Chapter 12, Problem 12.42AP

Interpretation Introduction

(a)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=110\left(98\%\right)$ is to be stated.

Concept introduction:

In mass spectroscopy, compounds can be identified on the basis of mass of compound. When the compound breaks into fragment then they can be distinguished from the other compounds. This technique is also used to differentiate the isotopes of compounds. This technique did not interact with electromagnetic radiation. Two peaks are used to identify the compound, first, the molecular ion peak which is the mass of the compound and second, the base peak which is the most abundant element peak. It may be same or different.

Answer to Problem 12.42AP

The peak at $m/z=110\left(98\%\right)$ is observed due to bromide isotope $\left[{}^{81}\text{Br}\right]$ that is $\left[\text{M}+2\right]$.

Explanation of Solution

The molecule ethyl bromide contains carbon, hydrogen, bromide atoms. In this molecule carbon and hydrogen mainly exist in one isotope form ${}^{12}\text{C}$ with natural abundance $\left(0.989\right)$ and ${}^1\text{H}$ with natural abundance $\left(0.999\right)$. The bromine exists in two isotopes ${}^{79}\text{Br}$ and ${}^{81}\text{Br}$ with abundance $\left(0.505\right)$ and $\left(0.495\right)$. The bromine isotopes affects the relative abundance. When the ethyl bromide undergoes fragmentation it loses electron and gives radical cation as shown below.

Figure 1

The peak at $m/z=110\left(98\%\right)$ is observed due to $\left[{}^{81}\text{Br}\right]$ that is $\left[\text{M}+2\right]$.

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=110\left(98\%\right)$ is due to $\left[{}^{81}\text{Br}\right]$ that is $\left[\text{M}+2\right]$ as shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=108\left(100\%\right)$ is to be stated.

Concept introduction:

In mass spectroscopy, compounds can be identified on the basis of mass of compound. When the compound breaks into fragment then they can be distinguished from the other compounds. This technique is also used to differentiate the isotopes of compounds. This technique did not interact with electromagnetic radiation. Two peaks are used to identify the compound, first, the molecular ion peak which is the mass of the compound and second, the base peak which is the most abundant element peak. It may be same or different.

Answer to Problem 12.42AP

The peak at $m/z=108\left(100\%\right)$ is observed due to $\left[{}^{79}\text{Br}\right]$ that is $\left[\text{M}\right]$ the molecular ion peak.

Explanation of Solution

The mass of the compound, $\text{mass}=108$, gives the molecular ion peak when it gives away one electron. The peak at $m/z=108\left(100\%\right)$ is due to $\left[{}^{79}\text{Br}\right]$ as shown below.

Figure 2

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=108\left(100\%\right)$ is due to $\left[{}^{79}\text{Br}\right]$ molecular ion peak as shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=81\left(5\%\right)$ is to be stated.

Concept introduction:

In mass spectroscopy, compounds can be identified on the basis of mass of compound. When the compound breaks into fragment then they can be distinguished from the other compounds. This technique is also used to differentiate the isotopes of compounds. This technique did not interact with electromagnetic radiation. Two peaks are used to identify the compound, first, the molecular ion peak which is the mass of the compound and second, the base peak which is the most abundant element peak. It may be same or different.

Answer to Problem 12.42AP

The peak at $m/z=81\left(5\%\right)$ is due to bromonium cation of $\left[{}^{81}\text{Br}\right]$.

Explanation of Solution

When the ethyl bromide $\left[\text{M}+2\right]$ loses ethyl radical it release bromonium cation $\left[{\text{Br}}^{+}\right]$ whose $\text{mass}=81$. This gives peak at $m/z=81\left(5\%\right)$ when $\left[{}^{81}\text{Br}\right]$ is present. It is shown below.

Figure 3

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=81\left(5\%\right)$ is due to bromonium cation $\left[{\text{Br}}^{+}\right]$ of $\left[{}^{81}\text{Br}\right]$ isotope as shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=79\left(5\%\right)$ is to be stated.

