# Ch12 - 272"Solutions Chapter J J Ethers Epoxides and...

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"Solutions Chapter J J: Ethers, Epoxides, and Sulfides Problem 11.46 The Sharpless epoxidation is used when a single enantiomer product is required. Predict the structure of the predominant product of the follOWing transformation. " ~ . # . """ C2 .... C2 M ." " "(-)-Diethyl tartrate . l _ 9""'_ .~. Ti(O-iPr)4 "." 0 . OH ten-Butyl hydroperoxide f ,.z - OH (~~< > ~+<> Using (-)-diethyl t;artrate wi!! lead to the stereoisomer shown. Chapter 12: Infrared Spectroscopy Solutions 273 CHAPTER 12 Solutions. to the Problems £r,oblem 12. 1 Which is higher in energy? (a) Infrared radiation of 1715 cm- t or 2800 cm '? The higher the wavenumber, the higher the energy. As a result, 2800 cm-! is higher energy than 1705 cm-!. (b) Radiofrequency radiation of 300 MHz or 60 Hz? Energy is directly proportional to frequency, so 300 MHz is higher energy than 60 Hz. £roblem 12.2 Without doing the calculation, which l)1ember of each pair dD you expc;ct to occur at the higher frequency? (a) C=O or C=C stretching? The atomic weight of 0 is' slightly larger than that of C. However, the C=O bond' is much stronger than C=C and, hence, has a substantially larger force constant. Thus, C=O stretching occurs at higher frequency. (b) C=O or CoO stretching? Double bonds have higher force constants than single bonds, so the C=O bond will have a stretching frequency that occurs at a higher frequency than CoO. (e) C=C or C=O stretching? Triple bonds have higher force constants than double bonds, so the C.C bond.will have a stretching frequency that occurs at a higher frequericy than C=O. (d) C-H orC-C1 stretching? Assuming that C-H and C-CI have similar force constants, then the C-H will have an absorbance at a higher frequency because the atoJTIic weight of H is much smaller than CI. Problem l2.3 A compound shows strong, very broad IR absorption in the region 3300-3600 cm- t and strong, sharp absorption at 1715 cm:'. What functional group accounts for both of these abs.orptions? The very broad IR absorption centered at ;3300-3600 em'! corresponds to O-H stretching and the strong absorption at 1715 em" corresponds to a carbonyl C=O stretch. The functional group responsible for these absorptions is the -COOH group. . Problem 12.4- Propanoic acid and methyl ethanoate are constitutional isomers. Show how to distinguish between them by lR spectroscopy. . Propanoic acid Methyl ethanQate (Methyl acetate) , The key difference between these constitutional iSOmers is the OH group of propanoic acid that is not present in methyl ethanoate. The -OH group will show a strong, broad O-H stretch in the IR spectrum of propanoic acid between 2500-3300 em'! that will be absent in the spectrum of methyl ethanoate. Also, the strong C=O stretching absorption will be between 1700-1725 cm'l for propanoic acid, but 1735~1800 em'! for methyl ethanoate. Both spectra will have C-H stretching absorptions near 3000 cm-! and CoO stretching absorptions near 1100 cm-!. Humphead wrasse

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Ch12 - 272"Solutions Chapter J J Ethers Epoxides and...

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