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

Question: Predict the relative intensities of the M and M+2 peaks for the following molecules: (a) CH3CH2Cl, (b) CH3CH2Br, (c) BrCH2CH2Br, (d) CH3CH2SH. Answer: The relative intensities for each molecule are as follows: (a) M : M+2 ≈ 3.1 : 1, (b) M : M+2 ≈ 1.03 : 1, (c) M : M+2 ≈ 1 : 2.89, (d) M : M+2 ≈ 22.5 : 1.

Step-by-step solution

Identifying isotopic abundances

In each molecule, identify the elements' isotopes with significant natural abundances and their respective probabilities. For instance, chlorine (\({^{35}\mathrm{Cl}}\), 75.77%) and \({^{37}\mathrm{Cl}}\) (24.23%), bromine (\({^{79}\mathrm{Br}}\), 50.69%) and \({^{81}\mathrm{Br}}\) (49.31%), and sulfur ( \({^{32}\mathrm{S}}\), 95.02%) and \({^{34}\mathrm{S}}\) (4.21%).

(a) CH3CH2Cl: Calculating probabilities

Calculate the probability of M and M+2 peaks. For M, all atoms are the most abundant isotopes (\({^{12}\mathrm{C}}\), \({^{1}\mathrm{H}}\), \({^{35}\mathrm{Cl}}\)). For M+2, only the \({^{37}\mathrm{Cl}}\) isotope contributes since carbon and hydrogen do not have significant abundances at M+2. Probability of M = (0.999) * (0.999) * (0.999) * (0.7577) = 0.7547 Probability of M+2 = 0.2423 (Due to \({^{37}\mathrm{Cl}}\)) Relative intensities: M : M+2 ≈ 3.1 : 1

(b) CH3CH2Br: Calculating probabilities

Calculate the probability of M and M+2 peaks. For M, all atoms are the most abundant isotopes (\({^{12}\mathrm{C}}\), \({^{1}\mathrm{H}}\), \({^{79}\mathrm{Br}}\)). For M+2, only the \({^{81}\mathrm{Br}}\) isotope contributes. Probability of M = (0.999) * (0.999) * (0.999) * (0.5069) = 0.5063 Probability of M+2 = 0.4931 (Due to \({^{81}\mathrm{Br}}\)) Relative intensities: M : M+2 ≈ 1.03 : 1

(c) BrCH2CH2Br: Calculating probabilities

Calculate the probability of M and M+2 peaks. For M, all 2 Bromines (\({^{79}\mathrm{Br}}\)) and other most abundant isotopes (\({^{12}\mathrm{C}}\), \({^{1}\mathrm{H}}\)). For M+2 we have two scenarios: (1) both Br atoms are M+2 isotopes, namely (\({^{81}\mathrm{Br}}\)), (2) one Br atom is M+2 isotope(\({^{81}\mathrm{Br}}\)) and another is M isotope (\({^{79}\mathrm{Br}}\)). Probability of M = (0.999) * (0.999) * (0.999) * (0.999) * (0.5069) * (0.5069) = 0.2568 Case 1: Probability of both M+2 isotopes = (0.4931) * (0.4931) = 0.2432 Case 2: Probability of one M+2 isotope and one M isotope = 2 * (0.5069) * (0.4931) = 0.4996 Probability of M+2 = Case 1 + Case 2 = 0.7428 Relative intensities: M : M+2 ≈ 1 : 2.89

(d) CH3CH2SH: Calculating probabilities

Calculate the probability of M and M+2 peaks. For M, all atoms are the most abundant isotopes (\({^{12}\mathrm{C}}\), \({^{1}\mathrm{H}}\), \({^{32}\mathrm{S}}\)). For M+2, only the \({^{34}\mathrm{S}}\) isotope contributes. Probability of M = (0.999) * (0.999) * (0.999) * (0.999) * (0.9502) = 0.9492 Probability of M+2 = 0.0421 (Due to \({^{34}\mathrm{S}}\)) Relative intensities: M : M+2 ≈ 22.5 : 1