Ailurophile03
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why is splitting pattern if CH septumpletName:  image.jpg
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Ailurophile03
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Kian Stevens 😊
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Kian Stevens
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The CH proton is adjacent to six protons: there are two methyl groups, with each carbon having three protons, whilst the C=O carbon has no protons

Since there are six adjacent protons, the n+1 rule states that the CH environment will have a multiplicity of n+1, where n is the number of adjacent protons

Hence, the multiplicity is 6+1 = 7 (obviously) and so the splitting pattern is a septet
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Ailurophile03
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(Original post by Kian Stevens)
The CH proton is adjacent to six protons: there are two methyl groups, with each carbon having three protons, whilst the C=O carbon has no protons

Since there are six adjacent protons, the n+1 rule states that the CH environment will have a multiplicity of n+1, where n is the number of adjacent protons

Hence, the multiplicity is 6+1 = 7 (obviously) and so the splitting pattern is a septet
thank u very much
these nmr questions are driving me insane
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David Getling
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(Original post by Ailurophile03)
thank u very much
these nmr questions are driving me insane
This is something you absolutely must try and master. Unless things have changed recently you might well get a high mark question (12 to 16 marks) combining NMR with maybe some IR or mass spectroscopy.
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Kian Stevens
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(Original post by Ailurophile03)
thank u very much
these nmr questions are driving me insane
There's not really much revision, so to speak, involved in NMR (or any spectral analysis for that matter)

It's essentially just a case of 'practise makes perfect', and so you'll only improve when you do more questions and become familiar with actually interpreting spectra

For example, being able to determine an environment's multiplicity is an absolutely essential aspect of NMR interpretation, but the only way you'll become more able to do it is simply by practise
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