Butanone o 70 2063 275 352 problem 5 the normal

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Unformatted text preview: O 3 The first would have a peak for the methyl group in the 0- 50 region and one for the CCl3 group at a very large chemical shift because of the three chlorine atoms. The second isomer would have two peaks in the 50- 100 region, not that far apart. The second structure looks better but it would be easily confirmed by proton NMR as the first structure would have one peak only but the second would have two peaks for different CHs. The solvent is indeed the second structure 1,1,2- trichloroethane. 4 Solutions Manual to accompany Organic Chemistry 2e Two of the peaks (45.3 and 95.6) in the paint thinner are much the same as those for this compound (chemical shifts change slightly in a mixture as the two compounds dissolve each other). The other compound has a carbonyl group at 206.3 and three saturated carbon atoms, two close to the carbonyl group (larger shifts) and one further away. Butanone fits the bill perfectly. You were not expected to decide which CH2 group belongs to which molecule—that can be found out by running a spectrum of pure butanone. butanone O 7.0 206.3 27.5 35.2 PROBLEM 5 The ‘normal’ O–H stretch in the infrared (i.e. without hydrogen bonding) comes at about 3600 cm–1. What is the reduced mass (µ) for O–H? What happens to the reduced mass when you double the mass of each atom in turn, i.e. what is µ for O–D and what is µ for S–H? In fact, both O–D and S–H stretches come at about 2,500 cm –1. Why? Purpose of the problem To get...
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