NMR Spectroscopy Questions and Answers

Answer: A

A vicinal coupling averaged over free rotation lands at 6 to 8 Hz, which is why ethyl groups show that spacing so consistently. Rigid systems depart from it widely: a trans coupling across a double bond reaches 12 to 18 Hz and a cis one 6 to 12, while a dihedral angle near ninety degrees gives almost zero. Long range four bond couplings are the ones in the 0 to 1 range, and the larger figures quoted belong to couplings involving fluorine or phosphorus. The expected value is what makes a departure from it informative.

Answer: B

The Karplus relation gives the three bond coupling constant as a cosine function of the dihedral angle between the two coupled protons, with maxima near zero and one hundred and eighty degrees and a minimum near ninety. It is the basis for deducing ring conformation and peptide backbone geometry from coupling constants. Electronegativity affects shifts, molecular size affects relaxation through the correlation time, and exchange affects linewidth, but none of those is Karplus. It is the clearest case of NMR yielding geometry rather than just connectivity.

Answer: C

Both hold, and the second is the reason. By the Karplus relation a dihedral angle near one hundred and eighty gives a coupling of ten to thirteen hertz, whereas sixty degrees gives only two to five. Measuring the coupling therefore tells you whether a substituent is axial or equatorial, which is a routine conformational assignment in sugar and steroid chemistry. The argument works only because the ring holds the angle fixed; free rotation would average the two values into a single intermediate figure.

Answer: D

First order rules hold only while the separation between coupled signals is large compared with the coupling between them, conventionally a ratio above about ten. As the two approach, intensities distort, extra lines appear and the simple n plus one description fails, giving the leaning AB pattern and worse. Raising the field increases the separation in hertz while leaving the coupling alone, and so pushes a spectrum back towards first order. That is one of the strongest practical arguments for a higher field instrument.

Answer: A

As two coupled signals approach one another, the lines nearer the partner gain intensity and those further away lose it, so each multiplet leans towards its partner like a roof. It is a useful diagnostic, since it tells you which multiplets are coupled to each other even in a crowded region. The centre of gravity of the multiplet moves slightly but the coupling constant is still read from the line spacing, which does not change. Recognising leaning saves a great deal of guesswork in an aromatic region.

Answer: B

The area under a peak is proportional to the number of nuclei contributing to it, so a three to two to one pattern of integrals identifies a methyl, a methylene and a single proton. Only ratios are meaningful unless a known standard is added. Relaxation times, coupling constants and shielding are read from other features entirely. Integration is quantitative in proton NMR because the relaxation of protons is fast enough for full recovery between pulses under ordinary conditions, which is exactly what fails for carbon.

Answer: C

Two effects spoil it: carbon relaxation times vary from under a second to well over a minute, so a short repetition time saturates the slow ones, and broadband proton decoupling enhances protonated carbons more than quaternary ones through the nuclear Overhauser effect. The result is that peak areas do not track carbon counts. Quantitative carbon work uses inverse gated decoupling and a long relaxation delay, often with a paramagnetic relaxation agent. Weakness of signal is a nuisance rather than the reason, and decoupling removes splitting, not area.

Answer: D

Irradiating the whole proton range while acquiring carbon collapses all carbon to proton coupling, so each distinct carbon gives one sharp line. That both simplifies the spectrum and concentrates the intensity of each multiplet into a single peak, which with the added nuclear Overhauser enhancement is a large sensitivity gain. All carbons remain visible, though quaternary ones are weaker for want of nearby protons and of that enhancement. The shift range is a property of shielding and is unaffected. The price of decoupling is the loss of multiplicity information, which DEPT then restores.

Answer: A

DEPT-135 sorts carbons by the number of attached protons: methine and methyl point up, methylene points down, and carbons with no attached proton do not appear at all. Comparing it with the ordinary decoupled spectrum identifies the quaternary carbons by difference, and a DEPT-90 showing only methines separates CH from CH3. This is the routine substitute for the older off-resonance decoupled spectrum, and it is the fastest way to convert a list of carbon shifts into a count of hydrogens on each carbon.

Answer: B

DEPT works by transferring polarisation from attached protons, so quaternary carbons, carbonyls and substituted aromatic ring junctions give no signal. Their absence is itself diagnostic, since subtracting the DEPT from the full decoupled spectrum leaves exactly the non-protonated carbons. Methyls are among the strongest signals in DEPT, aromatic CH appears normally, and proximity to oxygen or nitrogen affects shift rather than visibility. The experiment trades completeness for information about hydrogen count.