nmr+instructions

Therefore this compound has a carbonyl somewhere also

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Unformatted text preview: l groups in the molecule. You may not be able to get a lot of information, but you can definitely see carbonyls, aromatic rings, and the like. IR has a few distinct peaks, and 1700 is one of them. Just check the nearest IR table and you'll see 1700 is a C=O peak. Therefore this compound has a carbonyl somewhere. Also note that the carbonyl is a C=O double bond. You know that the compound only has 1 DUS, so the carbonyl must be it. There are no more rings/multiple bonds in the structure. Now you can look at the NMR. The easiest way to do this is to make it like a puzzle - every peak represents a certain piece. You can tell from the integration and the shift what the peak is, and the multiplicity tells you what's nearby....
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This document was uploaded on 04/06/2014.

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