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376 Chapter 19 CARBOXYLIC ACIDS H 3.4(d) F = = 3328 cm⁻¹) via hydroboration-oxidation; A = (c) For G, = 16 + 2 = 18; degrees of unsaturation = (18 2 14) = 2 π bonds or rings. IR: = 1742 cm -1 is C=0, very possibly an ester because of the high value and the large number of oxygens in the formula. NMR: Only three signals, with integrations of 4, 4, and 6 H. The molecule must be symmetrical, 0 with pieces such as 2 (δ = 3.7), 2 (δ = 2.4), and two more equivalent (δ = 1.7). The splitting between the upfield signals suggests that the sets of CH₂'s are so a reasonable answer is 1. CH₂Cl₂ CHO CO₂H (d) 2. H₂O H₂O CHO G CO₂H OH MgBr 1. (CH₃CH₂)₂O CO₂H 1. (CH₃CH₂)₂O (e) How about 2. Mg 2. H₂O CO₂H CH₂OH 1. (CH₃CH₂)₂O (f) (CH₃CH₂)₂O CH₂ 1. CH₂Cl₂ 2. Zn. H₂O LiAIH, (CH₃CH₂)₂O OH 1 33. (a) (alkanoyl chloride) O (b) (mixed anhydride)