CHEMISTRY 2814/01 Chains, Rings and Spectroscopy

Similar documents
Afternoon Time: 1 hour 30 minutes

CHEMISTRY 2812/01 Chains and Rings

THIS IS A NEW SPECIFICATION

CHEMISTRY 2814/01 Chains, Rings and Spectroscopy

Thursday 26 January 2012 Afternoon

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

Thursday 19 January 2012 Afternoon

Tuesday 19 June 2012 Afternoon

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

Tuesday 14 June 2016 Afternoon

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

Wednesday 23 May 2012 Afternoon

CHEMISTRY. Chains, Rings and Spectroscopy. OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced GCE. 1 hour 30 minutes

Wednesday 19 June 2013 Morning

F334. CHEMISTRY B (SALTERS) Chemistry of Materials ADVANCED GCE. Friday 24 June 2011 Morning. Duration: 1 hour 30 minutes

Thursday 10 January 2013 Morning

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

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

Tuesday 2 June 2015 Afternoon

Friday 26 May 2017 Morning

F324. CHEMISTRY A Rings, Polymers and Analysis ADVANCED GCE. Friday 24 June 2011 Morning PMT. Duration: 1 hour

Wednesday 16 January 2013 Morning

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

Cherry Hill Tuition A Level Chemistry OCR (A) Paper 12

Thursday 10 January 2013 Morning

THIS IS A NEW SPECIFICATION

CHEMISTRY 2815/01 Trends and Patterns

Friday 27 May 2016 Morning

Thursday 23 May 2013 Morning

surname number OXFORD CAMBRIDGE AND RSA EXAMINATIONS ADVANCED GCE F324 CHEMISTRY A Rings, Polymers and Analysis

Cherry Hill Tuition A Level Chemistry OCR (A) Paper 10 THIS IS A NEW SPECIFICATION

Tuesday 13 June 2017 Afternoon

Cherry Hill Tuition A Level Chemistry OCR (A) Paper 1 THIS IS A NEW SPECIFICATION

F321. CHEMISTRY A Atoms, Bonds and Groups ADVANCED SUBSIDIARY GCE. Thursday 13 January 2011 Morning PMT. Duration: 1 hour

F331. CHEMISTRY B (SALTERS) Chemistry for Life ADVANCED SUBSIDIARY GCE. Monday 23 May 2011 Afternoon. Duration: 1 hour 15 minutes

Surname. Number OXFORD CAMBRIDGE AND RSA EXAMINATIONS ADVANCED SUBSIDIARY GCE F332/TEST CHEMISTRY B (SALTERS) Chemistry of Natural Resources

* * MATHEMATICS 4721 Core Mathematics 1 ADVANCED SUBSIDIARY GCE. Monday 11 January 2010 Morning QUESTION PAPER. Duration: 1 hour 30 minutes.

THIS IS A NEW SPECIFICATION

Tuesday 10 June 2014 Morning

Candidate number. Centre number

F331. CHEMISTRY B (SALTERS) Chemistry for Life ADVANCED SUBSIDIARY GCE. Thursday 13 January 2011 Morning. Duration: 1 hour 15 minutes

Monday 14 January 2013 Afternoon

CHEMISTRY (SALTERS) 2849/01 Chemistry of Materials

CHEMISTRY (SALTERS) 2848/01 Chemistry of Natural Resources

CHEMISTRY 2816/01 Unifying Concepts in Chemistry

THIS IS A NEW SPECIFICATION

THIS IS A NEW SPECIFICATION

Surname. Number OXFORD CAMBRIDGE AND RSA EXAMINATIONS ADVANCED SUBSIDIARY GCE F321 CHEMISTRY A. Atoms, Bonds and Groups

CHEMISTRY Chains and Rings. OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary GCE

Friday 23 June 2017 Morning

* * ADDITIONAL MATHEMATICS 6993 FREE-STANDING MATHEMATICS QUALIFICATION ADVANCED LEVEL. Friday 5 June 2009 Afternoon. Duration: 2 hours.

