http://i47.tinypic.com/sv3za0.gif
That is the answer but I am confused because I thought that compound would just lose a methyl group and the m/z would be 71.. what is the correct rule for fragmentation?
That is the answer but I am confused because I thought that compound would just lose a methyl group and the m/z would be 71.. what is the correct rule for fragmentation?
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The key is "base peak" meaning that it would be the peak with 100% abundance.
The compound shown (2,3-dimethylbutane):
H3C........CH3
.......\......./
.......CH-CH
......./.......\
H3C........CH3
(dots used as placeholders only)
is in essence two isopropyl groups (C3H7, m/z = 43), so fragmentation will result in a very high abundance of the ion having m/z of 43. The isopropyl radical cation is very stable and is favored over formation of other fragments, and thus it would be the base peak.
The mass spectrum does have a peak at m/z 71, but it isn't the base peak.
http://webbook.nist.gov/cgi/cbook.cgi?ID…
The compound shown (2,3-dimethylbutane):
H3C........CH3
.......\......./
.......CH-CH
......./.......\
H3C........CH3
(dots used as placeholders only)
is in essence two isopropyl groups (C3H7, m/z = 43), so fragmentation will result in a very high abundance of the ion having m/z of 43. The isopropyl radical cation is very stable and is favored over formation of other fragments, and thus it would be the base peak.
The mass spectrum does have a peak at m/z 71, but it isn't the base peak.
http://webbook.nist.gov/cgi/cbook.cgi?ID…