---
title: "NMR in Depth: Pulses, Relaxation and 2D"
book: "University Chemistry — Year 3"
subject: chemistry
language: en
chapter: 8
exercises: 12
source: https://one-course.com/books/chemistry/4/en/chapter/8-nmr-in-depth-pulses-relaxation-and-2d
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 8 — NMR in Depth: Pulses, Relaxation and 2D

A hospital MRI scanner and the NMR spectrometer of a chemistry department work on the same principle. Each places hydrogen nuclei in a strong magnetic field, tips their magnetisation with a short pulse of radio waves, and listens to the faint signal it induces as it precesses and relaxes. The scanner turns the signal into an image of soft tissue; the spectrometer into a list of chemical shifts and couplings, and, with several pulses in a row, into two-dimensional maps that show which atom is bonded to which. This chapter explains what the pulses do, why the signal fades, and how two-dimensional spectra are read — the method by which most new organic structures are now solved.

**You already know.**

The Year 1 volume interpreted ${}^1$H NMR spectra: chemical shift, shielding, equivalent protons, integration, spin–spin coupling, coupling constants and multiplets. The Year 2 volume added ${}^{13}$C NMR with broadband decoupling and DEPT, whose pulse sequence it took as given; the pulses are explained here. [Chapter 2](https://one-course.com/books/chemistry/4/en/chapter/2-many-electron-atoms-and-term-symbols#ch-b3-many-electron-atoms) treated spin as an angular momentum. From physics: a magnetic moment in a field has energy $-\boldsymbol\mu\cdot\mathbf B$, and populations of levels follow the Boltzmann factor $\eu^{-\Delta E/kT}$, developed in [Chapter 10](https://one-course.com/books/chemistry/4/en/chapter/10-statistical-thermodynamics-partition-functions#ch-b3-partition-functions).

![Left: a hospital MRI scanner. Right: an NMR laboratory; each superconducting magnet stands in its cryostat, refilled with liquid helium and liquid nitrogen from the dewars on the left. Photograph Shandchem, CC BY 2.0.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-advanced-nmr/img-eacbcc6419c7.jpg)

![Left: a hospital MRI scanner. Right: an NMR laboratory; each superconducting magnet stands in its cryostat, refilled with liquid helium and liquid nitrogen from the dewars on the left. Photograph Shandchem, CC BY 2.0.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-advanced-nmr/img-b21e496c2d1d.jpg)

*Left: a hospital MRI scanner. Right: an NMR laboratory; each superconducting magnet stands in its cryostat, refilled with liquid helium and liquid nitrogen from the dewars on the left. Photograph Shandchem, CC BY 2.0.*

## 8.1 Nuclear spins in a magnetic field

**Definition 8.1 (Nuclear spin quantum number, gyromagnetic ratio).**

A nucleus has a spin angular momentum of quantum number $I$, its *nuclear spin quantum number* ($I = 0$ for $\ce{^{12}C}$, $\ce{^{16}O}$; $\frac12$ for $\ce{^1H}$, $\ce{^{13}C}$, $\ce{^{15}N}$, $\ce{^{19}F}$, $\ce{^{31}P}$; 1 for $\ce{^2H}$, $\ce{^{14}N}$), and a magnetic moment $\boldsymbol\mu = \gamma\hat{\mathbf
I}$ proportional to it; $\gamma$ is the *gyromagnetic ratio*. Along a field $B_0$ (axis $z$), $\hat I_z$ takes the values $m_I\hbar$, $m_I = -I, \dots, I$.

**Proposition 8.2 (Zeeman levels).**

In a field $B_0$ the levels are $E_m = -m_I\gamma\hbar B_0$; the transitions $\Delta m_I = \pm1$ absorb at the frequency

$$
\nu_0 = \frac{\gamma B_0}{2\pi}.
$$

**Proof.** $E = -\boldsymbol\mu\cdot\mathbf B = -\gamma B_0\hat I_z$, of [eigenvalues](https://one-course.com/books/chemistry/4/en/chapter/1-quantum-mechanics-for-chemists-model-systems#def-b3-quantum-model-systems-operator) $-\gamma B_0m_I\hbar$. Adjacent levels differ by $\gamma\hbar B_0 = h\nu_0$. ∎

**Definition 8.3 (Larmor frequency).**

$\nu_0 = \gamma B_0/2\pi$ is the *Larmor frequency* of the nucleus: the frequency of the resonance, and also the frequency at which its magnetic moment precesses about the field.

