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Edexcel paper 2 chem unofficial mark scheme Watch

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    Chemistry Paper 2: Advanced Organic and Physical Chemistry

    Q 1 This is a question about alkanes.

    (a) What is the reaction mechanism when ethane and chlorine react in UV light. [1 mark]

    A electrophilic addition
    B electrophilic substitution
    C free radical addition
    D free radical substitution

    (b) What is the name of this alkane? [1 mark]

    A 2-ethyl-3-propylpentane
    B 4-ethyl-3-methylheptane
    C 3-methyl-4-propylhexane
    D 4-methyl-3-propylhexane

    (c) Alkanes are obtained by processing crude oil.

    (i) Explain why different alkanes in crude oil can be separated by fractional distillation. [2 marks]
    Different bp, as move away from heat source,condense at different temps and form diff fractions

    (ii) Complete the equation for the cracking of octane to produce ethene and only one other organic compound. State symbols are not required. [1 mark]
    c8h18---c2h4+c6h14

    (iii) Write the equation for the reforming of hexane into cyclohexane, using displayed formulae for the organic compounds. State symbols are not required. [1 mark]
    draw hexane-----cyclohexane +h2
    Q 2 Diamond, graphene and graphite are different forms of carbon.

    (a) The structural feature that graphene and graphite have in common is that the carbon atoms are arranged in [1 mark]

    A layers with each atom bonded to four others
    B hexagonal and pentagonal rings within a layer
    C hexagonal rings within a layer
    D a three-dimensional structure

    (b) The bond angles within a layer of graphene and a layer of graphite are [1 mark]

    A 90 and 109.5
    B all 109.5
    C 109.5 and 120
    D all 120

    (c) One way in which diamond differs from graphene and graphite is that only diamond has [1 mark]

    A a high melting temperature
    B a precise molecular formula
    C poor electrical conductivity
    D a giant structure

    Q 3 This is a question about halogenoalkanes and related compounds.

    (a) Explain the trend in reactivity of the primary chloro-, bromo- and iodoalkanes with aqueous hydroxide ions. [2 marks]
    Chloro least reactive-highest bond enthalpy, iodo most bond weakes
    (b) In aqueous sodium hydroxide, 1-bromoethane reacts to produce ethanol.

    (i) Write the mechanism for this reaction, including all relevant curly arrows, lone pairs and dipoles. Include the transition state. [4 marks]
    Your classic sn2 mechanism, hydroxide to delta pos carbon in first step and c-br bond breaks, 2 partial bonds in intermediate with square brackets and then finish by showing alcohol

    (ii) Give the reagents that are used to test that bromide ions are formed in this reaction mixture. Include the result of the test. [2 marks]
    AgNo3+hno3- cream ppt of agbr
    (c) The halogenoalkane 2-bromobutane reacts with ethanolic potassium hydroxide to produce a mixture of alkenes. Draw the skeletal formulae of all the alkenes that could be produced. [3 marks]
    Draw
    (d) Explain why ethene has a boiling temperature of -104 C, whereas ethanol has a boiling temperature of 78 C. [3 marks]
    Similar VDW, ehtanol has permanent dipoles and hydrogen bonding-more energy to break bonds

    Q 4 Traditionally, high-flying aircraft and Formula 1 racing cars have had their tyres inflated with nitrogen gas instead of air. Recently, this practice has been extended to some other cars.

    (a) A car tyre is filled with nitrogen gas to a volume of 8.98 dm3 and a pressure of 207 kPa at 20 C.

    (i) Using the Ideal Gas Equation, calculate the mass of nitrogen gas, in grams, present in the car tyre under these conditions. Give your answer to an appropriate number of significant figures. [3 marks]
    21.97
    (ii) During a car journey, the tyres become warm. Use the Ideal Gas Equation to deduce the effect that this has on the pressure in the tyres. [1 mark]

    (b) One reason for the use of nitrogen gas in car tyres is that less gas is lost from the tyres during use because nitrogen molecules are larger than oxygen molecules. A suggested explanation for this is that nitrogen atoms are larger than oxygen atoms.
    Explain why a nitrogen atom is larger than an oxygen atom. [2 marks]
    Lower nuclear number, weaker elec attraction, bigger atmoic radius so larger etc

    Q 5 This is a question about catalytic converters in car exhaust systems.

