CHERRY HILL TUITION OCR A CHEMISTRY A2 PAPER 19 MARK SCHEME ANNOTATIONS MUST BE USED CH 3 CH 3 CH 3 H + correct products
|
|
- Eileen Sharp
- 5 years ago
- Views:
Transcription
1 ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME ALLW Kekulé structures throughout 1 (a) ANNTATINS MUST BE USED ALLW skeletal ALLW + + N 2 R N 2 ALLW 1st curly arrow from the ring R from within the ring to any part of the N + 2 including the + charge + + D NT ALLW intermediate with broken ring less than halfway down: N 2 N 2 N 2 curly arrow from ring to N 2 + correct intermediate curly arrow from bond back to reform ring 1 mark for intermediate correct products 4 + N 2 orseshoe must have open end towards N 2 ALLW Kekulé mechanism: 1 mark for curly arrow N + 2 N 2 N 2 ALLW double bonds shown in other Kekulé arrangement IF has been omitted completely (ie benzene shown), D NT AWARD intermediate mark R products mark (max 2) IF N 2 is shown in incorrect position in intermediate or product, D NT AWARD intermediate mark but award other marks (max 3) 1
2 ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME 1 (b ) ALLW any correct unambiguous structures N 2 ALLW N 2 Note: connectivity is NT being assessed in this part 2 N N 2 N (c) 1st stage isomer: isomer 3 product: 2 N N 2 reagents: Sn AND (conc) l equation: ANNTATINS MUST BE USED ALLW structure of isomer 3 shown separately R in equation ALLW structure of product shown separately R in equation ALLW correct name (3,5-diaminomethylbenzene) IGNRE incorrect name D NT ALLW 6 3 (N 2 ) 2 ALLW Zn + l/ 2 + metal catalyst/lial 4 /Na in ethanol IGNRE NaB 4 ALLW Sn and l followed by Na D NT ALLW Sn and l and Na 12 [] 4 2 IF isomer 3 R product are given in equation but not shown previously then credit here 2 N N 2 2 N N 2 Also credit reagents here if shown (eg above arrow) ALLW correct structural R displayed R skeletal formula ALLW combination of formulae as long as unambiguous 2
3 ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME (c) (i) 2nd stage organic compound: 2 6 D NT ALLW molecular formula ALLW name of compound: propanedioic acid R propane-1,3-dioic acid ALLW absence of e after propan ALLW acyl dichloride: l 2 l ALLW cyclic acid anhydride of propanedioic acid: 2 type of polymer: polyamide ALLW Nylon or Kevlar D NT ALLW polypeptide D NT ALLW amide Total 12 3
4 ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME 2 (a) 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 (b) (i) methanol R ethanol AND renewable 1 BT points required for the mark ALLW correct structural R displayed R skeletal formula D NT ALLW molecular formulae ALLW easy/cheap to manufacture/produce as alternative for renewable/from plants/from fermentation/burns more easily/efficiently (b) (ii) equilibrium shifts to right 1 ALLW equilibrium shifts in forward direction ALLW more products form ALLW greater yield R fully reacts R goes to completion D NT ALLW improves atom economy 4
5 ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME 2 (c) ALLW correct structural R displayed R skeletal formula ALLW combination of formulae as long as unambiguous D NT ALLW molecular formulae ( 2 ) ALLW further esterification, ie ( 2 ) ALLW linear formula for anhydride, ie 2 2 If incorrect carboxylic acid/anhydride/alcohol is used, ALLW EF for second equation 5
6 ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME 2 (d) A B Mark A, B and independently ie 2 column column R R R R R R R R R R R R Total 8 A can be any of the alternatives in the 1st B can be any of the alternatives in the 2nd can be any of the alternatives in the 3rd column ALLW correct structural R displayed R skeletal formula ALLW combination of formulae as long as unambiguous D NT ALLW molecular formulae ALLW correct names for A, B and For B accept diester For, IGNRE n R brackets (even if wrong); ALLW solid side bonds Minimum is one correct repeat unit. Polymer must be open at both ends 6
7 ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME 3 (a) observation: silver R Ag ALLW black R grey type of reaction: oxidation ALLW redox organic product: 3 3 ALLW correct structural R displayed R skeletal formula ALLW combination of formulae as long as unambiguous D NT ALLW molecular formulae ALLW carboxylate, 3 (b) R R intermediate R products (+ ANNTATINS MUST BE USED ALLW mechanism showing curly arrows from lone pair on and of intermediate 1 mark for curly arrow from to of = 1 mark for correct dipole on = AND curly arrow from double bond to 1 mark for correct intermediate with negative charge on AND curly arrow from to of AND curly arrow from to of 1 mark for correct organic product 4 Dipole not required on D NT ALLW incorrect dipole on ALLW 1 mark for correct intermediate with charge on AND curly arrow from to + IGNRE missing D NT ALLW incorrect second product 7
8 ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME 3 (c) reagent: 2 D NT ALLW EF from incorrect reagent, eg 2,4-DNP observation: decolourised R orange to colourless D NT ALLW goes clear ALLW red/orange/yellow/brown in any combination organic product: 3 3 ALLW organic product from reaction of one of the double bonds only, ie 3 3 R ALLW correct structural R displayed R skeletal formula ALLW combination of formulae as long as unambiguous D NT ALLW molecular formulae ALTERNATIVE reagents For 1st mark, ALLW 2 R l 2 R I 2 R l R R I R 2 For 2nd mark, there must be a statement of no change R no observation or similar that implies there is no visible change EXEPT for I 2 which has an observation of decolourised R brown to colourless Total 10 For 3rd mark, correct organic product must be shown that could be from reaction of both or one of the double bonds. 8
