Questions: Molecular Spectroscopy for Structure Determination

Question 1

A compound shows a strong C=O stretch at 1715 cm⁻¹ in IR and a singlet at ~9.5 ppm in ¹H NMR. Which structural feature do these two observations together establish?

• A. An ester functional group, because the carbonyl stretch and the singlet indicate O-C=O
• B. A ketone flanked by two alkyl groups, because 1715 cm⁻¹ is the classic ketone carbonyl
• C. An aldehyde (CHO group), because the ~9.5 ppm singlet is the diagnostic aldehyde C–H signal
• D. A carboxylic acid, because the broad O–H and the carbonyl together indicate COOH

Answer: (C)

Both ketones and aldehydes absorb near 1715 cm⁻¹, so IR alone cannot distinguish them. The ~9.5 ppm singlet in ¹H NMR is the definitive aldehyde C–H signal, which is absent in ketones (which have no H on the carbonyl carbon). This is a classic example of why combining techniques is essential: IR narrowed the candidates to carbonyl-containing compounds, and NMR resolved the ambiguity.

Question 2

A compound shows no significant UV-Vis absorption above 220 nm. What can be concluded about its electronic structure?

• A. It likely contains an extended conjugated π system spanning several double bonds
• B. It must contain an aromatic ring with strong electron-withdrawing substituents
• C. It contains little or no conjugation — probably only isolated double bonds or fully saturated bonds
• D. It is a protein, since the absence of UV absorption is characteristic of biological macromolecules

Answer: (C)

Extended conjugation (multiple conjugated double bonds, aromatic rings, carbonyl-alkene systems) shifts UV absorption to longer wavelengths and higher intensities. Absorption only below 220 nm is characteristic of molecules with no significant conjugation — isolated C=C bonds absorb around 170–190 nm (often inaccessible in solution), and saturated bonds absorb even further into the vacuum UV. The absence of absorption above 220 nm essentially rules out extended π systems.

Question 3

UV-Vis spectroscopy can determine the complete connectivity of atoms in an organic molecule.

• T. True
• F. False

Answer: False

UV-Vis reports only on the electronic structure — specifically how electrons are delocalized through conjugated or aromatic systems. It cannot reveal the sequence of atoms, the number of distinct chemical environments, or the arrangement of saturated portions of the molecule. Connectivity mapping requires NMR spectroscopy (through chemical shifts, integration, and coupling patterns). UV-Vis is useful for confirming conjugation and aromatic character, but it is structurally silent about the rest of the molecule.

Question 4

IR spectroscopy identifies functional groups by detecting characteristic vibrational frequencies of specific bonds.

• T. True
• F. False

Answer: True

Different types of bonds vibrate at characteristic frequencies because bond strength (force constant) and atomic masses determine the vibrational frequency. A C=O bond vibrates near 1715 cm⁻¹, an O–H near 3200–3550 cm⁻¹, a C–H near 2850–3000 cm⁻¹, and so on. The IR spectrum is essentially a fingerprint of which bond types — and therefore which functional groups — are present in the molecule.

Question 5

Why must IR, NMR, and UV-Vis spectroscopy be used together rather than relying on just one technique for unambiguous structure determination?

Model answer: Each technique reveals a different and complementary aspect of molecular structure. IR identifies which functional groups are present (what types of bonds exist) but says little about connectivity. NMR maps the carbon-hydrogen framework — how many distinct environments exist, how many protons are in each, and which atoms are neighbors — but may not uniquely distinguish certain functional groups. UV-Vis reveals the electronic structure (the extent of conjugation and aromatic character) but is silent about saturated parts of the molecule. Only when all three data sets are consistent with a single proposed structure can you be confident in the determination.

The analogy in the Explainer is apt: each spectrum is a witness with partial testimony. IR may tell you a carbonyl is present and NMR may tell you there is no aldehyde proton, together ruling out an aldehyde and supporting a ketone or ester. UV-Vis can then confirm whether that carbonyl is conjugated with a double bond (an enone absorbs near 250 nm) or isolated (a simple ketone absorbs weakly near 280 nm). The convergence of all three is what transforms spectroscopic data into a definitive structural assignment.