orientation compound B N 2 4 ALLW any 4-amino-1,3-dimethylbenzene drawn in any orientation EF amine of incorrect compound A (e.g. position of N 2 or lack of methyl sticks/groups) compound + N N ALLW diazonium chloride salt of 1,3-dimethylbenzene EF diazonium salt/compound of incorrect compound B IGNRE l ion allow N + N not allow + N 2 1

20 F324 Mark Scheme June 2010 Question Expected Answers Marks Additional Guidance ALLW if + charge is floating between the two Ns only if it is closer to the correct N allow not allow + + N N N N compound D ALLW any of - ALLW in place of 2

21 F324 Mark Scheme June 2010 Question Expected Answers Marks Additional Guidance ii Equation to show formation of N + 2 ion mark 1 N S S N 2 ALLW N S S N 2 N S 4 S N N 2 If N 2 is in correct position do not penalise even if compound A in b(i) is not in correct position + N 2 mark 4 curly arrow from bond back to reform ring AND correct products N 2 + N ALLW mark 2 curly arrow must be from 1,3- dimethylbenzene to N 2 + and EF for marks 3 and 4 D NT ALLW intermediate -ring must be more than 1 / 2 way up + N 2 mark 2 curly arrow from ring to + N 2 mark S 4 mark 3 intermediate with ring broken in the correct place 2 S 4 Link to compound A in part (i) cannot score full marks [in b(i) & b(ii)] if N 2 is not adjacent to a methyl ALLW 3 s shown ALLW S S 4 iii 2 1 No other correct response Total 13 3

22 F324 Mark Scheme June 2010 Question Expected Answers Marks Additional Guidance 2 a i ALLW for both marks Ester group must be displayed to get both marks and must contain 4 s ALLW for one mark ALLW for one mark 2 2 ALLW Kekulé structure / ( 2 ) 2 ALLW one mark if end bonds missing ALLW 1 mark if the 2 2 is drawn skeletally ALLW for ALLW 1 mark if repeat unit shows a displayed ester group and contains a benzene ring and two other carbons D NT ALLW ii ALLW Kekulé structure/ ( 2 ) 2 2 for ester groups 6 4 if already penalised in a(i) 4

23 F324 Mark Scheme June 2010 Question Expected Answers Marks Additional Guidance b i ALLW any order of elements ALLW or = ii ALLW / 2 ALLW Penalise incorrect bond linkage in 2b(ii) only. Do not penalise elsewhere on the paper ALLW ( 2 ) 2 c i - (Na + ) 2 ALLW any of the following for 1 mark + Na - or D NT ALLW any other response - Na + or 2 - Na + ii (PGA is) (bio)degradable R photodegradable R hydrolysed (but hydrocarbon based polymers are nonbiodegradable) ne of (bio)degradable R photodegradable R hydrolysed must be spelt correctly if one spelt correctly and another incorrectly spelt ALLW mark 1 ALLW broken down by bacteria (must be spelt correctly) ALLW degrade as alternative to degradable ALLW undergoes hydrolysis as alternative to hydrolysed IGNRE any additional information if the additional information is correct e.g. biodegradable and doesn t produce toxic gases D NT ALLW any additional information if the additional information is incorrect e.g. biodegradable and can be recycled Total 9 5

24 F324 Mark Scheme June 2010 Question Expected Answers Marks Additional Guidance 3 a Tollens test AND silver precipitate/mirror is the aldehyde react with 2,4-DNP() and orange precipitate must be the ketone 2,4-DNP() AND orange precipitate is either aldehyde R ketone ALLW carbonyl R = Tollens test & silver ppt/mirror is the aldehyde Alternative approaches Tollens test AND silver precipitate/mirror is the aldehyde react with carbonate/ hydrogencarbonate/ Na/Mg and fizzes/ bubbles/ effervesces/ gas evolved must be the (carboxylic) acid 2,4-DNP() and no orange precipitate is the (carboxylic) acid Tollens test & silver ppt/mirror is the aldehyde 4 ALLW ammoniacal AgN 3 / Ag + (N 3 ) 2 / Ag + (N 3 ) ALLW acidified dichromate R Fehlings as an alternative to Tollens observation turn green R red precipitate respectively ALLW acidified manganagate(vii) and observation as either brown precipitate/decolourised/pale pink ALLW Brady s (reagent) ALLW orange/red/yellow for colour of the 2,4-DNP() precipitate ALLW solid/crystals in place of precipitate IGNRE any reference to melting points ALLW Pl 5 as a test for the acid observation would be white fumes (of l) ALLW detection of (carboxylic) acid by reacting with an alcohol to make an ester but no mark for the observation. D NT ALLW detection of (carboxylic) acid by p or indicator Please annotate, use ticks to show where marks are awarded b Peak in range (cm 1 ) or (around) 3000 shows [need wavenumber (or range) and bond] 1 D NT ALLW single peak quoted within range other than 3000 (cm 1 ) for D NT ALLWrange (cm 1 ) IGNRE any reference to - or = 6

