13-CNMR peaks

[SaintJude]

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How do I get the answer? Read EK and still have no clue...

 

[milski]

4,4? You are looking for differently connected carbons. For example, in the first compound you have one in the methyl group on top, one at the connection to the cyclic group, then the next two are the same - they're both next to the "connected" C from the cycle and to another two carbons which are the same, and then you have two more carbons at the bottom which also happen to be the same. The total is 4. When I talk about same carbons, it is the 'sameness' that we use to determine chirality centers - same atoms connected to the carbon in any direction.

 

[SaintJude]

Yes you are right. So we're only trying to identify the number of chemically equivalent groups. The reason it's 4 for the right hand molecule is b/c:

1st peak--for two carbons in the methyl substituents --located most upfield, right?
2nd peak--for the 1, 2 carbons that are attached to the methyl groups
3rd peaks--for the two "ortho" carbons to the methyl
4th peaks--for the 4,5 carbons --most downfield?

 

[milski]

The groups are correct. I think the upfield/downfield logic is correct too but I'm somewhat marginally good at these.


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[BerkReviewTeach]

How do I get the answer?

The best answer can be found by going to the AAMC topic list and reading through it to see that 13C NMR is not tested. They removed it years ago. If you are using used books, it's a good idea to match the AAMC list to the table of contents in the book you're using.

I hope that doesn't come across as a smart aleck answer, because I'm serious about checking that list. There are many used books out there, including BR books from as recently as a couple years ago, that have 13C NMR and shouldn't. It was removed from BR organic last year.

There are other topics such as benzene chemistry, alkene chemistry, and ether chemistry that still appear in several review books and shouldn't. There's is nothing to gain by studying something that's not on that list and hasn't appeared on the MCAT in at least five years.

 

[MedPR]

Yes you are right. So we're only trying to identify the number of chemically equivalent groups. The reason it's 4 for the right hand molecule is b/c:

1st peak--for two carbons in the methyl substituents --located most upfield, right?
2nd peak--for the 1, 2 carbons that are attached to the methyl groups
3rd peaks--for the two "ortho" carbons to the methyl
4th peaks--for the 4,5 carbons --most downfield?

Edit, nevermind. I was talking about IR

 

[milski]

What a coincidence, I was just rereading about spectroscopy 🙂

Stiffer bonds (double bonds, triple bonds, etc) resonate at higher frequencies = shorter wavelength = upfield (right).

Heavier atoms resonate at lower frequencies = higher wavelength = downfield (left).

Based on those rules, I think carbons 1 and 2 on the benzene will be the peak most upfield, followed by carbons 3 and 6, then carbons 4 and 5, and the methyl groups will be last since they have the longest bonds.

What about the benzene ring current thing? I don't even remember the name but in general the benzene ring was supposed to be very de-shielded. Or does that apply only to H NMRs? I have a final on this in a week and I'm still sort of shaky about it all. 😱