(a) Interpretation: The mass spectrum fragmentation of ethyl bromide at m / z = 110 ( 98 % ) 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.

Question

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Answer 1

Question

Chapter 12, Problem 12.42AP

Interpretation Introduction

(a)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=110(98%)$ 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.

Expert Solution

Answer to Problem 12.42AP

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 1

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 2

Figure 1

The peak at $m/z=110(98%)$ is observed due to $[81Br]$ that is $[M+2]$ .

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=108(100%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=108(100%)$ is observed due to $[79Br]$ that is $[M]$ the molecular ion peak.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 3

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 4

Figure 2

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=81(5%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=81(5%)$ is due to bromonium cation of $[81Br]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 5

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 6

Figure 3

Conclusion

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

Interpretation Introduction

(d)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=79(5%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=79(5%)$ is due to bromonium cation of $[79Br]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 7

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 8

Figure 4

Conclusion

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

Interpretation Introduction

(e)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=29(61%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=29(61%)$ is due to ethyl cation $[C2H5+]$

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 9

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethyl cation $[C2H5+]$ with $mass=29$ . Therefore, peak at $m/z=29(61%)$ is observed. It is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 10

Figure 5

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=29(61%)$ is due to ethyl cation $[C2H5+]$ as shown in Figure 5.

Interpretation Introduction

(f)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=28(25%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=28(25%)$ is due to ethyl radical cation $[C2H4•+]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 11

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethyl radical cation $[C2H4•+]$ with $mass=28$ . Therefore, peak at $m/z=28(25%)$ is observed. It is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 12

Figure 6

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=28(25%)$ is due to ethyl radical cation $[C2H4•+]$ as shown below in Figure 6.

Interpretation Introduction

(g)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=27(53%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=28(25%)$ is due to ethene cation $[CH=CH+]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 13

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 14

Figure 7

Conclusion

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

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Answer 2

Question

Chapter 12, Problem 12.42AP

Interpretation Introduction

(a)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=110(98%)$ 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.

Expert Solution

Answer to Problem 12.42AP

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 1

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 2

Figure 1

The peak at $m/z=110(98%)$ is observed due to $[81Br]$ that is $[M+2]$ .

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=108(100%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=108(100%)$ is observed due to $[79Br]$ that is $[M]$ the molecular ion peak.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 3

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 4

Figure 2

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=81(5%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=81(5%)$ is due to bromonium cation of $[81Br]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 5

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 6

Figure 3

Conclusion

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

Interpretation Introduction

(d)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=79(5%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=79(5%)$ is due to bromonium cation of $[79Br]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 7

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 8

Figure 4

Conclusion

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

Interpretation Introduction

(e)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=29(61%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=29(61%)$ is due to ethyl cation $[C2H5+]$

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 9

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethyl cation $[C2H5+]$ with $mass=29$ . Therefore, peak at $m/z=29(61%)$ is observed. It is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 10

Figure 5

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=29(61%)$ is due to ethyl cation $[C2H5+]$ as shown in Figure 5.

Interpretation Introduction

(f)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=28(25%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=28(25%)$ is due to ethyl radical cation $[C2H4•+]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 11

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethyl radical cation $[C2H4•+]$ with $mass=28$ . Therefore, peak at $m/z=28(25%)$ is observed. It is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 12

Figure 6

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=28(25%)$ is due to ethyl radical cation $[C2H4•+]$ as shown below in Figure 6.

Interpretation Introduction

(g)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=27(53%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=28(25%)$ is due to ethene cation $[CH=CH+]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 13

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 14

Figure 7

Conclusion

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

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Answer 3

Question

Chapter 12, Problem 12.42AP

Interpretation Introduction

(a)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=110(98%)$ 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.

Expert Solution

Answer to Problem 12.42AP

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 1

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 2

Figure 1

The peak at $m/z=110(98%)$ is observed due to $[81Br]$ that is $[M+2]$ .

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=108(100%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=108(100%)$ is observed due to $[79Br]$ that is $[M]$ the molecular ion peak.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 3

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 4

Figure 2

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=81(5%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=81(5%)$ is due to bromonium cation of $[81Br]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 5

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 6

Figure 3

Conclusion

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

Interpretation Introduction

(d)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=79(5%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=79(5%)$ is due to bromonium cation of $[79Br]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 7

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 8

Figure 4

Conclusion

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

Interpretation Introduction

(e)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=29(61%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=29(61%)$ is due to ethyl cation $[C2H5+]$

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 9

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethyl cation $[C2H5+]$ with $mass=29$ . Therefore, peak at $m/z=29(61%)$ is observed. It is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 10

Figure 5

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=29(61%)$ is due to ethyl cation $[C2H5+]$ as shown in Figure 5.

