Comprehensive Study Notes

Recall that the orbital angular momentum $\mathbf{L}$ of an electron has a magnetic moment $-\left(e / 2 m{e}\right) \mathbf{L}$ associated with it [Eq. (6.128)], where $-e$ is the electron charge. It is natural to suppose that there is also a magnetic moment $\mathbf{m}{S}$ associated with the electronic spin angular momentum $\mathbf{S}$. We might guess that $\mathbf{m}{S}$ would be $-\left(e / 2 m{e}\right)$ times $\mathbf{S}$. Spin is a relativistic phenomenon, however, and we cannot expect $\mathbf{m}{S}$ to be related to $\mathbf{S}$ in exactly the same way that $\mathbf{m}{L}$ is related to $\mathbf{L}$. In fact, Dirac's relativistic treatment of the electron gave the result that (in SI units)

\(
\begin{equation}
\mathbf{m}{S}=-g{e} \frac{e}{2 m_{e}} \mathbf{S} \tag{10.55}
\end{equation}
\)

where Dirac's treatment gave $g{e}=2$ for the electron $g$ factor $g{e}$. Theoretical and experimental work subsequent to Dirac's treatment has shown that $g{e}$ is slightly greater than $2 ; g{e}=2(1+\alpha / 2 \pi+\cdots)=2.0023$, where the dots indicate terms involving higher powers of $\alpha$ and where the fine-structure constant $\alpha$ is

\(
\begin{equation}
\alpha \equiv \frac{e^{2}}{4 \pi \varepsilon_{0} \hbar c}=0.00729735257 \tag{10.56}
\end{equation}
\)

$g{e}$ has been measured with extraordinary accuracy [D. Hanneke et al., Phys. Rev. A, 83, 052122 (2011); available at arxiv.org/abs/1009.4831]: $g{e}=2.002319304361$. [Some workers prefer to omit the minus sign in (10.55) and use the convention that $g_{e}$ is negative; see www.stanford.edu/group/Zarelab/publinks/zarepub642.pdf.]

The magnitude of the spin magnetic moment of an electron is (in SI units)

\(
\begin{equation}
\left|\mathbf{m}{S}\right|=g{e} \frac{e}{2 m{e}}|\mathbf{S}|=g{e} \sqrt{\frac{3}{4}} \frac{e \hbar}{2 m_{e}} \tag{10.57}
\end{equation}
\)

The ferromagnetism of iron is due to the electron's magnetic moment.

The two possible orientations of an electron's spin and its associated spin magnetic moment with respect to an axis produce two energy levels in an externally applied magnetic field. In electron-spin-resonance (ESR) spectroscopy, one observes transitions between these two levels. ESR spectroscopy is applicable to species such as free radicals and transition-metal ions that have one or more unpaired electron spins and hence have a nonzero total electron spin and spin magnetic moment.

NMR Spectroscopy

Many atomic nuclei have a nonzero spin angular momentum I. Similar to (10.4) and (10.5), the magnitude of $\mathbf{I}$ is $[I(I+1)]^{1 / 2} \hbar$, where the nuclear-spin quantum number $I$ can be $0, \frac{1}{2}, 1$, and so on, and the $z$ component of $\mathbf{I}$ has the possible values $M{I} \hbar$, where $M{I}=-I,-I+1, \ldots, I$. Some $I$ values are: 0 for every nucleus with an even number of protons and an even number of neutrons (for example, ${ }{8}^{16} \mathrm{O}$ and ${ }{6}^{12} \mathrm{C}$ ); $\frac{1}{2}$ for ${ }{1}^{1} \mathrm{H},{ }{6}^{13} \mathrm{C},{ }{9}^{19} \mathrm{~F}$, and ${ }{15}^{31} \mathrm{P} ; 1$ for ${ }{1}^{2} \mathrm{H}$ and ${ }{7}^{14} \mathrm{~N} ; \frac{3}{2}$ for ${ }{5}^{11} \mathrm{~B},{ }{11}^{23} \mathrm{Na}$ and ${ }{17}^{35} \mathrm{Cl}$. If $I \neq 0$, the nucleus has a spin magnetic moment $\mathbf{m}{I}$ given by an equation similar to (10.55):

\(
\begin{equation}
\mathbf{m}{I}=g{N}\left(e / 2 m_{p}\right) \mathbf{I} \equiv \gamma \mathbf{I} \tag{10.58}
\end{equation}
\)

where $m{p}$ is the proton mass and the nuclear $g$ factor $g{N}$ has a value characteristic of the nucleus. The quantity $\gamma$, called the magnetogyric ratio of the nucleus, is defined by $\gamma \equiv \mathbf{m}{I} / \mathbf{I}=g{N} e / 2 m{p}$. Values of $I, g{N}$, and $\gamma$ for some nuclei are