**Example 8.4 (The common nuclei).**

For a spin-$\frac12$ nucleus of moment $\mu$, $\gamma/2\pi = \mu/(Ih) = 2\mu/h$. With the measured moments: $\ce{^1H}$ $42.58\,\mathrm{MHz}/\mathrm{T}$, $\ce{^{19}F}$ $40.08\,\mathrm{MHz}/\mathrm{T}$, $\ce{^{31}P}$ $17.25\,\mathrm{MHz}/\mathrm{T}$, $\ce{^{13}C}$ $10.71\,\mathrm{MHz}/\mathrm{T}$, $\ce{^{15}N}$ $-4.32\,\mathrm{MHz}/\mathrm{T}$. A “$400\,\mathrm{MHz}$ spectrometer” has $B_0 = 9.4\,\mathrm{T}$: its protons resonate at $400.2\,\mathrm{MHz}$, its carbons at $100.7\,\mathrm{MHz}$.

**Proposition 8.5 (A tiny population difference).**

For spin $\frac12$ at temperature $T$, the excess of nuclei in the lower level is

$$
\frac{N_\alpha - N_\beta}{N} \approx \frac{\gamma\hbar B_0}{2kT}.
$$

**Proof.** $N_\beta/N_\alpha = \eu^{-\Delta E/kT} \approx 1 - \Delta E/kT$ since $\Delta E \ll kT$; so $(N_\alpha - N_\beta)/(N_\alpha + N_\beta) \approx \Delta E/2kT$, with $\Delta E = \gamma\hbar B_0$. ∎

For protons at $9.4\,\mathrm{T}$ and $300\,\mathrm{K}$ the excess is $3.2 \times 10^{-5}$: only three nuclei in a hundred thousand contribute. Since both the excess and the voltage induced in the coil grow with $\gamma$ and $B_0$, the signal grows roughly as $\gamma^3B_0^2$: the reason for ever stronger magnets, and for the low sensitivity of nuclei with small $\gamma$ and low natural abundance such as $\ce{^{13}C}$.

## 8.2 Pulses and the free induction decay

**Definition 8.6 (Net magnetisation, rotating frame).**

The *net magnetisation* $\mathbf M$ of a sample is the sum of its nuclear magnetic moments per unit volume; at equilibrium it is $M_0$ along $\mathbf B_0$. The *rotating frame* is a set of axes $x'$, $y'$, $z$ turning about $z$ at the frequency of the radio waves applied; in it the precession at $\nu_0$ is seen slowed to the offset $\nu_0 -
\nu_{\mathrm{rf}}$.

**Proposition 8.7 (Precession).**

A magnetisation in a field obeys $\dd\mathbf M/\dd t = \gamma\,\mathbf M\times\mathbf B$; in a static field $B_0\mathbf e_z$, its transverse component turns about $z$ at the angular frequency $\omega_0 = \gamma B_0$ while $M_z$ stays constant.

**Partial proof.** The equation is the classical torque equation for a magnetic moment carrying angular momentum, admitted from physics. With $\mathbf B = B_0\mathbf e_z$: $\dot M_x = \gamma
B_0M_y$, $\dot M_y = -\gamma B_0M_x$, $\dot M_z = 0$, whose solution is $M_x + \iu M_y \propto
\eu^{-\iu\gamma B_0t}$: a rotation at $\omega_0$ (clockwise seen from $+z$ for $\gamma > 0$). ∎

**Definition 8.8 (Radiofrequency pulse, flip angle).**

A *radiofrequency pulse* is a short burst of radio waves at $\nu_{\mathrm{rf}} \approx \nu_0$, whose magnetic field $B_1$, fixed in the [rotating frame](#def-b3-advanced-nmr-magnetisation) along $x'$, turns $\mathbf M$ about $x'$. The angle turned is the *flip angle* $\theta = \gamma B_1t_p$ for a pulse of duration $t_p$: a $90^\circ$ pulse brings $\mathbf M$ into the $x'y'$ plane, a $180^\circ$ pulse inverts it.

![The magnetisation in the rotating frame. A 90 pulse about x' turns M from z to y' (with > 0 the rotation follows the left-hand rule about B_1, the usual convention); a 180 pulse inverts it.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-advanced-nmr/fig-ac9d9940739e.svg)

*The magnetisation in the [rotating frame](#def-b3-advanced-nmr-magnetisation). A $90^\circ$ pulse about $x'$ turns $\mathbf M$ from $z$ to $y'$ (with $\gamma > 0$ the rotation follows the left-hand rule about $\mathbf B_1$, the usual convention); a $180^\circ$ pulse inverts it.*

**Definition 8.9 (Free induction decay).**

After a $90^\circ$ pulse, the transverse magnetisation precesses at the [Larmor frequency](#def-b3-advanced-nmr-larmor) and induces an oscillating voltage in the coil around the sample, which dies away as the magnetisation relaxes: the *free induction decay* (FID).

