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    So there's a question with the H1 NMR Spectrum:

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    The mark scheme says the peak at 7.2 is from a benzene ring with 5 H's. Doesn't the big number of peaks mean there should be a lot of adjacent protons? But the mark scheme says the final structure is:
    Attachment 517273517275
    But there are no adjacent protons on the benzene ring?
    What am I missing?;(
    Thanks!
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    The fine structure at 7.2 is characteristic of benzene protons, the splitting doesnt come from adjacent protons for reasons you'll learn if you do degree level chemistry. All you need to know is that when you get a high ppm jagged multiplet, you can bet your bottom dollar that there arr benzene protons.
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    The spitting is because the benzene ring is substituted - by having the OC(O)CH(CH3)2 group, your hydrogens on the aromatic ring aren't all magnetically equivalent; you've got three different sets of hydrogens - ortho, meta and para to the substituted group. All of these will still come out around the same position, but will couple with each other, causing the splitting and a multiplet at ~7 corresponding to the aromatic ring.
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    (Original post by Spudgunhimself)
    The fine structure at 7.2 is characteristic of benzene protons, the splitting doesnt come from adjacent protons for reasons you'll learn if you do degree level chemistry. All you need to know is that when you get a high ppm jagged multiplet, you can bet your bottom dollar that there arr benzene protons.
    Ahh okay, thanks also I was wondering how the spectrum says there are 5 protons in the benzene environment but there are like 3 different proton environments on the benzene?
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    (Original post by Stiff Little Fingers)
    The spitting is because the benzene ring is substituted - by having the OC(O)CH(CH3)2 group, your hydrogens on the aromatic ring aren't all magnetically equivalent; you've got three different sets of hydrogens - ortho, meta and para to the substituted group. All of these will still come out around the same position, but will couple with each other, causing the splitting and a multiplet at ~7 corresponding to the aromatic ring.
    Right, thank you!
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    (Original post by white_o)
    Ahh okay, thanks also I was wondering how the spectrum says there are 5 protons in the benzene environment but there are like 3 different proton environments on the benzene?
    In benzene there's just one environment because all of the protium nuclei are magnetically equivalent. This will give you a peak area of 6 and no splitting.

    When you start adding substituents to the ring you're changing how the protium nuclei couple with each other. This causes new environments to be formed in the ring. These give slightly different signals, resulting in the splitting patterns.

    The relative peak area tells you how many protium atoms are attached to the ring, but to see where those atoms are you need to look at the splitting patterns. If there's only one substituent it's not necessary though because there's only one way it could be structured. But if you had something like Dimethylbenzene you'd see different splitting depending on where the substituents are. So 1,4-Dimethylbenzene for example, would have a doublet of doublets. 1,3 Dimethylbenzene on the other hand would just be plain nasty. I expect you'd get a doublet of doublets or a triplet, a singlet and two doublets all mushed up together in that small fuzzy mess of peaks. I'm not even going to think about 1,2-dimethylbenzene right now, it's too late for me to think this hard. The peak areas would be 4 for all three of them though.

    Edit: I just spent half an hour tapping my phone instead of sleeping and I just noticed that you already got an answer to your second question. Whoops.
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    (Original post by white_o)
    So there's a question with the H1 NMR Spectrum:

    Name:  chem 2.png
Views: 132
Size:  19.0 KB

    The mark scheme says the peak at 7.2 is from a benzene ring with 5 H's.
    It looks like Q5c on June 13 OCR A F324 (look it up), but it isn't as the OCR MS mentions the peak at 7.3.
    Which exam board have you taken the Q from?

    If it were OCR A, the spec says: "Candidates will be expected to identify aromatic protons from chemical shift values but will not be expected to analyse their splitting patterns." i.e. stop panicking.
 
 
 
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