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Practice Problem 19.11
Outlined below is a synthesis of a compound used in perfumes, called lily
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Chapter 19 Solutions
Organic Chemistry
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- : Treatment of (CHa)CHCH(OH)CH,CH3 with TSOH affords two products (M and N) with molecular formula CgH12. The 'H NMR spectra of M and N are given below. Propose structures for M and N and draw a mechanism to explain their formation. 1H NMR of M 3H 1H NMR of N 3H 3H 3 H 1H 3 H 2 H 2H 2H 8 7 6 4 1 0 9 8. 2 1 ppm ppm 4.arrow_forwardThree compounds, A, B, and C, have the same molecular formula, C5H6. In the presence of a platinum catalyst, all three compounds absorb 3 molar equivalents of hydrogen and yield pentane. Compounds B and C give a precipitate when treated with ammoniacal silver nitrate; compound A give no reaction. Compounds A and B show an absorption maximum near 250 nm. Compound C shows no absorption maximum beyond 200 nm. Propose a structure for A, B, and C.arrow_forwardIsoerythrogenic acid, C18H26O2, is an acetylic fatty acid that turns a vivid blue on exposure to UV light. On Catalytic hydrogenation over a palladium catalyst, five molar equivalents of hydrogen are absorbed, and stearic acid, CH3(CH2)16CO2H, is produced. Ozonolysis of isoerythrogenic acid yields the following products: formaldehyde, CH2O, malonic acid, HO2CCH2CO2H, adipic acid, HO2C(CH2)4CO2H, and the aldehyde carboxylic acid, OHC(CH2)6CO2H. Provide a structure for isoerythrogenic acid.arrow_forward
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