(b) (i) Compound B AND M + /molecular ion peak (at m/z) = 124

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Candice Strickland
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1 Answer Mark Guidance 1 (a) (Relative) solubility (in stationary phase) 1 ALLW how well the compound dissolves IGNRE retention time AND partition D NT ALLW adsorption R absorption (b) (i) Compound B AND M /molecular ion peak (at m/z) = ALLW Mr = 124 IGNRE compound B because m/z = 124 ALLW C 7 H 8 2 = 124 R C 7 H 8 2 = 124 ALLW peak at (m/z =) 109 due to HC 6 H 4 ALLW peak at (m/z =) 109 due to loss of CH 3 IGNRE reference to other peaks in the spectrum (ii) Compound (B) is less soluble in the stationary phase/ liquid 1 RA Answer refers to the first compound to emerge from the column ALLW compound (B) is more soluble in mobile phase/gas ALLW compound interacts less with stationary phase/liquid R compound interacts more with mobile phase/gas IGNRE compound adsorbs less IGNRE compound is not very soluble (comparison needed) IGNRE volatility R reactivity

2 Answer Mark Guidance (c) (i) reagent = K 2 Cr 2 7 AND H 2 S 4 3 ALLW acidified dichromate ALLW H /any acid IGNRE concentration of acid ALLW Na 2 Cr 2 7 /Cr /(potassium R sodium) dichromate((vi)) ALLW acidified Mn 4 - ALLW Tollens reagent/ammoniacal silver nitrate IGNRE conditions compound C = ALLW correct structural R displayed R skeletal formulae R a combination of above as long as unambiguous ALLW ECF from incorrect compound C Check positions of H groups ALLW esterification of phenol group ester =

3 Answer Mark Guidance (ii) curly arrow from H to C δ 3 ANNTATE ANSWER WITH TICKS AND CRSSES ETC curly arrow must come from lone pair on H or negative charge on H dipole AND curly arrow from C= bond to curly arrow must come from the bond, not the carbon atom correct intermediate AND curly arrow to H curly arrow must come from lone pair on or negative charge on and go to H or positive charge on H Where circles have been placed round charges, this is for clarity only and does not indicate a requirement ALLW correct structural R displayed R skeletal formulae R a combination of above as long as unambiguous ALLW for second stage IF H 2 is used it MUST show the curly arrow from the negative charge or lone pair on the oxygen atom of the intermediate to H in H 2 AND from the H bond to the in H 2. Dipole not required on water molecule Penalise missing H on intermediate only IGNRE product already given credit in part (i)

4 Answer Mark Guidance (d) CH 3 H 2 Br 2 CH 3 H 2HBr 1 ALLW correct structural R displayed R skeletal formulae R a combination of above as long as unambiguous ALLW disubstitution at any positions on benzene ring Br Br Total 10

5 Answer Mark Guidance 2 (a) FIRST react all with Tollens reagent AND silver mirror/ppt/solid (formed) with compound D R with Fehling s/benedict s solutions AND (brick-red/orange) solid/precipitate (formed) with compound D NTE: eliminates D 4 ALLW ammonia silver nitrate for reagent ALLW black solid/ppt ALLW the aldehyde gives a silver mirror ALLW solid R crystals R ppt as alternatives for precipitate ALLW correct structural R displayed R skeletal formulae R combination of above as long as unambiguous [] D NT ALLW molecular formulae for organic structures THEN react C and E with H 2 S 4 /H AND K 2 Cr 2 7 / Cr /Na 2 Cr 2 7 AND colour change R green colour with compound C R no change R no reaction R no green colour with compound E IGNRE all references to 2,4-dinitrophenylhydrazine/Brady s ACCEPT acidified dichromate ALLW blue/green blue IGNRE equation for oxidation of D ALLW equation for partial oxidation 2[] 2

6 Answer Mark Guidance ALLW alternative sequences e.g. FIRST react all with H 2 S 4 AND K 2 Cr 2 7 colour change with C and D eliminates E At least one correct equation and structure of one product from either reaction required for the second mark. NB several possible products for the oxidation of D 2 (b) THEN react C and D with Tollens distinguishes between C and D 4 ALLW correct structural R displayed R skeletal formulae R combination of above as long as unambiguous First curly arrow must come from either a lone pair on H or negative charge on H curly arrow from H to C (δ) of correct C= group dipole correct AND curly arrow from C= bond to (δ ) IF aldehyde reduced R both carbonyls reduced D NT AWARD first mark (second, third and fourth marks can be awarded ECF) IGNRE lack of C H if entirely skeletal IGNRE curly arrows in second stage correct intermediate with negative charge on Apply ecf to error in structure e.g. CH 2 missing from the chain or CH/-CH instead of CH correct product IGNRE other products

