ChaPTeR 9 AlKYNES SOLUTIONS TO TEXT PROBLEMS 9.1 The reaction is an acid-base process; water is the proton donor. Two separate proton-transfer steps are involved Carbide ion Acetylide ion Hydroxide ion -O2 9.2 a triple bond may connect C-I and C-2 or C-2 and c-3 in an unbranched chain of five carbons CH3CH2CH2C≡ CHCH, CH2C≡CCH3 One of the C-h isomers has a branched carbon chain H3 209 Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
__ __ __ 209 CHAPTER 9 ALKYNES SOLUTIONS TO TEXT PROBLEMS 9.1 The reaction is an acid–base process; water is the proton donor. Two separate proton-transfer steps are involved. 9.2 A triple bond may connect C-1 and C-2 or C-2 and C-3 in an unbranched chain of five carbons. One of the C5H8 isomers has a branched carbon chain. 3-Methyl-1-butyne CH3CHC CH CH3 1-Pentyne CH3CH2CH2C CH 2-Pentyne CH3CH2C CCH3 H CCH Acetylene H O Hydroxide ion C C H Water Acetylide ion H H O Carbide ion C H Acetylide ion O H Water Hydroxide ion C H H O C C Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
210 ALKYNES 9.3 The bonds become shorter and stronger in the series as the electronegativity increases. NH HO N(3.0) O(3.5) F(4.0) Bond distance(pm) N-H(101) O—H(95) F-H(92) Bond dissociation energy (kJ/mol) H(435) )-H(497)FH(568) Bond dissociation energy (kcal/mol): N-H(104) O-H(119) F-H(136) 9. 4(b) A proton is transferred from acetylene to ethyl anion. CHCH CH, CH3 Ethyl anio (stronger acid) ker base)(weaker acid (pK26) (pKa=62) The position of equilibrium lies to the right. Ethyl anion is a very powerful base and depro- tonates acetylene quantitatively. (c) Amide ion is not a strong enough base to remove a proton from ethylene. The equilibrium lies ( weaker acid) ( weaker base) (d) Alcohols are stronger acids than ammonia; the position of equilibrium lies to the right CH,CCCH,O2H+NH, CHC≡CCI Amide ion (weaker base) on ammonia (weaker acid 9.5 (b) The desired alkyne has a methyl group and a butyl group attached to a-c=C-unit. Two alkylations of acetylene are therefore required: one with a methyl halide, the other with a butyl NaNH. N HC≡CH2CH2B NaNH2, NH, CH2C≡CH 2. CHCH, CH, CH, CHC≡CCH2CH2CHCH It does not matter whether the methyl group or the butyl group is introduced first; the order of steps shown in this synthetic scheme may be inverted (c) An ethyl group and a propyl group need to be introduced as substituents on a-C=C-unit. As in part(b), it does not matter which of the two is introduced first. HC≡CH 1. NaNH,, NH L NaN CHCh.CH Br CH3CH CHC=CH2CHCHB CH2CH,CHC≡CCH,CH Acetylene I-Pentyne 3-Heptyne Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
210 ALKYNES 9.3 The bonds become shorter and stronger in the series as the electronegativity increases. NH3 H2O HF Electronegativity: N (3.0) O (3.5) F (4.0) Bond distance (pm): N@H (101) O@H (95) F@H (92) Bond dissociation energy (kJ/mol): N@H (435) O@H (497) F@H (568) Bond dissociation energy (kcal/mol): N@H (104) O@H (119) F@H (136) 9.4 (b) A proton is transferred from acetylene to ethyl anion. The position of equilibrium lies to the right. Ethyl anion is a very powerful base and deprotonates acetylene quantitatively. (c) Amide ion is not a strong enough base to remove a proton from ethylene. The equilibrium lies to the left. (d) Alcohols are stronger acids than ammonia; the position of equilibrium lies to the right. 