solns21b-ps8-Mar11

solns21b-ps8-Mar11 - Problem Set #8 Answer Key (Winter...

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Problem Set #8 Answer Key (Winter 20 11 ) Problem 1 Recall the strategies offered up in the ps# 5 solutions for analyzing the IR spectra. The strategy for these problems is almost the same, except now, on most of them, we have an H-NMR spectrum to help narrow down our final structure. H-NMR spectra give us information about the total number of hydrogens and the neighboring environment that each hydrogen is in. In ps#5 we had a problem where we had to choose between an α or β chlorine; as we saw, that can be a difficult choice at times with just an IR spectrum to look at, but if we had been given the H-NMR spectrum as well, the choice would have been obvious. H- NMR also gives us some functional group identification (mostly alkane groups), but it generally isn’t as clear-cut as in the IR spectrum. Additionally, we’re given UV absorption and mass spec data to help out; we'll discuss strategy on these data as they come up. a) From the IR spectrum we have a carbonyl group and a typical alkane stretching type peak (3000-2850), and from NMR we have a classic ethyl group pattern (triplet ~1 and quadruplet ~ 2.4, 3:2 ratio, leaning toward each other due to geometric proximity). Since there are no other peaks in the H-NMR, we know that all of the hydrogens in the molecule are found in ethyl groups. The MW=86, and there is really only one way to combine a carbonyl with ethyl groups to get to 86. diethylketone Double bonds absorb in the UV region, and a weak absorption around 280 nm is typical of a carbonyl. In a mass spec experiment, think of the molecule being shattered into smaller fragments. You can easily envision this molecule shattering into an ethyl fragment (29 amu) and also an ethyl-carbonyl fragment (57 amu). b) This molecule has one degree of unsaturation. IR shows a carbonyl peak and alkane stretching peak. With two oxygens we should also see a C-O peak and think in terms of a possible O=C–O group. With no carboxylic acid peak, we should think of a possible ester. NMR is curious and will take some thought. The molecule has only two kinds of hydrogens at a 3:1 ratio. For molecules with small molecular weights, the H-NMR generally gives an exact count of the hydrogens.
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solns21b-ps8-Mar11 - Problem Set #8 Answer Key (Winter...

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