Chapter 22, Problem 22.FE

(a)

Interpretation Introduction

Interpretation:

A composition for each molecule and the expected isotopic peak intensities has to be suggested and calculated.

Explanation of Solution

Form the given intensity we can identified a compound for ${\text{M}}^{+·}=94$ is phenol.

The molecular formula for phenol is ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\text{O}$ structure of phenol is

Now,

$\begin{array}{l}\text{Rings + double}\text{bonds }\text{=}\text{c-h/2+n/2+1}\\ =6-6/2+0/2+1\\ =4\end{array}$

The expected intensity for $\text{M+1}$ from the given table 21-2:

$\begin{array}{l}\text{1}\text{.08(6) + 0}\text{.012(6) + 0}\text{.038(1) = 6}\text{.59\%}\\ \text{ C}\text{H}\text{O}\end{array}$

Observed intensity of $\text{M+1 = 68/999 = 6}\text{.8\%}$

Expected intensity of $\text{M+2 = 0}\text{.0058(6)(5) + 0}\text{.205(1) = 0}\text{.38\%}$

Observed intensity of $\text{M+2 = 0}\text{.3\%}$

(b)

Interpretation Introduction

Interpretation:

A composition for each molecule and the expected isotopic peak intensities has to be suggested and calculated.

Explanation of Solution

Form the given intensity we can identified a compound for ${\text{M}}^{+·}=156$ is bromophenol.

The molecular formula for bromophenol is ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{Br}$ structure of phenol is

Now,

$\begin{array}{l}\text{Rings + double}\text{bonds }\text{=}\text{c-h/2+n/2+1}\\ =6-6/2+0/2+1\\ =4\end{array}$

Here, h includes $\text{H+Br}$

The expected intensity for

$\begin{array}{l}\text{M+1}=\text{1}\text{.08(6) + 0}\text{.012(5) = 6}\text{.54\%}\\ \text{C}\text{H}\end{array}$

Observed intensity of $\text{M+1 = 45/566 = 8}\text{.1\%}$

Expected intensity of $\text{M+2 = 0}\text{.0058(6)(5) + 97}\text{.3(1) = 97}\text{.5\%}$

Observed intensity of $\text{M+2 = 520/566 = 91}\text{.9\%}$

The isotopic partner of $\text{M+2}$ is $\text{M+3}$( ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{}^{\text{81}}\text{Br}$) and which has ${}^{\text{81}}\text{Br}$ plus either one ${}^{\text{13}}\text{C}\text{or}\text{one}{}^{\text{1}}\text{H}\text{.}$

Hence, the expected intensity of $\text{M+3}$ is $1.08(6)+0.012(5)=6.54\%$

The predicted intensity of ( ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{}^{\text{81}}\text{Br}$) at $\text{M+2}$ is $\text{0}\text{.0654(97}\text{.3) = 6}\text{.4\% }\text{of}{\text{M}}^{+·}$

Observed intensity of $\text{M+3}$ is $35/566=6.2\%.$

(c)

Interpretation Introduction

Interpretation:

A composition for each molecule and the expected isotopic peak intensities has to be suggested and calculated.

Explanation of Solution

From the given data ${\text{M}}^{+·}=224$ we can assign these value for the following compound and which has the correct structure shown below.  There is no some other way to assign the isomeric structure from the given data.

The molecular formula is ${\text{C}}_{\text{7}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{Cl}}_{\text{3}}$

$\begin{array}{l}\text{Rings + double}\text{bonds }\text{=}\text{c-h/2+n/2+1}\\ =7-6/2+0/2+1\\ =5\end{array}$

The expected intensity for

$\begin{array}{l}\text{M+1}=\text{1}\text{.08(7) + 0}\text{.012(3) + 0}\text{.038(2) = 7}\text{.67\%}\\ \text{C}\text{H}\text{O}\end{array}$

Observed intensity of $\text{M+1 = 63/791 = 8}\text{.0\%}$

Expected intensity of $\text{M+2 = 0}\text{.0058(7)(6) + 0}\text{.205(2) + 32}\text{.0(3) = 96}\text{.7\%}$

Observed intensity of $\text{M+2 = 754/791 = 95}\text{.4\%}$

The $\text{M+3}$ peak is the isotopic partner of ${\text{C}}_{\text{7}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{\text{35}}{\text{Cl}}_{\text{2}}{}^{\text{37}}\text{Cl}$ at $\text{M+2}\text{.}$

$\text{M+2}$ has ${}^{\text{37}}\text{Cl}$ plus either one ${}^{\text{13}}\text{C}\text{or}\text{one}{}^{\text{1}}{\text{H or }}^{17}\text{O}\text{.}$

Expected intensity of $\text{M+3}$ is $1.08(7)+0.012(3)+0.038(2)=7.67\%$

Predicted intensity of ( ${\text{C}}_{\text{7}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{\text{35}}{\text{Cl}}_{\text{2}}{}^{\text{37}}\text{Cl}$) at $\text{M+2}$ is $\text{32}\text{.0(3) = 96}\text{.0\% }\text{of}{\text{M}}^{+·}$

Expected intensity of $\text{M+3=7}\text{.67\% of 96}\text{.0\%=7}{\text{.4\% of M}}^{+·}$

Observed intensity $=60/791=7.6\%.$

For $\text{M+4}$ which is completely composed of ${\text{C}}_{\text{7}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{\text{35}}{\text{Cl}}^{\text{37}}{\text{Cl}}_2$ along with little amount of ${\text{C}}_{\text{7}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{\text{35}}{\text{Cl}}_{\text{2}}{}^{\text{37}}\text{Cl}\text{.}$

Other formulas like ${}^{\text{12}}{\text{C}}_{\text{6}}{}^{\text{13}}{\text{CH}}_{\text{3}}{}^{\text{16}}{\text{O}}^{\text{17}}{\text{O}}^{\text{35}}{\text{Cl}}_{\text{2}}{}^{\text{37}}\text{Cl}$ also add up to $\text{M+4,}$ which has two minor isotopes of ${}^{\text{13}}\text{C}\text{and}{}^{\text{17}}\text{O}$ but there less likely to occur.