9.5 (b) The desired alkyne has a methyl group and a butyl group attached to a @C>C@ unit. Two alkylations of acetylene are therefore required: one with a methyl halide, the other with a butyl halide. It does not matter whether the methyl group or the butyl group is introduced first; the order of steps shown in this synthetic scheme may be inverted. (c) An ethyl group and a propyl group need to be introduced as substituents on a @C>C@ unit. As in part (b), it does not matter which of the two is introduced first. 1. NaNH2, NH3 2. CH3CH2CH2Br Acetylene 1-Pentyne 3-Heptyne HC CH 1. NaNH2, NH3 2. CH3CH2Br CH3CH2CH2C CH CH3CH2CH2C CCH2CH3 1. NaNH2, NH3 2. CH3Br Acetylene Propyne 2-Heptyne HC CH 1. NaNH2, NH3 2. CH3CH2CH2CH2Br CH3C CH CH3C CCH2CH2CH2CH3 Amide ion (stronger base) NH2 Ammonia (weaker acid) Ka1036 (pKa 36) NH3 2-Butyn-1-olate anion (weaker base) CH3C CCH2O 2-Butyn-1-ol (stronger acid) Ka 10161020 (pKa 1620) CH3C CCH2O H Ammonia (stronger acid) Ka 1036 (pKa 36) NH3 Vinyl anion (stronger base) CH2 CH Amide ion (weaker base) NH2 Ethylene (weaker acid) Ka 1045 (pKa 45) CH2 CH H CH3CH3 Ethane (weaker acid) Acetylide ion (weaker base) HC C Ethyl anion (stronger base) CH2CH3 Acetylene (stronger acid) HC C H Ka 1026 (pKa 26) Ka 1062 (pKa 62) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
ALKYNES 211 9.6 Both l-pentyne and 2-pentyne can be prepared by alkylating acetylene. All the alkylation steps involve nucleophilic substitution of a methyl or primary alkyl halide HC≡CH CH2CH2C≡CH CH3CH2C≡CCH Acetylene 2-Pentyne A third isomer, 3-methyl-1-butyne, cannot be prepared by alkylation of acetylene, because it re quires a secondary alkyl halide as the alkylating agent. The reaction that takes place is elimination. not substitution HC≡C:+CH2CHCH HC≡CH+CH,=CHCH Br n ion bromide 9.7 Each of the dibromides shown yields 3, 3-dimethyl-l-butyne when subjected to double dehydro- halogenation with strong base 1.3Na (CH3)3CCCH, or(CH3)3CCHaCHBI CH3)3 CCHCH2Br 2.O (CH3)3CC≡CH 2.2-Dibromo-3,3- 1. 1-Dibromo-3.3. 1. 2-Dibromo-3,3 3,3-Dimethyl-1-butyr dimethylbutane 9.8 (b) The first task is to convert 1-propanol to propene CH3 CH,CH,OH CH3CHECH Propene After propene is available, it is converted to 1, 2-dibromopropane and then to propyne as described in the sample solution for part(a). (c) Treat isopropyl bromide with a base to effect dehydrohalogenation (CH3)2CHBr CHaCH-CH Isopropyl bromide Next, convert propene to propyne as in parts(a)and (b) (d) The starting material contains only two carbon atoms, and so an alkylation step is needed at some point. Propyne arises by alky lation of acetylene, and so the last step in the synthesis HC≡CH CHC≡CH The designated starting material, 1, l-dichloroethane, is a geminal dihalide and can be used to prepare acetylene by a double dehydrohalogenation NaNd CH_ CHCI2 2HO HC≡CH L. 1-Dichloroethane Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
