19.1: Theory of Nuclear Magnetic Resonance

As is the case with other forms of optical spectroscopy, the signal in nuclear magnetic resonance (NMR) spectroscopy arises from a difference in the energy levels occupied by the nuclei in the analyte. In this section we develop a general theory of nuclear magnetic resonance spectroscopy that draws on quantum mechanics and on classical mechanics to explain these energy levels.

Quantum Mechanical Description of NMR

The quantum mechanical description of an electron is given by four quantum numbers: the principal quantum number, \(n\), the angular momentum quantum number, \(l\), the magnetic quantum number, \(m_l\), and the spin quantum number, \(m_s\). The first three of these quantum numbers tell us something about where the electron is relative to the nucleus and something about the electron's energy. The last of these four quantum numbers, the spin quantum number, tells us something about the ability of an electron to interact with an applied magnetic field. An electron has possible spins of +1/2 or of –1/2, which we often refer to as spin up, using an upwards arrow, \(\uparrow\), to represent it, or as spin down, using a downwards arrow, \(\downarrow\), to represent it.

A nucleus, like an electron, carries a charge and has a spin quantum number. The overall spin, \(I\), of a nucleus is a function of the number of protons and neutrons that make up the nucleus. Here are three simple rules for nuclear spin states:

• If the number of neutrons and the number of protons are both even numbers, then the nucleus does not have a spin; thus, ¹²C, with six protons and six neutrons, has no overall spin and \(I = 0\).
• If the number of neutrons plus the number of protons is an odd number, then the nucleus has a half-integer spin, such as 1/2 or 3/2; thus, ¹³C, with six protons and seven neutrons, has an overall spin of \(I = 1/2\); this also is true for ¹H.
• If the number of neutrons and the number of protons are both odd numbers, then the nucleus has an integer spin, such as 1 or 2; thus, ²H, with one proton and one neutron, has an overall spin of \(I = 1\).

Note

Predicting that ¹³C has a spin of \(I = 1/2\), but that ¹²⁷I has a spin of \(I = 3/2\) and that ¹⁷O has a spin of \(I = 5/2\) is not trivial. A periodic table that provides spin states for elements is available here.

The total number of spin states—that is, the total number of possible orientations of the spin—is equal to \((2 \times I) + 1\). To be NMR active, a nucleus must have at least two spin states so that a change in spin states, and, therefore, a change in energy, is possible; thus, ¹²C, for which there are \((2 \times 0) + 1 = 1\) spin states, is NMR inactive, but ¹³C, for which there are \((2 \times 1/2) + 1 = 2\) spin states with values of \(m = +1/2\) and of \(m = -1/2\), is NMR active, as is ²H for which there are \((2 \times 1) + 1 = 3\) spin states with values of \(m = +1/2\), \(m = 0\), and \(m = -1/2\). As our interest in this chapter is in the NMR spectra for ¹H and for ¹³C, we will limit ourselves to considering \(I = 1/2\) and spin states of \(m = +1/2\) and of \(m = -1/2\).

Energy Levels in an Applied Magnetic Field

Suppose we have a large population of ¹H atoms. In the absence of an applied magnetic field the atoms are divided equally between their possible spin states: 50% of the atoms have a spin of +1/2 and 50% of the atoms have a spin of –1/2. Both spin states have the same energy, as is the case on the left side of Figure \(\PageIndex{1}\), and neither absorption nor emission occurs.

Figure \(\PageIndex{1}\). Energy level diagram for a proton, ¹H, in the absence (left) and in the presence (right) of an applied magnetic field.

In the presence of an applied magnetic field, as on the right side of Figure \(\PageIndex{1}\), the nuclei are either aligned with the magnetic field with spins of \(m = +1/2\), or aligned against the magnetic field with spins of \(m = -1/2\). The energies in these two spin states, \(E_\text{lower}\) and \(E_\text{upper}\), are given by the equations

\[E_\text{lower} = - \frac{\gamma h}{4 \pi}B_0 \label{nmr1} \]

\[E_\text{upper} = + \frac{\gamma h}{4 \pi}B_0 \label{nmr2} \]

where \(\gamma\) is the magnetogyric ratio for the nucleus, \(h\) is Planck's constant, and \(B_0\) is the strength of the applied magnetic field. The difference in energy, \(\Delta E\), between the two states is

\[\Delta E = E_\text{upper} - E_\text{lower} = + \frac{\gamma h}{4 \pi}B_0 - \left( - \frac{\gamma h}{4 \pi}B_0 \right) = \frac{\gamma h}{2 \pi}B_0 \label{nmr3} \]

Substituting Equation \ref{nmr3} into the more familiar equation \(\Delta E = h \nu\) gives the frequency, \(\nu\), of electromagnetic radiation needed to effect a change in spin state as

\[\nu = \frac{\gamma B_0}{2 \pi} \label{nmr4} \]

