17.44A carbonyl group is evident from the strong infrared absorption at 1710 cm1. Since all the 1H NMRsignals are singlets, there are no nonequivalent hydrogens in a vicinal or “three-bond”relationship.The three-proton signal at 2.1 ppm, and the 2-proton signal at2.3 ppm can be understood asarising from a unit. The intense 9-proton singlet at 1.0 ppm is due to the three equiva-lent methyl groups of a (CH3)3C unit. The compound is 4,4-dimethyl-2-pentanone.17.45The molecular formula of compounds A and B (C6H10O2) indicates an index of hydrogen deficiencyof 2. Because we are told the compounds are diketones, the two carbonyl groups account for all theunsaturations.The1H NMR spectrum of compound A has only two peaks, both singlets, at2.2 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 forThe simplicity of the spectrum can be understood if we are dealing with a symmetric diketone.
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