Bruice Chapter 13 Problem #4:
The Problem states:
“Identify the hydrocarbon that has a molecular ion with an m/z value of 128, a base peak with an m/z value of 43, and significant peaks with m/z values of 57, 71 and 85.”
This is a mass spectrum fragmentation problem. The first thing to realize is that there are only hydrogen and carbon atoms in this problem that make up our molecule. With that being the case and knowing that our molecular ion (the peak with the greatest m/z value) is 128, we can start by subtracting a rational amount of carbons (assuming each carbon comes with two hydrogen atoms). From here we can determine that there are 9 carbons, since 9*12 is 108 and we can fill in with hydrogen atoms. It is also useful to refer to the formula CnH2n+2. For this problem our molecule has a formula of C9H20. Due to this hydrocarbon combination we know that there is no ring or double bonds present.
Since the base peak must have the greatest relative abundance to form a more stable carbocation we must have a base peak of 43. And to do this we know that we need 43, which is 3 carbon and 7 hydrogen atoms. This fragmentation can be seen in the first fragmentation in the image above.
To make the expected fragments at m/z 57, 85 and 71 it is understood that there would be a need for 6 carbon and 13 hydrogen atoms for the m/z 85 fragment. This is the step is the next one provided. The m/z 57 fragment has the cleavage at the same location, however the cation is on said fragment. For the last fragment it is easy to see that there would need to be 5 carbon atoms and 11 hydrogen atoms. The m/z 71 fragment is shown at the last step of the image.
One last point of consideration. It would have been possible to get the mentioned fragments by building a 2-methyloctane molecule instead of a 2, 6-dimethylheptane molecule however doing so would have produced m/z fragments at 29 and 99 which are not mentioned in the problem.