1559T_ch13_247-25811/03/0520:12Pa9e247 EQA 13 Alkynes:The Carbon-Carbon Triple Bond Now that the erties of carbon-carbon double bonds have been examined in detail.it's time to have a brief look at their relatives,carbon-carbon triple bonds.Not surprisingly,what you will find will be very similar to what you've swill make up the main portion of triple strong bases to and synthetically useful cas of carbanions called alkynyl anions. Outline of the Chapter 13-1 Nomenclature 13-2 Structure and Bonding 13-3 Spectroscopy of the Alkynes An unusual effect in the NMR due to electron motion. 13-4 Preparation of Alkynes by Double Elimination 135 Preparation of Alkynes by Alkylation of Alkynyl Anions A synthesis of internal alkynes starting from terminal alkynes 13-6,13-7,13-8 Reactions of Alkynes 13-9 Alkenyl Halides 13-10 Ethyne as an Industrial Starting Material 13-11 Naturally Occurring and Physiologically Active Alkynes Keys to the Chapter 13-1and13-2.Non The alkyne functional ar nclere,5rctwria0dteahneasimcarTtusaocns roup has a geom merism is possible:and in naming the alkynes it is necessary only to indicate the position of the triple bond very strong.polarized carbon-hydrogen bond,making it relatively acidic.Section 13-5 examines some conse quences of that acidity. 247
13 Alkynes: The Carbon–Carbon Triple Bond Now that the properties of carbon–carbon double bonds have been examined in detail, it’s time to have a brief look at their relatives, carbon–carbon triple bonds. Not surprisingly, what you will find will be very similar to what you’ve just seen. Addition reactions will make up the main portion of triple bond chemistry. An important feature is the hydrogen attached to triply bonded carbon: It is unusually acidic, allowing ready removal by strong bases to form a new and synthetically useful class of carbanions called alkynyl anions. Outline of the Chapter 13-1 Nomenclature 13-2 Structure and Bonding 13-3 Spectroscopy of the Alkynes An unusual effect in the NMR due to electron motion. 13-4 Preparation of Alkynes by Double Elimination 13-5 Preparation of Alkynes by Alkylation of Alkynyl Anions A synthesis of internal alkynes starting from terminal alkynes. 13-6, 13-7, 13-8 Reactions of Alkynes 13-9 Alkenyl Halides A brief description of these compounds, and an introduction to reactions of organic compounds catalyzed by transition metals. 13-10 Ethyne as an Industrial Starting Material 13-11 Naturally Occurring and Physiologically Active Alkynes Keys to the Chapter 13-1 and 13-2. Nomenclature, Structure, and Bonding The alkyne functional group has a geometry simpler than that of the alkenes: It is linear. Thus, no cis-trans isomerism is possible; and in naming the alkynes it is necessary only to indicate the position of the triple bond. When the triple bond is at the end of a chain, it is said to be terminal, and it thus possesses a OCqCOH unit. This hydrogen is characterized by unexpectedly high-field absorption in the proton NMR spectrum and by a very strong, polarized carbon–hydrogen bond, making it relatively acidic. Section 13-5 examines some consequences of that acidity. 247 1559T_ch13_247-258 11/03/05 20:12 Page 247
