[nate]

Can anyone explain to me why ¹³C NMR imaging only "sees" 6 carbons when viewing 2,3-Dimethylpentane and not 7.  I know that the two Me groups aren't viewed as the same and the first and fifth C cannot be viewed as the same.  I see no plane of symmetry unless I am just looking at it incorrectly.  Thanks for any help.

[mike]

Are you looking at a particular spectrum of 2,3-dimethylpentane? Maybe yours has the methyl groups overlapped?

[nate]

In this example, it is simply the stick depiction of the molecule.

[Mitch]

here is why

[nate]

Thanks, that helps.  Where did you get that info from?  Our textbooks didn't go that indepth to it and what you showed me would have helped make sense out of a somewhat confusing subject.  And if you have any tips on figuring it out when you don't have that type of information would be helpful also.

[Mitch]

I have a program called chemdraw that can make a prediction on what the peaks should look like. I would worry more about learning the material solid, afterwards you can cheat like the rest of us do.

[nate]

Thanks.  I am working on it.  Seems like everytime I think I have it down, I get tripped up on something like this.  Thanks again.

[mike]

Here is one where the methyl groups are separated..

[sdekivit]

2,3-dimethylpentaan has two equivalen CH3-groups, namely the first CH3 is chemically equivalent to that on the 2nd carbon.

--> the peak intensity of this CH3-group wopuld be twice as that of the other carbonpeaks.

[Organishe]

Mike (or anyone else that can answer), is the reason the two carbons labeled 3 show up as two peaks because they are diastereotopic due to the stereocenter at carbon 6?  If not, then why, since they should be magnetically equivalent (excluding stereoisomerism).

thanks

[Winga]

Mike (or anyone else that can answer), is the reason the two carbons labeled 3 show up as two peaks because they are diastereotopic due to the stereocenter at carbon 6?  If not, then why, since they should be magnetically equivalent (excluding stereoisomerism).

thanks

The two carbon3s experience different chemical environments due to the adjacent stereocenter as you mentioned (try to rotate the *C-C bond).
As these two carbon3s are chemically non-equivalent, so they are also magnetically non-equivalent.

[Winga]

2,3-dimethylpentaan has two equivalen CH3-groups, namely the first CH3 is chemically equivalent to that on the 2nd carbon.

--> the peak intensity of this CH3-group wopuld be twice as that of the other carbonpeaks.

The peak intensity does not proportion to the ratio of no. of carbons from each set in ¹³C NMR spectrum. It is because the relaxation time of ¹³C is much longer than proton's and non-equivalent ¹³Cs will have different relaxation time. (and more reasons...)

For time saving, practically, we will not wait for all ¹³Cs to relax before next excitation. Therefore, the peak intensity does not relate to their ratio.

The intensity for ¹³C NMR spectrum is related to the no. of adjacent protons (for ¹³C NMR) attached to ¹³Cs due to Nuclear Overhauser Effect (NOE).
More the adjacent protons, higher the intensity. (for ¹³C NMR)

The peak intensity of ¹³C spectrum predicted by ChemDraw is not reliable, but chemical shift.

[sdekivit]

then you have other NMR-basics then me. Because two CH3-groups of the same chemical environment are present, these show as one peak, but with higher intensity.

It's the same with H-NMR, but here there is a direct relation between amount of H's and peak intensity. With C-NMR this is indeed less --> but peak intensity for these CH3-groups is higher.

it's just the same with the following compound:

H3C - CH2 - CO - O - CH2 - CH(CH3)2

[Winga]

then you have other NMR-basics then me. Because two CH3-groups of the same chemical environment are present, these show as one peak, but with higher intensity.

It's the same with H-NMR, but here there is a direct relation between amount of H's and peak intensity. With C-NMR this is indeed less --> but peak intensity for these CH3-groups is higher.

it's just the same with the following compound:

H3C - CH2 - CO - O - CH2 - CH(CH3)2

The chemical environment of two methyl groups (H or C) should not be the same as there is no plane of symmetry in this molecule.

Why there is one peak?
Obviously, their chemical shifts are too close to each other that give you one peak.

It is true that the peak intensity increases as the no. of equivalent carbons increases, but not proportion to the others.

e.g. if the molecule has 4 equivalent quaternary cabons, and only one primary carbon (-CH₃), the ratio of peak intensity is not 4:1, maybe the peak intensity of 1^o C is much higher than that of quaternary cabons due to NOE (effect).

The compound that you mentioned, the intensity of these two carbons is the highest as they are primary and have the highest number. However, you cannot find the ratio of different sets of carbons.

e.g. The ratio of these 2 carbons and the carbonyl carbon that you find out from the peak intensity is not 2:1, the carbonyl carbon (quanternary) has a very very small instensity.

For proton NMR, as the relaxation time of protons is very very short compare to ¹³C, all the protons are being relaxed after each excitation. (no need to wait)

For ¹³C NMR, assume there is 80% of one set of carbons relaxed after an excitation and before next excitation, the other sets of carbons will not have the same amount to be relaxed, e.g. 60%, 70%, 90%. Moreover, there will be less and less ¹³Cs (each set) are being excited, 80% -> 64% -> 51.2% (not true in real cases ^^).

[movies]

You should also account for the low natural abundance of ¹³C relative to that of ¹H.  If you have a sample containing a high level of ¹³C then you can integrate the spectrum and get reasonable numbers, but that is because the NOEs that Winga mentioned are responsible for a smaller percentage of the observed carbon signal.

Also, with respect to the spectrum that Mitch posted, the ChemDraw NMR prediction program can't predict splitting of diastereotopic signals, that's why the two "3" carbons look the same.  In reality, these would have slightly different signals as shown in the spectrum that Mike posted.