334/01 CHEMISTRY CH4. A.M. MONDAY, 23 January (1 hour 40 minutes)

Transcription

1 Candidate Name Centre Number Candidate Number WELSH JOINT EDUCATION COMMITTEE General Certificate of Education Advanced CYD-BWYLLGOR ADDYSG CYMRU Tystysgrif Addysg Gyffredinol Uwch 334/01 CHEMISTRY CH4 A.M. MONDAY, 23 January 2006 (1 hour 40 minutes) FOR EXAMINER S USE ONLY Section Question A Mark ADDITIONAL MATERIALS CJ*(334/01) In addition to this examination paper, you will need: a calculator; an 8 page answer book; a Data Sheet which contains a Periodic Table supplied by WJEC. Refer to it for any relative atomic masses you require. INSTRUCTIONS TO CANDIDATES Write your name, centre number and candidate number in the spaces at the top of this page. Section A Answer all questions in the spaces provided. Section B Answer both questions in Section B in a separate answer book, which should then be placed inside this question-and-answer book. Candidates are advised to allocate their time appropriately, between Section A (35 marks) and Section B (40 marks). INFORMATION FOR CANDIDATES The number of marks is given in brackets at the end of each question or part-question. The maximum mark for this paper is 75. Your answers must be relevant and must make full use of the information given to be awarded full marks for a question. You are reminded that marking will take into account the Quality of Written Communication used in your answers. No certificate will be awarded to a candidate detected in any unfair practice during the examination. B 4 5 TOTAL MARK

2 2 Examiner only Arholwr yn unig SECTION A Answer all the questions in the spaces provided. 1. (a) Adrenalin, a hormone which acts as a stimulant in the body, has the structure shown below. H H HO C C N H HO H O H CH 3 (i) Name two functional groups in the molecule. [2] I.... II.... (ii) On the diagram, identify a chiral centre in the molecule, using an asterisk, *. (iii) Draw the structure of the product formed when adrenalin is treated with I. excess aqueous sodium hydroxide, II. aqueous acid, H + (aq). (334-01)

3 3 Examiner only Arholwr yn unig (b) Adrenalin can be synthesised in several stages from a compound called 3,4-dihydroxybenzenecarbaldehyde, the structure of which is given below. HO C O HO H (i) During the first stage, the compound is reacted with hydrogen cyanide, HCN. Classify this reaction. (ii) Theoretically, 1 mole of 3,4-dihydroxybenzenecarbaldehyde can be converted to 1 mole of adrenalin. Calculate the maximum yield of adrenalin, in grams, which can be obtained from 8 42 g of 3,4-dihydroxybenzenecarbaldehyde. (Give your answer correct to 3 significant figures.) [3] Total [9] (334-01) Turn over.

4 4 Examiner only Arholwr yn unig 2. (a) Functional groups in aliphatic compounds behave differently to functional groups attached to benzene rings. For each of the following pairs of compounds, describe a chemical test that can be used to distinguish between them. The reagent(s) and condition(s) used, as well as the observations for each compound, are required. (i) 1-Chlorobutane and chlorobenzene. [2] (ii) Ethanol and phenol. [2] (iii) Ethylamine and phenylamine. [2] (334-01)

5 5 Examiner only Arholwr yn unig (b) A hydrocarbon of molecular formula C 8 H 10 has the NMR spectrum shown below. Simplified peak of peak area 5 Peak area 2 Peak area Chemical shift (δ) / ppm (i) Deduce the structure of the compound, giving your reasoning. [4] (ii) Draw the structure of one isomer of the hydrocarbon, C 8 H 10. Total [11] (334-01) Turn over.

6 6 Examiner only Arholwr yn unig 3. (a) Alkanes, alkenes and benzene can be halogenated, but by different mechanisms. (i) When propene reacts with hydrogen bromide, there are two possible products, although one is dominant. Name or give the structure of both possible products in the spaces below.... and... (ii) Draw the mechanism for the reaction between propene and hydrogen bromide, which leads to the formation of the dominant product. [3] (iii) Explain why one of the possible products is dominant. (iv) Explain why benzene does not readily undergo addition reactions. [2] (v) Name the type of mechanism that occurs when methane reacts directly with chlorine. (vi) Explain why ethane can also form during the direct chlorination of methane. [2] (334-01)

7 7 Examiner only Arholwr yn unig (b) Many organohalogens are used as herbicides. One of the first herbicides to be developed, 2,4-dichlorophenoxyethanoic acid, was used to defoliate forests in the Vietnam War. Its structure is shown below. OCH 2 COOH Cl Cl (i) State the reagent(s) and condition(s) needed to substitute a chlorine atom into a benzene ring. [2] (ii) Give a chemical test (apart from the use of indicators) to show that the herbicide above contains a carboxylic acid group. [2] Reagent(s)... Observation(s)... (c) Give another large-scale use of a named organohalogen compound of your choice Total [15] Total Section A [35] (334-01) Turn over.