15597.ch13247-25811/02/0520:12Page248 EQA 248.Chapter 13 ALKYNES:THE CARBON-CARBON TRIPLE BOND 13-3 Spectroscopy of the Alkynes signals are often easy to spot because of their characteristic splitting pattems.which arise from long-range cou 风gnR 13-4.Preparation of Alkynes by Double Elimination route to alkynes involves removal of two moles of hydrogen halide by strong base from a you now d in the following way: -CH2-CHBrKCCo Alkane Haloalkane -CH-CH-- CHB:CHBr- -C=C- Alken Alkyn Preparation of Alkynes from Alkynyl Anior ibility of nucleophilic carbanic naalynes (Sction -).ie variety of ine alkynes y be made ry y.v the scheme R-Cec-H →R一C=C-E whereE"is the electrophilic carbon ina primary haloalkane.a strained cyclic ether.ora carbonyl compound 13-6,13-7,and 13-8.Reactions of Alkynes aeingle addition to mae alkee and theadinalan erivThmec anisms.stereochemistry and regiochemistry.are essentially analogous to those you've already seen.So.you ck to uct.This flexibility contributes to the usefuln s of alkynes in synthesis
13-3. Spectroscopy of the Alkynes The high-field position of the hydrogen resonance is surprising at first, but it is the logical result of the cylindrical symmetry of the triple bond, which allows electrons to rotate in a tight circle about its axis. These alkyne signals are often easy to spot because of their characteristic splitting patterns, which arise from long-range coupling to the neighboring nuclei across the triple bond (see, for instance, the alkynyl hydrogen triplet in Fig. 13-5). Other NMR properties of these compounds are pretty normal. Diagnostic bands for CqC and qCOH in the IR spectra of terminal alkynes complement the NMR data, especially when the latter become more complex as a result of overlapping signals. Take note, however, of the weakness or even absence of an IR band for internal CqC triple bonds. 13-4. Preparation of Alkynes by Double Elimination The elimination route to alkynes involves removal of two moles of hydrogen halide by strong base from a dihaloalkane. The most common kind of sequence involves forming the dihaloalkane by addition of halogen to an alkene double bond. Because you already know how to make alkenes, you now have access to alkynes in the following way: 13-5. Preparation of Alkynes from Alkynyl Anions The other major alkyne preparation is based on the easy accessibility of nucleophilic carbanions from terminal alkynes (Section 13-2). So, a wide variety of internal alkynes may be made from any terminal alkyne, via the scheme where “E” is the electrophilic carbon in a primary haloalkane, a strained cyclic ether, or a carbonyl compound. 13-6, 13-7, and 13-8. Reactions of Alkynes Just as you saw with alkenes, alkynes are subject to a variety of addition reactions. These can occur in two stages: a single addition to make an alkene, and then a second addition to give an alkane derivative. The mechanisms, stereochemistry and regiochemistry, are essentially analogous to those you’ve already seen. So, you can look back to the reactions of the previous chapter as points of reference. Only an occasional detail or two will be different. When you choose to stop the addition reaction at the alkene stage, there is often the possibility of picking reagents and conditions to allow for specific formation of either the trans or cis alkene product. This flexibility further contributes to the usefulness of alkynes in synthesis. A special note should be made concerning hydration reactions of alkynes. Remember that hydration of alkenes leads to alcohols. Alkynes can be hydrated, too. Markovnikov addition is achieved with an aqueous Strong base Any electrophile “E” R Nucleophile R C C H C C R C C E Addition, e.g., Br2 Double elimination, e.g., NaNH2 CH CH CHBr CHBr Alkene 1,2-Dihaloalkane C C Alkyne Halogenation, e.g., Br2, hv Elimination, e.g., KOC(CH3)3 CH2 CH2 CH2 CHBr Alkane Haloalkane 248 • Chapter 13 ALKYNES: THE CARBON–CARBON TRIPLE BOND 1559T_ch13_247-258 11/03/05 20:12 Page 248
