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I went through achiever and it say there are two singlets in this nmr.
In case you didn't know, there are two methyl group para to each other on a benzene.
I am guessing since the H of the aromatic is equivalent*all four of them, there is no splitting because there is no nonequivalent. Nonequivalent only split. Is this correct?
In case you didn't know, there are two methyl group para to each other on a benzene.
I am guessing since the H of the aromatic is equivalent*all four of them, there is no splitting because there is no nonequivalent. Nonequivalent only split. Is this correct?