Part 1 presented a challenge to determine an experiment to distinguish two very similar products from each other, namely 3-methyl-5-(pyridin-2-yloxy)pyridine and 5′-methyl-2H-1,3′-bipyridin-2-one. The products have identical formula weights and the LC/MS and 1H NMR are too similar to draw any conclusion from.
The first step is to determine what is different between the two products and then identify an experiment specifically designed to focus on that difference. The obvious difference between the two products is the position of the oxygen atom—an ester group verse a carbonyl group. An FT-IR experiment, as commented by the reader Felipe A., can be used to sort out the products.
Other experiments can include the use of reducing agents, 15N NMR, 1H -13C HMBC, 1D NOE, 1H-1H TOCSY, MS2, etc. Note free water, acids and sample concentration can inhibit the use of some of these experiments.
A 13C NMR experiment may appear to be another good choice when trying to identify a carbonyl group. However, the carbonyl is part of a conjugated system and so the 13C chemical shift is expected around 160 ppm, which also happens to be expected for the 13C chemical shift of the O-C=N group on the other product.