Paper Reference. Advanced Unit Test 6B (Synoptic) Thursday 18 June 2009 Morning Time: 1 hour 30 minutes

Similar documents
Paper Reference. Friday 16 January 2009 Morning Time: 1 hour

Paper Reference. (including synoptic assessment) Monday 23 January 2006 Morning Time: 1 hour 30 minutes

Paper Reference. Thursday 12 June 2008 Afternoon Time: 1 hour 30 minutes

Paper Reference. Advanced Unit Test 6B (Synoptic) Thursday 24 January 2008 Morning Time: 1 hour 30 minutes

Paper Reference. Wednesday 6 June 2007 Morning Time: 1 hour

Paper Reference. (including synoptic assessment) Thursday 12 June 2008 Afternoon Time: 1 hour 30 minutes

Paper Reference. (including synoptic assessment) Thursday 11 June 2009 Afternoon Time: 1 hour 30 minutes

Paper Reference. Advanced Unit Test 6B (Synoptic) Monday 25 June 2007 Morning Time: 1 hour 30 minutes

Paper Reference. Paper Reference(s) 6246/02 Edexcel GCE Chemistry Advanced Unit Test 6B (Synoptic)

Paper Reference. Unit Test 6B (Synoptic) Thursday 25 January 2007 Afternoon Time: 1 hour 30 minutes

Paper Reference. Tuesday 23 January 2007 Morning Time: 1 hour 30 minutes

Paper Reference. Thursday 18 January 2007 Morning Time: 1 hour

Paper Reference. Tuesday 22 January 2008 Morning Time: 1 hour 30 minutes

Paper Reference. Paper Reference(s) 6246/02 Edexcel GCE Chemistry Advanced Unit Test 6B (Synoptic)

Paper Reference. Thursday 23 June 2005 Afternoon Time: 1 hour 30 minutes

Paper Reference. Monday 18 June 2007 Afternoon Time: 1 hour 30 minutes

Paper Reference. Sample Assessment Material Time: 1 hour

GCE AS/A level 1092/01 CHEMISTRY CH2

London Examinations IGCSE

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

London Examinations IGCSE

Paper Reference. Paper Reference(s) 6256/01 Edexcel GCE Chemistry (Nuffield) Advanced Level Unit Test 6 (Synoptic)

London Examinations GCE

Advanced Subsidiary Unit 1: The Core Principles of Chemistry

Advanced Subsidiary Unit 1: The Core Principles of Chemistry

PhysicsAndMathsTutor.com. Advanced Subsidiary Unit 1: The Core Principles of Chemistry

London Examinations IGCSE

Paper Reference. Wednesday 18 January 2006 Morning Time: 1 hour

Advanced Subsidiary Unit 1: The Core Principles of Chemistry

International Advanced Level Chemistry Advanced Subsidiary Unit 2: Application of Core Principles of Chemistry

CHEM2. General Certificate of Education Advanced Subsidiary Examination January Unit 2 Chemistry in Action (JAN12CHEM201) PMT

Unit 4: General Principles of Chemistry I Rates, Equilibria and Further Organic Chemistry (including synoptic assessment)

London Examinations IGCSE

London Examinations IGCSE

London Examinations IGCSE

Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level

Advanced Subsidiary Unit 1: The Core Principles of Chemistry

Advanced Subsidiary Unit 2: Application of Core Principles of Chemistry

Tuesday 9 October 2018

PhysicsAndMathsTutor.com. Advanced Subsidiary Paper 2: Core Organic and Physical Chemistry

3.2.8 Haloalkanes. Nucleophilic Substitution. 267 minutes. 264 marks. Page 1 of 36

London Examinations IGCSE

Paper Reference. Sample Assessment Material Time: 2 hours

International Advanced Level Chemistry Advanced Subsidiary Unit 3: Chemistry Laboratory Skills I

*AC112* *20AC11201* Chemistry. Assessment Unit AS 1 [AC112] FRIDAY 26 MAY, MORNING. assessing Basic Concepts in Physical and Inorganic Chemistry

TIME 1 hour 30 minutes, plus your additional time allowance.

