Uncovering Chemical Structure with NMR: Analysis of C10H14 Hydrocarbon

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SUMMARY

The discussion focuses on the analysis of the hydrocarbon C10H14 using NMR spectroscopy. Key findings indicate the presence of a benzene ring (C6H5) and a side chain consisting of a CH2-CH3 group, leading to the conclusion that the compound is methylcumene (C6H5-CH(CH3)-CH2-CH3). The NMR peaks at 7 ppm, 2.5 ppm, 1.5 ppm, 1.23 ppm, and 0.75 ppm correspond to specific hydrogen environments that confirm this structure. The analysis highlights the importance of interpreting NMR data accurately to deduce molecular structures.

PREREQUISITES
  • Understanding of NMR spectroscopy and chemical shifts
  • Familiarity with interpreting NMR peak patterns (multiplets, quartets, singlets)
  • Knowledge of organic chemistry nomenclature and structure
  • Experience with hydrocarbon structures and functional groups
NEXT STEPS
  • Study the principles of NMR spectroscopy, focusing on chemical shifts and splitting patterns
  • Learn how to interpret NMR spectra for complex organic molecules
  • Research the structural characteristics of methylcumene and its derivatives
  • Explore software tools for visualizing NMR data and molecular structures
USEFUL FOR

Chemists, organic chemistry students, and researchers involved in molecular structure determination using NMR spectroscopy will benefit from this discussion.

nautica
This should be easy, but I am missing something.

5 hydrogens at 7 ppm seeing what appears to be an overlap of several
1 hydrogen at 2.5 ppm seeiing 5
2 hydrogen at 1.5 ppm seeing 4
3 hydrogen at 1.23 ppm seeing 1
3 hydrogen at .75ppm seeing 2

The formula is C10H14

It appears I have a benzene ring but I am unable to get the side chain, which I am asumming there is only 1 and with the 3 seeing 2 and 2 seeing 3, there should be a CH2CH3.

Thanks
nautica
 
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Hello,

Your notation is not familiar with me, but I will try to predict the formula from your data.

1) The 7-ppm-centered peak system is due to a benzene ring, a plain phenyl substituent causes a characteristic multiplet. So we have a C6H5 for now.

2) A pentet at 2,5 ppm (I think you want to mean this by saying "seeing 5") having one proton means there is a branching CH group, one methyl and one ethyl; a total of five protons seen with only one proton on hand is possible this way.

3) A quartet at 1,5 ppm having two protons indicate a CH2-CH3 group (on the left).

4) A singlet peak at 1,23 ppm with three protons clearly shows that this group is a CH3.

5) A doublet at 0,75 ppm with three protons is indicative of CH2-CH3 group (on the right).

So the molecule you are seeking is methylcumene, C6H5-CH(CH3)-CH2-CH3, which can far more easily seen from the attachment .zip file containing a tif image file.

Regards
chem_tr
 
Great, thanks. That is what I had drawn, but for some reason it seemed like something wasn't right.

thanks again
nautica
 

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