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    [QUOTE=golddust&lipgloss;31608027]
    (Original post by dan1993)


    i could marry you for this, i'm chuffed that the paper's finally here!
    thankyou ! :kitty: xoxoxoxoxoxo
    I second this
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    Here's the F334 paper hope it helps

    F334Jan11.pdf
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    (Original post by dan1993)
    Here's the F334 paper hope it helps

    F334Jan11.pdf
    thanks so much!

    p.s do you have the mark scheme for this?
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    MS_F334_Jan11.pdf
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    How is everybody's revision going?
    I was just wondering, can anyone explain the 'ratios' in n.m.r. spectroscopy? I understand the theory of splitting, and sharing the splitting patterns etc. but I don't quite understand the ratios. Are they indicating towards the splitting patterns of each of the protons, or am I just making things up?
    Thanks to anyone that can help!
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    (Original post by brendan.)
    How is everybody's revision going?
    I was just wondering, can anyone explain the 'ratios' in n.m.r. spectroscopy? I understand the theory of splitting, and sharing the splitting patterns etc. but I don't quite understand the ratios. Are they indicating towards the splitting patterns of each of the protons, or am I just making things up?
    Thanks to anyone that can help!
    Do you mean the ratios of how each peak is split?

    I think its a standard thing - so if a peak is split into three, it will have two smaller peaks and one taller. But I don't think we need to understand anything about why. I don't think its ever been explained to me anyway...
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    (Original post by twelve)
    Do you mean the ratios of how each peak is split?

    I think its a standard thing - so if a peak is split into three, it will have two smaller peaks and one taller. But I don't think we need to understand anything about why. I don't think its ever been explained to me anyway...
    Ah, no sorry! At least I don't think it's that.
    Take for instance the 2011 paper above ^^, Q5(c) is about nmr, and in the mark scheme it mentions ratios of intensities/areas.

    It seems the post with the paper and mark scheme have gone now..
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    (Original post by brendan.)
    Ah, no sorry! At least I don't think it's that.
    Take for instance the 2011 paper above ^^, Q5(c) is about nmr, and in the mark scheme it mentions ratios of intensities/areas.

    It seems the post with the paper and mark scheme have gone now..
    The ratio of the areas of the peaks, for example 3:2:3:4, gives the ratio of the protons in the different environments.

    So for the ratio above, for every 3 protons in the first environment, there are 2 in the second, 3 in the third and four in the fourth. The height of the peak(the area) tells you how many protons are in each environment. BUT its only a ratio, so the actualy numbers of protons could be something like 6:4:6:8 or 15:10:15:20
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    When doing calculations, to what number of significant figures do we put our final answer if we're not told. I usually put it as 3 but i was just doing a paper and the markscheme said to put it to 2 significant figures. I'm confused
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    (Original post by tkoki1993)
    When doing calculations, to what number of significant figures do we put our final answer if we're not told. I usually put it as 3 but i was just doing a paper and the markscheme said to put it to 2 significant figures. I'm confused
    Its either the same number of significant as the data in the question is give, or to one more significant figure... I can never remember which - sorry!
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    (Original post by twelve)
    Its either the same number of significant as the data in the question is give, or to one more significant figure... I can never remember which - sorry!
    ahhh that makes sense... why dont they just tell us that!! lol
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    Has anyone done the June 2007 paper? Question 3 and parts of question 5 are tricky.
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    For nmr, if there is a quartet (split into four peaks), that means there are 3 Hydrogens on the adjacent carbon atom.

    However, what if there is a central carbon atom in a molecule with one carbon adjacent to it on the left, and one carbon adjacent to it on the right.

    If this had a quartet splitting pattern, does this mean:-
    1.) Each carbon next to it has 3 hydrogens, making a total of 6 hydrogens
    2.) The sum of hydrogens on each side is 3 (e.g. 2 Hydrogens on 1 carbon, and 1 on the other)
    3.) 1 carbon atom has three Hydrogens on it, the other we do not know
    4.) Or is there another reason

    Thanks for your help
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    (Original post by Adam9)
    For nmr, if there is a quartet (split into four peaks), that means there are 3 Hydrogens on the adjacent carbon atom.

    However, what if there is a central carbon atom in a molecule with one carbon adjacent to it on the left, and one carbon adjacent to it on the right.

    If this had a quartet splitting pattern, does this mean:-
    1.) Each carbon next to it has 3 hydrogens, making a total of 6 hydrogens
    2.) The sum of hydrogens on each side is 3 (e.g. 2 Hydrogens on 1 carbon, and 1 on the other)
    3.) 1 carbon atom has three Hydrogens on it, the other we do not know
    4.) Or is there another reason

    Thanks for your help
    I always thought it only considered 1 of the carbons next to it.

    Just had a quick google though and I typed in nmr spectrum for Propane
    and it came up with this: http://www.muhlenberg.edu/depts/chem.../Image1452.gif

    As you can see, the peak with a higher shift is split into 7 which suggests there is 6 carbons 'next door'. Therefore the answer to your question is that the splitting is for the total number of Hydrogens on all of the adjacent carbons.

    I think...
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    (Original post by brendan.)
    Ah, no sorry! At least I don't think it's that.
    Take for instance the 2011 paper above ^^, Q5(c) is about nmr, and in the mark scheme it mentions ratios of intensities/areas.

    It seems the post with the paper and mark scheme have gone now..
    Basically the ratio of the area beneath the peak tells you the ratio of the number of protons in each environment. Say for example you had methanol you would have two peaks one with an area of 3H and another with an area of 1H.
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    Hardest part of this paper is going to be memorising every reaction condition and reagent.
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    If anyone is interested the UMS marks for the January F335 paper they are available here.
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    Does anyone know that if in the exam if they asked you to circle to chromophore of the molecule ( that obviously they show you ) do you circle just the two benzene rings with the N=N bond in between it, or do you also circle the N and O and C=O (if this is shown attached to the benzene ring) becuase technically since all of them extends the delocalisation of dye they should be part of the chromophore but in some question they dont circle them. So are we supposed to include that in our "chromophore" or not?.. Thanks
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    (Original post by Prydie)
    If anyone is interested the UMS marks for the January F335 paper they are available here.
    Hey looked at the UMS marks for the f335 chemistry by deign paper, does that mean that is we get 80/120 looking at the raw mark which is actually 67% that it is worth an A ?
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    (Original post by pav)
    Does anyone know that if in the exam if they asked you to circle to chromophore of the molecule ( that obviously they show you ) do you circle just the two benzene rings with the N=N bond in between it, or do you also circle the N and O and C=O (if this is shown attached to the benzene ring) becuase technically since all of them extends the delocalisation of dye they should be part of the chromophore but in some question they dont circle them. So are we supposed to include that in our "chromophore" or not?.. Thanks
    My teacher said just the benzene rings and the N-N triple bond.

    But when there was a question about would the colour of this molecules change if you add a different side group, the answer was accepted as yes or no - so yes would be correct if you said that it was because the chromophore as changed, and no would be correct if you said that the chromophore wouldn't be changed. So I think its one of those where it would be accepted either way...
 
 
 
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