1559T_ch13_247-25811/03/0520:12Pa9e249 EQA Solutions to Problems.249 stable and isomerize to carbonyl compounds in a reaction called automerism c-c way to the nearby cadetails concerning the process will be upcom wher 13-.Alkenyl Halides onds in alkenyl halides do not ate in the usual substitution reaction they may be converted into carbon metal bonds,thus allowing formation of lithium and Grignard reagents that y09 metals.The Heck reaction utilizes this property to enable the coupling of the halogen-bearing carbon of an nded carbons c a second alkene,gr ienes as products.W stream mechanistic organic chemistry.However.take note of the fact that such processe s now make un an important part of the so-called synthetic toolbox of the pharmaceutical and medicinal chemist:Reac ctions such apuneAndbgospnodhoo6eopKag2pet2nw Solutions to Problems Br 21.(a)C (H入 22.(a)3-Chloro-3-methyl-1-butyne (b)2-Methyl-3-butyn-2-ol (c)4-Propyl-5-hexyn-1-ol (d)trans-3-Penten-1-yne (e)E-5-Methyl-4-(1-methylbutyl)-4-hepten-2-yne (f)cis-1-Ethenyl-2-ethynylcyclopentane 23.Bond strengths:cthynecthene -H bond uses the sp orbital from carbon.which to the carbon nucleus This effect shifts the bond polarity.which follows the same order:CHgreatest in ethyne.In tur.the greater bond polarity
acidic Hg(II) catalyst. Anti-Markovnikov addition occurs via a modified hydroboration–oxidation sequence. Both initially give vinylic alcohols (or enols) as products, but these are kinetically and thermodynamically unstable and isomerize to carbonyl compounds in a reaction called tautomerism. The latter is thermodynamically favorable because of the very strong carbon–oxygen double bond that is formed. It is kinetically rapid because the enol OOH bond is acidic and readily deprotonated, allowing the proton eventually to find its way to the nearby carbon. More details concerning the process will be upcoming when carbonyl compounds are discussed. Note that the hydration of alkynes is a new synthesis of aldehydes and ketones. 13-9. Alkenyl Halides Although the carbon–halogen bonds in alkenyl halides do not participate in the usual substitution reactions, they may be converted into carbon–metal bonds, thus allowing formation of lithium and Grignard reagents that can be used in the synthesis of allylic alcohols by exposure to ketones and aldehydes. The alkenyl halide carbon–halogen bond is also reactive in a variety of transformations involving transition metals. The Heck reaction utilizes this property to enable the coupling of the halogen-bearing carbon of an alkenyl halide to one of the double-bonded carbons of a second alkene, giving rise to dienes as products. We won’t be exploring this kind of chemistry in much depth, partly because it falls outside the scope of mainstream mechanistic organic chemistry. However, take note of the fact that such processes now make up an important part of the so-called synthetic toolbox of the pharmaceutical and medicinal chemist: Reactions such as these have made the syntheses of many types of compounds of therapeutic value much easier than was the case before their discovery and development. Solutions to Problems 21. (a) (b) (c) 22. (a) 3-Chloro-3-methyl-1-butyne (b) 2-Methyl-3-butyn-2-ol (c) 4-Propyl-5-hexyn-1-ol (d) trans-3-Penten-1-yne (e) E-5-Methyl-4-(1-methylbutyl)-4-hepten-2-yne (f) cis-1-Ethenyl-2-ethynylcyclopentane 23. Bond strengths: ethyne ethene ethane. Ethyne COH bond uses the sp orbital from carbon, which overlaps best with the 1s orbital on hydrogen. The high (50%) s character strongly attracts the bonding electrons to the carbon nucleus. This effect shifts these electrons closer to carbon, thereby enhancing the bond polarity, which follows the same order: COH greatest in ethyne. In turn, the greater bond polarity HO Br Cl O H C C H O C C Solutions to Problems • 249 1559T_ch13_247-258 11/03/05 20:12 Page 249
