Science & Fun | Home | Introduction into the ^{1}H NMR Spectroscopy | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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Lösung:
Reasoning:Starting with the composition C_{3}H_{5}Cl_{3}, we can suggest the following isomers:
Spectrum (A):A spectrum containing two doublets and one doublet of quartets (8 individual lines) can only be generated by a molecule of the following structure:
since only in this case will the signal of H^{M} be split due to coupling with H^{A} (n_{A} = 1, and therefor M_{M} = 2) into a doublet. The same holds true for H^{X}. Finally, under these circumstances the proton H^{A} will couple to both H^{M} and H^{X} (n_{X} = 1 and n_{M} = 3, which leads to M_{A} = 4 · 2 = 8), splitting the signal into a doublet of quartets. From these considerations follows that spectrum (A) has to belong to isomer 2. Spectrum (B):The fact that we observe only two singulet signals in this spectrum indicates that there is no coupling between the two groups of equivalent protons. This limits the possible structures to the type
In our case that would be the isomer 4. Now continue to the next problem. |