[interminable]

I've been reading the section of my textbook on Mass Spectrometry, more specifically amines. In one example, it's mentioned that for diethylamine, the most intense peak is at m/z = 58, due to the loss of a methyl group. I've figured that this occurs by alpha-cleavage. Then the textbook states that this fragment can undergo further fragmentation to produce an m/z = 30 peak. The same idea also occurs with triethylamine.

I can't seem to figure out how the fragmentation occurs to result in a m/z = 30...I've been moving electrons around and breaking bonds, but the closest I can get is 29. I also figure that since this fragment is even, it should have the one nitrogen in it, with the nitrogen rule being backwards for fragments.

I found this website: http://www.chemistry.ccsu.edu/glagovich/teaching/316/ms/amine.html
which refers to beta-hydrogen transfer, but there's one carbon I can't account for in their scheme leading to the m/z 30 peak. Assuming this is indeed what happens, does this just occur twice for triethylamine?

Kind of a related problem for diisopropylamine - I can't account for the peak at m/z = 44. If this idea of beta-hydrogen transfer is correct, then I can account for it, but I can't find explanation of this type of transfer in my textbook, or on any other website.

[KritikalMass]

I can't seem to figure out how the fragmentation occurs to result in a m/z = 30...

I found this website: http://www.chemistry.ccsu.edu/glagovich/teaching/316/ms/amine.html

Going from 58 to 30, and judging from that website, I'd say 2 carbons come off along with 4 hydrogens. Think the process is the exact same whether it is diethylamine or dipropylamine.

Will have to think about the diisopropylamine for a few.