June 29, 2012
by Arvin Moser, Team Manager, Application Scientists, ACD/Labs
A subtle clue can guide an elucidator to choose one scenario over another.
Given the 13C NMR spectrum below, it is apparent that either the C at 147 ppm or the CH3 at 29 ppm is an impurity based on the limit of the carbon count. Since both carbons lack any 1H-13C HMBC correlations, a good starting point is to choose the CH3 as the impurity and the quaternary C as belonging to the compound of interest. The reason for this is that it is reasonable for a quaternary C to lack any HMBC correlations while this is less common for a CH3 group.
We cannot make a decisive ruling that either C atom is an impurity. It is important to note that each scenario should be fully evaluated to ensure no possibility is overlooked.
Thank you TLP for your comment.