9 ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME 4 (a) (i) l( ) + 3N 3 2 N( ) + N N 4 l 1 ALLW use of two N 3 : l( ) + 2N 3 2 N( ) + N l ALLW products as above R 2 N( ) + N 4 l ALLW use of one N 3 : l( ) + N 3 2 N( ) l ALLW products as above R 2 N( ) + l For alternatives below, for N 4 l, ALLW N 4 + l R N l for l, ALLW + l R + + l for 2 N( ) + N 4 + ALLW 2 N( ) N 4 + R 2 N( )N 4 ALLW R in equation in place of (either or both sides) ALLW correct structural R displayed R skeletal formula ALLW combination of formulae as long as unambiguous D NT ALLW molecular formulae (a) (ii) N ALLW correct structural R displayed R skeletal formula ALLW combination of formulae as long as unambiguous 1 ALLW product from carboxylate ion as nucleophile: 2 N 9
10 ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME 4 (b) (i) R D NT ALLW any structure containing R (except in ) 1 (b) (ii) 2 2 ALL bond linkages must be correct, eg the chiral must be linked to the of the, the of the 2 and the N of the N 2 (connectivity is being tested) 2 N N 2 2 The 2nd mark is for the mirror image of an amino acid. This could be any amino acid EXEPT glycine D NT penalise connectivity more than once ALLW R in equation in place of 2 (either or both sides) Each structure must have four central bonds, with at least two wedges, one in; one out For bond into paper, accept: 4 (c) Disadvantages Any two from: (one stereoisomer might have harmful) side effects reduces the (pharmacological) activity/effectiveness cost R difficulty in separating stereoisomers 2 max ANNTATINS MUST BE USED ALLW optical isomer R enantiomers as alternative for stereoisomers ALLW a response that implies an increased dose Synthesis of a single optical isomer Any two from: using enzymes or bacteria using a chiral catalyst R transition metal complex/transition metal catalyst using chiral synthesis R chiral starting material R natural amino acid 2 max Total 8 ALLW biological catalyst ALLW 'chiral pool' R L-amino acids R D-sugars 10
11 ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME 5 (a) (i) Adsorption (onto the stationary phase) ALLW adsorbtion or adsorb(s) or adsorbed spelled correctly at least once Quality of Written ommunication 1 D NT ALLW anything that begins with ab... Adsorption must be spelled correctly (a) (ii) ALLW any value in the range IGNRE significant figures D NT ALLW fraction/percent as final answer (a) (iii) Spot may contain more than one compound/component 1 ALLW compounds have similar R f values/adsorptions R compounds have not (fully) separated R B is spread over a large region R compounds are similar IGNRE retention times 5 (b) (i) G separates the components/compounds ALLW chromatography for G ALLW they have different retention times (ii) (iii) (iv) AND MS is compared to a database/reference 1 nerol and geraniol AND they are stereoisomers R primary alcohols 1 stereoisomers have the same structural formula AND different 3D arrangements 1 1 ALLW MS analyses compounds/gives structural information/gives different mass spectra ALLW (uses) fragmentation patterns/fragments/peaks/parts of the compound D NT ALLW MS identifies compounds (in question) D NT ALLW molecular ion alone/m r etc. ompounds AND reason required for the mark ALLW they are E/Z isomers R cis-trans isomers ALLW straight-chain alcohols R unsaturated alcohols BT points required for the mark ALLW different arrangements in space ircle must include the correct = double bond AND must not extend further than the adjacent atoms in the main chain, ie limit is: 11
12 ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME 5 (c) (b) (v) * * * orrectly calculates amount of myrcene = 34/136 R 0.25 (mol) orrectly calculates 60% yield of menthol = /100 R 0.15 (mol) 2 orrectly calculates mass of menthol = = 23.4 (g) 3 ALL TREE chiral centres required for 2 marks ANY TW chiral centres required for 1 mark If more than three asterisks are shown, mark incorrect asterisk(s) first ANNTATINS MUST BE USED ALLW amount of myrcene 60/100 ALLW amount of menthol 156 ALLW alternative approach based on reacting masses (using same EF principles as above): correctly calculates mass of myrcene that could be obtained from 34 g myrcene: mass = /136 = 39 (g) 156 ; % of 39 g = 39 60/100 = 23.4 (g) ALLW final answer to 2 or more significant figures correctly rounded orrect answer of 23.4 (g) with no working scores all 3 marks Total 12 12
13 ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME 6 (a) ANNTATINS MUST BE USED a singlet for position 2 R a singlet because it has no adjacent s A triplet for positions 4 and 6 R a triplet because it has 2 adjacent s A quintet for position 5 R a quintet because it has four adjacent s 3 ALLW a response that implies a single peak R no splitting ALLW a response that implies a splitting into three D NT ALLW implications of more than one triplet ALLW pentet R a response that implies a splitting into five R multiplet ALLW 1 mark for singlet and triplet and quintet/pentet/multiplet with no identification of protons Any suggestion that the oxygens cause a splitting scores a maximum of 2 marks. All 3 remaining splitting patterns correct 2 marks. Any 2 correct 1 mark. IF number labels for protons in diagram are not identified, ALLW identification by chemical shifts for 2 marks max: singlet at AND a triplet at quintet/pentet/multiplet at lear and unambiguous identification of the protons other than by position number should be credited, ie 2 between two oxygens Quality of Written ommunication singlet R triplet R quintet R pentet R multiplet (see Guidance) must be spelled correctly at least once 13
