__ __ __ 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 C 5 H 8 isomers has a branched carbon chain. 3-Methyl-1-butyne CH 3 CHC CH CH 3 1-Pentyne CH 3 CH 2 CH 2 C CH 2-Pentyne CH 3 CH 2 C CCH 3 H11001 CCHH Acetylene H O H11002 Hydroxide ion H11001 C H11002 CH Acetylide ionWater HHO H11001H11001 Carbide ion CH Acetylide ion HO H11002 Hydroxide ionWater HHOC H11002 C H11002 C H11002 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 3 H 2 OHF 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 depro- tonates 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. NaNH 2 , NH 3 2. CH 3 CH 2 CH 2 Br Acetylene 1-Pentyne 3-Heptyne CHHC 1. NaNH 2 , NH 3 2. CH 3 CH 2 Br CH 3 CH 2 CH 2 C CH CH 3 CH 2 CH 2 C CCH 2 CH 3 1. NaNH 2 , NH 3 2. CH 3 Br Acetylene Propyne 2-Heptyne CHHC 1. NaNH 2 , NH 3 2. CH 3 CH 2 CH 2 CH 2 Br CH 3 C CH CH 3 C CCH 2 CH 2 CH 2 CH 3 H11001 Amide ion (stronger base) NH 2 H11002 Ammonia (weaker acid) K a 10 H1100236 (pK a 36) NH 3 2-Butyn-1-olate anion (weaker base) CH 3 C CCH 2 O H11002 H11001 2-Butyn-1-ol (stronger acid) K a H11015 10 H1100216 H1100210 H1100220 (pK a H11015 16H1100220) HCH 3 C CCH 2 O H11001 Ammonia (stronger acid) K a 10 H1100236 (pK a 36) NH 3 Vinyl anion (stronger base) CH 2 CHH11001 Amide ion (weaker base) NH 2 H11002H11002 Ethylene (weaker acid) K a H11015 10 H1100245 (pK a H11015 45) HCH 2 CH CH 3 CH 3 Ethane (weaker acid) H11001H11001 Acetylide ion (weaker base) HC C H11002H11002 Ethyl anion (stronger base) CH 2 CH 3 Acetylene (stronger acid) HC C H K a 10 H1100226 (pK a 26) K a H11015 10 H1100262 (pK a H11015 62) 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 re- quires 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 dehydro- halogenation 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. NaNH 2 , NH 3 2. H 2 O CH 3 CHCl 2 1,1-Dichloroethane HC CH Acetylene 1. NaNH 2 , NH 3 2. CH 3 Br HC CH Acetylene CH 3 CCH Propyne (CH 3 ) 2 CHBr NaOCH 2 CH 3 PropeneIsopropyl bromide CH 3 CH CH 2 CH 3 CH 2 CH 2 OH H 2 SO 4 heat Propene1-Propanol CH 3 CH CH 2 1. 3NaNH 2 2. H 2 O or or(CH 3 ) 3 CCCH 3 Br Br 2,2-Dibromo-3,3- dimethylbutane (CH 3 ) 3 CCH 2 CHBr 2 1,1-Dibromo-3,3- dimethylbutane Br (CH 3 ) 3 CCHCH 2 Br 1,2-Dibromo-3,3- dimethylbutane (CH 3 ) 3 CC CH 3,3-Dimethyl-1-butyne H11001H11001 E2 CH 3 CHCH 3 Br Isopropyl bromide HC CH Acetylene CH 2 CHCH 3 Propene HC Acetylide ion C H11002 1. NaNH 2 , NH 3 2. CH 3 CH 2 Br Acetylene 1-Butyne 2-Pentyne CHHC 1. NaNH 2 , NH 3 2. CH 3 Br CH 3 CH 2 C CH CH 3 CH 2 C CCH 3 Acetylene 1-Pentyne CHHC CHCH 3 CH 2 CH 2 C 1. NaNH 2 , NH 3 2. CH 3 CH 2 CH 2 Br ALKYNES 211 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 CH 3 CH 2 CH 2 Br could be carried out to give 4-octyne (CH 3 CH 2 CH 2 C>CCH 2 CH 2 CH 3 ), 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. NaNH 2 , NH 3 2. CH 3 CH 2 CH 2 CH 2 Br 2-Heptyne CCH 2 CH 2 CH 2 CH 3 CH 3 CCHCH 3 C Propyne CC H H 3 C H CH 2 CH 2 CH 2 CH 3 (E)-2-Heptene CH 2 CH 2 CH 2 CH 3 CC H H 3 C H (Z)-2-Heptene H 2 Lindlar Pd Li, NH 3 1. Na, NH 3 2. H 3 O H11001 C(CH 2 ) 7 CO 2 HCH 3 (CH 2 ) 7 C Stearolic acid CC H CH 3 (CH 2 ) 7 H (CH 2 ) 7 CO 2 H Elaidic acid 2H 2 Pt C(CH 2 ) 7 CO 2 HCH 3 (CH 2 ) 7 C Stearolic acid Stearic acid CH 3 (CH 2 ) 16 CO 2 H H 2 Lindlar Pd C(CH 2 ) 7 CO 2 HCH 3 (CH 2 ) 7 C Stearolic acid CC H CH 3 (CH 2 ) 7 H (CH 2 ) 7 CO 2 H Oleic acid NaNH 