Chapter_13_Solutions-1

Chapter_13_Solutions-1 - P1: OXT/SRB P2: xxx Printer: Bind...

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Unformatted text preview: P1: OXT/SRB P2: xxx Printer: Bind Rite JWCL234-13 JWCL234-Solomons-v1 December 15, 2009 17:7 CONFIRMING PAGES 13 CONJUGATED UNSATURATED SYSTEMS SOLUTIONS TO PROBLEMS 13.1 The following two allylic radicals are possible, differing only because of the isotopic label. Together they allow for four constitutional isomers with respect to the 13 C label. (In the absence of the isotopic label, only one constitutional isomer (as a racemic mixture) would be possible.) Br H22841 H22841 H22841 H22841 H22841 H22841 + Br Br ( H22841 = 13 C labeled position) + Br 13.2 (a) + + (b) because it represents a 2 carbocation. + (c) Cl Cl and (racemic) 13.3 (a) (b) + + + (c) 278 P1: OXT/SRB P2: xxx Printer: Bind Rite JWCL234-13 JWCL234-Solomons-v1 December 15, 2009 17:7 CONFIRMING PAGES CONJUGATED UNSATURATED SYSTEMS 279 + + + (d) O H O H (e) + + + O H Br Br- + (f ) (g) + + + + + ( j) (h) (i) O O- (minor)- S + S + O- O N + O- O- 2 + N O N + O- 13.4 (a) + because the positive charge is on a tertiary carbon atom rather than a primary one (rule 8). (b) + because the positive charge is on a secondary carbon atom rather than a primary one (rule 8). (c) + N because all atoms have a complete octet (rule 8b), and there are more covalent bonds (rule 8a). (d) O OH because it has no charge separation (rule 8c). P1: OXT/SRB P2: xxx Printer: Bind Rite JWCL234-13 JWCL234-Solomons-v1 December 15, 2009 17:7 CONFIRMING PAGES 280 CONJUGATED UNSATURATED SYSTEMS (e) because the radical is on a secondary carbon atom rather than a primary one (rule 8). (f ) NH 2 N because it has no charge separation (rule 8c). 13.5 In resonance structures, the positions of the nuclei must remain the same for all structures (rule 2). The keto and enol forms shown differ not only in the positions of their electrons, but also in the position of one of the hydrogen atoms. In the enol form, it is attached to an oxygen atom; in the keto form, it has been moved so that it is attached to a carbon atom. 13.6 (a) (3 Z )-Penta-1,3-diene, (2 E ,4 E )-2,4-hexadiene, (2 Z ,4 E )-hexa-2,4-diene, and 1,3- cyclohexadiene are conjugated dienes. (b) 1,4-Cyclohexadiene and 1,4-pentadiene are isolated dienes. (c) Pent-1-en-4-yne (1-penten-4-yne) is an isolated enyne. 13.7 The formula, C 6H8, tells us that A and B have six hydrogen atoms less than an alkane. This unsaturation may be due to three double bonds, one triple bond and one double bond, or combinations of two double bonds and a ring, or one triple bond and a ring. Since both A and B react with 2 mol of H 2 to yield cyclohexane, they are either cyclohexyne or cyclohexa- dienes. The absorption maximum of 256 nm for A tells us that it is conjugated. Compound B, with no absorption maximum beyond 200 nm, prossesses isolated double bonds. We can rule out cyclohexyne because of ring strain caused by the requirement of linearity of the C C system. Therefore, A is 1,3-cyclohexadiene; B is 1,4-cyclohexadiene. A has three signals in its 13 C NMR spectrum. With its higher symmetry,C NMR spectrum....
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Chapter_13_Solutions-1 - P1: OXT/SRB P2: xxx Printer: Bind...

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