Chapter 21, Problem 21.25P

(a)

Interpretation Introduction

Interpretation:

Plausible formula for ${\text{C}}_n{\text{H}}_x{\text{O}}_y{\text{N}}_z$ has to be suggested when nominal mass is 79 and ratio of ${\text{M}}^{+}+{\text{1/M}}^{+}$ peak is $5.9\text{ \%}$.

Concept Introduction:

Nominal mass may be defined as mass of most abundant isotope of a particular ion or molecule. For instance, carbon exist as two isotopes $\text{C-}12$ and $\text{C-}13$. $\text{C-}12$ is abundant as much as $98.9\text{ \%}$ while abundance of $\text{C-}13$ is only $\text{1}\text{.1 \%}$. So clearly nominal mass of carbon is $12\text{ g}$.

Explanation of Solution

Odd nominal mass of 79 suggests there are odd numbers of $\text{N}$ atoms. So it can be 1, 3 or any odd number.

Natural abundance of $\text{C-}13$ is $\text{1}\text{.1 \%}$. Hence number of $\text{C}$ atoms is calculated as follows:

  $\begin{array}{c}\frac{{\text{M}}^{+}+1}{{\text{M}}^{+}}=n\left(\text{1}\text{.1 \%}\right)\\ n=\frac{5.9\text{ \%}}{\text{1}\text{.1 \%}}\\ =5.3\\ \approx 5\end{array}$

Hence number of $\text{C}$ atom is 5.

Since each $\text{C}$ has mass of 12, so five $\text{C}$ atoms contribute a mass of 60 to overall formula ${\text{C}}_n{\text{H}}_x{\text{O}}_y{\text{N}}_z$. This leaves 19 mass that indicates possibility of one $\text{N}$ atom and remainare five $\text{H}$ atoms. So with no $\text{O}$ atom, the plausible formula of compound is ${\text{C}}_5{\text{H}}_5\text{N}$.

(b)

Interpretation Introduction

Interpretation:

Plausible formula for ${\text{C}}_n{\text{H}}_x{\text{O}}_y{\text{N}}_z$ has to be suggested when nominal mass is 123 and ratio of ${\text{M}}^{+}+{\text{1/M}}^{+}$ peak is $\text{6}\text{.1 \%}$.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Odd nominal mass of 123 suggests there are odd numbers of $\text{N}$ atoms. So it can be 1, 3 or any odd number.

Natural abundance of $\text{C-}13$ is $\text{1}\text{.1 \%}$. Hence number of $\text{C}$ atoms is calculated as follows:

  $\begin{array}{c}\frac{{\text{M}}^{+}+1}{{\text{M}}^{+}}=n\left(\text{1}\text{.1 \%}\right)\\ n=\frac{\text{6}\text{.1 \%}}{\text{1}\text{.1 \%}}\\ =5.5\\ \approx 5\text{ or 6}\end{array}$

Hence number of $\text{C}$ atom is 5 or 6.

Since each $\text{C}$ has mass of 12 so six $\text{C}$ atoms are assumed in formula then they contribute a mass of 72 to overall formula ${\text{C}}_n{\text{H}}_x{\text{O}}_y{\text{N}}_z$. This leaves 51 as mass that indicates possibility of only one $\text{N}$ atom. If molecule has 3 $\text{N}$ atoms then it would have contributed 41 and then no possibility of $\text{O}$ atom exists.

With 1$\text{N}$, and 2 $\text{O}$ and remainder 5 $\text{H}$ atoms, contribution of 51 gets satisfied and thus one plausible formula is ${\text{C}}_6{\text{H}}_5{\text{O}}_2\text{N}$.

When five $\text{C}$ atoms are assumed in formula, they contribute a mass of 60 to overall formula ${\text{C}}_n{\text{H}}_x{\text{O}}_y{\text{N}}_z$. This leaves 63 as mass that indicates possibility of 3 $\text{N}$ atoms and this would have contributed 42 and then one possibility of 1 $\text{O}$ atom and remainder five $\text{H}$ atoms. Thus, another possible formula is ${\text{C}}_5{\text{H}}_5{\text{ON}}_3$.

(c)

Interpretation Introduction

Interpretation:

Plausible formula for ${\text{C}}_n{\text{H}}_x{\text{O}}_y{\text{N}}_z$ has to be suggested when nominal mass is 148 and ratio of ${\text{M}}^{+}+{\text{1/M}}^{+}$ peak is $\text{7}\text{.4 \%}$.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Even nominal mass of 148 suggests there are even numbers of $\text{N}$ atoms. So it can be 2, 4 or any even number.

Natural abundance of $\text{C-}13$ is $\text{1}\text{.1 \%}$. Hence number of $\text{C}$ atoms is calculated as follows:

  $\begin{array}{c}\frac{{\text{M}}^{+}+1}{{\text{M}}^{+}}=n\left(\text{1}\text{.1 \%}\right)\\ n=\frac{\text{7}\text{.4 \%}}{\text{1}\text{.1 \%}}\\ =6.7\\ \approx 7\text{ or }6\end{array}$

Hence number of $\text{C}$ atom is 7 or 6.

Plausible formulas with 6 $\text{C}$ atoms that satisfy nominal mass of 148 include ${\text{C}}_6{\text{H}}_{12}{\text{O}}_4$ and ${\text{C}}_6{\text{H}}_{16}{\text{O}}_2{\text{N}}_{\text{2}}$.

Plausible formulas with 7 $\text{C}$ atoms that satisfy nominal mass of 148 include ${\text{C}}_7{\text{H}}_4{\text{O}}_2{\text{N}}_{\text{2}}$ and ${\text{C}}_7{\text{O}}_4$.

(d)

Interpretation Introduction

Interpretation:

Plausible formula for ${\text{C}}_n{\text{H}}_x{\text{O}}_y{\text{N}}_z$ has to be suggested when nominal mass is 168 and ratio of ${\text{M}}^{+}+{\text{1/M}}^{+}$ peak is $\text{12}\text{.5 \%}$.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Even nominal mass of 168 suggests there are even numbers of $\text{N}$ atoms. So it can be 2, 4 or any even number.

Natural abundance of $\text{C-}13$ is $\text{1}\text{.1 \%}$. Hence number of $\text{C}$ atoms is calculated as follows:

  $\begin{array}{c}\frac{{\text{M}}^{+}+1}{{\text{M}}^{+}}=n\left(\text{1}\text{.1 \%}\right)\\ n=\frac{\text{12}\text{.5 \%}}{\text{1}\text{.1 \%}}\\ =11.36\\ \approx 11\text{ or 12}\end{array}$

Hence number of $\text{C}$ atom is 11 or 12.

Plausible formulas with 11 $\text{C}$ atoms that satisfy nominal mass of 168 includes ${\text{C}}_{11}{\text{H}}_{20}\text{O}$, ${\text{C}}_{11}{\text{H}}_4{\text{O}}_2$ or ${\text{C}}_{11}{\text{H}}_8{\text{N}}_2$.

Plausible formulas with 12 $\text{C}$ atoms that satisfy nominal mass of 168 include ${\text{C}}_{12}{\text{H}}_{24}$ and ${\text{C}}_{12}{\text{H}}_8\text{O}$.