2 NH 3 H 2 Pt BrCH 2 (CH 2 ) 4 CH 3 HC CH Acetylene HC CNa Sodium acetylide HC CCH 2 (CH 2 ) 4 CH 3 1-Octyne CH 3 CH 2 CH 2 (CH 2 ) 4 CH 3 Octane 1. NaNH 2 , NH 3 2. H 2 O 1. NaNH 2 , NH 3 2. CH 3 Br H 2 SO 4 heat Br 2 CH 3 CH 2 OH Ethyl alcohol BrCH 2 CH 2 Br 1,2-Dibromoethane HC CH Acetylene CH 3 CCH PropyneEthylene H 2 CCH 2 212 ALKYNES 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 @CO 2 H groups was once part of a @C>C@ unit. The two fragments CH 3 (CH 2 ) 4 CO 2 H and HO 2 CCH 2 CH 2 CO 2 H account for only 10 of the original 16 carbons. The full complement of carbons can be accommodated by assuming that two molecules of CH 3 (CH 2 ) 4 CO 2 H are formed, along with one molecule of HO 2 CCH 2 CH 2 CO 2 H. The starting alkyne is therefore deduced from the ozonolysis data to be as shown: CH 3 (CH 2 ) 4 CO 2 HHO 2 C(CH 2 ) 4 CH 3 HO 2 CCH 2 CH 2 CO 2 H CH 3 (CH 2 ) 4 C C(CH 2 ) 4 CH 3 CCH 2 CH 2 C H11001H11001 Carbocation 2-Butanone CH 3 CH 2 CCH 3 O Water O H H Hydronium ion H O H H H11001 CCH 3 CH 3 CH 2 O H H11001 H11001 Water O H H Carbocation CCH 3 CH 3 CH 2 OH H11001 CH 3 CH 2 CCH 3 OH H11001 Hydronium ion 2-Buten-2-ol CCH 3 CH 3 CH OH OH H H H11001 CCH 3 H11001 H 2 OCH 3 C 2-Butyne 2-Butanone CH 3 CCH 2 CH 3 O 2-Buten-2-ol (enol form) CHCH 3 CH 3 C OH CHHC Acetylene CH 3 CHCl 2 1,1-Dichloroethane CH 3 CHBr 2 1,1-Dibromoethane 2HCl 1. NaNH 2 , NH 3 2. H 2 O CHClH 2 C Vinyl chloride CH 3 CHCl 2 1,1-Dichloroethane HCl ALKYNES 213 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 CH 3 CH 2 CHCH 2 C CH 2 CH 3 CH 2,5-Dimethyl-3-hexyne is CH 3 CHC CH 3 CCHCH 3 CH 3 4-Octyne is CH 3 CH 2 CH 2 C CCH 2 CH 2 CH 3 3-Octyne is CH 3 CH 2 C CCH 2 CH 2 CH 2 CH 3 2-Octyne is CH 3 C CCH 2 CH 2 CH 2 CH 2 CH 3 1-Octyne is HC CCH 2 CH 2 CH 2 CH 2 CH 2 CH 3 CH 3 CC CH 3 CH 3 CCCH 3 is 2,2,5,5-tetramethyl-3-hexyne CH 3 CH 3 324651 CH 3 CH 2 CH 2 CH 2 CHCH 2 CH 2 CH 2 CH 2 CH 3 is 4-butyl-2-nonyne C CCH 3 45 6 7 8 9 321 (Parent chain must contain the triple bond.) CCH 2 23 CH 2 C 113 is cyclotridecyne CH 2 CH 2 CH 2 C CH is 5-cyclopropyl-1-pentyne 543 2 1 CH 3 C H 3 C CCHCHCH 3 is 4,5-dimethyl-2-hexyne CH 3 564321 CH 3 CH 2 C CCH 3 is 2-pentyne 54321 CH 3 CH 2 CH 2 C CH is 1-pentyne 543 2 1 CH 3 CC CH 3 CH 3 CH 3,3-Dimethyl-1-butyne CH 3 CHC CH 3 CCH 3 4-Methyl-2-pentyne CH 3 CH 2 CHC CH 3 CH 3-Methyl-1-pentyne CH 3 CHCH 2 C CH 3 CH 4-Methyl-1-pentyne CH 3 CH 2 CH 2 CH 2 CCH CH 3 CH 2 CH 2 C CCH 3 CH 3 CH 2 C CCH 2 CH 3 1-Hexyne 2-Hexyne 3-Hexyne 214 ALKYNES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (g) (h) 9.20 Ethynylcyclohexane has the molecular formula C 8 H 12 . All the other compounds are C 8 H 14 . 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. CH 3 CH 2 C CCH 2 CH 3 3-Hexyne CH 2 CH 2 C C H11002 BrCH 2 CH 3 H11001prepared from CH 2 CH 3 CH 2 CH 2 CH 2 CH 1-Hexene CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 I 1-Iodohexane KOC(CH 3 ) 3 DMSO CHHC Acetylene CHCH 3 CH 2 CH 2 CH 2 C 1-Hexyne C H11002 Na H11001 HC NaNH 2 NH 3 CH 3 CH 2 CH 2 CH 2 Br 1-Hexene 1,2-Dibromohexane CHCH 3 CH 2 CH 2 CH 2 CCH 3 CH 2 CH 2 CH 2 CH CH 2 1-Hexyne Br 2 CCl 4 1. NaNH 2 , NH 3 2. H 2 O CH 3 CH 2 CH 2 CH 2 CHCH 2 Br Br 1. NaNH 2 , NH 3 2. H 2 O CH 3 CH 2 CH 2 CH 2 CH 2 CHCl 2 1,1-Dichlorohexane CHCH 3 CH 2 CH 2 CH 2 C 1-Hexyne HC CCHCH 2 CHCO H11002 CH 3 H11001 NH 3 O CH 3 CH 2 CHCH 2 CHCO H11002 CH 3 H11001 NH 3 O H 2 Pt C 7 H 11 NO 2 Homoleucine H 2 Pt or or 3-Ethyl-1-hexyne 4-Ethyl-1-hexyne 4-Ethyl-2-hexyne 3-Ethylhexane 3-Ethyl-3-methyl-1-pentyne is CH 3 CH 2 CC CH 2 CH 3 CH 3 CH CEthynylcyclohexane is CH ALKYNES 215 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The desired intermediate, 1-butyne, is available by halogenation followed by dehydrohalo- genation 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 C 10 H 19 Br. 