15597.ch13247-25811/02/0520:12Page250 EQA 250.Chapter 13 ALKYNES:THE CARBON-CARBON TRIPLE BOND with the enhanced stability of the conjugate Re evet,that cleavage (to C- nd H ( 2 ond strength e greatest and the bond length smallest in propyne,again as a result of the to alkynes.alkenes,nd alkanes.the three compound types differ in the hybidation of the AC d strengths vary in the same way: CH,C=NH*>CH;CH=NH2*>CH,CH2NH2* A-PA-Pe entyne.Cyclopentene has the most bond enero Notice(Section 13-2)that heats of hydrogenaion or a 35 kcal molond.whercas ths nomal ns for alencs 27.(a)3-Heptyne>1-heptyne (internal more stable than terminal) (b)Stability decreases from left to right.The first two.isomers of propynvlcyclopentane.follow the Imore stable than terminal."The las and has a strain energy in excess of 20 kcal mol 28.Degrees of unsaturation are calculated for each compound. ( H.4 s 2 CHCH adding up to CH Two C atoms are all that are left to account for.To get two degrees of unsaturation,make a triple bond between them (2bonds): 2 CH:CH,-and-C=C-CH CH,-C=C-CH,CH 8=0.9(化3D→CH,next to CH, 8=1.3(m.4HD? =1.5(quintet,2H)CH2.with CH2 groups on both sides 8=1.7 (t.small /H)Aha!How about -Split by- H-C=C-CH2 (Compare Figure 13-5.) =2.2(m.2H)Perhaps the CHa referred to here? cClm The mple So far,you have CH-CH
in ethyne contributes to the acidity of the hydrogen (together with the enhanced stability of the conjugate base, the ethynyl anion, also a result of hybridization effects). It may seem paradoxical to you that the strongest COH bond is the easiest one to deprotonate. Remember, however, that bond strength relates to homolytic cleavage (to C and H), whereas acidity refers to heterolytic cleavage (to C and H). 24. The bond strength should be greatest and the bond length smallest in propyne, again as a result of the sp (50% s character) orbital at C2. 25. In analogy to alkynes, alkenes, and alkanes, the three compound types differ in the hybridization of the nitrogen atom. Acid strengths vary in the same way: CH3CqNH CH3CHPNH2 CH3CH2NH2 26. Stability order is cyclopentene 1,4-pentadiene 1-pentyne. Cyclopentene has the most bonds, which are generally stronger than bonds. 1,4-Pentadiene and 1-pentyne both have two bonds, but the alkyne is of higher energy. Notice (Section 13-2) that heats of hydrogenation for alkynes are 65–70 kcal mol1 , or 32.5–35 kcal mol1 per bond, whereas the normal range for alkenes is 27–30 kcal mol1 (Section 11-7). 27. (a) 3-Heptyne 1-heptyne (internal more stable than terminal) (b) Stability decreases from left to right. The first two, isomers of propynylcyclopentane, follow the order “internal more stable than terminal.” The last, cyclooctyne, despite being internal, is less stable than either as a result of bond angle strain. Make a model: The alkyne carbons cannot have 180° bond angles. The compound has actually been made, but it doesn’t hang around very long and has a strain energy in excess of 20 kcal mol1 . 28. Degrees of unsaturation are calculated for each compound. (a) Hsat 12 2 14; degrees of unsaturation (14 10)/2 2 bonds or rings. NMR looks like an ethyl group: CH3 (t, 1.0) next to CH2 (q, 2.0). Because the molecule has 10 H’s, there must be two equivalent ethyl groups, 2 CH3CH2O, adding up to C4H10. Two C atoms are all that are left to account for. To get two degrees of unsaturation, make a triple bond between them (2 bonds): 2 CH3CH2O and OCqCO F CH3CH2OCqCOCH2CH3 (b) Hsat 14 2 16; degrees of unsaturation (16 12)/2 2 bonds or rings. IR: terminal OCqCH. NMR: 0.9 (t, 3 H) F CH3, next to CH2 1.3 (m, 4 H) F ? 