**Theorem 8.10 (From the FID to the spectrum).**

The Fourier transform of a decaying oscillation $\eu^{-t/T_2}\cos(2\pi\nu t)$ ($t \ge 0$) is, near $\nu$, a Lorentzian line of full width at half maximum $1/(\pi T_2)$.

**Proof.** Write the signal $\eu^{-t/T_2}\eu^{2\pi\iu\nu t}$ and integrate: $\int_0^\infty\eu^{-t/T_2}\eu^{2\pi\iu(\nu - f)t}\dd t = \frac{1}{1/T_2 - 2\pi\iu(\nu - f)}$, whose real part is $\frac{T_2}{1 + 4\pi^2T_2^2(f - \nu)^2}$. It is maximal at $f = \nu$ and falls to half when $2\pi T_2|f - \nu| = 1$, i.e. at $f - \nu = \pm1/(2\pi T_2)$: a full width $1/(\pi T_2)$. ∎

![Left: the FID of two lines (offsets 100 and 250\, Hz, T_2 = 0.10\, s), beating as the two frequencies interfere inside an exponential envelope (dashed). Right: its Fourier transform, two Lorentzian lines 3.2\, Hz wide.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-advanced-nmr/fig-02c38652c5ae.svg)

*Left: the FID of two lines (offsets 100 and $250\,\mathrm{Hz}$, $T_2 = 0.10\,\mathrm{s}$), beating as the two frequencies interfere inside an exponential envelope (dashed). Right: its Fourier transform, two Lorentzian lines $3.2\,\mathrm{Hz}$ wide.*

**Proposition 8.11 (Signal averaging).**

Adding $N$ FIDs multiplies the signal by $N$ and the random noise by $\sqrt N$: the signal-to-noise ratio grows as $\sqrt N$.

**Proof.** The signal adds coherently. The noise of independent acquisitions has zero mean and variance $\sigma^2$ each; the variance of a sum of $N$ independent terms is $N\sigma^2$ (the propagation law of the Year 2 volume), so its standard deviation is $\sqrt N\sigma$. ∎

DEPT, which sorts ${}^{13}$C signals by the number of attached hydrogens, applies a sequence of pulses to both protons and carbons separated by delays of $1/2J$: the large proton magnetisation is transferred to the carbon through the one-bond coupling, which multiplies the carbon signal by about $\gamma_H/\gamma_C \approx 4$, and the final proton pulse angle weights CH, $\ce{CH2}$ and $\ce{CH3}$ differently. The same idea — moving magnetisation between nuclei through their couplings — underlies the two-dimensional experiments below.

## 8.3 Relaxation

**Definition 8.12 (Longitudinal and transverse relaxation times).**

The return of $M_z$ to $M_0$ after a perturbation is exponential, with the *longitudinal relaxation time* $T_1$ (spin–lattice relaxation: energy given to the surroundings). The decay of the transverse magnetisation to zero is exponential with the *transverse relaxation time* $T_2 \le T_1$ (spin–spin relaxation: loss of phase coherence between spins).

**Proposition 8.13 (Inversion recovery).**

After a $180^\circ$ pulse, $M_z(t) = M_0(1 - 2\eu^{-t/T_1})$; it passes through zero at $t_{\mathrm{null}} = T_1\ln2$.

**Proof.** The relaxation equation $\dd M_z/\dd t = (M_0 - M_z)/T_1$ with $M_z(0) = -M_0$ has the solution $M_0 - 2M_0\eu^{-t/T_1}$, which vanishes when $\eu^{-t/T_1} = \frac12$. ∎

**Method 8.14 (Measuring T1T_1T1​ and setting a quantitative recycle delay).**

1. Run the sequence $180^\circ$ – $\tau$ – $90^\circ$ –acquire for a series of delays $\tau$ , waiting at least $5T_1$ between scans.
2. Fit each signal to $M_0(1 - 2\eu^{-\tau/T_1})$ , or read the null: $T_1 =  t_{\mathrm{null}}/\ln2$ .
3. For quantitative integration with $90^\circ$ pulses, wait at least $5T_1$ of the slowest nucleus between scans: then $\eu^{-5} < 1\,\%$ of the magnetisation is missing.

