Análise Espectroscópica de Compostos Orgânicos

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32 Organic Chemistry Solutions Manual The diol has no C=O group in the ¹³C NMR or infrared and has a molecular ion in the mass spectrum at two mass units higher than the other two products. Distinguishing them is more tricky. The hydroxy-ketone has a conjugated carbonyl group (about 1680 cm⁻¹ in the infrared) but the hydroxy-aldehyde is not conjugated (about 1730 cm⁻¹ in the infrared). The chemical shift of the C-OH carbon atoms in the 100-150 p.p.m. region will also be different because the benzene ring is D. H. Williams and I. Fleming (1995). Spectroscopic methods in next to this atom in the hydroxy-ketone. Calculations from tables in Williams and Fleming suggest organic chemistry (5th edn). about 80 p.p.m. for the hydroxy-ketone and about 60 p.p.m. for the hydroxy-aldehyde. The mass McGraw-Hill, London. spectra will also be different - simple α-cleavage gives quite different fragments. 0 Problem 5 The triketone shown here is called 'ninhydrin' and is used for the detection of amino acids. It exists in aqueous solution as a monohydrate. Which of the three ketones is hydrated and why? 0 Purpose of the problem To start you thinking about why some carbonyl groups are more stable than others. 0 Suggested solution oH The two ketones next to the benzene ring are conjugated with it and thereby stabilized though they OH are also destabilized by the middle carbonyl group - two electron-withdrawing groups next to each other is a bad thing. The central carbonyl group has no stabilization from the benzene ring and a 0 double dose of destabilization from its neighbours. ninhydrin hydrate Problem 6 0 This hydroxy-ketone shows no peaks in its infrared spectrum between 1600 and 1800 cm⁻¹ but it does show a broad absorption at 3000 to 3400 cm⁻¹. In the NMR spectrum, there Ho are no peaks above 150 p.p.m. but there is a peak at 110 p.p.m. Suggest an explanation. Purpose of the problem Revision of Chapter 3 with a reaction from this chapter. Suggested solution The evidence shows that there is no carbonyl group in this molecule but that there is an OH group. The peak at 110 p.p.m. looks at first sight like an alkene but that is not possible (try to draw any alkene structures and you will see why) so it must be an unusual saturated carbon atom (perhaps one with two oxygen atoms). You might also argue that an alcohol and a ketone could react to give a hemiacetal, and that, of course, is what it is. The compound exists as the stable cyclic hemiacetal - stable because of the ring size. 110 p.p.m 0 OH OH H