instructor_supplement_25

instructor_supplement_25 - Instructor Supplemental...

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Instructor Supplemental Solutions to Problems © 2010 Roberts and Company Publishers Chapter 25 The Chemistry of the Aromatic Heterocycles Solutions to In-Text Problems 25.1 (b) 25.2 (b) 2-Bromo-4-nitropyrrole (d) 8-Methoxyquinoline 25.4 (c) Because the dipole vector of pyrrole is directed away from the nitrogen, and because the resultant of the two carbon-chlorine bond dipoles is also directed away from the nitrogen, the dipole moment of 3,4-dichloropyrrole should be greater than that of pyrrole: 25.6 (b) Imidazole has one nitrogen that has the electronic character of the nitrogen in pyrrole, and one that is like the nitrogen in pyridine, except that it is somewhat more basic than the nitrogen of pyridine (Study Problem 25.1 on text p. 1225). Because of its basicity, imidazole can accept hydrogen bonds from water and, like pyridine, it is very soluble in water. 25.7 (b) Because the pyridine nitrogen is the more basic nitrogen, it is also the more nucleophilic nitrogen. Consequently, alkylation occurs on this nitrogen. 25.9 (b) This is a Friedel–Crafts acylation; both the methyl group and the ring direct substitution to the open “para” position (5-position).
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INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 25 2 (d) This is essentially a “benzylic” bromination: a free-radical bromination at a carbon adjacent to an aromatic ring (Sec. 17.2, text p. 793). 25.12 Substitution at the 4-position gives a carbocation that is stabilized by resonance interaction with an unshared electron pair on the oxygen. Substitution at the 3-position gives a carbocation that is not stabilized by such an interaction. 25.14 4-Bromopyridine undergoes nucleophilic aromatic substitution by phenolate ion. The anionic intermediate is stabilized because negative charge is delocalized onto the nitrogen, as shown in Eq. 25.42b on text p. 1238 (with Y = Br, : Nuc = OPh). In the analogous substitution reaction of 3-bromopyridine, negative charge cannot be delocalized onto the nitrogen in the anionic intermediate; consequently, the intermediate is less stable and the reaction doesn’t occur. 25.15 (b) 25.16 (b) In this case, the pyridine ring is nitrated on the 3-position by direct nitration. 25.17 (b) The bromine at the 4-position is displaced by ammonia for the reasons discussed in the solution to Problem 25.14 in this supplement.
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3 25.18 (b) Begin with the imine product of Eq. 25.55, text p. 1243, formed from pyridoxal phosphate and the amino group of serine. (We abbreviate the structure of pyridoxal phosphate as in the text.) The first step is a b -elimination that produces formaldehyde and a resonance-stabilized “carbanion” in which the charge is essentially neutralized by delocalization of electrons onto the nitrogen of the pyridinium ring. (Be sure to show this!) This “carbanion” is then protonated to give an imine, which, in turn, hydrolyzes to glycine and pyridoxal phosphate. 25.20
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instructor_supplement_25 - Instructor Supplemental...

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