9.6 Both 1-pentyne and 2-pentyne can be prepared by alkylating acetylene. All the alkylation steps involve nucleophilic substitution of a methyl or primary alkyl halide. A third isomer, 3-methyl-1-butyne, cannot be prepared by alkylation of acetylene, because it requires a secondary alkyl halide as the alkylating agent. The reaction that takes place is elimination, not substitution. 9.7 Each of the dibromides shown yields 3,3-dimethyl-1-butyne when subjected to double dehydrohalogenation with strong base. 9.8 (b) The first task is to convert 1-propanol to propene: After propene is available, it is converted to 1,2-dibromopropane and then to propyne as described in the sample solution for part (a). (c) Treat isopropyl bromide with a base to effect dehydrohalogenation. Next, convert propene to propyne as in parts (a) and (b). (d) The starting material contains only two carbon atoms, and so an alkylation step is needed at some point. Propyne arises by alkylation of acetylene, and so the last step in the synthesis is The designated starting material, 1,1-dichloroethane, is a geminal dihalide and can be used to prepare acetylene by a double dehydrohalogenation. 1. NaNH2, NH3 2. H2O CH3CHCl2 1,1-Dichloroethane HC CH Acetylene 1. NaNH2, NH3 HC CH 2. CH3Br Acetylene CH3C CH Propyne (CH3)2CHBr NaOCH2CH3 Isopropyl bromide Propene CH3CH CH2 CH3CH2CH2OH H2SO4 heat 1-Propanol Propene CH3CH CH2 1. 3NaNH2 2. H2O (CH or or 3)3CCCH3 Br Br 2,2-Dibromo-3,3- dimethylbutane (CH3)3CCH2CHBr2 1,1-Dibromo-3,3- dimethylbutane Br (CH3)3CCHCH2Br 1,2-Dibromo-3,3- dimethylbutane (CH3)3CC CH 3,3-Dimethyl-1-butyne E2 CH3CHCH3 Br Isopropyl bromide HC CH Acetylene CH2 CHCH3 Propene HC Acetylide ion C 1. NaNH2, NH3 2. CH3CH2Br Acetylene 1-Butyne 2-Pentyne HC CH 1. NaNH2, NH3 2. CH3Br CH3CH2C CH CH3CH2C CCH3 Acetylene 1-Pentyne HC CH CH3CH2CH2C CH 1. NaNH2, NH3 2. CH3CH2CH2Br ALKYNES 211 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
212 ALKYNES (e) The first task is to convert ethyl alcohol to acetylene. Once acetylene is prepared it can be alkylated with a methyl halide CH CHOH H2 SO4 L NaNH,, NH3 1NNH2NH,CH3C≡CH 2. CH, Br Ethyl alcohol 1. 2-Dibromoethane Propyne 9.9 The first task is to assemble a carbon chain containing eight carbons. Acetylene has two carbon atoms and can be alkylated via its sodium salt to 1-octyne. Hydrogenation over platinum converts BrCH,(CH,), CH HC≡CH HC≡CNa HCECCH,(CH2)4CH3-Pt CH_ CH2CH_(CH2)4CH lene I-Octyne Octane Alternatively, two successive alkylations of acetylene with CH CH, CH, Br could be carried out to ive 4-octyne(CH3 CH,CH, C=CCH,CH, CH3), which could then be hydrogenated to octane. 9.10 Hydrogenation over Lindlar palladium converts an alkyne to a cis alkene. Oleic acid therefore has the structure indicated in the following equation H, (CH,).CO,H CHa(CH,)7CEC(CH,)7CO,H Lindlar P Stearolic acid Oleic acid Hydrogenation of alkynes over platinum leads to alkanes CHa(CH2)7CEC(CH2),CO,H CH3(CH,),6CO,H Stearolic acid Stearic acid 9.11 Alkynes are converted to trans alkenes on reduction with sodium in liquid ammonia. CH(CH,)7CEC(CH2)7CO,H 2.H2O (CH,,CO,H Stearolic acid Elaidic acid 9.12 The proper double-bond stereochemistry may be achieved by using 2-heptyne as a reactant in the final step. Lithium-ammonia reduction of 2-heptyne gives the trans alkene; hydrogenation over Lindlar palladium gives the cis isomer. The first task is therefore the alky lation of propyne to H,C CH,CH,CH, CH 1. NaNH2. NH, (E)-2-Heptenc CHC≡CH 2. CH,CH2CH2CH, Br CH, CECCH, CH,CH,CH Propyne 2-Heptyne HC CH,CH, CH,CH Lindlar pd (Z)-2-Heptene Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(e) The first task is to convert ethyl alcohol to acetylene. Once acetylene is prepared it can be alkylated with a methyl halide. 