MS 914 - Mass Spec Practice
Submitted by
Matt on August 6, 2011.
Textbook and Chapter: Carey and Giuliano 8th Ed. (2010) , Chapter 13
Keywords: alpha cleavage, heterolytic cleavage, homolytic cleavage, mass spec
Description: Goes over alpha-cleavage and the McLafferty rearrangement in mass spectrometry.
Problem # 657
The mass spec of chlorocyclohexane shows a peak at m/z = 83.
Use curved arrows to show the heterolytic cleavage that accounts for this fragment.
Solution Heteroatoms (atoms that are not C or H) are always the most likely atoms to lose an electron during mass spec, and this will be no exception. Mass spec will blow off an electron from the chlorine, leaving behind the molecular ion (M+ ). But the molecular ion will then fragment.
The problem told us it was a heterolytic cleavage, in which both electrons form the bond go towards the positive charge. Because two electrons are involved, we use curved arrows as usual.
This results in a neutral chlorine radical and a carbocation with formula C6 H11 , for a total mass of 83 (6 x 12 + 11 x 1). Note that the chlorine radical doesn't give an MS peak because it is neutral.
Problem Details MendelSet practice problem #
657 submitted by
Matt on July 17, 2011.
Problem # 658
The mass spec of methyl ethyl ether shows peaks at m/z = 45 and 59.
Use hooks to show the alpha cleavages that result in these two fragments.
Solution MS ionization will knock off an electron from the heteroatom (atom that's not C or H), in this case, the oxygen, leaving behind a positively charge compound. This is the molecular ion (M+ ).
Oxygen usually undergoes homolytic cleavage- the bond splits and each atom gets one electron. Since only one electorn is involed, we use hooks instead of the usual curved arrows.
If your professor is into mass spec cleavage mechanisms, I suggest you practice this mechanism, called an alpha cleavage (see image below).
Problem Details MendelSet practice problem #
658 submitted by
Matt on July 17, 2011.
Problem # 659
The mass spec of 4-nonanone shows peaks at m/z = 58, 71, 86, 99.
Using curved arrows or hooks, show how each of these fragments can form via alpha cleavage or the McLafferty rearrangement. (and draw the structure of the indicated species in the appropriate box).
Solution Note that I drew the McLafferty rearrangements using arrows (2 electrons moving at once). Some textbooks use hooks instead, but the results are the same.
In most ungraduate organic chemistry courses, being able to draw an alpha cleavage is much more important than a being able to draw a McLafferty rearrangement (which tends to only show up on bonus problems).
Problem Details MendelSet practice problem #
659 submitted by
Matt on July 17, 2011.
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