1559r.eh16.291-30311/3/0510:02Page299 Soutionsto Problems29 CH.C CH,O CH;O 41 Cls.CHC. Mg.THF HCo 42.The synthesis of URB597 was extremely challenging.more so for the positions of its substituents in meta pos e to aromatic substitution reactions performed on biphenyl cannot serve as an entry to the preparation of this important compound. 43.(a)NMR shows two sets of benzene hydrog ns in a 2:3 integration ratio,and the IR bands at 685 and 735 cmindicate a monosubstituted benzene.This analysis leads to the only logical answer, bromobenzene,CsHsBr. -Br.which is made from benzene +Br2 and FeBrs. (b)NMR shows three benzene hydrogens.H triplet andaH doublet what you would expect fo ydrogens inc poso nd oa Ch CH 蓝故男。 ount?You to six carbons Putting the s the data NH2 plicated (d)Similar,but the NMR(two doublets,each 2 H)and the IR (820 cm)indicate that this is the nAume any orhopra mixtures can be readily separated to give the para productSolutions to Problems • 299 41. 42. The synthesis of URB597 was extremely challenging, more so for the positions of its substituents than for their structure. Both groups are located in meta positions relative to the bond between the rings. Problem 36 revealed that phenyl rings as substituents are ortho, para directing. Therefore electrophilic aromatic substitution reactions performed on biphenyl cannot serve as an entry to the preparation of this important compound. 43. (a) NMR shows two sets of benzene hydrogens in a 2 : 3 integration ratio, and the IR bands at 685 and 735 cm1 indicate a monosubstituted benzene. This analysis leads to the only logical answer, bromobenzene, C6H5Br, which is made from benzene Br2 and FeBr3. (b) NMR shows three benzene hydrogens, a 1 H triplet and a 2 H doublet; what you would expect for three hydrogens in adjacent positions around a ring, with the end ones equivalent: CHOCHOCH. What else is there? A 2 H lump at
 4.5 ppm together with two bands in the IR at 3382 and 3478 cm1 strongly suggest an NH2 group. How about an atom count? You’re up to six carbons for the benzene ring, five H and an N. The only formula that will fit is C6H5Br2N. Putting the NH2 on the benzene ring and then attaching the two Br’s one each side of it fits the data: 2,6-dibromobenzenamine is the answer. (c) The NH2 is again present, but the NMR now shows four benzene hydrogens in a complicated pattern. The IR helps: The band at 745 cm1 is in the range for ortho disubstituted rings. 2-Bromobenzenamine is the answer: C6H6BrN. (d) Similar, but the NMR (two doublets, each 2 H) and the IR (820 cm1 ) indicate that this is the para isomer of C: 4-bromobenzenamine: Synthesis. (Assume any ortho para mixtures can be readily separated to give the para product in good yield.) Br NH2 Br NH2 Br Br H NH2 H H Br, CH3O Cl CH3O CH3O MgCl anisyl alcohol Cl2, CHCl3, 0° Mg, THF 1. HCHO 2. H, H2O 1559T_ch16_291-303 11/3/05 10:02 Page 299