[jamesrb]

So the format for an upcoming test in my Organic Chemistry II class is that we will be asked to deduce the structure of an unknown compound. We will not be given the chemical or empirical formula. We will be shown four graphs with certain pieces of information:

Mass spec (M+ will be given to us)
IR
HNMR (ppm and integration ratios will be given to us)
CNMR (ppm and multiplicity will be given to us)

I think what I am having the most trouble with is understanding what I need to do with the integration values for the HNMR spectra and the multiplicity for the CNMR spectra and how these can help me determine the structure.

Starting with CNMR and multiplicity. The multiplicity tell you how many protons are attached to that carbon, correct? Would a carbon with a triplet multiplicity would be a CH2?

Regarding HNMR spectroscopy; each peak tells you the different types of hydrogen in a compound, or how many hydrogen are in their own unique chemical environment. How the integration values tell us the ratio of each type of hydrogen there are. So if I have a spectra with 3 peaks at 1.78, 3.91 and 4.85 I would have a compound that has 3 unique hydrogen atoms. Now say if the integration for each of those peaks is say 1 for 1.78, 2 for 3.91 and 1 for 4.85 I would have a ratio of 1:2:1 for the hydrogen atoms.

Is there anyway that I can use the CNMR spectra and multiplicity plus the HNMR spectra and integration to figure out an empirical or molecular formula? I am going to try to get a scanned copy of a practice problem later today.

[discodermolide]

Lots of questions here.
You are correct with the ¹³C a CH₂ will be a triplet in the off-resonance spectrum.
Your integration is WAY off.
The integration of a signal supplies the number of protons under each signal. In your example. integral of 1 for peak at 1.78 means 1 proton, integral of 2 for peak 3.91 means 2 protons, and 1 for 4.85 means 1 proton.
Do not get it confused with coupling (J values).
All it does is tell you how many protons are giving that signal.
The ¹³ C and ¹H tell you what magnetic environment the atoms are in. i.e. a OCH₃ signal will be different from a O=C(CH₃)₂ signal. In both sorts of spectra.
So you can see the bits that make up the ,molecule. The MS and IR give you info as to the mass and functional groups respectively. Put it all together and see what comes out.
See for an example http://www.chemicalforums.com/index.php?topic=62628.0

[jamesrb]

So if I see a signal with an integration of 1 but there are 5 peaks total, 1 big and 4 "sub peaks" (2 to the right of the big peak, 2 to the left), I should assume that it is the signal for 1 hydrogen being split by 4 neighboring hydrogen?

[discodermolide]

Yes. The integration gives the area under the signal which is directly proportional to the number of protons.
See:

http://www.wfu.edu/~ylwong/chem/nmr/h1/integration.html

[jamesrb]

I have the spectra for a practice problem and have posted them below along with my explanation of how I worked through the problem. I would appreciate any feedback or tips and tricks anyone feels like providing.

MS & IR spectra:


HNRM & CNMR:


So the molecular ion is labeled for me at 99. This is odd, so the compound has an odd number of nitrogen. Looking at the IR spectra I see what I believe to be two weak absorptions that indicate an N-H bond(s) at ~3300-3400 cm^-1. I also believe that the strong absorption at ~1700 is an indication that we have a C=O double bond present. Moving on to the CNMR spectra I can see that we must have at least 5 carbons. The carbon at 17.7 ppm is a triplet and therefore is a CH₂. Then the 29.34 ppm carbon is a quartet thus it is CH₃. 30.7 carbon is a triplet so it's CH₂. Same for the carbon at 49.4, another triplet, CH₂. Finally the carbon at 174.7 is a singlet. Since carbon never bonds with another carbon and calls it a day, is it safe to assume this is my C=O double bond? It has to be bonded to something, and that is waaay up at 174.7 ppm, hence that makes me think it is my C=O. So now all told we have account for:

CH₂, CH₂, CH₂, CH₃, C=O, and 1 N-H. Adding that up, we have 100. That is not the weight of my molecular ion, it is the my M+1. I have gone over my budget and am a bit confused. Should I not include the N-H, meaning the N is attached to one of my already accounted for H atoms? Taking away one H would put he at 99.

The integrations for my hydrogen atoms are 1, 1, 1.5 and 1. We can't have 1.5 hydrogen, and looking at what I have so far I would say the actual number of hydrogen atoms is a multiple of 2. 2+2+3+2=9. So we have the formula C₅H₉NO.

Calculating my degrees of unsaturation (C-(H/2)-(X/2)+(N/2)+1) = (5-(9/2)-(0/2)+(1/2)+1) = 2. So I could have 2 double bonds, or 2 rings, 1 ring and 1 double bond, or 1 triple bond.

 Trying to put the compound together proved to difficult for me and I had to give up. I looked at the answer and it turns out it was N-Methyl-2-pyrrolidone:
http://en.wikipedia.org/wiki/Methylpyrrolidone

Other then getting stumped at the end and not being able to put all the pieces together, if anyone has any input to add, any oversights I have made or noticed any faulty logic I followed, I would greatly appreciate the input. I am going to do be doing these all night, I have 3-4 more to do. I would love any tips on telltale IR absorptions I missed. I can see that most of what I didn't bother with on the IR spectra was in the fingerprint region, something I know little about and don't feel comfortable messing around with.

EDIT: After looking at the structure of N-Methyl-2-pyrrolidone I think I may have messed up. Is there no N-H bond in that compound? I figured there was...

[jamesrb]

MS & IR spectra:

HNMR & CNMR:


Fortunately I am given the mass percent composition on this one, therefore I can come up with an empirical formula. Assuming 100g of a sample
54.53% carbon
9.15% hydrogen
36.32% oxygen

54.53/12=4.54
9.15/1=9.15
36.32/16=2.27

Dividing by 2.27:
4.54/2.27=2
9.15/2.27=4
2.27/2.27=1

(C₂H₄O₁)_n

Looking at the IR spectra... I don't recognize a damn thing, other then C_sp³-H. Looking at my CNMR spectra I can see I have 2(or some multiple of 2) carbons, a quartet(CH₃) and a doublet(CH). HNMR tells me the same thing, 4 hydrogen or some multiple of that, right? 3 of one type, 1 of another. I'm looking all around, at the IR, CNMR and HNMR to give me an idea of how the O is.

I had to cheat and look again. It turns out it is 2,4,6-trimethyl-1,3,5-trioxane (paraldehyde):
http://en.wikipedia.org/wiki/Paraldehyde

I came up with Acetaldehyde: CH₃-CHO:
http://en.wikipedia.org/wiki/Acetaldehyde
Looking at the fingerprint region I can now see the C-O bond at 1100 but is there anything I could have done to recognize that it was paraldehyde?

[jamesrb]

More to come later.

[discodermolide]

The second problem: you can see it is not acetaldehyde by the missing aldehyde signal at about 9.5. The spectrum is very simple therefore you have a symmetrical molecule. You can see that the CH and the CH₃ are coupled. From the chemical shift of the CH you can say it is next to an electron withdrawing atom, probably oxygen, so you have a CH₃-CH-(O)₂ group in your molecule. (  O-CH(CH₃)-O group). You have the empirical formula.

I think you are on the right track with your analysis of the data. You just need to have more practice in recognising what the chemical shifts mean. But that will come.