8 8 SECTION B Answer both questions in the separate answer book provided. 4. (a) Study the reaction scheme shown below. HBr NH 3 HNO 2 A C 5 H 11 Br B C 2 Cr 2 O 7 /H + D (i) All the compounds are straight chained isomers. Compound A shows geometric isomerism. Compounds B and C each contain a chiral centre. Compound D gives an orange-yellow solid with 2,4-dinitrophenylhydrazine but does not give a silver precipitate with Tollens reagent. Use all the information to give the structures of compounds A-D, showing your reasoning in each case. [8] (ii) State the reagent(s) and condition(s) needed for the conversion of D C. (b) A compound E contains carbon, hydrogen and oxygen only. It has a molar mass of g mol 1. Quantitative analysis of compound E shows that it contains 39 97% carbon and 6 73% hydrogen by mass. Calculate both the empirical and molecular formulae of compound E. [3] (334-01)

9 9 (c) Carboxylic acids may be converted into several important derivatives. Give the reagent(s) and necessary condition(s) for the conversion of (i) propanoic acid to propanamide, [2] (ii) propanoic acid to ethyl propanoate. [2] (d) State how the infrared spectrum of propanoic acid would differ from that of ethyl propanoate, by using the characteristic infrared absorption frequencies given in the Data Sheet. (e) Explain, in detail, why propanoic acid is more acidic than propan-1-ol. [3] Total [20] Turn over for Question 5 (334-01) Turn over.

10 10 5. (a) Give one example of each of the following processes, giving necessary reactants, reaction conditions and balanced chemical equations: (i) elimination; [3] (ii) electrophilic substitution; [3] (iii) nucleophilic substitution. [3] (b) α-amino acids are very important in biochemistry since they are the monomers that make up the natural polymers called proteins. (i) Write down the graphic (full structural) formula of aminoethanoic acid. (ii) (iii) Write down the structural formula of the peptide link that forms when two aminoethanoic acid molecules react. Aminoethanoic acid, methyl ethanoate and propanoic acid have similar relative molecular masses. State and explain the trend in the melting temperatures of these compounds. [4] (c) Poly(phenylethene), commonly known as polystyrene, and nylon are both important synthetic polymers, but are different types of polymer. (i) Draw the structure of the monomer used in the production of poly(phenylethene) and name this type of polymerisation. [2] (ii) Draw the structure of a monomer used in the production of nylon 6,6. (iii) Give two differences between these two types of polymerisation. [2] Total [20] Section B Total [40] (334-01)

11 CH4 Section A Q.1 (a) (i) Alcohol / hydroxyl / hydroxy (not hydroxide) (1) Amine / amino (1) Phenol (1) (accept any 2) [2] (ii) (iii) I (accept O instead of -ONa) II (b) (i) Nucleophilic addition (ii) Mr (3,4-dihyd ) = (½) Mr (adrenalin) = (½) (accept 138 and 183) moles 3,4-dihyd mass adrenalin =

12 = 11 2 g (3 sig.figs) Total [9]

13 Q.2 (a) (i) Heat with aqueous NaOH followed by HNO3 and AgNO3 (aq) 1-chlorobutane gives white ppt., chlorobenzene no reaction (ii) Add iron(iii) chloride solution ethanol no reaction, phenol gives purple colour (accept add bromine, phenol gives white precipitate / iodoform test) (accept diazonium salt, phenol gives yellow/orange ppt) (accept heat with acidifies Cr2O7 2, ethanol orange to green) (iii) Add NaNO2(aq)/HCl(aq) below 5 ºC (accept HNO2) ethylamine gives bubbles (of nitrogen) phenylamine forms a solution (accept add bromine, phenylamine gives white precipitate) (accept diazonium salt, phenylamine gives yellow precipitate) (b) (i) δ 1.2 is CH3 group (attached to R) / triplet suggests next to CH2 quadruplet suggests next to CH3 δ 7.2 is C6H5 group (so all the protons are equivalent) / singlet suggests no proton on adjacent carbon (Accept use of peak areas) Compound is

