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    (Original post by DragonRider17)
    I got around 23% I think (it was denitely in the 20s), I thought it was too low, but the next question did say "the yield is low," so I think it was nearly right. If not, only one mark lost becuase of a slip somewhere.
    Yep I remember getting 23.something% what two things did you mention for the next question?
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    (Original post by Mango88)
    Yep I remember getting 23.something% what two things did you mention for the next question?
    I said that some of the benzoic acid would be lost in the recrysalisation (as some will remain dissolved in the water) and that some of the methylbenzene would not be oxidised to benzoic acid. I wasn't too happy with that second point, hopefull they're not going to be extremely picky.
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    The yield was definitely 24.3% i rounded to 24%
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    (Original post by Chemistryxx77)
    I don't want to bring either of you down but it was revealed in the high resolution NMR that the doublet at 9.8 indicated an aldehyde. And another peak said somewhere between 2.0 and 3.0 there was a multi something number of peaks which means either "CH3-C=O(- R)" or "-CH2-C=O(-R)" so it had to have a ketone in there no matter what and the C=O was not part of the aldehyde. So it had to be an aldehyde with a ketone or possibly an ester.
    Correct me if i'm wrong though.
    There was only one oxygen in the molecular formula though..
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    (Original post by Formless)
    The yield was definitely 24.3% i rounded to 24%
    I messed my yield up because I stupidly used the Mr of methylbenzene rather than benzoic acid - probably 2 marks down the drain
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    (Original post by Chemistryxx77)
    I don't want to bring either of you down but it was revealed in the high resolution NMR that the doublet at 9.8 indicated an aldehyde. And another peak said somewhere between 2.0 and 3.0 there was a multi something number of peaks which means either "CH3-C=O(- R)" or "-CH2-C=O(-R)" so it had to have a ketone in there no matter what and the C=O was not part of the aldehyde. So it had to be an aldehyde with a ketone or possibly an ester.
    Correct me if i'm wrong though.
    I Think you may be wrong:/ I had a aldehyde with a methyl group as someone else said further up - CH3CH(CH3)COH

    There were three peaks in total which suggested there was 3 hydrogen environments first of all which lead me away from it being a ketone as there was 4 carbons in the molecular formula so the compound must have had two environments which were equivalent and I couldn't think of a ketone which structure applied to those rules.

    I agree with you on the peak at 9.8 ppm indicating a aldehyde, but it had a doublet splitting pattern, and due to the n+1 rule that must mean its neighboring carbon bonded to one other hydrogen (CH).

    The best clue though is the peak split into 7. This must mean again due to n+1 it has neighboring carbons with a total of 6 hydrogen atoms, so that peak was referring to the carbon bonded to the two CH3 groups.

    Thats what i got anyway xD
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    A whole 7 marks on the benzene mechanism was an absoulte god sent imo
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    (Original post by sumeyyatontus)
    I messed my yield up because I stupidly used the Mr of methylbenzene rather than benzoic acid - probably 2 marks down the drain
    Ah unlucky, i think you can still get one mark somewhere.

    Any suggestions on how many marks ill loose for saying it was Butanal, got all the m/z stuff right, chemical environment, all the ppm and group it corresponds to. Just messed up on the peaks.
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    Hai
    This is really bugging me, i know it was only one mark-but what did sucrose hydrolyse to? It says in the question there's an aldehyde group in both glucose and sucrose produced, and i can't fathom how this could be (especially after doing biology...)
    Thanks!
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    (Original post by Formless)
    Ah unlucky, i think you can still get one mark somewhere.

    Any suggestions on how many marks ill loose for saying it was Butanal, got all the m/z stuff right, chemical environment, all the ppm and group it corresponds to. Just messed up on the peaks.
    I think it was only one mark for getting the structure, so as long as you got everything else right, you've only lost one mark
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    (Original post by Hedd11)
    Hai
    This is really bugging me, i know it was only one mark-but what did sucrose hydrolyse to? It says in the question there's an aldehyde group in both glucose and sucrose produced, and i can't fathom how this could be (especially after doing biology...)
    Thanks!
    I have no idea, and have just realised I forgot about the aldehyde group so must have written the wrong reaction, greeeaaaat. That's one mark gone
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    There were no double ocsigens though? 😫 Ah this question is so difficult!
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    (Original post by Hedd11)
    Hai
    This is really bugging me, i know it was only one mark-but what did sucrose hydrolyse to? It says in the question there's an aldehyde group in both glucose and sucrose produced, and i can't fathom how this could be (especially after doing biology...)
    Thanks!
    Was this the question before the one about why sucrose, glucose and fructose are soluble?
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    (Original post by sumeyyatontus)
    Paper wasn't too bad - think I'm looking at around 62/63 marks
    .

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    (Original post by ZAM-H)
    Ermmm that paper was disgusting... Ppl found the paper hard and you're just saying how you think you got 60+. Just leave.

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    Wtf lol 😂😂 someone's bitter....
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    (Original post by Mango88)
    So what should the structure have been?
    Well according to the NMR spectrum with ppm and the data sheet
    it should be.

    CH3-(C=O)-(C=O-H)
    ^Ketone ^Aldehyde
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    (Original post by Chemistryxx77)
    Well according to the NMR spectrum with ppm and the data sheet
    it should be.

    CH3-(C=O)-(C=O-H)
    ^Ketone ^Aldehyde
    But the molecular formula only contained one oxygen
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    (Original post by sumeyyatontus)
    Wtf lol 😂😂 someone's bitter....
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    (Original post by DragonRider17)
    Apologies if I don't quite understand what you're saying, but I'll show how I think ours works. Using CH3-CH(CH3)-CHO, you have the one proton on the aldehyde, with chemical shift of about 9.8 and area 1, which is a doublet as it is next to the single proton on the CH group. Then this CH proton is next to several other protons (2 CH3 groups and the CHO group), so is multi-split and has area 1. Basically for a peak to be multi-split it needs to be next to several carbons with protons. Those examples you have given would have no splitting as those protons are not next to any other protons. Then the two methyl groups are in identical envronments so produce one peak with area 6. It is also a doublet as each CH3 group is next to the CH group in the centre of the molecule. This seems to match the NMR prefectly and has Mr 72, which the mass spec confirmed was the case.
    That's fine i'm not here to hate I just want to figure it out.
    Ah but the aldehyde is a doublet because you have one H and the oxygen counts as 0 but the n+1 rule means one so therefore two peaks.
    I have no idea about the multi splitting peaks.
    There was no area of peak 6.
    They're not examples though those where the actual figures for the NMR given in ppm.
    Also for two environments to make six they both must be adjacent to a carbon
    e.g.
    (H3C)CH2(CH3) will give a peak area of 6.
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    (Original post by sumeyyatontus)
    There was only one oxygen in the molecular formula though..
    I know which is why even I was confused however, I remember two things which may be why this is the case.
    1) I'm pretty sure it said "suggest" a formula.
    2) And I though why would they say suggest ? And it occurred to me that they never said the sample was pure. Which means that some peaks where representing impure compounds parts of it.
    Some could be ...
    CH3C=O+
    or
    Ch3Ch2C=O-H+
 
 
 
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