14.9: Characteristics of ¹³C NMR Spectroscopy

¹³C NMR Spectroscopy - Signal Averaging and FT-NMR

In some ways, it’s surprising that carbon NMR is even possible. After all, ¹²C, the most abundant carbon isotope, has no nuclear spin and can’t be seen by NMR. Carbon-13 is the only naturally occurring carbon isotope with a nuclear spin, but its natural abundance is only 1.1%. Thus, only about 1 of every 100 carbon atoms in an organic sample can be observed by NMR. The problem of low abundance has been overcome, however, by the use of signal averaging and Fourier-transform NMR (FT–NMR). Signal averaging increases instrument sensitivity, and FT–NMR increases instrument speed.

The low natural abundance of ¹³C means that any individual NMR spectrum is extremely “noisy.” That is, the signals are so weak that they are cluttered with random background electronic noise, as shown in Figure \(\PageIndex{1}\)a. If, however, hundreds or thousands of individual runs are added together by a computer and then averaged, a greatly improved spectrum results (Figure \(\PageIndex{1}\)b). Background noise, because of its random nature, increases very slowly as the runs are added, while the nonzero signals stand out clearly. Unfortunately, the value of signal averaging is limited when using the method of NMR spectrometer operation described in Section 13.2, because it takes about 5 to 10 minutes to obtain a single spectrum. Thus, a faster way to obtain spectra is needed if signal averaging is to be used.

Figure \(\PageIndex{1}\): Carbon-13 NMR spectra of 1-pentanol, CH₃CH₂CH₂CH₂CH₂OH. Spectrum (a) is a single run showing the high level of background noise. Spectrum (b) is an average of 200 runs.

 

In the method of NMR spectrometer operation described in Section 13.2, the rf frequency is held constant while the strength of the magnetic field is varied so that all signals in the spectrum are recorded sequentially. In the FT–NMR technique used by modern spectrometers, however, all the signals are recorded simultaneously. A sample is placed in a magnetic field of constant strength and is irradiated with a short pulse of rf energy that covers the entire range of useful frequencies. All ¹H or ¹³C nuclei in the sample resonate at once, giving a complex, composite signal that is mathematically manipulated using so-called Fourier transforms and then displayed in the usual way. Because all resonance signals are collected at once, it takes only a few seconds rather than a few minutes to record an entire spectrum.

Combining the speed of FT–NMR with the sensitivity enhancement of signal averaging is what gives modern NMR spectrometers their power. Literally thousands of spectra can be taken and averaged in a few hours, resulting in sensitivity so high that a ¹³C NMR spectrum can be obtained from less than 0.1 mg of sample and a ¹H spectrum can be recorded from only a few micrograms.

One further question needs to be answered before moving forward with our discussion of ¹³C NMR. Why is spin–spin splitting seen only for ¹H NMR? Why is there no splitting of carbon signals into multiplets in ¹³C NMR? After all, you might expect that the spin of a given ¹³C nucleus would couple with the spin of an adjacent magnetic nucleus, either ¹³C or ¹H.

No coupling of a ¹³C nucleus with nearby carbons is seen because their low natural abundance makes it unlikely that two ¹³C nuclei will be adjacent. No coupling of a ¹³C nucleus with nearby hydrogens is seen because ¹³C spectra are normally recorded using broadband decoupling. At the same time that the sample is irradiated with a pulse of rf energy to cover the carbon resonance frequencies, it is also irradiated by a second band of rf energy covering all the hydrogen resonance frequencies. This second irradiation makes the hydrogens spin-flip so rapidly that their local magnetic fields average to zero and no coupling with carbon spins occurs.

Factors on ¹³C NMR Spectra

At its simplest, ¹³C NMR makes it possible to count the number of different carbon atoms in a molecule. Look at the ¹³C NMR spectra of 1-pentanol (Figure \(\PageIndex{1}\)) shown previously and ethyl acetate (Figure \(\PageIndex{2}\)). In each case, a single sharp resonance line is observed for each different carbon atom.

Most ¹³C resonances are between 0 and 220 ppm downfield from the TMS reference line, with the exact chemical shift of each ¹³C resonance dependent on that carbon’s electronic environment within the molecule. Figure \(\PageIndex{1}\) shows the correlation of chemical shift with environment.

Figure \(\PageIndex{1}\): Chemical shift correlations for ¹³C NMR.

 

The factors that determine chemical shifts are complex, but it’s possible to make some generalizations from the data in Figure \(\PageIndex{1}\). One trend is that a carbon atom’s chemical shift is affected by the electronegativity of nearby atoms. Carbons bonded to oxygen, nitrogen, or halogen absorb downfield (to the left) of typical alkane carbons. Because electronegative atoms attract electrons, they pull them away from neighboring carbon atoms, causing those carbons to be deshielded and to resonate at a lower field.

