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Concept explainers
(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
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Answer to Problem 12.42AP
The peak at m/z=110(98%) is observed due to bromide isotope [81Br] that is [M+2].
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
Figure 1
The peak at m/z=110(98%) is observed due to [81Br] that is [M+2].
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.
(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.
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Answer to Problem 12.42AP
The peak at m/z=108(100%) is observed due to [79Br] that is [M] the molecular ion peak.
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.
Figure 2
The mass spectrum fragmentation of ethyl bromide at m/z=108(100%) is due to [79Br] molecular ion peak as shown in Figure 2.
(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.
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Answer to Problem 12.42AP
The peak at m/z=81(5%) is due to bromonium cation of [81Br].
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.
Figure 3
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.
(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.
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Answer to Problem 12.42AP
The peak at m/z=79(5%) is due to bromonium cation of [79Br].
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.
Figure 4
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.
(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.
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Answer to Problem 12.42AP
The peak at m/z=29(61%) is due to ethyl cation [C2H5+]
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.
Figure 5
The mass spectrum fragmentation of ethyl bromide at m/z=29(61%) is due to ethyl cation [C2H5+] as shown in Figure 5.
(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.
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Answer to Problem 12.42AP
The peak at m/z=28(25%) is due to ethyl radical cation [C2H4•+].
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.
Figure 6
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.
(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.
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Answer to Problem 12.42AP
The peak at m/z=28(25%) is due to ethene cation [CH=CH+].
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.
Figure 7
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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Chapter 12 Solutions
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