| 20-00-253P66 | | Title | End-of-Chapter Problem 66 | | Caption | Proton and carbon-13 NMR spectra associated with end-of-chapter Problem 66. | | Notes | Terpineol (C10H18O) is an optically active compound with one asymmetric carbon which is part of a six-membered ring, and which is located on the opposite side of this ring to a methyl group attached to this ring. Reaction of terpineol with one equivalent of hydrogen generates an optically inactive compound with the molecular formula C10H20O. Dehydrating this latter compound with acid, followed by ozonolysis, followed by workup under reducing conditions, produces acetone and a final product compound whose proton and carbon-13 NMR spectra are shown in the figure. Determine the structure of terpineol. | | Keywords | end-of-chapter, problem, 66, proton, carbon-13, NMR | |