Concept introduction:

In mass spectroscopy, compounds can be identified on the basis of mass of compound. When the compound breaks into fragment then they can be distinguished from the other compounds. This technique is also used to differentiate the isotopes of compounds. This technique did not interact with electromagnetic radiation. Two peaks are used to identify the compound, first, the molecular ion peak which is the mass of the compound and second, the base peak which is the most abundant element peak. It may be same or different.

Answer to Problem 12.42AP

The peak at $m/z=79\left(5\%\right)$ is due to bromonium cation of $\left[{}^{79}\text{Br}\right]$.

Explanation of Solution

When the ethyl bromide $\left[\text{M}\right]$ loses ethyl radical it release bromonium cation $\left[{\text{Br}}^{+}\right]$ with $\text{mass}=79$. This gives peak at $m/z=79\left(5\%\right)$ when $\left[{}^{79}\text{Br}\right]$ is present. It is shown below.

Figure 4

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=79\left(5\%\right)$ is due to bromonium cation $\left[{\text{Br}}^{+}\right]$ of $\left[{}^{79}\text{Br}\right]$ isotope as shown in Figure 4.

Interpretation Introduction

(e)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=29\left(61\%\right)$ is to be stated.

Concept introduction:

In mass spectroscopy, compounds can be identified on the basis of mass of compound. When the compound breaks into fragment then they can be distinguished from the other compounds. This technique is also used to differentiate the isotopes of compounds. This technique did not interact with electromagnetic radiation. Two peaks are used to identify the compound, first, the molecular ion peak which is the mass of the compound and second, the base peak which is the most abundant element peak. It may be same or different.

Answer to Problem 12.42AP

The peak at $m/z=29\left(61\%\right)$ is due to ethyl cation $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{}^{+}\right]$

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethyl cation $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{}^{+}\right]$ with $\text{mass}=29$. Therefore, peak at $m/z=29\left(61\%\right)$ is observed. It is shown below.

Figure 5

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=29\left(61\%\right)$ is due to ethyl cation $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{}^{+}\right]$ as shown in Figure 5.

Interpretation Introduction

(f)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=28\left(25\%\right)$ is to be stated.

Concept introduction:

In mass spectroscopy, compounds can be identified on the basis of mass of compound. When the compound breaks into fragment then they can be distinguished from the other compounds. This technique is also used to differentiate the isotopes of compounds. This technique did not interact with electromagnetic radiation. Two peaks are used to identify the compound, first, the molecular ion peak which is the mass of the compound and second, the base peak which is the most abundant element peak. It may be same or different.

Answer to Problem 12.42AP

The peak at $m/z=28\left(25\%\right)$ is due to ethyl radical cation $\left[{\text{C}}_{\text{2}}{\text{H}}_4{}^{•+}\right]$.

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethyl radical cation $\left[{\text{C}}_{\text{2}}{\text{H}}_4{}^{•+}\right]$ with $\text{mass}=28$. Therefore, peak at $m/z=28\left(25\%\right)$ is observed. It is shown below.

Figure 6

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=28\left(25\%\right)$ is due to ethyl radical cation $\left[{\text{C}}_{\text{2}}{\text{H}}_4{}^{•+}\right]$ as shown below in Figure 6.

Interpretation Introduction

(g)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=27\left(53\%\right)$ is to be stated.

Concept introduction:

In mass spectroscopy, compounds can be identified on the basis of mass of compound. When the compound breaks into fragment then they can be distinguished from the other compounds. This technique is also used to differentiate the isotopes of compounds. This technique did not interact with electromagnetic radiation. Two peaks are used to identify the compound, first, the molecular ion peak which is the mass of the compound and second, the base peak which is the most abundant element peak. It may be same or different.

Answer to Problem 12.42AP

The peak at $m/z=28\left(25\%\right)$ is due to ethene cation $\left[\text{CH}={\text{CH}}^{+}\right]$.

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethene cation $\left[\text{CH}={\text{CH}}^{+}\right]$ with $\text{mass}=27$. Therefore, peak at $m/z=27\left(53\%\right)$ is observed. It is shown below.

Figure 7

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=27\left(53\%\right)$ is due to ethene cation $\left[\text{CH}={\text{CH}}^{+}\right]$ as shown in Figure 7.