CHAPTER 10 CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS SOLUTIONS TO TEXT PROBLEMS 10.1 As noted in the sample solution to part(a), a pair of electrons is moved from the double bond toward the positively charged carbon. (b) H,C →HC-C=CH C(CH3)2 C(CH3) 10.2 For two isomeric halides to yield the same carbocation on ionization, they must have the same carbon skeleton. They may have their leaving group at a different location, but the carbocations must become equivalent by allylic resonance. Cl CH 3-Bromo-I 3-Chloro-3. methylcyclohexane Not an allylic carbocation 4-Bromo.I 230 Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
CHAPTER 10 CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS SOLUTIONS TO TEXT PROBLEMS 10.1 As noted in the sample solution to part (a), a pair of electrons is moved from the double bond toward the positively charged carbon. (b) (c) 10.2 For two isomeric halides to yield the same carbocation on ionization, they must have the same carbon skeleton. They may have their leaving group at a different location, but the carbocations must become equivalent by allylic resonance. CH3 Br CH3 4-Bromo-1- methylcyclohexene Not an allylic carbocation CH3 CH3 Cl CH3 CH3 Br 3-Bromo-1- methylcyclohexene 3-Chloro-3- methylcyclohexene C(CH3)2 C(CH3)2 H2C CH2 CH3 C H2C CH2 CH3 C 230 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS 231 Not an allylic carbocation 5-chloro CH CH ot an 10.3 The allylic hydrogens are the ones shown in the structural formulas. I-Methylcyclohexene 2, 3, 3-Trimethyl-1-butene 1-Octene 10.4 The statement of the problem specifies that in allylic brominations using N-bromosuccinimide the active reagent is Br,. Thus, the equation for the overall reaction is H Br2 B 3-Bromocyclohexene Hydrogen The propagation steps are analogous to those of other free-radical brominations. An allylic hydrogen is removed by a bromine atom in the first step ·Br: H+H—Br Cyclohexene Bromine Hydrogen Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
10.3 The allylic hydrogens are the ones shown in the structural formulas. (b) (c) (d) 10.4 The statement of the problem specifies that in allylic brominations using N-bromosuccinimide the active reagent is Br2. Thus, the equation for the overall reaction is The propagation steps are analogous to those of other free-radical brominations. An allylic hydrogen is removed by a bromine atom in the first step. H H Cyclohexene Bromine atom H 2-Cyclohexenyl radical Hydrogen bromide Br H Br Br2 H H Cyclohexene Bromine Br H 3-Bromocyclohexene Hydrogen bromide HBr 1-Octene H H 2,3,3-Trimethyl-1-butene CH3 H H H H CH3 1-Methylcyclohexene CH3 1-Bromo-3- methylcyclohexene Not an allylic carbocation CH3 Br CH3 5-Chloro-1- methylcyclohexene Not an allylic carbocation CH3 Cl CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS 231 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
232 CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS The allylic radical formed in the first step abstracts a bromine atom from Br, in the second propa Br 2-Cyclohexenyl 3-Bromocyclohexene Bromine 10.5 Write both resonance forms of the allylic radicals produced by hydrogen atom abstraction from the CH CH CH, (CH3)3CC (CH3)3CC →(CH3)3CC H 2, 3,3-Trimethyl-l-butene Both resonance forms are equivalent, and so 2, 3, 3-trimethyl-l-butene gives a single bromide on treatment with N-bromosuccinimide(NBS) (CH3)3CC (CH3)3CC=CH2 CH CHB ethyl-l-butene Hydrogen atom abstraction from 1-octene gives a radical in which the unpaired electron is delocalized between two nonequivalent positions CH=CHCH,(CH2),CH,CH2=CHCH(CH2),CH, CH,CH=CH(CH,)4CH3 1-Octene Allylic bromination of 1-octene gives a mixture of products NBS CH2CHCH,(CH,)4CH3 CH2=CHCH(CH2)4CH3 BrCH- CH=CH(CH,)4CH3 3-Bromo-1-octene I-Bromo-2-octene(cis and trans) 10.6(b) All the double bonds in humulene are isolated, because they are separated from each other by one or more spcarbon atoms CH HC Humulene Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
