Chapter G, Problem G.1P

Interpretation Introduction

Interpretation:

Mass spectra of $\text{butylcyclopentane}$ and $\text{tert-butylcyclopentane}$ were acquired. Spectrum A exhibited significant mass peaks at m/z values of $\text{126}$, $\text{97}$, $\text{83}$, $\text{69}$, $\text{55}$, and $\text{41}$. Spectrum B exhibited significant peaks at m/z values of $\text{111}$, $\text{69}$, $\text{57}$, and $\text{41}$. 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 ${\text{M}}^{\text{+}}$ is more accurately depicted as ${\text{M}}^{\text{+ •}}$. The major difference in the two spectra are the peaks at m/z values $\text{126}$, $\text{97}$, and $\text{83}$ for spectrum A and at the m/z value of 111 for spectrum B.

Answer to Problem G.1P

The structures for $\text{butylcyclopentane}$ and $\text{tert-butylcyclopentane}$ and possible fragmentation ions are shown below:

Explanation of Solution

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

Conclusion

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