17.44 A carbonyl group is evident from the strong infrared absorption at 1710 cm 2 1 . Since all the 1 H NMR signals are singlets, there are no nonequivalent hydrogens in a vicinal or “ three-bond ” relationship. The three-proton signal at d 2.1 ppm, and the 2-proton signal at d 2.3 ppm can be understood as arising from a unit. The intense 9-proton singlet at d 1.0 ppm is due to the three equiva-lent methyl groups of a (CH 3 ) 3 C unit. The compound is 4,4-dimethyl-2-pentanone. 17.45 The molecular formula of compounds A and B (C 6 H 10 O 2 ) indicates an index of hydrogen de f ciency of 2. Because we are told the compounds are diketones, the two carbonyl groups account for all the unsaturations. The 1 H NMR spectrum of compound A has only two peaks, both singlets, at d 2.2 and 2.8 ppm. Their intensity ratio (6:4) is consistent with two equivalent methyl groups and two equivalent meth-ylene groups. The chemical shifts are appropriate for The simplicity of the spectrum can be understood if we are dealing with a symmetric diketone. The correct structure is
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This note was uploaded on 09/28/2011 for the course CHM 2210 taught by Professor Reynolds during the Fall '01 term at University of Florida.