CHERRY HILL TUITION OCR A CHEMISTRY A2 PAPER 19 MARK SCHEME ANNOTATIONS MUST BE USED CH 3 CH 3 CH 3 H + correct products

Similar documents
1 (a) (CH 3 CO) 2 O + CH 3 CH(OH)CH 3 CH 3 COOCH(CH 3 ) 2 + CH 3 COOH

ALLOW CO 2 and CO 2 H CH 3

ANNOTATIONS MUST BE USED

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

Question Answer Marks Guidance 1 (a) M1 EITHER in words: (pyruvic acid forms) hydrogen bonds with water

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

Question Answer Mark Guidance 1 (a) monomers join/bond/add/react/form polymer/form chain AND another product/small molecule e.g.

F324 Mark Scheme January 2010 F324 Rings, Polymers and Analysis. Question Expected Answers Marks Additional Guidance 1 (a)

Where circles have been placed round charges, this is for clarity only and does not indicate a requirement. ALLOW delocalised carboxylate.

2 Answer all the questions.

Allow CONH- or - COHN - 1(a)(i) Mark two halves separately

Nitrogen Compounds - MS 1. (a) (i) is an amine and a carboxylic acid / contains both NH2 and COOH functional groups (1) AW 1

Question Answer Mark Guidance 1 (a) (i) M1 p-orbitals overlap (to form pi/π-bonds) 4 ANNOTATE ANSWER WITH TICKS AND CROSSES ETC

GCE. Chemistry. Mark Scheme for June Advanced GCE 2814/01 Chains, Rings and Spectroscopy. Oxford Cambridge and RSA Examinations

PMT GCE. Chemistry A. Advanced GCE Unit F324: Rings, Polymers and Analysis. Mark Scheme for January 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

Practice paper Set 1 MAXIMUM MARK 100. Final. H432/02 Mark Scheme Practice 1. A Level Chemistry A H432/02 Synthesis and analytical techniques

4 ALLOW sugar from equation

Acceptable sequence of stages are: chlorination. nitration, reduction, chlorination nitration. nitration, chlorination, reduction, reduction

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

DO NOT ALLOW any reference to spatial/space

flowers, leaves and roots of roses rose oil heat

Tuesday 19 June 2012 Afternoon

CHERRY HILL TUITION OCR (SALTERS) CHEMISTRY A2 PAPER Answer all the questions. O, is formed in the soil by denitrifying bacteria. ...

Advanced GCE Chemistry A

8 (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

1 hour 45 minutes plus your additional time allowance

The amide or peptide link is found in synthetic polyamides and also in naturally-occurring proteins.

THIS IS A NEW SPECIFICATION

Final. Mark Scheme. Chemistry CHEM4. (Specification 2420) Unit 4: Kinetics, Equilibria and Organic Chemistry

Question Answer Mark Guidance 1 (a) (i) 4 Please use annotations (rate) molecules/particles in smaller volume OR increases concentration

Subject: Chains, Rings and Spectroscopy Code: Session: June Year: Final Mark Scheme


PMT GCE. Chemistry A. Advanced GCE F324. Mark Scheme for June Oxford Cambridge and RSA Examinations

A mass spectrometer can be used to distinguish between samples of butane and propanal. The table shows some precise relative atomic mass values.

UNIT 4 REVISION CHECKLIST CHEM 4 AS Chemistry

Candidates answer on the question paper. Time: 1 hour 15 minutes Additional Materials: Data Sheet for Chemistry (Inserted) Scientific calculator

(07) 3 (e) Calculate the ph of this buffer solution at 298 K. Give your answer to 2 decimal places

F324: Rings, Polymers and Analysis Carbonyl Compounds

CHEMISTRY 2814/01 Chains, Rings and Spectroscopy

Surname. Number OXFORD CAMBRIDGE AND RSA EXAMINATIONS ADVANCED GCE F324 CHEMISTRY A. Rings, Polymers and Analysis

Question Answer Mark Guidance 1 (a)

Thursday 26 January 2012 Afternoon

AS Demonstrate understanding of the properties of selected organic compounds

2 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.

, by reacting CH 3 with ethanoic anhydride, (CH 3

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

Alcohols. Nomenclature. 57 minutes. 57 marks. Page 1 of 9

3.4 A2 Unit F324: Rings, Polymers and Analysis

2 Answer all the questions. 1 Alkenes and benzene both react with bromine but alkenes are much more reactive.

SPECIMEN MATERIAL v1.0. A-LEVEL Chemistry. Paper 2: Organic and Physical Chemistry Mark scheme. 7405/2 Specimen Paper (set 2) Version 1.

Organic Chemistry SL IB CHEMISTRY SL

Answer Marks Guidance

Name/CG: 2012 Term 2 Organic Chemistry Revision (Session II) Deductive Question

Acceptable Answers Reject Mark. Acceptable Answers Reject Mark. ALLOW Iron/Fe or Zn/Zinc for tin Conc for concentrated. Acceptable Answers Reject Mark

Page 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

GCE A level 1094/01 CHEMISTRY CH4

Amides, Amino acids and Chirality

2 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.

COCH 3. + HCl H 5 C 6. Not molecular formulae Not allow C. phenylethanone Ignore number 1 in name but penalise other numbers

Section 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.

CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME. ±½ a square

Candidate number. Centre number

Exam 1 (Monday, July 6, 2015)

Morning. This document consists of 15 printed pages, 1 blank page and a Data Sheet for Chemistry.