Interpretation Introduction

(f)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=28(25%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=28(25%)$ is due to ethyl radical cation $[C2H4•+]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 11

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethyl radical cation $[C2H4•+]$ with $mass=28$ . Therefore, peak at $m/z=28(25%)$ is observed. It is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 12

Figure 6

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=28(25%)$ is due to ethyl radical cation $[C2H4•+]$ as shown below in Figure 6.

Interpretation Introduction

(g)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=27(53%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=28(25%)$ is due to ethene cation $[CH=CH+]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 13

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 14

Figure 7

Conclusion

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

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Answer 4

Question

Chapter 12, Problem 12.42AP

Interpretation Introduction

(a)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=110(98%)$ 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.

Expert Solution

Answer to Problem 12.42AP

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 1

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 2

Figure 1

The peak at $m/z=110(98%)$ is observed due to $[81Br]$ that is $[M+2]$ .

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=108(100%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=108(100%)$ is observed due to $[79Br]$ that is $[M]$ the molecular ion peak.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 3

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 4

Figure 2

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=81(5%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=81(5%)$ is due to bromonium cation of $[81Br]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 5

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 6

Figure 3

Conclusion

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

Interpretation Introduction

(d)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=79(5%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=79(5%)$ is due to bromonium cation of $[79Br]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 7

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 8

Figure 4

Conclusion

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

Interpretation Introduction

(e)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=29(61%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=29(61%)$ is due to ethyl cation $[C2H5+]$

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 9

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethyl cation $[C2H5+]$ with $mass=29$ . Therefore, peak at $m/z=29(61%)$ is observed. It is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 10

Figure 5

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=29(61%)$ is due to ethyl cation $[C2H5+]$ as shown in Figure 5.

Interpretation Introduction

(f)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=28(25%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=28(25%)$ is due to ethyl radical cation $[C2H4•+]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 11

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethyl radical cation $[C2H4•+]$ with $mass=28$ . Therefore, peak at $m/z=28(25%)$ is observed. It is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 12

Figure 6

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=28(25%)$ is due to ethyl radical cation $[C2H4•+]$ as shown below in Figure 6.

Interpretation Introduction

(g)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=27(53%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=28(25%)$ is due to ethene cation $[CH=CH+]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 13

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 14

Figure 7

Conclusion

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

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Answer 5

Chapter 12, Problem 12.42AP

Interpretation Introduction

(a)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=110(98%)$ 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.

Expert Solution

Answer to Problem 12.42AP

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 1

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 2

Figure 1

The peak at $m/z=110(98%)$ is observed due to $[81Br]$ that is $[M+2]$ .

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=108(100%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=108(100%)$ is observed due to $[79Br]$ that is $[M]$ the molecular ion peak.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 3

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 4

Figure 2

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=81(5%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=81(5%)$ is due to bromonium cation of $[81Br]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 5

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 6

Figure 3

Conclusion

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

Interpretation Introduction

(d)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=79(5%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=79(5%)$ is due to bromonium cation of $[79Br]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 7

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 8

Figure 4

Conclusion

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

Interpretation Introduction

(e)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=29(61%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=29(61%)$ is due to ethyl cation $[C2H5+]$

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 9

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethyl cation $[C2H5+]$ with $mass=29$ . Therefore, peak at $m/z=29(61%)$ is observed. It is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 10

Figure 5

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=29(61%)$ is due to ethyl cation $[C2H5+]$ as shown in Figure 5.

Interpretation Introduction

(f)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=28(25%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=28(25%)$ is due to ethyl radical cation $[C2H4•+]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 11

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethyl radical cation $[C2H4•+]$ with $mass=28$ . Therefore, peak at $m/z=28(25%)$ is observed. It is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 12

Figure 6

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=28(25%)$ is due to ethyl radical cation $[C2H4•+]$ as shown below in Figure 6.

Interpretation Introduction

(g)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=27(53%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=28(25%)$ is due to ethene cation $[CH=CH+]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 13

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 14

Figure 7

Conclusion

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

Answer 6

Chapter 12, Problem 12.42AP

Interpretation Introduction

(a)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=110(98%)$ 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.

Expert Solution

Answer to Problem 12.42AP

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 1

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 2

Figure 1

The peak at $m/z=110(98%)$ is observed due to $[81Br]$ that is $[M+2]$ .