    (a) When petrol is burnt in a car engine, pollutant gases including carbon monoxide and nitrogen monoxide are formed.

    (i) Write the equation for the reaction between these two polluting gases that takes place on the surface of a catalytic converter. State symbols are not required. [1 mark]
    2NO+2CO---2co2 +N2

    (ii) Describe the stages by which the reaction in (a)(i) occurs in a catalytic converter. [3 marks]
    catalyst was adsorption, bond weakening then desorption-

    (b) Which area in the Maxwell-Boltzmann distribution diagram represents the increase in the number of particles with sufficient energy to react in the presence of a catalyst? [1 mark]

    A area A
    B area B
    C area A - area B
    D area A + area B

    (c) In the UK, the exhaust emissions of a petrol-fuelled vehicle must be less than 1.00g of carbon monoxide per kilometre.
    What is the maximum number of carbon monoxide molecules that can be emitted per kilometre for a vehicle to meet this regulation? [1 mark]

    A 1.37 x 1022
    B 2.15 x 1022
    C 6.02 x 1023
    D 1.69 X 1025

    Q 6 This is a question about polymerisation.

    (a) But-1-ene and cyclohexene both form addition poymers. Draw a section of each polymer, showing two repeat units. [2 marks]
    Draw

    (b) Deduce the two monomers needed to produce the polyamide shown. [2 marks]
    From diagra, put on cl/ oh for carboxyl and extra h for amine

    (c) Deduce the single monomer that could be used to produce the polyamide shown. [1 mark]
    Fix both ends with amine and cocl group

    (d) PLA is a biodegradable polyester which is made from 2- hydroxypropanoic acid, CH3CH(OH)COOH.

    (i) Draw the two enantiomers of 2-hydroxypropanoic acid. [2 marks]
    Draw so central carbon oh above, far right wedges was cooh, below was h, far left was ch3 then o=invert ch3 and cooh
    (ii) State how separate samples of these two enantiomers could be distinguished in a laboratory. [1 mark]
    Polarimeter- different drection of rotation of ppl
    (iii) Biodegradable polyesters breakdown naturally. State why this is an advantage. [1 mark]
    Less pollutants/energy costs to incinerate etc

    Q 7 Antifebrin was the trade name for N-phenylethanamide which was used as a painkiller until paracetamol was discovered.

    (a) Some of the following reagents can be used to produce Antifebrin from benzene.

    - Aluminium chloride
    - Ammonia, concentrated
    - Benzene
    - Ethanal
    - Ethanoic acid
    - Ethanol
    - Ethanoyl chloride
    - Hydrochloric acid, concentrated
    - Hydrochloric acid, dilute
    - Iron
    - Nitric acid, concentrated
    - Nitric acid, dilute
    - Propanone
    - Sodium chloride
    - Sulfuric acid, concentrated
    - Tin

    Selecting from only these reagents, devise a three-step synthetic pathway to convert benzene into Antifebrin. You should include the structures of the two intermediate compounds and the reaction conditions. [5 marks]
    Conc hno3 and conc h2so4, followed by tin and conc hcl followed by ethanoyl chloride
    (b) What is the number of peaks in a C-13 NMR spectrum of Antifebrin? [1 mark]

    A 5
    B 6-----------right
    C 7
    D 8

    (c) Paracetamol is structurally similar to Antifebrin, but has a hydroxy group attached directly to the benzene ring.
    The bromination of the benzene ring in paracetamol occurs much more readily compared to the bromination of benzene.
    Explain this increased reactivity. [2 marks] Oxygen increased density as lone pair incoroporate, greater nucleophile, more suscpetible to electrophilic attack
    (d) A tablet with a total mass of 500 mg contained 3.10 x 10-3 mol of paracetamol.
    Calculate the percentage by mass of paracetamol in the tablet, quoting your answer to an appropriate number of significant figures. [2 marks]
    93.62
    Q 8 Phenylethanone is an ingredient in many types of chewing gum.

