Carbon-13 NMR

Core Concepts

In this article, we will explore what ¹³C-NMR is, how it occurs, and its uses as it relates to compound and structure determination.

Related Articles

• NMR Spectroscopy
• Spectral Analysis of Organic Compounds
• Mass Spectrometry and Mass Spectrometers
• What is an Isotope
• The Critical Element Carbon

What is ¹³C-NMR?

¹³C-NMR, also known as Carbon NMR, is a form of spectroscopy that is used to identify unique carbons in a compound. The technique involves using radio frequency waves to determine the locations of carbon within an organic compound. In brief, radio waves affect the nuclear spin of certain carbon atoms, resulting in an energy transition that can be measured. For more details on the physics behind NMR, see our article on the topic.

However, not all carbon isotopes can be analyzed through NMR. Carbon has two major isotopes – ¹²C and ¹³C, which have a relative abundance of 98.9% and 1.1% respectively. ¹²C is invisible in NMR, while ¹³C is NMR active as a result of its unique nuclear spin. ¹³C atoms and Hydrogen atoms are coupled to one another, which may introduce barriers to NMR; decoupling occurs via the addition of a second radio frequency.

What does a ¹³C-NMR spectrum look like?

For each unique carbon atom in a compound, a single signal will appear on the NMR spectrum. Unique carbon atoms are carbons not related to another by symmetry.

Each signal shows as a vertical peak. The peak size is often proportionate to the number of Carbons represented, though simplified spectrums may have a uniform peak height of one. Unlike in ¹H NMR, ¹³C does not involve any “splitting”. Since ¹³C are rare relative to ¹²C, the probability of two ¹³C occurring in the same molecule is low, thus eliminating the signal interferance observed in ¹H NMR.

The bottom axis of the spectrum measures the signal’s chemical shift (unit: ppm). Higher frequencies are further left, or “downfield,” while lower frequencies are further right, or “upfield”. The carbon’s chemical shift is relative to its proximity to various functional groups and electronegative atoms, with the general trend that more electron-rich carbons have higher ppm. Consider the following table:

Carbons attached to another carbon atom with attached electropositive atoms will have a lower chemical shift. Therefore, we consider these carbon atoms’ signals to be upfield. Carbons directly attached to oxygen or pi bonded to another carbon atom will have a higher chemical shift. Therefore, we consider these carbon atoms’ signals to be downfield.

How do we use a ¹³C-NMR spectrum to identify a compound’s structure?

Typically, drawing a structure from a ¹³C-NMR spectrum occurs in concert with the use of mass spectrometry data or a known formula. The example below will demonstrate how to complete a structural drawing with either supplemental data.

Problem: A compound’s mass spectrum gives a molecular ion peak of 106 m/z and M+2 peak of 108 m/z. The two peaks have a ratio of roughly 3:1. Below is the given Carbon NMR spectrum.

Solution: First, we count the number of unique carbon atoms. As the ¹³C-NMR spectrum has four separate peaks of equal size, we can conclude that there are four unique carbon atoms.

Next, we work with the mass spectrum data. From the molecular ion peak, we know the molecular weight. Further, we can identify the presence of a chlorine atom, resulting from the molecular ion and M+2 peak ratio of 3:1. We subtract the mass of chlorine’s lower isotope (35 amu) from the total mass of 106 amu to give 71 amu. 

From here, we notice a peak around 205 ppm. When checking with the table, we can identify a ketone/aldehyde group. We then subtract the mass of oxygen from the remaining mass: 71 amu – 16 amu = 55 amu.

Lastly, we can compute the number of remaining atoms using the 55 amu. While a structure can have four unique carbon peaks but more than four total carbons (i.e.1,8-dichlorooctane), the remaining mass of 55 amu indicates that the structure can only contain four maximum carbons. Four carbon atoms amount to 48 amu, so 7 amu remain to account for the remaining atoms. Therefore, we know the compound has 7 hydrogen atoms (7 amu/1 amu = 7).

From our calculations, we therefore have the formula: C₄H₇OCl.

*At this point, the processes, whether using a formula or mass spectrum with ¹³C-NMR data, intersect.*

We can now build our structure with the calculated formula. 

As identified above, there is a carbon with a double bonded oxygen. Because the range for this functional group is 190-210 ppm, and the chemical shift provided is 205 ppm, we can likely conclude the group is a ketone, rather than an aldehyde which will produce a signal that is further upfield due to an additional hydrogen. 

The next signal we can identify is -CH₂Cl, which comes from the identification of the Cl atom. This signal is around 45 ppm, which leaves two more signals at 20 and 30 ppm.

Given the remaining signals and possible atoms, we can thus conclude there is a -CH₂– group and -CH₃ group. Because the -CH₃ signal is at about 20 ppm, the most downfield in the range, we can conclude that it is bonded to the ketone group. Further, the -CH₂– signal is at about 30 ppm, also very downfield, so it must be between two downfield-shifting groups. From these two conclusions we can draw our final structure, seen below.

How do we draw a ¹³C-NMR spectrum given a compound’s structure?

We can draw a spectrum using a ¹³C-NMR table and a line angle (or similar) structure. Follow along for the process.

Structure Review

From the above structure, we must first identify the number of unique carbons. The disubstituted benzene ring contains six carbons; of these six, there are four unique carbons.

We can identify unique carbons by drawing a line of symmetry vertically down the molecule. The carbon connected to the chlorine atom (1) is unique as there are no other carbons on the structure with the same exact atom connectivity; the same can be said for the carbon connected to the bromine atom (4). The two carbons directly attached to the bromine-attached carbon are the same due to their relation by symmetry (3). Both carbons are double bonded to another carbon and attached to another carbon and a hydrogen. 

Further, the two carbons directly attached to the chlorine-attached carbon are also the same by symmetry (2). Therefore, they constitute one unique carbon signal. However, they are unique from the other two remaining carbons due to their increased proximity to the chlorine atom, and their decreased proximity to the bromine atom.

Using this knowledge, we can draw the ¹³C-NMR for this compound. The unique carbons are numbered for clarity.

¹³C-NMR Spectrum Considerations and Drawing

Carbons that are double bonded with another carbon will show chemical shifts within the range of 100 and 160 ppm. We can determine the approximate shifts within this range using electronegative atom proximity. Carbon 1 is directly bonded to Cl, which produces a more downfield chemical shift than Br. Therefore, carbon 1’s signal will be more downfield than carbon 4.

Carbon 2 and 3 are not directly bonded to an electronegative atom, so they will be more upfield than signal 1 and 4. However, they follow the same pattern, so carbon 2 will be slightly further downfield than carbon 3.

Thus, below is the approximate spectrum we can draw for this compound.