[onemanlan]

I have had a pretty hard time grasping the material in Organic Chemistry 2 right now and 13C NMR is no exception by ant means. We have a homework question that goes like so:

Compound F, a hydrocarbon, with M+ = 96 and a M+1/M+ ratio is 7.92% in it's mass spec, undergoes reaction with HBr to yield compound G. Propose the structures for F and G, whose 13C NMR data is giving below. Be sure to indicate which piece of spectral data are reveals to you for partial credit.

Compound F:
Broadband Decoupled 13C NMR: 27.6, 29.3, 32.2, 132.4
DEPT - 135: Positive peak 132.4, Negative Peaks: 27.6, 29.3, 32.2

Compound G:
Broadband Decoupled 13C NMR: 25.1, 27.7, 39.9, 56.0
DEPT - 135: Positive peak: 56.0, Negative peaks: 25.1, 27.7, 39.9

We're given a sheet of 13C NMR trends, but it's given in the typical range spectrum that generally overlap one another. What I do not get is how to put numbers as accurate to one decimal place into the given ranges.

So lets give what I do know about these:

13C NMR - Tells us the number of carbons involved and what bonds the carbons might have
DEPT 135 - Positive peaks can be CH or CH3. Negative peaks are CH2's.

Compound F:
M+ = 96
M+1/M+ ratio = 7.76
27.6 - R₂CH₂ or RCH₃
29.3 - R₂CH₂
32.2 - R₃CH or R₄C
132.4 - R₂C=CR₂

Compound G:
25.1
27.7
39.9
56.0


I'm really not sure how to even go about this problem any further. I see 4 peaks on the 13NMR, but I can't quite get it through my head how you get M+ = 96 out of that. I'm further confused about how to interpret the DEPT 135 data into a better structure. A general since of direction would be helpful, because I feel like I'm shooting for a bull's eye in the dark.

[Kryolith]

Welcome to the forum 

Compound F:
Broadband Decoupled 13C NMR: 27.6, 29.3, 32.2, 132.4
DEPT - 135: Positive peak 132.4, Negative Peaks: 27.6, 29.3, 32.2

13C NMR - Tells us the number of carbons involved and what bonds the carbons might have

It tells you the number of chemical equivalent carbons!

DEPT 135 - Positive peaks can be CH or CH3. Negative peaks are CH2's.

And what kind of carbon atoms give no peak?

27.6 - R₂CH₂ or RCH₃

No. Read your own post about DEPT 135.

29.3 - R₂CH₂

OK.

32.2 - R₃CH or R₄C

No. Read your own post again and try to answer my question above.

132.4 - R₂C=CR₂

OK.

Now try to figure out the formula of the compound.

[onemanlan]

I don't quite know what makes carbons chemically equivalent. I have a general idea of how to distinguish between chemically equal protons and carbons, but it doesn't really mean a whole lot to me when visualizing it because I do not know how to put that data into the works to help build the molecule.

Quaternary carbons do not peak on DEPT 135*.

So that means that 27.6 cannot be a R₃CH because it's peak is negative on the DEPT 135.
32.2 could not be R₄C because there would be no peak.

Does the M+1/M+ ratio not allow us to find the # of carbons in the compound by dividing the percentage by the 1.11% for the 13C isotope. In this case it would say there are 7 carbons involved. 3 of which do not show up on the DEPT 135 meaning they're quaternary...? I'm just speculating at this point.

This is proving to be much more difficult than I expected. I'm still working on it right now. Thanks for the help and I'll take as much as I can get.

Thank you for the forum welcome too.

[Kryolith]

Quaternary carbons do not peak on DEPT 135*.

So that means that 27.6 cannot be a R₃CH because it's peak is negative on the DEPT 135.
32.2 could not be R₄C because there would be no peak.

Does the M+1/M+ ratio not allow us to find the # of carbons in the compound by dividing the percentage by the 1.11% for the 13C isotope. In this case it would say there are 7 carbons involved.

You're absolutely right. Can you tell the formula now?

