10 Ch06
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10 Ch06

Course Number: CHEM 1570, Fall 2008

College/University: Cornell

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CHAPTER 6 The Reactions of Alkynes An Introduction to Multistep Synthesis Important Terms acetylide ion the conjugate base of a terminal alkyne RC C - aldehyde a compound with a carbonyl group that is bonded to an alkyl group and to a hydrogen (or bonded to two hydrogens). O R C or H H O C H alkylation reaction alkyne carbonyl group a reaction that adds an alkyl group to a reactant. a hydrocarbon that...

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6 CHAPTER The Reactions of Alkynes An Introduction to Multistep Synthesis Important Terms acetylide ion the conjugate base of a terminal alkyne RC C - aldehyde a compound with a carbonyl group that is bonded to an alkyl group and to a hydrogen (or bonded to two hydrogens). O R C or H H O C H alkylation reaction alkyne carbonyl group a reaction that adds an alkyl group to a reactant. a hydrocarbon that contains a triple bond. a carbon doubly bonded to an oxygen. O C enol an ,-unsaturated alcohol. OH C geminal dihalide internal alkyne keto-enol tautomers a compound with two halogen atoms bonded to the same carbon. an alkyne with its triple bond not at the end of the carbon chain. a ketone and its isomeric ,-unsaturated alcohol. O RCH2CR keto OH RCH enol CR ketone a compound with a carbonyl group that is bonded to two alkyl groups. O R C R -complex polymer radical anion a complex formed between an electrophile and a triple bond. a large molecule made by linking up many small molecules called monomers. a species with a negative charge and an unpaired electron. 184 Chapter 6 retrosynthetic analysis or retrosynthesis tautomerization tautomers terminal alkyne vinylic cation vinylic radical working backward (on paper) from a target molecule to available starting materials. interconversion of tautomers. 185 isomers that are in rapid equilibrium. The keto and enol tautomers differ in the location of a double bond and a hydrogen. an alkyne with its triple bond at the end of the carbon chain. a compound with a positive charge on a vinylic carbon. a compound with an unpaired electron on a vinylic carbon. 186 Chapter 6 Solutions to Problems 1. The molecular formula of a noncyclic hydrocarbon with 14 carbons and no bonds is C14 H30 (C n H 2n+2 ) . Because a compound has two fewer hydrogens for every ring and bond, a compound with one ring and 4 bonds (2 triple bonds) would have 10 fewer hydrogens than the C n H 2n +2 formula. Thus, the molecular formula is C14 H 20 . CH3 2. a. ClCH2 CH2 C CCH 2 CH3 c. CH3CHC CH3 CH e. HC CCH2CCH3 CH3 b. d. HC CCH 2Cl f. CH3 C CCH 3 3. 