Infrared Spectroscopy Questions and Answers

Answer: C

The reduced mass of a carbon-hydrogen pair is close to hydrogen's own, far smaller than any bond between heavy atoms, and the frequency goes as one over its square root. The force constant is ordinary. Lone pairs and bond length are not the mechanism. That every bond to hydrogen appears at high wavenumber, whatever it is bonded to, follows from the mass alone and is the first pattern to notice in the group frequency region. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: D

A carbon-oxygen double bond has a large force constant, which raises the frequency, but both atoms are heavy, which lowers it, and the two effects land the band near 1700. The large dipole change makes the band intense rather than shifting it, which is a separate and equally useful property. That the strongest and most recognisable band in organic infrared owes its position to stiffness and its intensity to polarity is worth keeping distinct. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: A

A triple bond is stiffer, so its force constant is larger and the band appears near 2200 against a double bond's 1650. The reduced mass is unchanged between them, since the same atoms are involved, which isolates the force constant as the cause. Bond order rising from one to two to three walks the band up the spectrum in steps, and that progression is one of the most useful regularities the technique offers. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: B

Doubling the light atom's mass changes the reduced mass and hence the frequency:

\[ \begin{aligned} \frac{\nu_{D}}{\nu_{H}} &= \sqrt{\frac{\mu_{H}}{\mu_{D}}} \\ &\approx \frac{1}{1.36} \end{aligned} \]

giving about 2130 reciprocal centimetres. Halving would be right only if the frequency went as one over the mass rather than its square root. Deuterium substitution is used deliberately to confirm which bands involve hydrogen, since the shift is large and predictable.

Answer: C

Replacing hydrogen roughly divides the frequency by 1.36, so a band that moves is one involving hydrogen and a band that does not is not. Intensity is little changed, preparation is generally harder, and the force constant is essentially unaffected because the electrons are the same. Changing one quantity in a relation and seeing which observations respond is the general way to assign a spectrum, and mass is the easiest quantity to change. The shared properties explain why these belong to one family, and the differences explain why more than one exists.

Answer: D

Intensity follows how much the dipole moment changes during the vibration, so polar bonds such as carbonyl and hydroxyl give strong bands and near-symmetric ones give weak. Force constant and reduced mass set the position rather than the height, and atom count sets how many bands there are. That position and intensity carry independent information, one about stiffness and mass and the other about polarity, is what lets a band be identified from both together.

Answer: A

In a symmetrically substituted alkene the two ends are equivalent, so stretching produces almost no change in dipole moment and the band is very weak. The force constant and frequency are perfectly ordinary, near 1650, and masking by other bands is not the reason. That the absence of an expected band is itself evidence - here, of symmetry - is a useful habit, and it recurs throughout structural spectroscopy. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: B

Above about 1500 reciprocal centimetres each functional group absorbs at a characteristic wavenumber almost regardless of the rest of the molecule, so the region reports which groups are present. Compound identity is settled by the fingerprint region, concentration by band intensity, and molecular weight by mass spectrometry. Two regions answering two different questions is what makes an infrared spectrum useful both for classification and for identification.

Answer: C

Below about 1500 reciprocal centimetres the bands arise from skeletal and bending modes involving the whole molecule, so the pattern is complex and highly specific but hard to assign band by band. Functional groups are read above it. A region that is uninterpretable in detail but unique as a whole is exactly what an identification by comparison needs.

Answer: D

Bending and skeletal vibrations involve many atoms moving together, so a band belongs to the molecule as a whole rather than to one bond and shifts unpredictably with any structural change. The bands are often strong, resolution is adequate, and solvent absorption is a separate problem. That the very complexity which prevents assignment is what makes the region diagnostic is the point worth grasping. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.