![Relaxation (model values). The inverted longitudinal magnetisation recovers through zero at T_1 2; the transverse magnetisation decays faster, with T_2.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-advanced-nmr/fig-1978201cfdd1.svg)

*Relaxation (model values). The inverted longitudinal magnetisation recovers through zero at $T_1\ln2$; the transverse magnetisation decays faster, with $T_2$.*

**Definition 8.15 (Spin echo).**

In the sequence $90^\circ$–$\tau$–$180^\circ$–$\tau$, spins that drifted apart in phase because of field inhomogeneities are refocused at time $2\tau$, producing a *spin echo*; its amplitude decays with the true $T_2$, free of the inhomogeneity of the magnet.

**Definition 8.16 (Nuclear Overhauser effect).**

Saturating or perturbing one spin changes, through their dipolar relaxation, the intensity of the signal of another spin close in space: the *nuclear Overhauser effect* (NOE). Its build-up rate is proportional to $r^{-6}$, so it is seen only between nuclei less than about $0.5\,\mathrm{nm}$ apart, whether bonded or not.

## 8.4 Two-dimensional NMR

**Definition 8.17 (Two-dimensional spectrum, cross peak, diagonal peak).**

A *two-dimensional spectrum* is recorded by repeating a pulse sequence with an incremented delay $t_1$ before the acquisition time $t_2$; a double Fourier transform gives a map of intensity against two frequencies. A *cross peak* at $(\delta_1,
\delta_2)$ shows that the nuclei at $\delta_1$ and $\delta_2$ are connected (by a coupling or by proximity); a *diagonal peak* at $(\delta, \delta)$ belongs to a single nucleus.

**Definition 8.18 (COSY, HSQC, HMBC, NOESY).**

*COSY* (correlation spectroscopy) correlates protons coupled to each other, usually through three bonds (H–C–C–H). *HSQC* (heteronuclear single-quantum coherence) correlates each carbon with the protons directly bonded to it. *HMBC* (heteronuclear multiple-bond correlation) correlates carbons with protons two or three bonds away, across quaternary carbons and heteroatoms. *NOESY* correlates protons close in space through the [nuclear Overhauser effect](#def-b3-advanced-nmr-noe).

![Pulse sequences (time to the right; filled boxes: pulses, labelled with their flip angles in degrees; blue: the recorded signal). The spin echo peaks at 2 after the first pulse.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-advanced-nmr/fig-d8a9871b1a8a.svg)

*Pulse sequences (time to the right; filled boxes: pulses, labelled with their [flip angles](#def-b3-advanced-nmr-pulse) in degrees; blue: the recorded signal). The [spin echo](#def-b3-advanced-nmr-echo) peaks at $2\tau$ after the first pulse.*

**Method 8.19 (Solving a structure with two-dimensional spectra).**

1. From the formula, the ${}^1$ H integrals and the ${}^{13}$ C/DEPT spectrum, list the CH, $\ce{CH2}$ , $\ce{CH3}$ and quaternary carbons.
2. [HSQC](#def-b3-advanced-nmr-experiments) : attach each proton signal to its carbon.
3. [COSY](#def-b3-advanced-nmr-experiments) : join the protonated carbons into spin systems (chains of neighbours).
4. [HMBC](#def-b3-advanced-nmr-experiments) : join the spin systems across quaternary carbons, carbonyls and heteroatoms, where [COSY](#def-b3-advanced-nmr-experiments) is silent.
5. [NOESY](#def-b3-advanced-nmr-experiments) : fix relative configurations and conformations from through-space contacts.
6. Check every signal against the proposed structure.

![Two-dimensional spectra of ethyl butanoate, CH3CH2CH2COOCH2CH3, drawn at its measured shifts. COSY: black, diagonal peaks; red, cross peaks between protons on neighbouring carbons — two spin systems, ethyl and propyl, with no cross peak across the ester oxygen. Right: HSQC peaks (squares) join each proton to its carbon; HMBC peaks (circles) show the OCH_2 protons (4.13\, ppm) and the CH_2C=O protons (2.28\, ppm) both reaching the carbonyl carbon at 173.6\, ppm.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-advanced-nmr/fig-add44526fbe2.svg)

*Two-dimensional spectra of ethyl butanoate, $\ce{CH3CH2CH2COOCH2CH3}$, drawn at its measured shifts. [COSY](#def-b3-advanced-nmr-experiments): black, [diagonal peaks](#def-b3-advanced-nmr-two-d); red, [cross peaks](#def-b3-advanced-nmr-two-d) between protons on neighbouring carbons — two spin systems, ethyl and propyl, with no [cross peak](#def-b3-advanced-nmr-two-d) across the ester oxygen. Right: [HSQC](#def-b3-advanced-nmr-experiments) peaks (squares) join each proton to its carbon; [HMBC](#def-b3-advanced-nmr-experiments) peaks (circles) show the OCH$_2$ protons ($4.13\,\mathrm{ppm}$) and the CH$_2$C=O protons ($2.28\,\mathrm{ppm}$) both reaching the carbonyl carbon at $173.6\,\mathrm{ppm}$.*