1.5 (quintet, 2H) F CH2, with CH2 groups on both sides 1.7 (t, small J, H) F Aha! How about So far, you have CH3OCH2O and OCH2OCqCH, or C5H8; you need C2H4 more. The simplest way is CH3CH2CH2CH2CH2CqCH, 1-heptyne. HCC Split by CH2 (Compare Figure 13-5.) 2.2 (m, 2H) Perhaps the CH2 referred to here? 250 • Chapter 13 ALKYNES: THE CARBON–CARBON TRIPLE BOND 1559T_ch13_247-258 11/03/05 20:12 Page 250
1559T_ch13_247-25811/03/0520:26Pa9e251 ⊕ EQA Solutions to Problems.251 (e)The molecular formula is CsHO.How?Briefly Carbon.71.4%of 84 60:60/12 (atomic mass of C)=5 Hydrogen.9.6%of 84 =8:8/1 (atomic mass of H)=8 Oxygen,19%(the remainder)of 84=16:16/16(atomic mass of ) Double check using exact masses from Table 11-5 512.00000)+81.00783)+15.9949=84.05754 plest splittin suggest =1.8 (broad s,1 HOH,broad singlet gives it away 8=3.7 (t.2 H)CH2.next OH (chemical shift tells you this).and also next to another CH2 (triplet splitting tells you that) 6=19(t=CH(na owness of spliting is typical)."long-range"coupled to a CH2o the other side of the triple bond. thar'srsn the formula s just put them together:HO-CH CH CH2 29 =C-H of terminal alkyne has vc-3300 cm (a)D-C=CCH2CH2CH2CH2CH2C=C-D (b)C=C-D(vc-p) 30.(a)CH,CH2CH(CH])C=CH (b)CH,OCH,CH,CH2C=CCH, (after aqueous work-up) (R=CHCHCH,-) Mes R (d)The reverse of the meso compound,this gives @ 31.(a)frans-3-octene.via two sequential one-electron reductions as described in Section 13-6: 入入入 (b)Upon sodium/liquid amn duction of a triple bond.th f the doubl electron gives the alkyne radical anion,in which the two substituents on the original alkyne carbons
(c) The molecular formula is C5H8O. How? Briefly: Carbon, 71.4% of 84 60; 60/12 (atomic mass of C) 5 Hydrogen, 9.6% of 84 8; 8/1 (atomic mass of H) 8 Oxygen, 19% (the remainder) of 84 16; 16/16 (atomic mass of O) 1 Double check using exact masses from Table 11-5: 5(12.00000) 8(1.00783) 15.9949 84.05754 Hsat 10 2 12; degrees of unsaturation (12 8)/2 2 bonds or rings. IR: OCqCO stretch shows at 2100 cm1 , broad band from 3200–3500 cm1 suggest OOOH. NMR, focus on the signals with the simplest splitting patterns first: 1.8 (broad s, 1 H) F OH, broad singlet gives it away 3.7 (t, 2 H) F CH2, next OH (chemical shift tells you this), and also next to another CH2 (triplet splitting tells you that) 1.9 (t, 1 H) F CqCH (narrowness of splitting is typical), “long-range” coupled to a CH2 on the other side of the triple bond. Let’s see what we know so far. We have figured out that the molecule contains the two pieces HOOCH2OCH2O and OCH2OCqCH. Add them up and you get C5H8O: that’s all there is in the formula, so just put them together: HOOCH2OCH2OCH2OCqCH. The two middle CH2 groups are responsible for the two signal sets that we didn’t bother to try to interpret because they were more complicated. Try to figure out on your own why they look the way they do. 29. qCOH of terminal alkyne has ˜COH 3300 cm1 . (a) DOCqCCH2CH2CH2CH2CH2CqCOD (b) CqCOD (˜COD) (c) Before reaction, m1 is H (mass 1) and m2 is C9H11 (mass 119). Rewrite the Hooke’s law equation as ˜ 2 k2 f(m1 m2)/m1m2. So (3300)2 k2 f(120/119), or k2 f 1.1 107 . Because k and f are assumed to be constant, use this value for k2 f to predict ˜ 2 for the product. Now m1 is D (mass 2), so ˜ 2 (1.1 107 )(122/240) 5.6 106 and