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 1o
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 13
C, all the protons are being relaxed after each excitation. (no need to wait)
C 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 13
Cs (each set) are being excited, 80% -> 64% -> 51.2% (not true in real cases ^^).