339 CHAPTER 9 ALKYNES H ydrocarbons that contain a carbon–carbon triple bond are called alkynes. Non- cyclic alkynes have the molecular formula C n H 2nH110022 . Acetylene (HCPCH) is the simplest alkyne. We call compounds that have their triple bond at the end of a carbon chain (RCPCH) monosubstituted, or terminal, alkynes. Disubstituted alkynes (RCPCRH11032) are said to have internal triple bonds. You will see in this chapter that a car- bon–carbon triple bond is a functional group, reacting with many of the same reagents that react with the double bonds of alkenes. The most distinctive aspect of the chemistry of acetylene and terminal alkynes is their acidity. As a class, compounds of the type RCPCH are the most acidic of all sim- ple hydrocarbons. The structural reasons for this property, as well as the ways in which it is used to advantage in chemical synthesis, are important elements of this chapter. 9.1 SOURCES OF ALKYNES Acetylene was first characterized by the French chemist P. E. M. Berthelot in 1862 and did not command much attention until its large-scale preparation from calcium carbide in the last decade of the nineteenth century stimulated interest in industrial applications. In the first stage of that synthesis, limestone and coke, a material rich in elemental car- bon obtained from coal, are heated in an electric furnace to form calcium carbide. Calcium carbide is the calcium salt of the doubly negative carbide ion ( ). Car- bide dianion is strongly basic and reacts with water to form acetylene: CPC H11002H11002 Calcium oxide (from limestone) CaO Carbon (from coke) 3C Carbon monoxide CO 1800–2100°C CaC 2 Calcium carbide H11001H11001 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEM 9.1 Use curved arrows to show how calcium carbide reacts with water to give acetylene. Beginning in the middle of the twentieth century, alternative methods of acetylene production became practical. One of these is based on the dehydrogenation of ethylene. The reaction is endothermic, and the equilibrium favors ethylene at low temperatures but shifts to favor acetylene above 1150°C. Indeed, at very high temperatures most hydro- carbons, even methane, are converted to acetylene. Acetylene has value not only by itself but is also the starting material from which higher alkynes are prepared. Natural products that contain carbon–carbon triple bonds are numerous. Two exam- ples are tariric acid, from the seed fat of a Guatemalan plant, and cicutoxin, a poiso- nous substance isolated from water hemlock. Diacetylene (HCPC±CPCH) has been identified as a component of the hydro- carbon-rich atmospheres of Uranus, Neptune, and Pluto. It is also present in the atmo- spheres of Titan and Triton, satellites of Saturn and Neptune, respectively. 9.2 NOMENCLATURE In naming alkynes the usual IUPAC rules for hydrocarbons are followed, and the suffix -ane is replaced by -yne. Both acetylene and ethyne are acceptable IUPAC names for HCPCH. The position of the triple bond along the chain is specified by number in a manner analogous to alkene nomenclature. PROBLEM 9.2 Write structural formulas and give the IUPAC names for all the alkynes of molecular formula C 5 H 8 . When the ±CPCH group is named as a substituent, it is designated as an ethynyl group. Propyne HCPCCH 3 1-Butyne HCPCCH 2 CH 3 2-Butyne CH 3 CPCCH 3 4,4-Dimethyl-2-pentyne (CH 3 ) 3 CCPCCH 3 Tariric acid CH 3 (CH 2 ) 10 CPC(CH 2 ) 4 COH O X Cicutoxin HOCH 2 CH 2 CH 2 CPC±CPCCH?CHCH?CHCH?CHCHCH 2 CH 2 CH 3 W OH Ethylene CH 2 ?CH 2 Hydrogen H 2 HCPCH Acetylene H11001 heat H11001H11001 Water 2H 2 O Ca(OH) 2 Calcium hydroxide HCPCH AcetyleneCalcium carbide Ca 2H11001 C ? C 2H11002 340 CHAPTER NINE Alkynes Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 9.3 PHYSICAL PROPERTIES OF ALKYNES Alkynes resemble alkanes and alkenes in their physical properties. They share with these other hydrocarbons the properties of low density and low water-solubility. They are slightly more polar and generally have slightly higher boiling points than the corre- sponding alkanes and alkenes. 9.4 STRUCTURE AND BONDING IN ALKYNES: sp HYBRIDIZATION Acetylene is linear, with a carbon–carbon bond distance of 120 pm and carbon–hydro- gen bond distances of 106 pm. Linear geometries characterize the H±CPC±C and C±CPC±C units of ter- minal and internal triple bonds, respectively as well. This linear geometry is responsible for the relatively small number of known cycloalkynes. Figure 9.1 shows a molecular model for cyclononyne in which the bending of the C±CPC±C unit is clearly evi- dent. Angle strain destabilizes cycloalkynes to the extent that cyclononyne is the small- est one that is stable enough to be stored for long periods. The next smaller one, cyclooc- tyne, has been isolated, but is relatively reactive and polymerizes on standing. In spite of the fact that few cycloalkynes occur naturally, they gained recent atten- tion when it was discovered that some of them hold promise as anticancer drugs. (See the boxed essay Natural and “Designed” Enediyne Antibiotics following this section.) An sp hybridization model for the carbon–carbon triple bond was developed in Section 1.18 and is reviewed for acetylene in Figure 9.2. Figure 9.3 maps the electro- static potential in ethylene and acetylene and shows how the second H9266 bond in acety- lene causes a band of high electron density to encircle the molecule. H C C H 120 pm 106 pm106 pm 180° 180° 9.4 Structure and Bonding in Alkynes: sp Hybridization 341 FIGURE 9.1 Molecular model of cyclononyne, showing bending of bond angles associated with triply bonded carbons. This model represents the structure obtained when the strain energy is minimized according to molecular mechanics and closely matches the structure determined ex- perimentally. Notice too the degree to which the staggering of bonds on adjacent atoms governs the overall shape of the ring. Examples of physical proper- ties of alkynes are given in Appendix 1. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website At this point, it’s useful to compare some structural features of alkanes, alkenes, and alkynes. Table 9.1 gives some of the most fundamental ones. To summarize, as we progress through the series in the order ethane → ethylene → acetylene: 1. The geometry at carbon changes from tetrahedral → trigonal planar → linear. 