Tuesday 4 June 2013 Afternoon

surname number OXFORD CAMBRIDGE AND RSA EXAMINATIONS ADVANCED SUBSIDIARY GCE F322 CHEMISTRY A Chains, Energy and Resources

Friday 10 June 2016 Afternoon Time allowed: 1 hour 30 minutes

Candidate number. Centre number

B278A MATHEMATICS C (GRADUATED ASSESSMENT) MODULE M8 SECTION A GENERAL CERTIFICATE OF SECONDARY EDUCATION. Monday 8 March 2010 Morning WARNING

Tuesday 14 June 2016 Afternoon

Surname. Number OXFORD CAMBRIDGE AND RSA EXAMINATIONS ADVANCED SUBSIDIARY GCE F322 CHEMISTRY A. Chains, Energy and Resources

Monday 19 June 2017 Morning

Friday 10 June 2016 Afternoon

G484. PHYSICS A The Newtonian World ADVANCED GCE. Thursday 27 January 2011 Afternoon. Duration: 1 hour

CHEMISTRY Chains, Rings and Spectroscopy. OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced GCE. 1 hour 30 minutes

THIS IS A NEW SPECIFICATION

A Level Chemistry B (Salters) H433/02 Scientific literacy in chemistry. Practice paper Set 1 Time allowed: 2 hours 15 minutes

THIS IS A NEW SPECIFICATION

Monday 18 June 2012 Morning

Thursday 16 June 2016 Morning

Tuesday 4 June 2013 Afternoon

CHEMISTRY (SALTERS) 2849/01 Chemistry of Materials

MATHEMATICS 4725 Further Pure Mathematics 1

Monday 19 June 2017 Morning

Tuesday 22 January 2013 Afternoon

F322: Chains, Energy and Resources Alkenes

Wednesday 16 January 2013 Morning

1 hour 45 minutes plus your additional time allowance

G484. PHYSICS A The Newtonian World ADVANCED GCE. Monday 27 June 2011 Morning. Duration: 1 hour

Wednesday 16 January 2013 Morning

CHEMISTRY 2815/06 Transition Elements

This document consists of 12 printed pages and a Data Sheet for Chemistry.

MATHEMATICS 4722 Core Mathematics 2

THIS IS A NEW SPECIFICATION MODIFIED LANGUAGE

SPECIMEN. Date Morning/Afternoon Time allowed: 1 hour 30 minutes. AS Level Chemistry B (Salters) H033/02 Chemistry in depth Sample Question Paper

MATHEMATICS 4728 Mechanics 1

ADVANCED GCE 2854/01 CHEMISTRY (SALTERS) Chemistry by Design THURSDAY 19 JUNE 2008 Time: 2 hours *CUP/T42936*

CHEMISTRY 2811/01 Foundation Chemistry

2 Answer all the questions.

2 Answer all the questions. 1 PLA is a biodegradable polymer made from corn starch. The structure of part of a PLA chain is shown below.

Surname. Number OXFORD CAMBRIDGE AND RSA EXAMINATIONS ADVANCED SUBSIDIARY GCE F331 CHEMISTRY B (SALTERS) Chemistry for Life

F332/TEST. CHEMISTRY B (SALTERS) Chemistry of Natural Resources ADVANCED SUBSIDIARY GCE. Friday 27 May 2011 Afternoon. Duration: 1 hour 45 minutes

Candidate number. Centre number

Wednesday 21 May 2014 Afternoon

CHEMISTRY (SALTERS) 2853

Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level

Wednesday 13 June 2012 Morning

CHEMISTRY Chains and Rings. OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary GCE

Thursday 19 January 2012 Afternoon

MATHEMATICS 4723 Core Mathematics 3

* * MATHEMATICS (MEI) 4757 Further Applications of Advanced Mathematics (FP3) ADVANCED GCE. Wednesday 9 June 2010 Afternoon PMT

Monday 20 June 2016 Morning

Wednesday 8 June 2016 Afternoon

Transcription:

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: Data Sheet for hemistry (inserted) Other Materials Required: Scientific calculator Monday 1 February 2010 Morning Duration: 1 hour 30 minutes * 2 8 1 4 0 1 * INSTRUTIONS TO 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, however additional Examiner s Use Only: paper may be used if necessary. 1 INFORMATION FOR ANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 90. You will be awarded marks for the quality of written communication where this is indicated in the question. You may use a scientific calculator. A copy of the Data Sheet for hemistry is provided as an insert with this question paper. You are advised to show all the steps in any calculations. This document consists of 16 pages. Any blank pages are indicated. 2 3 4 5 6 7 Total [F/100/3426] D (LEO/SW) 19368/2 OR is an exempt harity Turn over

2 Answer all the questions. 1 4-Methylphenylamine is manufactured from benzene in three stages, as shown below. 6 6 stage 1 stage 2 stage 3 3 6 5 3 6 4 NO 2 benzene methylbenzene 4-methylnitrobenzene 4-methylphenylamine (a) In the box above draw the displayed formula of 4-methylphenylamine. [1] (b) In this question, one mark is available for the quality of use and organisation of scientific terms. State the type of reaction, the reagents and conditions and write a balanced equation for stage 2 and stage 3. Describe, with the aid of curly arrows, the mechanism for stage 2. Include an equation for the formation of the electrophile.

3... [11] Quality of Written ommunication [1] (c) 4-Methylphenylamine can be converted into the azo dye, shown below, via a diazonium ion. O N 3 N (i) Draw the structure of the diazonium ion. [1] (ii) Deduce the number of carbon and the number of hydrogen atoms in the azo dye. number of atoms... number of atoms...... [2] [Total: 16] Turn over

2 Glutamic acid and glycine are both -amino acids that occur widely in living organisms. The structures of the amino acids are shown below. 4 2 N OO 2 N OO ( 2 ) 2 OO glutamic acid glycine (a) State the general formula of an -amino acid... [1] (b) Glutamic acid has optical isomers but glycine does not. Explain why. Include suitable 3-D diagrams in your answer. (c) Amino acids form different ions at different p values. (i) Draw the structure of the ion formed by glycine at p 2. [4] [1]

(ii) Draw the structure of the ion formed by glutamic acid at p 14. 5 [2] (d) Glutamic acid and glycine can react to form a mixture of dipeptides. Draw two dipeptides formed when one molecule of glutamic acid reacts with one molecule of glycine. (e) In the presence of a suitable catalyst, glutamic acid is esterified by an excess of ethanol to form a compound with molecular formula 9 17 NO 4. [3] (i) Identify a suitable catalyst for this reaction.... [1] (ii) Draw the structure of the compound with molecular formula 9 17 NO 4. [2] [Total: 14] Turn over

3 Ecoflex is a biodegradeable plastic made by combining three different monomers. A short section of the Ecoflex polymer is shown below. 6 O O O O O ( 2 ) 4 O ( 2 ) 4 O ( 2 ) 4 O (a) Draw a circle around an ester link in the section of Ecoflex shown above. [1] (b) One of the monomers used to make Ecoflex is OO( 2 ) 4 OO. (i) Give the systematic name of OO( 2 ) 4 OO.... [1] (ii) Draw the structures of the other two monomers used to make Ecoflex. (c) The monomer OO( 2 ) 4 OO can be made in the laboratory by oxidising compound X, 6 10 O 2. (i) Draw the displayed formula and the skeletal formula of compound X. [2] [2]

(ii) 7 Write a balanced equation for the oxidation of compound X to form OO( 2 ) 4 OO. Use [O] to represent the oxidising agent.... [1] (iii) Explain how infra-red spectra of compound X and OO( 2 ) 4 OO could be used to confirm that the oxidation had taken place.... [1] (d) Poly(but-1-ene) can be made from the monomer but-1-ene, 3 2 = 2. Draw a section of poly(but-1-ene) to show two repeat units. (e) Ecoflex and poly(but-1-ene) are both polymers but the type of polymerisation reaction used to make each one is different. State the type of polymerisation reaction used to make each polymer. ecoflex... poly(but-1-ene)... [1] [1] (f) Describe how the structure of atactic poly(but-1-ene) differs from isotactic poly(but-1-ene)... [2] [Total: 12] Turn over