232 CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS The allylic radical formed in the first step abstracts a bromine atom from Br2 in the second propagation step. 10.5 Write both resonance forms of the allylic radicals produced by hydrogen atom abstraction from the alkene. Both resonance forms are equivalent, and so 2,3,3-trimethyl-1-butene gives a single bromide on treatment with N-bromosuccinimide (NBS). Hydrogen atom abstraction from 1-octene gives a radical in which the unpaired electron is delocalized between two nonequivalent positions. Allylic bromination of 1-octene gives a mixture of products 10.6 (b) All the double bonds in humulene are isolated, because they are separated from each other by one or more sp3 carbon atoms. Humulene CH3 CH3 H3C CH3 CH2 CHCH2(CH2)4CH3 BrCH2CH CH(CH2) CH2 CHCH(CH2 4CH3 )4CH3 Br NBS 1-Octene 3-Bromo-1-octene 1-Bromo-2-octene (cis and trans) CH2 CHCH2(CH2)4CH3 1-Octene CH2 CHCH(CH2)4CH3 CH2CH CH(CH2)4CH3 NBS (CH3)3CC CH2 CH3 2,3,3-Trimethyl-1- butene (CH3)3CC CH2 CH2Br 2-(Bromomethyl)-3,3- dimethyl-1-butene (CH3)3CC CH2 CH3 2,3,3-Trimethyl-1-butene (CH3)3CC CH2 CH2 (CH3)3CC CH2 CH2 3-Bromocyclohexene Bromine atom 2-Cyclohexenyl radical Bromine H H Br Br Br Br Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS 233 (c) The C-1 and C-3 bonds of cembrene are conjugated with each other. CH (CH3)2CH CH The double bonds at C-6 and C-10 are isolated from each other and from the conjugated diene (d) The sex attractant of the dried-bean beetle has a cumulated diene system involving C-4, C-5, and C-6. This allenic system is conjugated with the C-2 double bond CH3(CH2)CH, CH=C=CHCH=CHCO, CH 10.7 The more stable the isomer, the lower its heat of combustion. The conjugated diene is the most stable and has the lowest heat of combustion the cumulated diene is the least stable and has the highest heat of combustion HC H,C=CHCH, CH=CH, H,C=C=CHCH, CH3 (E)-1, 3-Pentadiene 1. 4-Pentadiene Most stabl 3186 kJ/mol (68.9 kcal/mol) 3251 (761.6 kcal/mol) (777.1 kcal/mol) 10.8 Compare the mirror-image forms of each compound for superposability. For 2-methyl-2, 3- pentadiene 2-methyl-2, 3-pentadiene H-C H,C. CH3 H-C H C unit demonstrates that the reference structure and its mirror image are superposable he Rotation of the mirror image 180 around an axis passing through the three carbons of th H C. CH CH H CH Mirror image 2-Methyl-2, 3-pentadiene is an achiral allene Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(c) The C-1 and C-3 double bonds of cembrene are conjugated with each other. The double bonds at C-6 and C-10 are isolated from each other and from the conjugated diene system. (d) The sex attractant of the dried-bean beetle has a cumulated diene system involving C-4, C-5, and C-6. This allenic system is conjugated with the C-2 double bond. 10.7 The more stable the isomer, the lower its heat of combustion. The conjugated diene is the most stable and has the lowest heat of combustion. The cumulated diene is the least stable and has the highest heat of combustion. 10.8 Compare the mirror-image forms of each compound for superposability. For 2-methyl-2,3- pentadiene, Rotation of the mirror image 180° around an axis passing through the three carbons of the C?C?C unit demonstrates that the reference structure and its mirror image are superposable. 2-Methyl-2,3-pentadiene is an achiral allene. C Rotate 180 Mirror image C H CH3 H3C CH3 C Reoriented mirror image C C H3C H H3C CH3 C 2-methyl-2,3-pentadiene and Reference structure C C H3C H H3C CH3 C Mirror image C C H CH3 H3C CH3 C H2C CHCH2CH CH2 H2C C CHCH2CH3 (E)-1,3-Pentadiene Most stable 3186 kJ/mol (761.6 kcal/mol) 1,4-Pentadiene 3217 kJ/mol (768.9 kcal/mol) 1,2-Pentadiene Least stable 3251 kJ/mol (777.1 kcal/mol) C C H H3C H CH CH2 CH3(CH2)6CH2CH C CHCH CHCO2CH3 6 54 3 2 1 (CH3)2CH CH3 CH3 CH3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Cembrene CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS 233 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