Experiment 8 Optical Isomers. In this experiment you will be given the opportunity to see the 3-dimensional aspects of

2 Answer all the questions. 1 Nitrogen monoxide is formed when nitrogen and oxygen from the air combine. (g) + O 2

A Level Chemistry A H432/02 Synthesis and analytical techniques. Practice paper Set 1 Time allowed: 2 hours 15 minutes

Chem 14C Lecture 1 Spring 2017 Final Exam Part B Page 1

2.8 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

CHEM1102 Worksheet 4 Answers to Critical Thinking Questions Model 1: Infrared (IR) Spectroscopy

3 Answer all the questions.

abc Mark Scheme Chemistry 5421 General Certificate of Education 2005 examination - June series CHM4 Further Physical and Organic Chemistry

(a) Name the alcohol and catalyst which would be used to make X. (2)

* * Cambridge International Examinations Cambridge International Advanced Level

The mechanism of the nitration of methylbenzene is an electrophilic substitution.

Monday 19 June 2017 Morning Time allowed: 2 hours 15 minutes

4. NMR spectra. Interpreting NMR spectra. Low-resolution NMR spectra. There are two kinds: Low-resolution NMR spectra. High-resolution NMR spectra

IGNORE Just benzene has a delocalised ring Benzene does not have C=C double bonds Any references to shape/ bond angles. Acceptable Answers Reject Mark

Afternoon Time: 1 hour 30 minutes

Name this organic reagent and state the conditions for the preparation. Reagent... Conditions (3)

Tuesday 19 June 2012 Afternoon

CHEM J-8 June Complete the following table. Make sure you give the name of the starting material where indicated. REAGENTS/ CONDITIONS

CHEM 203. Midterm Exam 1 October 31, 2008 ANSWERS. This a closed-notes, closed-book exam. You may use your set of molecular models

THE UNIVERSITY OF CALGARY FACULTY OF SCIENCE MIDTERM EXAMINATION CHEMISTRY 353

klm Mark Scheme Chemistry 2421 General Certificate of Education Kinetics, Equilibria and Organic Chemistry 2010 examination - January series

Unit 2. Answers to examination-style questions. Answers Marks Examiner s tips. 1 (a) heat energy change at constant pressure

Organic and Biochemical Molecules. 1. Compounds composed of carbon and hydrogen are called hydrocarbons.

F322: Chains, Energy and Resources Basic Concepts

This document consists of 16 printed pages and a Data Sheet for Chemistry.

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

GCE Chemistry A. Mark Scheme for June Unit H432A/02: Synthesis and analytical techniques. Advanced GCE. Oxford Cambridge and RSA Examinations

CHEMISTRY (SALTERS) Chemistry of Materials. OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced GCE

AQA A2 CHEMISTRY TOPIC 4.10 ORGANIC SYNTHESIS AND ANALYSIS TOPIC 4.11 STRUCTURE DETERMINATION BOOKLET OF PAST EXAMINATION QUESTIONS

Calculate a rate given a species concentration change.

Tuesday 14 June 2016 Afternoon

Pearson Edexcel Level 3 GCE Chemistry Advanced Paper 2: Advanced Organic and Physical Chemistry

CHEM4 Kinetics, Equilibria and Organic Chemistry Mark scheme

Question Answer Marks Guidance 1 (a) (+)5 1 ALLOW 5+ OR V OR Cr 5+ 1 (b) For equations, IGNORE any state symbols; ALLOW multiples

Q1. The following pairs of compounds can be distinguished by simple test tube reactions.

Transcription:

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

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 2 2 1 (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

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

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: 1.2.3 R spaces: 1 2 3 D NT ALLW 123 2 (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

ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME 2 (c) 2 + 2 2 2 + 2 ALLW correct structural R displayed R skeletal formula ALLW combination of formulae as long as unambiguous D NT ALLW molecular formulae ( 2 ) 2 + 2 2 2 + 2 2 ALLW further esterification, ie ( 2 ) 2 + 2 2 2 2 2 + 2 ALLW linear formula for anhydride, ie 2 2 If incorrect carboxylic acid/anhydride/alcohol is used, ALLW EF for second equation 5

ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME 2 (d) A B Mark A, B and independently ie 2 column 2 2 2 2 2 2 2 2 column R R R 2 2 3 R R R 2 2 5 5 2 5 2 R R R 3 2 2 2 R R R 3 3 3 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

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

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

ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME 4 (a) (i) l( ) + 3N 3 2 N( ) + N + 4 + N 4 l 1 ALLW use of two N 3 : l( ) + 2N 3 2 N( ) + N + 4 + 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 4 + + 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

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

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) 0.2 1 ALLW any value in the range 0.1 0.3 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

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 = 0.25 60/100 R 0.15 (mol) 2 orrectly calculates mass of menthol = 0.15 156 = 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 = 34 156/136 = 39 (g) 156 ; 136 60% 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

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 3.3 4.2 AND a triplet at 3.3 4.2 quintet/pentet/multiplet at 0.7 2.0 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

ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME F324 Mark Scheme January 2011 6 (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 6 4 + / 6 5 2 + ALLW molecular mass R relative molecular mass ALLW 6 4 / 6 5 2 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 ~125 140 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

ERRY ILL TUITIN R A EMISTRY A2 PAPER 19 MARK SEME Mark Scheme January 2011 6 (b) Number of peaks for other three isomers matched to structures: Any 2 correct for 2 marks 1 correct for 1 mark 2 5 4 peaks 3 peaks 6 peaks ALLW carbon environments for peaks orrect structure shown: 6 Total 9 15 15