15597ch13247-25911/03/0520:12Page247 EQA 13 Alkynes:The Carbon-Carbon Triple Bond what you've just seen.Addition tant feat Outline of the Chapter 13-1 Nomenclature 风 13-2 Structure and Bonding 133 Secroaie2gatekneoermmim 13-4 Preparation of Alkynes by Double Elimination 13-5 Preparation of Alkynes by Alkylation of Alkynyl Anions A synthesis of intemal alkynes starting from terminal alkynes. 13-6,13-7,13-8 Reactions of Alkynes 13-9 Alkenyl Halides 众物dp 13-10 Ethyne as an Industrial Starting Material 13-11 Naturally Occurring and Physiologically Active Alkynes Keys to the Chapter 13-1and13-2 d Bo ding erminal,and it thus posse H unit very str by unexpec5w防and b 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
1559T_ch13_247-25811/03/0520:12Pa9e248 EQA 248.Chapter 13 ALKYNES:THE CARBON-CARBON TRIPLE BOND egAaBect 13-3 py of thee is surprising at first.but it is the logical result of e gen re drical symmetry of the triple bond,which allows electrons to rotate in a tight circle about its axis.These alkyne nd is normal.Diagnostic bands for CCand 13-4. Preparation of Alkynes by Double Elimination to an alkene double bond.Because you already know how to make alkenes.you now have access to alkynes in the following way: eg Brs 一CH,-CHBr- 、cga The other major alky w达出 R-C=-C-H M →R-C=C-E where"E"is the electrophilic carbon ina primary haloalkane.a strained cyclic ether.or a carbonyl compound. y 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 mech- anisms,stere ustry and reg emistry,are essentally analogous to tho you ve alre o.you tions to allow for spec formation of cither the trans or cis alkene prod- ctions of akynes.Remember that hydration of alkenes leads to alcohols.Alkynes can be hydrated,too.Markovnikov addition is achieved with an agucous
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
15597.ch13247-25811/03/0520:12Pag0249 EQA Solutions to Problems249 stable and isomerize to carbonyl compounds in a reaction calledmr -c0- eropey compounds are discussed.Note that the hydration of alkynes is a new synthesis of aldehydes and 13-9.Alkenyl Halides ed in ynt sis of allylic alcohols by exposure to ketones and aldehydes. e ca s invo alkenyl halide to one of the double-bonded carbons of a second alkene.givi rise to diene as products.We won'tbe this kind of chemist y in muc cau eit falls outside the scope of main o the the pharmaceutical and medicinal chemist:Reactions such Solutions to Problems Br 21.(a)C1= (b)= 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-methylbutyI)-4-hepten-2-yne (f)cis-1-Ethenyl-2-ethynylcyclopentane 23.Bond strengths:ethyne>ethene>ethane.Ethyne C-H bond uses the sp orbital from carbon,which overlaps best with
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
1559T_ch13_247-25811/03/0520:12Pa9e250 ⊕ EQA 250.Chapter 13 ALKYNES:THE CARBON-CARBON TRIPLE BOND s to the acidity of the hyd gen(together with the enhanced stability of the conjugate It may se atoyou that the stronges nd is the molytic greatest and the bond length smallest in propyne.again asa result of th 25ehmXahm ths vary in the anes. he three compound types differ in the hybridization of the CHC=NH*CH,CH=NH2*>* -pentyne both have twobonds.but the n 13-2)tha t heats of hydroge 1a0 normal (Section 11-7). 27.(a)3-Heptyne1-heptyne (internal more stable than terminal) (b)Stability decreases from left to right.The first two,isomers of propynylcyclopentane,follow the more s spite eing inter 18bond angles.The compound has b ng around very long 28.Degrees of unsaturation are calculated for each compound. (a) H=12+2=14: be two equivalent ethyl groups.2 CHCHa- -adding up to C4Ho.Two C atoms are all that are lef to account for.To get two degrees of unsaturation,make a triple bond between them (2 bonds) 2 CH:CH,-and-C=C-CH,CH,-C=C-CH,CH =0.9 (t.3 H)CHa.next to CH2 8=1.3(m,4H田→? =1.5(quintet,2H)CH2,with CH2 groups on both side 8=1.7 (t,small J.H)Aha!How about 一spy H-C=C-CH2-(Compare Figure 13-5.) =2.2 (m,2H)Perhaps the CHa referred to here?
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
1559Tch13247-25911/03/0520:26Page251 EQA Solutions to Problems251 (e)The molecular formula is CHO.How?Briefy: Carbon.71.4%of 84 60:60/12 (atomic mass of C)=5 Hydrogen.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: 512.00000)+8(1.00783)+15.9949=84.05754 H.a =10+2=12;degrees of unsaturation =(12-8)2=2 m bonds or rings.IR:-C=C- stretch shows at 2100cm lwith the simplest splitting pattems frs NMK.focus on =1.8 (broad s.1 H)OH,broad singlet gives it away =3.7(tH)CH next OH (chemical shift tells you this),and also next to another CH (trinlet splitting tells you that) the two pieces e in the CHCCH.The two middle CHa groups are 29.=C-H of terminal alkyne has vc-H~3300 cm (a)D-C=CCH,CH,CH,CH,CH,C=C-D (b)C=C-DV) ☒ 30.(a)CH;CH2CH(CHa)C=CH (b)CH.OCH.CH.CH.C=CCH (after aqucous work-up) 99 H (c) C-C (R=CH:CHCH2-) H RH H (anti) R (d)The reverse of the meso compound,this gives =C @ R 31.(a)trans-3-octene.via two sequential one-electon reductions as described in Section 13-6 入入 (b)Upon sodium/liquid ammonia reduction of a triple bond,the trans stereochemistry of the double iginal alkyne cathon
(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