3 of which do not show up on the DEPT 135 meaning they're quaternary...? I'm just speculating at this point.

How many DEPT 135 signals would you get for e.g. butane, which doesn't contain any quaternary carbon atoms?

[onemanlan]

It took me a little while, but I think I might have the answer for compound F.

H₂C=CH-C(CH₃)₂-HC=CH₂

So when I was looking at this I figured that since the decoupled 13C NMR spectrum shows 4 chemically equivalent carbons. Through the M+1/M+ ratio I found that there were 7 carbons in the molecule. If it has 7 C we can find that C₇H₁₂ = 94 and that the degrees of unsaturation = 2. 2 degrees means two double bonds or one triple bond. Since triple bond didn't show up in the 13C NMR spectrum it rules that one out. From there I pieced it together. With the other information I had regarding the chemical shifts. I hope it's right. Thanks for your help anyway.

[macman104]

Won't always have your compound, but the following is a great site for spectroscopy:

http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/cre_index.cgi?lang=eng

[Kryolith]

Great! A link to a database is what you need to solve spectras on your own 

[Kryolith]

It took me a little while, but I think I might have the answer for compound F.

H₂C=CH-C(CH₃)₂-HC=CH₂

This compound would have 3 DEPT 135 signals (positive or negative?) and 4 ¹³C-{¹H} signals. Try to understand why.

...means two double bonds or one triple bond...

C₃H₆ must not be propene, and what about the degree of unsaturation of benzene?

EDIT: Sorry for answering my own post, I clicked "Quote" but wanted to edit my previous post and insert the quote...

[macman104]

Great! A link to a database is what you need to solve spectras on your own 

I didn't say that.  However, it's a great tool for verifying your spectra, or looking at similar spectra if you are truly stuck.  Also, if you really are stuck, I found it extremely helpful to look at spectra I couldn't solve and work backwards to understand how the peaks came about.  Jumping to conclusions is awesome 

[onemanlan]

It took me a little while, but I think I might have the answer for compound F.

H₂C=CH-C(CH₃)₂-HC=CH₂

This compound would have 3 DEPT 135 signals (positive or negative?) and 4 ¹³C-{¹H} signals. Try to understand why.

Oh, well I must have not been thinking, but it would only show 3 DEPT 135 signals. 2 positive bands for the CH's and CH₃'s, then 1 negative band down for the CH₂'s. It would show for ¹³ signals because there are 4 chemically equivalent carbons in the molecule.

I had my test on this today. I'm not sure how well I did, but I'm not giving up on this problem. I need to learn this stuff. It's though, but it's sort of fun piecing molecules together based on bits and pieces of information.

[Kryolith]

I hope you were successful at your test. 

So can you tell the structure now? And what about the meaning of the degree of unsaturation?

[onemanlan]

To tell you the truth even after that test was said and done I don't understand ¹³C NMR very well at all. I have a better idea of what chemical equivalence is and how it works, but my teacher didn't go over carbon NMR too much. It's kind of a bummer too because my book says it's easier than Proton NMR, but I think thats way, way easier to interpret than former.

As for DU I think it saved my ass on the test. Being able to figure out DU and then propose a proper molecular structure based on a few fragments of data made the test much easier. Several problems we had contained IR, H NMR, and ¹³C NMR.

DU = [(2x + 2) - y)]/2 where x = # C and y = # H.

DU = 0 -> Alkane chain
DU = 1 -> One double bond or ring structure
DU = 2 -> Two double bonds, one triple bond, or two ring structures, double bond in ring structure...
DU = 3 -> Two triple bonds?
DU = 4 -> Benzene

[Kryolith]

To tell you the truth even after that test was said and done I don't understand ¹³C NMR very well at all.

Practice makes perfect 

DU = 2 -> Two double bonds, one triple bond, or two ring structures, double bond in ring structure...
DU = 3 -> Two triple bonds?
DU = 4 -> Benzene

DU = 2 -> "two ring structure" is called bicyclic
DU = 3 -> e.g. one triple and one double bond
DU = 4 -> e.g. two triple bonds