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Unformatted text preview: CHEM 344 Spectroscopy Practice Problem Set I Mass Spectrometry ANSWER KEY 1) Draw the molecular ions and calculate the m/z value of the following compounds. Any of the above representations of the benzene and 1pentene molecular ion are correct. For each molecule, structure A shows the unpaired electron and the + charge delocalized within the system, while structures B and C show localized electrons and + charges. 2) Draw the major fragments you would expect from the molecular ions of the following compounds. Calculate the m/z value of the molecular ion and the fragment ions in each case. For alkanes, we can't specify the location of the unpaired electron or charge with any certainty so the bracket convention is OK. Branched alkanes fragment either side of the branching point(s) to give 2o or 3o carbocations. A carbonyl compound think cleavage either side of carbonyl group! This carbonyl compound (2pentanone) has protons attached to the carbon atom, so we also need to consider the McLafferty rearrangement. Remember that the McLafferty rearrangement produces a radical cation and a neutral molecule (an alkene). From Q1, we know that there are 2 localized structures to draw for the benzene molecular ion. From there, we can push single electrons around to produce reasonable looking fragments. The alkyl chain is also a likely site of electron removal. Again, bracket convention is OK. Loss of CH3 radical leads to a substituted benzyl species, which will rearrange to give an aromatic tropylium carbocation. There is also a McLaffertytype rearrangement product with m/z =92 see if you can work out what it might be (hint: it originates from one of the "localized" molecular ions of benzene). 3) Carbonyl compound C8H8O: IHD = [(16 + 2) 8] x 0.5 = 5 An IHD of 5 usually means a benzene ring (4) and a double bond (1). The double bond is due to the C=O group. That takes care of at least 7 carbon atoms (6 in the benzene ring and 1 for the carbonyl group). We can also assign 5 hydrogen atoms to the benzene ring. 105 77 120 43 15 The base peak is m/z=105, which is [M]15 and suggests loss of a CH3 radical. So far we know that the molecule contains a carbonyl group, and that there are phenyl and (probably) methyl groups present in the molecule We know that cleavage is typical for carbonyl compounds, so if we predict that the methyl and phenyl groups are attached either side of the carbonyl group, we would expect to see peaks at m/z = 105 and m/z= 43 due to acylium cations. And we do! The peak at m/z = 77 is typical of a phenyl cation, produced by loss of a molecule of CO from the phenylacylium cation (m/z =105). This confirms the IHD evidence for a benzene ring. Finally, a low abundance peak at m/z=15 confirms the presence of a methyl unit. This leads us to conclude that the unknown is acetophenone. 4) Pr = Propyl group, CH3CH2CH2. The oxonium cation m/z = 31 is the base peak for 1butanol since loss of a Pr radical is more favorable than loss of an H radical. The single electron can be more readily stabilized in a longer chain (e.g. by further fragmentation or rearrangement). Similarly, m/z = 45 is the base peak for 2butanol. Note that the peak m/z =59 in tertbutanol is not the same as the m/z =59 peak observed for fragmentation of 2butanol. ...
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- Spring '11