Uncovering Chemical Structure with NMR: Analysis of C10H14 Hydrocarbon

[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

 

[chem_tr]

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

 

[nautica]

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

thanks again
nautica