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Topic: Combined spectroscopy problem  (Read 9081 times)

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Re: Combined spectroscopy problem
« Reply #15 on: June 10, 2021, 01:35:53 PM »
Sorry for the delay, this is what I've managed to do so far:

The M+2/M intensity is about 1/3 so Cl is present.
The M peak is odd so it contains N too (teacher said that if there's N it's only one for these exercises).
Then I divided 171 by 13 and I got C13H15 formula, then I needed to substract the Cl and the N so it is a CH2 for the N and C2H11 for the Cl.
This gives C10H2 remaining, so there should be more H than the initial guess. My guess was 6 or a multiple of 6 but I don't see how to "guess" it now. Do I just try 12, if it doesn't work 18 and so on? Or just add 6H to that formula? Maybe my initial guess is wrong but adding the H in the NMR spectra is 6 no matter what you do.

I'm stuck with the formula, I think if I manage to get it the remaining part should be easier.
Never, ever do that that division by 13 to find the C count. Instead, measure the ratio of the intensity of the molecular ion at [M+1] to the molecular ion [M].
Deduce approximate C atom count from this ratio.
You have a MW = 171, so N =odd count (1 atom) and Cl =1 atom.
171 - 14 N - 35Cl  = 122; no evidence of S or P or other halogen.
Set up a table as shown below,

C   H    O  = 122

10  2    0
9   14   0
8   26   0
8   10   1

Etc until you get a combination of numbers of C, H and O atoms that fits your given data.

Did you look up the "Number of rings & double bond" equivalents?
If not, you should.


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