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Reasoning:Starting with the composition C3H5Cl3, 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 HM be split due to coupling with HA (nA = 1, and therefor MM = 2) into a doublet. The same holds true for HX.
Finally, under these circumstances the proton HA will couple to both HM and HX (nX = 1 and nM = 3, which leads to MA = 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.