NMR Spectroscopy Questions and Answers

Answer: C

The aldehyde proton sits at 9 to 10 ppm, deshielded both by the electronegative oxygen and by the anisotropy of the carbonyl double bond it is attached to. That position is nearly diagnostic, since little else in an ordinary organic spectrum appears there. The 2 to 3 region holds protons next to a carbonyl rather than on it, 5 to 6 holds alkenes, and 12 to 13 is the carboxylic acid region, a broad line that vanishes on shaking with heavy water. Learning the half dozen landmark regions makes most spectra readable at a glance.

Answer: D

The hydroxyl proton exchanges with deuterium, so the signal disappears and a new HOD peak grows. The deuterium exchange test is the standard way of identifying exchangeable protons on alcohols, amines, thiols and acids. It does not sharpen the line, since the proton is simply gone, and neither shifts nor doubles it. The position of an exchangeable proton is also strongly dependent on concentration, temperature and solvent, which is why it is never quoted as tightly as a carbon-bound proton and why the exchange test is preferred to a table lookup.

Answer: A

Both are true and the second explains the first. When the hydroxyl proton hops between molecules faster than the coupling constant, the spin state of its partner averages out and the multiplet collapses to a single broad line. Drying the sample or working in dry dimethyl sulfoxide slows the exchange enough that the expected coupling reappears, which is both the proof of the mechanism and a useful piece of technique. Exchange broadening of this kind is a general phenomenon, seen also in amines and in hindered rotation.

Answer: B

The dominant shielding term for carbon is the paramagnetic contribution from low lying excited states involving its p orbitals, which varies enormously with hybridisation and substitution. Hydrogen has only an s valence orbital, so its shifts depend on the smaller diamagnetic term and span a narrow range. Mass does not enter the Larmor equation as such, decoupling changes multiplicity rather than shift, and the gyromagnetic ratio governs frequency and sensitivity. The wide carbon range is why carbon spectra are so much less crowded than proton ones.

Answer: C

Ketone and aldehyde carbonyl carbons resonate near 200 ppm, the far downfield end of the ordinary carbon range, and esters and acids come a little higher field near 170. Aromatic and alkene carbons cluster around 110 to 150, carbons bearing oxygen around 60 to 90, and simple alkyl carbons below 40. A peak beyond 190 is therefore strong evidence for a ketone or aldehyde. These broad regions carry most of the structural information in a carbon spectrum, since coupling is normally removed by decoupling.

Answer: D

Coupling is transmitted through the bonding electrons, the spin of one nucleus polarising the electrons and that polarisation reaching the neighbour. Because the route is through bonds, the coupling constant falls away sharply beyond three bonds in saturated systems and carries information about connectivity and geometry. The direct through space dipolar interaction averages to zero in a freely tumbling liquid, chemical exchange destroys coupling rather than creating it, and spinning is an instrumental matter. Coupling is what turns a list of shifts into a structure.

Answer: A

The n plus one rule gives four lines, a quartet, with intensities 1:3:3:1 from the binomial coefficients. The three neighbouring spins can be arranged in eight ways that fall into four distinct total spin values, and the degeneracies of those give the intensity pattern. This is the ethyl group signature: the methyl appears as a triplet from two neighbours and the methylene as a quartet from three. The rule holds only when the neighbours are equivalent to each other and the shift difference is large compared with the coupling.

Answer: B

A triplet arises from two equivalent neighbours whose spins can be both up, one up and one down in two ways, or both down, so the intensities follow 1:2:1. The pattern is the row of Pascal's triangle for n = 2, and the quartet from three neighbours follows as 1:3:3:1. Equal intensities would imply three inequivalent lines rather than a genuine multiplet, and the other ratios given belong to no simple case. Reading intensities as well as line counts is what distinguishes a real triplet from three accidentally overlapping singlets.

Answer: C

The coupling constant is a property of the molecule, transmitted through its bonds, so it is unchanged in hertz at any field. Chemical shift differences in hertz do scale with field, which is why multiplets that overlap at low field separate cleanly at high field. Expressed in ppm the coupling appears to shrink as the field rises, which is precisely why coupling constants are always quoted in hertz. Running a sample at two fields and seeing which separations move is the standard test for distinguishing coupling from shift.

Answer: D

The separation in ppm is 0.06, and multiplying by the operating frequency converts it to hertz:

\[ J = 0.06 \times 500 = 30 \]

so the coupling is 30 Hz, which is unusually large and would suggest a coupling to phosphorus or fluorine rather than to another proton. Ordinary proton to proton couplings run from about 2 to 18 Hz. The arithmetic matters more than the chemistry here: a separation quoted in ppm means nothing until the field is known.