14 ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME F324 Mark Scheme January (b) ANY 5 marks plus correct structure (in box) ANNTATINS MUST BE USED Molecular ion/m + peak at (m/z of) 106 Fragment peak at 91 is / ALLW molecular mass R relative molecular mass ALLW 6 4 / ALLW peak at 91 represents loss of Molecular formula is 8 10 (or implied, ie any one of the structures below) 2 5 ALLW correct structural R displayed R skeletal formula ALLW combination of formulae as long as unambiguous ALLW a correct name eg a dimethylbenzene ALL FUR structures needed for 1 mark ALLW correct names 13 NMR spectrum shows 5 environments Peak near 20 is a attached at another carbon, R peaks at ~ for aromatic s ALLW NMR spectrum shows five different types of carbon D NT ALLW NMR spectrum has five peaks the mark is for realising what the peaks show, not for just describing the spectrum 14 14
15 ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME Mark Scheme January (b) Number of peaks for other three isomers matched to structures: Any 2 correct for 2 marks 1 correct for 1 mark peaks 3 peaks 6 peaks ALLW carbon environments for peaks orrect structure shown: 6 Total
1 (a) (CH 3 CO) 2 O + CH 3 CH(OH)CH 3 CH 3 COOCH(CH 3 ) 2 + CH 3 COOH
Question er Mark Guidance 1 (a) ( 3 ) 2 + 3 () 3 3 ( 3 ) 2 + 3 1st mark orrect structure of ester: 3 ( 3 ) 2 2nd mark Equation contains correct formulae for ( 3 ) 2, 3 () 3 AND 3 2 ALLW correct structural
More informationALLOW CO 2 and CO 2 H CH 3
CERRY ILL TUITI CR A CEMISTRY A PAPER 1 MARK SCEME 1 (a) (i) The p R point at which the zwitterion exists 1 ALLW p/point at which there is no overall/net charge IGRE p/point at which there is no charge/
More informationANNOTATIONS MUST BE USED
1 (a) ATATIS MUST BE USED ALLW skeletal ALLW R ALLW 1st curly arrow from the ring R from within the ring to any part of the 2 including the charge H D T ALLW intermediate with broken ring less than halfway
More information(b) (i) Compound B AND M + /molecular ion peak (at m/z) = 124
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
More informationQuestion Answer Marks Guidance 1 (a) M1 EITHER in words: (pyruvic acid forms) hydrogen bonds with water
1 (a) M1 EITER in words: (pyruvic acid forms) hydrogen bonds with water 2 R correctly labelled diagram showing hydrogen bond between pyruvic acid and water FR M1 only: if use diagram ALLW a labelled hydrogen
More informationPMT GCE. Chemistry A. Advanced GCE. Unit F324: Rings, Polymers and Analysis. Mark Scheme for January Oxford Cambridge and RSA Examinations
GCE Chemistry A Advanced GCE Unit F324: Rings, Polymers and Analysis Mark Scheme for January 2012 xford Cambridge and RSA Examinations CR (xford Cambridge and RSA) is a leading UK awarding body, providing
More informationQuestion Answer Mark Guidance 1 (a) monomers join/bond/add/react/form polymer/form chain AND another product/small molecule e.g.
Question Answer Mark Guidance (a) monomers join/bond/add/react/form polymer/form chain AND another product/small molecule e.g. 2 /l IGNRE two when referring to monomers, ie (two) monomers... (b) (i) QW
More informationF324 Mark Scheme January 2010 F324 Rings, Polymers and Analysis. Question Expected Answers Marks Additional Guidance 1 (a)
F324 Rings, Polymers and Analysis Question Expected Answers Marks Additional Guidance 1 (a) ALLW 6 6 + Br 2 6 5 Br + Br + Br 2 Br + Br (b) (i) White precipitate R white solid R white crystals 1 D T ALLW
More informationWhere circles have been placed round charges, this is for clarity only and does not indicate a requirement. ALLOW delocalised carboxylate.
Question (a) Answer Mark Guidance Where circles have been placed round charges, this is for clarity only and does not indicate a requirement (i) ALLOW correct structural OR displayed OR skeletal formulae
More information2 Answer all the questions.
ERRY ILL TUITI R A EMISTRY A2 PAPER 17 2 Answer all the questions. 1 A chemist was investigating the reactions of benzene, phenol and cyclohexene with bromine. She found that they all reacted with bromine
More informationAllow CONH- or - COHN - 1(a)(i) Mark two halves separately
ERRY ILL TUITIN AQA EMISTRY A2 PAPER 27 MARK SEME Question Marking Guidance (a)(i) 2 2 N N 2 Mark Allow N- or - N - N 6 Mark two halves separately 2 N 6 omments lose each for missing trailing bonds at
More informationNitrogen Compounds - MS 1. (a) (i) is an amine and a carboxylic acid / contains both NH2 and COOH functional groups (1) AW 1
itrogen ompounds - MS. (a) (i) is an amine and a carboxylic acid / contains both and functional groups () AW (c) R( ) () R Does not fit the formula because and are not attached to the same carbon () AW
More informationQuestion Answer Mark Guidance 1 (a) (i) M1 p-orbitals overlap (to form pi/π-bonds) 4 ANNOTATE ANSWER WITH TICKS AND CROSSES ETC
Question Answer Mark Guidance (a) (i) M p-orbitals overlap (to form pi/π-bonds) 4 ANNOTATE ANSWER WIT TICKS AND CROSSES ETC IGNORE p-orbitals overlap to form sigma bonds M2 π-bond(s) are delocalised in
More informationGCE. Chemistry. Mark Scheme for June Advanced GCE 2814/01 Chains, Rings and Spectroscopy. Oxford Cambridge and RSA Examinations
GE hemistry Advanced GE 2814/01 hains, Rings and Spectroscopy Mark Scheme for June 2010 xford ambridge and RSA Examinations R (xford ambridge and RSA) is a leading UK awarding body, providing a wide range
More informationPMT GCE. Chemistry A. Advanced GCE Unit F324: Rings, Polymers and Analysis. Mark Scheme for January Oxford Cambridge and RSA Examinations
GE hemistry A Advanced GE Unit F324: Rings, Polymers and Analysis Mark Scheme for January 2013 xford ambridge and RSA Examinations R (xford ambridge and RSA) is a leading UK awarding body, providing a
