Questions: Infrared Spectroscopy: Functional Group Identification

Question 1

A compound shows a strong carbonyl absorption at 1735 cm⁻¹ and no broad O–H stretch anywhere in the spectrum. Which functional group is most consistent with this data?

• A. Ketone — the carbonyl near 1715 cm⁻¹ is close enough
• B. Ester — carbonyl stretches at 1735–1750 cm⁻¹ with no O–H
• C. Carboxylic acid — the carbonyl at ~1710 cm⁻¹ is in the right region
• D. Amide — nitrogen donation shifts the carbonyl to higher frequency

Answer: (B)

Ester carbonyls absorb at 1735–1750 cm⁻¹ and have no O–H stretch, matching both observations. Ketones absorb near 1715 cm⁻¹, carboxylic acids near 1710 cm⁻¹ and always show a broad O–H (2500–3300 cm⁻¹), and amide carbonyls appear near 1650 cm⁻¹ — the lowest of all carbonyls because nitrogen lone pair donation weakens the C=O bond, not strengthens it.

Question 2

A spectrum shows a very broad, flat O–H absorption spanning 2500–3300 cm⁻¹ and a carbonyl stretch at ~1710 cm⁻¹. Which functional group best explains both absorptions together?

• A. Alcohol — O–H and a separate carbonyl in the molecule
• B. Carboxylic acid — the broad O–H and ~1710 cm⁻¹ carbonyl are diagnostic
• C. Aldehyde — aldehydes show a C–H stretch near 2720 cm⁻¹ which could be confused with O–H
• D. Ketone with intramolecular hydrogen bonding

Answer: (B)

A carboxylic acid has both features: the O–H stretch is characteristically very broad (spanning 2500–3300 cm⁻¹) due to strong hydrogen bonding, and the C=O absorbs near 1710 cm⁻¹. An alcohol shows a narrower O–H (3200–3550 cm⁻¹) and no carbonyl. The broad, flat O–H spanning over 800 cm⁻¹ is the distinguishing signature of a carboxylic acid.

Question 3

The fingerprint region (500–1500 cm⁻¹) is too complex to yield any structural information and should be ignored in routine IR interpretation.

• T. True
• F. False

Answer: False

The fingerprint region is complex precisely because its overall pattern is unique to each molecule, making it valuable for confirming identity by comparison with a reference spectrum. While individual peaks in this region are difficult to assign, the region as a whole is not uninformative. The standard practice is to use functional group bands (above 1500 cm⁻¹) for identification and the fingerprint region for confirmation.

Question 4

Hydrogen bonding in alcohols causes the O–H stretch to appear at higher frequency and as a sharper, more distinct peak compared to a non-hydrogen-bonded O–H group.

• T. True
• F. False

Answer: False

Hydrogen bonding does the opposite. It weakens the O–H bond slightly (since the hydrogen is partially shared with a neighboring oxygen), lowering the stretching frequency to 3200–3550 cm⁻¹ instead of the free O–H value near 3600 cm⁻¹. It also broadens the peak significantly — the variable strengths of hydrogen bonds in a bulk sample smear the absorption over a wider range of frequencies. A free (non-hydrogen-bonded) O–H would appear sharper and at higher frequency.

Question 5

Why can't IR spectroscopy alone determine the complete molecular structure of an unknown organic compound?

Model answer: IR identifies which functional groups are present or absent but cannot determine the carbon skeleton, the number of carbon atoms, or the connectivity between fragments. Different compounds can share functional groups and produce overlapping bands. The fingerprint region can confirm identity only when a reference spectrum is available. A complete structure determination requires combining IR (functional groups) with NMR (connectivity and hydrogen environment) and mass spectrometry (molecular weight and fragmentation pattern).

This is the practical limitation that makes IR a first-pass diagnostic tool rather than a structure-solving tool. For example, two carboxylic acids will both show broad O–H and a carbonyl near 1710 cm⁻¹; only NMR distinguishes them. The power of IR is negative information: ruling out functional groups whose bands are absent, which efficiently narrows the structure space before moving to other techniques.