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Why does the C=O for a vinyl ester have a higher frequency than the C=O for an...
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You are talking about Infrared Spectroscopy, usually abbreviated to IR. This is an analytical instrumentation technique that involves exposing a chemical sample to numerous wavelengths of light across the infrared portion of the electromagnetic spectrum and measuring how well the sample absorbs the different wavelengths of light. The computer performs a series of mathematical calculations called Fourier transform that converts the measurements into a spectrum where the percent transmittance (the opposite of absorbance) is plotted versus the frequency of light. The frequency is usually given in units called wavenumbers, or reciprocal centimeters. So the baseline is at the top of the spectrum and the peaks are bands that dip down lower in the spectrum (a lower % transmittance means a higher % absorbance).
The carbonyl peak is one of the most distinctive peaks in an IR spectrum. The carbon oxygen double bond strongly absorbs infrared radiation to give a very strong peak. The exact position of the carbonyl peak depends on the chemical environment of the carbonyl itself in the molecule. Carbonyls in ketones, aldehydes, esters, and amides all give different values for their carbonyl peaks. In addition, the chemical environment around the carbonyl also affects its position. In your example, you have two different esters, an aliphatic one and a vinyl one. The aliphatic one is surrounded by only carbon carbon single bonds. The vinyl one is adjacent to a carbon carbon double bond. The pi electrons in the double bond are in resonance with the pi electrons in the carbonyl bond, thus the total of four pi electrons are delocalized over four different atoms (3 carbons and one oxygen). This added delocalization of the pi electrons alters the IR absorbance characteristics of the carbonyl group, and that is why the vinyl ester has a slightly different frequency than the aliphatic ester.
Posted by ncchemist on October 20, 2013 at 3:53 PM (Answer #1)
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