In nuclear-magnetic-resonance (NMR) spectroscopy, one observes transitions between nuclear-spin energy levels in an applied magnetic field. The sample (most commonly a dilute solution of the compound being studied) is placed between the poles of a strong magnet. The energy of interaction between an isolated nuclear-spin magnetic moment $\mathbf{m}{I}$ in an external magnetic field $\mathbf{B}$ is given by Eq. (6.131) as $E=-\mathbf{m}{I} \cdot \mathbf{B}$. Using (10.58) for $\mathbf{m}_{I}$ and taking the $z$ axis as coinciding with the direction of $\mathbf{B}$, we have

\(
E=-\mathbf{m}{I} \cdot \mathbf{B}=-\gamma\left(I{x} \mathbf{i}+I{y} \mathbf{j}+I{z} \mathbf{k}\right) \cdot(B \mathbf{k})=-\gamma B I_{z}
\)

We convert this classical expression for the energy into a Hamiltonian operator by replacing the classical quantity $I{z}$ by the operator $\hat{I}{z}$. Thus, $\hat{H}=-\gamma B \hat{I}{z}$. Let $\left|M{I}\right\rangle$ denote the function that is simultaneously an eigenfunction of the operators $\hat{I}^{2}$ (for the square of the magnitude of the nuclear-spin angular momentum) and $\hat{I}_{z}$. We have

\(
\begin{equation}
\hat{H}\left|M{I}\right\rangle=-\gamma B \hat{I}{z}\left|M{I}\right\rangle=-\gamma B M{I} \hbar\left|M_{I}\right\rangle \tag{10.59}
\end{equation}
\)

Therefore, (10.59) gives the energy levels of the isolated nuclear spin in the applied magnetic field as

\(
E=-\gamma \hbar B M{I}, \quad M{I}=-I,-I+1, \ldots, I
\)

In NMR spectroscopy, the sample is exposed to electromagnetic radiation that induces transitions between nuclear-spin energy levels. The selection rule turns out to be $\Delta M_{I}= \pm 1$. The NMR transition frequency $\nu$ is found as follows:

\(
\begin{align}
& h \nu=|\Delta E|=|\gamma| \hbar B\left|\Delta M{I}\right|=|\gamma| \hbar B \
& \nu=(|\gamma| / 2 \pi) B=\left|g{N}\right|\left(e / 4 \pi m_{p}\right) B \tag{10.60}
\end{align}
\)

The value of $\gamma$ differs greatly for different nuclei, and in any one experiment, one studies the NMR spectrum of one kind of nucleus. The most commonly studied nucleus is ${ }^{1} \mathrm{H}$, the proton. The second most studied nucleus is ${ }^{13} \mathrm{C}$. The ${ }^{13} \mathrm{C}$ isotope occurs in $1 \%$ abundance in carbon.

Equation (10.60) is for a nucleus isolated except for the presence of the external magnetic field $\mathbf{B}$. For a nucleus present in a molecule, we also have to consider the contribution of the molecular electrons to the magnetic field felt by each nucleus. In most ground-state molecules, the electron spins are all paired and there is no electronic orbital angular momentum. With no electronic spin or orbital angular momentum, the electrons do not contribute to the magnetic field experienced by each nucleus. However, the application of the external applied field $\mathbf{B}$ perturbs the molecular electronic wave function, thereby producing an electronic contribution to the magnetic field at each nucleus. This electronic contribution is proportional to the magnitude of the external field $B$ and is usually in the opposite direction to $B$. Therefore, the magnetic field experienced by nucleus $i$ in a molecule is $B-\sigma{i} B=\left(1-\sigma{i}\right) B$, where the proportionality constant $\sigma_{i}$ is called the screening constant or shielding constant for nucleus $I$ and is much less than 1. Equation (10.60) becomes for a nucleus in a molecule

\(
\begin{equation}
\nu{i}=(|\gamma| / 2 \pi)\left(1-\sigma{i}\right) B \tag{10.61}
\end{equation}
\)

The value of $\sigma{i}$ is the same for nuclei that are in the same electronic environment in the molecule. For example, in $\mathrm{CH}{3} \mathrm{CH}{2} \mathrm{OH}$, the three $\mathrm{CH}{3}$ protons have the same $\sigma{i}$, and the two $\mathrm{CH}{2}$ protons have the same $\sigma{i}$. (A Newman projection of ethanol, which has a staggered conformation, shows that two of the three $\mathrm{CH}{3}$ hydrogens are closer to the OH group than is the third $\mathrm{CH}_{3}$ hydrogen, but the low barrier to internal rotation in ethanol allows the three methyl hydrogens to be rapidly interchanged at room temperature, thereby making the electronic environment the same for these three hydrogens.)