predicted ˜COD 2366 cm1 . The discrepancy of about 10% is typical and due to changes in k and f. 30. (a) CH3CH2CH(CH3)CqCH (b) CH3OCH2CH2CH2CqCCH3 (after aqueous work-up) (c) (d) The reverse of the meso compound, this gives 31. (a) trans-3-octene, via two sequential one-electron reductions as described in Section 13-6: (b) Upon sodium/liquid ammonia reduction of a triple bond, the trans stereochemistry of the double bond that normally results is determined in the first two steps of the mechanism. Addition of one electron gives the alkyne radical anion, in which the two substituents on the original alkyne carbons Z Cl C H C R R E Cl C H C R R ) CH3 (R CH3CHCH2 Rotate Meso Cl C H R Cl C H R H Cl R C Cl C H R OCH3 (anti) Solutions to Problems • 251 1559T_ch13_247-258 11/03/05 20:26 Page 251
15597ch13247-25911/03/0520:12Pag0252 EQA 252.Chapter 13 ALKYNES:THE CARBON-CARBON TRIPLE BOND cause angle,torsional,an other strain effects combine to raise the energy of its trans counterpart H is-C dectenyl Reduced strain) The cis-cyclooctenyl radical is formed preferentially,and ultimately goes on via reduction by a second electron to give cis-cyclooctene. 32.All products are those obtained after aqueous work-up. CH(CH3)2 (a)CH;CH2C=CCHa (b) C C HO、C=CCH OH (c) (d) -CH-C=CCH OH (e)CH,CH-CHz-C=CCH, )H- CH;C=C H 33.Method (d)is the only one that will give a high yield of the correct molecule.Methods (b)and (c)will Method (a)is totally bogus 34.In most cases the answer given is just one of several correct ones. CCIC=CCH.CI 0 (b)HC=CLi CHaCHaCCHa-product
can exist in either a cis or a trans arrangement. In the reduction of an acyclic alkyne, the trans radical anion is the more stable because it suffers from less steric hindrance than the cis isomer. Beginning with cyclooctyne, however, the reverse is the case: The cis radical anion is more stable because angle, torsional, and other strain effects combine to raise the energy of its trans counterpart: The cis-cyclooctenyl radical is formed preferentially, and ultimately goes on via reduction by a second electron to give cis-cyclooctene. 32. All products are those obtained after aqueous work-up. (a) CH3CH2CqCCH3 (b) (c) (d) (e) CH3CHOCH2OCqCCH3 (f) 33. Method (d) is the only one that will give a high yield of the correct molecule. Methods (b) and (c) will give some of the target molecule, together with some regioisomeric alkynes. Method (e), and SN2 with a basic nucleophile on a secondary halide, works to a certain extent but gives much of the E2 product as well. Method (a) is totally bogus. 34. In most cases the answer given is just one of several correct ones. (a) O B (b) HCqCLi CH3CH2CCH3 88n product CH3CH2CH2Br, DMSO 1. NaNH2, NH3 2. CH3CH2Br, DMSO HC CLi CCH HC 2CH2CH3 product CH3C CH3 Blocks top C HO H C CCH3 OH CH HO C CCH3 C CH3 CH(CH3)2 CH3 H Via E2; haloalkane too hindered for SN2 C Na Na H H cis-Cyclooctyne radical anion (Reduced strain) trans-Cyclooctyne radical anion (Severely strained) cis-Cyclooctenyl radical Cyclooctyne (Moderately strained: angles around triple bond are distorted from 180) trans-Cyclooctenyl radical H NH2 H NH2 252 • Chapter 13 ALKYNES: THE CARBON–CARBON TRIPLE BOND OH A 1559T_ch13_247-258 11/03/05 20:12 Page 252
1559T_ch13_247-25811/03/0520:12Pa9e253 ⊕ EQA Solutions to Problems.253 HC=CLi HC-CHCHs-product (d)Reactin of the basic alkynyln witha tertiary Instead.use efm9aaoecnlonhubeeiydiaioegeaioneueacegemcTaesheupteboad 1.Mg OH (CH cc,(C.CC-Cods 35.Priority of D is higher than H.but lower than anything else.Structure is CH2CH3 CH;C=C- y p CH2CHs (S)D- -Br+LiC=CCHs Dtso product CH H D 36.(a C=0 D (b) c-C-# (e)CH:CI-CH2 D BrCH、 (d)CH,CICHs (e) H 0 (g)CH,CClCHI2 (h)CH3CCH O CH-CH-CH 37.In the structures below R=cyclohexyl. R D C=C (b) C=C. (c)RCI=CHR (E and Z) D D R Br (e C-C (f) Br II 0 (g)RCCICCIR (h)and (i)RCCH2R
