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122 |
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Answers:
Reasoning:
In this example we will try to find the correct answer by eliminating all
the false ones. To do so we will compare the spectra of all possible
isomers with the actual spectrum of our mystery molecule.
| |
Isomer |
Signals |
Groups |
d [ppm] |
Irel. |
Multiplicity |
| 1. |
CH3-CH2-CH2-CH2-OH |
5 |
|
|
|
|
| 2. |
 |
4 |
(CH3)2
>CH-
CH2
-OH |
» 1
»1.8
»3.6
up to 5.2 |
6 1 2 1 |
2 21(7*3) 4(2*2) 3 |
| 3. |
 |
5 |
|
|
|
|
| 4. |
 |
2 |
|
|
|
|
| 5. |
CH3-CH2-O-CH2-CH3 |
2 |
|
|
|
|
| 6. |
CH3-O-CH2-CH2-CH3 |
4 |
CH3 CH2
CH3-O- CH2-O- |
»1
»1.5
»3.6
»3.8 |
3 2 3 2 |
3 12(3*4) 1 3 |
| 7. |
 |
3 |
|
|
|
|
The spectra of only two of the seven possible isomers have the same number
of lines as the experimental spectrum. Picking isomer 2 of these
two options is the rather obvious solution, once you consider the
relative intensities and the respective multiplicities of the signals.
Now on to one last problem of this type!
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