We might thus expect the ${ }^{1} \mathrm{H}$ NMR spectrum of ethanol to show three peaks-one for the $\mathrm{CH}{3}$ protons, one for the $\mathrm{CH}{2}$ protons, and one for the OH proton, with the relative intensities of these peaks being 3:2:1. However, there is an additional effect, called spinspin coupling, in which the nuclear spins of the protons on one carbon affect the magnetic field experienced by the protons on an adjacent carbon. Different possible orientations of the proton spins on one carbon produce different magnetic fields at the protons of the adjacent carbon, thereby splitting the NMR transition of the protons at the adjacent carbon. For example, the two $\mathrm{CH}_{2}$ proton nuclei have the following four possible nuclear-spin orientations:

where the up and down arrows represent $M{I}=\frac{1}{2}$ and $M{I}=-\frac{1}{2}$, respectively. [Actually, because of the indistinguishability of identical particles, the middle two spin states in (10.62) must be replaced by linear combinations of these two states, to give nuclear-spin states that are analogous to the electron spin functions (10.22) to (10.25).] The middle two spin states in (10.62) have the same effect on the magnetic field felt by the $\mathrm{CH}{3}$ protons, so the four spin states in (10.62) produce three different magnetic fields at the $\mathrm{CH}{3}$ protons, thereby splitting the $\mathrm{CH}{3}$ proton NMR absorption line into three closely spaced lines of relative intensities $1: 2: 1$, corresponding to the number of $\mathrm{CH}{2}$ proton spin states that produce each magnetic field. One might expect the $\mathrm{CH}_{2}$ protons to also split the OH proton NMR line. However, even a trace of water present in the ethanol will catalyze a rapid exchange of the OH proton between different ethanol molecules, thereby eliminating the splitting of the OH line.

A similar analysis of the possible $\mathrm{CH}{3}$ proton orientations (Prob. 10.22) shows that the $\mathrm{CH}{3}$ protons split the ethanol $\mathrm{CH}{2}$ proton NMR line into four lines of relative intensities 1:3:3:1. The general rule is that a group of $n$ equivalent protons on an atom splits the NMR line of protons on an adjacent atom into $n+1$ lines. The spin-spin splitting (which is transmitted through the chemical bonds) is too weak to affect the NMR lines of protons separated by more than three bonds from the protons doing the splitting. In the proton NMR spectrum of $\mathrm{CH}{3} \mathrm{CH}{2} \mathrm{C}(\mathrm{O}) \mathrm{H}$, the $\mathrm{CH}{3}$ protons split the $\mathrm{CH}{2}$ proton line into 4 lines, and each of these is split into two lines by the $\mathrm{C}(\mathrm{O}) \mathrm{H}$ proton, so the $\mathrm{CH}{2}$ NMR line is split into 8 lines. (The intermolecular proton exchange in ethanol prevents the OH proton from splitting the $\mathrm{CH}_{2}$ proton NMR absorption.)

Nuclei with $I=0$ (for example, ${ }^{12} \mathrm{C},{ }^{16} \mathrm{O}$ ) don't split proton NMR peaks. It turns out that nuclei with $I>\frac{1}{2}$ (for example, ${ }^{35} \mathrm{Cl},{ }^{37} \mathrm{Cl},{ }^{14} \mathrm{~N}$ ) generally don't split proton NMR peaks. The ${ }^{19} \mathrm{~F}$ nucleus has $I=\frac{1}{2}$ and does split proton NMR peaks. Also, a quantummechanical analysis shows that the spin-spin interactions between equivalent protons don't affect the NMR spectrum.

The treatment just given (called a first-order analysis) is actually an approximation that is valid provided that the spin-spin splittings are much smaller than all the NMR frequency differences between chemically nonequivalent nuclei. In very large molecules, there will likely be chemically nonequivalent nuclei that are in only slightly different electronic environments, so the NMR frequency differences between them will be quite small and the first-order analysis will not hold. By increasing the strength of the applied magnetic field, one increases the NMR frequency differences between chemically nonequivalent nuclei, thereby tending to make the spectrum first-order, which is easier to analyze. Also, the signal strength is increased as the field is increased. Therefore, people try and use as high a field as is feasible. Current NMR research spectrometers have fields that correspond to proton NMR frequencies in the range 300 to 1000 MHz .

NMR spectroscopy is the premier structural research tool in organic chemistry, and special NMR techniques allow the structures of small proteins to be determined with the aid of NMR.