8 4 In 1893, a chemist discovered an organic acid which became known as Feist s acid, 6 6 O 4. (a) 2.50 10 3 mol of Feist s acid was completely burnt. alculate the volume of carbon dioxide and mass of water that would be formed at r.t.p. 1 mol of a gas occupies 24 dm 3 at r.t.p. volume of carbon dioxide =... cm 3 (b) Two possible structures for Feist s acid are shown below. mass of water =... g [4] OO OO OO OO structure E structure F The n.m.r. spectrum of Feist s acid is shown below. absorption of energy 1 the number by each peak shows the relative area under the peak 1 peak Y 1 peak Z peak X 12 11 10 9 8 7 6 5 4 3 2 1 0 / ppm

9 (i) Identify the type of protons responsible for peak X.... [1] (ii) Explain how D 2 O can be used to help confirm which protons are responsible for peak X.... [2] (iii) Use peaks Y and Z to decide whether structure E or structure F is correct for Feist s acid. Explain your reasoning and state how the n.m.r. spectrum for the other structure would differ.... [3] [Total: 10] Turn over

10 5 Neral, 10 16 O, is a naturally occurring compound found in many plants. It has a sweet lemon smell and is used extensively in perfumery. The skeletal formula of neral is shown below. O (a) Name the two functional groups in neral......... and........ [2] (b) Neral has stereoisomers. (i) What is meant by the term stereoisomers?... [1] (ii) The stereoisomerism in neral is caused by one of the double bonds. On the skeletal formula above, circle the double bond that gives rise to stereoisomerism. Explain why this double bond causes stereoisomerism.... [3] (c) Neral is used in scented candles. When burnt it produces carbon dioxide and water vapour. Write a balanced equation for the complete combustion of neral... [1]

11 (d) Neral can be used in the laboratory to produce a variety of other compounds. Neral can be reduced to geraniol, shown below, which contributes to the smell of roses. O (i) Identify a suitable reducing agent for this reduction of neral.... [1] (ii) Write a balanced equation for this reduction of neral. Use [] to represent the reducing agent.... [1] (e) Neral can be converted into a hydroxycarboxylic acid using a two stage process. O stage 1 N O stage 2 OO O (i) Describe, with the aid of curly arrows, the mechanism in stage 1. Show any relevant dipoles and lone pairs of electrons. [5] (ii) State the type of reaction involved in stage 2. Identify the reagent and conditions used. type of reaction... reagent and conditions... [3] [Total: 17] Turn over

12 6 ompounds A, B and all have the same relative molecular mass. The mass spectrum of compound A is shown below. 100 80 relative intensity 60 40 20 0 10 15 20 25 30 35 40 45 m / e 50 55 60 65 70 75 ompound A reacts with 2,4-dinitrophenylhydrazine to give an orange precipitate but does not give a positive result with Tollens reagent. ompound B is a saturated hydrocarbon and the n.m.r. of compound B contains just one peak. The infra-red spectrum of compound contains a broad absorption between 2500 3300 cm 1 as well as an absorption between 1680 1750 cm 1. Identify compounds A, B and. Show all of your reasoning. compound A compound B compound

13 [10] [Total: 10] Turn over

7 In this question, one mark is available for the quality of spelling, punctuation and grammar. 14 yclohexene reacts with bromine at room temperature. Benzene reacts with bromine only in the presence of a halogen carrier. Phenol rapidly decolourises bromine at room temperature. Explain the different reactivities of bromine with cyclohexene, benzene and phenol. In your answer, include balanced equations and state the reaction type in each case.

15... [10] Quality of Written ommunication [1] [Total: 11] END OF QUESTION PAPER

16 PLEASE DO NOT WRITE ON TIS PAGE opyright Information OR is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. OR 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 OR 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 (www.ocr.org.uk) after the live examination series. If OR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OR 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. OR 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.