[Ranowa]

I'm trying to identify the structure of a C10H14O compound given H and C13 NMR readouts. I've got a signal with a ton of splitting for 6Hs, which means there has to be some equivalent Hs in there, but I have ten different signals for Carbons. How can I have equivalent hydrogens if there aren't any equivalent carbons? The compound has a ton of resonance structures that cause lots of splitting in the H readout; could one Carbon involved in two resonance structures somehow be sending two signals? Alternatively, while their are no equivalent Cs, I do have two signals coming in with .5 ppm difference. Right now I'm thinking it's two methyl groups and they're giving me the 6H signal, and there's just some very small difference between the two groups- is that even possible, or would any difference at all make the Hs come in as two different signals?

Edit: picture of both NMR readouts is my next post further down

[Babcock_Hall]

Individual resonance forms do not exist; only the blend of all the forms does.  I am not sure what you mean.

[billnotgatez]

@Ranowa
There is a feature in the edit/posting screen that lets you attach txt, jpg, gif, pdf, and png  files.
Just click on the plus next to
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[Ranowa]

Here's a picture of my NMR readout- everything handwritten is just my speculations, except for when I changed the original molecular formula (there was a typo) and when I labelled one of the H's signal splitting as 'singlet but special'. Apparently there's some long/more complex signal splitting going on there we haven't studied; professor told it is indicative of H-CpibondC-CH2. 

[Dan]

Here's a picture of my NMR readout- everything handwritten is just my speculations, except for when I changed the original molecular formula (there was a typo) and when I labelled one of the H's signal splitting as 'singlet but special'. Apparently there's some long/more complex signal splitting going on there we haven't studied; professor told it is indicative of H-CpibondC-CH2.

You have 9 signals in the ¹H - the footnote (*) says that the 6H multiplet is 3 overlapping signals.

You have 10 signals in the ¹³C. Remember, not all the carbons are necessarily bonded to hydrogens.

[Babcock_Hall]

You might calculate the degrees of unsaturation as a starting point.

[Vidya]

Rightly said sart with calculating the degrees of unsaturations ... it has a high degrees of unsaturation ..check for signals in H1NMR in the range of 7-8ppm ...aromatic ring yes or No ..Go step by step ...we will help you