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) CH 3 CH 2 CH 2 CH 2 CCH NH 3 Li CH 3 CH 2 CH 2 CH 2 CH CH 2 1-Hexyne 1-Hexene CH 3 CH 2 CH 2 CH 2 C 1-Hexyne H11001CH H 2 Lindlar Pd CH 3 CH 2 CH 2 CH 2 CH 1-Hexene CH 2 CH 3 CH 2 CH 2 CH 2 C 1-Hexyne H11001CH 2H 2 Pt CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 Hexane Br BrCH 2 CH(CH 2 ) 7 CH 3 1,2-Dibromodecane KOH ethanol–water CC Br H H (CH 2 ) 7 CH 3 (E)-1-Bromo-1-decene H11001 CC H Br H (CH 2 ) 7 CH 3 (Z)-1-Bromo-1-decene H 2 C 2-Bromo-1-decene Br Br BrCH 2 CH(CH 2 ) 7 CH 3 1,2-Dibromodecane KOH ethanol–water C(CH 2 ) 7 CH 3 HC HCCH Acetylene NaNH 2 NH 3 C H11002 H11001 Na CH 3 CH 2 Br HC 1-Butyne CCH 2 CH 3 1. NaNH 2 , NH 3 2. H 2 O CH 3 CH 2 C CH 1-Butyne CH 3 CH 2 CH 2 CHCl 2 1,1-Dichlorobutane CH 3 CH 2 CCH 3 CH 2 C CH 1-Butyne NaNH 2 NH 3 C H11002 H11001 Na CH 3 CH 2 Br 3-Hexyne CH 3 CH 2 C CCH 2 CH 3 Br 2 1. NaNH 2 , NH 3 CH 3 CH 2 CHCH 2 BrH11001 2. H 2 O CH 3 CH 2 C CH 1-Butyne Br CH 3 CH 2 CH CH 2 1-Butene 216 ALKYNES 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 CH 3 CH 2 C CCH 2 CH 3 H 2 H11001 CC H CH 3 CH 2 CH 2 CH 3 H (Z)-3-Hexene 3-Hexyne Pt Hexane CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 CH 3 CH 2 C CCH 2 CH 3 2H 2 H11001 CH 3 CH 2 CH 2 CH 2 CCH 1-Hexyne 1. O 3 2. H 2 O Pentanoic acid CH 3 CH 2 CH 2 CH 2 COH O Carbonic acid H11001 HOCOH O CH 3 CH 2 CH 2 CH 2 CCH 1-Hexyne H 2 O, H 2 SO 4 HgSO 4 2-Hexanone CH 3 CH 2 CH 2 CH 2 CCH 3 O CH 3 CH 2 CH 2 CH 2 CCH 1-Hexyne Cl 2 (2 mol) 1,1,2,2-Tetrachlorohexane CH 3 CH 2 CH 2 CH 2 CCHCl 2 Cl Cl CH 3 CH 2 CH 2 CH 2 CCH 1-Hexyne Cl 2 (1 mol) CC Cl CH 3 CH 2 CH 2 CH 2 H Cl (E)-1,2-Dichloro-1-hexene CH 3 CH 2 CH 2 CH 2 CCH 1-Hexyne 2,2-Dichlorohexane CH 3 CH 2 CH 2 CH 2 CCH 3 Cl Cl HCl (2 mol) CH 3 CH 2 CH 2 CH 2 CCH 1-Hexyne 2-Chloro-1-hexene CH 3 CH 2 CH 2 CH 2 CCH 2 Cl HCl (1 mol) Sodium 1-hexynide tert-Butyl bromide CH 3 CH 2 CH 2 CH 2 C (CH 3 ) 3 CBr H11002 Na H11001 C H11001 CH 3 CH 2 CH 2 CH 2 CCH 1-Hexyne 2-Methylpropene H11001 (CH 3 ) 2 CCH 2 CH 3 CH 2 CH 2 CH 2 C CCH 2 CH 2 CH 2 CH 3 5-DecyneSodium 1-hexynide 1-Bromobutane CH 3 CH 2 CH 2 CH 2 CCH 3 CH 2 CH 2 CH 2 Br H11002 Na H11001 C H11001 CH 3 CH 2 CH 2 CH 2 CCH NH 3 NaNH 2 1-Hexyne Sodium 1-hexynide CH 3 CH 2 CH 2 CH 2 C H11002 Na H11001 C ALKYNES 217 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, CH 3 C>CCH 2 CH 2 CH 2 CH 3 . Two regioisomeric enols are formed, each of which gives a different ketone. H11001 O CH 3 CCH 2 CH 2 CH 2 CH 2 CH 3 CH 3 C CCH 2 CH 2 CH 2 CH 3 2-Heptyne CH 3 C OH CHCH 2 CH 2 CH 2 CH 3 2-Hepten-2-ol 2-Heptanone O CH 3 CH 2 CCH 2 CH 2 CH 2 CH 3 CH 3 CH OH CCH 2 CH 2 CH 2 CH 3 2-Hepten-3-ol 3-Heptanone H 2 O, H 2 SO 4 HgSO 4 3-Hexyne 1. O 3 2. H 2 O Propanoic acid 2CH 3 CH 2 COH O CH 3 CH 2 C CCH 2 CH 3 3-Hexyne H 2 O, H 2 SO 4 HgSO 4 3-Hexanone CH 3 CH 2 CCH 2 CH 2 CH 3 O CH 3 CH 2 C CCH 2 CH 3 3-Hexyne Cl 2 (2 mol) CH 3 