Join TSR now and get all your revision questions answeredSign up now

OCR Chemistry F322~ 4th June 2013~ AS Chemistry Watch

  • View Poll Results: How did you find the exam
    Hard A= 72
    42
    15.22%
    It was okay A= 76
    99
    35.87%
    Easy A= 79
    75
    27.17%
    Very Easy A=82
    60
    21.74%

    Offline

    0
    ReputationRep:
    does anyone know what was full UMS in the Jan 2013 paper? Think I dropped 4 marks.
    Offline

    0
    ReputationRep:
    UNOFFICIAL MARKSHEME - TO BE CONTRIBUTED...

    Q1.
    a) i) C10H22 (1)
    ii) 2-methylnonane (2)
    iii) Fewer points of contact, less VDWs, less energy required to break them. (2)

    b) i) C15H32 à C10H22 + C5H10 (1)
    ii) Carbon to carbon bond can be broken in a variety of places. (1)

    c) i) Draw an octagon (1)
    ii) Cyclic hydrocarbons burn more efficiently as fuels (1)

    Q2.
    a) i) E & H (1)
    ii) H (1)
    iii) F (1)

    b) i) C5H9OH (1)
    ii) 2-methylpentan-3-ol (1)

    c) A series of organic compounds with the same functional group, each differing by CH2. (2)

    d) 1st box -> Ketone, same as diagram but change –OH to =O (4)

    2nd box -> Ester, same as diagram minus the H on –OH, add ethanoic acid but minus H on the carboxyl.

    4th & 3rd box -> Alkene, same as diagram minus –OH and put a double bond in the ring. H20 in other box due to dehydration.
    Q3.
    a) Alkene and Ester (2)

    b) Double carbon bond within chain. (1)

    c) Orange to Colourless (1)

    d) i) Same structural formula, but different arrangement of atoms in space. (1)

    ii) Ester group on opposite side of the double carbon bond (pointing downwards). (1)

    e) i) The enthalpy change when 1 mol of a substance reacts completely with excess oxygen, under standard conditions. (2)

    ii) Q = mcT, (50x4.18x(20.2-54))/1000 = (-)7.06kJ (2)

    iii) n = m/Mr 1.34/(12x17)+32+32 = 5x10^-3mol (2)

    iv) 5x10^-3 mol 7.06kJ
    1mol 7.06 x 200 = 1412kJ (depending on rounding) (3)

    v) Incomplete combustion, carbon particulates were created. (1)

    f) C6H12O6 à 2C2H5OH + 2CO2
    Yeast + 37 degrees Celsius. (3)

    Q4.
    a) i) Curly arrows from bond in IBr to Br. From double bone to I.
    Next molecule: I bonded to one carbon, leaving Br- ion with lone pair, and a carbocation. Arrow from lone pair to carbocation. (3)

    ii) Electrophilic Addition (1)

    iii) Same as last molecule, but switch I and Br around. (1)

    b) i) Ultraviolet radiation (1)





    ii) Free Radical Substitution (7)

    Initiation
    IBr I + Br (2 radicals formed due to UV, bond breaks homolytically)

    Propagation
    I + CH4 CH3I + H
    Br + CH4 HBr + Ch3

    Termination
    CH3 + H CH4
    Br + H HBr
    I + CH3 CH3
    Q5.
    a) Products – Reactants
    (4 x +54) (4 x -394)
    (5 x -20) (12 x -242)

    +116 – (-4480) = 4596 kJmol-1 (3)

    b) i) n = 240/24 (N2O) = 10 mol
    1 mol à 164kJ
    10 mol à 1640kJ (2)

    ii) For 2N2O = 164
    for N20 = 82kJmol-1 (1)

    iii) 447 – 164 = +283kJmol-1 (1)

    c) O3 split by UV (2)
    O3 à O2 + O
    O2 + O à O3

    d) NO + O3 à NO2 + O2 (2)
    O + O3 à 2O2

    Q6.
    a) i) 256(+) (1)
    ii) S8 (1)
    iii) S4(+) (1)

    b) (33/100 x 194) + (34/100 x 195) + (25/100 x 196) + (8/100 x 198) = 195.2 (2)
    c) Breathalyser (1)

    d) C-O peak at “” Carbonyl (5)
    C=O peak at “” Aldehyde, ketone, ester

    C 66.7/12 H 11.1/1 O 22.2/16
    5.558 11.1 1.3875
    /smallest (1.3875)
    4:8:1
    C4H8O Mr = 72

    c-c-c-c=o c-c-(=o)c-c Aldehyde (Butanal) or Ketone (Butanone)