    One method for the production of phenylethanone involves the reaction of benzene with ethanoyl chloride, CH3COCl.

    (a) (i) Write the equation for the formation of the electrophile, CH3CO+, from ethanoyl chloride using the catalyst aluminium chloride. [1 mark]
    Ch3cocl+alcl3----alcl4- + ch3co+
    (ii) Complete the diagram, including curly arrows, to show the mechanism for the reaction between this electrophile and benzene to produce phenylethanone. Include the regeneration of the catalyst. [4 marks]
    Draw mechanism, half clown face, regenerate ring etc bla bla
    (b) Phenylethanone can be distinguished from its structural isomer, phenylethanal, in a number of different ways.

    (i) Which would react with phenylethanone but not with phenylethanal? [1 mark]

    A acidified sodium dichromate(VI)
    B alkaline iodine solution
    C Fehling's solution
    D Tollens' reagent

    (ii) Give the steps to show how 2,4-dinitrophenylhydrazine could be used to distinguish between phenylethanone and phenylethanal. [4 marks]
    2,4 DNPH derivates, orange crystalline forms, recrystallise ppt, find mp using capillary tube method- compare with database of known bradys reagent derivatives

    *(iii) Compare and contrast the high resolution proton NMR spectra of phenylethanone and phenylethanal.
    You should use the Data Booklet. [6 marks]

    Diff proton nmr, one had 5 one had 4 intensity was like 3 to 2 to 2 to 1 and 1 to 2 to 2 to 2 to 1
    find chemical shifts from data booklet and state splitting patterns


    (c) The compound 1-phenylethanol can be formed from phenylethanone.
    Give the reagent and conditions that would be used to form 1- phenylethanol. [2 marks]
    Lialh4 in dry ether

    Q 9 This question is about reaction kinetics.

    (a) Compound A decomposes in a first order reaction.
    Calculate the time it takes for the mass of A to decrease from 600g to 37.5g if the decomposition has a constant half-life of 14 minutes. [1 mark] 56 mins

    (b) The 'initial rates' method was used to investigate the orders of reaction with respect to reactants X, Y and Z. The table shows the results obtained.

    (i) Calculate the orders with respect to X, Y and Z. [3 marks]

    X .....1............
    Y ......2...........
    Z ....0.............

    (ii) Give the rate equation for the reaction and hence calculate the rate constant, k, to an appropriate number of significant figures. Include units in your answer. [4 marks]
    K was 241.11 mol-2dm6/s

    (c) The kinetics of the 'bromine clock' were investigated and the rate equation was found to be

    Rate = k[BrO3-][Br-][H+]2

    (i) What is the overall reaction order? [1 mark]

    A First
    B Second
    C Third
    D Fourth

    (ii) Calculate the concentration of bromide ions required to produce a reaction rate of 4.08 x 10-3 mol dm-3 s-1 at 298K given that
    0.255

    k = 8.00 dm9 mol-3 s-1
    [BrO3-] = 0.200 mol dm-3
    [H+] = 0.100 mol dm-3 [2 marks]

    (d) The rate constant for the reaction between bromoethane and aqueous hydroxide ions was determined at five different temperatures.
    The results are given in the table.

    Complete the data in the table and use them to plot a graph of ln k against 1/T and hence determine the activation energy, Ea, in kJ mol-1.