2. The C±C and C±H bonds become shorter and stronger. 3. The acidity of the C±H bonds increases. All of these trends can be accommodated by the orbital hybridization model. The bond angles are characteristic for the sp 3 , sp 2 , and sp hybridization states of carbon and don’t require additional comment. The bond distances, bond strengths, and acidities are related to the s character in the orbitals used for bonding. s Character is a simple concept, being nothing more than the percentage of the hybrid orbital contributed by an s orbital. Thus, an sp 3 orbital has one quarter s character and three quarters p, an sp 2 orbital has one third s and two thirds p, and an sp orbital one half s and one half p. We then use this information to analyze how various qualities of the hybrid orbital reflect those of its s and p contributors. Take C±H bond distance and bond strength, for example. Recalling that an elec- tron in a 2s orbital is, on average, closer to the nucleus and more strongly held than an 342 CHAPTER NINE Alkynes (c)(b)(a) Ethylene Acetylene FIGURE 9.2 The carbon atoms of acetylene are connected by a H9268 H11001 H9266 H11001 H9266 triple bond. Both carbon atoms are sp-hybridized, and each is bonded to a hydrogen by an sp–1s H9268 bond. The H9268 component of the triple bond arises by sp–sp overlap. Each carbon has two p orbitals, the axes of which are perpendicular to each other. One H9266 bond is formed by overlap of the p orbitals shown in (b), the other by overlap of the p orbitals shown in (c). Each H9266 bond contains two electrons. FIGURE 9.3 Electro- static potential maps of eth- ylene and acetylene. The re- gion of highest negative charge (red) is associated with the H9266 bonds and lies be- tween the two carbons in both. This electron-rich re- gion is above and below the plane of the molecule in eth- ylene. Because acetylene has two H9266 bonds, its band of high electron density encircles the molecule. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website electron in a 2p orbital, it follows that an electron in an orbital with more s character will be closer to the nucleus and more strongly held than an electron in an orbital with less s character. Thus, when an sp orbital of carbon overlaps with a hydrogen 1s orbital to give a C±H H9268 bond, the electrons are held more strongly and the bond is stronger and shorter than electrons in a bond between hydrogen and sp 2 -hybridized carbon. Sim- ilar reasoning holds for the shorter C±C bond distance of acetylene compared to eth- ylene, although here the additional H9266 bond in acetylene is also a factor. The pattern is repeated in higher alkynes as shown when comparing propyne and propene. The bonds to the sp-hybridized carbons of propyne are shorter than the corre- sponding bonds to the sp 2 hybridized carbons of propene. An easy way to keep track of the effect of the s character of carbon is to associ- ate it with electronegativity. As the s character of carbon increases, so does that carbon’s apparent electronegativity (the electrons in the bond involving that orbital are closer to carbon). The hydrogens in C±H bonds behave as if they are attached to an increasingly more electronegative carbon in the series ethane → ethylene → acetylene. PROBLEM 9.3 How do bond distances and bond strengths change with elec- tronegativity in the series NH 3 , H 2 O, and HF? The property that most separates acetylene from ethane and ethylene is its acidity. It, too, can be explained on the basis of the greater electronegativity of sp-hybridized carbon compared with sp 3 and sp 2 . H 106 pm 146 pm 121 pm CCCH 3 Propyne C CH 3 H HH 134 pm 151 pm 108 pm C Propene 9.4 Structure and Bonding in Alkynes: sp Hybridization 343 TABLE 9.1 Structural Features of Ethane, Ethylene, and Acetylene Feature Systematic name Molecular formula C±C bond distance, pm C±H bond distance, pm H±C±C bond angles C±C bond dissociation energy, kJ/mol (kcal/mol) C±H bond dissociation energy, kJ/mol (kcal/mol) Hybridization of carbon s character in C±H bonds Approximate acidity as measured by K a (pK a ) Structural formula Ethyne C 2 H 2 120 106 180° 820 (196) 536 (128) sp 50% 10 H1100226 (26) Acetylene C CHH Ethene C 2 H 4 134 110 121.4° 611 (146) 452 (108) sp 2 33% 10 H1100245 (45) Ethylene C H HH H C Ethane C 2 H 6 153 111 111.0° 368 (88) 410 (98) sp 3 25% 10 H1100262 (62) Ethane C H H H H H H C How do the bond dis- tances of molecular models of propene and propyne compare with the experimental values? Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 9.5 ACIDITY OF ACETYLENE AND TERMINAL ALKYNES The C±H bonds of hydrocarbons show little tendency to ionize, and alkanes, alkenes, and alkynes are all very weak acids. The ionization constant K a for methane, for exam- ple, is too small to be measured directly but is estimated to be about 10 H1100260 (pK a 60). H H H H C Methane H H11001 Proton H11001 H H H C H11002 Methide anion (a carbanion) 344 CHAPTER NINE Alkynes NATURAL AND “DESIGNED” ENEDIYNE ANTIBIOTICS B eginning in the 1980s, research directed toward the isolation of new drugs derived from natural sources identified a family of tumor-inhibitory antibiotic substances characterized by novel structures containing a CPC±C?C±CPC unit as part of a 9- or 10-membered ring. With one double bond and two triple bonds (-ene H11001 di- H11001 -yne), these compounds soon became known as enediyne antibiotics. The simplest member of the class is dynemicin A*; most of the other enediynes have even more complicated structures. Enediynes hold substantial promise as anti- cancer drugs because of their potency and selectivity. Not only do they inhibit cell growth, they have a greater tendency to kill cancer cells than they do nor- mal cells. The mechanism by which enediynes act in- volves novel chemistry unique to the CPC±C?C±CPC unit, which leads to a species that cleaves DNA and halts tumor growth. The history of drug development has long been based on naturally occurring substances. Often, how- ever, compounds that might be effective drugs are produced by plants and microorganisms in such small amounts that their isolation from natural