NMR Spectroscopy Questions and Answers

Answer: C

With a thirty degree pulse most of the longitudinal magnetisation is left in place, so the next pulse can follow quickly without waiting five T1, and more scans accumulate in a given time. Signal per scan is lower than at ninety degrees, but signal per unit time is higher, which is what matters. Decoupling is still applied, and the Overhauser enhancement is still present and welcome. The Ernst angle formalises the best choice for a given relaxation time and repetition rate.

Answer: D

Inversion recovery applies a one hundred and eighty degree pulse, waits a variable time and then reads the magnetisation with a ninety degree pulse; plotting intensity against delay gives an exponential from which T1 follows, and the delay at which the signal passes through zero is about seventy per cent of T1. Linewidth measures T2 star instead, scan counting measures nothing, and comparing fields would confound relaxation with other field dependent effects. A T1 measurement is the honest way to choose a relaxation delay for quantitative work.

Answer: A

Sensitivity in proton NMR is high enough that a few milligrams in half a millilitre gives a usable single scan spectrum, so concentration is the deciding factor. Boiling point matters only for the practicalities of handling, molar mass affects linewidth for very large molecules but not the general case, and colour is irrelevant to a magnetic technique, which is one of its advantages over optical spectroscopy. When a sample is scarce, the choices are more scans, a smaller tube of matched volume, or a cryoprobe.

Answer: B

Cooling the receiver coil and preamplifier to around twenty kelvin cuts their thermal noise and gives a signal to noise gain of three to four times, which is equivalent to running ten to sixteen times as many scans. The sample itself stays at its own temperature. Field is set by the magnet, shift range by shielding, and relaxation by the molecule. For a scarce natural product a cryoprobe often turns an impossible experiment into an overnight one.

Answer: C

The small proportion of ordinary chloroform left in the deuterated solvent gives a peak at 7.26 ppm that is commonly used to reference the spectrum when no TMS has been added. Its position is tabulated for every common solvent. The lock follows deuterium, not this proton signal; its concentration is not known accurately enough for quantitation; and shimming is judged from lineshape and lock level rather than from this peak. Referencing to the residual peak is convenient and adequate for almost all routine reporting.

Answer: D

Both are true and the reason is correct. Coupling between equivalent nuclei exists physically but produces no change in the observed transitions, so the signal remains a singlet. That is why the six protons of ethane, or the two of a symmetric methylene, give one unsplit line. The moment the equivalence is broken, by a stereocentre elsewhere in the molecule for instance, geminal coupling of ten to fifteen hertz appears and the methylene becomes an AB quartet, which often surprises the newcomer.

Answer: A

When a molecule contains a stereocentre, the two protons of a nearby methylene sit in different environments and have different chemical shifts and couple to each other, giving an AB pattern rather than a singlet. It is a frequent puzzle in the spectra of chiral compounds and a strong clue that a stereocentre is present. Vicinal coupling, exchange and accidental coincidence are all different situations. Replacing each proton in turn and asking whether the products are diastereomers is the formal test.

Answer: B

Shift is a property of the electronic environment, so electronegativity, ring and carbonyl anisotropy and hydrogen bonding all move a peak, sometimes by several ppm in the last case. Accumulating more scans improves signal to noise and changes nothing about position. Keeping clear which observables depend on the molecule and which on the experiment is the difference between interpreting a spectrum and merely describing it: shift, coupling and integral belong to the sample, while noise, resolution and artefacts belong to the instrument.

Answer: C

DMSO dissolves sugars, amides and salts that chloroform will not touch, and because it accepts hydrogen bonds strongly it slows hydroxyl exchange enough that OH protons appear as resolved, coupled multiplets, which is genuinely informative. It does give a residual peak, a quintet at 2.50 ppm from the incompletely deuterated methyl, it boils high and is difficult to remove, and its lock signal is unremarkable. The trade is solubility and information against recovery of the sample.

Answer: D

Broad underlying humps come from species that tumble slowly, such as polymers, aggregates or particulate matter, whose short T2 gives very wide lines. Filtering the sample or diluting it often removes them. A phase error deforms individual peaks rather than adding a featureless hump, truncation gives sinc wiggles beside sharp peaks, and excess TMS gives a large sharp line at zero. Distinguishing a broad genuine signal from a baseline fault is a matter of seeing whether it survives a change of processing.