Solution Found!
The mass spectrum of 2-bromopentane shows many fragments.
Chapter 15, Problem 15.5(choose chapter or problem)
The mass spectrum of 2-bromopentane shows many fragments. (a) One fragment appears at M79. Would you expect a signal at M77 that is equal in height to the M79 peak? Explain. (b) A fragment appears at M15. Would you expect a signal at M13 that is equal in height to the M15 peak? Explain. (c) One fragment appears at M29. Would you expect a signal at M27 that is equal in height to the M29 peak? Explain.
Questions & Answers
QUESTION:
The mass spectrum of 2-bromopentane shows many fragments. (a) One fragment appears at M79. Would you expect a signal at M77 that is equal in height to the M79 peak? Explain. (b) A fragment appears at M15. Would you expect a signal at M13 that is equal in height to the M15 peak? Explain. (c) One fragment appears at M29. Would you expect a signal at M27 that is equal in height to the M29 peak? Explain.
ANSWER:Step 1 of 3
The equal mix of the two naturally occurring isomers of bromine \(\left({ }^{78} \mathrm{Br}\right.\) and \(\left.{ }^{81} \mathrm{Br}\right)\) that gives rise to two equal rise peaks: separated by two \(\mathrm{m} / \mathrm{z}\) units.
(a)
In this fragment, the bromine atom has already been lost, and so we would not expect a signal at \(\mathrm{M}-77\) that is equal in height to the \(\mathrm{M}-79\) peak.