sources is not practical. If the structure is relatively simple, chem- ical synthesis provides an alternative source of the drug, making it more available at a lower price. Equally important, chemical synthesis, modification, or both can improve the effectiveness of a drug. Building on the enediyne core of dynemicin A, for example, Professor Kyriacos C. Nicolaou and his associates at the Scripps Research Institute and the University of Cali- fornia at San Diego have prepared a simpler analog that is both more potent and more selective than dynemicin A. It is a “designed enediyne” in that its structure was conceived on the basis of chemical rea- soning so as to carry out its biochemical task. The de- signed enediyne offers the additional advantage of being more amenable to large-scale synthesis. OH OH O OOH CH 3 C C OCH 3 COH O C C O HN Dynemicin A “Designed” enediyne O N S HOCH 2 CH 2 O O 2H11001 H11002 OO H11002 C C C C O *Learning By Modeling contains a model of dynemicin A, which shows that the CPC±C?C±CPC unit can be incorporated into the molecule without much angle strain. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The conjugate base of a hydrocarbon is called a carbanion. It is an anion in which the negative charge is borne by carbon. Since it is derived from a very weak acid, a car- banion such as H11002 : CH 3 is an exceptionally strong base. In general, the ability of an atom to bear a negative charge is related to its elec- tronegativity. Both the electronegativity of an atom X and the acidity of H±X increase across a row in the periodic table. Using the relationship from the preceding section that the effective electronega- tivity of carbon in a C±H bond increases with its s character (sp 3 H11021 sp 2 H11021 sp), the order of hydrocarbon acidity behaves much like the preceding methane, ammonia, water, hydrogen fluoride series. The acidity increases as carbon becomes more electronegative. Ionization of acetylene gives an anion in which the unshared electron pair occupies an orbital with 50% s character. In the corresponding ionizations of ethylene and ethane, the unshared pair occupies an orbital with 33% (sp 2 ) and 25% (sp 3 ) s character, respectively. Terminal alkynes (RCPCH) resemble acetylene in acidity. Although acetylene and terminal alkynes are far stronger acids than other hydro- carbons, we must remember that they are, nevertheless, very weak acids—much weaker than water and alcohols, for example. Hydroxide ion is too weak a base to convert acety- lene to its anion in meaningful amounts. The position of the equilibrium described by the following equation lies overwhelmingly to the left: Because acetylene is a far weaker acid than water and alcohols, these substances are not suitable solvents for reactions involving acetylide ions. Acetylide is instantly converted to acetylene by proton transfer from compounds that contain hydroxyl groups. H11001 Acetylene (weaker acid) K a H11005 10 H1100226 pK a H11005 26 HHCC Hydroxide ion (weaker base) OH H11002 Acetylide ion (stronger base) H C C H11002 H11001 Water (stronger acid) K a H11005 1.8 H11003 10 H1100216 pK a H11005 15.7 H OH (CH 3 ) 3 CCPCH 3,3-Dimethyl-1-butyne K a H11005 3 H11003 10 H1100226 (pK a H11005 25.5) HHCC Acetylene Proton H H11001 H11001 H spC C H11002 Acetylide ion CH 3 CH 3 Ethane K a H11015 10 H1100262 pK a H11015 62 (weakest acid) CH 2 ?CH 2 Ethylene H11015 10 H1100245 H11015 45 HCPCH Acetylene H11005 10 H1100226 H11005 26 (strongest acid) H11021 H11021 CH 4 Methane K a H11015 10 H1100260 pK a H11015 60 (weakest acid) NH 3 Ammonia H1101510 H1100236 H1101536 H 2 O Water 1.8 H11003 10 H1100216 15.7 HF Hydrogen fluoride 3.5 H11003 10 H110024 3.2 (strongest acid) H11021 H11021H11021 9.5 Acidity of Acetylene and Terminal Alkynes 345 The electrostatic poten- tial map of (CH 3 ) 3 CCPCH on Learning By Modeling clearly shows the greater positive char- acter of the acetylenic hydrogen relative to the methyl hydrogens. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Amide ion is a much stronger base than acetylide ion and converts acetylene to its conjugate base quantitatively. Solutions of sodium acetylide (HCPCNa) may be prepared by adding sodium amide (NaNH 2 ) to acetylene in liquid ammonia as the solvent. Terminal alkynes react similarly to give species of the type RCPCNa. PROBLEM 9.4 Complete each of the following equations to show the conjugate acid and the conjugate base formed by proton transfer between the indicated species. Use curved arrows to show the flow of electrons, and specify whether the position of equilibrium lies to the side of reactants or products. (a) (b) (c) (d) SAMPLE SOLUTION (a) The equation representing the acid–base reaction between propyne and methoxide ion is: Alcohols are stronger acids than acetylene, and so the position of equilibrium lies to the left. Methoxide ion is not a strong enough base to remove a proton from acetylene. Anions of acetylene and terminal alkynes are nucleophilic and react with methyl and primary alkyl halides to form carbon–carbon bonds by nucleophilic substitution. Some useful applications of this reaction will be discussed in the following section. 9.6 PREPARATION OF ALKYNES BY ALKYLATION OF ACETYLENE AND TERMINAL ALKYNES Organic synthesis makes use of two major reaction types: 1. Functional group transformations 2. Carbon–carbon bond-forming reactions Both strategies are applied to the preparation of alkynes. In this section we shall see how to prepare alkynes while building longer carbon chains. By attaching alkyl groups to acetylene, more complex alkynes can be prepared. CH 3 CPC±H Propyne (weaker acid) H11001H11001 Propynide ion (stronger base) CH 3 CPC H11002 Methoxide ion (weaker base) OCH 3 H11002 Methanol (stronger acid) H±OCH 3 CH 3 CPCCH 2 OH H11001 NH 2 H11002 CH 2 ?CH 2 H11001 NH 2 H11002 HCPCH H11001 H 2 CCH 3 H11002 CH 3 CPCH H11001 OCH 3 H11002 H11001 Acetylene (stronger acid) K a H11005 10 H1100226 pK a H11005 26 HHCC Amide ion (stronger base) NH 2 H11002 Acetylide ion (weaker base) H C C H11002 H11001 Ammonia (weaker acid) K a H11005 10 H1100236 pK a H11005 36 H NH 2 346 CHAPTER NINE Alkynes Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Reactions that attach alkyl groups to molecular