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Chapter 14: Problem 12

The molecular ions of both \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{~S}\) and \(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}\) appear at \(m / z 102\) in low- resolution mass spectrometry. Show how determination of the correct molecular formula can be made from the appearance and relative intensity of the \(M+2\) peak of each compound.

Short Answer

Expert verified
Answer: \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{S}\) has a higher M+2 peak intensity compared to \(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}\).

Step by step solution

01

Identify the isotope contributions to the M+2 peak

There are three primary isotopes that can contribute to an M+2 peak in organic molecules: \(^{13}\mathrm{C}\), \(^{2}\mathrm{H}\) (D), and \(^{34}\mathrm{S}\). Oxygen has no significant isotope contributing to the M+2 peak. The natural abundance of these isotopes is: - \(^{13}\mathrm{C}\): 1.1% - \(^{2}\mathrm{H}\) (D): 0.015% - \(^{34}\mathrm{S}\): 4.4%
02

Calculate the probability of each isotopic contribution for both molecular formulas

For \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{S}\): - Probability of \(^{13}\mathrm{C}\): 5 atoms * 1.1% = 5.5% - Probability of \(^{2}\mathrm{H}\) (D): 10 atoms * 0.015% = 0.15% - Probability of \(^{34}\mathrm{S}\): 1 atom * 4.4% = 4.4% For \(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}\): - Probability of \(^{13}\mathrm{C}\): 6 atoms * 1.1% = 6.6% - Probability of \(^{2}\mathrm{H}\) (D): 14 atoms * 0.015% = 0.21% - Probability of \(^{34}\mathrm{S}\): 0 atoms * 4.4% = 0%
03

Calculate the relative intensity of the M+2 peak for both molecular formulas

For \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{S}\): The M+2 peak intensity = 5.5% + 0.15% + 4.4% = 10.05% For \(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}\): The M+2 peak intensity = 6.6% + 0.21% + 0% = 6.81%
04

Compare the M+2 peak intensities to determine the molecular formula

By comparing the relative intensity of the M+2 peak in the mass spectra of the given molecules, we find that the compound with a higher M+2 peak intensity corresponds to the molecular formula \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{S}\), while the compound with the lower M+2 peak intensity corresponds to \(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}\). This allows us to distinguish between the two molecular formulas and determine which compound has the corresponding molecular formula.

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Most popular questions from this chapter

Show how the compounds with the molecular formulas \(\mathrm{C}_{6} \mathrm{H}_{9} \mathrm{~N}\) and \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NO}\) can be distinguished by the \(m / z\) ratio of their molecular ions in high- resolution mass spectrometry.

Both \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}\) and \(\mathrm{C}_{7} \mathrm{H}_{14}\) have the same nominal mass, namely 98 . Show how these compounds can be distinguished by the \(m / z\) ratio of their molecular ions in highresolution mass spectrometry.

The following is the mass spectrum of compound \(\mathrm{C}, \mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\). Compound \(\mathrm{C}\) is infinitely soluble in water, undergoes reaction with sodium metal with the evolution of a gas, and undergoes reaction with thionyl chloride to give a water-insoluble chloroalkane.

For primary amines with no branching on the carbon bearing the nitrogen, the base peak occurs at \(m / z 30\). What cation does this peak represent? How is it formed? Show by drawing contributing structures that this cation is stabilized by resonance.

Determine the probability of the following in a natural sample of ethane. (a) One carbon in an ethane molecule is \({ }^{13} \mathrm{C}\). (b) Both carbons in an ethane molecule are \({ }^{13} \mathrm{C}\). (c) Two hydrogens in an ethane molecule are replaced by deuterium atoms.
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