    You should include the value and units of the gradient of the line.
    82-87 range probably

    The Arrhenius equation can be expressed as ln k = - Ea/R x 1/T + constant [7 marks
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    (Original post by glad-he-ate-her)
    Organic synthesis- conc no3, conc h2so4 +alcl3 goes to nitrobenzene then add tin and conc hcl to phenylamine then add ethanoyl chloride ( with no halogen carrier) to desired product
    Ea was 84000 nearabouts joules, and divide for kj
    mass of nitrogen was 21.97
    Orders were x-1 y-2 z-0 k was 241.11
    Diff proton nmr, one had 5 one had 4 intensity was like 3 to 2 to 2 to 1 and 1 to 2 to 2 to 2 to 1
    find chemical shifts from data booklet and state splitting patterns
    24 DNPH derivates, orange crystalline forms, recrystallise ppt, find mp using capillary tube method- compare with database of known bradys reagent derivatives
    catalyst was adsrorbption, bond weakening then desrorption- explain those terms
    graphie and graphene was single layer of hexagons, bond angle 120 and poor electrical conductivity was answer
    reforming equation was just make cyclic and balance with hydrogen
    2NO+2CO---2co2 +N2 was an answer
    conc of bromine was 0.255
    paracetomal percent was 93.6

    Agree with that
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    Can you do the synthesis by adding ethanoic acid? And water would be removed for a condensation reaction?
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    (Original post by NatoHeadshot)
    Can you do the synthesis by adding ethanoic acid? And water would be removed for a condensation reaction?
    As opposed to ethanoyl chloride for the last step?
    Yes but its less efficient and yield is less but youd still get the marks
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    (Original post by glad-he-ate-her)
    Organic synthesis- conc no3, conc h2so4 +alcl3 goes to nitrobenzene then add tin and conc hcl to phenylamine then add ethanoyl chloride ( with no halogen carrier) to desired product
    Ea was 84000 nearabouts joules, and divide for kj
    mass of nitrogen was 21.97
    Orders were x-1 y-2 z-0 k was 241.11
    Diff proton nmr, one had 5 one had 4 intensity was like 3 to 2 to 2 to 1 and 1 to 2 to 2 to 2 to 1
    find chemical shifts from data booklet and state splitting patterns
    24 DNPH derivates, orange crystalline forms, recrystallise ppt, find mp using capillary tube method- compare with database of known bradys reagent derivatives
    catalyst was adsrorbption, bond weakening then desrorption- explain those terms
    graphie and graphene was single layer of hexagons, bond angle 120 and poor electrical conductivity was answer
    reforming equation was just make cyclic and balance with hydrogen
    2NO+2CO---2co2 +N2 was an answer
    conc of bromine was 0.255
    paracetomal percent was 93.6
    You don't need AlCl3 for 1st step in synthesis as h2so4 is the catalyst.

    Also, you sure it was 84,000J - I got ~840,000J which was similar to an answer on a practice question I did
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    (Original post by 153179)
    You don't need AlCl3 for 1st step in synthesis as h2so4 is the catalyst.