4. To answer this question, you need to have the information in Section 6.2. a. 1-bromo-1,3-pentadiene b. 1-hepten-5-yne c. 4-hepten-1-yne HC CCH2CH2CH2CH3 1-hexyne butylacetylene CH3CH2CHC CH CH3C CCH2CH2CH3 CH3CH2C CCH2CH3 2-hexyne methylpropylacetylene CH3CHCH2C CH 3-hexyne diethylacetylene CH3CHC CCH3 CH3 3-methyl-1-pentyne sec-butylacetylene CH3 4-methyl-1-pentyne isobutylacetylene CH3 CH3CC CH CH3 4-methyl-2-pentyne isopropylmethylacetylene CH3 3,3-dimethyl-1-butyne tert-butylacetylene 5. 6. a. 5-bromo-2-pentyne b. 6-bromo-2-chloro-4-octyne a. 1-hepten-4-yne b. 4-methyl-1,4-hexadiene c. 5-vinyl-5-octen-1-yne pentane 1-pentene c. 1-methoxy-2-pentyne d. 3-ethyl-1-hexyne d. 3-butyn-1-ol e. 1,3,5-heptatriene f. 2,4-dimethyl-4-hexen-1-ol 1-pentyne 7. Chapter 6 8. a. sp2sp2 b. sp2sp3 c. spsp2 d. spsp3 e. spsp f. sp2sp2 g. sp2sp3 h. spsp3 i. sp2sp 187 9. The less stable reactant will be the more reactive reactant, if the less stable reactant has the more stable transition state, or if the less stable reactant has the less stable transition state and the difference in the stabilities of the reactants is greater than the difference in the stabilities of the transition states. Br Br 10. a. CH2 Br CCH3 Br d. HC CCH3 Br Br Br b. CH3CCH3 Br Br e. CH3CH2CCH3 Br Br Br c. CH3C Br CCH3 f. CH3CCH2CH2CH3 + CH3CH2CCH2CH3 Br Br 11. a. H3C C Br C Br CH3 Only anti addition occurs, because the intermediate is a cyclic bromonium ion. Therefore the product has the trans configuration. 12. Because the alkyne is not symmetrical, two ketones will be obtained. O CH3CH2CCH2CH2CH2CH3 and O CH3CH2CH2CCH2CH2CH3 CCH2CH3 13. a. CH3C CH b. CH3CH2C c. HC C The best answer for "b" is 3-hexyne, because it would form only the desired ketone. 2-Hexyne would form two different ketones, so only half of the product would be the desired ketone. OH Because the ketone has identical substituents bonded to the carbonyl carbon, it has only one enol tautomer. 14. a. CH2 CCH3 188 Chapter 6 OH OH and CH3CH2C H C CH2CH2CH3 E HO C CH3CH2 E OH OH and CH3C C CH2CH3 CH3CH2 Z Because each enol has identical groups on one of its sp2 carbons, E and Z isomers are not possible for either enol. O H HO C C H H3C Z CH2CH3 C OH CHCH2CH3 CH2CH2CH3 E and Z isomers are possible for each of these enols. b. CH3CH H CCH2CH2CH3 OH C C H3C c. CH2 C O 15. a. (1) CH3CH2CCH3 O b. (1) CH3CH2CCH3 O (2) CH3CH2CH2CH O (2) CH3CH2CCH3 O c. (1) CH3CH2CH2CCH3 O and CH3CH2CCH2CH3 O (2) CH3CH2CH2CCH3 and CH3CH2CCH2CH3 major product because there is less steric hindrance to addition of BH3 at the 2-position 16. Ethyne (acetylene) An alkyne can form an aldehyde only if the OH group adds to a terminal sp carbon. When water adds to a terminal alkyne, the proton adds to the terminal sp carbon. Therefore, the only way the OH group can add to a terminal sp carbon is if there are two terminal sp carbons in the alkyne. In other words, the alkyne must be ethyne. a. CH3CH2CH2C 1-pentyne CH or CH3CH2C CCH3 H2 Pt/C 17. CH3CH2CH2CH2CH3 2-pentyne