Spectroscopic Techniques Questions and Answers

Answer: A

Correct answer: 2200 reciprocal centimetres

The stretching frequency rises with bond strength and falls with reduced mass, so a triple bond absorbs well above a double bond. A carbon-carbon triple bond appears near 2200 and a nitrile near 2250; a carbon-carbon double bond near 1650. The band is often weak, since a symmetrically substituted alkyne has little dipole change, and in a perfectly symmetrical one it vanishes altogether.

Answer: B

A small fraction of the scattered photons gain or lose a vibrational quantum, and the shift rather than the absolute frequency carries the information. The much stronger unshifted Rayleigh scattering must be filtered out.

Answer: C

Correct answer: polarisability

Raman scattering depends on the incident field inducing a dipole in the molecule, so the mode is active only if the polarisability - the ease with which the electron cloud is distorted - changes during the vibration. This is the complementary rule to the infrared one, and in a centrosymmetric molecule the two are mutually exclusive: a mode active in one is inactive in the other, which is why the two techniques are used together.

Answer: D

Correct answer: a weak Raman scatterer

Water absorbs infrared radiation very strongly, which makes aqueous infrared spectroscopy difficult, but it is a poor Raman scatterer because the O-H stretch involves little change in polarisability. Aqueous solutions can therefore be examined directly, in ordinary glass, without drying or solvent substitution - which is why Raman is so useful in biological and electrochemical work.

Answer: A

Correct answer: shielding by the surrounding electrons

The electrons around a nucleus circulate in the applied field and generate a small opposing field, so the nucleus experiences slightly less than the applied strength and resonates at a slightly different frequency. The extent of that shielding depends on the electron density, and hence on the chemical environment - which is why the shift identifies the group. Coupling to neighbouring spins is the separate phenomenon that splits the signal.

Answer: B

Correct answer: independent of the instrument’s field strength

The frequency difference between a signal and the reference is proportional to the applied field, so it would be a different number on every instrument. Dividing by the operating frequency removes that dependence:

A proton at 7.26 ppm is at 7.26 ppm on a 60 MHz instrument and on a 600 MHz one, which is what makes published shift tables usable at all.

Answer: C

Correct answer: tetramethylsilane

TMS is used because its twelve protons are equivalent and give a single sharp line; silicon is less electronegative than carbon so they are unusually shielded and the signal falls at one end of the range, clear of almost everything else; it is chemically inert and volatile, so it is easily removed. Its signal is defined as zero, and shifts are quoted relative to it.

Answer: D

Coupling through the intervening bonds means each neighbour's two spin states shift the observed resonance slightly, splitting it into a multiplet. The separation, the coupling constant, is independent of the field strength.

Answer: A

Correct answer: quartet

The multiplicity is \( n + 1 \) where \( n \) is the number of equivalent neighbours, so three neighbours give four lines - a quartet - with intensities 1:3:3:1 from Pascal’s triangle. This is why the classical ethyl group shows a quartet for the methylene protons, which have three methyl neighbours, and a triplet for the methyl protons, which have two.

Answer: B

Correct answer: protons giving rise to it

Every proton in a given environment contributes equally to the signal, so the integrated area is proportional to how many there are. The integral therefore gives the ratio of protons in each environment - three to two to one for ethanol, for instance - which together with the chemical shifts and the splitting patterns is usually enough to settle the structure.