    Q7.
    a) N (1)
    b) i) S (1)
    ii) Nucleophilic Substitution
    OH- arrow to back of last carbon bonded to Br. Arrow from C-Br to Br. C(delta +) and Br(delta -) OH replace Br, Br- left. (4)

    iii) Increase in temp, increase Ek, increase rate of reaction. (1)

    c) But-2-ene + HCL à R (2)
    d) HCFs (1)
    Q8.
    a) –C-C-C-C— with CN on 2nd and 4th Carbon, H’s on rest. (1)
    b) Addition reaction, no waste products, 1 product. (1)

    c) Le Chatelier’s Principle – When a dynamic equilibrium is subject to change, the position of the equilibrium will shift to minimise the change.

    Increase temp à
    Shifts to left, endothermic side of reaction, reduce temp.

    Increase pressure à
    Shifts to left, less moles of gas, reduce pressure.

    Absense of catalyst à
    Will not shift equilibriums position. (5)



    d) Theoretical = 220 mol
    Actual = 11.13x10^3/(12x3)+3+14 = 210
    210/220 x 100 = 95.5% (2)

    e) Boltzmann for temp increase à shift to the right, Ea same position, but more area thus more molecules reach activation energy. Increase in temp increase Ek, more frequent and successful collision thus increases rate of reaction.

    Boltmann for catalyst à shift Ea to left, lowers Ea so more molecules can reach it, more successful collision, provide alternate route, increasing rate of reaction.

    Graphs on y-axise: number of molecules, x-axis: energy
    (7)
    Offline

    1
    ReputationRep:
    For the functional group would I get marks for put c-o and c double bond o?


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by Liberty.)
    I've attempted one, but considering it's my own answers simple-d down even more it's going to be heavily flawed.

    Please list improvements and I'll change it

    I'll write it up all fancy-pancy once the improvements are finished tomorrow after my darn physics exam.

    P.S I had no idea what I was doing with the Ozone stuff

    Thank you Q 3d, they are two isomers, the second box would have the same molecule but with the C=C bond in a different place

    Also for the 240dm^3 one you multiply by 5 not by 10 since the original equation had 2moles in it
    Offline

    0
    ReputationRep:
    (Original post by Adil16)
    I put exactly that however in the heat of the moment I may have got my left and right mixed up :-( hope I didn't lol
    I put that too, however I put without catalyst that it is unchanged becasue the rate of the forward and reverse reaction are exactly the same.
    Offline

    0
    ReputationRep:
    (Original post by Liberty.)
    I've attempted one, but considering it's my own answers simple-d down even more it's going to be heavily flawed.

    Please list improvements and I'll change it

    I'll write it up all fancy-pancy once the improvements are finished tomorrow after my darn physics exam.

    P.S I had no idea what I was doing with the Ozone stuff
    THANK YOU! feel better about this now these answers seem similar
    Offline

    0
    ReputationRep:
    (Original post by Farringtonn)
    For the functional group would I get marks for put c-o and c double bond o?


    Posted from TSR Mobile
    I put ketones and alkenes, and I am not even sure if thats right!
    Offline

    2
    ReputationRep:
    (Original post by TheFootyKing19)
    Found it ok . Just some weird questioms

    Hopefully did good!

