NMR spectrum

[NA19]

---

Members don't see this ad.

The answer is B. Can someone explain?

 

[The Kraken]

Carbonyl pulls more electron density than a double bond, making II correct. What is your question? Why only two or why two is correct?

 

[monkeyvokes]

1 is most deshielded because it's nearest to oxygen

carbon two's neighbors 👎 each have one proton, n+1 =3 (a triplet)

 

[NA19]

I'm confused about why I and III are incorrect.

 

[monkeyvokes]

I'm confused about why I and III are incorrect.

does my explanation help
?

 

[The Kraken]

I'm confused about why I and III are incorrect.

N+1 rules applies to sp3 carbons, not sp2 for answer 3. The NMR of the compound technically has aromatic hydrogens with different coupling constants but that's not really mcat level nmr analysis. I would have said 6 at the mcat level, but they arent exactly the same...

http://www.sas.upenn.edu/~genette/NMRcoupling.pdf

 

[monkeyvokes]

N+1 rules applies to sp3 carbons, not sp2 for answer 3. The NMR of the compound technically has aromatic hydrogens with different coupling constants but that's not really mcat level nmr analysis. I would have said 6 at the mcat level, but they arent exactly the same...

http://www.sas.upenn.edu/~genette/NMRcoupling.pdf

Why doesn't n+1 apply to carbon 2?
As I understand it, the coupling constant J is for aromatic hydrogens. Ortho hydrogens split peaks more than meta hydrogens. If you have two ortho hydrogens, they can split the peak by the same distance twice and form a triplet.
I don't think any of that applies to carbon 2, does it?

 

[The Kraken]

Why doesn't n+1 apply to carbon 2?
As I understand it, the coupling constant J is for aromatic hydrogens. Ortho hydrogens split peaks more than meta hydrogens. If you have two ortho hydrogens, they can split the peak by the same distance twice and form a triplet.
I don't think any of that applies to carbon 2, does it?

Regarding choice III) Carbon 2 is an sp2 carbon and iirc n+1 does not apply for sp2 hybridized atoms.

Regarding choice I on why there are more than 6 nonequiv H's) Look up coupling.

 

[monkeyvokes]

Regarding choice III) Carbon 2 is an sp2 carbon and iirc n+1 does not apply for sp2 hybridized atoms.

Regarding choice I on why there are more than 6 nonequiv H's) Look up coupling.

I agree there are more than six, i labeled them in the image above.
I was under the impression that n+1 only didn't apply to sp2 hybridized atoms if they were in a benzene ring (and perhaps other aromatics). I have always used n+1 for alkenes and never had a problem. can't seem to find your position in google but that doesn't mean i'm correct.

 

[NA19]

monkeyvokes: Position 4 can't have an H, the carbon already has four bonds, doesn't it? So then how are there 6 non-equivalent H's?

Also, why don't sp2 hybridized C's following the n+1 rule? What rule do they follow instead?

 

[NA19]

Can anyone help with this?

 

[gettheleadout]

This question was addressed here: http://forums.studentdoctor.net/showthread.php?t=982246

I'm aware of stereochemical effects on J-values for coupling, and I would not have expected that "the pi-conjugation of the molecule slows rotation of the phenyl ring, eliminating its symmetry." Honestly that sounds like BS.

 

[NA19]

Okay, so III is wrong because complex splitting is going on, right? Which means you will get a doublet of doublets...aka a multiplet. I still don't understand why saying that there are 6 non-equivalent Hs is wrong. Is it because of the direction of the carbonyl upwards which is spatially closer to the Hs higher up on the benzene ring?

 

[JingleChips]

Okay, so III is wrong because complex splitting is going on, right? Which means you will get a doublet of doublets...aka a multiplet. I still don't understand why saying that there are 6 non-equivalent Hs is wrong. Is it because of the direction of the carbonyl upwards which is spatially closer to the Hs higher up on the benzene ring?

According to the answer key posted in the other thread, it says because the whole molecule is conjugated (test it yourself - start pushing electrons from the far end of the phenyl), the rotation of the phenyl ring is slowed and the symmetry is eliminated. I don't think the MCAT will be testing you on this....

 

[gettheleadout]

According to the answer key posted in the other thread, it says because the whole molecule is conjugated (test it yourself - start pushing electrons from the far end of the phenyl), the rotation of the phenyl ring is slowed and the symmetry is eliminated. I don't think the MCAT will be testing you on this....

Agree wholeheartedly here, this wouldn't even be tested in my organic courses (and never was).