Chapter 22, Problem 22.41P

Interpretation Introduction

Interpretation:

The expected intensities of $\text{M,}\text{M+1,}\text{and}\text{M+2}$ for each of the given species has to be predicted.

Explanation of Solution

The expected intensities for the given species can be calculated as follows.

The intensities for the species $37:3$ which is having the molecular formula of $\left[{\text{MNH}}_{\text{4}}{}^{\text{+}}\right]{\text{= C}}_{\text{37}}{\text{H}}_{\text{72}}\text{ON}$ can be calculated as

$\begin{array}{l}\text{X+1 = 0}{\text{.12n}}_{\text{H}}\text{ + 1}{\text{.08 n}}_{\text{C}}\text{ + 0}{\text{.369 n}}_{\text{N}}\text{+ 0}{\text{.038 n}}_{\text{O}}\\ \text{=}\text{0}\text{.12(72) + 1}\text{.08}\left(\text{37}\right)\text{ + 0}\text{.369}\left(\text{1}\right)\text{+ 0}\text{.038}\left(\text{1}\right)\\ \text{=}\text{41}\text{.2\% (Observer = 35}\text{.8\%)}\\ \\ \text{X+2 = 0}{\text{.0058 n}}_{\text{C}}{\text{(n}}_{\text{C}}\text{-1) + 0}{\text{.205n}}_{\text{O}}\\ \text{=0}\text{.0058(37)(36)+0}\text{.205(1)}\\ \text{=}\text{7}\text{.9\%}\text{(Observed=}\text{7}\text{.0\%)}\end{array}$

The intensities for the species $37:3$ which is having the molecular formula of ${\left[\text{MH}\right]}^{+}{\text{= C}}_{\text{37}}{\text{H}}_{\text{69}}\text{O}$ can be calculated as

$\begin{array}{l}\text{X+1 = 0}{\text{.12n}}_{\text{H}}\text{ + 1}{\text{.08 n}}_{\text{C}}\text{ + 0}{\text{.038 n}}_{\text{O}}\\ \text{=}\text{0}\text{.12(69) + 1}\text{.08}\left(\text{37}\right)\text{ + 0}\text{.038}\left(\text{1}\right)\\ \text{=}\text{40}\text{.8\% (Observer = 23}\text{.0\%)}\\ \\ \text{X+2 = 0}{\text{.0058 n}}_{\text{C}}{\text{(n}}_{\text{C}}\text{-1) + 0}{\text{.205n}}_{\text{O}}\\ \text{=0}\text{.0058(37)(36)+0}\text{.205(1)}\\ \text{=}\text{7}\text{.9\%}\text{(Observed=}8.\text{0\%)}\end{array}$

The intensities for the species $37:3$ which is having the molecular formula of ${\left[{\text{MNH}}_4\right]}^{+}{\text{= C}}_{\text{37}}{\text{H}}_{\text{74}}\text{ON}$ can be calculated as

$\begin{array}{l}\text{X+1 = 0}{\text{.12n}}_{\text{H}}\text{ + 1}{\text{.08 n}}_{\text{C}}\text{ + 0}{\text{.369 n}}_{\text{N}}\text{+ 0}{\text{.038 n}}_{\text{O}}\\ \text{=}\text{0}\text{.12(74) + 1}\text{.08}\left(\text{37}\right)\text{ + 0}\text{.369}\left(\text{1}\right)\text{+ 0}\text{.038}\left(\text{1}\right)\\ \text{=}\text{41}\text{.3\% (Observer = 40}\text{.8\%)}\\ \\ \text{X+2 = 0}{\text{.0058 n}}_{\text{C}}{\text{(n}}_{\text{C}}\text{-1) + 0}{\text{.205n}}_{\text{O}}\\ \text{=0}\text{.0058(37)(36)+0}\text{.205(1)}\\ \text{=}\text{7}\text{.9\%}\text{(Observed=}3.\text{7\%)}\end{array}$

The intensities for the species $37:3$ which is having the molecular formula of ${\left[\text{MH}\right]}^{+}{\text{= C}}_{\text{37}}{\text{H}}_{\text{71}}\text{O}$ can be calculated as

$\begin{array}{l}\text{X+1 = 0}{\text{.12n}}_{\text{H}}\text{ + 1}{\text{.08 n}}_{\text{C}}\text{ + 0}{\text{.038 n}}_{\text{O}}\\ \text{=}\text{0}\text{.12(71) + 1}\text{.08}\left(\text{37}\right)\text{ + 0}\text{.038}\left(\text{1}\right)\\ \text{=}40.8\text{\% (Observed = 33}\text{.4\%)}\\ \\ \text{X+2 = 0}{\text{.0058 n}}_{\text{C}}{\text{(n}}_{\text{C}}\text{-1) + 0}{\text{.205n}}_{\text{O}}\\ \text{=0}\text{.0058(37)(36)+0}\text{.205(1)}\\ \text{=}\text{7}\text{.9\%}\text{(Observed=}8.\text{4\%)}\end{array}$

Conclusion

The expected intensities of $\text{M,}\text{M+1,}\text{and}\text{M+2}$ for each of the given species was predicted.