Expected intensity of $\text{M+4}$ from ${\text{C}}_{\text{7}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{\text{35}}{\text{Cl}}_{\text{2}}{}^{\text{37}}\text{Cl}$ is $\text{5}\text{.11(3)(2) = 30}{\text{.7\% of M}}^{+·}.$

The contribution from ${\text{C}}_{\text{7}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{\text{35}}{\text{Cl}}_2{}^{\text{37}}\text{Cl}$ on basis of the predicted intensity of ${\text{C}}_{\text{7}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{\text{35}}{\text{Cl}}_2{}^{\text{37}}\text{Cl}$ at $\text{M+2}\text{.}$

The predicted intensity of ${\text{C}}_{\text{7}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{\text{35}}{\text{Cl}}_2{}^{\text{37}}\text{Cl}$ is $\text{32}\text{.0(3) = 96}\text{.0\%}\text{of}{\text{M}}^{+·}.$

The predicted intensity from ${\text{C}}_7{\text{CH}}_{\text{3}}{}^{\text{16}}{\text{O}}^{\text{18}}{\text{O}}^{\text{35}}{\text{Cl}}_{\text{2}}{}^{\text{37}}\text{Cl}\text{at }\text{M+4}$ is $\text{0}\text{.205}\left(\text{2}\right)\text{= 0}\text{.410\% of 96}\text{.0\% = 0}\text{.4\%}\text{.}$

The total expected intensity of $\text{M+4}$ is $\text{30}\text{.7\%+ 0}\text{.4\% = 31}{\text{.1\% of M}}^{+·}.$

Observed intensity $=264/791=33.4\%.$

Expected intensity of $\text{M+5}$ from ${}^{\text{12}}{\text{C}}_{\text{6}}{}^{\text{13}}{\text{CH}}_{\text{3}}{\text{O}}_2{}^{\text{35}}{\text{Cl}}^{\text{37}}{\text{Cl}}_2$ and ${}^{\text{12}}{\text{C}}_7{\text{H}}_2{}^2{\text{HO}}_2{}^{\text{35}}{\text{Cl}}^{\text{37}}{\text{Cl}}_2$ and ${\text{C}}_7{\text{H}}_{\text{3}}{}^{\text{16}}{\text{O}}^{\text{17}}{\text{O}}^{\text{35}}{\text{Cl}}^{\text{37}}{\text{Cl}}_{\text{2}}$ which is based on the predicted intensity of ${\text{C}}_{\text{7}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{\text{35}}{\text{Cl}}^{\text{37}}{\text{Cl}}_2$ at $\text{M+4}\text{.}$ $\text{M+5}$ should have $1.08(7)+0.012(3)+0.038(2)=7.7\%\text{of}{\text{C}}_7{\text{H}}_{\text{3}}{}^{\text{16}}{\text{O}}^{\text{17}}{\text{O}}^{\text{35}}{\text{Cl}}^{\text{37}}{\text{Cl}}_{\text{2}}$ $\text{at M+4 =7}\text{.7\% of 30}\text{.7\%=2}\text{.4\%}\text{.}$

Observed intensity $=29/791=3.7\%.$

(d)

Interpretation Introduction

Interpretation:

A composition for each molecule and the expected isotopic peak intensities has to be suggested and calculated.

Explanation of Solution

From the given data ${\text{M}}^{+·}=154$ we can assign these value for the following compound and which has the correct structure shown below.  The observed $\text{M+2}\text{is 12/122 = 9}\text{.8\%,}$ which indicates that the compound is having two sulfur atoms.  The composition ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}{\text{O}}_{\text{2}}{\text{S}}_{\text{2}}$ is exact match for this compound with two sulfur atoms and having molecular mass of 154.

The structure for this compound can be shown as follows.

Now let’s find the following things,

$\begin{array}{l}\text{Rings + double}\text{bonds }\text{=}\text{c-h/2+n/2+1}\\ =4-10/2+0/2+1\\ =0\end{array}$

The expected intensity for $\text{M+1}$:

$\begin{array}{l}\text{1}\text{.08(4) + 0}\text{.012(10) + 0}\text{.038(2) + 0}\text{.801(2) = 6}\text{.12\%}\\ \text{ C}\text{H}\text{O}\text{S}\end{array}$

Observed intensity of $\text{M+1 = 9/122 = 7}\text{.4\%}$

Expected intensity of $\text{M+2 = 0}\text{.0058(4)(3) + 0}\text{.205(2)+4}\text{.52(2) = 9}\text{.52\%}$

Observed intensity of $\text{M+2 = 12/122=9}\text{.8\%}\text{.}$