Mass Spectroscopy, Example #1

Analysis: C₅H₁₂O MW = 88.15 Help with Analysis

Mass Spectrum

Mass Spectrum

Correlation Table Common Fragments Interpret MS

Give Structure Next Example Return to MS Home

C₅H₁₂O MW = 88.15

From the molecular formula, the compound has "no degrees of unsaturation" (no double bonds, carbonyls or rings).

Return to Top

The spectrum shows a small molecular ion and a small m-1 peak, suggesting the presence of an alcohol (it cannot be an aldehyde since there are no degrees of unsaturation). The m-15 peak represents loss of a methyl group and the m-17 is consistent with loss of a hydroxy radical. For an alcohol, the base peak is often formed by expulsion of an alkyl chain to give the simple oxonium ion R^'CR^''

OH⁺; to generate the observed m/e = 45, R^' must be CH₃ and R^' a H.

Fragmentation of Common Functional Groups
Common Ions and Fragments

Return to Mass Spectrum

Structure:

IUPAC Name: 2-pentanol

MS Fragments:

The spectrum shows a small molecular ion and a small peak resulting from loss of a hydrogen atom from the alcohol. Loss of a methyl group gives the m-15 peak and loss of hydroxy radical gives the secondary carbocation at m-17. The base peak, not surprisingly, is formed by expulsion of the alkyl chain to give the simple oxonium ion at m/e = 45.

Return to Top

Correlation Table	Common Fragments	Interpret MS
Give Structure	Next Example	Return to MS Home