CH 2 C CCH 2 CH 3 3,3,4,4-Tetrachlorohexane CH 3 CH 2 C Cl CCH 2 CH 3 Cl Cl Cl 3-Hexyne Cl 2 (1 mol) CH 3 CH 2 C CCH 2 CH 3 CC Cl CH 3 CH 2 CH 2 CH 3 Cl (E)-3,4-Dichloro-3-hexene 3-Hexyne HCl (2 mol) CH 3 CH 2 C CCH 2 CH 3 3,3-Dichlorohexane CH 3 CH 2 CCH 2 CH 2 CH 3 Cl Cl 3-Hexyne HCl (1 mol) CH 3 CH 2 C CCH 2 CH 3 CC Cl CH 3 CH 2 CH 2 CH 3 H (Z)-3-Chloro-3-hexene 3-Hexyne Li NH 3 CH 3 CH 2 C CCH 2 CH 3 CC H CH 3 CH 2 CH 2 CH 3 H (E)-3-Hexene 218 ALKYNES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 9.29 The alkane formed by hydrogenation of (S)-3-methyl-1-pentyne is achiral; it cannot be optically active. The product of hydrogenation of (S)-4-methyl-1-hexyne is optically active because a stereogenic center is present in the starting material and is carried through to the product. Both (S)-3-methyl-1-pentyne and (S)-4-methyl-1-hexyne yield optically active products when their triple bonds are reduced to double bonds. 9.30 (a) The dihaloalkane contains both a primary alkyl chloride and a primary alkyl iodide functional group. Iodide is a better leaving group than chloride and is the one replaced by acetylide. (b) Both vicinal dibromide functions are converted to alkyne units on treatment with excess sodium amide. (c) The starting material is a geminal dichloride. Potassium tert-butoxide in dimethyl sulfoxide is a sufficiently strong base to convert it to an alkyne. (d) Alkyl p-toluenesulfonates react similarly to alkyl halides in nucleophilic substitution reactions. The alkynide nucleophile displaces the p-toluenesulfonate leaving group from ethyl p-toluenesulfonate. H11001 Phenylacetylide ion C C H11002 Ethyl p-toluenesulfonate OS O O CH 2 CH 3 CH 3 1-Phenyl-1-butyne C CCH 2 CH 3 KOC(CH 3 ) 3 , DMSO heat CCH 3 Cl Cl 1,1-Dichloro-1- cyclopropylethane CHC Ethynylcyclopropane BrCH 2 CHCH 2 CH 2 CHCH 2 Br 1,2,5,6-Tetrabromohexane 1. excess NaNH 2 , NH 3 2. H 2 O HC CCH 2 CH 2 C CH Br Br 1,5-Hexadiyne ClCH 2 CH 2 CH 2 CH 2 CH 2 CH 2 IH11001NaC CH ClCH 2 CH 2 CH 2 CH 2 CH 2 CH 2 C CH Sodium acetylide 1-Chloro-6-iodohexane 8-Chloro-1-octyne H 2 Pt H H 3 C CH 2 C CH 3 CH 2 CH (S)-4-Methyl-1-hexyne CH 2 CH 2 CH 3 H H 3 C CH 3 CH 2 (S)-3-Methylhexane C C H 2 Pt CH 3 CH 2 CHCH 2 CH 3 CH 3 3-Methylpentane (does not have a stereogenic center; optically inactive) H H 3 C CCH CH 3 CH 2 (S)-3-Methyl-1-pentyne C ALKYNES 219 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (e) Both carbons of a unit are converted to carboxyl groups on ozonolysis. ( f ) Ozonolysis cleaves the carbon–carbon triple bond. O < (g) Hydration of a terminal carbon–carbon triple bond converts it to a @CCH 3 group. (h) Sodium-in-ammonia reduction of an alkyne yields a trans alkene. The stereochemistry of a double bond that is already present in the molecule is not altered during the process. (i) The primary chloride leaving group is displaced by the alkynide nucleophile. H11001 O O(CH 2 ) 8 Cl 8-Chlorooctyl tetrahydropyranyl ether CCH 2 CH 2 CH 2 CH 3 NaC Sodium 1-hexynide O O(CH 2 ) 8 C CCH 2 CH 2 CH 2 CH 3 9-Tetradecyn-1-yl tetrahydropyranyl ether (Z)-13-Octadecen-3-yn-1-ol CC H CH 3 CH 2 CH 2 CH 2 H CH 2 (CH 2 ) 7 C CCH 2 CH 2 OH (3E,13Z)-3,13-Octadecadien-1-ol CC H CH 3 CH 2 CH 2 CH 2 H CH 2 (CH 2 ) 7 CC H H CH 2 CH 2 OH 1. Na, NH 3 2. H 2 O CH 3 CHCH 2 CC CH CH 3 CH 3 OH 3,5-Dimethyl-1-hexyn-3-ol CH 3 CHCH 2 C CCH 3 CH 3 CH 3 HO O 3-Hydroxy-3,5-dimethyl-2-hexanone H 2 O, H 2 SO 4 HgO 1. O 3 2. H 2 O H11001 HOCOH O Carbonic acid1-Ethynylcyclohexanol OH C CH 1-Hydroxycyclohexane- carboxylic acid OH COH O 1. O 3 2. H 2 O Cyclodecyne HOC(CH 2 ) 8 COH O O Decanedioic acid ( CO 2 H)CC 220 ALKYNES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ( j) Hydrogenation of the triple bond over the Lindlar catalyst converts the compound to a cis alkene. 