1)
We can, in fact, write four processes, where one produces an atom, another produces a radical and the others both produce molecules. These can be represented as follows:
C2H<r
H++C2Hj       C2HJ- + H2
2)
CH^OH+>
m/z 32
CH2OH+ + H*
m/z 31
CH3+ + OH* m/z 15
CHO+ + H2
m/z 29
(a) m/z 46 (molecular ion)
(b) m/z 31 (base peak)
(c) CH3CH2OFT"--> CH.OH' + CH^
Tlie base peak has the same formula as that in the methanol spectrum. It is formed by C — C bond cleavage in the case of ethanol.
(d) In methanol, m/z 29 was formulated as CHCK which arose from the fragmentation:
CH2OH~--> CHO + H:
This may also happen with ethanol, but the m/z 29 ion may also be formulated as QH^, which may be formed from the fragmentation:
QHgOH •--> QH^ + OH-
This second process is analogous to:
CH.OH---> CH^ + OH"
in methanol.
(e) The steps are the loss of a hydrogen atom from the molecular ion (m/z 46) to give m/z 45, followed by further loss of a hydrogen molecule from m/z 45 to give m/z 43.
This can be represented as
CH3CH2OH+»--> OTCHOFT + H*
CHjCCV + H2
The fragmentations exactly parallel those of methanol.
(f) m/z 18 is H20*, i.e.
C2H5OH+*--> H2Ol'+ C2H4
(g) The fragmentation pattern is as follows:
C2HsOH+-
CH2OH+ + CH^ C2HJ + OH'
H20+-+C2H4 CHO+ + H2 CH3CHOH+ + H'
CH3CO+ + H2