## 8.5 Other nuclei and dynamics

Fluorine-19 and phosphorus-31, spin $\frac12$, 100 % abundant and with large $\gamma$, are almost as easy to observe as protons; their wide ranges of shifts make them precise probes of their environment (drugs, phosphates, ligands such as phosphines in [Chapter 20](https://one-course.com/books/chemistry/4/en/chapter/20-organometallic-chemistry-bonding-and-ligands#ch-b3-organometallic-bonding)). Nitrogen-15 is rare and weak, and is usually observed indirectly through the protons bonded to it. Coupling between different nuclei is removed by decoupling, as for ${}^{13}$C in the Year 2 volume.

**Definition 8.20 (Coalescence temperature).**

When a nucleus exchanges between two environments of shifts separated by $\Delta\nu$ (in hertz), its two lines broaden as the exchange speeds up and merge into one at the *coalescence temperature* $T_c$.

**Proposition 8.21 (Rate at coalescence).**

For two equally populated sites without coupling, the exchange rate constant at coalescence is $k_c = \pi\Delta\nu/\sqrt2$.

**Proof.** *Admitted at this level.* ∎

The derivation, from the Bloch equations with exchange, is treated in more advanced courses. With the Eyring equation ([Chapter 12](https://one-course.com/books/chemistry/4/en/chapter/12-theories-of-reaction-rates#ch-b3-rate-theories)) the rate gives the barrier: dynamic NMR measures rotations about amide bonds, ring flips and ligand exchanges with barriers of 40 to $100\,\mathrm{kJ}/\mathrm{mol}$.

**Example 8.22 (An MRI image).**

In magnetic resonance imaging a field gradient makes the [Larmor frequency](#def-b3-advanced-nmr-larmor) of the water protons depend on position, so that the Fourier transform of the signal is a map of where they are. Contrast comes from relaxation: the repetition time and echo delays are chosen to weight the image by $T_1$ or by $T_2$, which differ between tissues; contrast agents containing gadolinium shorten the $T_1$ of nearby water ([Chapter 24](https://one-course.com/books/chemistry/4/en/chapter/24-bioinorganic-chemistry#ch-b3-bioinorganic)).

**In the lab — Preparing a spectrometer.**

The sample, dissolved in a deuterated solvent, is lowered into the probe. The spectrometer locks on the deuterium signal to compensate the slow drift of the field; the probe is tuned and matched to the frequency of each nucleus; the field is made homogeneous over the sample by adjusting the shim coils until the lines are narrow. The magnet’s field is permanent: steel tools, credit cards and pacemakers are kept outside the marked safety line.

**History — From resonance to two dimensions.**

Felix Bloch and Edward Purcell detected nuclear magnetic resonance in bulk matter in 1946 (Nobel Prize in Physics, 1952). Richard Ernst introduced pulsed Fourier-transform NMR in 1966 and, following an idea of Jean Jeener, two-dimensional spectroscopy in the 1970s (Nobel Prize in Chemistry, 1991).

## 8.6 Exercises

**Exercise 8.1 ★.**

Compute the Larmor frequencies of $\ce{^1H}$, $\ce{^{13}C}$ and $\ce{^{19}F}$ at $14.1\,\mathrm{T}$ ($\gamma/2\pi = 42.58$, 10.71 and $40.08\,\mathrm{MHz}/\mathrm{T}$).

**Solution of Exercise 8.1.**

$\nu_0 = (\gamma/2\pi)B_0$: $\ce{^1H}$ $600.4\,\mathrm{MHz}$, $\ce{^{13}C}$ $151.0\,\mathrm{MHz}$, $\ce{^{19}F}$ $565.1\,\mathrm{MHz}$.

**Exercise 8.2 ★.**

Compute the relative excess of protons in the lower level at $9.4\,\mathrm{T}$ and $300\,\mathrm{K}$. By how much does it change at $4\,\mathrm{K}$?

**Solution of Exercise 8.2.**

$h\nu_0/2kT = 6.626\times10^{-34} \times 400.2\times10^6/(2 \times 1.381\times10^{-23} \times 300)
= 3.2 \times 10^{-5}$. At $4\,\mathrm{K}$ it is 75 times larger, $2.4 \times 10^{-3}$ (the approximation still holds).