9.31 Ketones such as 2-heptanone may be readily prepared by hydration of terminal alkynes. Thus, if we had 1-heptyne, it could be converted to 2-heptanone. Acetylene, as we have seen in earlier problems, can be converted to 1-heptyne by alkylation. 9.32 Apply the technique of reasoning backward to gain a clue to how to attack this synthesis problem. A reasonable final step is the formation of the Z double bond by hydrogenation of an alkyne over Lindlar palladium. The necessary alkyne 9-tricosyne can be prepared by a double alkylation of acetylene. It does not matter which alkyl group is introduced first. The alkyl halides are prepared from the corresponding alcohols. CH 3 (CH 2 ) 12 OH 1-Tridecanol CH 3 (CH 2 ) 12 Br 1-Bromotridecane HBr or PBr 3 CH 3 (CH 2 ) 7 OH 1-Octanol CH 3 (CH 2 ) 7 Br 1-Bromooctane HBr or PBr 3 1. NaNH 2 , NH 3 2. CH 3 (CH 2 ) 7 Br 1. NaNH 2 , NH 3 2. CH 3 (CH 2 ) 12 Br C(CH 2 ) 12 CH 3 CH 3 (CH 2 ) 7 C 9-Tricosyne CHCH 3 (CH 2 ) 7 C 1-Decyne CHHC Acetylene H 2 Lindlar Pd CC H CH 3 (CH 2 ) 7 H (CH 2 ) 12 CH 3 (Z)-9-Tricosene C(CH 2 ) 12 CH 3 CH 3 (CH 2 ) 7 C 9-Tricosyne HC C(CH 2 ) 4 CH 3 CH 3 CH 2 CH 2 CH 2 CH 2 BrHC C H11002 Na H11001 H11001 NaNH 2 NH 3 HC CH HC C H11002 Na H11001 H 2 O H 2 SO 4 , HgSO 4 CH 3 C(CH 2 ) 4 CH 3 O 2-Heptanone HC C(CH 2 ) 4 CH 3 1-Heptyne O O(CH 2 ) 8 C CCH 2 CH 2 CH 2 CH 3 9-Tetradecyn-1-yl tetrahydropyranyl ether O O(CH 2 ) 8 CC HH CH 2 CH 2 CH 2 CH 3 (Z)-9-Tetradecen-1-yl tetrahydropyranyl ether H 2 Lindlar Pd ALKYNES 221 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 9.33 (a) 2,2-Dibromopropane is prepared by addition of hydrogen bromide to propyne. The designated starting material, 1,1-dibromopropane, is converted to propyne by a double dehydrohalogenation. (b) As in part (a), first convert the designated starting material to propyne, and then add hydrogen bromide. (c) Instead of trying to introduce two additional chlorines into 1,2-dichloropropane by free- radical substitution (a mixture of products would result), convert the vicinal dichloride to propyne, and then add two moles of Cl 2 . (d) The required carbon skeleton can be constructed by alkylating acetylene with ethyl bromide. Addition of 2 mol of hydrogen iodide to 1-butyne gives 2,2-diiodobutane. (e) The six-carbon chain is available by alkylation of acetylene with 1-bromobutane. 1. NaNH 2 , NH 3 2. CH 3 CH 2 CH 2 CH 2 Br HC CCH 2 CH 2 CH 2 CH 3 1-Hexyne HC CH Acetylene H11001HC CCH 2 CH 3 1-Butyne 2HI Hydrogen iodide 2,2-Diiodobutane CH 3 CCH 2 CH 3 I I NaNH 2 NH 3 CH 3 CH 2 Br HC CCH 2 CH 3 1- Butyne HC C H11002 Na H11001 Sodium acetylide HC CH Acetylene Propyne CH 3 CCH 1. NaNH 2 , NH 3 2. H 2 O 1,2-Dichloropropane CH 3 CHCH 2 Cl Cl 1,1,2,2-Tetrachloropropane CH 3 CCHCl 2 Cl Cl 2Cl 2 Propyne CH 3 CCH 1,2-Dibromopropane 1. NaNH 2 , NH 3 2. H 2 O CH 3 CHCH 2 Br Br 2,2-Dibromopropane CH 3 CCH 3 Br Br 2HBr Propyne CH 3 CCH 1,1-Dibromopropane CH 3 CH 2 CHBr 2 1. NaNH 2 , NH 3 2. H 2 O H11001 2HBr Hydrogen bromide Propyne CH 3 CCH 2,2-Dibromopropane CH 3 CCH 3 Br Br 222 ALKYNES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The alkylating agent, 1-bromobutane, is prepared from 1-butene by free-radical (anti- Markovnikov) addition of hydrogen bromide. Once 1-hexyne is prepared, it can be converted to 1-hexene by hydrogenation over Lindlar palladium or by sodium–ammonia reduction. ( f ) Dialkylation of acetylene with 1-bromobutane, prepared in part ( f ), gives the necessary ten- carbon chain. Hydrogenation of 5-decyne yields decane. (g) A standard method for converting alkenes to alkynes is to add Br 2 and then carry out a double dehydrohalogenation. (h) Alkylation of the triple bond gives the required carbon skeleton. Hydrogenation over the Lindlar catalyst converts the carbon–carbon triple bond to a cis dou- ble bond. H 2 Lindlar Pd C CCH 3 1-(1-Propynyl)cyclohexene (Z)-1-(1-Propenyl)cyclohexene C CH H H 3 C CCH C CCH 3 1. NaNH 2 , NH 3 2. CH 3 Br 1-Ethynylcyclohexene 1-(1-Propynyl)cyclohexene Br 2 NaNH 2 NH 3 (CH 3 ) 13 CH CH Cyclopentadecene (CH 2 ) 13 CH CH Br Br 1,2-Dibromocyclopentadecane (CH 2 ) 13 CC Cyclopentadecyne 5-Decyne 2H 2 Pt Decane CH 3 (CH 2 ) 3 CH 2 CH 2 (CH 2 ) 3 CH 3 CH 3 (CH 2 ) 3 C C(CH 2 ) 3 CH 3 1. NaNH 2 , NH 3 2. CH 3 CH 2 CH 2 CH 2 Br CHCH 3 CH 2 CH 2 CH 2 C 1-Hexyne CH 3 CH 2 CH 2 CH 2 C CCH 2 CH 2 CH 2 CH 3 5-Decyne HC CH Acetylene 1. NaNH 2 , NH 3 2. CH 3 CH 2 CH 2 CH 2 Br 1-Hexyne CH 3 CH 2 CH 2 CH 2 CCH 1-Hexene CH 3 CH 2 CH 2 CH 2 CH CH 2 H 2 , Lindlar Pd or Na, NH 3 CH 3 CH 2 CH 2 CH 2 Br 1-Bromobutane H11001 1-Butene CH 3 CH 2 CH CH 2 Hydrogen bromide HBr peroxides ALKYNES 223 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (i) The stereochemistry of meso-2,3-dibromobutane is most easily seen with a Fischer projection: Recalling that the addition of Br 2 to alkenes occurs with anti stereochemistry, rotate the sawhorse diagram so that the bromines are anti to each other: Thus, the starting alkene must be trans-2-butene. trans-2-Butene is available from 2-butyne by metal-ammonia reduction: 9.34 Attack this problem by first planning a synthesis of 4-methyl-2-pentyne from any starting material in a single step. Two different alkyne alkylations suggest themselves: Isopropyl bromide is a secondary alkyl halide and cannot be used to alkylate accord- ing to reaction (a). A reasonable last step is therefore the alkylation of (CH 3 ) 2 CHC>CH via reac- tion of its anion with methyl iodide. The next question that arises from this analysis is the origin of (CH 3 ) 2 CHC>CH. One of the available starting materials is 1,1-dichloro-3-methylbutane. It can be converted to (CH 3 ) 2 CHC>CH by a double dehydrohalogenation. The complete synthesis is therefore: 9.35 The reaction that produces compound A is reasonably straightforward. Compound A is 14-bromo-1- tetradecyne. H11001 Sodium acetylide NaC CH Br(CH 2 ) 12 Br 1,12-Dibromododecane CHBr(CH 2 ) 12 C Compound A (C 14 H 25 Br) (CH 3 ) 2 CHCH 2 CHCl 2 1. NaNH 2 , NH 3 2. H 2 O 1. NaNH 2 2. CH 3 I (CH 3 ) 2 CHC CCH 3 1,1-Dichloro-3-methylbutane 4-Methyl-2-pentyne3-Methyl-1-butyne (CH 3 ) 2 CHC CH CH 3 C C H11002 4-Methyl-2-pentyne CH 3 C CCH(CH 3 ) 2 CH 3 C C H11002 from and BrCH(CH 3 ) 2 (a) CH 3 Ifrom and(b) CCH(CH 3 ) 2 H11002 C Br Br H H CH 3 H 3 C meso-2,3-Dibromobutanetrans-2-Butene CC