Mass spectra and infrared evidence

What you’ll learn

• Infer molecular mass and plausible fragments from mass spectra.
• Use infrared data to identify functional groups.
• Combine independent evidence without overclaiming a unique structure.

Understand the idea

A molecular ion can identify relative molecular mass when its charge is known, while fragment ions provide clues to connectivity. The highest-intensity peak is the base peak and is not necessarily the molecular ion. Only charged fragments are detected; a proposed fragmentation must conserve atoms and account for the accompanying neutral species.

Infrared absorption occurs when a vibration produces a changing molecular dipole. Use the supplied absorption ranges to identify bonds such as O–H, N–H, C=O, C=C and C–X. An alcohol O–H absorption is commonly broad around 3200–3600 $\mathrm{cm^{-1}}$; an acid O–H band is broader and can extend roughly 2500–3300 $\mathrm{cm^{-1}}$. Carbonyl absorptions commonly occur in the region around 1700 $\mathrm{cm^{-1}}$, with the exact value depending on the group.

Combine constraints rather than choosing a structure from one peak. A carbonyl band can occur in an aldehyde, ketone or carboxylic acid. Molecular mass, formula, other absorptions and chemical tests help distinguish them. A fingerprint-region match with a trusted reference supports identification, but absence of one weak peak should be interpreted cautiously.

Key formulae

$$m/z=\frac{\text{ion mass}}{\text{charge number}}$$

$$\widetilde\nu=\frac1\lambda\quad(\lambda\text{ in cm gives }\widetilde\nu\text{ in }\mathrm{cm^{-1}})$$

$$\mathrm{CH_3CH_2OH}^{+\bullet}\rightarrow\mathrm{CH_2OH^+}+\mathrm{CH_3}^{\bullet}$$

$$m/z(\mathrm{CH_2OH^+})=12+3+16=31$$

Worked example

A compound has molecular formula C₃H₆O and molecular-ion m/z 58. It has a strong carbonyl IR absorption, no broad O–H absorption and a positive aldehyde test. Suggest its structure.

1. The formula mass is 3(12) + 6(1) + 16 = 58, agreeing with the molecular ion.
2. A carbonyl group with no broad O–H band suggests an aldehyde or ketone rather than an alcohol.
3. The positive aldehyde test selects the three-carbon aldehyde rather than the ketone.

Answer: Propanal, $\mathrm{CH_3CH_2CHO}$.