More informationPMT 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
More informationPractice paper Set 1 MAXIMUM MARK 100. Final. H432/02 Mark Scheme Practice 1. A Level Chemistry A H432/02 Synthesis and analytical techniques
432/02 Mark Scheme Practice 1 Practice paper Set 1 A Level hemistry A 432/02 Synthesis and analytical techniques MARK SEME Duration: 2 hours 15 minutes MAXIMUM MARK 100 Final This document consists of
More information4 ALLOW sugar from equation
Question Expected Answers Marks Additional Guidance 1 (a) method 1: fermentation of sugars or carbohydrates R reaction with yeast with sugar or carbohydrates 6 12 6 2 2 5 + 2 2 4 ALLW sugar from equation
More informationAcceptable sequence of stages are: chlorination. nitration, reduction, chlorination nitration. nitration, chlorination, reduction, reduction
Question er Mark Guidance 1 (a) (i) Response requires three stages Acceptable sequence of stages are: chlorination nitration, reduction, chlorination nitration nitration, chlorination, reduction, reduction
More informationGCE Chemistry A. Mark Scheme for June Unit F324: Rings, Polymers and Analysis. Advanced GCE. Oxford Cambridge and RSA Examinations
GCE Chemistry A Unit F324: Rings, Polymers and Analysis Advanced GCE Mark Scheme for June 2017 Oxford Cambridge and RSA Examinations OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing
More informationDO NOT ALLOW any reference to spatial/space
Question Answer Mark Guidance 1 (a) (i) (compounds or molecules having the) same molecular 1 ALLOW different structure OR different displayed formula but different structural formulae formula OR different
More informationflowers, leaves and roots of roses rose oil heat
1 Rose oil can be extracted from the flowers, leaves and roots of roses using the apparatus below. coolant flowers, leaves and roots of roses rose oil heat (a) The rose oil contains a mixture of compounds,
More informationTuesday 19 June 2012 Afternoon
Tuesday 19 June 2012 Afternoon A2 GCE CEMISTRY A F324 Rings, Polymers and Analysis *F314730612* Candidates answer on the Question Paper. CR supplied materials: Data Sheet for Chemistry A (inserted) ther
More informationCHERRY HILL TUITION OCR (SALTERS) CHEMISTRY A2 PAPER Answer all the questions. O, is formed in the soil by denitrifying bacteria. ...
2 Answer all the questions. 1 itrous oxide gas, 2, is formed in the soil by denitrifying bacteria. (a) Give the systematic name for nitrous oxide. ne model of the bonding in nitrous oxide includes a dative
More informationAdvanced GCE Chemistry A
Advanced GE hemistry A Unit F4 Rings, Polymers and Analysis Medium banded andidate Style Answer Introduction R has produced these candidate style answers to support teachers in interpreting the assessment
More information8 (c) (i) H 2 /Ni or H 2 /Pt or Sn/HCl or Fe/HCl (conc or dil or neither) allow dil H 2 SO 4 ignore mention of NaOH
Polymers Answers EM4 - AQA GE hemistry Mark Scheme 200 January series 8 (c) (i) 2 /i or 2 /Pt or Sn/l or e/l (conc or dil or neither) allow dil 2 S 4 ignore mention of a ot ab 4 ot LiAl 4 ot a/ 2 5 not
More information1 hour 45 minutes plus your additional time allowance
GE A Level 1094/01 EMISTRY 4 P.M. TUESDAY, 14 June 2016 1 hour 45 minutes plus your additional time allowance Surname Other Names entre Number andidate Number 2 WJE BA Ltd. J*(S16-1094-01)MLP 2 For Examiner
More informationThe amide or peptide link is found in synthetic polyamides and also in naturally-occurring proteins.
ERRY ILL TUITIN AQA EMISTRY A2 PAPER 27 10 14 4 (a) (i) The amide or peptide link is found in synthetic polyamides and also in naturally-occurring proteins. Draw the repeating unit of the polyamide formed
More informationTHIS IS A NEW SPECIFICATION
TIS IS A EW SPEIFIATI ADVAED GE EMISTRY A Rings, Polymers and Analysis F324 * E / 1 5371* andidates answer on the Question Paper R Supplied Materials: Data Sheet for hemistry A (inserted) ther Materials
More informationFinal. Mark Scheme. Chemistry CHEM4. (Specification 2420) Unit 4: Kinetics, Equilibria and Organic Chemistry
ERRY ILL TUITI AQA EMISTRY A2 PAPER 30 MARK SEME Question : /A Question 2: /A Question 3: /A General ertificate of Education (A-level) June 203 hemistry EM4 (Specification 2420) Unit 4: Kinetics, Equilibria
More informationQuestion Answer Mark Guidance 1 (a) (i) 4 Please use annotations (rate) molecules/particles in smaller volume OR increases concentration
ERRY ILL TUITIN R (SALTERS) EMISTRY A2 PAPER 28 MARK SEME 1 (a) (i) 4 Please use annotations (rate) molecules/particles in smaller volume R increases concentration ALLW molecules/particles closer together
More informationSubject: Chains, Rings and Spectroscopy Code: Session: June Year: Final Mark Scheme
Subject: hains, Rings and Spectroscopy ode: : e : Mark Scheme MAXIMUM MARK 90 ADVIE T EXAMINERS N TE ANNTATIN F SRIPTS 1. Please ensure that you use the final version of the Mark Scheme. You are advised
More informationMark Scheme Page 1 of 8 Unit ode 2814 Session Jan Year 2004 Qu. Expected answers: Marks: 1 (a) (i) (relative) molecular mass / M r (ii) right / highest m /e / highest mass / second highest mass etc AW
More informationPMT GCE. Chemistry A. Advanced GCE F324. Mark Scheme for June Oxford Cambridge and RSA Examinations
GE hemistry A Advanced GE F324 Mark Scheme for June 200 xford ambridge and RSA Examinations R (xford ambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the
More informationA mass spectrometer can be used to distinguish between samples of butane and propanal. The table shows some precise relative atomic mass values.
