Chapter 10, Problem 51P

In the mass spectrum of the following compounds, which would be more intense: the peak at m/z = 57 or the peak at m/z = 71?

1. a. 3-methylpentane
2. b. 2-methylpentane

Interpretation Introduction

Interpretation

The tallest peak should be identified.

Explanation of Solution

Mass fragmentation of 3-methyl pentane: In this molecule will be more adopted to lose an ethyl radical forming a secondary carbocation and a primary radical than a methyl radical forming secondary carbocation and methyl radical.  In addition, 3-methylpentane ha undergoes for two path ways to lose an ethyl radical,

Therefore, the peak at $\text{m/z=57}$ would be intense than the peak at $\text{m/z=57}$ , the corresponding fragmentations has shown below.

Mass fragmentation of 2-methyl pentane: The 2-methylpentane has undergoes for pathways to lose a methyl radical in each pathway and it cannot form a secondary carbocation by losing an ethyl radical.  Loss of an ethyl radical would form a primary carbocation and a primary radical. Therefore, it will be more adopt to lose a methyl radical than an ethyl radical, the corresponding peak at $\text{m/z=71}$ would be more intense than the peak at $\text{m/z=57}\text{.}$

                    Figure 1

Conclusion

The 3-methyl pentane molecular peak at the m/e=71 has more intense then the peak at $m/z=57.$