Chemistry Assessment Unit AS 1

Advanced Subsidiary Paper 2: Core Organic and Physical Chemistry

Advanced Subsidiary Unit 1: The Core Principles of Chemistry

Advanced Subsidiary Unit 2: Application of Core Principles of Chemistry. Wednesday 17 January 2018 Morning Time: 1 hour 30 minutes

CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level

CHEMISTRY Unit 2 Energy, Rate and Chemistry of Carbon Compounds

London Examinations IGCSE

Advanced Subsidiary Unit 1: The Core Principles of Chemistry. Wednesday 11 January 2017 Morning Time: 1 hour 30 minutes

For the element X in the ionic compound MX, explain the meaning of the term oxidation state.

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level

Paper Reference. London Examinations IGCSE. Foundation Tier. Tuesday 10 November 2009 Afternoon Time: 1 hour 30 minutes

Advanced Subsidiary Unit 1: The Core Principles of Chemistry

London Examinations IGCSE

Unit 4: General Principles of Chemistry I Rates, Equilibria and Further Organic Chemistry (including synoptic assessment)

Advanced Subsidiary Unit 1: The Core Principles of Chemistry

Haloalkanes. Isomers: Draw and name the possible isomers for C 5 H 11 Br

Candidate Name Centre Number Candidate Number. Energy, Rate and Chemistry of Carbon Compounds

Advanced Subsidiary Unit 3B: Chemistry Laboratory Skills I Alternative

B410U20-1 S17-B410U20-1. CHEMISTRY AS component 2 Energy, Rate and Chemistry of Carbon Compounds

Advanced Subsidiary Unit 1: The Core Principles of Chemistry

Unit 2 Foundation Physical and Inorganic Chemistry

CHEM5 (JUN13CHEM501) General Certificate of Education Advanced Level Examination June Unit 5 Energetics, Redox and Inorganic Chemistry

B410U10-1 S17-B410U10-1. CHEMISTRY AS component 1 The Language of Chemistry, Structure of Matter and Simple Reactions

London Examinations IGCSE

pent-2-ene CH 3CH = CHCH 2CH 3 3-methylbut-1-ene (CH 3) 2CHCH = CH 2 2-methylbut-2-ene (CH 3) 2C = CHCH 3 2-methylbut-1-ene H 2C = C(CH 3)CH 2CH 3

SPECIMEN. Date Morning/Afternoon Time allowed: 1 hour 30 minutes. AS Level Chemistry A H032/01 Breadth in chemistry Sample Question Paper PMT

Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level

CHEM2. General Certificate of Education Advanced Subsidiary Examination June Unit 2 Chemistry in Action. Friday 27 May pm to 3.

Chemistry Assessment Unit A2 1

Advanced Subsidiary Unit 1: The Core Principles of Chemistry

Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level

CHEM2. General Certificate of Education Advanced Subsidiary Examination January Unit 2 Chemistry in Action

CHEM5. (JAN12CHEM501) WMP/Jan12/CHEM5. General Certificate of Education Advanced Level Examination January 2012

Advanced Subsidiary Unit 1: The Core Principles of Chemistry

PhysicsAndMathsTutor.com. International Advanced Level Chemistry Advanced Subsidiary Unit 2: Application of Core Principles of Chemistry

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level

Cherry Hill Tuition A Level Chemistry OCR (A) Paper 9 THIS IS A NEW SPECIFICATION

M08/4/CHEMI/SP2/ENG/TZ2/XX CHEMISTRY. Thursday 8 May 2008 (afternoon) Candidate session number. 1 hour 15 minutes INSTRUCTIONS TO CANDIDATES

Advanced Subsidiary Unit 3: Chemistry Laboratory Skills I

NANDI CENTRAL DISTRICT JOINT MOCK 2013

SPECIMEN. Candidate Number

AQA Level 1/2 Certificate SCIENCE: DOUBLE AWARD

Advanced Subsidiary Unit 1: The Core Principles of Chemistry

CHEM2. (JUN14CHEM201) WMP/Jun14/CHEM2/E9w. General Certificate of Education Advanced Subsidiary Examination June Unit 2 Chemistry in Action

Chemistry Assessment Unit AS 2

CHEM5. (JAN13CHEM501) WMP/Jan13/CHEM5. General Certificate of Education Advanced Level Examination January 2013

Q1. Pentanenitrile can be made by reaction of 1-bromobutane with potassium cyanide.

CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level

Chemistry Assessment Unit A2 1

Candidate number. Centre number

Advanced Unit 6B: Chemistry Laboratory Skills II Alternative

A-level CHEMISTRY 7405/1. Paper 1: Inorganic and Physical Chemistry. SPECIMEN MATERIAL v1.2

CHEM2. (JAN13CHEM201) WMP/Jan13/CHEM2. General Certificate of Education Advanced Subsidiary Examination January Unit 2 Chemistry in Action

CHEM4. (JUN15CHEM401) WMP/Jun15/CHEM4/E5. General Certificate of Education Advanced Level Examination June 2015

Transcription:

Centre No. Paper Reference Surname Initial(s) Candidate No. 6 2 4 6 0 2 Signature Paper Reference(s) 6246/02 Edexcel GCE Chemistry Examiner s use only Team Leader s use only Advanced Unit Test 6B (Synoptic) Thursday 18 June 2009 Morning Time: 1 hour 30 minutes Question Number 1 2 3 4 Blank Materials required for examination Nil Items included with question papers Nil Candidates may use a calculator. Instructions to Candidates In the boxes above, write your centre number, candidate number, your surname, initial(s) and signature. Check that you have the correct question paper. The paper reference is shown above. Answer Section A in the spaces provided in this question paper. Answer TWO questions in Section B in the spaces provided in this question paper. Indicate which question you are answering by marking the box ( ). If you change your mind about a question, put a line through the box ( ) and then mark your new question with a cross ( ). Do not use pencil. Use blue or black ink. Show all the steps in any calculations and state the units. Information for Candidates The marks for individual questions and parts of questions are shown in round brackets: e.g.. The total mark for this paper is 50. There are 20 pages in this question paper. Any pages are indicated. A Periodic Table is printed on the back cover of this question paper. Advice to Candidates You are reminded of the importance of clear English and careful presentation in your answers. You will be assessed on your Quality of Written Communication in this paper. This publication may be reproduced only in accordance with Edexcel Limited copyright policy. 2009 Edexcel Limited. Printer s Log. No. N33913A W850/R6246/57570 7/7/7/3/ *N33913A0120* Total Turn over

SECTION A Answer ALL parts of this question in the spaces provided. 1. Ascorbic acid, vitamin C, is an essential nutrient for many animals. It is a reducing agent which acts as an antioxidant in biological systems. It is very soluble in water and its concentration (e.g. in fruit juice) can be found as follows. Procedure A known volume of the solution containing ascorbic acid is acidified with dilute sulphuric acid and about 1 g of solid potassium iodide is dissolved in the mixture. 2 3 cm 3 of starch solution is added. The mixture is titrated with a standard solution of potassium iodate(v), KIO 3, until the appearance of a permanent blue colour. The following reactions occur. Iodate(V) ions react with iodide ions to form iodine. As soon as iodine is produced, it is reduced to iodide ions by the ascorbic acid. When the ascorbic acid has been used up, the next iodine to be produced by the iodate(v) ions reacts with the starch to give a blue colour. 2 *N33913A0220*

(a) (i) Use the data below to show that iodine can be reduced by ascorbic acid. State the molar ratio in which they react. I 2 + 2e 2I E = +0.54 V OH OH O O CHCH 2 OH O O CHCH 2 OH + 2H + + 2e E = 0.13 V O O HO OH ascorbic acid (ii) The reaction that produces the iodine is IO 3 + 5I - + 6H + 3I 2 + 3H 2 O In an experiment as described on page 2, 25.00 cm 3 of a solution containing ascorbic acid required a mean titre of 20.83 cm 3 of a potassium iodate(v) solution of concentration 0.0100 mol dm 3. Calculate the mass of ascorbic acid (molar mass 176 g mol 1 ) present in 25.00 cm 3 of the solution. (4) *N33913A0320* 3 Turn over

(b) (i) Explain how the intermolecular forces between ascorbic acid and water molecules form. Hence explain why it is very soluble in water. (3) (ii) Draw the structure of the molecule produced by heating ascorbic acid under reflux with potassium dichromate(vi) and sulphuric acid. Assume that the ring structure remains intact. (1) Q1 (Total 10 marks) TOTAL FOR SECTION A: 10 MARKS 4 *N33913A0420*