This is called the Larmor frequency for the nucleus. For example, if the magnet has a field strength of 11.74 Tesla, then the frequency needed to effect a change in spin state for ¹H, for which \(\gamma\) is \(2.68 \times 10^8 \text{ rad}\text{ T}^{-1} \text{s}^{-1}\), is

\[\nu = \frac{(2.68 \times 10^8 \text{rad} \text{ T}^{-1}\text{s}^{-1})(11.74 \text{ T})}{2 \pi} = 5.01 \times 10^8 \text{ s}^{-1} \nonumber \]

or 500 MHz, which is in the radio frequency (RF) range of the electromagnetic spectrum. This is the Larmor frequency for ¹H.

Population of Spin States

The relative population of the upper spin state, \(N_\text{upper}\), and of the lower spin state, \(N_\text{lower}\), is given by the Boltzmann equation

\[\frac{N_\text{upper}}{N_\text{lower}} = e^{- \Delta E/k T} \label{nmr5} \]

where \(k\) is Boltzmann's constant (\(1.38066 \times 10^{-23} \text{ J/K}\)) and \(T\) is the temperature in Kelvin. Substituting in Equation \ref{nmr3} for \(\Delta E\) gives this ratio as

\[\frac{N_\text{upper}}{N_\text{lower}} = e^{-\gamma h B_0/2 \pi k T} \label{nmr6} \]

IF we place a population of ¹H atoms in a magnetic field with a strength of 11.74 Tesla, the ratio \(\frac{N_\text{upper}}{N_\text{lower}}\) at 298 K is

\[\frac{N_\text{upper}}{N_\text{lower}} = e^{-\frac{(2.68 \times 10^{8} \text{ rad} \text{ s}^{-1})(6.626 \times 10^{-34} \text{ Js})(11.74 \text{ T})}{(2 \pi)(1.38 \times 10^{-23} \text{ JK}^{-1})(298 \text{ K})}} = 0.99992 \nonumber \]

If this ratio is 1:1, then the probability of absorption and emission are equal and there is no net signal. In this case, the difference in the populations is on the order of 8 per 100,000, or 80 per 1,000,000, or 80 ppm. The small difference in the two populations means that NMR is less sensitive than many other spectroscopic methods.

Classical Description of NMR

To understand the classical description of an NMR experiment we draw upon Figure \(\PageIndex{2}\). For simplicity, let's assume that in the population of nuclei available to us, there is an excess of just one nucleus with a spin state of +1/2. In Figure \(\PageIndex{2}a\), we see that the spin of this nucleus is not perfectly aligned with the applied magnetic field, \(B_0\), which is aligned with the z-axis; instead the nucleus precesses around the z-axis at an angle of theta, \(\Theta\). As a result, the net magnetic moment along the z-axis, \(\mu_z\), is less than the magnetic moment, µ, of the nucleus. The precession occurs with an angular velocity, \(\omega_0\), of \(\gamma B_0\).

Figure \(\PageIndex{2}\). Illustration that shows in (a) the magnetic moment, µ, of a nucleus whose spin is aligned with the applied magnetic field, \(B_0\). Application of the magnetic field, \(B_1\) flips the spin of the nucleus, as shown in (b).

If we apply a source of radio frequency (RF) electromagnetic radiation along the x-axis such that its magnetic field component, \(B_1\), is perpendicular to \(B_0\), then it will generate its own angular velocity in the xy-plane. When the angular velocity of the precessing nucleus matches the angular velocity of \(B_1\), absorption takes place and the spin flips, as seen in Figure \(\PageIndex{2}b\).

Relaxation

When the magnetic field \(B_1\) is removed, the nucleus returns to its original state, as seen in Figure \(\PageIndex{2}a\), a process called relaxation. In the absence of relaxation, the system is saturated with equal populations of the two spin states and absorption approaches zero. This process of relaxation has two separate mechanisms: spin-lattice relaxation and spin-spin relaxation.

In spin-lattice relaxation the nucleus in its higher energy spin state, Figure \(\PageIndex{2}b\), returns to its lower energy state spin state, Figure \(\PageIndex{2}a\), by transferring energy to other species present in the sample (the lattice in spin-lattice). Spin-lattice relaxation is characterized by first-order exponential decay with a characteristic relaxation time of \(T_1\) that is a measure of the average time the nucleus remains in its higher energy spin state. Smaller values for \(T_1\) result in more efficient relaxation.

If two nuclei of the same type, but in different spin states, are in close proximity to each other, they can trades places in which the nucleus in the higher energy spin state gives up its energy to the nucleus in the lower energy spin state. The result is a decrease in the average life-time of an excited state. This is called spin-spin relaxation and it is characterized by a relaxation time of \(T_2\).

