Uncovering Chemical Structure with NMR: Analysis of C10H14 Hydrocarbon

  • Thread starter nautica
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In summary, the formula for the molecule described in the conversation is C10H14, specifically methylcumene. There is a benzene ring with a CH2-CH3 group branching off on the left and a CH3 group on the right. The proton peaks at 2.5, 1.5, 1.23, and 0.75 ppm correspond to the different groups in the molecule.
  • #1
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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  • #2
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
 
  • #3
Great, thanks. That is what I had drawn, but for some reason it seemed like something wasn't right.

thanks again
nautica
 

FAQ: Uncovering Chemical Structure with NMR: Analysis of C10H14 Hydrocarbon

What is NMR and how does it work?

NMR, or nuclear magnetic resonance, is a technique used to analyze the structure of molecules. It involves placing a sample in a strong magnetic field and then applying radio frequency energy to the sample. This causes the nuclei of the atoms in the sample to resonate, producing a unique signal that can be used to determine the chemical structure of the molecule.

Why is NMR important in chemistry?

NMR is important in chemistry because it allows scientists to identify the molecular structure of a compound, which is crucial for understanding its properties and behavior. It is a non-invasive and non-destructive technique, making it a valuable tool in the study of chemical structures.

What information can be obtained from NMR spectroscopy?

NMR spectroscopy can provide information about the number and types of atoms in a molecule, the connectivity of atoms, and the local environment of atoms within the molecule. It can also reveal the presence of functional groups and the overall symmetry of the molecule.

How is NMR used to analyze C10H14 hydrocarbon?

NMR can be used to analyze C10H14 hydrocarbon by first preparing a sample of the compound and placing it in a strong magnetic field. The sample is then subjected to radio frequency energy, causing the nuclei of the atoms in the compound to resonate and produce a spectrum. This spectrum can then be analyzed to determine the structure of the hydrocarbon.

What are the advantages of using NMR for analyzing chemical structures?

The advantages of using NMR for analyzing chemical structures include its non-destructive nature, its ability to provide detailed information about the molecular structure, and its versatility in analyzing a wide range of compounds. It is also a relatively fast and accurate technique, making it a valuable tool for chemists in research and industrial settings.

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