BLANK PAGE NOW TURN OVER FOR SECTION B *N33913A0520* 5 Turn over

SECTION B Answer any TWO questions from this section in the spaces provided. If you answer Question 2 put a cross in this box. 2. Double bonds are very important in synthetic chemistry, usually being much more reactive than single bonds. (a) The C=C and C=O bonds have the same electronic structure. (i) Draw dot and cross diagrams of ethene, H 2 C=CH 2, and of methanal, H 2 C=O, to show that this is so. Show outer electrons only. (ii) C=C and C=O bonds undergo different types of addition reaction. State the type of reaction in each case and explain the difference. (3) (1) 6 *N33913A0620*

(b) Propenenitrile, H 2 C=CHCN, is polymerised to make Orlon. Assuming the presence of a radical initiator, R, show the polymerisation mechanism as far as a radical containing two propenenitrile units. Mechanistic arrows must not be used. *N33913A0720* 7 Turn over

(c) Propenenitrile, H 2 C=CHCN, is also used to make Nylon. Propenenitrile is first reduced to hexanedinitrile, NC(CH 2 ) 4 CN, which is then reduced further to give 1,6-diaminohexane, H 2 N(CH 2 ) 6 NH 2. (i) Suggest a suitable reagent and conditions for the reduction of hexanedinitrile to 1,6-diaminohexane. (ii) Explain whether, or not, your suggested method would be suitable for use on an industrial scale. (1) (iii) Nylon-6,6 is made from 1,6-diaminohexane and hexanedioyl dichloride, ClOC(CH 2 ) 4 COCl. Draw the repeating unit of the polymer and identify the other molecule produced when the polymer is formed. Repeating unit: Identity of other molecule produced... 8 *N33913A0820*

(d) Polyvinyl alcohol (CH 2 CH(OH) ) n is a water-soluble polymer. All attempts to make its monomer vinyl alcohol, H 2 C=CHOH, have failed, the product invariably being CH 3 CHO. Use the bond enthalpies below to calculate the enthalpy change for the conversion of H 2 C=CHOH to CH 3 CHO and suggest why vinyl alcohol does not exist. H 2 C=CHOH CH 3 CHO bond C=C C O O H C H C C C=O bond enthalpy / kj mol 1 612 360 463 412 348 743 QUESTION 2 CONTINUES ON THE NEXT PAGE *N33913A0920* 9 Turn over

(e) Give the mechanism for the reaction between ethanal and hydrogen cyanide, HCN, in the presence of cyanide ions. (3) 10 *N33913A01020*

(f) The C=N bond also has the same electronic structure as C=C and C=O. (i) By analogy with the reaction of the C=O bond with Grignard reagents, suggest the structural formula of the compound that is formed when methylmagnesium CH 3 CH 3 bromide reacts with the compound C N and the product H is then hydrolysed. (ii) By analogy with compounds containing the C=C group, suggest in terms of the electronic structure and the bond angle around the nitrogen atom, why the CH 3 CH 3 nitrogen compound C N has a stereoisomer. H (iii) Draw the structure of the other stereoisomer. (1) (Total 20 marks) (1) Q2 *N33913A01120* 11 Turn over

If you answer Question 3 put a cross in this box. 3. Water is one of the most important chemicals we use; this question is about various aspects of the chemistry of water. (a) (i) Draw the shape of a water molecule, marking the bond angle on your diagram. Explain why the molecule has this shape and is polar overall. (3) (ii) Explain why the polarity of the water is important in dissolving ionic compounds. 12 *N33913A01220*

(iii) Draw a fully labelled Hess s Law cycle to show the enthalpy changes involved in dissolving potassium bromide, KBr, in water. Use it, and the data below, to determine the enthalpy of solution. ΔH latt (KBr) 670 kj mol 1 ΔH hydration (K + ) 322 kj mol 1 ΔH hydration (Br ) 335 kj mol 1 (3) QUESTION 3 CONTINUES ON THE NEXT PAGE *N33913A01320* 13 Turn over

