EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
2nd Edition
ISBN: 9780393630817
Author: KARTY
Publisher: W.W.NORTON+CO. (CC)
Question
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Chapter G, Problem G.1P
Interpretation Introduction

Interpretation:

Mass spectra of butylcyclopentane and tert-butylcyclopentane were acquired. Spectrum A exhibited significant mass peaks at m/z values of 126, 97, 83, 69, 55, and 41. Spectrum B exhibited significant peaks at m/z values of 111, 69, 57, and 41. Each spectrum with its compound is to be matched.

Concept introduction:

The ionization process causes the loss of a single electron from a closed-shell, uncharged molecule, so the molecular ion must have an unpaired electron, making it a free radical. More specifically, it also carries a positive charge, so the molecular ion is a radical cation, and M+ is more accurately depicted as M+ •. The major difference in the two spectra are the peaks at m/z values 126, 97, and 83 for spectrum A and at the m/z value of 111 for spectrum B.

Expert Solution & Answer
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Answer to Problem G.1P

The structures for butylcyclopentane and tert-butylcyclopentane and possible fragmentation ions are shown below:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter G, Problem G.1P

Explanation of Solution

The peak in Spectrum A at m/z = 97 is from the loss of an ethyl radical fragment, and this is only possible for butylcyclopentane. The peak in Spectrum B at m/z = 111 is from the loss of methyl radical, which is more likely to occur in tert-butylcyclopentane owing to the stability of the tertiary cation. Therefore, Spectrum A is from butylcyclopentane and Spectrum B is from tert-butylcyclopentane. The M+ peak for B is absent, the 3o carbocation that is produced upon fragmentation is relatively a stable cation.

Conclusion

Spectrum A is for butylcyclopentane and Spectrum B is for tert-butylcyclopentane. Each spectrum is matched with its compound on the basis of stability of possible fragmentation ions.

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