Question

Carboxylic acids often give a strong fragment ion at \(m / z(M-17)\). What is the likely structure of this cation? Show by drawing contributing structures that it is stabilized by resonance.

Short answer

Answer: The fragment ion formed from carboxylic acids at \(m/z (M-17)\) is R-COO^+. Its resonance stabilization is achieved through the delocalization of the positive charge between the two oxygen atoms in the ion. This stabilization is demonstrated by two resonance-contributing structures, where the positive charge alternates between the two oxygen atoms linked to the central carbon atom.

Step-by-step solution

Understand the carboxylic acid structure

A carboxylic acid has the general structure R-COOH, where R is an alkyl or aryl group and COOH represents the carboxyl functional group.

Determine the fragment ion with \(m / z(M-17)\)

Given that the fragment ion has a mass of \(M-17\), we can deduce that the ion is formed by the loss of 17 mass units from the carboxylic acid molecule. As the molecular weight of OH is 17 (Oxygen: 16, Hydrogen: 1), we can assume that the loss of OH from the carboxyl group forms the cation. The new structure after losing OH will be R-COO^+. This is the structure of the cation.

Draw the resonance-contributing structures

Resonance involves the delocalization of electrons within a molecule or ion. The resonance-contributing structures represent different possible arrangements of electrons in the molecule and depict the actual molecule's stability. For the R-COO^+ ion, we can draw the following resonance-contributing structures: Structure 1: Single bond between C and O, with one of O having a positive charge and a lone pair of electrons. [ R-C=O ... O^+ ] Structure 2: Double bond between C and O with a positive charge on the other O after sharing its lone pair of electrons. [ R-C+O ... O= ] The actual structure of R-COO^+ is a hybrid of these resonance-contributing structures. The positive charge is delocalized between the two oxygen atoms, which stabilizes the ion.