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OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD watch

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    (Original post by RayMasterio)
    I have a question.

    During polymerisation of an amino acid, how do you know if you're forming a dipeptide or if you're polymerising the amino acid itself?
    a dipeptide would have closed bonds and the NH2 and COOH groups at the end, but a polymer would have open bonds and NH/CO at each end?
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    "for carbon-13 soectra the carbon peak from CDCl3 are removed from the spectrum" then whats the point of using it? isnt it so it won't produce a signal?


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    no CDCL3 is used as a solvent because you need your sample to be a solution in NMR but its removed for C13 NMR because there is a carbon in it and it produces a peak
    (Original post by ranz)
    "for carbon-13 soectra the carbon peak from CDCl3 are removed from the spectrum" then whats the point of using it? isnt it so it won't produce a signal?


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    Anyone who's done 2015 NMR question, why is there a peak on the 13C spectrum furthest to the left at around 168-169ppm? What is it representing? It doesn't seem to add up...
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    guys for nmr, do u guys first look at splitting, integeration etc and then look at chemical shifts,
    or do u guys do it simultaneously?
    thanks!


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    (Original post by Yazmin123)
    no CDCL3 is used as a solvent because you need your sample to be a solution in NMR but its removed for C13 NMR because there is a carbon in it and it produces a peak
    ah i see okk! thanks


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    (Original post by bulkybros100)
    Anyone who's done 2015 NMR question, why is there a peak on the 13C spectrum furthest to the left at around 168-169ppm? What is it representing? It doesn't seem to add up...
    That's exactly my thought!
    I asked my teacher and this was his response
    The answer is in the hydrogen NMR. Look at the structure and identify the peaks due to the 4 hydrogens given in the molecule. The aromatic is size 7 so there must be another aromatic ring with 4 hydrogens on it. Then look at the info about a peak H-C=N. it is size three so must be a methyl group. The answer should fall out easily. The C13 spectrum is really just used to make sure your answer has 14 carbon environments

    If anyone has better insight into this, it'd be great
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    (Original post by suibster)
    That's exactly my thought!
    I asked my teacher and this was his response
    The answer is in the hydrogen NMR. Look at the structure and identify the peaks due to the 4 hydrogens given in the molecule. The aromatic is size 7 so there must be another aromatic ring with 4 hydrogens on it. Then look at the info about a peak H-C=N. it is size three so must be a methyl group. The answer should fall out easily. The C13 spectrum is really just used to make sure your answer has 14 carbon environments

    If anyone has better insight into this, it'd be great
    I'm also interested, it confused me when doing the question for the first time, was always thinking about a C=O


    Although using all of the other info on the proton NMR and 7 environments, you can kinda figure it out. it confused me in the time conditions of the mock we did though
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    guys for a q like this how would u know whether 2 ch3 attached to same carbon or its an overall symmetrical compound?


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    (Original post by ranz)
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    guys for a q like this how would u know whether 2 ch3 attached to same carbon or its an overall symmetrical compound?


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    The top one would be a singlet, because it's symmetrical. The bottom one wouldn't be a singlet, but a multiplet because the hydrogen is adjacent to 6 protons. But the 2xCh3 would produce a doublet because they're adjacent to a carbon with 1 proton
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    (Original post by RayMasterio)
    The top one would be a singlet, because it's symmetrical. The bottom one wouldn't be a singlet, but a multiplet because the hydrogen is adjacent to 6 protons. But the 2xCh3 would produce a doublet because they're adjacent to a carbon with 1 proton
    but hwat if the compound was ch3ch3ch3 is symmetrical and would also produce a multiplet?


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    (Original post by itsConnor_)
    I'm also interested, it confused me when doing the question for the first time, was always thinking about a C=O


    Although using all of the other info on the proton NMR and 7 environments, you can kinda figure it out. it confused me in the time conditions of the mock we did though
    ^^My thoughts exactly, especially when the Proton NMR had a peak in the C=O region 2.4ppm as well !

    God I HATE Chemistry - so inconsistent and illogical.
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    (Original post by ranz)
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    guys for a q like this how would u know whether 2 ch3 attached to same carbon or its an overall symmetrical compound?


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    Think about it simply. The top one, the CH3s are in the same environment, right? Therefore the top one is a singlet (adjacent to no hs) (let's pretend X is a ketone). Whereas the bottom is a multiplet (2xCH3) as it isn't in the same environment :/ Get it?
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    (Original post by bulkybros100)
    Anyone who's done 2015 NMR question, why is there a peak on the 13C spectrum furthest to the left at around 168-169ppm? What is it representing? It doesn't seem to add up...
    I just did this question earlier and had the same problem, I think its because of the C=N carbon environment, but it doesn't have a chemical shift value on the insert, so its purely there for confusion
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    (Original post by bulkybros100)
    Anyone who's done 2015 NMR question, why is there a peak on the 13C spectrum furthest to the left at around 168-169ppm? What is it representing? It doesn't seem to add up...
    That's the shift for the C connected to the N via the C=N bond, the shift for which is not listed on the data sheet. You don't need to use any information from the 13C spectrum to do the question, other than the fact that there are 14 peaks.
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    (Original post by marioman)
    That's the shift for the C connected to the N via the C=N bond, the shift for which is not listed on the data sheet. You don't need to use any information from the 13C spectrum to do the question, other than the fact that there are 14 peaks.
    that seems a bit ridiculous? On Tuesday should I be thinking about ignoring peaks that aren't listed on our data sheet?
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    (Original post by itsConnor_)
    that seems a bit ridiculous? On Tuesday should I be thinking about ignoring peaks that aren't listed on our data sheet?
    Read the question carefully - in June 2015 it tells you to "use your answer to (c)" (14 peaks) and "the chemical shifts and splitting patterns in the 1H NMR spectrum". Although it doesn't say it outright then it's heavily hinting that the main clues to the structure are in the 1H spectrum rather than the 13C spectrum.
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    (Original post by ranz)
    but hwat if the compound was ch3ch3ch3 is symmetrical and would also produce a multiplet?


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    Yes, with an RPA of 9
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    My problem with the June 2015 paper was which environment caused the peak. Like the NH bond never crossed my mind, i thought it was a carbonyl compound because it has the same chemical shifts.

    How would you overcome this problem?
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    Can anyone explain this question? Like how do you know the position of the double bond in the third monomer?
 
 
 
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