    Also, you sure it was 84,000J - I got ~840,000J which was similar to an answer on a practice question I did
    Oh yeah i dont even think i put it thinking back.
    Pretty certain on 84000, few of my friends got the same
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    (Original post by glad-he-ate-her)
    Oh yeah i dont even think i put it thinking back.
    Pretty certain on 84000, few of my friends got the same
    I got that the first time but it was all about the 10-x on 1/T which was where the trap lies
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    Guys I think on average I have 140-150 (so 70-75 ish per paper) on the lower bound for my total score over the total 300 how well would you think I have to do in paper 3 to confirm the a* or is it unreasonable now? /:
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    Very high grade boundaries for this paper because it was the easiest paper you'll ever see. 81/90 A*?
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    (Original post by WhiteScythe)
    Very high grade boundaries for this paper because it was the easiest paper you'll ever see. 81/90 A*?
    Yeah,for an A I would say 70-75.Hopefully they will be lower though since many people found them difficult and this is the first year of now exams,remember tsr is full of probably topics students who students really hard.
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    (Original post by WhiteScythe)
    Very high grade boundaries for this paper because it was the easiest paper you'll ever see. 81/90 A*?
    78/79 Id say for a star, 70 for an A
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    (Original post by glad-he-ate-her)
    Organic synthesis- conc no3, conc h2so4 goes to nitrobenzene then add tin and conc hcl to phenylamine then add ethanoyl chloride ( with no halogen carrier) to desired product
    Ea was 84000 nearabouts joules, and divide for kj
    mass of nitrogen was 21.97
    Orders were x-1 y-2 z-0 k was 241.11
    Diff proton nmr, one had 5 one had 4 intensity was like 3 to 2 to 2 to 1 and 1 to 2 to 2 to 2 to 1
    find chemical shifts from data booklet and state splitting patterns
    24 DNPH derivates, orange crystalline forms, recrystallise ppt, find mp using capillary tube method- compare with database of known bradys reagent derivatives
    catalyst was adsrorbption, bond weakening then desrorption- explain those terms
    graphie and graphene was single layer of hexagons, bond angle 120 and poor electrical conductivity was answer
    reforming equation was just make cyclic and balance with hydrogen
    2NO+2CO---2co2 +N2 was an answer
    conc of bromine was 0.255
    paracetomal percent was 93.6
    I messed up the order for the reaction rate question, do I still get ecf marks?
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    Nice markscheme! I've added it to the unofficial markscheme list here
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    (Original post by glad-he-ate-her)
    78/79 Id say for a star, 70 for an A
    That sounds just about right.
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    (Original post by bobby147)
    I messed up the order for the reaction rate question, do I still get ecf marks?
    Yes, look at old mark schemes, they say ecf from part 1 in making rate equation
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    (Original post by glad-he-ate-her)
    Organic synthesis- conc no3, conc h2so4 goes to nitrobenzene then add tin and conc hcl to phenylamine then add ethanoyl chloride ( with no halogen carrier) to desired product
    Ea was 84000 nearabouts joules, and divide for kj
    mass of nitrogen was 21.97
    Orders were x-1 y-2 z-0 k was 241.11
    Diff proton nmr, one had 5 one had 4 intensity was like 3 to 2 to 2 to 1 and 1 to 2 to 2 to 2 to 1
    find chemical shifts from data booklet and state splitting patterns
    24 DNPH derivates, orange crystalline forms, recrystallise ppt, find mp using capillary tube method- compare with database of known bradys reagent derivatives
    catalyst was adsrorbption, bond weakening then desrorption- explain those terms
    graphie and graphene was single layer of hexagons, bond angle 120 and poor electrical conductivity was answer
    reforming equation was just make cyclic and balance with hydrogen
    2NO+2CO---2co2 +N2 was an answer
    conc of bromine was 0.255
    paracetomal percent was 93.6
    I got the exact same on everything except for the activation energy where I got 85.3kjmol-1 but I used the largest and lowest value on the table to workout gradient and then times it by 8.31. but must have made an arithmetic error somewhere.
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    (Original post by StudentOfHell)
    I got the exact same on everything except for the activation energy where I got 85.3kjmol-1 but I used the largest and lowest value on the table to workout gradient and then times it by 8.31. but must have made an arithmetic error somewhere.
    I used the table values as well tbh, i got something in that range, their answer will be a range anyway as gradient value will vary
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    (Original post by glad-he-ate-her)
    Oh yeah i dont even think i put it thinking back.
    Pretty certain on 84000, few of my friends got the same
    I got + 84.3 kjmol so your bang on target.
    Did you remember to convert into kilojoules?
    I got the right number of peaks for the nmr question,but my confidence is so low in nmr with cyclic compounds I didn't proceed .
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    i got 94.2% for the paracetamol question...
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    (Original post by StudentOfHell)
    I got the exact same on everything except for the activation energy where I got 85.3kjmol-1 but I used the largest and lowest value on the table to workout gradient and then times it by 8.31. but must have made an arithmetic error somewhere.
    No, its always a range,so you got full marks.
 
 
 
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