9.9 The first task is to assemble a carbon chain containing eight carbons. Acetylene has two carbon atoms and can be alkylated via its sodium salt to 1-octyne. Hydrogenation over platinum converts 1-octyne to octane. Alternatively, two successive alkylations of acetylene with CH3CH2CH2Br could be carried out to give 4-octyne (CH3CH2CH2C>CCH2CH2CH3), which could then be hydrogenated to octane. 9.10 Hydrogenation over Lindlar palladium converts an alkyne to a cis alkene. Oleic acid therefore has the structure indicated in the following equation: Hydrogenation of alkynes over platinum leads to alkanes. 9.11 Alkynes are converted to trans alkenes on reduction with sodium in liquid ammonia. 9.12 The proper double-bond stereochemistry may be achieved by using 2-heptyne as a reactant in the final step. Lithium–ammonia reduction of 2-heptyne gives the trans alkene; hydrogenation over Lindlar palladium gives the cis isomer. The first task is therefore the alkylation of propyne to 2-heptyne. 1. NaNH2, NH3 2. CH3CH2CH2CH2Br 2-Heptyne CH3C CH CH3C CCH2CH2CH2CH3 Propyne C C H H3C H CH2CH2CH2CH3 (E)-2-Heptene CH2CH2CH2CH3 C C H H3C H (Z)-2-Heptene H2 Lindlar Pd Li, NH3 1. Na, NH3 2. H3O C(CH2 ) CH3(CH2)7C 7CO2H Stearolic acid C C H CH3(CH2)7 H (CH2)7CO2H Elaidic acid 2H2 Pt C(CH2) CH3(CH2)7C 7CO2H Stearolic acid Stearic acid CH3(CH2)16CO2H H2 Lindlar Pd C(CH2) CH3(CH2)7C 7CO2H Stearolic acid C C H CH3(CH2)7 H (CH2)7CO2H Oleic acid NaNH2 NH3 H2 Pt BrCH2(CH2)4CH3 HC CH Acetylene HC CNa Sodium acetylide HC CCH2(CH2)4CH3 1-Octyne CH3CH2CH2(CH2)4CH3 Octane 1. NaNH2, NH3 2. H2O 1. NaNH2, NH3 2. CH3Br H2SO4 heat Br2 CH3CH2OH Ethyl alcohol BrCH2CH2Br 1,2-Dibromoethane HC CH Acetylene CH3C CH Ethylene Propyne H2C CH2 212 ALKYNES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
213 9. 13 (b) Addition of hydrogen chloride to vinyl chloride gives the geminal dichloride 1, 1 H C=CHCI CH3CHCI2 Vinyl chloride L.I-Dichloroethane (c) Since 1, I-dichloroethane can be prepared by adding 2 mol of hydrogen chloride to acetylene Is shown in the sample solution to part(a), first convert 1, l-dibromoethane to acetylene by dehydrohalogenation NaNH, NH, CH CHBI HC≡CH 2HCI 2.H,O CH3CHCI 1.l-Dibromoethane 1.l-Dichloroethane 9. 14 The enol arises by addition of water to the triple bond. CH3C=CHCH3 CHC≡CCH3+H2O CHaCCH,CH3 2-Butyne 2-Buten-2-ol (enol form) 2-Butanone The mechanism described in the textbook Figure 9.6 is adapted to the case of 2-butyne hydration as H CH3CH=CCH3 O: CH,-CCH CH CHCCH Hydronium 2-Buten-2-ol CH CH,-CCH3 +:O CH2CH2CCH3+H一 Carbocation Water 2-Butanone Hydronium ion 9. 15 Hydration of 1-octyne gives 2-octanone according to the equation that immediately precedes this problem in the text. Prepare 1-octyne as described in the solution to Problem 9.9, and then carry out its hydration in the presence of mercury(f) sulfate and sulfuric acid Hydration of 4-octyne gives 4-octanone. Prepare 4-octyne as described in the solution to Problem 9.9 9.16 Each of the carbons that are part of-CO, H groups was once part of a-C=C-unit. The two fragments CH3 (CH,)CO, H and HO, CCH,CH, CO, H account for only 10 of the original 16 carbons The full complement of carbons can be accommodated by assuming that two molecules of CHa(CH,)CO, H are formed, along with one molecule of HO, CCH, CH,CO, H. The starting alkyne is therefore deduced from the ozonolysis data to be as shown: CH,(CH2),C#,_ CH,C3C(CH)CH, CHa(CH2)4CO,H HO, CCH,CH,CO,H HO, C(CH,)4CH3 Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