Chapter 6 b. CH3C CCH3 H2 Lindlar catalyst 189 H3C C H CH3CH2 C CH3 H H C CH3 Na NH3 2-butyne c. CH3CH2C CCH3 Na NH3 liq C H 2-pentyne d. CH3CH2CH2CH2C 1-hexyne CH H2 Lindlar catalyst or CH3CH2CH2CH2CH CH2 18. The base used to remove a proton must be stronger than the base that is formed as a result of proton removal. A terminal alkyne has a pKa ~ 25. Therefore, the base used to remove a proton from a terminal alkyne must be a stronger base than the terminal alkyne. In other words, any base whose conjugate acid has a pKa greater than 25 can be used. The electronegativities of carbon atoms decrease in the order: sp > sp2 > sp3. The more electronegative the carbon atom, the less stable it will be with a positive charge. a. CH3 CH 2 + 19. b. H 2 C CH + 20. The reaction will not favor products because the carbanion that would be formed is a stronger base than the amide ion. (Recall that the equilibrium favors reaction of the strong and formation of the weak; Section 1.19) CH3 CH3 pK a > 60 weaker acid weaker base + - NH 2 CH3 C H 2 - + NH3 pK a = 36 stronger base stronger acid - 21. a. CH3 CH 2 CH 2 CH 2 - - > CH3 CH 2 CH CH > CH3 CH 2 C C- b. 22. 23. NH 2 > CH3 C C- > CH3 CH2 O - > F - Solved in the text. a. HC CH - 1. NH 2 2. CH 3CH 2CH 2Br CH3CH2CH2C CH b. HC CH - 1. NH2 2. CH3Br - 1. NH2 2. CH3Br CH3C CH H2 /Lindlar catalyst or Na/NH3 liq - 1. NH2 2. CH3Br CH3CH CH2 H3C C H C H CH3 c. HC CH CH3C CH CH3C CCH3 H2 Lindlar catalyst 190 Chapter 6 O 1. disamylborane 2. HO, H2O2, H2O HBr - d. product of a e. product of b CH3CH2CH2CH2CH CH3CHCH3 Br Cl f. HC CH 1. NH2 2. CH3Br - CH3C CH excess HCl CH3CCH3 Cl Cl Cl 24. a. CH3CH2CCH3 Cl Cl b. CH3CH2CH2CCH2CH3 Cl Cl + CH3CH2CCH2CH2CH2CH3 Cl c. CH3CH2CH2CCH2CH2CH3 Cl 25. a. CH3C CCH2CH2CH3 CCHCH2CH2CH3 CH2CH3 g. BrC h. CCH2CH2CH3 b. CH3CH2C HC CC 2 Br c. CH3 C CH i. CH3 CH 2 C CCH 2 CH3 CH3 d. CH2 CHC CH CH3 CCCH3 CH3 C CH CH3 j. CH3CC CH3 e. CH3 C C k. CH3 CH3 CCHCH2CH3 f. CH3CC CH3 + l. CH3C CCH2CHCHCH3 CH3 26. CH3CH2C CH2 + electrophile Cl nucleophile - CH3CH2C Cl CH2 Chapter 6 CH3C CH + H Br CH3C + 191 CH2 + Br- nucleophile electrophile - CH3C C H + NH2 electrophile nucleophile CH3C C - + NH3 27. a. 5-bromo-2-hexyne b. 5-methyl-2-octyne O c. 5,5-dimethyl-2-hexyne d. 6-chloro-2-methyl-3-heptyne O RCH CH2 e. 1,6-cyclooctadiene f. 1,6-dimethyl-1,3-cylohexadiene 28. RCH2CH3 RCH CH2 RCOOH excess H2, Pd/C RCHCH3 Br H2/Lindlar catalyst or Na/NH3 1. H2/Lindlar catalyst or Na/NH3 2. HBr Br excess HBr RC CH H2O/H2SO4 + HgSO4 RCCH3 Br 1 equiv HBr 1. disiamylborane 2. HO-, H2O2, H2O O RCCH3 O RC Br CH2 RCH2CH 29. Al named only one correctly. a. 4-methyl-2-hexyne b. 7-bromo-3-heptyne HC CCH2CH2CH2CH2CH3 1-heptyne pentylacetylene CH3CH2CH2CHC CH c. correct d. 2-pentyne CH3C CCH2CH2CH2CH3 CH3CH2C CCH2CH2CH3 30. 