14 (ii) Total [11] Q.3 (a) (i) 1-bromopropane (½) and 2-bromopropane (½) (accept structures) (ii) [3] (iii) Secondary carbocation is more stable than primary carbocation therefore secondary bromoalkane is dominant. (accept stability due to inductive effect of methyl groups) (iv) Benzene contains delocalised π electron system (which is very stable) addition destroys the delocalised π cloud (v) (Free) radical substitution / photochlorination (vi) During the reaction methyl radicals form; two of these radicals can combine to give ethane (b) (i) React with chlorine at room temperature (½) / absence of light (½) in the presence of aluminium chloride (ii) Add Na2CO3 / NaHCO3 bubbles (of CO2 form) which turn lime water cloudy (accept warm with alcohol/conc H2SO4, sweet smelling ester formed) (c) DDT in pesticides / PVC in window frames etc. Total [15] Total Section A [35]

15 Section B Q4(a)(i) is to be assessed for Quality of Written Communication. Candidates must satisfy the following criteria to be awarded the marks designated as [L]. Selection of a form and style of writing appropriate to the question Organisation of the relevant information clearly and coherently Use of legible text with adequate spelling, punctuation and grammar Q.4 (a) (i) Give 4 marks for reasons and 4 marks for structures. Reasons e.g. Since A shows geometric isomerism it must be pent-2- ene. Since B is formed from a bromoalkane and ammonia it is an amine. Since D reacts with 2,4-dnph but not with Tollens reagent it is a ketone. Since C is oxidised to D it must be a secondary alcohol. (Penalise once only if functional group is on third carbon) [4] (ii) Sodium tetrahydridoborate(iii)/nabh4 (½) in water/ethanol/ methanol (½) (accept lithium tetrahydridoaluminate(iii) in ether)

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17 (b) C : H : O : 6.73 : Empirical formula = CH2O Molecular formula = C3H6O3 (c) (i) Phosphorus(V) chloride/ PCl3/ SOCl2 (to form acid chloride) followed by ammonia (accept ammonia/ammonium carbonate (to form ammonium salt) followed by strong heat) (ii) Ethanol (Reflux with) concentrated (½) sulphuric acid (½) (d) Only infrared spectrum of propanoic acid would contain absorption between 2500 and 3550 cm -1 (due to O H bond). (Need comparison and reference to wavelengths) (e) Propanoic acid loses proton easier than propan-1-ol since the negative charge on the resulting propanoate ion is delocalised therefore ion left after losing proton is stabilised. (accept converse.) Total [20]

18 Q.5 (a) (i) Alcohol to alkene Heat with concentrated sulphuric acid at 180 C/Al2O3/H3PO4) C2H5OH C2H4 + H2O (accept bromoalkane to alkene, hot conc. NaOH in ethanol) (ii) Nitration of benzene Heat with concentrated HNO3 and H2SO4 (at 50 C) C6H6 + NO2 + C6H5NO2 + H + (accept chlorination of benzene, Cl2 and AlCl3 at room temp.) (iii) Hydrolysis of halogenoalkane Heat with aqueous NaOH C4H9Br + OH C4H9OH + Br (accept addition of CN, reflux with KCN in ethanol or addition of NH3, heat with excess of NH3) (b) (i) (ii) (iii) (If full structure given, it must all be correct) Trend in melting temperatures is aminoethanoic acid > propanoic acid > methyl ethanoate. Aminoethanoic acid highest since it exists as a zwitterions/shows ionic bonding.

19 Propanoic acid next since it can form hydrogen bonding between molecules. Methyl ethanoate lowest since it only has van der Waals forces between molecules.

20 (c) (i) Addition polymerisation (ii) (accept diacyl chloride) (accept shortened structural formulae) (accept condensation polymerisation forms small molecule/water whereas addition polymerisation does not) (iii) In addition polymerisation only polymer is formed, in condensation polymerisation small molecule is also formed. Addition polymerisation uses only one type of monomer, condensation polymerisation uses two different monomers. Addition polymerisation uses an initiator, no other compound needed for condensation polymerisation. (accept any 2) Total [20] Total Section B [40]