234 CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS Comparison of the mirror-image forms of 2-chloro-2,3-pentadiene reveals that they are not superposable. 2-Chloro-2, 3-pe ne is a chiral allene -Chloro-2, 3-pentadiene CH Rotate180° HC H Reference structure Mirror image 10.9 Both starting materials undergo B-elimination to give a conjugated diene system. Two minor prod ucts result. both of which have isolated double bonds H,C=CHCH, CCH, CH X=OH 3-Methyl-S-hexen-3-ol X= Br 4-Bromo-4-methyl-1-hexene Faster Slower CH CH H,C=CHCH=CCH,CH3 H,C=CHCH,C=CHCH3 H,C=CHCH, CCHaCH3 2-Ethyl-1, 4-pentadiene (mixture of E and Z isom (mixture of E and Z isomers: major product) 10.10 The best approach is to work through this reaction mechanistically. Addition of hydrogen halides always proceeds by protonation of one of the terminal carbons of the diene system. Protonation of C-1 gives an allylic cation for which the most stable resonance form is a tertiary carbocation. Pro- tonation of C-4 would give a less stable allylic carbocation for which the most stable resonance form is a secondary carbocation H,C 尤C-CH=CH2 H-C H,C=CCH=CH, (CH3),CCH=CH, H3C thyl-13-butadiene =CH-CH, H,C Under kinetically controlled conditions the carbocation is captured at the carbon that bears the great est share of positive charge, and the product is the tertiary chloride. Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
Comparison of the mirror-image forms of 2-chloro-2,3-pentadiene reveals that they are not superposable. 2-Chloro-2,3-pentadiene is a chiral allene. 10.9 Both starting materials undergo -elimination to give a conjugated diene system. Two minor products result, both of which have isolated double bonds. 10.10 The best approach is to work through this reaction mechanistically. Addition of hydrogen halides always proceeds by protonation of one of the terminal carbons of the diene system. Protonation of C-1 gives an allylic cation for which the most stable resonance form is a tertiary carbocation. Protonation of C-4 would give a less stable allylic carbocation for which the most stable resonance form is a secondary carbocation. Under kinetically controlled conditions the carbocation is captured at the carbon that bears the greatest share of positive charge, and the product is the tertiary chloride. H2C CH2 CH3 CCH (CH3)2CCH CH2 Cl HCl C CH CH2 H3C H3C C CH CH2 H3C H3C 2-Methyl-1,3-butadiene 3-Chloro-3-methyl-1- butene (major product) X OH 3-Methyl-5-hexen-3-ol X Br 4-Bromo-4-methyl-1-hexene CHCH2CCH2CH3 X H2C CH3 H2C CHCH CCH2CH3 CH3 H2C CHCH2C CHCH3 CH3 H2C CHCH2CCH2CH3 CH2 4-Methyl-1,3-hexadiene (mixture of E and Z isomers; major product) 4-Methyl-1,4-hexadiene (mixture of E and Z isomers; minor product) 2-Ethyl-1,4-pentadiene (minor product) Faster Slower Reference structure C C H3C H Cl CH3 C C C H3C H H3C Cl C Mirror image C C H CH3 Cl CH3 C Rotate 180 Reoriented mirror image and 2-Chloro-2,3-pentadiene 234 CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
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
238 CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS (b) Write the carbon skeleton corresponding to 2, 2,6, 6-tetramethylheptane Compounds of molecular formula CuH2o have two double bonds or one triple bond. The only compounds with the proper carbon skeleton are the alkyne and the allene shown. (CH,)3CC=CCH, C(CH3)3(CH3)3CCH=C==CHC(CH3)3 10.20 The dienes that give 2, 4-dimethylpentane on catalytic hydrogenation must have the same carbon skeleton as that alkane 2. 4-Dimethyl- 2. 4-Dimethylpentane 1. 3-pentadiene 2,3-pentadiene conjugated diene isolated diene cumulated diene 10.21 The important piece of information that allows us to complete the structure properly is that the ant repellent is an allenic substance. The allenic unit cannot be incorporated into the ring, because the three carbons must be collinear. The only possible constitution is therefore CH CH HO C=CHCCH 10.22 (a) Allylic halogenation of propene with N-bromosuccinimide gives allyl bromide M-bromosuccinimid H,C=CHCH3 CCI heat H,C=CHCH, Br Allyl bromide (b) Electrophilic addition of bromine to the double bond of propene gives 1, 2-dibromopropane HC=CHCH BrChcHCh Propene 1, 2-Dibromopropane (c) 1, 3-Dibromopropane is made from allyl bromide from part(a) by free-radical addition of HC=CHCHBr des BrCH2CH2CH2BI Allyl bromid Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(b) Write the carbon skeleton corresponding to 2,2,6,6-tetramethylheptane. Compounds of molecular formula C11H20 have two double bonds or one triple bond. The only compounds with the proper carbon skeleton are the alkyne and the allene shown. 