2-heptyne butylmethylacetylene CH3CH2CHCH2C CH 3-heptyne ethylpropylacetylene CH3CHCH2CH2C CH CH3 3-methyl-1-hexyne CH3CH2CHC CCH3 CH3 4-methyl-1-hexyne CH3CHCH2C CCH3 CH3 5-methyl-1-hexyne isopentylacetylene CH3CHC CCH2CH3 CH3 4-methyl-2-hexyne sec-butylmethylacetylene CH3 5-methyl-2-hexyne isobutylmethylacetylene CH3 2-methyl-3-hexyne ethylisopropylacetylene 192 Chapter 6 CH3 CCH3 CH3CC CH3 CH3CCH2C CH CH3 CH3CH2CC CH CH3 4,4-dimethyl-2-pentyne tert-butylmethylacetylene CH2CH3 CH3CH2CHC CH CH3 4,4-dimethyl-1-pentyne CH3 CH3CHCHC CH3 CH CH3 3,3-dimethyl-1-pentyne tert-pentylacetylene 3-ethyl-1-pentyne 3,4-dimethyl-1-pentyne 31. Addition of Br2 to the triple bond in the presence of H2O forms an enol that immediately tautomerizes to a ketone. O CH3CH2CH2C CH Br2 H2O CH3CH2CH2C OH CHBr CH3CH2CH2CCH2Br O 32. a. CH3CH2CH2C b. CH 1. disiamylborane - 2. HO, H O , H O 2 2 2 CH3CH2CH2CH2CH CH3CH2CH CH2 1. BH3 - 2. HO, H2O2, H2O CH3CH2CH2CH2OH O c. CH3CH2CH2C CCH2CH2CH3 + H2O H2SO4 CH3CH2CH2CCH2CH2CH2CH3 This symmetrical alkyne will give the greatest yield of the desired ketone. Because the reactant is not a terminal alkyne, the reaction can take place without the mercuric ion catalyst. 33. 34. a. H 2 , Lindlar catalyst b. Na, NH3 (liq) c. excess H2 , Pt / C a. 1-octen-6-yne b. cis-3-hexen-1-ol c. 1,5-octadiyne d. 5-chloro-1,3-cyclohexadiene e. 1-methyl-1,3,5-heptatriene 35. The molecular formula of the hydrocarbon is C32 H56 . C n H 2n+2 = C32 H 66 With one triple bond, two double bonds, and one ring, the degree of unsaturation is 5. Therefore, the compound is missing 10 hydrogens from C n H 2n +2 . Chapter 6 O 193 36. a. CH2 Br CCH3 Br e. CH3CCH3 O i. CH3 CH C 2 b. CH3CCH3 Br f. CH3CH2CH j. CH3 C C - c. BrHC Br Br CCH3 Br g. CH3 CH 2 CH3 k. CH3C CCH2 CH2 CH2 CH2 CH3 d. BrCHCCH3 Br h. CH3CH C 2 O H CH3 C C H 37. a. CH3CH CCH3 Br Br e. CH3CCH2CH3 O i. H3C b. CH3CH2CCH3 Br Br f. CH3CCH2CH3 j. no reaction c. CH3C Br Br CCH3 g. CH3 CH 2 CH 2 CH3 k. no reaction Br CCH3 Br CH H3C CH3 C C H CH3CHCH CH3 H2O H2SO4 d. CH3C Br h. H H2 38. a. 1. CH3CHC CH3 Lindlar catalyst CH2 OH CH3CCH2CH3 CH3 H2O H CH3CCH2CH3 CH3 + CH3CCHCH3 CH3 + 194 Chapter 6 2. CH3CHC CH3 CH H2 Lindlar catalyst CH3CHCH CH3 CH2 1. BH3 2. H2O2, HO-, 2 CH3CHCH2CH2OH CH3 b. 3-Methyl-2-butanol will be a minor product obtained from both 1. and 2. 3-Methyl-2-butanol will be obtained from 1. because water can attack the secondary carbocation before it has a chance to rearrange to the tertiary carbocation. 3-Methyl-2-butanol will be obtained from 2. because in the second step of the synthesis boron can also add to the other sp2 carbon; it will be a minor product because the transition state for its formation is less stable than the transition state leading to the major product. Because a carbocation is not formed as an intermediate, a carbocation rearrangement cannot occur. (The proton cannot add to the other sp2 carbon in the second step of part 1. because that would require the formation of a primary carbocation in the second step of the synthesis. Primary carbocations are so unstable that they can never be formed.) OH CH3CHCH2CHCH3 3-methyl-2-butanol 39. Three of the names are correct. a. 3-heptyne c. correct b. 5-methyl-3-heptyne d. 6,7-dimethyl-3-octyne only c and e are keto-enol tautomers. O H2O, H2SO4 HgSO4 1. NaNH2 2. CH3CH2Br CH3 e. correct f. correct 40. 