9.13 (b) Addition of hydrogen chloride to vinyl chloride gives the geminal dichloride 1,1- dichloroethane. (c) Since 1,1-dichloroethane can be prepared by adding 2 mol of hydrogen chloride to acetylene as shown in the sample solution to part (a), first convert 1,1-dibromoethane to acetylene by dehydrohalogenation. 9.14 The enol arises by addition of water to the triple bond. The mechanism described in the textbook Figure 9.6 is adapted to the case of 2-butyne hydration as shown: 9.15 Hydration of 1-octyne gives 2-octanone according to the equation that immediately precedes this problem in the text. Prepare 1-octyne as described in the solution to Problem 9.9, and then carry out its hydration in the presence of mercury(II) sulfate and sulfuric acid. Hydration of 4-octyne gives 4-octanone. Prepare 4-octyne as described in the solution to Problem 9.9. 9.16 Each of the carbons that are part of @CO2H groups was once part of a @C>C@ unit. The two fragments CH3(CH2)4CO2H and HO2CCH2CH2CO2H account for only 10 of the original 16 carbons. The full complement of carbons can be accommodated by assuming that two molecules of CH3(CH2)4CO2H are formed, along with one molecule of HO2CCH2CH2CO2H. The starting alkyne is therefore deduced from the ozonolysis data to be as shown: CH3(CH2)4CO2H HO2C(CH2) HO2CCH2CH2CO2H 4CH3 CH3(CH2)4C C(CH2) CCH2CH2C 4CH3 Carbocation 2-Butanone CH3CH2CCH3 O Water O H H Hydronium ion H O H H CH3CH2 CCH3 O H Water O H H Carbocation CH3CH2 CCH3 OH CH3CH2CCH3 OH Hydronium ion 2-Buten-2-ol CH3CH CCH3 OH O H H H CH3C CCH3 H2O 2-Butyne 2-Butanone CH3CCH2CH3 O 2-Buten-2-ol (enol form) CH3C CHCH3 OH HC CH Acetylene CH3CHCl2 1,1-Dichloroethane CH3CHBr2 1,1-Dibromoethane 2HCl 1. NaNH2, NH3 2. H2O H CHCl 2C Vinyl chloride CH3CHCl2 1,1-Dichloroethane HCl ALKYNES 213 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
214 ALKYNES 9.17Three isomers have unbranched carbon chains: CH2CH2CH2CH2C≡CHCH3CH2CH2C≡CCH3CH2CHC≡CCH2CH3 1-Hexyne 2-Hexyne Next consider all the alkynes with a single methyl branch CH3CHCH2C≡CHCH3CH2CHC≡ CH CH3CHO≡CCH2 H3 4-Methyl-l-pentyne 3-Methyl-l-pentyne 4-Methyl-2-pentyne One isomer has two methyl branches. None is possible with an ethyl branch CH2CC≡CH 3.3-Dimethyl-1-butyne 9. 18(a) CH CH, CHaCECH is 1-pentyne (b) CH CH,CECCH, is 2-pentyne (c) CH_ C=CCHCHCH3 is 4, 5-dimethyl-2-hexyne H3C CH (d) H, CH,CH, CECH is 5-cyclopropyl-l-pentyne CECCHos is cyclotridecyne (e) (f) CHCH, CH,CH, CHCH,CH, CH,CH, CH3 is 4-butyl-2-nonyne (Parent chain must contain the triple bond. H3 (g) CH CC=CCCH3 is 2, 2,5,5-tetramethyl-3-hexyne CH3 CH 9. 19 (a) 1-Octyne is HCECCH, CH,CH,CH, CH, CH (b) 2-Octyne is CH C=CCH,CH,CH,CH,CH (c) 3-Octyne is CH CH,C=CCH, CH, CH, CH3 (d) 4-Octyne is CH_CH,CH,C=CCH, CH,CH3 (e) 2, 5-Dimethyl-3-hexyne is CH CHCECCHCH () 4-Ethyl-1-hexyne is CH3 CH2CHCH,CECH CHCH Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