fragments are called alkylation reactions. One way in which alkynes are prepared is by alkylation of acetylene. Alkylation of acetylene involves a sequence of two separate operations. In the first one, acetylene is converted to its conjugate base by treatment with sodium amide. Next, an alkyl halide (the alkylating agent) is added to the solution of sodium acetylide. Acetylide ion acts as a nucleophile, displacing halide from carbon and forming a new carbon–carbon bond. Substitution occurs by an S N 2 mechanism. The synthetic sequence is usually carried out in liquid ammonia as the solvent. Alterna- tively, diethyl ether or tetrahydrofuran may be used. An analogous sequence using terminal alkynes as starting materials yields alkynes of the type RCPCRH11032. Dialkylation of acetylene can be achieved by carrying out the sequence twice. As in other nucleophilic substitution reactions, alkyl p-toluenesulfonates may be used in place of alkyl halides. PROBLEM 9.5 Outline efficient syntheses of each of the following alkynes from acetylene and any necessary organic or inorganic reagents: (a) 1-Heptyne (b) 2-Heptyne (c) 3-Heptyne SAMPLE SOLUTION (a) An examination of the structural formula of 1-heptyne reveals it to have a pentyl group attached to an acetylene unit. Alkylation of acetylene, by way of its anion, with a pentyl halide is a suitable synthetic route to 1-heptyne. 1. NaNH 2 , NH 3 2. CH 3 CH 2 Br 1. NaNH 2 , NH 3 2. CH 3 Br 2-Pentyne (81%) CCH 2 CH 3 CH 3 C Acetylene CHHC 1-Butyne CCH 2 CH 3 HC Sodium acetylide CNaHC H11001 1-Bromobutane CH 3 CH 2 CH 2 CH 2 Br NH 3 1-Hexyne (70–77%) CHCH 3 CH 2 CH 2 CH 2 C Alkyne CRHC Sodium acetylide CNaHC H11001H11001 Alkyl halide RX Sodium halide NaX via CHC R H11002 X Acetylene CHHC Sodium acetylide CNaHCH11001H11001 Sodium amide NaNH 2 Ammonia NH 3 Acetylene HHCC Monosubstituted or terminal alkyne RHCC Disubstituted derivative of acetylene RRH11032CC 9.6 Preparation of Alkynes by Alkylation of Acetylene and Terminal Alkynes 347 NaNH 2 NH 3 CH 3 Br 4-Methyl-1-pentyne CH(CH 3 ) 2 CHCH 2 C 5-Methyl-2-hexyne (81%) CCH 3 (CH 3 ) 2 CHCH 2 CCNa(CH 3 ) 2 CHCH 2 C Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The major limitation to this reaction is that synthetically acceptable yields are obtained only with methyl halides and primary alkyl halides. Acetylide anions are very basic, much more basic than hydroxide, for example, and react with secondary and ter- tiary alkyl halides by elimination. The desired S N 2 substitution pathway is observed only with methyl and primary alkyl halides. PROBLEM 9.6 Which of the alkynes of molecular formula C 5 H 8 can be prepared in good yield by alkylation or dialkylation of acetylene? Explain why the prepa- ration of the other C 5 H 8 isomers would not be practical. A second strategy for alkyne synthesis, involving functional group transformation reactions, is described in the following section. 9.7 PREPARATION OF ALKYNES BY ELIMINATION REACTIONS Just as it is possible to prepare alkenes by dehydrohalogenation of alkyl halides, so may alkynes be prepared by a double dehydrohalogenation of dihaloalkanes. The dihalide may be a geminal dihalide, one in which both halogens are on the same carbon, or it may be a vicinal dihalide, one in which the halogens are on adjacent carbons. Double dehydrohalogenation of a geminal dihalide Double dehydrohalogenation of a vicinal dihalide The most frequent applications of these procedures are in the preparation of terminal alkynes. Since the terminal alkyne product is acidic enough to transfer a proton to amide anion, one equivalent of base in addition to the two equivalents required for double Vicinal dihalide R H X C H X CRH11032 H11001H11001 2NH 3 AmmoniaSodium amide 2NaNH 2 H11001 2NaX Sodium halideAlkyne C C RH11032R Geminal dihalide R H H C X X CRH11032 H11001H11001 2NH 3 AmmoniaSodium amide 2NaNH 2 H11001 2NaX Sodium halideAlkyne C C RH11032R E2 HC C H11002 Acetylide H CH 3 CH 3 CH 2 C Br tert-Butyl bromide HC CH Acetylene H11001 CH 2 CH 3 CH 3 C 2-Methylpropene H11001 Br H11002 Bromide HCPCH Acetylene HCPCNa Sodium acetylide HCPCCH 2 CH 2 CH 2 CH 2 CH 3 1-Heptyne NaNH 2 NH 3 CH 3 CH 2 CH 2 CH 2 CH 2 Br 348 CHAPTER NINE Alkynes Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website dehydrohalogenation is needed. Adding water or acid after the reaction is complete con- verts the sodium salt to the corresponding alkyne. Double dehydrohalogenation of a geminal dihalide Double dehydrohalogenation of a vicinal dihalide Double dehydrohalogenation to form terminal alkynes may also be carried out by heating geminal and vicinal dihalides with potassium tert-butoxide in dimethyl sulfoxide. PROBLEM 9.7 Give the structures of three isomeric dibromides that could be used as starting materials for the preparation of 3,3-dimethyl-1-butyne. Since vicinal dihalides are prepared by addition of chlorine or bromine to alkenes (Section 6.14), alkenes, especially terminal alkenes, can serve as starting materials for the preparation of alkynes as shown in the following example: PROBLEM 9.8 Show, by writing an appropriate series of equations, how you could prepare propyne from each of the following compounds as starting mate- rials. You may use any necessary organic or inorganic reagents. (a) 2-Propanol (d) 1,1-Dichloroethane (b) 1-Propanol (e) Ethyl alcohol (c) Isopropyl bromide SAMPLE SOLUTION (a) Since we know that we can convert propene to propyne by the sequence of reactions all that remains to completely describe the synthesis is to show the preparation of propene from 2-propanol. Acid-catalyzed dehydration is suitable. (CH 3 ) 2 CHOH 2-Propanol CH 3 CH?CH 2 Propene H H11001 heat CH 3 CH?CH 2 Propene CH 3 CHCH 2 Br W Br 1,2-Dibromopropane CH 3 CPCH Propyne Br 2 1. NaNH 2 , NH 3 2. H 2 O Br 2 3-Methyl-1-butyne (52%) CH(CH 3 ) 2 CHC 1,2-Dibromo-3-methylbutane (CH 3 ) 2 CHCHCH 2 Br Br 3-Methyl-1-butene (CH 3 ) 2 CHCH CH 2 1. NaNH 2 , NH 3 2. H 2 O 3NaNH 2 NH 3 H 2 O 1-Decyne (54%) CHCH 3 (CH 2 ) 7 C Sodium salt of alkyne product (not isolated) CNaCH 3 (CH 2 ) 7 C 1,2-Dibromodecane CH 3 (CH 2 ) 7 CHCH 2 Br Br 3NaNH 2 NH 3 H 2 O 3,3-Dimethyl- 1-butyne (56–60%) CH(CH 3 ) 3 CC 1,1-Dichloro-3,3- dimethylbutane (CH 3 ) 3 CCH 2 CHCl 2 Sodium salt of alkyne product (not isolated) CNa(CH 3 ) 3 CC 9.7 Preparation of Alkynes by Elimination Reactions 349 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 9.8 REACTIONS OF ALKYNES We have already discussed one important chemical property of alkynes, the acidity of acetylene and terminal alkynes. In the remaining sections of this chapter several other reactions of alkynes will be explored. Most of them will be similar to reactions of alkenes. Like alkenes, alkynes undergo addition reactions. We’ll begin with a reaction familiar to us from our study of alkenes, namely, catalytic hydrogenation. 