Infrared spectroscopy — common mistakes

Examiners often test your ability to distinguish between an alcohol and a carboxylic acid. The key is the C=O peak. An alcohol has a broad O-H peak but NO C=O peak. A carboxylic acid has BOTH a very broad O-H peak AND a strong, sharp C=O peak.

Pay as much attention to the absence of peaks as their presence. If you are trying to decide between a ketone and an alcohol, the lack of a broad O-H peak is definitive evidence against the alcohol structure.

The broadness of the O-H peak is caused by hydrogen bonding. In a bulk sample of an alcohol or carboxylic acid, molecules are hydrogen-bonded to varying degrees. This creates a wide range of O-H bond environments, each with a slightly different vibrational frequency, smearing the absorption over a broad range of wavenumbers.

Yes, it is excellent for distinguishing between functional group isomers (e.g., propanal and propanone vs. prop-2-en-1-ol) because they have different functional groups that absorb at different frequencies. However, it is generally not used to distinguish between positional isomers (e.g., propan-1-ol and propan-2-ol) or stereoisomers, as their key functional groups are the same and their spectra are often very similar in the functional group region.

The C=O absorption for both aldehydes and ketones falls in a similar range (~1700-1740 cm⁻¹) and they cannot be reliably distinguished by this peak alone. However, an aldehyde also has characteristic C-H stretches for the aldehyde proton at ~2720 cm⁻¹ and ~2820 cm⁻¹. The presence of these weak but distinct peaks, in addition to the C=O peak, confirms an aldehyde.

No. You will be provided with a Data Booklet in the exam that contains a table of characteristic infrared absorption values for common bonds. Your task is to learn how to use that data to interpret a given spectrum, not to memorise the numbers themselves. You should, however, be familiar with the general regions and shapes (e.g., 'broad O-H', 'strong, sharp C=O').