CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS 235 10.11 The two double bonds of 2-methyl-1,3-butadiene are not equivalent, and so two different products of direct addition are possible, along with one conjugate addition product. H,C=CCH=CH, BrCH,CCH=CH, H,C=CCHCH, Br BrCH,C==CHCH, Br CH3 CH H3 3.4-Dibromo-3. 1.4-Dibromo.2. methyl-l-but (conjugate addition) 10.12 The molecular formula of the product, CIoHOCIO, is that of a 1: 1 Diels-Alder adduct between 2-chloro-1, 3-butadiene and benzoquinone H 2-Chloro-1.3. CIoHgCiO, 10.13"Unravel "the Diels-Alder adduct as described in the sample solution to part(a). H C≡N is prepared from Diels-Alder adduct (c) o is prepared from Dienophile 10.14 Two stereoisomeric Diels-Alder adducts are possible from the reaction of 1, 3-cyclopentadic methyl acrylate. In one stereoisomer the CO, CH3 group is syn to the HC=CH bridge, and is the endo isomer. In the other stereoisomer the CO, CH3 group is anti to the HC=CH bridge and called the exo isomer HC=CHCOCH H OCH Methyl acrylate Endo isomer(75%) Exo isomer (25%) ( Stereoisomeric forms of bicyclo[2. 2. 1 ]hept-5-ene-2-ca Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
10.11 The two double bonds of 2-methyl-1,3-butadiene are not equivalent, and so two different products of direct addition are possible, along with one conjugate addition product. 10.12 The molecular formula of the product, C10H9ClO2, is that of a 1:1 Diels–Alder adduct between 2-chloro-1,3-butadiene and benzoquinone. 10.13 “Unravel” the Diels–Alder adduct as described in the sample solution to part (a). (b) (c) 10.14 Two stereoisomeric Diels–Alder adducts are possible from the reaction of 1,3-cyclopentadiene and methyl acrylate. In one stereoisomer the CO2CH3 group is syn to the bridge, and is called the endo isomer. In the other stereoisomer the CO2CH3 group is anti to the bridge and is called the exo isomer. 1,3-Cyclopentadiene O OCH3 C H Endo isomer (75%) O OCH3 C H Exo isomer (25%) (Stereoisomeric forms of methyl bicyclo[2.2.1]hept-5-ene-2-carboxylate) Methyl acrylate H2C CHCOCH3 O HC CH HC CH O is prepared from O CH O 3 Diene CH3 Dienophile O O O Diels–Alder adduct Diene C N C N is prepared from Dienophile (cyano groups are cis) H N C H N C C C O O Benzoquinone O O H H Cl Cl 2-Chloro-1,3- C10H9ClO2 butadiene 3,4-Dibromo-3- methyl-1-butene (direct addition) 3,4-Dibromo-2- methyl-1-butene (direct addition) 2-Methyl-1,3- butadiene CCH CH2 CH3 H2C Br CH3 BrCH2CCH CH2 1,4-Dibromo-2- methyl-2-butene (conjugate addition) BrCH2C CH3 CHCH2Br Br2 Br CH3 H CCHCH2Br 2C CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS 235 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
236 CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS 10.15 An electrophile is by definition an electron-seeker: When an electrophile attacks ethylene, it inter acts with the T orbital because this is the orbital that contains electrons. The Orbital of ethylene is unocal 10.16 Analyze the reaction of two butadiene molecules by the Woodward-Hoffmann rules by examining e symmetry properties of the highest occupied molecular orbital(HOMO)of one diene and the lowest unoccupied molecular orbital (LUMO)of the other. Soop Antibonding/ LUMO This reaction is forbidden by the Woodward-Hoffmann rules. Both interactions involving the ends of the dienes need to be bonding for concerted cycloaddition to take place. Here, one is bonding and the other is antibonding 10.17 Dienes and trienes are named according to the IUPAC convention by replacing the -ane ending of the alkane with -adiene or -atriene and locating the positions of the double bonds by number. The stereoisomers are identified as E or Z according to the rules established in Chapter 5 (a) 3, 4-0ctadiene: CH CH,CH=C=CHCH2,CH3 H3C H (b) (E, E)-3,5-Octadiene CHCH (c)(Z, 2)-1, 3-Cyclooctadiene (d) (Z, Z)-1, 4-Cyclooctadiene (e)(E, E)-1, 5-Cyclooctadiene H,C HH CH3 (f)(2E, 42, 6E)-2, 4, 6-0ctatriene (g) 5-Allyl-1, 3-cyclopentadiene CH,CH=CH H,C=CH (h) trans-1, 2-Divinylcycloprop CH=CH, H, C=CCH=CCH (i 2,4-Dimethy l-1, 3-pentadiene: CH Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