9.9 HYDROGENATION OF ALKYNES The conditions for hydrogenation of alkynes are similar to those employed for alkenes. In the presence of finely divided platinum, palladium, nickel, or rhodium, two molar equivalents of hydrogen add to the triple bond of an alkyne to yield an alkane. PROBLEM 9.9 Write a series of equations showing how you could prepare octane from acetylene and any necessary organic and inorganic reagents. Substituents affect the heats of hydrogenation of alkynes in the same way they affect alkenes. Alkyl groups release electrons to sp-hybridized carbon, stabilizing the alkyne and decreasing the heat of hydrogenation. Alkenes are intermediates in the hydrogenation of alkynes to alkanes. The heat of hydrogenation of an alkyne is greater than twice the heat of hydrogenation of the derived alkene. The first hydrogenation step of an alkyne is therefore more exother- mic than the second. Noting that alkenes are intermediates in the hydrogenation of alkynes leads us to consider the possibility of halting hydrogenation at the alkene stage. If partial hydro- genation of an alkyne could be achieved, it would provide a useful synthesis of alkenes. In practice it is a simple matter to convert alkynes to alkenes by hydrogenation in the presence of specially developed catalysts. The one most frequently used is the Lindlar catalyst, a palladium on calcium carbonate combination to which lead acetate and quino- line have been added. Lead acetate and quinoline partially deactivate (“poison”) the cat- alyst, making it a poor catalyst for alkene hydrogenation while retaining its ability to catalyze the addition of hydrogen to alkynes. Alkane RCH 2 CH 2 RH11032 Alkyne CRH11032RC Alkene CHRH11032RCH H 2 catalyst H 2 catalyst H11002H9004H° (hydrogenation) 1-Butyne 292 kJ/mol (69.9 kcal/mol) CHCH 3 CH 2 C 2-Butyne 275 kJ/mol (65.6 kcal/mol) CCH 3 CH 3 C Alkane RCH 2 CH 2 RH11032 Alkyne CRH11032RC Hydrogen 2H 2 H11001 Pt, Pd, Ni, or Rh Hydrogen 2H 2 H11001 4-Methyl-1-hexyne CHCH 3 CH 2 CHCH 2 C CH 3 3-Methylhexane (77%) CH 3 CH 2 CHCH 2 CH 2 CH 3 CH 3 Ni 350 CHAPTER NINE Alkynes The high energy of acetylene is released when it is mixed with oxygen and burned in an oxyacetylene torch. The temperature of the flame (about 3000°C) exceeds that of any other hydrocarbon fuel and is higher than the melting point of iron (1535°C). The structure of quinoline is shown on page 430. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website In subsequent equations, we will not specify the components of the Lindlar palladium catalyst in detail but will simply write “Lindlar Pd” over the reaction arrow. Hydrogenation of alkynes to alkenes yields the cis (or Z) alkene by syn addition to the triple bond. PROBLEM 9.10 Oleic acid and stearic acid are naturally occurring compounds, which can be isolated from various fats and oils. In the laboratory, each can be prepared by hydrogenation of a compound known as stearolic acid, which has the formula CH 3 (CH 2 ) 7 CPC(CH 2 ) 7 CO 2 H. Oleic acid is obtained by hydrogenation of stearolic acid over Lindlar palladium; stearic acid is obtained by hydrogenation over platinum. What are the structures of oleic acid and stearic acid? 9.10 METAL–AMMONIA REDUCTION OF ALKYNES A useful alternative to catalytic partial hydrogenation for converting alkynes to alkenes is reduction by a Group I metal (lithium, sodium, or potassium) in liquid ammonia. The unique feature of metal–ammonia reduction is that it converts alkynes to trans (or E) alkenes whereas catalytic hydrogenation yields cis (or Z) alkenes. Thus, from the same alkyne one can prepare either a cis or a trans alkene by choosing the appropriate reac- tion conditions. PROBLEM 9.11 Sodium–ammonia reduction of stearolic acid (see Problem 9.10) yields a compound known as elaidic acid. What is the structure of elaidic acid? PROBLEM 9.12 Suggest efficient syntheses of (E)- and (Z)-2-heptene from propyne and any necessary organic or inorganic reagents. The stereochemistry of metal–ammonia reduction of alkynes differs from that of catalytic hydrogenation because the mechanisms of the two reactions are different. The mechanism of hydrogenation of alkynes is similar to that of catalytic hydrogenation of alkenes (Sections 6.1 and 6.3). A mechanism for metal–ammonia reduction of alkynes is outlined in Figure 9.4. Na NH 3 (E)-3-Hexene (82%) C CH 3 CH 2 H H CH 2 CH 3 C 3-Hexyne CH 3 CH 2 C CCH 2 CH 3 CH 3 (CH 2 ) 3 C C(CH 2 ) 3 CH 3 5-Decyne H 2 Lindlar Pd (Z)-5-Decene (87%) C CH 3 (CH 2 ) 3 H (CH 2 ) 3 CH 3 H C C OH CH 1-Ethynylcyclohexanol H11001 H 2 Hydrogen Pd/CaCO 3 lead acetate, quinoline 1-Vinylcyclohexanol (90–95%) CH 2 C OH H 9.10 Metal–Ammonia Reduction of Alkynes 351 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The mechanism includes two single-electron transfers (steps 1 and 3) and two proton transfers (steps 2 and 4). Experimental evidence indicates that step 2 is rate- determining, and it is believed that the observed trans stereochemistry reflects the dis- tribution of the two stereoisomeric alkenyl radical intermediates formed in this step. The more stable (E)-alkenyl radical, in which the alkyl groups R and RH11032 are trans to each other, is formed faster than its Z stereoisomer. Steps 3 and 4, which follow, are fast, and the product distribution is determined by the E–Z ratio of radicals produced in step 2. 