Butane and propanal are compounds with M r = 58.0, calculated using data from your Periodic Table. (a) A mass spectrometer can be used to distinguish between samples of butane and propanal. The table shows
More informationUNIT 4 REVISION CHECKLIST CHEM 4 AS Chemistry
UNIT 4 REVISION CHECKLIST CHEM 4 AS Chemistry Topic 4.1 Kinetics a) Define the terms: rate of a reaction, rate constant, order of reaction and overall order of reaction b) Deduce the orders of reaction
More informationCandidates answer on the question paper. Time: 1 hour 15 minutes Additional Materials: Data Sheet for Chemistry (Inserted) Scientific calculator
SPECIMEN Advanced GCE CEMISTRY A F324 QP Unit F324: Rings, Polymers and Analysis Specimen Paper Candidates answer on the question paper. Time: 1 hour 15 minutes Additional Materials: Data Sheet for Chemistry
More information(07) 3 (e) Calculate the ph of this buffer solution at 298 K. Give your answer to 2 decimal places
7 3 (e) An acidic buffer solution is formed when 10.0 cm3 of 0.125 mol dm 3 aqueous sodium hydroxide are added to 15.0 cm3 of 0.174 mol dm 3 aqueous HX. The value of Ka for the weak acid HX is 3.01 10
More informationF324: Rings, Polymers and Analysis Carbonyl Compounds
F324: Rings, Polymers and Analysis 4.1.2 arbonyl ompounds 1. ydroxyethanal, 2, is sometimes referred to as the first sugar as it is the simplest possible molecule that contains both an aldehyde group and
More informationCHEMISTRY 2814/01 Chains, Rings and Spectroscopy
TIS IS A LEGAY SPEIFIATION ADVANED GE EMISTRY 2814/01 hains, Rings and Spectroscopy * OE / 1 9368* andidates answer on the Question Paper A calculator may be used for this paper OR Supplied Materials:
More informationSurname. Number OXFORD CAMBRIDGE AND RSA EXAMINATIONS ADVANCED GCE F324 CHEMISTRY A. Rings, Polymers and Analysis
Candidate Forename Centre Number Candidate Surname Candidate Number OXFORD CAMBRIDGE AND RSA EXAMINATIONS ADVANCED GCE F324 CHEMISTRY A Rings, Polymers and Analysis WEDNESDAY 27 JANUARY 2010: Morning DURATION:
More informationQuestion Answer Mark Guidance 1 (a)
Question Answer Mark Guidance 1 (a) 1 ALLW correct structural R displayed R skeletal 3 3 formula R mixture of the above 3 D NT ALLW molecular formula Br Br ALLW dichloro or diiodo compound instead of the
More informationThursday 26 January 2012 Afternoon
Thursday 26 January 2012 Afternoon A2 GE EMISTRY A F324 Rings, Polymers and Analysis *F314460112* andidates answer on the Question Paper. R supplied materials: Data Sheet for hemistry A (inserted) ther
More informationAS Demonstrate understanding of the properties of selected organic compounds
No Brain Too Small EMISTRY AS 91165 Demonstrate understanding of the properties of selected organic compounds ollated Flow hart Type Questions / types of reaction (2017) (a) (i) omplete the following reaction
More information2 Answer all the questions. CH(NH 2. )COOH, R is CH [1] (ii) Draw the structures of the ions formed by alanine at ph 6.0 and at ph 1.5.
2 Answer all the questions. 1 This question looks at the properties and chemistry of some α-amino acids. The general formula of an α-amino acid is R(N 2 ). (a) In the α-amino acid alanine, 3 (N 2 ), R
More information, by reacting CH 3 with ethanoic anhydride, (CH 3
1 A chemist prepares and analyses some esters. (a) The chemist prepares an ester of propan-2-ol, H 3 H(H)H 3, by reacting H 3 H(H)H 3 with ethanoic anhydride, (H 3 ) 2. Using structural formulae, write
More informationPMT. GCE Chemistry A. Unit F324: Rings, Polymers and Analysis. Advanced GCE. Mark Scheme for June Oxford Cambridge and RSA Examinations
GCE Chemistry A Unit F324: Rings, Polymers and Analysis Advanced GCE Mark Scheme for June 2015 Oxford Cambridge and RSA Examinations OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing
More informationAlcohols. Nomenclature. 57 minutes. 57 marks. Page 1 of 9
..0 Alcohols Nomenclature 57 minutes 57 marks Page of 9 M. (a) % O =.6 % () If % O not calculated only M available C O () = 5. =.5 =.5 Ratio: : 0: ( C 0 O) () If arithmetic error in any result lose M If
More information3.4 A2 Unit F324: Rings, Polymers and Analysis
3.4 A2 Unit F324: Rings, Polymers and Analysis This unit builds upon the chemical concepts that have been developed during AS Chemistry. This unit consists of three teaching modules: Module 1: Rings, Acids
More information2 Answer all the questions. 1 Alkenes and benzene both react with bromine but alkenes are much more reactive.
2 Answer all the questions. 1 Alkenes and benzene both react with bromine but alkenes are much more reactive. (a) Explain the relative resistance to bromination of benzene compared with alkenes. In your
More informationSPECIMEN MATERIAL v1.0. A-LEVEL Chemistry. Paper 2: Organic and Physical Chemistry Mark scheme. 7405/2 Specimen Paper (set 2) Version 1.