(c) To get the triple bond one carbon away from the alcohol carbon, use the ring opening of an oxacyclopropane by an alkynyl anion. (d) Reaction of the basic alkynyl anion with a tertiary halide would give elimination. Instead, use a tertiary Grignard reagent to make the necessary carbon–carbon bond by addition to an aldehyde. An elimination–halogenation–double-dehydrohalogenation sequence generates the triple bond. 35. Priority of D is higher than H, but lower than anything else. Structure is Best bond to make is marked (arrow). Synthesis of the optically active product could be achieved if the enantiomerically pure haloalkane could be obtained. 36. (a) (b) (c) CH3CIPCH2 (d) CH3CI2CH3 (e) (f ) O O B B (g) CH3CCl2CHI2 (h) CH3CCH3 (i) CH3CH2CH 37. In the structures below R cyclohexyl. (a) (b) (c) RCICHR (E and Z) (d) RCI2CH2R (e) (f ) II O AA B (g) RCCl2Cl2R RCClCClR (h) and (i) RCCH2R C R I Cl R C C R Br Br R C C R D D R C C D R D R C C H CH3 Cl I C C H CH3 Br Br C C H CH3 D D C C CH3 H D D C H (S)-D CCH3 CH2CH3 Br product LiC SN2 DMSO H CH3C C CH2CH3 D (CH3)3CCl (CH3)3CCHCH3 OH (CH3)3CCH CH2 product 2. CH3CH O 1. Mg 2. LDA, THF 1. PBr3 2. NaNH2, NH3 1. Br2, CCl4 HC O CLi H2C CHCH3 product Solutions to Problems • 253 1559T_ch13_247-258 11/03/05 20:12 Page 253
1559r.ah13247-25811/02/0520:12Page254 EQA 254.Chapter 13 ALKYNES:THE CARBON-CARBON TRIPLE BOND 38.In these,"racemic"means a racemic mixture ofR.R and stereoisomers. R 0 R (a)meso-RCHDCHDR (a)racemic RCHDCHDR (b)racemic RCDCDR (b)meso-RCDCDR Br Br BrBr (rucemie (e)rcm9品 H OH H OH a ki fo HO OH HO OH 39.The only high-yield prec ne with both good vield an selectivin 3-hepaol()dare the problem is elimination from 3.4-dichlorohep obtained as minor products.Finally,addition of chlorine to trans-3-heptene gives 3.4- 4o.acH,Cc=G1盟 product (b)CH,CH.CH.CHC=CH prodoct 9aH,C=cCH也s4CH H C=C BrCCproduct H (d)CC=CCH,CH CH: HE-CH CH、Br (e)CHsC=CCH, h-e CH Mainly (f)CH.CH-CH.C=CCH,CH.CH,product
38. In these, “racemic” means a racemic mixture of R,R and S,S stereoisomers. (a) meso-RCHDCHDR (a) racemic RCHDCHDR (b) racemic RCDCDR (b) meso-RCDCDR AA AA Br Br Br Br (c) racemic (c) racemic (d) (d) (e) meso- (e) racemic 39. The only high-yield precursor is 3-heptyne (g), which gives cis-3-heptene with both good yield and selectivity upon hydrogenation over Lindlar’s catalyst. Eliminations of 3-chloroheptane (a) with base and 3-heptanol (d) with acid are the poorest choices because both will give regioisomeric and stereoisomeric mixtures of cis- and trans-2- and 3-heptenes. Similar eliminations of 4-chloroheptane (b) and 4-heptanol (e) are better because at least the regioisomer problem is gone: Only cis- and trans-3-heptenes can form. Double elimination from 3,4-dichloroheptane (c) gives mostly 3-heptyne (g), but other unsaturated regioisomers are obtained as minor products. Finally, addition of chlorine to trans-3-heptene gives 3,4-dichloroheptane. 40. (a) (b) (c) (d) (e) (f) CH3CH2CH2C CCH2CH2CH3 product HgSO4, H2SO4, H2O CH3C CCH3 HBr product Cl2, CCl4 C C H CH3 CH3 Br Mainly CH3C CCH3 product H2, Pd-BaSO4, quinoline, CH3CH2OH Br2, CCl4 C C H CH3 CH3 H CH3C CCH3 product Na, NH3 Br2, CCl4 C C H CH3 CH3 H CH3CH2CH2CH2C 2HI CH product CH3CH2C 1. HCl 2. HBr CH product D R D R HO OH D D R R HO OH D R R D O D D R R O D R D R H OH D D R R H OH C R R D D C C R R D D C 254 • Chapter 13 ALKYNES: THE CARBON–CARBON TRIPLE BOND 1559T_ch13_247-258 11/03/05 20:12 Page 254
