Chapter 13, Problem 13.25P

Interpretation Introduction

(a)

Interpretation: The expected molecular ion for the given compound is to be determined.

Concept introduction: Molecular mass is the sum of the atomic weights of each constituent element multiplied by the number of atoms of that element. The molecular weight of compound is equal to the mass of the molecular ion.

Answer to Problem 13.25P

The structure of the given compound is,

The expected molecular ion peak of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ is observed at $m\text{/}z=78$.

Explanation of Solution

The molecular formula of given compound is ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$.

The $m\text{/}z$ value of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ is calculated as,

$\begin{array}{rll}{\text{Mass of C}}_{\text{6}}{\text{H}}_{\text{6}}& =& \left(6\times \text{Mass of C atom}+6\times \text{Mass of H atom}\right)\\ & =& \left(\text{6}\times \text{12}+\text{6}\times \text{1}\right)\\ & =& \left(\text{72 + 6}\right)\\ & =& \text{ 78 g/mol}\end{array}$

The molecular weight of the compound is equal to the mass of the molecular ion.

Hence, the expected molecular ion peak of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ is observed at $m\text{/}z=78$.

Conclusion

The expected molecular ion peak of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ is observed at $m\text{/}z=78$.

Interpretation Introduction

(b)

Interpretation: The expected molecular ion for the given compound is to be determined.

Concept introduction: Molecular mass is the sum of the atomic weights of each constituent element multiplied by the number of atoms of that element. The molecular weight of compound is equal to the mass of the molecular ion.

Answer to Problem 13.25P

The structure of the given compound is,

The expected molecular ion peak of ${\text{C}}_{10}{\text{H}}_{\text{16}}$ compound is observed at $m\text{/}z=136$

Explanation of Solution

The molecular formula of given compound is ${\text{C}}_{10}{\text{H}}_{\text{16}}$.

The $m\text{/}z$ value of ${\text{C}}_{10}{\text{H}}_{\text{16}}$ is calculated as,

$\begin{array}{rll}{\text{Mass of C}}_{10}{\text{H}}_{\text{16}}& =& \left(10\times \text{Mass of C atom}+16\times \text{Mass of H atom}\right)\\ & =& \left(10\times \text{12}+1\text{6}\times \text{1}\right)\\ & =& \left(\text{120}+\text{16}\right)\\ & =& \text{ 136 g/mol}\end{array}$

The molecular weight of the compound is equal to the mass of the molecular ion.

Hence, the expected molecular ion peak of ${\text{C}}_{10}{\text{H}}_{\text{16}}$ compound is observed at $m\text{/}z=136$.

Conclusion

The expected molecular ion peak of ${\text{C}}_{10}{\text{H}}_{\text{16}}$ is observed at $m\text{/}z=136$.

Interpretation Introduction

(c)

Interpretation: The expected molecular ion for the given compound is to be determined.

Concept introduction: Molecular mass is the sum of the atomic weights of each constituent element multiplied by the number of atoms of that element. The molecular weight of compound is equal to the mass of the molecular ion.

Answer to Problem 13.25P

The structure of the given compound is,

The expected molecular ion peak of ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}\text{O}$ is observed at $m\text{/}z=86$.

Explanation of Solution

The molecular formula of given compound is ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}\text{O}$

The $m\text{/}z$ value of ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}\text{O}$ is calculated as,

$\begin{array}{rll}{\text{Mass of C}}_{\text{5}}{\text{H}}_{\text{10}}\text{O}& =& \left(5\times \text{Mass of C atom}+10\times \text{Mass of H atom }+\text{Mass of O atom}\right)\\ & =& \left(5\times \text{12+}10\times \text{1+16}\right)\\ & =& \left(\text{60+10+16}\right)\\ & =& \text{ 86 g/mol}\end{array}$

The molecular weight of the compound is equal to the mass of the molecular ion.

The expected molecular ion peak of ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}\text{O}$ is observed at $m\text{/}z=86$.

Conclusion

The expected molecular ion peak of ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}\text{O}$ is observed at $m\text{/}z=86$.

Interpretation Introduction

(d)

Interpretation: The expected molecular ion for the given compound is to be determined.

Concept introduction: Molecular mass is the sum of the atomic weights of each constituent element multiplied by the number of atoms of that element. The molecular weight of compound is equal to the mass of the molecular ion.

Answer to Problem 13.25P

The structure of the given compound is,

The larger molecular ion $\text{M}$ peak is observed at $m\text{/}z=106$ (${\text{C}}_5{\text{H}}_{11}{}^{35}\text{Cl}$) and small $\text{M}+\text{2}$ peak is observed at $m\text{/}z=108$ (${\text{C}}_5{\text{H}}_{11}{}^{37}\text{Cl}$).

Explanation of Solution

The molecular formula of given compound is ${\text{C}}_5{\text{H}}_{11}\text{Cl}$. Chlorine has two isotopes ${}^{\text{35}}\text{Cl}$ and ${}^{\text{37}}\text{Cl}$.

