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#1 2011-03-20 07:56:41

orgopete
Administrator

2-pentanone from mass spectrum

Yahoo!Answers in chemistry wrote:

The mass spectrum of an unknown ketone showed the molecular ion peak at m/z = 86. The relative abundances of the M to the M+1 peak were in the ratio of about 100 to 5. The other main peaks appeared at m/z = 71 and 43. Suggest a possible structure for this ketone and briefly explain your reasoning.

The answer is pentan-2-one but I'm not sure how they got it.

The ratio of 13C/12/C is about 0.01. From the spectrum, the ratio of M+1/M = 5/100. Therefore, 5x(13C/12C) = 5/100. So we know the compound contains 5 carbons. That would mean we have a pentane, but its mw is only 72 (C5H12). You might also be able to determine the number of oxygen atoms from the M+2 based on the natural abundance of 18O/16O. (These can be of dubious utility as many compounds will have virtually no M, M+1, and M+2 peaks. This is more common for electron impact than chemical ionization spectra.) Since M is even (86), the additional mw must come from an oxygen. However, C5H12O has a mw of 88, so it must be two hydrogens less. The possibilities are a ring or double bond. The double bond can be between two carbons or between a carbon and oxygen.

If it were an ring, the product would be an alcohol. It would lose water (18) and other cleavages. Not present in spectra, not an alcohol.

If the double bond were a C=C double bond, then we would see allyl losses (41) or something along those lines. Not found.

If the double bond is a C=O, then we would see alpha cleavage. Look for loss of the groups attached to the C=O. If aldehyde, 86-C4H9 = 29. If 3-pentanone, symmetrical, loss of C2H5 (29), peak at 67. If 2-pentanone, then loss of CH3 or C3H7, 71 and 43. Therefore it must be 2-pentanone.

       http://www.curvedarrow.com/chem/ZZ0D2A3B56.jpg

If you receive or look up experimental (electron impact) MS, you will often find the parent peak missing or very small. That can make identifying an unknown compound very difficult. That is also a reason that MS had been of limited use as an experimental tool. With the development of chemical ionization tools (and other developments), fragmentation is reduced and the utility of MS has grown exponentially. If you are in a lab and were trying to identify the compound, an NMR spectrum would be of much greater value combined with the MW, which could be obtain from a chemical ionization MS.

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