Mass Spectroscopy Homework
From textbook: 12.23, 12.24, 12.26
12.23 (a)
M+ = 86, means molecular weight of ketone must be 86
Formula which fits is C5H10O – there can only be 3 possiblities
H3C
H3C
O
H3C
O
CH3
O
CH3 H3C
(a)
(b)
CH3
(c)
The fragement of 71 – 86-71 = 15 – loss of a fragment of 15 – a CH3 group
- since ketones fragment by alpha cleavage – this means that 71 is a product of alpha
cleavage – where the ketone loses a CH3 group - it could be (b) or (c)
The fragment of 43 – loss of a fragment of 86-43=43 – C3H7 – so this is the product of
alpha cleavage where the ketone loses a C3H7 group – it could be (b) or (c)
Both (b) and (c) are therefore consistent with the data.
In theory, one would then look at the other peaks – e.g. (b) you would expect to see a
peak due to the McLafferty rearrangements
OH
H2C
CH3 - peak at 58. However the question does not explicitly say there is or is no
peak at 58 – so to be on the safe side (b) or (c).
12.23(b)
Alcohol with M+=88 – means molecular weight of 88. These are the 8 possiblities
H3C
H3C
O
H
CH3
A
H3C
OH
CH3
H3C
CH3
B
CH3
H3C
OH
CH3
H3C
C
D
CH3
H3C
H3C
H3C
OH
OH
CH3
H3C
OH
OH
OH
CH3
E
F
G
H
Fragment of 73 - 88-73= 15 – loss of a fragment of CH3 group – since alcohols
fragment by alpha cleavage, alcohol loses one CH3 group - where the CH3 is attached to
the alpha carbon – i.e. A, D, E, F, G can be excluded straightaway
Fragment of 70 –
88-70= 18 – loss of H2O – characteristic of alcohols
Fragment of 59 - 88-59=29 – loss of C2H5 group – again since alchols fragment by
alpha cleavage – the C2H5 group must be attached to the alpha carbon – only A and H
are possible.
Therefore H is the only alcohol consistent with the results.
12.24
H3C
+
CH3
H3C
CH3
CH3
Formula Weight
H3C
+
CH2
H3C
CH
Formula Weight
= 71.1402914
H3C
+
+
CH2
CH2
Formula Weight
= 57.1137114
= 43.0871314
Formula Weight
= 29.0605514
CH3
+
HC
CH3
Formula Weight
= 43.0871314
The M+ peak is the small one at 84.
The base peak is the one at 43.
The base peak is so because 43 – represents two possible fragments – and one of them is
quite stable because it is a formal secondary carbocation so should be more abundant.
12.26
(a)
Formula Weight
= 120.14852
Formula Weight
CH2
CH3
+
HO
O
Formula Weight
= 148.20168
Formula Weight
CH3
+
C
C
O
O
= 71.0972314
= 105.1134514
A
B (+ not shown)
C
D
A – molecular ion, B – Mclafferty rearrangement product (only one side of ketone)
C and D – alpha cleavage products
(b)
Formula Weight
= 110.19676
OH
OH
CH3
Formula Weight
A
CH3
+
C
Formula Weight
= 128.21204
B (+not shown)
= 99.1503914
C
A – molecular ion, B – dehydration product C – alpha cleavage
(c)
Formula Weight
= 99.17412
H
N
CH3
Molecular ion – if there is alpha cleavage – you will end up with 99 still.