25 F324 Mark Scheme June 2010 Question Expected Answers Marks Additional Guidance c Alternative approaches depending on whether or not the aldehyde is correct ALLW 3-methylbutanal, any correct unambiguous structure ALLW two marks for correct aldehyde with no explanation Doublet indicates adjacent is bonded to only 1 R (relative) peak area indicates 2 x 3 (in the same environment) Doublet indicates adjacent is bonded to only 1 AND (relative) peak area indicates 2 x 3 (in the same environment) ALLW doublet/peak at 0.9ppm due to R ALLW the splitting shows adjacent to /environment that contains 1 /proton ALLW 6 s/ protons in same environment D NT ALLW 6 s in same environment next to d i If aldehyde is correct ( 3 ) 2 2 If aldehyde is correct only need to explain doublet R peak areas If aldehyde identified is incorrect if aldehyde is incorrect must explain both doublet or peak areas e.g. 3 3 ALLW displayed/skeletal formulae would score two marks if the doublet and the peak areas were correctly explained ii ketone 3 There are 4 (different ) environments ALLW 2 s are in same environment/equivalent (therefore) it is ketone 2 / ALLW 3-methylbutan(-2-)one/ any correct unambiguous structure ALLW 2-methylbutan-3-one 3 ( responsible for peak at = 210 ppm) is =/carbonyl carbon Total 12 ALLW 7

26 F324 Mark Scheme June 2010 Question Expected Answers Marks Additional Guidance 4 a i The time (from the injection of the sample) for the component to leave the column 1 ALLW time from injection to detection ALLW time spent in column ALLW time taken to reach detector ii They have similar retention times 1 ALLW both are esters therefore partition/adsorption/retention times will be very similar ALLW EF if they describe R f values in part a(i) ALLW same retention times b i iii Butylbutanoate 1 ALLW butyl butanoate ALLW but-1-yl butanoate hydrocarbon chain must be correct for one mark ester group and ethyl group must be correct for one mark 2 D NT ALLW butanyl butanoate ALLW any correct unambiguous structure/ 3 ( 2 ) / 3 ( 2 ) ( 2 ) 4 () D NT ALLW etc ALLW 2 for ester l ALLW 1 mark for correct 2,4-decadiene structure e.g. ALLW 1 mark for correct ethyl oate structure e.g. or or 2 5 8

27 F324 Mark Scheme June 2010 Question Expected Answers Marks Additional Guidance ii 2 R R 2 R ALLW any orientation of the three fatty acids c 1. react phenylethanal with 2 S 4 /K 2 r 2 7 if either phenylethanoic acid or 2- phenyethanol not prepared automatically lose two marks 2. to get phenylethanoic acid/ mark 2 can be scored if dichromate is used without being acidified 3. react phenylethanal with NaB 4 4. to get 2-phenylethanol/ mark 3 must be correct to score mark 4 5. react phenylethanoic acid with 2-phenylethanol. If both already correctly named ALLW acid and alcohol 6. 2 S 4 if linked to the reaction of an alcohol and acid 7. reflux in either (1) or (5) or catalyst used in (5) QW must spell catalyst or reflux correctly 7 ALLW + & r or 2 S 4 /Na 2 r any other oxidising agent or other named acid please consult with TL ALLW LiAl 4 as alternative to NaB 4 phenylethanoic acid & phenylethanol must be unambiguously identified by either name or formula D NT ALLW or oxidised to form(a carboxylic) acid or reduced to form alcohol for marks 2 and 4 ALLW conc 2 S 4 D NT ALLW dilute or 2 S 4 (aq) D NT ALLW just acid catalyst D NT ALLW l, N 3 Please annotate, use ticks to show where marks are awarded Total 13 9

28 F324 Mark Scheme June 2010 Question Expected Answers Marks Additional Guidance 5 a i N 1 ALLW * in place of circle ALLW if circle extends to include ii Mark 1 production of a single isomer is more expensive/difficult R separation of the single isomer is expensive/difficult 4 Mark 2 one of the isomers is more (pharmacologically) active or one of the isomers might have adverse/harmful/nasty side effects IGNRE any reference to dosage ALLW one is more effective/works (better) Marks 3 and 4 problems are overcome by using: Enzymes/bacteria/biological catalyst hiral synthesis hiral catalyst or transition metal complex Start with a natural chiral molecule or chiral pool any D NT ALLW use naturally occurring isomer unless stated that it is a chiral compound D NT ALLW transition metal ion D NT ALLW pool synthesis hiral pool synthesis scores 1 (not 2) marks b i N N 2 1 ALLW N 2 ALLW epoxy ethane as 2 4, ( 2 ) 2, 2 2 ALLW product as ( 2 ) 2 N 2 D NT ALLW product as 2 7 N ii 2 2 N ALLW ( 2 ) 2 ALLW displayed/skeletal formula D NT ALLW molecular formula 10

29 F324 Mark Scheme June 2010 Question Expected Answers Marks Additional Guidance c i 2 2 N 3 + l - Must show l ion 1 ALLW 2 2 N 3 l if formula is correct and both charges not shown ALLW ( 2 ) 2 / any correct unambiguous structure D NT ALLW ions joined by covalent bonds ii 2 2 N 3 + S - Must show S ion 1 ALLW if formula is correct and both charges not shown ALLW ( 2 ) 2 / any correct unambiguous structure ALLW N 3 S 2-2 d i Both N 2 and are joined to the same 1 ALLW 2 N 2 or R(N 2 ) 2 R The 4 groups/atoms attached to the can be in any order but must be adjacent. ( ) not essential ii 2 2 N 2 + 2[] 2 N ALLW ( 2 ) 2 D NT ALLW molecular formula e i Question 5e is followed by two blank lined pages (15 and 16) which candidates can use instead of requesting additional paper. Please check to see whether or not pages 15 or 16 have been used 11

30 F324 Mark Scheme June 2010 Question Expected Answers Marks Additional Guidance e i Isomer F N 2 2 ALLW ( 2 ) 4 N 2 / ALLW any correct unambiguous structure of 1-aminobutan-4-ol Isomer G * * ALLW 3 ()(N 2 ) 3 ALLW any correct unambiguous structure of 2-aminobutan-3-ol. N 2 *notrequired Total 13 12

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