Continuous Wave NMR vs. Fourier Transform NMR

In Chapter 16 we learned that we can record an infrared spectrum by using a scanning monochromator to pass, sequentially, different wavelengths of IR radiation through a sample, obtaining a spectrum of absorbance as a function of wavelength. We also learned that we can obtain the same spectrum by passing all wavelengths of IR radiation through the sample at the same time using an interferometer, and then use a Fourier transform to convert the resulting interferogram into a spectrum of absorbance as a function of wavelength. Here we consider their equivalents for NMR spectroscopy.

Continuous Wave NMR

If we scan \(B_1\) while holding \(B_0\) constant—or scan \(B_0\) while holding \(B_1\) constant—then we can identify the Larmor frequencies where a particular nucleus absorbs. The result is an NMR spectrum that shows the intensity of absorption as a function of the frequency at which that absorption takes place. Because we record the spectrum by scanning through a continuum of frequencies, the method is known as continuous wave NMR. Figure \(\PageIndex{2}\) provides a useful visualization for this experiment.

Fourier Transform NMR

In Fourier transform NMR, the magnetic field \(B_1\) is applied as a brief pulse of radio frequency (RF) electromagnetic radiation centered at a frequency appropriate for the nucleus of interest and for the strength of the primary magnetic field, \(B_0\). The pulse typically is 1-10 µs in length and applied in the xy-plane. From the Heisenberg uncertainty principle, a short pulse of \(\Delta t\) results in a broad range of frequencies as \(\Delta f = 1/\Delta t\); this ensures that the pulse spans a sufficient range of frequencies such that the nucleus of interest to us will absorb energy and enter into an excited state.

Before we apply the pulse, the population of nuclei are aligned parallel to the applied magnetic field, \(B_0\), some with a spin of +1/2 and others with a spin of –1/2. As we learned above, there is a slight excess of nuclei with spins of +1/2, which we can represent as a single vector that shows their combined magnetic moments along the z-axis, \(\mu_z\), as shown in Figure \(\PageIndex{3}a\). When we apply a pulse of RF electromagnetic radiation with a magnetic field strength of \(B_1\), the spin states of the nuclei tip away from the z-axis by an angle that depends on the nucleus's magnetogyric ratio, \(\gamma\), the value of \(B_1\), and the length of the pulse. If, for example, a pulse of 5 µs tips the the magnetic vector by 45° (Figure \(\PageIndex{3}b\)), then a pulse of 10 µs will tip the magnetic vector by 90° degrees (Figure \(\PageIndex{3}c\)), so that it now lies completely within the xy-plane.

Figure \(\PageIndex{3}\). Illustration that shows (a) the net magnetic moment on the z-axis for a population of nuclei in an applied magnetic field with a field strength of \(B_0\). Application of a pulsed RF field, \(B_1\), tips the magnetic moment (b) away from the z-axis. The longer the pulse of RF, the greater the angle of tipping, as shown in (c).

At the end of the pulse, the nuclei begin to relax back to their original state. Figure \(\PageIndex{4}\) shows that this relaxation occurs both in the xy-plane (spin-spin relaxation) and along the z-axis (spin-lattice relaxation). If we were to trace the path of the magnetic vector with time, we would see that it follows a spiral-like motion as its contribution in the xy-plane decreases and its contribution along the z-axis increases. We measure this signal—called the free induction decay, or FID—during this period of relaxation.

Figure \(\PageIndex{4}\). Illustration showing the process of relaxation during an NMR experiment. The bottom row shows the net magnetic moment aligned with the z-axis, \(\mu_z\), as a function of time. The top row shows the net magnetic moment in the xy-plane with the figure on the top right showing the overall change in the net magnetic moment in three dimensions.

The FID for a system that consists of only one type of nucleus is the simple exponentially damped oscillating signal in Figure \(\PageIndex{5}a\). The Fourier transform of this simple FID gives the spectrum in Figure \(\PageIndex{5}b\) that has a single peak. A sample with a more than one type of nucleus yields a more complex FID pattern, such as that in Figure \(\PageIndex{5}c\), and a more complex spectrum, such as the two peaks in Figure \(\PageIndex{5}d\). Note that, as we learned in an earlier treatment of the Fourier transform in Chapter 7, a broader peak in the frequency domain results in a faster decay in the time domain.

Figure \(\PageIndex{5}\). Panels (a) and (b) show the FID signal for a sample that yields single peak and the result of taking the FT of the FID. Panels (c) and (d) show the FID signal for a sample that consists of two peaks and the result of taking the FT of the FID.

Figure \(\PageIndex{6}\) shows a typical pulse sequence highlighting the total cycle time and its component parts: the pulse width, the acquisition time during which the FID is recorded, and a recycle delay before applying the next pulse and beginning the next cycle.

Figure \(\PageIndex{1}\): Example of a pulse sequence.