9.17 Three isomers have unbranched carbon chains: Next consider all the alkynes with a single methyl branch: One isomer has two methyl branches. None is possible with an ethyl branch. 9.18 (a) (b) (c) (d) (e) ( f ) (g) 9.19 (a) (b) (c) (d) (e) ( f ) 4-Ethyl-1-hexyne is CH3CH2CHCH2C CH2CH3 CH 2,5-Dimethyl-3-hexyne is CH3CHC CH3 CCHCH3 CH3 4-Octyne is CH3CH2CH2C CCH2CH2CH3 3-Octyne is CH3CH2C CCH2CH2CH2CH3 2-Octyne is CH3C CCH2CH2CH2CH2CH3 1-Octyne is HC CCH2CH2CH2CH2CH2CH3 CH3CC CH3 CH3 CCCH3 is 2,2,5,5-tetramethyl-3-hexyne CH3 CH3 1 2 46 3 5 CH3CH2CH2CH2CHCH2CH2CH2CH2CH3 is 4-butyl-2-nonyne C CCH3 45 6 7 8 9 3 21 (Parent chain must contain the triple bond.) CCH2 2 3 CH2C 13 1 is cyclotridecyne CH2CH2CH2C CH is 5-cyclopropyl-1-pentyne 543 2 1 CH3C H3C CCHCHCH3 is 4,5-dimethyl-2-hexyne CH3 1 2 3 4 5 6 CH3CH2C CCH3 is 2-pentyne 5 4 3 21 CH3CH2CH2C CH is 1-pentyne 543 2 1 CH3CC CH3 CH3 CH 3,3-Dimethyl-1-butyne CH3CHC CH3 CCH3 4-Methyl-2-pentyne CH3CH2CHC CH3 CH 3-Methyl-1-pentyne CH3CHCH2C CH3 CH 4-Methyl-1-pentyne CH3CH2CH2CH2C CH CH3CH2CH2C CCH3 CH3CH2C CCH2CH3 1-Hexyne 2-Hexyne 3-Hexyne 214 ALKYNES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
215 (8) Ethynylcyclohexane is C≡CH CH3 (h) 3-Ethyl-3-methyl-l-pentyne is CH,CH,CC=CH CHCH 9.20 Ethynylcyclohexane has the molecular formula C&Hi. All the other compounds are CHi 9.21 Only alkynes with the carbon skeletons shown can give 3-ethy lhexane on catalytic hydrogenation 9.22 The carbon skeleton of the unknown acetylenic amino acid must be the same as that of homoleucine The structure of homoleucine is such that there is only one possible location for a carbon-carbon triple bond in an acetylenic precursor. HC≡ CCHCH,CHCO CH CHCHCHCHCO CHNI ChInO Homoleucine 9.23 (a) CH, CH-CH_CH2 CH- CHCI2 2.. NW, CH3 CH,CH_CHIC=CH hexyne 1. NaNH. NH CH CH, CH,CH, CH=CH2 CH CH, CHCH, CHCH, Br CH2CH2CH2CHC≡CH 2-Dibromohexane l-Hexyne HC≡CH HC≡ C Na* CH,CH2 CH- CH,Br CH2CH2CH2CHC≡CH (d) CH_CH__,CH,I DMSO CH3 CH,CH,CH,CHECH I-lodohexane -Hexene 1-Hexene is then converted to l-hexyne as in part(b). 9. 24 (a) Working backward from the final product, it can be seen that preparation of 1-butyne will allow the desired carbon skeleton to be constructed CHaCH,=CCH,CH3 prepared from CH,CH,CEC: BrCH,CH3 3-Hexyne Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(g) (h) 9.20 Ethynylcyclohexane has the molecular formula C8H12. All the other compounds are C8H14. 9.21 Only alkynes with the carbon skeletons shown can give 3-ethylhexane on catalytic hydrogenation. 9.22 The carbon skeleton of the unknown acetylenic amino acid must be the same as that of homoleucine. The structure of homoleucine is such that there is only one possible location for a carbon–carbon triple bond in an acetylenic precursor. 9.23 (a) (b) (c) (d) 1-Hexene is then converted to 1-hexyne as in part (b). 9.24 (a) Working backward from the final product, it can be seen that preparation of 1-butyne will allow the desired carbon skeleton to be constructed. CH3CH2C CCH2CH3 3-Hexyne CH2CH2C C prepared from BrCH2CH3 CH3CH2CH2CH2CH CH2 1-Hexene CH3CH2CH2CH2CH2CH2I 1-Iodohexane KOC(CH3)3 DMSO HC CH Acetylene CH3CH2CH2CH2C CH 1-Hexyne CNa HC NaNH2 NH3 CH3CH2CH2CH2Br 1-Hexene 1,2-Dibromohexane CH3CH2CH2CH2CH CH2 CH3CH2CH2CH2C CH 1-Hexyne Br2 CCl4 1. NaNH2, NH3 2. H2O CH3CH2CH2CH2CHCH2Br Br 1. NaNH2, NH3 2. H2O CH3CH2CH2CH2CH2CHCl2 1,1-Dichlorohexane CH3CH2CH2CH2C CH 1-Hexyne HC CCHCH2CHCO CH3 NH3 O CH3CH2CHCH2CHCO CH3 NH3 O H2 Pt C7H11NO2 Homoleucine H2 Pt or or 3-Ethyl-1-hexyne 4-Ethyl-1-hexyne 4-Ethyl-2-hexyne 3-Ethylhexane 3-Ethyl-3-methyl-1-pentyne is CH3CH2CC CH2CH3 CH3 CH Ethynylcyclohexane is CH C ALKYNES 215 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
