NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY
QUESTION 1 Answer all parts
a) Which isotopes of potassium (Z = 19) are NMR active ?
39K 40K 41K
b) Explain the definition of the chemical shift and calculate its 1H value of a
peak resonating at 500.138678 MHz vs the Si(CH3)4 reference resonating
at 500.133456 MHz. Propose an assignment for this proton.
c) The carbon – oxygen distance in 13C16O is slightly longer than in 13C17O due
to the mass dependence of the vibrational frequency and the anharmonicity of
the potential energy. Do you expect the same 13C chemical shift in these two
molecules ? Explain your answer.
d) Draw the 1H NMR spectrum of vinyl acetate.
e) Where is the bulk magnetization after a 90°(y) – 180°(x) – 90°(y) pulse
sequence ? Propose another pulse sequence that will produce the same effect.
f) The transverse magnetization (along x) obeys the following equation:
= − M#(t) dt
Explain the terms in this equation. What does the equation predict ?
g) What is the chemical shift anisotropy and what effect does it have on the
NMR spectrum of a static solid sample ?
h) The static 207Pb solid state NMR spectrum of Pb(NO3)2 is given below.
Estimate the isotropic chemical shift.
i) Consider the PF5 molecule in trigonal bipyramidal geometry. Are the
fluorines magnetically equivalent ? Explain your answer.
QUESTION 2 Answer all parts
a) The 1H decoupled 11B NMR spectrum of carborane nido-2,3-
(C2H5)2C2B4H6 (Figure 1i) is given in Figure 1ii.
Figure 1. (i) Chemical structure of nido-2,3-(C2H5)2C2B4H6. (ii) 1H decoupled 11B
NMR spectrum. (iii) 1H decoupled COSY 11B 11B NMR spectrum. Peaks at 0, -
3 and -46 ppm, in a 1:2:1 ratio, are detected in the 1D 11B NMR spectrum. The
peak at -12 ppm is an impurity and must not be considered. All spectra were
recorded at an external magnetic field B0 = 8.5 T and the 90° pulse was set to
I. What is the Larmor frequency ? How does this frequency depend on
the static external magnetic field ?
10 0 -10 -20 -30 -40 -50 ppm
10 0 -10 -20 -30 -40 -50
11B Chemical Shift / ppm
II. Calculate the 11B Larmor frequency in MHz of the spectrometer used
to record the data provided in Figure 1ii and iii. Assuming that this
Larmor frequency is the reference frequency, calculate the resonance
frequency in MHz of the 3 boron resonances of the carborane nido2,3-(C2H5)2C2B4H6 (Figure 1i). Give all your frequencies with 3
III. Deduce the frequency offset in kHz between these peaks. Using your
answer, comment whether or not the 11B spectrum is quantitative.
[9 marks] IV. Explain the origin of the chemical shift.
V. How does the chemical shift δ vary with the magnetic shielding σ ?
Provide an expression relating these 2 quantities. Predict qualitatively the
evolution of the chemical shift with the coordination number and propose a
preliminary assignment of the 11B NMR spectrum of nido-2,3-
b) The 1H decoupled two dimensional 11B 11B COSY NMR spectrum of nido2,3-(C2H5)2C2B4H6 is provided in Figure 1iii.
I. What is the basic principle of two dimensional NMR spectroscopy ?
Using a general pulse program, explain briefly how such spectrum is
II. What interaction does the COSY experiment probe ? Where does the
NMR signal of bonded spins appear ? Use your answer to explain the
spectrum in Figure 1iii and refine your preliminary assignment in
Question 2 a) V.
III. Confirm your assignment above from the signal integration.
IV. Predict the two dimensional 11B 11B NOESY NMR spectrum of nido2,3-(C2H5)2C2B4H6.
QUESTION 3 Answer all parts
a) The structure and 1H decoupled 13C solid state NMR spectra of crystalline
adamantane are given in Figure 2.
Figure 2. (i) Chemical structure of adamantane. 13C Cross Polarisation (CP)
Magic Angle Spinning (MAS) NMR spectra of (ii) adamantane and (iii) fully
deuterated adamantane (d18-adamantane). All spectra were recorded at a
Larmor frequency ν0 (
13C) = 100.74 MHz using a sample spinning frequency of
6.5 kHz and with 1H decoupling. In (ii), the shifts are at 38.5 and 29.4 ppm,
while in (iii) the multiplicities are centred at 37.1 and 28.2 ppm.
I. What does 1H decoupling refer to ?
II. Explain the 13C CP MAS NMR spectrum of adamantane (Figure 2ii).
Predict the corresponding spectrum without 1H decoupling.
III. Explain the 13C CP MAS NMR spectrum of d18-adamantane (Figure
2iii). Use your results to propose an assignment of the 13C CP MAS
NMR spectrum of adamantane.
IV. The dipolar coupling constant dij (in Hz) between two spins i and j is
obtained by the following expression:
d,- = 16µπ/ 2 hγr,-2,γExplain the terms of this equation and calculate the dipolar coupling
constant of a carbon and a deuterium separated by 1.1 Å.
V. Explain why no spinning sideband is observed in Figure 2iii.
[3 marks] b)
The following equation describes the time evolution of the z magnetization Mz:
/ exp − τ + M6
/ M6 τ =
M6(0) − M6 T<
I. Explain the terms of this equation.
II. Draw a sketch of the evolution of the z magnetization Mz for three
different values of T1 and justify your answer.
III. The inversion recovery experiment is not practical to measure very
long T1 as it requires extended acquisition times. An alternative
method consists of using the saturation recovery experiment: this
starts with the spin being irradiated for a long time so that it becomes
saturated (Mz = 0) followed by a delay τ, a 90° pulse and then
observation of the FID. Draw the corresponding pulse program.
IV. Show that the z-magnetization at the end of τ is given by:
/1 − exp− τ
M6 τ = M6T<
Make a sketch of Mz(τ) and predict its evolution.
[8 marks] V. Show that a plot of
ln S∝ − S(τ)
against τ provides values of T1 (S∞ andS(τ) are the NMR peak heights
at equilibrium and time τ, respectively).
VI. A plot of
ln S∝ − S(τ)
against τ for 13C gives the following data
Molecule adamantane d18-adamantane
Shifts / ppm 38.5 29.4 37.1 28.2
Gradient -0.0013 -0.001 -0.02 -0.01
Determine the T1 values and provide an explanation for their
Nucleus Gyromagnetic ratio γ
1H 26.7522⋅107 99.98 1/2
2H 4.1066⋅107 0.02 1
10B 2.8747⋅107 19.58 3
11B 8.5847⋅107 80.1 3/2
13C 6.7283⋅107 1.1 1/2
19F -25.1815⋅107 100 1/2
31P 10.8394⋅107 100 1/2
h = 6.626⋅10-34 m2 kg s-1
µ0 = 4 π ⋅10-7 H m-1 kB =
1.381⋅10-23 m2 kg s-2 K-1
R = 8.314 J mol-1 K
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