Question

Predict the relative intensities of the \(M\) and \(M+2\) peaks for the following. (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}\) (c) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{Br}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SH}\)

Short answer

Based on the calculations, the relative intensities of the M and M+2 peaks for each compound are: (a) CH3CH2Cl: M = 75%, M+2 = 25% (b) CH3CH2Br: M = 50%, M+2 = 50% (c) BrCH2CH2Br: M = 25%, M+2 = 50% (d) CH3CH2SH: M = 100%, M+2 = negligible

Step-by-step solution

Calculate the isotopic contribution

In each case, we will find the isotopic contribution of the elements or groups involved in the compound. The natural abundance of isotopes will be crucial for the analysis. For chlorine, we have: - 75% of \(^{35}\text{Cl}\) (mass number = 35) - 25% of \(^{37}\text{Cl}\) (mass number = 37) For bromine, we have: - 50% of \(^{79}\text{Br}\) (mass number = 79) - 50% of \(^{81}\text{Br}\) (mass number = 81)

Apply isotopic contribution to each compound

Now, we will analyze each compound separately. (a) For \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\), we have one Cl atom. The isotopic distribution is as follows: - M: 75% (contribution from \(^{35}\text{Cl}\)) - M+2: 25% (contribution from \(^{37}\text{Cl}\)) (b) For \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}\), we have one Br atom. The isotopic distribution is as follows: - M: 50% (contribution from \(^{79}\text{Br}\)) - M+2: 50% (contribution from \(^{81}\text{Br}\)) (c) For \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{Br}\), we have two Br atoms. The isotopic distribution is as follows: - M: 25% (contribution from two \(^{79}\text{Br}\) atoms) - M+2: 50% (contribution from one \(^{79}\text{Br}\) and one \(^{81}\text{Br}\) atom) (d) For \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SH}\), there are no significant isotopes present. The isotopic distribution is as follows: - M: 100% (no measurable M+2 peak) The relative intensities of the M and M+2 peaks for each compound are as follows: (a) M: 75%, M+2: 25% (b) M: 50%, M+2: 50% (c) M: 25%, M+2: 50% (d) M: 100%, M+2: negligible

Key concepts

Isotopic Distribution

In mass spectrometry, isotopic distribution refers to the pattern of peaks that emerge due to the presence of different isotopes in a sample. When you analyze a molecule, it's common to observe peaks at the molecular ion level (\(M\)) and subsequent peaks like the \(M+2\) peak, which arise from isotopes with additional neutrons. For accurate analysis, understanding isotopic distribution is vital.

Natural isotopic abundances define these distributions. For example:

• Chlorine has two stable isotopes: \(^{35}\text{Cl}\) (abundance: 75%) and \(^{37}\text{Cl}\) (abundance: 25%). This results in distinct \(M\) and \(M+2\) peaks.
• Bromine also has two common isotopes: \(^{79}\text{Br}\) and \(^{81}\text{Br}\), each with an equal 50% occurrence, influencing the peaks significantly.

Knowing these abundances helps predict the intensity and presence of mass spectrometry peaks in a given molecular compound.

Chlorine Isotopes

Chlorine's isotopes are vital in predicting the mass spectrometric profile of compounds containing this element.

The isotopic composition:

• \(^{35}\text{Cl}\) accounts for about 75% of chlorine in nature.
• \(^{37}\text{Cl}\) constitutes the remaining 25%.

This distribution affects the intensity of the \(M\) and \(M+2\) peaks. For instance, in a compound like \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\), the predominant \(M\) peak, representing the lighter \(^{35}\text{Cl}\) isotope, would appear at 75% intensity. Conversely, the \(M+2\) peak, corresponding to the heavier \(^{37}\text{Cl}\), has a 25% intensity.

This clear separation in peaks is a direct result of the chlorine isotopic distribution, distinguishing it from elements with less pronounced isotopic differences in their mass spectra.

Bromine Isotopes

Bromine's isotopes significantly impact its mass spectrometric profile, often more dramatically than chlorine due to even isotopic distribution.

The key isotopes in bromine are:

• \(^{79}\text{Br}\) with a 50% natural abundance.
• \(^{81}\text{Br}\), also with a 50% abundance.

This equal distribution leads to equal intensity peaks in mass spectrometry. Taking \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}\) as an example, the \(M\) and \(M+2\) peaks both appear with equal 50% intensity.

In molecules like \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{Br}\), which contain two bromine atoms, the \(M\) peak can drop to 25% and the \(M+2\) peak remains significant at 50%. This arises when one atom is \(^{79}\text{Br}\) and the other \(^{81}\text{Br}\), making bromine isotopes crucial in interpreting such spectra.

Relative Peak Intensities

Relative peak intensities in mass spectrometry offer insight into isotopic composition of a molecule. Observing \(M\) and \(M+2\) intensities can reveal specific isotopic contributions. Molecules with chlorine or bromine exhibit clear peak ratios due to their distinctive isotopic patterns.

Here's why relative intensities matter:

• For compounds like \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{Cl}\), the \(^{35}\text{Cl}\) and \(^{37}\text{Cl}\) isotopes result in peak ratios of 75% and 25%.
• Bromine-containing compounds, such as \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{Br}\), show equal \(M\) and \(M+2\) peaks due to equal abundances of \(^{79}\text{Br}\) and \(^{81}\text{Br}\).
• Compounds without significant isotopes have an almost exclusive \(M\) peak, like \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SH}\), showing no noticeable \(M+2\) peak.

Understanding these intensities helps predict and interpret mass spectra, essential for identifying or confirming the structure of the compounds involved.