(b) The following standard electrode potentials are required to answer this question. [Fe(H 2 O) 6 ] 3+ (aq) + e [Fe(H 2 O) 6 ] 2+ (aq) E = + 0.77 V [Fe(CN) 6 ] 3 (aq) + e [Fe(CN) 6 ] 4 (aq) E = + 0.36 V ½O 2 (g) + 2H + (aq) + 2e H 2 O(l) E = + 1.23 V (i) Write the overall equation for the oxidation of [Fe(H 2 O) 6 ] 2+ ions by molecular oxygen in acidic solution. State, with a reason, whether the reaction is feasible under standard conditions. (ii) State, with a reason, whether the oxidation by molecular oxygen of hexacyanoferrate(ii) ions, [Fe(CN) 6 ] 4, to hexacyanoferrate(iii) ions, [Fe(CN) 6 ] 3, is more or less likely than the similar oxidation involving the hexaaqua ions. (1) (c) Water reacts vigorously with silicon tetrachloride, SiCl 4, at room temperature. However, carbon tetrachloride, CCl 4, does not react with water. (i) Give the equation for the reaction between water and silicon tetrachloride. State symbols are not required. (1) 14 *N33913A01420*

(ii) Explain why carbon tetrachloride does not react with water. (3) (d) Solid magnesium chloride dihydrate, MgCl 2.2H 2 O, gives off hydrogen chloride when heated in a dry tube. Explain why this occurs and identify the residue............. (e) Primary halogenoalkanes are hydrolysed by water much more slowly than tertiary halogenoalkanes. State briefly how you could demonstrate the difference in rate, using test tube reactions and the compounds 1-bromobutane and 2-bromo-2-methylpropane......................... (3) Q3 (Total 20 marks) *N33913A01520* 15 Turn over

If you answer Question 4 put a cross in this box. 4. Sulphuric acid is one of the most important of all inorganic chemicals. (a) (i) Commercial concentrated sulphuric acid contains 98% H 2 SO 4 and 2% water, and has a density of 1800 g dm 3. Calculate its concentration in mol dm 3. (ii) In an aqueous solution of sulphuric acid the following equilibria exist. H 2 SO 4 (aq) + H 2 O(l) H 3 O + (aq) + HSO 4 (aq) K a very large HSO 4 (aq) + H 2 O(l) H 3 O + (aq) + SO 4 2 (aq) K a = 0.01 (K a values at 25 C) The ph of a 0.10 mol dm 3 solution of sulphuric acid at 25 C is 0.98. The strong monobasic acid, hydrochloric acid, of the same concentration has a ph of 1. Explain, in terms of the equilibria above, why the ph values are very similar. 16 *N33913A01620*

(b) Concentrated sulphuric acid is a strong acid and an oxidising agent and reacts in different ways with different halide ions. (i) Write the equation for the reaction of concentrated sulphuric acid with solid sodium chloride. State symbols are not required. (1) (ii) The reaction of concentrated sulphuric acid with solid sodium bromide gives hydrogen bromide, bromine, and sulphur dioxide. With sodium iodide the products include iodine, sulphur dioxide, sulphur and hydrogen sulphide. Use this information, and your answer to (i), to state and explain the trend in reducing ability of the halide ions. (3) (c) Chloroalkanes and bromoalkanes can be made by reacting an alcohol with the sodium halide in 50% aqueous sulphuric acid. This method cannot be used to make iodoalkanes. (i) State the reagents which are used to make 1-iodobutane from butan-1-ol. (ii) Give the formula of the compound formed from the reagents in (i) that then reacts with the alcohol. (1) QUESTION 4 CONTINUES ON THE NEXT PAGE *N33913A01720* 17 Turn over

(d) The mass spectrum of a primary halogenobutane is given below. 100 80 Relative intensity 60 40 20 0 20 30 40 50 60 70 80 90 100 110 120 130 140 m/e (i) Explain why the peak at m/e 29 is consistent with the compound being an unbranched halogenobutane. (1) (ii) The natural isotopic composition of the halogens is given below, with the percentage abundance of each isotope shown in brackets. chlorine-35 (75%) chlorine-37 (25%) bromine-79 (50%) bromine-81 (50%) iodine-127 (100%) Which halogenobutane generated the mass spectrum? Give TWO reasons for your choice. 18 *N33913A01820*

(e) 1-bromobutane can be converted into but-1-ene. (i) Give the reagent and the solvent needed for this conversion. (1) (ii) If HBr is added to 1-butene, the major product is 2-bromobutane. Give the mechanism for this reaction. (3) (iii) Explain why the major product is 2-bromobutane rather than 1-bromobutane. Q4 (Total 20 marks) TOTAL FOR SECTION B: 40 MARKS TOTAL FOR PAPER: 50 MARKS END *N33913A01920* 19

20 *N33913A02020*