114
Answer: |
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Reasoning:
All four candidates would give rise to three signals in their 1H-NMR spectra. The first and fourth molecule, however, can be eliminated right away based on the expectation ranges for their chemical shifts (Our spectrum contains a signal with d > 8ppm). Besides, the relative intensities of the signals in their spectra (6 : 1 : 1) diverge widely from what we see in our spectrum (3 : 3 : 2).The decision in favour of the third molecule is based on the expected values for the chemical shifts. It is important to keep in mind that for both options b) and c) the methylene groups give rise to triplets (m=3), the neighboring methyls to quartets (m=4), and the second methyl groups to singulets (m=1).
| Molecule | Equivalent protons | d [ppm] (expected) | d [ppm] (observed) | Multiplicities |
|---|---|---|---|---|
| b) CH3-CH2-CO-O-CH3 | -CH2-CH3
CH3-CH2-CO- -OCH3 |
1.9 ... 3.2 3.3 ... 4.1 |
4.1 2.0 |
4 1 |
| c) CH3-CO-O-CH2-CH3 | -CH2-CH3
-O-CH2-CH3 CH3-CO- |
3.3 ... 4.6 1.7 ...2.7 |
4.1 2.0 |
4 1 |
If you had any trouble with this problem, you might want to go back and have another look at the section on relative intensities!
Otherwise you might as well go ahead and tackle the next problem.