**Exercise 8.3 ★.**

A $90^\circ$ pulse lasts $10\,\text{µ}\mathrm{s}$ for protons. What is $\gamma B_1/2\pi$? How long is a $180^\circ$ pulse, and what does a $5\,\text{µ}\mathrm{s}$ pulse do?

**Solution of Exercise 8.3.**

$\theta = \gamma B_1t_p = \pi/2$ in $10\,\text{µ}\mathrm{s}$: $\gamma B_1/2\pi = 1/(4 \times
10\,\text{µ}\mathrm{s}) = 25\,\mathrm{kHz}$. A $180^\circ$ pulse lasts $20\,\text{µ}\mathrm{s}$; $5\,\text{µ}\mathrm{s}$ gives $45^\circ$.

**Exercise 8.4 ★.**

A line is $3.2\,\mathrm{Hz}$ wide at half height. What is $T_2$ (assuming a perfectly homogeneous field)?

**Solution of Exercise 8.4.**

$T_2 = 1/(\pi \times 3.2) = 0.10\,\mathrm{s}$.

**Exercise 8.5 ★★.**

Using signal $\propto\gamma^3$ per nucleus and the 1.1 % natural abundance of $\ce{^{13}C}$, compare the sensitivity of ${}^{13}$C and ${}^1$H NMR. How many more scans does a ${}^{13}$C spectrum need for the same signal-to-noise ratio?

**Solution of Exercise 8.5.**

$(10.71/42.58)^3 = 0.0159$, times 0.011: $1.75 \times 10^{-4}$, about 1/5700 of the proton signal. Since S/N $\propto\sqrt N$, equal S/N needs $5700^2 \approx 3 \times 10^{7}$ times more scans — impossible, which is why ${}^{13}$C spectra use more concentrated samples, decoupling and polarisation transfer.

**Exercise 8.6 ★★.**

In an inversion-recovery experiment the signal of a carbon vanishes at $\tau = 1.39\,\mathrm{s}$. Find $T_1$ and the recycle delay for quantitative work.

**Solution of Exercise 8.6.**

$T_1 = 1.39/\ln2 = 2.0\,\mathrm{s}$; recycle delay $\ge 5T_1 = 10\,\mathrm{s}$.

**Exercise 8.7 ★★.**

In the [COSY](#def-b3-advanced-nmr-experiments) spectrum of butanone, $\ce{CH3COCH2CH3}$, which [cross peaks](#def-b3-advanced-nmr-two-d) appear? Which carbons show [HMBC](#def-b3-advanced-nmr-experiments) peaks from the singlet methyl protons?

**Solution of Exercise 8.7.**

[COSY](#def-b3-advanced-nmr-experiments): one [cross peak](#def-b3-advanced-nmr-two-d), between the $\ce{CH2}$ quartet and the $\ce{CH3}$ triplet of the ethyl group; the $\ce{CH3CO}$ singlet couples to nothing. [HMBC](#def-b3-advanced-nmr-experiments) from the singlet methyl: the carbonyl carbon (two bonds) and the $\ce{CH2}$ carbon (three bonds).

**Exercise 8.8 ★★.**

Two methyl signals of an amide, $50\,\mathrm{Hz}$ apart at low temperature, coalesce at $330\,\mathrm{K}$. Compute the exchange rate at coalescence and the Gibbs energy of activation with $\Delta G^\ddagger = RT_c\ln(k_BT_c/hk_c)$.

**Solution of Exercise 8.8.**

$k_c = \pi \times 50/\sqrt2 = 111\,\mathrm{s}^{-1}$. $\Delta G^\ddagger = 8.314 \times 330 \times
\ln(6.88\times10^{12}/111) = 68\,\mathrm{kJ}/\mathrm{mol}$.

**Exercise 8.9 ★★.**

An NOE between protons H$_a$ and H$_b$ is four times stronger than between H$_a$ and H$_c$, at $0.30\,\mathrm{nm}$. Estimate the H$_a$–H$_b$ distance.

**Solution of Exercise 8.9.**

NOE $\propto r^{-6}$: $r_b = 0.30 \times 4^{-1/6} = 0.24\,\mathrm{nm}$.

**Exercise 8.10 ★★★.**

Show, by solving $\dd\mathbf M/\dd t = \gamma\mathbf M\times\mathbf B_1$ in the [rotating frame](#def-b3-advanced-nmr-magnetisation) with $\mathbf B_1$ along $x'$, that a pulse rotates $\mathbf M$ by $\gamma B_1t_p$ about $x'$.

