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The 13 C nucleus is present in only 1.08% natural abundance. Therefore, acquisition of a spectrum usually takes much longer than in 1 H NMR. The magnetogyric.

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Presentation on theme: "The 13 C nucleus is present in only 1.08% natural abundance. Therefore, acquisition of a spectrum usually takes much longer than in 1 H NMR. The magnetogyric."— Presentation transcript:

1 The 13 C nucleus is present in only 1.08% natural abundance. Therefore, acquisition of a spectrum usually takes much longer than in 1 H NMR. The magnetogyric ratio of the 13 C nucleus is about 1/4 that of the 1 H nucleus. Therefore, the resonance frequency in 13 C NMR is much lower than in 1 H NMR. (75 MHz for 13 C as opposed to 300 MHz for 1 H in a 7.04 Tesla field). At these lower frequencies, the excess population of nuclei in the lower spin state is reduced, which, in turn, reduces the sensitivity of NMR detection. Unlike 1 H NMR, the area of a peak is not proportional to the number of carbons giving rise to the signal. Therefore, integrations are usually not done. Each unique carbon in a molecule gives rise to a 13 C NMR signal. Therefore, if there are fewer signals in the spectrum than carbon atoms in the compound, the molecule must possess symmetry. When running a spectrum, the protons are usually decoupled from their respective carbons to give a singlet for each carbon atom. This is called a proton-decoupled spectrum. Introduction to C-13 NMR

2 Carbon-13 Chemical Shift Table C  C triple bonds

3 Alkane: 2-methylpentane

4 Alcohol: 2-hexanol

5 Alkyl Halide: 3-bromopentane

6 Alkene: 1-hexene

7 Aromatic Ring: eugenol

8 Carboxylic Acid: pentanoic acid

9 Ester: ethyl valerate

10 Amide: pentanamide

11 Ketone: 3-methyl-2-pentanone

12 Aldehyde: 2-methylpentanal

13 Symmetry in C-13 NMR Each unique carbon in a molecule gives rise to a 13 C NMR signal. Therefore, if there are fewer signals in the spectrum than carbon atoms in the compound, the molecule must possess symmetry. Examples:

14 Enantiotopic vs Diastereotopic CH 3 ’s * * * * *

15 Determine the number of signals in the proton-decoupled C-13 NMR spectrum of each of the following compounds:

16 ppm Carbon # Carbon-13 NMR Spectrum of Geraniol

17 T 1 and NOE Effects in C-13 NMR Because of unequal T 1 and NOE effects, peaks heights vary widely in C-13 NMR. This is why C-13 spectra are normally not integrated. CarbonT1 (sec)NOE CH CH 3

18 Carbon-13 Proton-Coupled Patterns

19 Carbon-13 Proton-Coupled Spectrum of Ethyl Phenylacetate Typical coupling constants for 13 C- 1 H one- bond couplings are between 100 to 250 Hz. Difficult to interpret C=O

20 DEPT Spectra Quaternary carbons (C) do not show up in DEPT.

21 Simulated DEPT Spectra of Ethyl Phenylacetate Normal C-13 spectrum DEPT-45 DEPT-90 DEPT-135

22 DEPT Spectra of Codeine

23 Predict the normal C-13, DEPT-90, and DEPT-135 spectra of ipsenol, whose structure appears below.

24 DEPT Spectra of Ipsenol Normal C-13 spectrum CDCl 3 DEPT-135 DEPT-90

25 Determine the number and appearance of the signals in the DEPT-45, DEPT 90, and DEPT 135 NMR spectrum of each of the following compounds:


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