9.11 ADDITION OF HYDROGEN HALIDES TO ALKYNES Alkynes react with many of the same electrophilic reagents that add to the carbon–car- bon double bond of alkenes. Hydrogen halides, for example, add to alkynes to form alkenyl halides. C R H RH11032 C (Z)-Alkenyl radical (less stable) C R HRH11032 C (E)-Alkenyl radical (more stable) 352 CHAPTER NINE Alkynes H11002 RC?CRH11032 H11001 H±NH 2 ±￡ RC?CHRH11032 H11001 NH 2 Overall Reaction: RCPCRH11032 H11001 2Na H11001 2NH 3 ±￡ RCH?CHRH11032 H11001 2NaNH 2 Alkyne Sodium Ammonia Step 1: Electron transfer from sodium to the alkyne. The product is an anion radical. RCPCRH11032 H11001 Na ±￡ RC?CRH11032 H11001 Na H11001 Alkyne Sodium H11002 Anion radical Sodium ion Step 2: The anion radical is a strong base and abstracts a proton from ammonia. H11002 Anion radical Alkenyl radical H11002 Amide ion Ammonia Step 3: Electron transfer to the alkenyl radical. Alkenyl radical Sodium Sodium ion Alkenyl anion Step 4: Proton transfer from ammonia converts the alkenyl anion to an alkene. H11002 H11002 RC?CHRH11032 H11001 Na ±￡ RC?CHRH11032 H11001 Na H11001 H 2 N±H H11001 RC?CHRH11032 ±￡ RCH?CHRH11032 H11001 H 2 N Ammonia Alkenyl anion Alkene Amide ion Trans alkene Sodium amide FIGURE 9.4 Mechanism of the sodium–ammonia reduction of an alkyne. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The regioselectivity of addition follows Markovnikov’s rule. A proton adds to the car- bon that has the greater number of hydrogens, and halide adds to the carbon with the fewer hydrogens. When formulating a mechanism for the reaction of alkynes with hydrogen halides, we could propose a process analogous to that of electrophilic addition to alkenes in which the first step is formation of a carbocation and is rate-determining. The second step according to such a mechanism would be nucleophilic capture of the carbocation by a halide ion. Evidence from a variety of sources, however, indicates that alkenyl cations (also called vinylic cations) are much less stable than simple alkyl cations, and their involve- ment in these additions has been questioned. For example, although electrophilic addi- tion of hydrogen halides to alkynes occurs more slowly than the corresponding additions to alkenes, the difference is not nearly as great as the difference in carbocation stabili- ties would suggest. Furthermore, kinetic studies reveal that electrophilic addition of hydrogen halides to alkynes follows a rate law that is third-order overall and second-order in hydrogen halide. Rate H11005 k[alkyne][HX] 2 This third-order rate dependence suggests a termolecular transition state, one that involves two molecules of the hydrogen halide. Figure 9.5 depicts such a termolecular process using curved arrow notation to show the flow of electrons, and dashed-line notation to H11001RC CH Alkyne slow fast Hydrogen halide H X Alkenyl cation RC CH 2 H11001 H11001 Halide ion X H11002 Alkenyl halide RC CH 2 X 1-Hexyne CHCH 3 CH 2 CH 2 CH 2 C Hydrogen bromide HBrH11001 2-Bromo-1-hexene (60%) CH 2 Br CH 3 CH 2 CH 2 CH 2 C Alkyne CRH11032RC Hydrogen halide HXH11001 Alkenyl halide X CRH11032RCH 9.11 Addition of Hydrogen Halides to Alkynes 353 H11002 H---X RCPCH ±￡ RC?CH 2 H11001 HX via: RCPCH H±X H±X X (a) (b) H11001 H---X δ δ FIGURE 9.5 (a), Curved ar- row notation and (b) tran- sition-state representation for electrophilic addition of a hydrogen halide HX to an alkyne. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website indicate the bonds being made and broken at the transition state. This mechanism, called Ad E 3 for addition-electrophilic-termolecular, avoids the formation of a very unstable alkenyl cation intermediate by invoking nucleophilic participation by the halogen at an early stage. Nevertheless, since Markovnikov’s rule is observed, it seems likely that some degree of positive character develops at carbon and controls the regioselectivity of addition. In the presence of excess hydrogen halide, geminal dihalides are formed by sequen- tial addition of two molecules of hydrogen halide to the carbon–carbon triple bond. The hydrogen halide adds to the initially formed alkenyl halide in accordance with Markovnikov’s rule. Overall, both protons become bonded to the same carbon and both halogens to the adjacent carbon. PROBLEM 9.13 Write a series of equations showing how you could prepare 1,1- dichloroethane from (a) Ethylene (b) Vinyl chloride (CH 2 ?CHCl) (c) 1,1-Dibromoethane SAMPLE SOLUTION (a) Reasoning backward, we recognize 1,1-dichloroethane as the product of addition of two molecules of hydrogen chloride to acetylene. Thus, the synthesis requires converting ethylene to acetylene as a key feature. As described in Section 9.7, this may be accomplished by conversion of ethylene to a vicinal dihalide, followed by double dehydrohalogenation. A suitable synthesis based on this analysis is as shown: Hydrogen bromide (but not hydrogen chloride or hydrogen iodide) adds to alkynes by a free-radical mechanism when peroxides are present in the reaction mixture. As in the free-radical addition of hydrogen bromide to alkenes (Section 6.8), a regioselectiv- ity opposite to Markovnikov’s rule is observed. 1-Hexyne CHCH 3 CH 2 CH 2 CH 2 C Hydrogen bromide HBrH11001 1-Bromo-1-hexene (79%) CHBrCH 3 CH 2 CH 2 CH 2 CH peroxides CH 2 ?CH 2 Ethylene BrCH 2 CH 2 Br 1,2-Dibromoethane CH 3 CHCl 2 1,1-Dichloroethane HCPCH Acetylene Br 2 1. NaNH 2 2. H 2 O 2HCl 3-Hexyne CCH 2 CH 3 CH 3 CH 2 C Hydrogen fluoride 2HFH11001 3,3-Difluorohexane (76%) F CH 3 CH 2 CH 2 CCH 2 CH 3 F Alkyne CRH11032RC H11001 Alkenyl halide X CRH11032RCH HX HX Geminal dihalide X RCH 2 CRH11032 X 354 CHAPTER NINE Alkynes For further discussion of this topic, see the article “The Electrophilic Addition to Alkynes” in the November 1993 edition of the Journal of Chemical Education (p. 873). Additional commen- tary appeared in the Novem- ber 1996 issue. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 9.12 HYDRATION OF ALKYNES By analogy to the hydration of alkenes, hydration of an alkyne is expected to yield an alcohol. The kind of alcohol, however, would be of a special kind, one in which the hydroxyl group is a substituent on a carbon–carbon double bond. This type of alcohol is called an enol (the double bond suffix -ene plus the alcohol suffix -ol). An important property of enols is their rapid isomerization to aldehydes or ketones under the condi- tions of their formation. The process by which enols are converted to aldehydes or ketones is called keto–enol isomerism (or keto–enol tautomerism) and proceeds by the sequence of pro- ton transfers shown in Figure 9.6. Proton transfer to the double bond of an enol occurs readily because the carbocation that is produced is a very stable one. The positive charge on carbon is stabilized by electron release from oxygen and may be represented in res- onance terms as shown on the following page. Alkyne CRH11032RC Water H 2 OH11001 Enol (not isolated) CRH11032 OH RCH slow fast RH11032 H11005 H; aldehyde RH11032 H11005 alkyl; ketone O RCH 2 CRH11032 9.12 Hydration of Alkynes 355 Overall Reaction: OH RCH CRH11032 ±￡ RCH 2 ±CRH11032 Enol O Step 1: The enol is formed in aqueous acidic solution. The first step of its transformation to a ketone is proton transfer to the carbon–carbon double bond. EnolHydronium ion Water Carbocation Step 2: The carbocation transfers a proton from oxygen to a water molecule, yielding a ketone WaterCarbocation Ketone RCH 2 ±CRH11032H11001 O ±￡ RCH 2 CRH11032 H11001 H±O H11001 OH H H O H11001 H H Hydronium ion O±H H11001 RCH CRH11032 O H11001 RCH±CRH11032 H11001 OH H H H H H H11001 OH Ketone (aldehyde if RH11032H11005H) FIGURE 9.6 Conversion of an enol to a ketone takes place by way of two solvent-mediated proton transfers. A proton is transferred to carbon in the first step, then removed from oxy- gen in the second. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Delocalization of an oxygen lone pair stabilizes the cation. All the atoms in B have octets of electrons, making it a more stable structure than A. Only six electrons are associated with the positively charged carbon in A. PROBLEM 9.14 Give the structure of the enol formed by hydration of 2-butyne, and write a series of equations showing its conversion to its corresponding ketone isomer. In general, ketones are more stable than their enol precursors and are the products actually isolated when alkynes undergo acid-catalyzed hydration. The standard method for alkyne hydration employs aqueous sulfuric acid as the reaction medium and mer- cury(II) sulfate or mercury(II) oxide as a catalyst. Hydration of alkynes follows Markovnikov’s rule; terminal alkynes yield methyl- substituted ketones. PROBLEM 9.15 Show by a series of equations how you could prepare 2-octanone from acetylene and any necessary organic or inorganic reagents. How could you prepare 4-octanone? Because of the regioselectivity of alkyne hydration, acetylene is the only alkyne structurally capable of yielding an aldehyde under these conditions. At one time acetaldehyde was prepared on an industrial scale by this method. Modern methods involve direct oxidation of ethylene and are more economical. 9.13 ADDITION OF HALOGENS TO ALKYNES Alkynes react with chlorine and bromine to yield tetrahaloalkanes. Two molecules of the halogen add to the triple bond. Acetylene CHHC Water H 2 OH11001 Vinyl alcohol (not isolated) CHOHCH 2 Acetaldehyde O CH 3 CH 1-Octyne CCH 2 CH 2 CH 2 CH 2 CH 2 CH 3 HC H 2 OH11001 H 2 SO 4 HgSO 4 2-Octanone (91%) O CH 3 CCH 2 CH 2 CH 2 CH 2 CH 2 CH 3 4-Octyne CCH 2 CH 2 CH 3 CH 3 CH 2 CH 2 2 OH11001 H H11001 , Hg 2H11001 4-Octanone (89%) O CH 3 CH 2 CH 2 CH 2 CCH 2 CH 2 CH 3 RCH 2 H11001 OH CRH11032 AB RCH 2 H11001 OH CRH11032 356 CHAPTER NINE Alkynes Mercury(II) sulfate and mer- cury(II) oxide are also known as mercuric sulfate and ox- ide, respectively. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A dihaloalkene is an intermediate and is the isolated product when the alkyne and the halogen are present in equimolar amounts. The stereochemistry of addition is anti. 9.14 OZONOLYSIS OF ALKYNES Carboxylic acids are produced when alkynes are subjected to ozonolysis. Ozonolysis is sometimes used as a tool in structure determination. By identifying the carboxylic acids produced, we can deduce the structure of the alkyne. As with many other chemical methods of structure determination, however, it has been superseded by spectroscopic methods. PROBLEM 9.16 A certain hydrocarbon had the molecular formula C 16 H 26 and contained two triple bonds. Ozonolysis gave CH 3 (CH 2 ) 4 CO 2 H and HO 2 CCH 2 CH 2 CO 2 H as the only products. Suggest a reasonable structure for this hydrocarbon. 9.15 SUMMARY Section 9.1 Alkynes are hydrocarbons that contain a carbon–carbon triple bond. Sim- ple alkynes having no other functional groups or rings have the general formula C n H 2nH110022 . Acetylene is the simplest alkyne. Section 9.2 Alkynes are named in much the same way as alkenes, using the suffix -yne instead of -ene. 1. O 3 2. H 2 O HOCOH O Carbonic acid1-Hexyne CHCH 3 CH 2 CH 2 CH 2 C Pentanoic acid (51%) CH 3 CH 2 CH 2 CH 2 CO 2 H H11001 CRH11032RC 1. O 3 2. H 2 O RCOH O HOCRH11032 O H11001 CH 3 CH 2 C CCH 2 CH 3 3-Hexyne Br 2 Bromine H11001 (E)-3,4-Dibromo-3-hexene (90%) C CH 3 CH 2 Br Br CH 2 CH 3 C Alkyne CRH11032RC H11001 Halogen (chlorine or bromine) 2X 2 Tetrahaloalkane X CRH11032 X RC X X Propyne CHCH 3 C H11001 Chlorine 2Cl 2 1,1,2,2-Tetrachloropropane (63%) Cl CH 3 CCHCl 2 Cl 9.15 Summary 357 Recall that when carbonic acid is formed as a reaction product, it dissociates to car- bon dioxide and water. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Section 9.3 The physical properties (boiling point, solubility in water, dipole moment) of alkynes resemble those of alkanes and alkenes. Section 9.4 Acetylene is linear and alkynes have a linear geometry of their X±CPC±Y units. The carbon–carbon triple bond in alkynes is com- posed of a H9268 and two H9266 components. The triply bonded carbons are sp- hybridized. The H9268 component of the triple bond contains two electrons in an orbital generated by the overlap of sp-hybridized orbitals on adja- cent carbons. Each to these carbons also has two 2p orbitals, which over- lap in pairs so as to give two H9266 orbitals, each of which contains two electrons. Section 9.5 Acetylene and terminal alkynes are more acidic than other hydrocarbons. They have a K a ’s for ionization of approximately 10 H1100226 , compared with about 10 H1100245 for alkenes and about 10 H1100260 for alkanes. Sodium amide is a strong enough base to remove a proton from acetylene or a terminal alkyne, but sodium hydroxide is not. Sections Table 9.2 summarizes the methods for preparing alkynes. 