Another trend is that sp³-hybridized carbons generally absorb from 0 to 90 δ, while sp² carbons absorb from 110 to 220 δ. Carbonyl carbons (C=O) are particularly distinct in ¹³C NMR and are always found at the low-field end of the spectrum, from 160 to 220 δ. Figure \(\PageIndex{2}\) shows the ¹³C NMR spectra of 2-butanone and para-bromoacetophenone and indicates the peak assignments. Note that the C=O carbons are at the left edge of the spectrum in each case.

Figure \(\PageIndex{2}\): Carbon-13 NMR spectra of (a) 2-butanone.

 

A second interesting point about both spectra in Figure \(\PageIndex{2}\) is that the peaks aren’t uniform in size. Some peaks are larger than others, even though they are one-carbon resonances (except for the two 2-carbon peaks of para-bromoacetophenone). This difference in peak size is a general feature of broadband-decoupled ¹³C NMR spectra, and explains why we can’t integrate ¹³C NMR spectra in the same way we integrate the resonances in a ¹H NMR spectrum. The local environment of each carbon atom determines not only its chemical shift but also the time it takes for the nuclei to return to their equilibrium state after receiving a pulse of rf radiation and flipping their spins. Quaternary carbons, regardless of their hybridization state or substituents, typically give smaller resonances than primary, secondary, or tertiary carbons.

 

Figure \(\PageIndex{2}\): Carbon-13 NMR spectra of (b) para-bromoacetophenone.

The ¹³C NMR spectrum of para-bromoacetophenone is interesting in several ways. Note particularly that only six carbon absorptions are observed, even though the molecule contains eight carbons. para-bromoacetophenone has a symmetry plane that makes ring carbons 4 and 4′, and ring carbons 5 and 5′ equivalent. (Remember from Section 2.4 that aromatic rings have two resonance forms.) Thus, the aromatic carbons show only four absorptions in the range 128 to 137 δ.

Worked Example \(\PageIndex{1}\): Predicting Chemical Shifts in ¹³C NMR Spectra

At what approximate positions would you expect ethyl acrylate, H₂C═CHCO₂CH₂CH₃, to show ¹³C NMR absorptions?

Strategy

Identify the distinct carbons in the molecule, and note whether each is alkyl, vinylic, aromatic, or in a carbonyl group. Then predict where each absorbs, using Figure \(\PageIndex{1}\) as necessary.

Solution

Ethyl acrylate has five chemically distinct carbons: two different C=C, one C=O, one O–C, and one alkyl C. From Figure \(\PageIndex{1}\), the likely absorptions are

 The actual absorptions are at 14.1, 60.5, 128.5, 130.3, and 166.0 δ.

Exercises

Exercise \(\PageIndex{1}\)

Predict the number of carbon resonance lines you would expect in the ¹³C NMR spectra of the following compounds:

1. Methylcyclopentane
2. 1-Methylcyclohexene
3. 1,2-Dimethylbenzene
4. 2-Methyl-2-butene
5.  
6. 

Answer

1. 4
2. 7
3. 4
4. 5
5. 5
6. 7

Exercise \(\PageIndex{2}\)

Propose structures for compounds that fit the following descriptions:

1. A hydrocarbon with seven lines in its ¹³C NMR spectrum
2. A six-carbon compound with only five lines in its ¹³C NMR spectrum
3. A four-carbon compound with three lines in its ¹³C NMR spectrum

Answer

1. 1,3-Dimethylcyclopentene
2. 2-Methylpentane
3. 1-Chloro-2-methylpropane

Exercise \(\PageIndex{3}\)

Classify the resonances in the ¹³C NMR spectrum of methyl propanoate, CH₃CH₂CO₂CH₃ :

Answer

−CH₃, 9.3 δ;  −CH₂− , 27.6 δ; C=O, 174.6 δ;− OCH₃, 51.4 δ

Exercise \(\PageIndex{4}\)

Which isomer of ortho, meta, or para xylene do you have based on the ¹³C NMR spectrum? 

Answer

There are 5 different carbons in the spectrum with four different aromatic carbons. p-Xylene would have 2 types of aromatic carbons and o-xylene would have 3 types. m-Xylene is the only one with 4 different types of aromatic carbons, which fits this spectrum. The methyl groups would all be similary, so not a point of difference.  

Exercise \(\PageIndex{5}\)

Propose a structure using the spectral data below for C₉H₁₀O. 

¹³C broadband decoupled spectrum: 

¹H NMR spectrum: Integration: 1 (doublet; J = 1 Hz):5 (multiplet):1 (quartet of doublets;J = 7 Hz and 1 Hz):3 (doublet; J = 7 Hz)

Answer