7 Answer Mark Guidance 2 (c) 1 Compound C D E Number of peaks all correct 2 (d) (i) pent-2-ene AND 3 ALLW correct structural R displayed R skeletal formulae R combination of above as long as unambiguous ALLW C 2 H 5 CH and CH 3 CH 2 (d) (ii) hexa-2,4-diene 1 ALLW correct structural R displayed R skeletal formulae R combination of above as long as unambiguous Total 13

8 er Marks Guidance 3 (a) (i) propane-1,2,3-triol 1 ALLW absence of e after propan ALLW 1,2,3-propanetriol ALLW absence of hyphens 1, 2 and 3 must be clearly separated: ALLW full stops: R spaces: D NT ALLW 123 IGNRE glycerol (ii) 3 ALLW correct structural R displayed R skeletal formulae R combination of above as long as unambiguous D NT ALLW cis structure ALLW R ne mark for decenoate salt R decenoic acid ne mark for hexanoate salt R hexanoic acid ne mark for BTH correct products shown as salts (with or without Na ) D NT ALLW (covalent bond) ALLW delocalised carboxylate IGNRE glycerol (b) one of the fatty acids is trans which may increase / cause / produce (the level of) bad /LDL cholesterol QWC cholesterol MUST be spelt correctly 2 ALLW one of the products is TRANS ALLW reduces (the level of) good /HDL cholesterol Total 6

9 Answer Marks Guidance 4 (a) (i) 1 ALLW correct structural R displayed R skeletal formulae R combination of above as long as unambiguous F = AND reagent NaBH 4 NB ne mark for BTH Wedge out of the paper is required i.e.( or or ) D NT ALLW dashed wedge on methyl group in this orientation ( or or or ) ALLW H (ii) Colour changes from orange to green / blue / green blue 1 (iii) to ensure carboxylic acid is formed R prevents formation of aldehyde R distillation only makes the aldehyde 1 (iv) (nucleophilic) addition 1 ALLW redox R reduction (b) 2,4-DNP(H) orange precipitate 2 ALLW Brady s (reagent) ALLW orange/red/yellow for colour of the 2,4-DNP(H) precipitate ALLW solid/crystals in place of precipitate IGNRE any reference to recrystallising/melting points

10 Answer Marks Guidance 4 (c) (i) ne of: 2 For bold wedge ALLW or or H For dashed wedge ALLW or or or R R H D NT ALLW any other representation of the structure, i.e. anything not skeletal ALLW open wedges ALLW isomers shown in any alternative correct orientation for one mark optical (isomerism) (ii) If answer = 63.5 award 3 marks moles of E used = 4.56/160(.0) / (mol) moles of G formed = 3.15/174(.0) / (mol) yield = / % / 63.5% mol is exact calculator value mol is to 3sf (calculator value ) IGNRE trailing numbers in this answer ALL ANSWERS MUST be to a minimum of 3sf, the final answer must be to 3 sf (calculator value gives %) (rounding of moles of G gives %) ALLW ecf from incorrect Mr or moles unless the yield is >100

11 Answer Marks Guidance (iii) 2 ALLW abbreviation of alkyl chain Wedge out of the paper is required i.e.( or or ) ther product = H 2 for first mark for second mark D NT ALLW dashed wedge on methyl group in this orientation ( or or or ) ALLW Be careful with orientation of lactone: ALLW Total 13

12 er Marks Guidance 5 (a) ( cis-isomer has Hs on same side R cis-isomer has branches on same side R cis-isomer has same groups on same side R cis-isomer has lowest priority groups on same side R cis-isomer has highest priority groups on same side 2 ALLW trans-isomer has Hs on opposite sides R trans-isomer has branches on opposite sides R trans-isomer has same groups on opposite sides D NT ALLW similar groups for same groups R trans-isomer has lowest priority groups on opposite sides R trans-isomer has highest priority groups on opposite sides For explanation, ALLW a clear diagram, ie: cis ALLW response in terms of packing, e.g. molecules/chains of trans-isomer pack close together R molecules/chains of cis-isomer do not pack closely together D NT ALLW carbon atoms for molecules/chains (ii) heart disease/strokes 1 ALLW any named heart/circulatory complaint e.g. atheroma, atheroscleros ALLW increase in bad cholesterol/ldl ALLW high in LDLs ALLW fat lining arteries ALLW high blood pressure ALLW hypertension IGNRE reference to HDLs and cholesterol on its own

13 er Marks Guidance (b) ( 27 1 (ii) 8 1 (c) ( alcohol IGNRE H R hydroxyl R hydroxy D NT ALLW phenol R hydroxide ester 2 IGNRE CR IF there is a list with more than two responses, mark wrong responses first, e.g. alcohol, ketone X, ether X zero marks alcohol, ester, methyl X 1 mark ester, hydroxide X, ketone X zero marks ester, hydroxyl I, ketone X 1 mark (ii) ensures correct chirality 1 ALLW enantiomer for optical isomer ALLW produces only one optical isomer ALLW stops need/cost/difficulty of separating optical isomers ALLW stops formation of the optical isomer which may have (harmful) side effects D NT ALLW lower doses/dosage needed D NT ALLW forms one stereoisomer (could be E/Z) D NT ALLW stereoselectivity

