a. If you have 1 million protons, how many of them will undergo a transition fro

Question

a. If you have 1 million protons, how many of them will undergo a transition from the low to high energy spin state if the proper amount of energy is added?

b. Now answer the same question for 1 million C-13 nuclei. This will require you to calculate the spin state population ratios for C-13, as you see done on the video for protons. You should find that the percentage difference between the low and high energy spin states are even smaller for C13.

c. How many times less sensitive is C-13 versus proton, assuming you have equal numbers of spin active nuclei?

d. Of course, you DON’T have equal numbers of spin active nuclei. If you have 1 million carbon nuclei (average, ordinary, natural carbon), how many nuclei actually undergo a transition from the low to high energy spin state if the proper amount of energy is added? (This answer here should be a surprise).

e. Finally, repeat the problem, but with the C=O stretch in the IR spectrum. If you had one million of these carbonyls at 1726 cm-1, how many would you see undergo the transition from the low to the higher energy vibrational state when the proper amount of energy is added?

f. That last calculation should really bring home the differences between NMR and the other spectroscopies. Contemplate it a moment, and realize what a miracle it is we can do NMR at all.

Explanation / Answer

a)

Most organic compounds exhibit proton resonances that fall within a 12 ppm range , and it is therefore necessary to use very sensitive and precise spectrometers to resolve structurally distinct sets of hydrogen atoms within this narrow range. In this respect it might be noted that the detection of a part-per-million difference is equivalent to detecting a 1 millimeter difference in distances of 1 kilometer.

c)

The low natural abundance of 13C has three principal consequences:

  1. It can be much more time-consuming to obtain 13C than 1H NMR spectra, especially if quantities or solubility is limited. Whereas an 1H spectrum on 1 mg of a typical organic compound can usually be obtained in 15-30 minutes of spectrometer time, it might take several hours to obtain a much lower quality 13C spectrum on the same sample. This is because, in addition to the 1% natural abundance of 13C (vs >99% for 1H), the Boltzman population difference between the and states is smaller (the enery gifference is only 1/4 as large), and the weaker magnetic dipole makes signals inherently harder to detect. Thus, unless your compound does not have useful 1H signals, you will usually measure many more 1H than 13C NMR spectra.

  Inherent sensitivity: 1.59% (1H = 100). Sens. = (C/H)3

  Actual sensitivity: 0.017% (1/5700)

  2. The effects of 13C nuclei on spectra of other nuclei (e.g., 1H, 19F, 31P) are very minor. Each proton signal is surrounded by 13C satellites separated by 1JC-H (typically 120-150 Hz), each with an intensity of 0.5% of the central peak. The central peak arises from the 98.9% of 12C which is NMR transparent. The 13C satellites can be readily detected for sharp peaks. In addition to being useful for measuring proton-carbon couplings, these satellites can sometimes be used for measurement of JHH between equivalent protons.

  3. Coupling between carbons (JCC) is not usually observed, because two adjacent 13C nuclei occur in only 1.1% of the carbons. There are thus 13C satellites on the carbon peaks (each about 0.5% of the intensity of the main peak), in the same way that there are 13C satellites on proton spectra. The couplings can be measured directly with some difficulty by accumulating many scans on a very concentrated sample, but a better way is to use one of the multi-pulse 2D experiments (e.g., INADEQUATE) which nulls the central peaks due to adjacent 12C atoms. Even so, large samples and long acquisition times are required.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.