NMR Spectroscopy Questions and Answers

Answer: A

Setting the square root of n equal to four gives:

\[ n = 4^{2} = 16 \]

so 16 scans are needed. The square root law is unforgiving: each further doubling of signal to noise costs four times the time already spent, so going from four times to eight times would take 64 scans. This is why weak samples are better addressed by a higher field or a cryoprobe than by simply running overnight, and why sample concentration is worth optimising before the experiment is started.

Answer: B

Appending zeros to the end of the free induction decay before transforming gives more data points per hertz, so the line is defined by more points and its position and shape are read more accurately. It adds no new information and therefore no genuine sensitivity, though it can make the spectrum look smoother. Phase errors are corrected by phasing, and lineshape by apodisation. The distinction between digital resolution, which zero filling improves, and true resolution, which the magnet and shims fix, is one worth keeping clear.

Answer: C

Multiplying the decay by a falling exponential weights the early, strong part of the record and suppresses the late, noisy tail, so noise falls but the effective decay is faster and the lines broaden. It is the standard apodisation for a weak carbon spectrum where sensitivity matters more than resolution. A rising function such as a Gaussian applied the other way sharpens lines at the cost of noise. No apodisation moves a peak or changes the phase, and the trade between resolution and sensitivity is a genuine one that cannot be evaded.

Answer: D

Digital resolution in hertz per point is one divided by the acquisition time, so a one second acquisition gives one hertz per point and a four second acquisition a quarter of that. Scans add signal but not resolution, the field sets dispersion rather than digitisation, and the pulse width sets the excitation bandwidth. To resolve a small coupling the acquisition must be long enough, which in turn requires the relaxation to be slow enough for signal to remain. Resolution and sensitivity again pull against each other.

Answer: A

Circulating electrons around a nucleus generate a small opposing field, so the nucleus experiences slightly less than the applied field and resonates at a slightly lower frequency. The more electron density, the greater the shielding and the further upfield the signal. Coupling splits a signal rather than moving it, inhomogeneity broadens every line equally, and populations govern intensity. Because shielding depends on the chemical environment, the shift is what makes NMR a structural tool rather than merely a count of nuclei.

Answer: B

Dividing the shift in hertz by the spectrometer frequency in megahertz gives a dimensionless number independent of the field, so a proton at 7.2 ppm is at 7.2 ppm on a 300 and on a 600 MHz instrument. In hertz it would be 2160 on one and 4320 on the other, and no table of values could be shared. Coupling constants deliberately stay in hertz precisely because they do not scale with field, and the contrast between the two is a standard way of telling a coupling from a shift difference: run the sample at another field.

Answer: C

Chemical shift is the offset in hertz divided by the operating frequency in megahertz:

\[ \delta = \tfrac{1200}{400} = 3.0 \]

giving 3.0 ppm, which is the region of protons next to a carbonyl or an amine. The same nucleus on a 600 MHz machine would be 1800 Hz from TMS and still 3.0 ppm. Converting confidently in both directions is needed whenever a coupling constant has to be extracted from a spectrum plotted on a ppm axis.

Answer: D

TMS gives a single sharp line from twelve equivalent, strongly shielded protons, placed at zero and upfield of nearly all organic resonances. Silicon is less electronegative than carbon, so the methyl protons are unusually shielded. It is chemically inert and volatile, so it is easily removed, and it does not react with the sample. Gyromagnetic ratio is a property of nuclei, not compounds, and TMS is in fact insoluble in water, where DSS or the residual solvent peak is used instead.

Answer: A

An electronegative neighbour draws electron density away, so the proton is deshielded and moves downfield to higher delta. The methyl protons of chloromethane appear near 3.0 ppm, of the bromide near 2.7 and of the iodide near 2.2, tracking electronegativity exactly. The effect falls off quickly with distance, being much weaker two bonds away and negligible at three. Broadening comes from exchange or relaxation, and splitting from coupling, so neither is the answer here. This single trend explains a large part of any routine proton spectrum.

Answer: B

The delocalised pi electrons circulate in the applied field and generate a field that opposes it inside the ring but reinforces it in the plane outside, so protons on the ring are strongly deshielded. A proton held above the ring, as in some cyclophanes, is shifted the other way and can even appear at negative delta, which is the clinching evidence for the explanation. Carbon electronegativity hardly changes with hybridisation on this scale, relaxation does not move peaks, and geometry relative to the field is not the point. Ring current shifts are the standard test for aromaticity in NMR.