14 er Marks Guidance (iii) ANNTATINS MUST BE USED 1st step reagent: functional groups: NaBH 4 alde yde forms an alcohol names required ALLW H 2 /Ni (catalyst) D NT ALLW LiAlH 4 (because LiAlH 4 reduces CH) IGNRE type of reaction or conditions IGNRE CH R H IGNRE carbonyl R hydroxyl R hydroxy D NT ALLW phenol R hydroxide 2nd step Marks NLY available from correct hydroxycarboxylic acid formed in 1st step reagent: functional groups: Acid R H (catalyst) alcoho and carboxylic acid / carboxyl group form an ester names required 4 ALLW named acid/correct formula IGNRE dilute/concentrated IGNRE H, CH, C, IGNRE hydroxyl R hydroxy D NT ALLW phenol R hydroxide Total 12

15 er Marks Guidance 6 (a) N 2 curly arrow from ring to N 2 N 2 H N 2 CH CH CH correct intermediate curly arrow from C-H bond back to reform ring correct products H M1 M2 M3 M Note: ALLW M1, M2 AND M3 for benzene R ANY substituted benzene compound For M4, credit NLY the correct products HN 3 H 2 S 4 N 2 H HS 4 H 2 S 4 R H 2 HS 4 ANNTATINS MUST BE USED Mark 1 (M1) ALLW curly arrow from the ring R from within the ring Mark 2 (M2) intermediate showing delocalisation over less than 6 carbons with the correct orientation BUT D NT ALLW intermediate with π system less than halfway up: H N 2 CH Mark 3 (M3) curly arrow from C H bond reforming π-delocalised ring in benzene ALLW Kekulé mechanism: N2 H N 2 N 2 HN 3 2H 2 S 4 N 2 H HS 4 H 2 S 4 H 3 2HS 4 6 H R HN 3 H 2 S 4 H 2 N 3 HS 4 AND H 2 N 3 N 2 H 2 H HS 4 H 2 S 4 CH CH CH ALLW double bonds shown in other Kekulé arrangement Mark 4 (M4) BTH correct products: 3-nitrobenzaldehyde AND H

16 er Marks Guidance (b) 2 C 6 H 5 CH KH C 6 H 5 CH 2 H C 6 H 5 CK R ALLW correct structural R displayed R skeletal formula 2 C 6 H 5 CH H C 6 H 5 CH 2 H C 6 H 5 C ALLW combination of formulae as long as unambiguous 1 mark for C 6 H 5 CH 2 H 1 mark for C 6 H 5 CK R C 6 H 5 CH R C 6 H 5 C ALLW use of NaH instead of KH throughout, i.e. 2 C 6 H 5 CH NaH C 6 H 5 CH 2 H C 6 H 5 CNa ALLW C 6 H 5 C K (c) 1 mark for complete fully correct balanced equation (i.e. as above) 3 C N H H ALLW correct structural R displayed R skeletal formula ALLW combination of formulae as long as unambiguous C C CH H H C C C C C H H H H 3

17 er Marks Guidance (d) ( ANNTATINS MUST BE USED H C 6 H 5 C C 6 H 5 C R C 6 H 5 C R IGNRE connectivity on H of product R - H H intermediate H organic product 1 mark for curly arrow from R to C of C= (lone pair not necessary) 1 mark for correct dipoles on C= AND curly arrow from double bond to 1 mark for correct intermediate with charge on 1 mark for correct product 4 Curly arrow MUST start from sign of R R from lone pair on R lone pair does not need to be shown on R IGNRE any curly arrows shown for stage 2 i.e. in intermediate (ii) H 3 C Li CH CH 2 CH 3 R H 3 C Li CH CH 2 CH 3 1 ALLW correct structural R displayed R skeletal formula ALLW combination of formulae as long as unambiguous IGNRE C 4 H 9 Li R C 4 H 9 Li Total 17

PMT GCE. Chemistry A. Advanced GCE Unit F324: Rings, Polymers and Analysis. Mark Scheme for June Oxford Cambridge and RSA Examinations

PMT GCE. Chemistry A. Advanced GCE Unit F324: Rings, Polymers and Analysis. Mark Scheme for June Oxford Cambridge and RSA Examinations GCE Chemistry A Advanced GCE Unit F324: Rings, Polymers and Analysis Mark Scheme for June 2013 xford Cambridge and RSA Examinations CR (xford Cambridge and RSA) is a leading UK awarding body, providing