Spectroscopic Techniques Questions and Answers

Answer: C

Correct answer: absorbs in the ultraviolet or visible region

A chromophore is the group responsible for the absorption - a carbonyl, a nitro group, an extended conjugated system - and it is what gives a compound its colour when the absorption falls in the visible. An auxochrome is the different thing described in the next question: a group that modifies a chromophore’s absorption without being the source of it.

Answer: D

Correct answer: shifting and intensifying its absorption

An auxochrome carries a lone pair that can conjugate with the chromophore, extending the delocalised system. That lowers the energy gap, so the absorption moves to longer wavelength - a bathochromic shift - and it usually increases the intensity as well, a hyperchromic effect. Hydroxyl, amino and halogen substituents all act this way, which is why phenol absorbs at longer wavelength than benzene.

Answer: A

Correct answer: longer wavelength

A bathochromic shift, also called a red shift, moves a band to longer wavelength and therefore to lower energy. The opposite, to shorter wavelength, is hypsochromic or blue. The intensity changes have their own names: hyperchromic for an increase and hypochromic for a decrease. Extending conjugation is the commonest cause of a bathochromic shift.

Answer: B

More delocalisation narrows the gap between the highest occupied and lowest unoccupied orbitals, so less energy is needed. Carry the conjugation far enough and the absorption reaches the visible and the compound is coloured.

Answer: C

The oscillating dipole is what couples to the electromagnetic field, so a symmetric stretch of a symmetrical molecule is silent. Such a vibration is instead Raman active, which is why the two techniques complement one another.

Answer: D

Correct answer: dipole moment

Infrared absorption requires the vibration to change the molecule’s dipole moment, since it is that oscillating dipole that couples to the electric field of the radiation. In the symmetric stretch of carbon dioxide both oxygen atoms move outward together, so the two bond dipoles remain equal and opposite and the net dipole stays zero throughout. The mode is therefore infrared inactive - though it is Raman active, since the polarisability does change.

Answer: A

Correct answer: \( 3N - 6 \)

A molecule of \( N \) atoms has \( 3N \) degrees of freedom. Three are taken by translation of the whole molecule and three by rotation about its principal axes, leaving

vibrational modes. A linear molecule has only two rotational degrees of freedom, since rotation about the molecular axis moves nothing, so it has \( 3N - 5 \) vibrations instead.

Answer: B

Correct answer: is characteristic of the molecule as a whole

Below about 1500 reciprocal centimetres the spectrum is dominated by skeletal bending and stretching modes that involve the whole molecular framework rather than any one group, so the pattern is complex and, like a fingerprint, effectively unique to the compound. It is therefore used for identification by comparison with a reference spectrum, while the region above 1500 is used to identify functional groups.

Answer: C

Correct answer: carbonyl group

The carbon-oxygen double bond is strong and highly polar, so its stretch is both high in frequency and very intense - typically 1700 to 1750 reciprocal centimetres, and one of the easiest bands to recognise. The exact position discriminates within the family: an aldehyde or ketone near 1715, an ester near 1740, an amide near 1650, and conjugation lowering each by 20 to 30.

Answer: D

Correct answer: hydrogen-bonded hydroxyl group

The O-H stretch appears in this region, and hydrogen bonding is what makes it broad: each molecule experiences a slightly different bonding environment, so the absorption is smeared over hundreds of reciprocal centimetres. A free hydroxyl in dilute solution gives a sharp band near 3600 instead, and the contrast between the two is a useful test for association.