1559T_ch13_247-25811/03/0520:12Pa9e255 EQA Solufionsto Problems255 CH ⊕ asistent with it the doubly deprotonated ethyne 43. →CH2=CHMgB ,人人 上H0 人 ,CH OH OH 人入个 CH)CO-K".CHCOH 上悠N
(g) (h) (i) 41. (a) (b) 42. Ca2 :CqC: , a calcium salt of ethyne, is consistent with its reaction with water to form HCqCH. One could call this material “calcium acetylide” or, perhaps “ethynediylcalcium,” with the “di” referring to the doubly deprotonated ethyne. 43. 44. (CH3)3CO K, (CH3)3COH HO Cl HO HO 1. Br2, CCl4, 2. NaNH2, NH3 PCC, CH2Cl2 product LiNH2, NH3 H2, Lindlar catalyst, CH3CH2OH HBr (1 equiv) Mg, THF HC CLi HC CH CH2 CHBr CH2 CHMgBr OH OH O 1. 2. H, H2O O 1. 2. H, H2O R3P, 100C Br 1% Pd(OCCH3)2 O Br COCH3 R3P, 100C O COCH3 O 1% Pd(OCCH3)2 O product H2, Pd-BaSO4, quinoline, CH3CH2OH 1. HC CLi, THF 2. H, H2O, O C CH THF HB 2 C CH ( ( 2 ( ( product H2O2, OH C C H H B HC CCHCH3 product H2, Pd-BaSO4, quinoline, CH3CH2OH 1. BH3, THF 2. H2O2, HO OH H2C CHCHCH3 OH Solutions to Problems • 255 1559T_ch13_247-258 11/03/05 20:12 Page 255
1559rch13247-25811/02/0520:12Pag0256 EQA 256.Chapter 13 ALKYNES:THE CARBON-CARBON TRIPLE BOND 45. 1.CH-so.CL.py* RCH2OH 2.Nal.HMPA 一RCHal LC=CDRCH-C=CH 1.M4T证 OH RCH.CH,CHCCH,RCH.CH.CH-C(CH)bermoee CHs 46.Work backward from the ozonolysis result. signed to alkenyl and alkynyl :bonds.r fom e on the bydrocaro Whe carbons are required by the molecular formula.The answer is CH-c c=C-H above,and (2)the NMR spectrum would show spin spn coupin(e.the methyl signal would bea doublet). 47.日 HO H HO H
45. *R is so hindered that this special sequence (tosylate n iodide) is necessary to allow SN2 reaction with alkynyl anion to be carried out. **The unhindered triple bond is hydroborated much faster than the hindered (trisubstituted) double bond in the R group. 46. Work backward from the ozonolysis result. Now look at the spectra: The IR shows bands at 1615 cm1 (CPC) and at 2110 cm1 (CqC). Notice, too, that absorptions at 3100 cm1 and 3300 cm1 are present, and may be assigned to alkenyl and alkynyl COH bonds, respectively. The NMR shows four types of hydrogens, with signals in an intensity ratio of 3 : 1 : 1 : 1. A total of six hydrogens are present in the unknown molecule, presumably a CH2 ( 1.9), an alkynyl H ( 2.8), and two alkenyl hydrogens ( 5.2 and 5.3). Can we combine this information with the ozonolysis data? The hydrocarbon shown above, which arises from hydrogenation of the unknown, has the formula C5H8. Our unknown must therefore be C5H6. What pieces have we identified? A CH3O, a OCqCOH, and two alkenyl H’s, adding up to C3H6; so two more carbons are required by the molecular formula. The answer is CH2 B CH3OCOCqCOH An isomeric possibility, CH3OCHPCHOCqCOH, cannot be correct because (1) hydrogenation over Lindlar catalyst would give a straight chain diene, CH3OCHPCHOCHPCH2, not the branched one shown above, and (2) the NMR spectrum would show spin–spin coupling (e.g., the methyl signal would be a doublet). 47. HgCl2 Cl H2O H H H H H H H HgCl HO H HgCl2 Cl O CH3CH H H HO H HgCl H H H HO HgCl O O O 2 HCH CH3C m CH ust have come from 1. O3, CH2Cl2 2. Zn, H, H2O CH3 C H. CH2CH2 C O OH RCH2OH RCH2CH2CH RCH2CH2CH2CCH3 RCH2CH2CH C(CH3)2 CH3 RCH2CH2CH2Br RCH2I RCH2C CH bergamotene 1. CH3 SO2Cl, py* 2. NaI, HMPA LiC CH, DMSO H2SO4, 1. NaBH4, CH3OH 2. PBr3 1. ( )2BH,**THF 2. H2O2, HO 1. Mg, THF 2. CH3CCH3 O 256 • Chapter 13 ALKYNES: THE CARBON–CARBON TRIPLE BOND 1559T_ch13_247-258 11/03/05 20:12 Page 256