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    (Original post by ali_t9001)
    *2[H]
    well, thank god I didn't make that mistake in the exam :P
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    (Original post by Ilkonebi)
    I put the same. Did you have to specify it was a sextet though or is multiplet enough?
    It was enough, you can use multiplet for 5 and higher.
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    (Original post by FluffyCommie)
    Pretty sure the only acceptable form would be:

    Aldehyde + 2[o] --> alcohol
    I did the same thing by putting an oh don't know if I will loose a mark and doesn't reduction involve 2H instead of 2O
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    (Original post by Michaelj99)
    thats what I did im so confused, it didn't say it was branched or anything either
    Not sure.... Both structures seem to have same molecular formula though.
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    (Original post by anndz3007)
    his/her answer was right i think, cus the middle C was attracted to N fom NH2 and the first one attract to O in OH no?/
    They used D20 (or something like that) and when looking at specta, you never include Hs from N nor O
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    (Original post by cr7alwayz)
    What I remember (there may be mistakes)

    octadec-9,12-enoic acid because it has two trans isomers? which increases level of bad cholesterol.

    Cis-trans isomerism requires different groups attached to each carbon in c=c bond.

    Alcohol and amide.functional groups

    condensation polymer was polyamide and used as fibres in clothing

    Measure Rf values and compare with known Rf values. 2 spots instead of 3 because compounds may not have separated fully

    Polylactic acid could breakdown because ester linkages could be hydrolyses and the stitches could be dissolved because lactic acid could form hydrogen bonds with water using its OH groups.

    Diester made of six carbons and two ester bonds (skeletal formula)
    Percentage yield mass = 2.66g

    formula for peak A was just molecular formula with positive charge
    peak B was CH3CH2NH2 (or something like that) and positive charge

    work out mr of Compound H Mr - 74.
    Compound H: CH3CH2COOH
    For Compound J: (CH3)3CH2OH
    where do you put the charge on A ?? i thought nothing break down so i just leave it neutral ??
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    (Original post by sakuraton)
    Got something like that except I got ethanoic acid instead of propanoic acid. 60 was the Mr

    COOH Mr is 45 so there is 15 as remaining which is CH3

    Am I right?

    Unless I calculated the Mr wrong
    Yeah the Total Mr was 74 so without cooh its 29 which is c2h5
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    (Original post by President J)
    They used D20 (or something like that) and when looking at specta, you never include Hs from N nor O
    yeah you not include it but it still affect the chemical shift no ??
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    My friend wrote his structures in pencil, will the pencil pick up for marking?
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    Nothing quiet like throwing away your past papers ect after an exam
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    (Original post by anndz3007)
    his/her answer was right i think, cus the middle C was attracted to N fom NH2 and the first one attract to O in OH no?/
    1. oh nh doesn't split 2. d2o is used
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    (Original post by anndz3007)
    where do you put the charge on A ?? i thought nothing break down so i just leave it neutral ??
    It's called "Molecular ion peak" and i remembered my teacher banging on about + charge so I just put the whole molecule and that whole thing has + charge
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    (Original post by swetu1)
    I did the same thing by putting an oh don't know if I will loose a mark and doesn't reduction involve 2H instead of 2O
    It does invole [H], you're right, that was just a stupid mistake in my comment, but at least i put [H] in my exam :3

    As far as i'm concerned you will probably lose the mark for putting OH-
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    (Original post by FluffyCommie)
    Pretty sure the only acceptable form would be:

    Aldehyde + 2[o] --> alcohol
    2[H]**

    (Original post by jamesgates1)
    Why did the secondary alcohol and ketone have a ch3 group off it? Why couldn't it be a straight chain?
    There was only 4 carbon environments on the C13 NMR spectra therefore you had to have 2 of the five C's in identical environments therefore you knew there was a CH(CH3)2

    (Original post by Michaelj99)
    thats what I did im so confused, it didn't say it was branched or anything either
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    (Original post by Michaelj99)
    thats what I did im so confused, it didn't say it was branched or anything either
    It wouldn't be branched at all, everybody just got a little over excited and over complicated it. It's literally ch3ch2ch2ch(oh)ch3. CNMR peaks are the distance of the carbons from the electronegative point. The Ch2 next to the Ch(oh) will have the same peak as the Ch3 on the other end of the Ch(oh) so there are only 4 peaks. We were right. The answer most people got was wrong because if you look at it carefully you'd see it only has 3 peaks.
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    Did anyone else fine that exam rather difficult? I would've much preferred a paper like last year's ...

    Very few standard questions and no big nmr question too

    Been getting 50-55/60 on past papers, I don't think I've got more that 40 looking back at all the mistakes I've made.
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    (Original post by Michaelj99)
    thats what I did im so confused, it didn't say it was branched or anything either
    if i remember correctly i think there wasn't enough carbon environments in the c nmr to be a straight chain? The number of environments was one less than the number of carbons in the compound. i could be wrong though
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    (Original post by jasonx2)
    Anyone got 2.86g for the mass of compound A?
    i got 2.86 but do you remember the questions
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    (Original post by Michaelj99)
    Does it have to have the methyl?! I just did it as straight pentanone :/
    i thought the same :s
    why is the methyl needed?
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    (Original post by SlippyDuck117)
    Did anyone else fine that exam rather difficult? I would've much preferred a paper like last year's ...

    Very few standard questions and no big nmr question too

    Been getting 50-55/60 on past papers, I don't think I've got more that 40 looking back at all the mistakes I've made.
    I feel the exact same. Actually though 2015 was easy but this was awful

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