H H 3 C CH 3 H 2-Butyne CH 3 C CCH 3 Br 2 Na, NH 3 H Br Br H CH 3 CH 3 Br Br H H CH 3 CH 3 CH 3 CH 3 H Br Br H which is equivalent to H Br Br H CH 3 CH 3 224 ALKYNES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Treatment of compound A with sodium amide converts it to compound B. Compound B on ozonol- ysis gives a diacid that retains all the carbon atoms of B. Compound B must therefore be a cyclic alkyne, formed by an intramolecular alkylation. Compound B is cyclotetradecyne. Hydrogenation of compound B over Lindlar palladium yields cis-cyclotetradecene (compound C). Hydrogenation over platinum gives cyclotetradecane (compound D). Sodium–ammonia reduction of compound B yields trans-cyclotetradecene. The cis and trans isomers of cyclotetradecene are both converted to O?CH(CH 2 ) 12 CH?O on ozonolysis, whereas cyclotetradecane does not react with ozone. 9.36–9.37 Solutions to molecular modeling exercises are not provided in this Study Guide and Solutions Man- ual. You should use Learning By Modeling for these exercises. SELF-TEST PART A A-1. Provide the IUPAC names for the following: (a) CH 3 C CH 3 CCHCH(CH 3 ) 2 (CH 2 ) 12 CC Na, NH 3 H H Compound E (C 14 H 26 ) (CH 2 ) 12 CC H 2 Pt Compound D (C 14 H 28 ) (CH 2 ) 12 CC H 2 Lindlar Pd HH Compound C (C 14 H 26 ) (CH 2 ) 12 CC 1. O 3 2. H 2 O HOC(CH 2 ) 12 COHBr(CH 2 ) 12 CCH OO NaNH 2 (CH 2 ) 11 C H 2 C Br C H11002 Na H11001 Compound A Compound B ALKYNES 225 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) (c) A-2. Give the structure of the reactant, reagent, or product omitted from each of the following reactions. (a) (b) (c) (d) (e) ( f ) (g) (h) A-3. Which one of the following two reactions is effective in the synthesis of 4-methyl-2-hexyne? Why is the other not effective? 1. 2. A-4. Outline a series of steps, using any necessary organic and inorganic reagents, for the prepa- ration of: (a) 1-Butyne from ethyl bromide as the source of all carbon atoms (b) 3-Hexyne from 1-butyne (c) 3-Hexyne from 1-butene (d) A-5. Treatment of propyne in successive steps with sodium amide, 1-bromobutane, and sodium in liquid ammonia yields as the final product ______. A-6. Give the structures of compounds A through D in the following series of equations. A NaNH 2 , NH 3 B H11001 D B C HBr, heat D CH 3 CH 2 CH 2 C CC(CH 3 ) 3 CH 3 CCH 2 CH(CH 3 ) 2 from acetylene O CH 3 ICH 3 CH 2 CHC CNa H11001 CH 3 CH 3 CCH 3 CH 2 CHCH 3 CNaH11001 Br CH 3 CH 2 CH 2 CHC ?CCH 2 CH 3 CH 3 1. O 3 2. H 2 O CH 3 C CCH 2 CH 3 ? Cl 2 (1 mol) CH 3 C CCH 2 CH 3 (E)-2-pentene ? (CH 3 ) 2 CHC CCH 2 CH 3 ? 1. NaNH 2 2. CH 3 CH 2 Br CH 3 C CCH 3 ? H 2 Lindlar Pd CH 3 CH 2 CH 2 CCH 3 CH 2 CH 2 CCH 3 CH ? O CH 3 CH 2 CH 2 CCH HCl (2 mol) ? CH 3 CH 2 CH 2 CCH HCl (1 mol) ? CH 3 CH 2 CH 2 CHCHC CH 2 CH 2 CH 3 CH 2 CH 3 CH 226 ALKYNES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A-7. What are the structures of compounds E and F in the following sequence of reactions? A-8. Give the reagents that would be suitable for carrying out the following transformation. Two or more reaction steps are necessary. PART B B-1. The IUPAC name for the compound shown is (a) 2,6-Dimethyl-3-octyne (b) 6-Ethyl-2-methyl-3-heptyne (c) 2-Ethylpropyl isopropyl acetylene (d ) 2-Ethyl-6-methyl-4-heptyne B-2. Which of the following statements best explains the greater acidity of terminal alkynes (RC>CH) compared