10.20 The dienes that give 2,4-dimethylpentane on catalytic hydrogenation must have the same carbon skeleton as that alkane. 10.21 The important piece of information that allows us to complete the structure properly is that the ant repellent is an allenic substance. The allenic unit cannot be incorporated into the ring, because the three carbons must be collinear. The only possible constitution is therefore 10.22 (a) Allylic halogenation of propene with N-bromosuccinimide gives allyl bromide. (b) Electrophilic addition of bromine to the double bond of propene gives 1,2-dibromopropane. (c) 1,3-Dibromopropane is made from allyl bromide from part (a) by free-radical addition of hydrogen bromide. H2C CHCH2Br Allyl bromide 1,3-Dibromopropane BrCH2CH2CH2Br HBr peroxides H2C CHCH3 Propene 1,2-Dibromopropane BrCH2CHCH3 Br2 Br H2C CHCH3 Propene Allyl bromide H2C CHCH2Br CCl4, heat N-bromosuccinimide HO CHCCH C 3 O CH3 CH3 HO CH3 or or 2,4-Dimethyl- 1,3-pentadiene conjugated diene (a) 2,4-Dimethyl- 1,4-pentadiene isolated diene 2,4-Dimethyl- 2,3-pentadiene cumulated diene (c) 2,4-Dimethylpentane (b) C H2 Pt (CH3)3CC CCH2C(CH3)3 (CH3)3CCH CHC(CH3) C 3 2,2,6,6-Tetramethyl-3-heptyne 2,2,6,6-Tetramethyl-3,4-heptadiene 238 CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS 239 (d) Addition of hydrogen chloride to allyl bromide proceeds in accordance with Markovnikov's rule H,C=CHCH, Br CH3 CHCH, Br Allyl bromide 1- Bromo-2-chloropropane (e) Addition of bromine to allyl bromide gives 1, 2, 3-tribromopropane H,C=CHCH, Br BrChChChaB Allyl bromide 1, 2, 3-Tribromopropane (f) Nucleophilic substitution by hydroxide on allyl bromide gives allyl alcohol H,C=CHCH.Br HC=CHCH.OH Allyl bromide Allyl alcohol (g) Alkylation of sodium acetylide using allyl bromide gives the desired 1-penten-4-yne H, C=CHCH. Br NaC=CH, HC=CHCHCECH Allyl bromide 1-Penten-4-yne (h) Sodium-ammonia reduction of 1-penten-4-yne reduces the triple bond but leaves the double bond intact. Hydrogenation over Lindlar palladium could also be used H_C=CHCH2 C=CH or H Lindlar Pd H2C=CHCH,CH-CH2 L4-Pentadiene 10.23 (a) The desired allylic alcohol can be prepared by hydrolysis of an allylic halide. Cyclopent can be converted to an allylic bromide by free-radical bromination with N-bromosuccinim NBS peroxides OH Cyclopentene 3-Bromocyclopentene 2-Cyclopenten-l-o (b) Reaction of the allylic bromide from part(a) with sodium iodide in acetone converts it to the 3-lodocyclopentene Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(d) Addition of hydrogen chloride to allyl bromide proceeds in accordance with Markovnikov’s rule. (e) Addition of bromine to allyl bromide gives 1,2,3-tribromopropane. ( f ) Nucleophilic substitution by hydroxide on allyl bromide gives allyl alcohol. (g) Alkylation of sodium acetylide using allyl bromide gives the desired 1-penten-4-yne. (h) Sodium–ammonia reduction of 1-penten-4-yne reduces the triple bond but leaves the double bond intact. Hydrogenation over Lindlar palladium could also be used. 10.23 (a) The desired allylic alcohol can be prepared by hydrolysis of an allylic halide. Cyclopentene can be converted to an allylic bromide by free-radical bromination with N-bromosuccinimide (NBS). (b) Reaction of the allylic bromide from part (a) with sodium iodide in acetone converts it to the corresponding iodide. 3-Bromocyclopentene 3-Iodocyclopentene Br NaI acetone I NBS heat, peroxides H2O Na2CO3 Br OH Cyclopentene 2-Cyclopenten-1-ol 3-Bromocyclopentene Na, NH3 1,4-Pentadiene H2C CHCH2CH CH2 1-Penten-4-yne H2C CHCH2C CH or H2, Lindlar Pd H2C CHCH2Br NaC CH Allyl bromide 1-Penten-4-yne H2C CHCH2C CH H2C CHCH2Br Allyl bromide Allyl alcohol H2C NaOH CHCH2OH H2C CHCH2Br Allyl bromide 1,2,3-Tribromopropane BrCH2CHCH2Br Br2 Br H2C CHCH2Br Allyl bromide 1-Bromo-2-chloropropane CH3CHCH2Br HCl Cl CONJUGATION IN ALKADIENES AND ALLYLIC SYSTEMS 239 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website