216 ALKYNES The desired intermediate, 1-butyne, is available by halogenation followed by dehydrohalo- genation of 1-butene I NaNH, NH, CH,CHLCH=CH, Br2 CHCH CHCH,Br 2HO CH2CHC≡CH 1-Butene Reaction of the anion of 1-butyne with ethyl bromide completes the synthesis CH2CH2C≡CH CH2CH2C≡ Cuzo CHCHC=CCh, CH 3-Hexyne (b) Dehydrohalogenation of 1, l-dichlorobutane yields 1-butyne. The synthesis is completed as in part(a). 1. NaNH2, NHs CH,CH,CH,CHCI 2H,o →CH3CH2C≡CH LI-Dichlorobutane NaNH, (c)HC≡CH HC≡C:Na+ CH, CHB, HC≡CCH2CH3 1-Butyne is converted to 3-hexyne as in part(a) 9.25 A single dehydrobromination step occurs in the conversion of 1, 2-dibromodecane to CioHjgBr Bromine may be lost from C-l to give 2-bromo-l-decene BrCH, CH(CH,)CH3 H,C=C(CH2),CH Br L2-Dibromodecane Loss of bromine from C-2 gives(E)-and(z)-l-bromo-1-decene (CH,_CH (CH2),CH3 BrCH, CH(CH,),CHa- =C B H 1. 2-Dibromodecane (Erl-Bromo-I-decene 9.26(a) CH, CH, CH,CH,CECH 2H2 CHa CH, CH,CH,,CH3 l-Hexyne (b) CH CH, CH, CH,C=CH H Lindlar p CH CH, CH, CH, CH=CH, 1-Hexene (c)CH3CH2CH2CH2C≡CH CH,CHCH,CH=CH 1-Hexyne 1-Hexene Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
The desired intermediate, 1-butyne, is available by halogenation followed by dehydrohalogenation of 1-butene. Reaction of the anion of 1-butyne with ethyl bromide completes the synthesis. (b) Dehydrohalogenation of 1,1-dichlorobutane yields 1-butyne. The synthesis is completed as in part (a). (c) 1-Butyne is converted to 3-hexyne as in part (a). 9.25 A single dehydrobromination step occurs in the conversion of 1,2-dibromodecane to C10H19Br. Bromine may be lost from C-1 to give 2-bromo-1-decene. Loss of bromine from C-2 gives (E)- and (Z)-1-bromo-1-decene. 9.26 (a) (b) (c) CH3CH2CH2CH2C CH NH3 Li CH3CH2CH2CH2CH CH2 1-Hexyne 1-Hexene CH3CH2CH2CH2C 1-Hexyne CH H 2 Lindlar Pd CH3CH2CH2CH2CH 1-Hexene CH2 CH3CH2CH2CH2C 1-Hexyne CH 2H 2 Pt CH3CH2CH2CH2CH2CH3 Hexane Br BrCH2CH(CH2)7CH3 1,2-Dibromodecane KOH ethanol–water C C Br H H (CH2)7CH3 (E)-1-Bromo-1-decene C C H Br H (CH2)7CH3 (Z)-1-Bromo-1-decene H2C 2-Bromo-1-decene Br Br BrCH2CH(CH2)7CH3 1,2-Dibromodecane KOH ethanol–water C(CH2)7CH3 HC CH HC Acetylene NaNH2 NH3 C Na CH3CH2Br HC 1-Butyne CCH2CH3 1. NaNH2, NH3 2. H2O CH3CH2C CH 1-Butyne CH3CH2CH2CHCl2 1,1-Dichlorobutane CH3CH2C CH CH3CH2C 1-Butyne NaNH2 NH3 C Na CH3CH2Br 3-Hexyne CH3CH2C CCH2CH3 Br2 1. NaNH2, NH3 CH3CH2CHCH2Br 2. H2O CH3CH2C CH 1-Butyne Br CH3CH2CH CH2 1-Butene 216 ALKYNES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