41. a. HC b. HC CH CH CH2 CHOH enol CH CH3CH H2 Lindlar catalyst CH3CH2C CH3CH2CH CH2 Br2 CH2Cl2 CH3CH2CHCH2Br Br O 1. NaNH2 2. CH3Br H2O, H2SO4 HgSO4 c. HC CH HC CCH3 CH3CCH3 Chapter 6 d. HC CH 1. NaNH2 2. CH3Br 195 HC CCH3 1. NaNH2 2. CH3CH2Br CH3C CCH2CH3 Na NH3 e. HC CH 1. NaNH2 2. CH3Br HC CCH3 1. NaNH2 2. CH3CH2Br CH3C CCH2CH3 Lindlar H2 catalyst f. HC CH 1. NaNH2 2. CH3Br HC CCH3 1. NaNH2 2. CH3CH2Br CH3C CCH2CH3 H2 Pd/C 42. a. Syn addition of H2 forms cis-2-butene. When Br2 adds to cis-2-butene, the threo pair of enantiomers is formed. CH3 H Br CH3 Br H Br H CH3 CH3 H Br b. Reaction with sodium and liquid ammonia forms trans-2-butene. And when Br2 adds to trans-2butene, a meso compound is formed. CH3 H H CH3 Br Br c. Anti addition of Cl2 forms trans-2,3-dichloro-2-butene. And when Br2 adds to trans-2,3-dichloro-2butene, a meso compound is formed. CH3 Cl Cl CH3 Br Br 196 Chapter 6 O O O 43. a. CH3CH2CCH3 b. CH3CH2CH2CCH3 1. NaNH2 2. CH3CH2CH2CH2Br c. O d. CH 44. a. HC CH CH3CH2CH2CH2C H2O H2SO4 HgSO4 CH O CH3CH2CH2CH2CCH3 H2 Lindlar catalyst or Na/NH3 liq b. HC CH 1. NaNH2 2. CH3CH2Br CH3CH2C CH CH3CH2CH HBr CH2 CH3CH2CHCH3 H2 Lindlar catalyst or Na/NH3 liq Br c. HC CH 1. NaNH2 2. CH3CH2CH2Br CH3CH2CH2C CH CH3CH2CH2CH H2O H2SO4 CH2 CH3CH2CH2CHCH3 OH O 1. disiamylborane 2. HO-, H2O2, H2O d. C CH CH2CH O e. C CH H2O, H2SO4 HgSO4 C H C C CH3 H CH3 f. C CCH3 H2 Lindlar catalyst Chapter 6 45. She can make 3-octyne by using 1-hexyne instead of 1-butyne. She would then need to use ethyl bromide (instead of butyl bromide) for the alkylation step: CH3 CH 2 CH 2 CH 2 C C 197 1. NaNH2 2. CH3CH2Br CH3 CH 2 CH 2 CH 2 C CCH 2 CH3 Or she could make the 1-butyne she needed by alkylating ethyne: HC CH 1. NaNH2 2. CH3CH2Br CH3CH2C CH 1. NaNH2 2. CH3CH2CH2CH2Br CH3CH2CH2CH2C CCH2CH3 46. a. Only one product is obtained from the hydroboration-oxidation of 2-butyne because the alkyne is symmetrical. Two different products can be obtained from hydroboration-oxidation of unsymmetrical 2-pentyne. CH3C CCH3 1. BH3, THF 2. HO-, H2O2, H2O CH3C CCH2CH3 1. BH3, THF 2. HO-, H2O2, H2O O CH3CH2CCH3 O O CH3CCH2CH3CH3 + CH3CH2CCH2CH3 b. Only one product will be obtained from hydroboration-oxidation of any symmetrical alkyne, such as 3-hexyne or 4-octyne. CH3 CH 2 C CCH 2 CH3 3-hexyne CH3 CH 2 CH 2 C CCH 2 CH 2 CH3 4-octyne 47. a. The first step forms a trans alkene. Syn addition to a trans alkene forms the threo pair of enantiomers. CH2CH3 H D D H