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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
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
Last edited by Kian Stevens; 1 year ago
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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
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
these nmr questions are driving me insane
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#6
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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