SPECIMEN MATERIAL v.0 A-LEVEL Chemistry Paper 2: Organic and Physical Chemistry Mark scheme 7405/2 Specimen Paper (set 2) Version.0 Mark schemes are prepared by the Lead Assessment Writer and considered,
More informationOrganic Chemistry SL IB CHEMISTRY SL
Organic Chemistry SL IB CHEMISTRY SL 10.1 Fundamentals of organic chemistry Understandings: A homologous series is a series of compounds of the same family, with the same general formula, which differ
More informationAnswer Marks Guidance
Question number (a) molecular formula: C 4 H 8 Answer Marks Guidance empirical formula: CH 2 This is a revision of earlier chapters. (b) (i) name of mechanism: electrophilic addition Remember that reactions
More informationName/CG: 2012 Term 2 Organic Chemistry Revision (Session II) Deductive Question
Name/G: 2012 Term 2 rganic hemistry Revision (Session II) Deductive Question 1(a) A yellow liquid A, 7 7 N 2, reacts with alkaline potassium manganate (VII) and on acidification gives a yellow solid B,
More informationAcceptable Answers Reject Mark. Acceptable Answers Reject Mark. ALLOW Iron/Fe or Zn/Zinc for tin Conc for concentrated. Acceptable Answers Reject Mark
(a)(i) (a)(ii) Concentrated nitric acid AND concentrated sulfuric acid concentrated nitric and sulfuric acids Concentrated HNO and concentrated H SO 4 Extra reagents To prevent multiple substitutions/
More informationPage 2. (1)-methylethyl ethanoate OR. Propan-2-yl ethanoate Ignore extra or missing spaces, commas or hyphens 1. (ii) M4 for 3 arrows and lp
M.(a) (i) (CH 3 ) 2 CHOH + (CH 3 CO) 2 O CH 3 COOCH(CH 3 ) 2 + CH 3 COOH Allow CH 3 CO 2 CH(CH 3 ) 2 and CH 3 CO 2 H Ignore (CH 3 ) 2 C in equation ()-methylethyl ethanoate OR Propan-2-yl ethanoate Ignore
More informationGCE A level 1094/01 CHEMISTRY CH4
Surname ther Names Centre 2 Candidate GCE A level 1094/01 CHEMISTRY CH4 P.M. MNDAY, 14 January 2013 1¾ hours ADDITINAL MATERIALS In addition to this examination paper, you will need: Data Sheet Periodic
More informationAmides, Amino acids and Chirality
R hemistry A 432 Amides, Amino Acids & hirality Amides, Amino acids and hirality aming of Amides The amide functional group consists of a carbonyl group bonded to the nitrogen of an amine. Like amines,
More information2 Answer all the questions. OH(g) ΔH = 91 kj mol 1 equation 1.1. (a) A pressure of between 50 and 100 atmospheres is used for this reaction.
2 Answer all the questions. 1 Methanol is added to ethanol to make the ethanol unfit to drink. Methanol can be made by the following reaction. CO(g) + 2H 2 (g) CH 3 OH(g) ΔH = 91 kj mol 1 equation 1.1
More informationCOCH 3. + HCl H 5 C 6. Not molecular formulae Not allow C. phenylethanone Ignore number 1 in name but penalise other numbers
Not molecular formulae Not allow C 6 H 5 CH CO M. (a) (i) CH COCl + C 6 H 6 C 6 H 5 COCH + HCl OR phenylethanone Ignore number in name but penalise other numbers AlCl can be scored in equation Allow RHS
More informationSection A. 1 at a given temperature. The rate was found to be first order with respect to the ester and first order with respect to hydroxide ions.
2 Section A Answer all questions in the spaces provided. Question 1:The N/Arate of hydrolysis of an ester X (HCOOCH2CH2CH3) was studied in alkaline 1 at a given temperature. The rate was found to be first
More informationCHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME. ±½ a square
Chemistry Advanced Level Paper 3 (9CH0/03) 1(a)(i) suitable scale and axes labelled including units (1) all points plotted correctly (1) line of best fit (1) Plotted points use at least half the available
More informationCandidate number. Centre number
Oxford Cambridge and RSA A Level Chemistry A H432/02 Synthesis and analytical techniques Practice paper Set 2 Time allowed: 2 hours 15 minutes *PRACTICE2016* You must have: the Data Sheet for Chemistry
More informationExam 1 (Monday, July 6, 2015)
Chem 231 Summer 2015 Assigned Homework Problems Last updated: Friday, July 24, 2015 Problems Assigned from Essential Organic Chemistry, 2 nd Edition, Paula Yurkanis Bruice, Prentice Hall, New York, NY,
More informationMorning. This document consists of 15 printed pages, 1 blank page and a Data Sheet for Chemistry.