    Wbu? Best of luck
    me too, overall a decent paper.

    good luck to you too! hope to see you back here on results day
    Offline

    0
    ReputationRep:
    (Original post by ninah1995162)
    I put ketones and alkenes, and I am not even sure if thats right!
    I think it was alkenes and ester

    Sent from my Nexus 4 using Tapatalk 4 Beta
    Offline

    0
    ReputationRep:
    In box 3 and 4 didn't it want the two possible isomers?
    Offline

    2
    ReputationRep:
    And waalllllaaaaaaaaaa!!!!!!!!!!!!!! !!!!

    Apologies on quality, I had to have it low to upload it
    Attached Images
  1. File Type: pdf UNOFFICIAL JUNE 2013 F322.pdf (216.9 KB, 1959 views)
    Offline

    0
    ReputationRep:
    (Original post by Legal drugdealer)
    Do you think they could be any lower?
    You know what?
    Possibly, yes.

    Minimum I would say would be 76?


    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by ché.)
    https://dl.dropboxusercontent.com/s/...rncq5urwq&dl=1

    here you go!
    This is the full paper that was another one uploading earlier...
    That will 10x better!
    :d


    posted from tsr mobile
    nooo i put decane instead of c10h22 on the first question. What is wrong with me.
    Offline

    2
    ReputationRep:
    (Original post by profit_master3)
    UNOFFICIAL MARKSHEME - TO BE CONTRIBUTED...
    Ooft, that looks awfully similar to something I made , ehe
    Offline

    0
    ReputationRep:
    (Original post by Sam_1996)
    In box 3 and 4 didn't it want the two possible isomers?
    Yeah, it said to put the organic products formed so H2O couldn't have been the other product.
    Offline

    2
    ReputationRep:
    And waalllllaaaaaaaaaa!!!!!!!!!!!!!! !!!!

    Apologies on quality, I had to have it low to upload it Apologies again
    Attached Images
  2. File Type: pdf UNOFFICIAL MARKSCHEME JUNE 2013 F322.pdf (216.9 KB, 390 views)
    Offline

    1
    ReputationRep:
    (Original post by profit_master3)
    UNOFFICIAL MARKSHEME - TO BE CONTRIBUTED...

    Q1.
    a) i) C10H22 (1)
    ii) 2-methylnonane (2)
    iii) Fewer points of contact, less VDWs, less energy required to break them. (2)

    b) i) C15H32 à C10H22 + C5H10 (1)
    ii) Carbon to carbon bond can be broken in a variety of places. (1)

    c) i) Draw an octagon (1)
    ii) Cyclic hydrocarbons burn more efficiently as fuels (1)

    Q2.
    a) i) E & H (1)
    ii) H (1)
    iii) F (1)

    b) i) C5H9OH (1)
    ii) 2-methylpentan-3-ol (1)

    c) A series of organic compounds with the same functional group, each differing by CH2. (2)

    d) 1st box -> Ketone, same as diagram but change –OH to =O (4)

    2nd box -> Ester, same as diagram minus the H on –OH, add ethanoic acid but minus H on the carboxyl.

    4th & 3rd box -> Alkene, same as diagram minus –OH and put a double bond in the ring. H20 in other box due to dehydration.
    Q3.
    a) Alkene and Ester (2)

    b) Double carbon bond within chain. (1)

    c) Orange to Colourless (1)

    d) i) Same structural formula, but different arrangement of atoms in space. (1)

    ii) Ester group on opposite side of the double carbon bond (pointing downwards). (1)

    e) i) The enthalpy change when 1 mol of a substance reacts completely with excess oxygen, under standard conditions. (2)

    ii) Q = mcT, (50x4.18x(20.2-54))/1000 = (-)7.06kJ (2)

    iii) n = m/Mr 1.34/(12x17)+32+32 = 5x10^-3mol (2)

    iv) 5x10^-3 mol 7.06kJ
    1mol 7.06 x 200 = 1412kJ (depending on rounding) (3)

    v) Incomplete combustion, carbon particulates were created. (1)

    f) C6H12O6 à 2C2H5OH + 2CO2
    Yeast + 37 degrees Celsius. (3)

    Q4.
    a) i) Curly arrows from bond in IBr to Br. From double bone to I.
    Next molecule: I bonded to one carbon, leaving Br- ion with lone pair, and a carbocation. Arrow from lone pair to carbocation. (3)

    ii) Electrophilic Addition (1)

    iii) Same as last molecule, but switch I and Br around. (1)

    b) i) Ultraviolet radiation (1)





    ii) Free Radical Substitution (7)