**Solution of Exercise 8.10.**

In the [rotating frame](#def-b3-advanced-nmr-magnetisation) on resonance only $\mathbf B_1 = B_1\mathbf e_{x'}$ remains: $\dot M_{x'} = 0$, $\dot M_{y'} = \gamma B_1M_z$, $\dot M_z = -\gamma B_1M_{y'}$ (components of $\gamma\mathbf M\times\mathbf B_1$). So $M_{x'}$ is constant and $(M_{y'}, M_z)$ turns at the angular rate $\gamma B_1$: after $t_p$, by the angle $\gamma B_1t_p$ about $x'$.

**Exercise 8.11 ★★★.**

Explain why the [spin echo](#def-b3-advanced-nmr-echo) refocuses the dephasing due to an inhomogeneous field but not that due to random interactions between spins. Which time constant does the echo amplitude measure?

**Solution of Exercise 8.11.**

A spin in a slightly stronger field gains phase at a constant extra rate during $\tau$; the $180^\circ$ pulse reverses its phase, and it loses exactly the same amount during the second $\tau$: all such spins realign at $2\tau$. Random spin–spin interactions fluctuate in time and do not repeat in the second interval: their dephasing is not undone. The echo amplitude decays with the true $T_2$.

**Exercise 8.12 ★★★.**

Propose the structure of a compound $\ce{C4H8O2}$ whose ${}^1$H spectrum shows a quartet (2H, 4.12 ppm), a singlet (3H, 2.05 ppm) and a triplet (3H, 1.26 ppm), and whose [HMBC](#def-b3-advanced-nmr-experiments) spectrum correlates the quartet protons with a carbon at 171 ppm (data of the exercise). Which other ester of the same formula would give a different [HMBC](#def-b3-advanced-nmr-experiments) pattern, and how?

**Solution of Exercise 8.12.**

Ethyl ethanoate, $\ce{CH3COOCH2CH3}$: $\ce{OCH2}$ quartet (4.12), $\ce{CH3CO}$ singlet (2.05), $\ce{CH3}$ triplet; the quartet protons see the carbonyl carbon three bonds away, through the ester oxygen. Methyl propanoate, $\ce{CH3CH2COOCH3}$, would show its quartet near 2.3 ppm (on a carbon near 27 ppm in [HSQC](#def-b3-advanced-nmr-experiments), not 60) and a singlet $\ce{OCH3}$ near 3.7 ppm whose [HMBC](#def-b3-advanced-nmr-experiments) peak to the carbonyl crosses the oxygen.

## 8.7 Problem: The Pineapple Ester in Two Dimensions

**Problem 8.1.**

Weekend problem — ethyl butanoate solved again, from two-dimensional spectra: spin systems from COSY, carbons from HSQC, the ester linkage from HMBC, and the delay needed for a quantitative carbon spectrum

An ester $\ce{C6H12O2}$ from the aroma of pineapple shows ${}^1$H signals at 4.13 (2H, q), 2.28 (2H, t), 1.66 (2H, m), 1.25 (3H, t) and 0.95 ppm (3H, t), and ${}^{13}$C signals at 173.6, 60.1, 36.3, 18.6, 14.3 and 13.7 ppm, recorded at $9.4\,\mathrm{T}$. Its 2D spectra are those of the figure of section 4. Data of the problem: the $T_1$ values of the carbons are about $20\,\mathrm{s}$ for the carbonyl and 2 to $6\,\mathrm{s}$ for the others.

**Part I — One dimension.**

1. Compute the degree of unsaturation of $\ce{C6H12O2}$ .
2. Which ${}^{13}$ C signal is the carbonyl? Which is the carbon bonded to oxygen?
3. How many protonated carbons are there? What would DEPT-135 show?
4. At $9.4\,\mathrm{T}$ , what are the Larmor frequencies of ${}^1$ H and ${}^{13}$ C?
5. The coupling of the quartet is $7.2\,\mathrm{Hz}$ . How wide, in ppm, is the quartet at $400\,\mathrm{MHz}$ ?
6. Why can the 1D spectra alone leave the order of the fragments open?

**Part II — [COSY](#def-b3-advanced-nmr-experiments).**

7. List the [cross peaks](#def-b3-advanced-nmr-two-d) of the [COSY](#def-b3-advanced-nmr-experiments) map.
8. Deduce the two spin systems.
9. Why is there no [cross peak](#def-b3-advanced-nmr-two-d) between 4.13 and 2.28 ppm?
10. Which proton signal is coupled to two others? Explain its multiplicity.
11. What would a [cross peak](#def-b3-advanced-nmr-two-d) between 0.95 and 1.25 ppm have meant?
12. Draw the two fragments.

