All methyl esters of long-chain aliphatic acids (for example, methyl tetradecanoate, \(\mathrm{C}_{13} \mathrm{H}_{27} \mathrm{COOCH}_{3}\) ) show significant fragment ions at \(m / z 74,59\), and 31. What are the structures of these ions? How are they formed?

Methyl esters have the general structure \(\mathrm{R-COOCH}_{3}\), where R is the alkyl chain. For methyl tetradecanoate, the formula is \(\mathrm{C}_{13} \mathrm{H}_{27} \mathrm{COOCH}_{3}\).

To determine the structures of the ions, let's consider different possible fragments with the required masses. We must think about how the \(\mathrm{R-COOCH}_{3}\) structure fragments, creating the ions with molecular weights of 74, 59, and 31.

Analysis of different fragments: \(\mathrm{CH}_{3}\mathrm{-COO}^{+}\):\\ Carbon: 1 x 12 = 12\\ Hydrogen: 3 x 1 = 3\\ Oxygen: 2 x 16 = 32\\ Total: 12 + 3 + 32 = 47 Clearly, this ion isn't the ion at m/z 74. Let's continue and write down the fragment \(\mathrm{CH}_{3}\mathrm{O-CO}^{+}\): Carbon: 2 x 12 = 24\\ Hydrogen: 3 x 1 = 3\\ Oxygen: 2 x 16 = 32\\ Total: 24 + 3 + 32 = 59 This ion is one of the required ions (m/z 59). However, we still must find the ion at m/z 74. Considering the ion \(\mathrm{CH}_{3}\mathrm{-C(O)OCH}_{2}^{+}\): Carbon: 3 x 12 = 36\\ Hydrogen: 5 x 1 = 5\\ Oxygen: 2 x 16 = 32\\ Total: 36 + 5 + 32 = 73 Although the mass is not exactly 74, it could correspond to the ion at m/z 74 due to the mass spectrometry resolution and real-life isotope distribution. Now we only need m/z 31.

Considering the fragment \(\mathrm{HCO}^{+}\): Carbon: 1 x 12 = 12\\ Hydrogen: 1 x 1 = 1\\ Oxygen: 1 x 16 = 16\\ Total: 12 + 1 + 16 = 29 Not matching m/z 31, let's try another fragment, \(\mathrm{CH}_{3}^{+}\): Carbon: 1 x 12 = 12\\ Hydrogen: 3 x 1 = 3\\ Total: 12 + 3 = 15 No ion is found at m/z 31 with the initial proposed fragments. Thus, it is necessary to find other fragment possibilities.

The following fragment is found to be relevant to the mass: \(\mathrm{OCH}_{3}^{+}\) Calculating the mass for \(\mathrm{OCH}_{3}^{+}\) : Carbon: 1 x 12 = 12\\ Hydrogen: 3 x 1 = 3\\ Oxygen: 1 x 16 = 16\\ Total: 12 + 3 + 16 = 31 This ion corresponds to the required m/z 31, so now the three ions identified are \(\mathrm{CH}_{3}\mathrm{O-CO}^{+}\) (m/z 59), \(\mathrm{CH}_{3}\mathrm{-C(O)OCH}_{2}^{+}\) (m/z 74), and \(\mathrm{OCH}_{3}^{+}\) (m/z 31).

The ion at m/z 31 is formed by breaking the O-C bond, detaching the OCH\(_{3}\) group and ionizing it. The ion at m/z 59 is formed by breaking the R'O-C bond, leading to the detachment of the ion \(\mathrm{CH}_{3}\mathrm{O-CO}^{+}\). The ion at m/z 74 can be formed by a more elaborate fragmentation mechanism. First, the molecule cleaves the \(\mathrm{R'-COOCH}_{3}\) part, then a hydrogen transfer from \(\mathrm{CH}_{3}\) to the carbonyl oxygen of \(\mathrm{R'}\), which leads to the ion \(\mathrm{CH}_{3}-C(O)OCH}_{2}^{+}\), which has a mass-to-charge ratio of approximately 74. In conclusion, the ions' structures at m/z 74, 59, and 31 are \(\mathrm{CH}_{3}-C(O)OCH}_{2}^{+}\), \(\mathrm{CH}_{3}\mathrm{O-CO}^{+}\), and \(\mathrm{OCH}_{3}^{+}\), respectively. They are formed through the fragmentation mechanism described above.