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Topic: help interpreting NMR?  (Read 4183 times)

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Offline zerofantasym

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help interpreting NMR?
« on: March 14, 2009, 11:15:14 AM »
Hi! I'm having trouble interpreting NMR spectrum. I think I know what the compound is but I cannot find an NMR for it online, so I have no idea if I got it right.

Here are the peaks:


1. 7.35 ppm: 5 H singlet
2. 4 ppm: 2H quartet
3. 3.5 ppm: 2H singlet
4. ~1.3 ppm: 3H triplet

I think I got that 7.35 is an aromatic 6-C ring; 4 and 1.3 is CH2-CH3. I think the 3.5 should be an alcohol or O-CH2-. I think that the compound is alpha-ethoxytoluene, but like I said,  I have no idea if it is right :S. Thank you!!!


Offline frenchy

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Re: help interpreting NMR?
« Reply #1 on: March 14, 2009, 03:53:39 PM »
Are you sure 7.35ppm is a singlet, if the ring is substituted once, should you not have 3 different signals?
You got a molecular formula?
PhD student in synthetic Inorganic and Supramolecular chemistry.

Offline zerofantasym

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Re: help interpreting NMR?
« Reply #2 on: March 14, 2009, 05:24:52 PM »
No I don't.
But yeah pretty sure it's a singlet.

Offline macman104

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Re: help interpreting NMR?
« Reply #3 on: March 15, 2009, 01:03:21 AM »
Are you sure 7.35ppm is a singlet, if the ring is substituted once, should you not have 3 different signals?
You got a molecular formula?
On lower field NMRs, if the aromatic peaks do not have a very wide separation (like the signlas PPM does not differ by much), then they may bleed together to look like a singlet.

Offline alphahydroxy

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Re: help interpreting NMR?
« Reply #4 on: March 15, 2009, 10:54:17 AM »
I think you're right with alpha-ethoxytoluene.

I thought it might be ethyl phenylacetate, but pulling up a spectrum for that indicates that it is not...

Offline Squirmy

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Re: help interpreting NMR?
« Reply #5 on: March 18, 2009, 01:43:53 AM »
I think your interpretation is pretty good, but...

In alpha-ethoxy toluene, both CH2's are deshielded by oxygen, but one is also deshielded by the ring. So, which one would you expect to be at higher ppm?

I think an IR would be helpful for this one...I'm guessing you'd see a nice C=O stretch.

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