ALKYNES 217 (d)CH3CHCH2CH2C≡CH CHCH2 CH..C≡C:Nat 1-Hexyne Sodium 1-hexynide CHa,CH,CH,C=C: Na CH, CH,CH,CH, Br CHa CH,CH,CH,C=CCH,CH,CH,CH Sodium l-hexynide 1-Bromobutane O) CH,CHCH,CH,CEC: Na +(CH3)3CBr CHaCH,CH, CH,CECH +(CH3)C=CH Sodium l-hexynide tert-Butyl l-Hexyne 2-Methylpropene bromide (g)CHCH2CH2CH2C≡CH CH3CHOCHICHC-CH 2-Chloro- I-hexene (h) CHi CH2 CH2 CECH(2 mol) CHa CH2 CH2CH2CCH3 1-Hexyne CH CHCHCH (i)CH3CH2CH2CH2C≡CH I-Hexyne (Er1, 2-Dichloro-I-hexene CH, CH,CH, CH2C≡CH-(2mo CHa CH,CH, CH,CCHCI l-Hexyne 1. 1. 2. 2-Tetrachlorohexane () CH3 CH, CH, CH,C=CH HO, H,,, CH,CH_CH,CH, CCH HgSOA 1-Hexyne 2-Hexanone (D)CH3CH2CH2CH2C≡CH CH.COH HOCOH Pentanoic acid Carbonic acid 9.27(a) CH,CH,C=CCH,CH3 2H, CH,,CH,CH,CH, CH3 3-Hexyne Hexane CHCH CHCH (b) CH,CH, C=CCH, CH, +H, (Z)3-Hexene Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(d) (e) ( f ) (g) (h) (i) ( j) (k) (l) 9.27 (a) (b) 3-Hexyne Lindlar Pd CH3CH2C CCH2CH3 H2 C C H CH3CH2 CH2CH3 H (Z)-3-Hexene 3-Hexyne Pt Hexane CH3CH2C CCH2CH3 CH3CH2CH2CH2CH2CH3 2H2 CH3CH2CH2CH2C CH 1-Hexyne 1. O3 2. H2O Pentanoic acid CH3CH2CH2CH2COH O Carbonic acid HOCOH O CH3CH2CH2CH2C CH 1-Hexyne H2O, H2SO4 HgSO4 2-Hexanone CH3CH2CH2CH2CCH3 O CH3CH2CH2CH2C CH 1-Hexyne Cl2 (2 mol) 1,1,2,2-Tetrachlorohexane CH3CH2CH2CH2CCHCl2 Cl Cl CH3CH2CH2CH2C CH 1-Hexyne Cl2 (1 mol) C C Cl CH3CH2CH2CH2 H Cl (E)-1,2-Dichloro-1-hexene CH3CH2CH2CH2C CH 1-Hexyne 2,2-Dichlorohexane CH3CH2CH2CH2CCH3 Cl Cl HCl (2 mol) CH3CH2CH2CH2C CH 1-Hexyne 2-Chloro-1-hexene CH3CH2CH2CH2C CH2 Cl HCl (1 mol) Sodium 1-hexynide tert-Butyl bromide CH3CH2CH2CH2C (CH3)3CBr Na C CH3CH2CH2CH2C CH 1-Hexyne 2-Methylpropene (CH3)2C CH2 CH3CH2CH2CH2C CCH2CH2CH2CH3 Sodium 1-hexynide 1-Bromobutane 5-Decyne CH3CH2CH2CH2C CH3CH2CH2CH2Br Na C CH3CH2CH2CH2C CH NH3 NaNH2 1-Hexyne Sodium 1-hexynide CH3CH2CH2CH2C Na C ALKYNES 217 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
218 ALKYNES () CH,CHC= CCH,CH31-A,→ CH,CH3 (E)3-Hexene CH CH2 (d)CHCH2C≡CCH2CH3 (I mol) CHCH (Z)-3-Chloro-3-hexene (e)CHCH2C≡CCH2CH2 (2 mol) CH3CH,,,CH 3-Hexyne 3.3-Dichlorohexane (f)CH3CH2C≡CCH2CH (I mol (E)-3, 4-Dichloro-3-hexer (g)CHCH2C≡CCH2CH CHCH,C—CCH2CH H.O. H (h) CH3=CCH2 CH3 Hgse CH,CHCCH,CH,CH 3-Hexyne (i)CH2CH2C≡CCH2CH2 2CH,. COH 9.28 The two carbons of the triple bond are similarly but not identically substituted in 2-heptyne CHaCECCH,,,CH3. Two regioisomeric enols are formed, each of which gives a different CHC≡CCH2CH2CH2CH CHa C=CHCH, CH, CH, CH3 CH,CH=CCH, CH,CH, CH3 2-Heptyne 2-Hepten-2-oI 2-Hepten-3-ol CHa CCH,CH,,: CH CHCCHCHCHCH Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(c) (d) (e) ( f ) (g) (h) (i) 9.28 The two carbons of the triple bond are similarly but not identically substituted in 2-heptyne, CH3C>CCH2CH2CH2CH3. Two regioisomeric enols are formed, each of which gives a different ketone. O CH3CCH2CH2CH2CH2CH3 CH3C CCH2CH2CH2CH3 2-Heptyne CH3C OH CHCH2CH2CH2CH3 2-Hepten-2-ol 2-Heptanone O CH3CH2CCH2CH2CH2CH3 CH3CH OH CCH2CH2CH2CH3 2-Hepten-3-ol 3-Heptanone H2O, H2SO4 HgSO4 3-Hexyne 1. O3 2. H2O Propanoic acid 2CH3CH2COH O CH3CH2C CCH2CH3 3-Hexyne H2O, H2SO4 HgSO4 3-Hexanone CH3CH2CCH2CH2CH3 O CH3CH2C CCH2CH3 3-Hexyne Cl2 (2 mol) CH3CH2C CCH2CH3 3,3,4,4-Tetrachlorohexane CH3CH2C Cl CCH2CH3 Cl Cl Cl 3-Hexyne Cl2 (1 mol) CH3CH2C CCH2CH3 C C Cl CH3CH2 CH2CH3 Cl (E)-3,4-Dichloro-3-hexene 3-Hexyne HCl (2 mol) CH3CH2C CCH2CH3 3,3-Dichlorohexane CH3CH2CCH2CH2CH3 Cl Cl 3-Hexyne HCl (1 mol) CH3CH2C CCH2CH3 C C Cl CH3CH2 CH2CH3 H (Z)-3-Chloro-3-hexene 3-Hexyne Li NH3 CH3CH2C CCH2CH3 C C H CH3CH2 CH2CH3 H (E)-3-Hexene 218 ALKYNES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website