NMR Spectroscopy Questions and Answers

Answer: A

A nucleus is NMR active only when its spin quantum number I is not zero, because only then does it carry a magnetic moment that can take up different orientations in a field. Carbon-12 and oxygen-16 have I = 0 and are silent, which is why natural abundance carbon work uses the 1.1 per cent carbon-13 instead. An even proton count does not decide the matter on its own, nuclear charge is irrelevant, and the electric dipole moment belongs to infrared and microwave spectroscopy rather than to magnetic resonance. Spin is the whole entry requirement.

Answer: B

The proton has I = 1/2, so it takes 2I + 1 = 2 orientations in a magnetic field and gives a single sharp transition. Spin-half nuclei have no electric quadrupole moment, so they are not broadened by the electric field gradients around them, and that is the second reason hydrogen-1, carbon-13, fluorine-19 and phosphorus-31 dominate routine work. Nuclei with I = 1 such as deuterium, or I = 3/2 such as chlorine-35, are quadrupolar and give broad lines. A spin of zero would give no signal at all.

Answer: C

The number of allowed orientations is 2I + 1, each labelled by a magnetic quantum number m running from +I to -I in integer steps. A spin-half nucleus therefore has two levels, deuterium with I = 1 has three, and a spin of 3/2 has four. The degeneracy of these levels is lifted only by the applied field, which is what Zeeman splitting means, and the selection rule for an NMR transition is a change of one in m. Counting the levels correctly is the first step in predicting how many lines a nucleus can give.

Answer: D

The Larmor equation says the resonance frequency is directly proportional to the applied field, the constant of proportionality being the gyromagnetic ratio of the nucleus divided by two pi. Doubling the field doubles the frequency, which is exactly why instruments are named by the proton frequency they reach: a 400 MHz spectrometer holds about 9.4 tesla. Neither an inverse nor a square root dependence appears anywhere in the treatment, and a field-independent frequency would make the whole technique useless for chemistry.

Answer: A

Multiplying the field by the gyromagnetic ratio in frequency units gives the answer directly:

\[ \nu = 42.58 \times 11.74 \approx 500 \]

so the instrument is a 500 MHz machine. The same arithmetic puts 400 MHz at 9.4 T and 600 MHz at 14.1 T, and those three fields cover most departmental spectrometers. Naming an instrument by its proton frequency rather than its field is a convenience: the frequency is what the chemist tunes and what sets the dispersion of the spectrum.

Answer: B

Two factors multiply together: carbon-13 is only 1.1 per cent of natural carbon, and its gyromagnetic ratio is about a quarter that of the proton. Sensitivity goes as the cube of the gyromagnetic ratio for a given field, so the combined penalty is roughly six thousand fold. Carbon-13 has I = 1/2 and so is certainly not silent, its relaxation is slow rather than fast, and its resonance sits comfortably in range. The remedy is many more scans, and the accumulation of signal is why a carbon spectrum takes minutes where a proton spectrum takes seconds.

Answer: C

Because the energy gap in NMR is minute compared with thermal energy, the excess in the lower level is only a few nuclei per million even at high field. That small excess is the entire source of the signal, and it is the fundamental reason NMR is insensitive next to infrared or ultraviolet spectroscopy. The populations are not equal, or there would be no net absorption at all, but they are very nearly so. Raising the field or lowering the temperature increases the excess, which is why high field magnets and cryoprobes are worth their cost.

Answer: D

Both parts are true and the second is the reason for the first. A bigger field widens the Zeeman gap, so the Boltzmann excess grows and the signal with it, and because chemical shift in hertz is proportional to field while coupling constants are not, crowded multiplets separate. That double benefit is why laboratories pay for higher field magnets despite the cost of the cryogens and the siting. The gain in sensitivity goes roughly as the field to the three halves power, and the gain in dispersion is simply linear.

Answer: A

The gyromagnetic ratio is the magnetic moment divided by the spin angular momentum, a fixed property of each isotope. It sets the resonance frequency in a given field through the Larmor equation and it sets the sensitivity, which scales as its cube. The proton has the largest value of the common nuclei, which is why it is the workhorse. Mass to charge ratio belongs to mass spectrometry, level populations follow from the Boltzmann law, and linewidth is a relaxation matter. Knowing the ratio for a nucleus lets you predict its frequency at any field.

Answer: B

Both statements hold. Zero spin means no nuclear magnetic moment and so no Zeeman levels to move between, and the two commonest isotopes of carbon and oxygen are exactly of that kind. The consequence is that organic NMR of carbon rests on the 1.1 per cent carbon-13, and that oxygen work uses the scarce oxygen-17 or is not attempted at all. Nuclei with an even number of both protons and neutrons are the ones that come out at I = 0, which is a useful rule for predicting which isotopes will be silent.