The $m\text{/}z$ value of ${\text{C}}_5{\text{H}}_{11}{}^{35}\text{Cl}$ is calculated as follows.

$\begin{array}{rll}{\text{Mass of C}}_5{\text{H}}_{11}{}^{35}\text{Cl}& =& \left(5\times \text{Mass of C atom}+11\times {\text{Mass of H atom + Mass of }}^{35}\text{Cl atom}\right)\\ & =& \left(5\times \text{12}+11\times \text{1+ 35 }\right)\\ & =& \left(\text{60+11+35}\right)\\ & =& \text{ 106 g/mol}\end{array}$

The $m\text{/}z$ value of ${\text{C}}_5{\text{H}}_{11}{}^{37}\text{Cl}$ is calculated as follows.

$\begin{array}{rll}{\text{Mass of C}}_5{\text{H}}_{11}{}^{37}\text{Cl}& =& \left(5\times \text{Mass of C atom}+11\times {\text{Mass of H atom + Mass of }}^{37}\text{Cl atom}\right)\\ & =& \left(5\times \text{12}+11\times \text{1+ 37 }\right)\\ & =& \left(\text{60+11+37}\right)\\ & =& \text{ 108 g/mol}\end{array}$

The molecular weight of the compound is equal to the mass of the molecular ion.

Hence, the larger molecular ion $\text{M}$ peak is observed at $m\text{/}z=106$ (${\text{C}}_5{\text{H}}_{11}{}^{35}\text{Cl}$) and small $\text{M}+\text{2}$ peak is observed at $m\text{/}z=108$ (${\text{C}}_5{\text{H}}_{11}{}^{37}\text{Cl}$).

Conclusion

The larger molecular ion $\text{M}$ is observed peak at $m\text{/}z=106$ (${\text{C}}_5{\text{H}}_{11}{}^{35}\text{Cl}$) and $\text{M}+\text{2}$ is observed small peak at $m\text{/}z=108$ (${\text{C}}_5{\text{H}}_{11}{}^{37}\text{Cl}$).

Interpretation Introduction

(e)

Interpretation: The expected molecular ion for the given compound is to be determined.

Concept introduction: Molecular mass is the sum of the atomic weights of each constituent element multiplied by the number of atoms of that element. The molecular weight of compound is equal to the mass of the molecular ion.

Answer to Problem 13.25P

The structure of the given compound is,

The larger molecular ion $\text{M}$ peak is observed at $m\text{/}z=192$ (${\text{C}}_8{\text{H}}_{17}{}^{79}\text{Br}$) and small $\text{M}+\text{2}$ peak is observed at $m\text{/}z=194$ (${\text{C}}_8{\text{H}}_{17}{}^{81}\text{Br}$).

Explanation of Solution

The molecular formula of given compound is ${\text{C}}_8{\text{H}}_{17}\text{Br}$. Bromine atom has two isotopes ${}^{79}\text{Br}$ and ${}^{81}\text{Br}$.

The $m\text{/}z$ value of ${\text{C}}_8{\text{H}}_{17}{}^{79}\text{Br}$ is calculated as follows.

$\begin{array}{rll}{\text{Mass of C}}_8{\text{H}}_{17}{}^{79}\text{Br}& =& \left(8\times \text{Mass of C atom}+17\times \text{Mass of H atom + Mass of Br atom}\right)\\ & =& \left(8\times \text{12}+17\times \text{1+ 79}\right)\\ & =& \left(\text{96 + 17+79}\right)\\ & =& \text{ 192}\end{array}$

The $m\text{/}z$ value of ${\text{C}}_8{\text{H}}_{17}{}^{81}\text{Br}$ is calculated as follows.

$\begin{array}{rll}{\text{Mass of C}}_8{\text{H}}_{17}{}^{81}\text{Br}& =& \left(8\times \text{Mass of C atom}+17\times \text{Mass of H atom + Mass of Br atom}\right)\\ & =& \left(8\times \text{12}+17\times \text{1+ 81}\right)\\ & =& \left(\text{96 + 17+81}\right)\\ & =& \text{ 194}\end{array}$

The molecular weight of the compound is equal to the mass of the molecular ion.

Hence, the larger molecular ion $\text{M}$ peak is observed at $m\text{/}z=192$ (${\text{C}}_8{\text{H}}_{17}{}^{79}\text{Br}$) and small $\text{M}+\text{2}$ peak is observed at $m\text{/}z=194$ (${\text{C}}_8{\text{H}}_{17}{}^{81}\text{Br}$).

Conclusion

The larger molecular ion $\text{M}$ peak is observed at $m\text{/}z=192$ (${\text{C}}_8{\text{H}}_{17}{}^{79}\text{Br}$) and small $\text{M}+\text{2}$ peak is observed at $m\text{/}z=194$ (${\text{C}}_8{\text{H}}_{17}{}^{81}\text{Br}$).