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H NMR and C NMR

In this problem the tertiary alcohol below was subjected to dehydration with phosphorous oxychloride. The potential products are shown below. The product was isolated and purified by distillation. An NMR of the major product exhibited two triplets only, one centered at 2.1δ and the other at 0.9δ . Indicate the product that seems to have been produced. Explain the spectrum.

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Response to Question 8:

In compound one, we have many different kinds of hydrogen atoms. Why is that? Practically every carbon that hydrogens are attached to are different because of the position of the -OH group. That one functional group makes the entire molecule totally unsymmetrical. As a result of this asymmetry, we would expect to see all kinds of peaks for all the different hydrogens.

In compound 2, again we do not have any symmetry. Not only that but we would clearly see that the lone hydrogen off the C=C would be expected to have a chemical shift different from all the rest of hydrogens attached to the sp3 hybridized carbons, right? Also, the two hydrogens on the CH2 group that is nearest that double bond (on the pentene ring) would be split into a complex pattern because they would be split by the single hydrogen on the double bond (into a doublet) and then split again by the two hydrogens on the nearest CH2 group into a triplet, giving rise to a complicated splitting pattern. Again, because of that double bond, the carbons ...

Solution Summary

The solution explains the spectrum and the resultant products of a dehydration of alcohol.

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