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Topic: Spectroscopy intagration  (Read 5034 times)

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Offline Daisy8tyle

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Spectroscopy intagration
« on: February 01, 2017, 10:43:13 PM »
Basically I have an unknown compound and need help figuring out the IR integration I get an integration of 1:12:6:8:10 but these numbers seem high for integration. What should my next step be?
I've posted two pictures and the second of a close up so you can clearly see if it's a triplet,doublet, etc.
I KNOW:
It's a Ketone
has a phenyl group
boiling point of about 157.5 celcius

Offline hypervalent_iodine

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Re: Spectroscopy intagration
« Reply #1 on: February 01, 2017, 11:09:06 PM »
Sorry to have to tell you, but your assignments on splitting and what you think is in there in no way match your spectra.

Your quartets are not quartets. The things you've labelled triplets, do not look like triplets. You have no sextet. The peaks are too broad and unresolved for you to give any definite splitting assignment.

The peak at 7.26 is chloroform, which is (I assume) the solvent you ran this in? You don't need to integrate it.

I see no aromatic peaks in there, so I do not think that there is a phenyl group. Or alternatively, whatever you took an NMR of is rife with impurity / grease, and your compound peaks are hidden.

What makes you think there is a ketone?


Offline Daisy8tyle

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Re: Spectroscopy intagration
« Reply #2 on: February 02, 2017, 12:41:27 AM »
I think a ketone is present b/c I have an IR spectrum that has a strong peak at 1716 and there isn't a peak at 2700 and 2800 to indicate an aldehyde. I forgot to account for the chloroform peak, thank you. Would an integration of 6:3:3:6 be better?

Offline Daisy8tyle

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Re: Spectroscopy intagration
« Reply #3 on: February 02, 2017, 12:54:08 AM »
Sorry to have to tell you, but your assignments on splitting and what you think is in there in no way match your spectra.

Your quartets are not quartets. The things you've labelled triplets, do not look like triplets. You have no sextet. The peaks are too broad and unresolved for you to give any definite splitting assignment.

The peak at 7.26 is chloroform, which is (I assume) the solvent you ran this in? You don't need to integrate it.

I see no aromatic peaks in there, so I do not think that there is a phenyl group. Or alternatively, whatever you took an NMR of is rife with impurity / grease, and your compound peaks are hidden.

What makes you think there is a ketone?

Sorry I got confused again I came up with an integration of 2H : 1 H: 1 H: 1 H : 2 H

Offline hypervalent_iodine

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Re: Spectroscopy intagration
« Reply #4 on: February 02, 2017, 01:17:22 AM »
Sorry to have to tell you, but your assignments on splitting and what you think is in there in no way match your spectra.

Your quartets are not quartets. The things you've labelled triplets, do not look like triplets. You have no sextet. The peaks are too broad and unresolved for you to give any definite splitting assignment.

The peak at 7.26 is chloroform, which is (I assume) the solvent you ran this in? You don't need to integrate it.

I see no aromatic peaks in there, so I do not think that there is a phenyl group. Or alternatively, whatever you took an NMR of is rife with impurity / grease, and your compound peaks are hidden.

What makes you think there is a ketone?

Sorry I got confused again I came up with an integration of 2H : 1 H: 1 H: 1 H : 2 H

The peak heights don't match that. Do you not have access to processing software to do the integration?

Do you have any sort of mass spec data on your compound? Can you post the IR?

Offline Borek

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Re: Spectroscopy intagration
« Reply #5 on: February 02, 2017, 03:49:34 AM »
figuring out the IR integration

Something I never tried to do.
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