alde123
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On the markscheme (http://www.ocr.org.uk/Images/176056-...lysis-june.pdf) for question 5ciii on this paper (http://www.ocr.org.uk/Images/175441-...d-analysis.pdf), I don't understand why the proton shift at 2.7 can't be the benzene ring, seeing as there is a benzene ring in the answer. I also don't get why the peaks at 120 on the Carbon NMR can't also represent a C=C double bond.

Please can someone explain this to me.

Thanks
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charco
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(Original post by alde123)
On the markscheme (http://www.ocr.org.uk/Images/176056-...lysis-june.pdf) for question 5ciii on this paper (http://www.ocr.org.uk/Images/175441-...d-analysis.pdf), I don't understand why the proton shift at 2.7 can't be the benzene ring, seeing as there is a benzene ring in the answer. I also don't get why the peaks at 120 on the Carbon NMR can't also represent a C=C double bond.

Please can someone explain this to me.

Thanks
Look at the integrals. There is only 1 proton in the signal at 2.7
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alde123
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(Original post by charco)
Look at the integrals. There is only 1 proton in the signal at 2.7
But in the datasheet it shows the hydrogen attached to a carbon which is attached to a benzene ring so isn't there still only one proton in that particular environment.
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charco
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(Original post by alde123)
But in the datasheet it shows the hydrogen attached to a carbon which is attached to a benzene ring so isn't there still only one proton in that particular environment.
All of the aromatic protons will appear at a similar delta value. The integral will show all of these protons. If the benzene ring is monosubstituted there must be an integral of five, if it's disubstituted an integral of four etc.

Your signal at 2.7 shows an integral of 1 - hence 1 proton. The only way it could be aromatic is if there are five substituents on the ring.
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alde123
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(Original post by charco)
All of the aromatic protons will appear at a similar delta value. The integral will show all of these protons. If the benzene ring is monosubstituted there must be an integral of five, if it's disubstituted an integral of four etc.

Your signal at 2.7 shows an integral of 1 - hence 1 proton. The only way it could be aromatic is if there are five substituents on the ring.
Thanks, I can see that now. So for the carbon NMR is that to do with the height of the peak to determine which structure on the data sheet is right?
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