Conclusion

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

Answer 7

Chapter 12, Problem 12.42AP

Interpretation Introduction

(a)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=110(98%)$ 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 8

Expert Solution

Answer to Problem 12.42AP

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 1

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 2

Figure 1

The peak at $m/z=110(98%)$ is observed due to $[81Br]$ that is $[M+2]$ .

Answer 13

Conclusion

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

Answer 14

Interpretation Introduction

(b)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=108(100%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=108(100%)$ is observed due to $[79Br]$ that is $[M]$ the molecular ion peak.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 3

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 4

Figure 2

Conclusion

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

Answer 15

Interpretation Introduction

(b)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=108(100%)$ 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 16

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=108(100%)$ is observed due to $[79Br]$ that is $[M]$ the molecular ion peak.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 3

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 4

Figure 2

Answer 21

Conclusion

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

Answer 22

Interpretation Introduction

(c)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=81(5%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=81(5%)$ is due to bromonium cation of $[81Br]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 5

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 6

Figure 3

Conclusion

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

Answer 23

Interpretation Introduction

(c)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=81(5%)$ 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 24

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=81(5%)$ is due to bromonium cation of $[81Br]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 5

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 6

Figure 3

Answer 29

Conclusion

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

Answer 30

Interpretation Introduction

(d)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=79(5%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=79(5%)$ is due to bromonium cation of $[79Br]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 7

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 8

Figure 4

Conclusion

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

Answer 31

Interpretation Introduction

(d)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=79(5%)$ 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 32

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=79(5%)$ is due to bromonium cation of $[79Br]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 7

Answer 33

Expert Solution

Answer 34

Expert Solution

Answer 35

Expert Solution

Answer 36

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 8

Figure 4

Answer 37

Conclusion

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

Answer 38

Interpretation Introduction

(e)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=29(61%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=29(61%)$ is due to ethyl cation $[C2H5+]$

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 9

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethyl cation $[C2H5+]$ with $mass=29$ . Therefore, peak at $m/z=29(61%)$ is observed. It is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 10

Figure 5

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=29(61%)$ is due to ethyl cation $[C2H5+]$ as shown in Figure 5.

Answer 39

Interpretation Introduction

(e)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=29(61%)$ 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 40

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=29(61%)$ is due to ethyl cation $[C2H5+]$

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 9

Answer 41

Expert Solution

Answer 42

Expert Solution

Answer 43

Expert Solution

Answer 44

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethyl cation $[C2H5+]$ with $mass=29$ . Therefore, peak at $m/z=29(61%)$ is observed. It is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 10

Figure 5

Answer 45

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=29(61%)$ is due to ethyl cation $[C2H5+]$ as shown in Figure 5.

Answer 46

Interpretation Introduction

(f)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=28(25%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=28(25%)$ is due to ethyl radical cation $[C2H4•+]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 11

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethyl radical cation $[C2H4•+]$ with $mass=28$ . Therefore, peak at $m/z=28(25%)$ is observed. It is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 12

Figure 6

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=28(25%)$ is due to ethyl radical cation $[C2H4•+]$ as shown below in Figure 6.

Answer 47

Interpretation Introduction

(f)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=28(25%)$ 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 48

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=28(25%)$ is due to ethyl radical cation $[C2H4•+]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 11

Answer 49

Expert Solution

Answer 50

Expert Solution

Answer 51

Expert Solution

Answer 52

Explanation of Solution

When the ethyl bromide breaks into fragment, it releases ethyl radical cation $[C2H4•+]$ with $mass=28$ . Therefore, peak at $m/z=28(25%)$ is observed. It is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 12

Figure 6

Answer 53

Conclusion

The mass spectrum fragmentation of ethyl bromide at $m/z=28(25%)$ is due to ethyl radical cation $[C2H4•+]$ as shown below in Figure 6.

Answer 54

Interpretation Introduction

(g)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=27(53%)$ 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.

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=28(25%)$ is due to ethene cation $[CH=CH+]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 13

Explanation of Solution

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

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 14

Figure 7

Conclusion

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

Answer 55

Interpretation Introduction

(g)

Interpretation:

The mass spectrum fragmentation of ethyl bromide at $m/z=27(53%)$ 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 56

Expert Solution

Answer to Problem 12.42AP

The peak at $m/z=28(25%)$ is due to ethene cation $[CH=CH+]$ .

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 12, Problem 12.42AP , additional homework tip 13

Answer 57

Expert Solution

Answer 58

Expert Solution

Answer 59

Expert Solution

Answer 60

Explanation of Solution

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