An IR spectrum shows a strong, sharp absorption at approximately 1735 cm⁻¹ and no broad O–H stretch. Which functional group is most consistent with this observation?
AAmide (R–CO–NR₂)
BCarboxylic acid (–COOH)
CEster (–COO–)
DConjugated ketone (ArCO–)
Esters absorb near 1735 cm⁻¹, which is the high end of the C=O range. Carboxylic acids absorb near 1710 cm⁻¹ and also show a broad O–H stretch (2500–3300 cm⁻¹). Amides absorb near 1680 cm⁻¹. Conjugated carbonyls are shifted to lower wavenumbers (~1670–1690 cm⁻¹). The absence of O–H rules out alcohols and carboxylic acids.
Question 2 True / False
If an IR spectrum shows no absorption in the 1680–1760 cm⁻¹ region, the molecule cannot contain any oxygen atoms.
TTrue
FFalse
Answer: False
The 1680–1760 cm⁻¹ region is specific to C=O stretching. Molecules with oxygen in ethers (C–O–C), alcohols (C–OH), or epoxides will show no carbonyl peak, yet clearly contain oxygen. The absence of a C=O peak eliminates aldehydes, ketones, esters, acids, and amides — but many oxygen-containing functional groups lack a carbonyl entirely.
Question 3 Short Answer
A chemist obtains an IR spectrum of an unknown compound and identifies a broad O–H absorption and a C=O stretch at ~1710 cm⁻¹. What additional technique would best confirm whether the compound is a carboxylic acid rather than a β-ketoalcohol, and why?
Think about your answer, then reveal below.
Model answer: ¹H NMR spectroscopy — it would show the distinctive downfield carboxylic acid proton (δ 10–12 ppm) and reveal the carbon skeleton connectivity, which IR cannot provide.
IR identifies functional groups by bond vibrations but cannot distinguish two compounds sharing the same groups arranged differently. NMR reveals the chemical environment of each hydrogen (and carbon), directly confirming whether the O–H and C=O belong to a –COOH unit or are separate groups on the molecule.