ADVANED GE 2814/01 EMISTRY hains, Rings and Spectroscopy TUESDAY 23 JANUARY 2007 Morning Additional materials: Scientific calculator Data Sheet for hemistry (Inserted) Time: 1 hour 30 minutes INSTRUTINS
More informationExperiment 8 Optical Isomers. In this experiment you will be given the opportunity to see the 3-dimensional aspects of
Experiment 8 Optical Isomers In this experiment you will be given the opportunity to see the 3-dimensional aspects of stereochemistry and optical isomers. Previously in class you were exposed to the concept
More information2 Answer all the questions. 1 Nitrogen monoxide is formed when nitrogen and oxygen from the air combine. (g) + O 2
2 Answer all the questions. 1 Nitrogen monoxide is formed when nitrogen and oxygen from the air combine. N 2 (g) + 2 (g) 2N(g) equation 1.1 Under normal atmospheric conditions, a further reaction occurs
More informationA Level Chemistry A H432/02 Synthesis and analytical techniques. Practice paper Set 1 Time allowed: 2 hours 15 minutes
A Level hemistry A 432/02 Synthesis and analytical techniques Practice paper Set 1 Time allowed: 2 hours 15 minutes You must have: the Data Sheet for hemistry A You may use: a scientific calculator a ruler
More informationChem 14C Lecture 1 Spring 2017 Final Exam Part B Page 1
Chem 14C Lecture 1 Spring 2017 Final Exam Part B Page 1 1. (2) Write the letter of the structure that best fits the following 13 C-NMR spectrum: 51 ppm (doublet), 44 ppm (triplet), 27 ppm (quartet), 26
More information2.8 EXTRA QUESTIONS MS. (iii) C 6 H 5 CHCl 2 / C 6 H 5 CCl 3 / C 6 H 5 CHCHC 6 H 5 / other correct possible answer (1) 1
.8 EXTRA QUESTIONS MS 1. (i) l l / ½ l l 1 (ii) l 6 5 6 5 l 6 5 l l 6 5 l (iii) 6 5 l / 6 5 l / 6 5 6 5 / other correct possible answer 1 [4]. (a) (i) (free)radical substitution 1 (both words required
More informationCHEM1102 Worksheet 4 Answers to Critical Thinking Questions Model 1: Infrared (IR) Spectroscopy
CEM1102 Worksheet 4 Answers to Critical Thinking Questions Model 1: Infrared (IR) Spectroscopy 1. See below. Model 2: UV-Visible Spectroscopy 1. See below. 2. All of the above. 3. Restricted to the identification
More information3 Answer all the questions.
CHERRY HILL TUITIN CR A CHEMISTRY A2 PAPER 18 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
More informationabc Mark Scheme Chemistry 5421 General Certificate of Education 2005 examination - June series CHM4 Further Physical and Organic Chemistry
Version.0 General Certificate of Education abc Chemistry 542 CM4 Further Physical and Organic Chemistry Mark Scheme 2005 examination - June series Mark schemes are prepared by the Principal Examiner and
More information(a) Name the alcohol and catalyst which would be used to make X. (2)
1 The chemical X is an ester with formula CH 3 COOC(CH 3 ) 3 which occurs in raspberries and pears. It can be prepared in the laboratory by refluxing ethanoic acid with an alcohol in the presence of a
More information* * Cambridge International Examinations Cambridge International Advanced Level
*3272601861* Cambridge International Examinations Cambridge International Advanced Level CHEMISTRY 9701/42 Paper 4 Structured Questions ctober/november 2014 2 hours Candidates answer on the Question Paper.
More informationThe mechanism of the nitration of methylbenzene is an electrophilic substitution.
Q1.Many aromatic nitro compounds are used as explosives. One of the most famous is 2-methyl-1,3,5-trinitrobenzene, originally called trinitrotoluene or TNT. This compound, shown below, can be prepared
More informationMonday 19 June 2017 Morning Time allowed: 2 hours 15 minutes
Oxford ambridge and RSA A Level hemistry A H432/02 Synthesis and analytical techniques Monday 19 June 2017 Morning Time allowed: 2 hours 15 minutes *6826116453* You must have: the Data Sheet for hemistry
More information4. NMR spectra. Interpreting NMR spectra. Low-resolution NMR spectra. There are two kinds: Low-resolution NMR spectra. High-resolution NMR spectra
1 Interpreting NMR spectra There are two kinds: Low-resolution NMR spectra High-resolution NMR spectra In both cases the horizontal scale is labelled in terms of chemical shift, δ, and increases from right
More informationIGNORE Just benzene has a delocalised ring Benzene does not have C=C double bonds Any references to shape/ bond angles. Acceptable Answers Reject Mark
1(a) All carbon to carbon bonds same length/ longer C-C and shorter C=C not present 1 IGNORE Just benzene has a delocalised ring Benzene does not have C=C double bonds Any references to shape/ bond angles
More informationAfternoon Time: 1 hour 30 minutes
ADVANED GE 2814/01 EMISTRY hains, Rings and Spectroscopy TURSDAY 12 JUNE 2008 Afternoon Time: 1 hour 30 minutes *UP/T68366* andidates answer on the question paper Additional materials (enclosed): Data
More informationName this organic reagent and state the conditions for the preparation. Reagent... Conditions (3)
1. (a) Propanoic acid, C 3 C 2 COO, can be prepared from carbon dioxide and an organic reagent. Name this organic reagent and state the conditions for the preparation. Reagent... Conditions...... (3) (b)
More informationTuesday 19 June 2012 Afternoon
Tuesday 19 June 2012 Afternoon A2 GCE CEMISTRY A F324 Rings, Polymers and Analysis *F314730612* Candidates answer on the Question Paper. CR supplied materials: Data Sheet for Chemistry A (inserted) ther
More informationCHEM J-8 June Complete the following table. Make sure you give the name of the starting material where indicated. REAGENTS/ CONDITIONS
CEM1102 2014-J-8 June 2014 Complete the following table. Make sure you give the name of the starting material where indicated. STARTIG MATERIAL REAGETS/ CDITIS STRUCTURAL FRMULA(S) F MAJR RGAIC PRDUCT(S)
More informationCHEM 203. Midterm Exam 1 October 31, 2008 ANSWERS. This a closed-notes, closed-book exam. You may use your set of molecular models
CEM 203 Midterm Exam 1 ctober 31, 2008 Your name: ANSWERS This a closed-notes, closed-book exam You may use your set of molecular models This exam contains 8 pages Time: 1h 30 min 1. / 15 2. / 16 3. /
More informationTHE UNIVERSITY OF CALGARY FACULTY OF SCIENCE MIDTERM EXAMINATION CHEMISTRY 353
12 TE UNIVERSITY F CALGARY FACULTY F SCIENCE MIDTERM EXAMINATIN CEMISTRY 353 FEBRUARY 28th, 2001 Time: 2 ours PLEASE WRITE YUR NAME, STUDENT I.D. NUMBER AND SECTIN NUMBER (01 for MWF lectures and 02 for
More informationklm Mark Scheme Chemistry 2421 General Certificate of Education Kinetics, Equilibria and Organic Chemistry 2010 examination - January series
Version.0: 02/200 klm General ertificate of Education hemistry 242 EM4 Kinetics, Equilibria and rganic hemistry Scheme 200 examination - January series schemes are prepared by the Principal Examiner and
More informationUnit 2. Answers to examination-style questions. Answers Marks Examiner s tips. 1 (a) heat energy change at constant pressure
(a) heat energy change at constant pressure This is in the spec but not so well known. Learn it. (b) N 2 (g) + ½O 2 (g) N 2 O(g) (c) (i) D = (bonds broken) (bonds made) = ½(945) + (3/2)(59) 3(278) = 23
More informationOrganic and Biochemical Molecules. 1. Compounds composed of carbon and hydrogen are called hydrocarbons.
