Infrared Spectroscopy Questions and Answers

Answer: A

Water's hydroxyl bands dominate the infrared spectrum and swamp the sample's, while its Raman scattering is weak and leaves the spectrum clear. It does dissolve salt cells, which is a separate practical nuisance. It certainly has Raman-active vibrations, just weak ones. That one technique tolerates the most important biological solvent and the other does not is a large part of why Raman is used for aqueous and biological samples at all.

Answer: B

The near infrared contains overtones and combination bands, which are weak and broad and so penetrate far into a sample - ideal for measuring moisture, protein or fat in grain or pharmaceuticals without preparation. Structural assignment needs the sharper fundamentals of the mid infrared, salts absorb far lower, and electronic transitions are ultraviolet. Weak absorption becoming an advantage because it permits deep penetration is a neat inversion of the usual priority.

Answer: C

Overtone and combination transitions are formally forbidden in a perfectly harmonic oscillator and gain intensity only through anharmonicity, so they are one or two orders of magnitude weaker than the fundamentals. Source and detector are actually better in the near infrared, and samples are thicker rather than thinner. A transition's intensity following from how far the real system departs from an idealised model is a pattern that recurs throughout spectroscopy.

Answer: D

Real bonds are anharmonic, so the vibrational levels converge and the first overtone falls a little short of twice the fundamental. Exactly twice would hold only for a perfect harmonic oscillator, which also would not permit the transition at all. The shortfall is small but systematic, and it is the same anharmonicity that gives the overtone whatever intensity it has - one departure from the ideal explaining both facts.

Answer: A

Both gases absorb in the mid infrared and both vary with the room, so their bands appear and change between the background and the sample scans. Dust, oxidation and detector damage are secondary. Purging with dry nitrogen removes the variability rather than merely the absorption, which is the point: a constant background divides out, and it is the change between scans that cannot be corrected.

Answer: B

Anharmonicity puts the overtone a little below twice the fundamental:

\[ \begin{aligned} \tilde{\nu}_{2} &\approx 2 \times 1700 - \Delta \\ &\approx 3300 \ \mathrm{cm^{-1}} \end{aligned} \]

giving about 3300 rather than exactly 3400. The shortfall is typically one or two per cent. That a weak band near twice a strong one is probably its overtone, rather than a new functional group, is worth checking before an unexpected feature is assigned.

Answer: C

The first overtone of a carbonyl near 1700 lands close to 3400, where a hydroxyl or amine stretch would also appear, and it is weak enough to be overlooked as an impurity. Amines do absorb nearby, which is a separate confusion, and background and moisture are instrumental rather than intrinsic. Checking whether a weak band is near twice a strong one is the quickest way to rule out an overtone before reporting a functional group.

Answer: D

Neither a mull nor a pellet has a defined thickness or a known sample concentration within it, so the Beer-Lambert law has no path length to work with. Bands do not shift, detectors are linear, and scattering is suppressed by the matching medium. Quantitative infrared work therefore uses solution cells of known path length or attenuated total reflectance with a fixed effective path, which is the practical consequence of this limitation.

Answer: A

A sealed liquid cell with a spacer of known thickness gives a defined path length, so the Beer-Lambert law applies exactly as it does in ultraviolet work. A pellet's thickness and a mull's are not reproducible, and the fingerprint region gives no internal standard. Quantitative work needs a defined path length whatever the spectral region, and the sampling technique has to be chosen with that in mind.

Answer: B

The out-of-plane bending bands fall between about 900 and 690, and their exact positions reveal how many adjacent hydrogens the ring carries and hence the substitution pattern. The 3100 region holds the aromatic stretches, 1600 to 1450 the ring modes, and the 2000 to 2500 region is nearly empty. A region whose bands answer a specific structural question, rather than merely announcing a group, is unusually valuable and worth learning as an exception.