Identify an unknown liquid from its IR spectrum
Five unlabelled spectra of common solvents: find the diagnostic bands, deduce the functional group, and name the compound.
Goal
Assign bands near 3300 (O–H), ~3000 (C–H), 1650–1750 (C=O), 1050–1250 (C–O) and below 900 (fingerprint/out-of-plane bends) to identify each sample.
Apparatus and reagents
FTIR spectrometer with an ATR accessory, dropper bottles of ethanol, acetone, ethyl acetate, acetic acid and toluene.
Procedure
- Sample A shows a very broad band near 3340 cm⁻¹ and a strong one near 1050 cm⁻¹ but no carbonyl — deduce the functional group.
- Samples B and C both show C=O: compare the carbonyl position (≈1715 vs ≈1740 cm⁻¹) and look for strong C–O bands.
- Sample D pairs a huge, very broad O–H band spanning 3300–2400 cm⁻¹ with C=O near 1710 cm⁻¹ — a carboxylic acid dimer signature.
- Sample E has C–H stretches just above 3000 cm⁻¹, ring bands at 1605/1495 cm⁻¹ and strong out-of-plane bends at 728/694 cm⁻¹ — an aromatic compound.
What to observe
- A broad O–H without C=O means an alcohol; O–H plus C=O means a carboxylic acid; C=O plus strong C–O bands means an ester.
- Carbonyl frequency tracks the environment: ketone ≈1715, carboxylic acid ≈1710, ester ≈1740 cm⁻¹ in these samples.
Explanation
Vibrational frequency follows Hooke’s law: ν̃ ∝ √(k/μ) — strong bonds and light atoms give higher wavenumbers. Answers: A = ethanol (broad O–H, C–O 1050), B = acetone (C=O 1715, no O–H), C = ethyl acetate (C=O 1742, strong C–O 1240/1048), D = acetic acid (very broad O–H + C=O 1712), E = toluene (aromatic C–H >3000, ring modes, 728/694 bends).
History of the experiment
Chemists behind it
Virtual experiment: a simplified model to build intuition. It does not replace real lab work or safety training; never repeat chemistry at home without supervision.