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Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread Watch

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    (Original post by Chemstar_456)
    Hey all, this is what I remember from the multiple choice:

    Type of reaction: Transesterification
    The hydrolysis of the palm oil would be: Triol (there was only 1)
    Radiation leading to bond breaking: UV (free radical formation)
    What the 2 mechanisms had in common: Both had bonds broken and formed
    Only 1 leads to: Elimination
    The one with the optical isomer: Aldehyde which was X
    The one with geometric isomer: Alkene Y (had C=C with E/Z isomerism)
    Which produced iodoform: Both W+X which is aldehyde and ketone
    Which is oxidised to form a carboxylic acid: Aldehyde X
    Which reacted with 2,4-DNP: W+X the carbonyls
    Gas used for carrier gas chromatography: Nitrogen
    To separate you use the: Stationary phase
    The repeat unit for the diol and dicarboxylic acid: A
    The monomer used for that polymer was: 2-hydroxybutanoic acid
    Ammonia and propanoyl chloride: C2H5CONH2
    The acid and base conjugate pairs was: A
    The pH of the buffer solution: 6.21
    The yield will increase by: reducing pressure and temperature
    Changes affecting Kp: Temperature change
    Iodine: High concentration of hexane, top layer

    Please correct me if I'm wrong
    Don't think any of them led to elimination. the cyanide one was addition and the bromine was substitution.

    I thought it was 3-hydroxypropanoic acid.
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    (Original post by Don Pedro K.)
    Don't think any of them led to elimination. the cyanide one was addition and the bromine was substitution.

    I thought it was 3-hydroxypropanoic acid.
    nope deffo 2-hydroxypropanoic acid
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    (Original post by Chemstar_456)
    What the 2 mechanisms had in common: Both had bonds broken and formed
    Only 1 leads to: Elimination
    Which is oxidised to form a carboxylic acid: Aldehyde X

    The monomer used for that polymer was: 2-hydroxybutanoic acid


    The pH of the buffer solution: 6.21
    I believe only 1 leads to racemic
    I think there was secondary alcohol which would've been oxidised to carboxylic acid too
    And it was 3-hydroxy because COOH carbon is the priority.
    pH I got as 4.03 or 4.02 around that

    (Original post by ramadeen)
    i think you're right actually. I just realised elimination means a double bond is formed, the whole time i was thinking one molecule gets eliminated lol. well thats one extra mark gone
    Yeah there was a question in IAL paper or something where we had to deduce the mechanism for elimination with OH but the step in the MCQ didn't look like that. Well I wouldn't worry now, it won't change anything, and the paper was tricky so gb will be forgiving. Gl with unit 5 tho
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    (Original post by ilovecake123)
    nope deffo 2-hydroxypropanoic acid
    I don't think so...Not sure xD

    (Original post by C0balt)
    I believe only 1 leads to racemic
    I think there was secondary alcohol which would've been oxidised to carboxylic acid too
    And it was 3-hydroxy because COOH carbon is the priority.
    pH I got as 4.03 or 4.02 around that



    Yeah there was a question in IAL paper or something where we had to deduce the mechanism for elimination with OH but the step in the MCQ didn't look like that. Well I wouldn't worry now, it won't change anything, and the paper was tricky so gb will be forgiving. Gl with unit 5 tho
    Secondary alcohols can't oxidise to carb acids
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    (Original post by Don Pedro K.)
    I don't think so...Not sure xD



    Secondary alcohols can't oxidise to carb acids
    actually on second thought it could be 3 hydroxy
    never mind LOL nothing we can do except smash unit 5
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    (Original post by Don Pedro K.)
    I don't think so...Not sure xD



    Secondary alcohols can't oxidise to carb acids
    Its 3 hydroxy look at the photo

    And oh my god that is right, flipping hell what happened to me lol it forms ketone of course

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    (Original post by C0balt)
    Its 3 hydroxy look at the photo

    And oh my god that is right, flipping hell what happened to me lol it forms ketone of course

    Posted from TSR Mobile
    yh that's what I got ! Pretty sure I got 19/18 on MC
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    (Original post by Don Pedro K.)
    yh that's what I got ! Pretty sure I got 19/18 on MC
    lol hopefully grade boundaries will be low
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    (Original post by C0balt)



    Yeah there was a question in IAL paper or something where we had to deduce the mechanism for elimination with OH but the step in the MCQ didn't look like that. Well I wouldn't worry now, it won't change anything, and the paper was tricky so gb will be forgiving. Gl with unit 5 tho
    yeah i hope the gb will be low. I reckon I got a low A on this one. Hopefully unit 5 goes better for all of us.
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    (Original post by C0balt)
    Its 3 hydroxy look at the photo

    And oh my god that is right, flipping hell what happened to me lol it forms ketone of course

    Posted from TSR Mobile
    No I think you're right, I don't think it was a secondary alcohol I think it was primary just the OH was on the right and the carbon had 2 other hydrogen atoms bound to it
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    (Original post by Don Pedro K.)
    Yeah it was substitution for the other one by SN2. I also put that a racemic mixture forms but I can't remember if the molecule was symmetric (in which case it wouldn't form a racemic mixture obvs) :/
    the molecule was indeed symmetric, there was a CH3 on either side of the C-Br, so I went for elimination. It asked what it leads to in that step, so i would say that it is true to say that in that step a Br- is eliminated......
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    boundary thoughts? cant see it being any more than 67-68/90 for an A and 73 for an A*??
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    (Original post by C0balt)
    Its 3 hydroxy look at the photo

    And oh my god that is right, flipping hell what happened to me lol it forms ketone of course

    Posted from TSR Mobile
    Nah you're right, there was a primary alcohol in there, sneaky edexcel lol nearly caught me out but i defo remember thinking abt it for a minute or two.
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    (Original post by Whizbox)
    No I think you're right, I don't think it was a secondary alcohol I think it was primary just the OH was on the right and the carbon had 2 other hydrogen atoms bound to it
    But there was also a secondary alcohol molecule with a methyl group wasn't there
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    (Original post by ughexams)
    the molecule was indeed symmetric, there was a CH3 on either side of the C-Br, so I went for elimination. It asked what it leads to in that step, so i would say that it is true to say that in that step a Br- is eliminated......
    Ahh okay haha so I lost that one then. What were the options for the one before that? I remember one of them being "attacks a planar site" or something which isn't right for the C-Br one but yeah
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    (Original post by C0balt)
    But there was also a secondary alcohol molecule with a methyl group wasn't there
    I can't remember, there was 2 which would be oxidised to a carboxylic acid though wasn't there?
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    what did people put for the question where it asks how you would know that the h2so4 used to precipitate out the acid was in excess?:ashamed2:
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    (Original post by ughexams)
    the molecule was indeed symmetric, there was a CH3 on either side of the C-Br, so I went for elimination. It asked what it leads to in that step, so i would say that it is true to say that in that step a Br- is eliminated......
    I don't think it was elimination
    AFAIK in elimination there should also be a double headed arrow from one of C-H bond to form C=C
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    (Original post by Whizbox)
    I can't remember, there was 2 which would be oxidised to a carboxylic acid though wasn't there?
    I can't remember lol Oh well whatever I probably got like 17/18 for MCQ and I hope full UMS is no higher than 80/90 lol
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    where was that link you put up for the grade boundaries ?

    (Original post by C0balt)
    I don't think it was elimination
    AFAIK in elimination there should also be a double headed arrow from one of C-H bond to form C=C
 
 
 
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