Experiment Reaction: An Aldol Reaction: trans-4-Nitrochalcone
Consider the product IR obtained in the lab and pictured below. Note that the sample is showing an unexpected band of OH (where? why?) but the relevant region is clear and accurate.
Identify and assign the carbonyl absorption band in your product spectrum: [ Select ] ["3361", "2360", "1656", "1607", "1513", "1333"] cm-1 [ Select ] ["stretch", "bend"] .
The acetophenone, which is the starting material, has an IR band assigned to its [ Select ] ["hydroxide", "carbonyl", "aromatic ring"] at 1681 cm-1. How this compare with the product carbonyl band that you found in this IR above? The carbonyl in your product is [ Select ] ["the same in cm-1", "at least 10 cm-1 higher", "at least 10 cm-1 lower"] from that of the acetophenone.
The reason for the difference in frequency value for the carbonyl, is that [ Select ] ["the double bond", "the OH", "the aldehyde"] creates an additional [ Select ] ["enolization", "conjugation", "localization"] which [ Select ] ["increases", "lowers"] the frequency of the C=O band.
The alpha, beta – unsaturated system in the final product results from the elimination of [ Select ] ["water", "nitro", "HCl"] in the last step of the reaction.