with monosubstituted alkenes (RCH?CH 2 )? (a) The sp-hybridized carbons of the alkyne are less electronegative than the sp 2 carbons of the alkene. (b) The two H9266 bonds of the alkyne are better able to stabilize the negative charge of the anion by resonance. (c) The sp-hybridized carbons of the alkyne are more electronegative than the sp 2 carbons of the alkene. (d ) The question is incorrect—alkenes are more acidic than alkynes. B-3. Referring to the following equilibrium (R H11005 alkyl group) (a) K H11021 1; the equilibrium would lie to the left. (b) K H11022 1; the equilibrium would lie to the right. (c) K H11005 1; equal amounts of all species would be present. (d ) Not enough information is given; the structure of R must be known. B-4. Which of the following is an effective way to prepare 1-pentyne? (d ) All these are effective. B-5. Which alkyne yields butanoic acid (CH 3 CH 2 CH 2 CO 2 H) as the only organic product on treat- ment with ozone followed by hydrolysis? (a) 1-Butyne (c) 1-Pentyne (b) 4-Octyne (d ) 2-Hexyne (a) 1. Cl 2 2. NaNH 2 , heat (b) 1. NaNH 2 2. CH 3 CH 2 CH 2 Br 1-Pentene Acetylene 1,1-Dichloropentane(c) 1. NaNH 2 , NH 3 2. H 2 O RCH 2 CH 3 CH11001 RC H11002 H11002 RCH 2 CH 2 CHH11001 RC CH 3 CHCH 2 C CCH(CH 3 ) 2 CH 2 CH 3 CCH CH 2 CH 2 OH 1. NaNH 2 , NH 3 2. CH 3 CH 2 Br H 2 O, H 2 SO 4 HgSO 4 CH 3 CH 2 CCH 2 CH 2 CH 3 Compound FCompound E O ALKYNES 227 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-6. Which of the following produces a significant amount of acetylide ion on reaction with acetylene? (a) Conjugate base of CH 3 OH (pK a 16) (b) Conjugate base of H 2 (pK a 35) (c) Conjugate base of H 2 O (pK a 16) (d) Both (a) and (c). B-7. Which of the following is the product of the reaction of 1-hexyne with 1 mol of Br 2 ? B-8. Choose the sequence of steps that describes the best synthesis of 1-butene from ethanol. (a) (1) NaC>CH; (c) (1) HBr, heat; (2) NaC>CH; (2) H 2 , Lindlar Pd (3) H 2 , Lindlar Pd (b) (1) NaC>CH; (d) (1) HBr, heat; (2) KOC(CH 3 ) 3 , DMSO; (2) Na, NH 3 (3) NaC>CH; (4) H 2 , Lindlar Pd B-9. What is (are) the major product(s) of the following reaction? (c) B-10. Which would be the best sequence of reactions to use to prepare cis-3-nonene from 1-butyne? (a) (1) NaNH 2 in NH 3 ; (2) 1-bromopentane; (3) H 2 , Lindlar Pd (b) (1) NaNH 2 in NH 3 ; (2) 1-bromopentane; (3) Na, NH 3 (c) (1) H 2 , Lindlar Pd; (2) NaNH 2 in NH 3 ; (3) 1-bromopentane (d) (1) Na, NH 3 ; (2) NaNH 2 in NH 3 ; (3) 1-bromopentane B-11. Which one of the following is the intermediate in the preparation of a ketone by hydration of an alkyne in the presence of sulfuric acid and mercury(II) sulfate? HO OH OH OH OH (b)(c)(d)(e)(a) (d) HC CCH 2 CH(CH 3 ) 2 (b) H11001 HC CHH 2 C CCH 3 CH 3 H 3 C H 3 C CH 3 (CH 3 ) 3 CC(a)CH (CH 3 ) 3 CBr H11001 ?HC C H11002 Na H11001 Br Br Br Br Br Br (a) (c) Br Br (b) Br Br (d) (e) 228 ALKYNES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-12. Which combination is best for preparing the compound shown in the box? (a) (b) (c) (d) C CH 3 CH 2 Br H CH 3 1. NaNH 2 , NH 3 CH2. BrCH 2 CH 2 CH 2 C C CH 3 CH 2 Br H H 3 C 1. NaNH 2 , NH 3 CH2. BrCH 2 CH 2 CH 2 C C CH 3 CH 2 CH 2 CH 2 CH 2 Br CH H CH 3 NaC C CH 3 CH 2 CH 2 CH 2 CH 2 Br H H 3 C CHNaC C CH 3 CH 3 CH 2 CH 2 CH 2 CH 2 CCH H ALKYNES 229 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website