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CHEM 213 Lecture Notes - Lecture 5: Hydrogen Bond, Ketone, Double Bond

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19 Sep 2016

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University of Victoria

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Double bond equivalents: ring unsaturation next week ideas. C=o stretch about same place quite easily distinguish ketone from aldehydes. Low enough so its separated from spy stretches. C=o around same place as aldekydes and ketones ~1710 two oxygens huge -o-h 3400-2400 almost covers all c-h stretches. Ester has no broad o-h feature not really carbonyl that does distinguishing. ~1690, which is below other c=o primary amides have two n-h peaks. Acids: fairly easy to spot from these two n-h stretches (symmetric/assymetric) amides can hydrogen bond, so they are a little bit broadened secondary amide replaced one hydrogen, thus, only one n-h stretch one peak n-h. May be minor contributor but it has c-o character. Thus c=o stretching frequency goes down a little bit. Doesn"t work for esters because there are two competing e ects. Pulling electrons entity away from the carbon strengthens the bond. Dont quite get to 1700 because esters start at 1735cm-1.

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Related Questions

Which of the following responses is correct regarding the mass spectra of the following compounds? The amide will have an even number for the parent ion peak while the amine will have an odd number. he amide has a degree of unsaturation equal to 2 and the amine has a degree of unsaturation equal to 1. Both of the compounds will have an odd number for the parent ion peak. Both the amide and the amine have a degree of unsaturation equal to 1. How would you distinguish between the following pair of compounds using IR spectroscopy? Choose the correct answer. Key features for the amide IR would be a C=O stretch at ~1700 cm-1. Key features for the amine would be a N-H stretch at ~ 3400 cm-1. Key features for the amide IR would be a C=O stretch at ~1700 cm -1 Key features for the amine would be two N-H stretches at ~3400 cm-1. Key features for the amide IR would be a C=O stretch at ~1700 cm-1 and a N-H stretch at ~3400 cm-1 Key features for the amine would be a N-H stretch at ~3400 cm-1. Key features for the amide IR would be a C=O stretch at ~3400 cm-1. Key features for the amine would be a N-H stretch at ~1700 cm-1. How would you distinguish between the following pair of compounds using 13C NMR spectroscopy? Choose the correct answer. The amide would show two resonances in the spectra, all below 80 ppm. The amine would show four resonances, all below 80 ppm. The amide would show two resonances in the spectra, one above 200 ppm (the carbonyl C) and one below 80 ppm (the CH3 carbons). The amine would show four resonances, all below 80 ppm. The amide would show two resonances in the spectra, one above 200 ppm (the carbonyl C) and one below 80 ppm (the CH3 carbons). The amine would show two resonances, both below 80 ppm. The amide would show a resonance between 160 and 175 ppm (the carbonyl C) and the rest below 80 ppm (the CH3 carbons). The amine would show two resonances, both below 80 ppm. How would you distinguish between the following pair of compounds using 1H NMR spectroscopy? Choose the correct answer. The amide would show two resonances, both of which would be singlets. The amine would show three resonances, all of which would be singlets. The amide would show two resonances, both below 3 ppm. The amine would show three resonances, one ranging from ~1-5 ppm for the N-H hydrogen, and two between 1 and 3 for the CH2 and CH3 hydrogens. The amide would show two resonances, both between 9 and 10 ppm. The amine would show three resonances, one ranging from ~1-5 ppm for the N-H hydrogen, and two between 1 and 3 for the CH2 and CH3 hydrogens. The amide would show two resonances one between 9 and 10 ppm for the aldehyde H and one between -1.5 and 3 ppm. The amine would show three resonances, one would be a singlet for the N-H hydrogen, and the CH2 and CH3 hydrogens would be a quartet and triplet, respectively. Which of the following structures is consistent with the NMR? 1H NMR