A compound with molecular formula C₄H₆O gives the infrared spectrum shown here. Identify the compound.

Infrared (IR) spectroscopy is the study of the interaction of the samplewith IR radiation which brings about change in vibrational-rotational level.I It is also called rotational-vibrational spectroscopy.

• It is a technique based on the vibrations of the atoms of a molecule.
• The waves are electromagnetic in nature and interact with polarity of the chemical bonds of the molecule.

If there is no polarity in the molecule then the infrared interaction is inactive and the molecule does not produce any IR spectrum

In the IR spectra of aldehyde, a peak usually appears around 2700 cm^-1. Simple ketone and aldehydes show a very strong band for the carbonyl group (C=O) that appears in the range of 1740-1710 cm^-1. This band is shifted to lower frequencies with conjugation to a C=C phenyl group. A very important doublet can be observed in the C-H stretch region for the aldehyde C-H near 2850 and 2750 cm^-1. The presence of this doublet allows aldehyde to distinguish from other carbonyl-containing compounds.

The ketone with the conjugated carbonyl group is expected to show a C=O stretch at 1685, whereas ketone with the isolated carbonyl group will show C=O stretch at 1720.

The methyl group usually shows a weak band at 1380 cm^-1 and a medium band at 1260 cm^-1.

The IR absorption at 1700 indicates that the compound has a carbonyl (C=O) group, and absorption at 1600 indicates that the compound has C=C.

The absorption band at 1380 indicates that the compound has methyl group. Since it is given in the molecular formula that the compound has four carbons and only one oxygen, we can conclude that the compound must be methyl vinyl ketone (but-3-en-2-one).

In the given IR spectra, the carbonyl (C=O) stretch is at 1700—lower frequency than expected value of ketone at 1720, because carbonyl (C=O) group develops single-bond character due to electron delocalization.