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Spectroscopy Problems
43
Copyright © 2017 Pearson Education, Inc.
The four major fragments have the same
m
>
z
values (57, 43, 29, and 15) as those formed by pentane, but
their relative intensities are different.
1
and 3 produce the most abundant fragments because they both form a secondary cation, and the stability
of the cation is more important than the stability of the radical in determining the most abundant fragments.
Therefore, we expect a base peak with
m
>
z
=
43
(because the secondary cation is accompanied by a
primary radical) and a less intense peak with
m
>
z
=
57
(because the secondary cation is accompanied by
a methyl radical).
Both spectra show a base peak at
m
>
z
=
43.
The major difference in the two spectra is the intensity of the
peak with
m
>
z
=
57.
The spectrum of isopentane should show a more intense peak because it is due to a
secondary cation, whereas the peak with
m
>
z
=
57
in the spectrum of pentane is due to a primary cation.
Thus,
pentane
gives the first mass spectrum and
isopentane
gives the second.
4.
The molecular ion for this compound has an
m
>
z
=
73.
The nitrogen rule states that if a molecular ion has
an odd value, then the structure must have an odd number of nitrogens. Therefore, we can eliminate the
alkane, the ketone, and the ether.
Now we can determine the molecular formula of the compound using the rule of 13. When we subtract 14
(the mass of nitrogen) from 73, we get 59.
59
13
=
4 carbons with 7 left over
Therefore, the molecular formula is
C
4
H
11
N.
Both amines given as possible structures have this formula.
To determine which of the amines is responsible for the spectrum, we can take clues from how ethers and
alcohols cleave and apply them to amines. Oxygen-containing species undergo
a
-cleavage. If nitrogen
behaved similarly, then we would expect a fragment to form by cleaving a
C
¬
C
bond alpha to the nitrogen
in each compound. Given the relative stability of this fragment, we can anticipate that it will be the base
peak of the mass spectrum.
CH
2
H
3
C
N
CH
3
-cleavage
CH
3
+
-cleavage
m/z
= 30
m/z
= 58
+
H
2
N CH
2
+
+
CH
2
CH
2
CH
3
CH
3
N
H
3
C
CH
2
CH
3
H
2
N
CH
2
CH
2
CH
2
CH
3
a
-Cleavage of
N,N
-dimethylethylamine gives a cation with
m
>
z
=
58.
a
-Cleavage of butylamine gives
a cation with
m
>
z
=
30.
The spectrum shows a base peak with
m
>
z
=
58,
indicating that the compound
that gives the spectrum is
N,N
-dimethylethylamine
.
N