Organic and Biochemical Molecules 1. Compounds composed of carbon and hydrogen are called hydrocarbons. 2. A compound is said to be saturated if it contains only singly bonded carbons. Such hydrocarbons
More informationF322: Chains, Energy and Resources Basic Concepts
F322: hains, Energy and Resources Basic oncepts 1. Some of the hydrocarbons in kerosene have the formula 10 22. (i) What is the name of the straight chain hydrocarbon with the formula 10 22? (ii) Draw
More informationThis document consists of 16 printed pages and a Data Sheet for Chemistry.
ADVANED GE 2814/01 EMISTRY hains, Rings and Spectroscopy MNDAY 18 JUNE 2007 Afternoon *UP/T24087* Additional materials: Scientific calculator Data Sheet for hemistry (Inserted) Time: 1 hour 30 minutes
More informationGCE Chemistry A. Mark Scheme for June Unit F324: Rings, Polymers and Analysis. Advanced GCE. Oxford Cambridge and RSA Examinations
GCE Chemistry A Unit F324: Rings, Polymers and Analysis Advanced GCE Mark Scheme for June 2016 Oxford Cambridge and RSA Examinations OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing
More informationGCE Chemistry A. Mark Scheme for June Unit H432A/02: Synthesis and analytical techniques. Advanced GCE. Oxford Cambridge and RSA Examinations
GE hemistry A Unit 432A/02: Synthesis and analytical techniques Advanced GE Mark Scheme for June 2017 Oxford ambridge and RSA Examinations 432A/02 Mark Scheme OR (Oxford ambridge and RSA) is a leading
More informationCHEMISTRY (SALTERS) Chemistry of Materials. OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced GCE
XFRD CAMBRIDGE AND RSA EXAMINATINS Advanced GCE CHEMISTRY (SALTERS) Chemistry of Materials 2849 Monday 19 JUNE 2006 Afternoon 1 hour 30 minutes Candidates answer on the question paper. Additional materials:
More informationAQA A2 CHEMISTRY TOPIC 4.10 ORGANIC SYNTHESIS AND ANALYSIS TOPIC 4.11 STRUCTURE DETERMINATION BOOKLET OF PAST EXAMINATION QUESTIONS
AQA A2 CHEMISTRY TOPIC 4.10 ORGANIC SYNTHESIS AND ANALYSIS TOPIC 4.11 STRUCTURE DETERMINATION BOOKLET OF PAST EXAMINATION QUESTIONS 1 1. Consider the following reaction sequence. CH 3 CH 3 CH 3 Step 1
More informationCalculate a rate given a species concentration change.
Kinetics Define a rate for a given process. Change in concentration of a reagent with time. A rate is always positive, and is usually referred to with only magnitude (i.e. no sign) Reaction rates can be
More informationTuesday 14 June 2016 Afternoon
xford ambridge and RSA Tuesday 14 June 2016 Afternoon A2 GE HEMISTRY A F324/01 Rings, Polymers and Analysis *5878609421* andidates answer on the Question Paper. R supplied materials: Data Sheet for hemistry
More informationPearson Edexcel Level 3 GCE Chemistry Advanced Paper 2: Advanced Organic and Physical Chemistry
Write your name here Surname Other names Pearson Edexcel Level 3 GCE Centre Number Candidate Number Chemistry Advanced Paper 2: Advanced Organic and Physical Chemistry Sample Assessment Materials for first
More informationCHEM4 Kinetics, Equilibria and Organic Chemistry Mark scheme
AQA Qualifications A-LEVEL EMISTRY EM4 Kinetics, Equilibria and rganic hemistry Mark scheme 2420 June 204 Version:. Final Mark schemes are prepared by the Lead Assessment Writer and considered, together
More informationQuestion Answer Marks Guidance 1 (a) (+)5 1 ALLOW 5+ OR V OR Cr 5+ 1 (b) For equations, IGNORE any state symbols; ALLOW multiples
Question Answer Marks Guidance 1 (a) (+)5 1 ALLW 5+ R V R Cr 5+ 1 (b) For equations, IGNRE any state symbols; ALLW multiples EXAMPLES ------------------------------------------------------------------------------------
More informationQ1. The following pairs of compounds can be distinguished by simple test tube reactions.
Q1. The following pairs of compounds can be distinguished by simple test tube reactions. For each pair of compounds, give a reagent (or combination of reagents) that, when added separately to each compound,
More information