    Initiation
    IBr I + Br (2 radicals formed due to UV, bond breaks homolytically)

    Propagation
    I + CH4 CH3I + H
    Br + CH4 HBr + Ch3

    Termination
    CH3 + H CH4
    Br + H HBr
    I + CH3 CH3
    Q5.
    a) Products – Reactants
    (4 x +54) (4 x -394)
    (5 x -20) (12 x -242)

    +116 – (-4480) = 4596 kJmol-1 (3)

    b) i) n = 240/24 (N2O) = 10 mol
    1 mol à 164kJ
    10 mol à 1640kJ (2)

    ii) For 2N2O = 164
    for N20 = 82kJmol-1 (1)

    iii) 447 – 164 = +283kJmol-1 (1)

    c) O3 split by UV (2)
    O3 à O2 + O
    O2 + O à O3

    d) NO + O3 à NO2 + O2 (2)
    O + O3 à 2O2

    Q6.
    a) i) 256(+) (1)
    ii) S8 (1)
    iii) S4(+) (1)

    b) (33/100 x 194) + (34/100 x 195) + (25/100 x 196) + (8/100 x 198) = 195.2 (2)
    c) Breathalyser (1)

    d) C-O peak at “” Carbonyl (5)
    C=O peak at “” Aldehyde, ketone, ester

    C 66.7/12 H 11.1/1 O 22.2/16
    5.558 11.1 1.3875
    /smallest (1.3875)
    4:8:1
    C4H8O Mr = 72

    c-c-c-c=o c-c-(=o)c-c Aldehyde (Butanal) or Ketone (Butanone)



    Q7.
    a) N (1)
    b) i) S (1)
    ii) Nucleophilic Substitution
    OH- arrow to back of last carbon bonded to Br. Arrow from C-Br to Br. C(delta +) and Br(delta -) OH replace Br, Br- left. (4)

    iii) Increase in temp, increase Ek, increase rate of reaction. (1)

    c) But-2-ene + HCL à R (2)
    d) HCFs (1)
    Q8.
    a) –C-C-C-C— with CN on 2nd and 4th Carbon, H’s on rest. (1)
    b) Addition reaction, no waste products, 1 product. (1)

    c) Le Chatelier’s Principle – When a dynamic equilibrium is subject to change, the position of the equilibrium will shift to minimise the change.

    Increase temp à
    Shifts to left, endothermic side of reaction, reduce temp.

    Increase pressure à
    Shifts to left, less moles of gas, reduce pressure.

    Absense of catalyst à
    Will not shift equilibriums position. (5)



    d) Theoretical = 220 mol
    Actual = 11.13x10^3/(12x3)+3+14 = 210
    210/220 x 100 = 95.5% (2)

    e) Boltzmann for temp increase à shift to the right, Ea same position, but more area thus more molecules reach activation energy. Increase in temp increase Ek, more frequent and successful collision thus increases rate of reaction.

    Boltmann for catalyst à shift Ea to left, lowers Ea so more molecules can reach it, more successful collision, provide alternate route, increasing rate of reaction.

    Graphs on y-axise: number of molecules, x-axis: energy
    (7)
    Is the question where you had to name the alcohol not 4-methyl pentan-3-ol I thought you had to name alcohols before methyl groups
    Offline

    0
    ReputationRep:
    (Original post by Williamhodds)
    nooo i put decane instead of c10h22 on the first question. What is wrong with me.
    Ah, man!?
    Sorry about that!



    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by joe1545)
    With the question where it was why does the reaction have 100% atom economy, even though it didnt as Hydrogen was being produced what did people say? I said because hydrogen is also a desired product as it can be sold on or used in process such as the hydrogenation of an Alkene, an example is the manufacture of margarine... Anyone?
    Sorry but that was the polymer question wasn't it? Because addition reactions only produce one product :/
    Offline

    0
    ReputationRep:
    Help

    did the Q=m x c x change in temp question say RISE IN TEMP OR DROP IN TEMP?

    PLEASE
 
 
 
Poll
If you won £30,000, which of these would you spend it on?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.