**Part III — [HSQC](#def-b3-advanced-nmr-experiments) and [HMBC](#def-b3-advanced-nmr-experiments).**

13. From the [HSQC](#def-b3-advanced-nmr-experiments) , assign each carbon to its protons.
14. Which two methyl carbons are distinguished only by the [HSQC](#def-b3-advanced-nmr-experiments) ?
15. Which [HMBC](#def-b3-advanced-nmr-experiments) peak links the ethyl fragment to the carbonyl, and through how many bonds?
16. Which [HMBC](#def-b3-advanced-nmr-experiments) peaks link the propyl fragment to the carbonyl?
17. Deduce the structure, and exclude its isomer propyl propanoate.
18. Why are [HMBC](#def-b3-advanced-nmr-experiments) peaks across four bonds or more usually absent?

**Part IV — A quantitative carbon spectrum.**

19. Why are the integrals of an ordinary ${}^{13}$ C spectrum not proportional to the number of carbons?
20. What fraction of its equilibrium magnetisation does the carbonyl recover if scans are repeated every $5\,\mathrm{s}$ with $90^\circ$ pulses (starting each time from zero)?
21. What fraction after $5T_1$ ?
22. Besides relaxation, which effect of proton decoupling distorts carbon integrals, and how is it suppressed?
23. With a delay of $5T_1$ , how long do 256 scans take?
24. State the result: the minimum recycle delay for a quantitative ${}^{13}$ C spectrum of this ester.

**Solution of Problem 8.1.**

**1.** $(2 \times 6 + 2 - 12)/2 = 1$: one C=O. **2.** 173.6 ppm: the ester carbonyl; 60.1 ppm: the $\ce{OCH2}$. **3.** Five; DEPT-135: $\ce{CH2}$ negative (60.1, 36.3, 18.6), $\ce{CH3}$ positive (14.3, 13.7), the carbonyl absent. **4.** $400.2\,\mathrm{MHz}$ and $100.7\,\mathrm{MHz}$. **5.** $3J = 21.6\,\mathrm{Hz} = 0.054\,\mathrm{ppm}$. **6.** The multiplicities show which groups are neighbours inside each fragment, but not how the fragments join across the oxygen and the carbonyl. **7.** 4.13–1.25; 2.28–1.66; 1.66–0.95 (and the symmetric ones). **8.** $\ce{OCH2CH3}$ (4.13, 1.25) and $\ce{CH2CH2CH3}$ (2.28, 1.66, 0.95). **9.** The $\ce{OCH2}$ and $\ce{CH2C=O}$ protons are separated by the oxygen and the carbonyl carbon: five bonds, no resolved coupling. **10.** 1.66 ppm, between $\ce{CH2}$ and $\ce{CH3}$: coupled to 2 + 3 = 5 protons with similar $J$, a sextet (a multiplet). **11.** Coupled methyls, hence two $\ce{CH3}$ on adjacent carbons (a $\ce{CH3-CH3}$ unit): impossible here. **12.** $\ce{-O-CH2-CH3}$ and $\ce{-CH2-CH2-CH3}$, plus the $\ce{C=O}$. **13.** 4.13/60.1; 2.28/36.3; 1.66/18.6; 1.25/14.3; 0.95/13.7. **14.** 14.3 (ethyl $\ce{CH3}$) and 13.7 ppm (butanoyl $\ce{CH3}$), only $0.6\,\mathrm{ppm}$ apart. **15.** 4.13 ppm to 173.6 ppm: H–C–O–C, three bonds. **16.** 2.28 ppm (two bonds) and 1.66 ppm (three bonds) to 173.6 ppm. **17.** $\ce{CH3CH2CH2C(=O)OCH2CH3}$, ethyl butanoate. Propyl propanoate would put an $\ce{OCH2}$ triplet (not a quartet) near 4.0 ppm and a quartet near 2.3 ppm. **18.** Four-bond couplings are usually below 1 Hz; the [HMBC](#def-b3-advanced-nmr-experiments) delay, tuned to 5–10 Hz, does not select them. **19.** Carbons relax at different rates and receive different nuclear Overhauser enhancements from decoupling. **20.** $1 - \eu^{-5/20} = 0.22$: the carbonyl is under-represented by a factor 4. **21.** $1 - \eu^{-5} = 0.993$. **22.** The NOE from proton decoupling, different for each carbon; it is removed by inverse-gated decoupling (decoupler on only during acquisition). **23.** $256 \times 100\,\mathrm{s} = 25\,600\,\mathrm{s}$, about 7 hours. **24.** **$5T_1$ of the carbonyl carbon: $100\,\mathrm{s}$ between scans.**