10.15 An electrophile is by definition an electron-seeker. When an electrophile attacks ethylene, it interacts with the orbital because this is the orbital that contains electrons. The * orbital of ethylene is unoccupied. 10.16 Analyze the reaction of two butadiene molecules by the Woodward–Hoffmann rules by examining the symmetry properties of the highest occupied molecular orbital (HOMO) of one diene and the lowest unoccupied molecular orbital (LUMO) of the other. This reaction is forbidden by the Woodward–Hoffmann rules. Both interactions involving the ends of the dienes need to be bonding for concerted cycloaddition to take place. Here, one is bonding and the other is antibonding. 10.17 Dienes and trienes are named according to the IUPAC convention by replacing the -ane ending of the alkane with -adiene or -atriene and locating the positions of the double bonds by number. The stereoisomers are identified as E or Z according to the rules established in Chapter 5. (a) 3,4-Octadiene: (b) (E,E)-3,5-Octadiene: (c) (Z,Z)-1,3-Cyclooctadiene: (d) (Z,Z)-1,4-Cyclooctadiene: (e) (E,E)-1,5-Cyclooctadiene: ( f ) (2E,4Z,6E)-2,4,6-Octatriene: (g) 5-Allyl-1,3-cyclopentadiene: (h) trans-1,2-Divinylcyclopropane: (i) 2,4-Dimethyl-1,3-pentadiene: H2C CCH3 CH3 CCH CH3 H H2C CH H CH CH2 H CH2CH CH2 H H H H H H3C CH3 H C C H CH3CH2 C H C H CH2CH3 H CH3CH2CH CHCH C 2CH2CH3 HOMO LUMO Antibonding Bonding 236 CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS 237 10.18 (a) H,C=CH(CH,)CH=CH2 (b)(CH ),C=CC=C(CH,)2 2,3, 4.5-Tetramethyl-2, 4-hexadiene (c)CH2=CH-CH—CH=CH CHECH CH3 3-lsopropenyl-1, 4-cyclohexadiene HH (1Z3E.5Z)-1. 6-Dichloro-1 3,5-hexatriene ()HC=C=CHCH=CHCH3 1.2. 4-Hexatriene HC、 CHECH, CHCH CH, CH3 (E)3-Ethyl-4-methyl-1, 3-hexadiene 10.19 (a) Since the product is 2, 3-dimethylbutane we know that the carbon skeleton of the starting ma ince 2,3-dimethylbutane is CH and the starting material is C6Hio, two molecules of H must have been taken up and the starting material must have two double bonds. The starting material can only be 2, 3-dimethy l-1,3-butadiene. H2C=C—C=CH2+2H2 CH3), CHCH(CH CH, CH Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
10.18 (a) (b) (c) (d) (e) ( f ) (g) (h) 10.19 (a) Since the product is 2,3-dimethylbutane we know that the carbon skeleton of the starting material must be Since 2,3-dimethylbutane is C6H14 and the starting material is C6H10, two molecules of H2 must have been taken up and the starting material must have two double bonds. The starting material can only be 2,3-dimethyl-1,3-butadiene. H2C C CH3 C CH2 CH3 2H2 (CH3)2CHCH(CH3)2 Pt C C C C C C (E)-3-Ethyl-4-methyl-1,3-hexadiene C C CH3CH2 H3C CH2CH3 CH CH2 (1E,5E,9E)-1,5,9-Cyclododecatriene H2C CHCH C CHCH3 1,2,4-Hexatriene H H H Cl H Cl H H (1Z,3E,5Z)-1,6-Dichloro-1,3,5-hexatriene CH2 CH3 H 3-Isopropenyl-1,4-cyclohexadiene CH2 CH CH CH CH2 CH CH2 3-Vinyl-1,4-pentadiene (CH3)2C C(CH3)2 CH3 CH3 CC 2,3,4,5-Tetramethyl-2,4-hexadiene H2C CH CH(CH2 2 )5CH 1,8-Nonadiene CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS 237 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website