Carbon-13 nmr problem

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  1. shawn_o1's Avatar
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    Carbon-13 nmr problem
    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN10.PDF

    Question 8 a) i) I just don't understand. I initially thought all three compounds had 2 peaks in their 13C spectrum, but the mark scheme says W has 3, X has 4 and Y has 2. How does W and X have any peaks other than d 110-160 for benzene and d 25-60 for C-N (single bond)?
  2. JMaydom's Avatar
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    Re: Carbon-13 nmr problem
    X only has 3 environments and hence 3 peaks in the carbon spectrum! Maybe it's a mark scheme error?
    The others only have 2 environments.

    You do use decoupled spectra for carbon? ie, there is never a splitting pattern for them?

    I'll run a simulation on the chem software i have just to check everything
  3. shawn_o1's Avatar
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    Re: Carbon-13 nmr problem
    (Original post by JMaydom)
    X only has 3 environments and hence 3 peaks in the carbon spectrum! Maybe it's a mark scheme error?
    The others only have 2 environments.

    You do use decoupled spectra for carbon? ie, there is never a splitting pattern for them?

    I'll run a simulation on the chem software i have just to check everything
    No, 13C peaks are never split.
  4. EierVonSatan's Avatar
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    Re: Carbon-13 nmr problem
    Determining the number of peaks is all about symmetry. The chemical environments are similar but not exactly the same - lets take W for example...

    (It's a bit annoying that it's a ring, it make it harder to explain in text.) It has a plane of symmetry between the two NO2 groups, cutting the molecule in half. You have three carbons on each side, one carbon is directly attached to the NO2 this is one environment, the next carbon along is further away from the nitro group (environment number 2) and the last is even further away (environment 3). So you predict it to have 3 peaks.
  5. TiTo20's Avatar
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    (Original post by shawn_o1)
    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN10.PDF

    Question 8 a) i) I just don't understand. I initially thought all three compounds had 2 peaks in their 13C spectrum, but the mark scheme says W has 3, X has 4 and Y has 2. How does W and X have any peaks other than d 110-160 for benzene and d 25-60 for C-N (single bond)?
    The markscheme is correct, the only trouble I have proving it is that I'm on my iPhone and I don't have my white board to help me prove. If no one else answers this before I'm done, I'll quickly take a picture when I'm finished with work =]

    Only tip I can give you without my stuff is look for symmetry, and that the distance between one carbon and a group, and another carbon and a group will give rise to different peaks.

    W has three peaks because of...

    C -- C -- C -- NO2 : 1 peak for the C furthest away from the NO2 group, and then +1 until then C next to the NO2 group.

    Hope this sort of helped!


    This was posted from The Student Room's iPhone/iPad App
  6. JMaydom's Avatar
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    Re: Carbon-13 nmr problem
    Sorry was in proton nmr mode when I wrote that: 3,4,2 makes sense
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