CH2CH3 D H CH2CH3 H D CH2CH3 b. The first step forms a cis alkene. Syn addition to a cis alkene forms the erythro pair of enantiomers, but since each asymmetric carbon is bonded to the same four groups, the product is a meso compound. CH2CH3 H H D D CH2CH3 198 48. Chapter 6 There are two reasons why hyperconjugation is less effective in stabilizing a positive charge on a vinylic cation than a positive charge on an alkyl cation. First, an sp2--s bond is stronger than an sp3--s bond, so the sp2 orbitals are less able to donate electron density to the adjacent positively charge carbon. Second, an sp2 carbon has 120 bond angles compared to the 109.5 bond angles of an sp3 carbon, so the orbitals of an sp2 carbon are farther away, which makes them less able to interact with the positively charged carbon. H H C H C H H a vinylic cation + H H C C + H an alkyl cation 49. (3E,6E)-3,7,11-trimethyl-1,3,6,10-dodecatetraene The configuration of the double bond at the 1-position and at the 10-position is not specified because isomers are not possible at those positions since there are 2 hydrogens on C-1 and two methyl groups on C-11. 50. a. HC CH - NH2 HC C- CH3Br HC CCH3 - NH2 -C CCH3 CH3CH2CH2CH2CH2Br H3C C H C CH2CH2CH2CH2CH3 H CH - H2 Lindar catalyst CH3CH2CH2CH2CH2C CCH3 b. HC NH2 HC C- CH3CH2Br HC CCH2CH3 - NH2 -C CCH2CH3 CH3CH2CH2Br CH3CH2 C H C H CH2CH2CH3 CH3CH2Br Na NH3(l) CH3CH2CH2C CCH2CH3 c. HC CH - NH2 HC C- HC CCH2CH3 - NH2 -C CCH2CH3 CH3CH2Br CH3CH2 CH3CH2CHCHCH2CH3 Br OH Br2 H2O CH2CH3 C C H H2 Lindar catalyst CH3CH2C CCH2CH3 H Chapter 6 Chapter 6 Practice Test 1. What reagents could be used to convert the given starting material into the desired product? O 199 a. C CH H CH2CH H C C CH3 b. C CCH3 2. Draw the enol tautomer of the following compound: O 3. Give the structure of a. sec-butyl isobutyl acetylene b. 2-methyl-1,3-cyclohexadiene 4. Indicate whether each of the following statements is true or false: a. A terminal alkyne is more stable than an internal alkyne. b. Propyne is more reactive than propene toward reaction with HBr. c. 1-Butyne is more acidic than 1-butene. d. An sp2 hybridized carbon is more electronegative than an sp3 hybridized carbon. T T T T F F F F 5. Give the systematic name for the following compounds: a. CH3CHC CH3 CCH2CH2Br b. CH3CHC CH3 CCH2CH2OH 200 6. Chapter 6 What alkyne would be the best reagent to use for the synthesis of the following ketone? O CH3CH2CH2CCH3 7. Rank the following compounds in order of decreasing acidity: (Label the most acidic compound #1.) NH3 CH3 C C CH3 CH3 H2 O CH3 CH C 2 8. 9. Give an example of a ketone that has two enol tautomers. Show how the target molecules could be prepared from the given starting materials. a. CH3CH2C CH CH3CH2CH2CH2CH2CH3 b. CH3CH2C CH CH3CH2CHCH2CH2CH3 Br O c. CH3CH2C CH CH3CH2CCH2CH2CH3

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