9.6–9.7 Section 9.8 Like alkenes, alkynes undergo addition reactions. Sections Table 9.3 summarizes reactions that reduce alkynes to alkenes and 9.9–9.10 alkanes. Sections Table 9.4 summarizes electrophilic addition to alkynes. 9.11–9.13 Section 9.14 Carbon–carbon triple bonds can be cleaved by ozonolysis. The cleavage products are carboxylic acids. PROBLEMS 9.17 Write structural formulas and give the IUPAC names for all the alkynes of molecular for- mula C 6 H 10 . 9.18 Provide the IUPAC name for each of the following alkynes: (a) CH 3 CH 2 CH 2 CPCH (b) CH 3 CH 2 CPCCH 3 (c) CH 3 C CH 3 CCHCH(CH 3 ) 2 1. O 3 2. H 2 O HOCCH 3 O Acetic acid2-Hexyne CCH 3 CH 3 CH 2 CH 2 C Butanoic acid O CH 3 CH 2 CH 2 COH H11001 Sodium amide NaNH 2 H11001 Ammonia NH 3 H11001 1-Butyne CHCH 3 CH 2 C Sodium 1-butynide CNaCH 3 CH 2 C 4,4-Dimethyl-2-pentyne 358 CHAPTER NINE Alkynes Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) (e) (f) (g) (CH 3 ) 3 CCPCC(CH 3 ) 3 CH 3 CH 2 CH 2 CH 2 CHCH 2 CH 2 CH 2 CH 2 CH 3 C CCH 3 CH 2 C CCH 2 CH 2 CH 2 CH 2 C CH Problems 359 TABLE 9.2 Preparation of Alkynes Reaction (section) and comments Alkylation of acetylene and terminal alkynes (Section 9.6) The acidity of acetylene and terminal alkynes permits them to be converted to their conjugate bases on treatment with sodium amide. These anions are good nucleophiles and react with methyl and primary alkyl halides to form carbon–carbon bonds. Secondary and tertiary alkyl halides cannot be used, because they yield only elimination products under these conditions. Double dehydrohalogenation of vicinal dihalides (Section 9.7) Dihalides in which the halogens are on adjacent carbons undergo two elimination processes analogous to those of geminal dihalides. Double dehydrohalogenation of geminal dihalides (Section 9.7) An E2 elimination reaction of a geminal dihalide yields an alkenyl halide. If a strong enough base is used, sodium amide, for example, a second elimination step follows the first and the alkenyl halide is converted to an alkyne. General equation and specific example Alkyne RCPCH H11001H11001 Sodium amide NaNH 2 Sodium alkynide RCPCNa Ammonia NH 3 Sodium alkynide RCPCNa H11001H11001 Primary alkyl halide RH11032CH 2 X Alkyne RCPCCH 2 RH11032 Sodium halide NaX Geminal dihalide RC±CRH11032 H W W H X W W X H11001H11001 Sodium amide 2NaNH 2 Alkyne RCPCRH11032 Sodium halide 2NaX 3,3-Dimethyl-1-butyne (CH 3 ) 3 CCPCH 4,4-Dimethyl-2- pentyne (96%) (CH 3 ) 3 CCPCCH 3 1. NaNH 2 , NH 3 2. CH 3 I Vicinal dihalide RC±CRH11032 H W W X H W W X H11001H11001 Sodium amide 2NaNH 2 Alkyne RCPCRH11032 Sodium halide 2NaX 1,1-Dichloro-3,3- dimethylbutane (CH 3 ) 3 CCH 2 CHCl 2 3,3-Dimethyl-1- butyne (56–60%) (CH 3 ) 3 CCPCH 1. 3NaNH 2 , NH 3 2. H 2 O 1,2-Dibromobutane CH 3 CH 2 CHCH 2 Br W Br 1-Butyne (78–85%) CH 3 CH 2 CPCH 1. 3NaNH 2 , NH 3 2. H 2 O Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 9.19 Write a structural formula or build a molecular model of each of the following: (a) 1-Octyne (b) 2-Octyne (c) 3-Octyne (d) 4-Octyne (e) 2,5-Dimethyl-3-hexyne (f) 4-Ethyl-1-hexyne (g) Ethynylcyclohexane (h) 3-Ethyl-3-methyl-1-pentyne 9.20 All the compounds in Problem 9.19 are isomers except one. Which one? 9.21 Write structural formulas for all the alkynes of molecular formula C 8 H 14 that yield 3-ethylhexane on catalytic hydrogenation. 360 CHAPTER NINE Alkynes TABLE 9.3 Conversion of Alkynes to Alkenes and Alkanes Reaction (section) and comments Hydrogenation of alkynes to alkanes (Section 9.9) Alkynes are completely hydrogenated, yielding alkanes, in the presence of the customary metal hydrogenation catalysts. Metal-ammonia reduction (Section 9.10) Group I metals—sodium is the one usually employed—in liquid ammonia as the solvent convert alkynes to trans alkenes. The reaction proceeds by a four-step sequence in which electron-transfer and proton-transfer steps alternate. Hydrogenation of alkynes to alkenes (Section 9.9) Hydrogenation of alkynes may be halted at the alkene stage by using special catalysts. Lindlar palladium is the metal catalyst employed most often. Hydrogenation occurs with syn stereochemistry and yields a cis alkene. General equation and specific example Alkyne RCPCRH11032H11001 Hydrogen 2H 2 Alkane RCH 2 CH 2 RH11032 metal catalyst Alkyne RCPCRH11032H11001 Hydrogen H 2 Cis alkene C?C RH11032 H R H ± ± ± ± Lindlar Pd cis-2-Heptene (59%) C?C CH 2 CH 2 CH 2 CH 3 H H 3 C H ± ± ± ± 2-Heptyne CH 3 CPCCH 2 CH 2 CH 2 CH 3 H 2 Lindlar Pd trans-2-Hexene (69%) C?C H CH 2 CH 2 CH 3 H 3 C H ± ± ± ± 2-Hexyne CH 3 CPCCH 2 CH 2 CH 3 Na NH 3 2H 2 , Pt Cyclodecyne Cyclodecane (71%) Alkyne RCPCRH11032H11001 Sodium 2Na H11001 Ammonia 2NH 3 H11001 Sodium amide 2NaNH 2 Trans alkene C?C H RH11032 R H ± ± ± ± Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 9.22 An unknown acetylenic amino acid obtained from the seed of a tropical fruit has the mo- lecular formula C 7 H 11 NO 2 . On catalytic hydrogenation over platinum this amino acid yielded homoleucine (an amino acid of known structure shown here) as the only product. What is the structure of the unknown amino acid? Homoleucine CH 3 CH 2 CHCH 2 CHCO H11002 CH 3 H11001 NH 3 O Problems 361 TABLE 9.4 Electrophilic Addition to Alkynes Reaction (section) and comments Addition of hydrogen halides (Section 9.11) Hydrogen halides add to alkynes in accordance with Markovnikov’s rule to give alkenyl halides. In the presence of 2 eq of hydrogen halide, a second addition occurs to give a geminal dihalide. Halogenation (Section 9.13) Addition of 1 equivalent of chlorine or bromine to an alkyne yields a trans dihaloalkene. A tetrahalide is formed on addition of a second equivalent of the halogen. Acid-catalyzed hydration (Section 9.12) Water adds to the triple bond of alkynes to yield ketones by way of an unstable enol intermediate. The enol arises by Markovnikov hydration of the alkyne. Enol formation is followed by rapid isomerization of the enol to a ketone. General equation and specific example Alkyne RCPCRH11032 Alkenyl halide RCH?CRH11032 W X HX HX Geminal dihalide RCH 2 CRH11032 X W W X Propyne CH 3 CPCH H11001 Hydrogen bromide 2HBr 2,2-Dibromo- propane (100%) CH 3 CCH 3 Br W W Br Propyne CH 3 CPCH H11001 Chlorine 2Cl 2 1,1,2,2-Tetrachloro- propane (63%) CH 3 CCHCl 2 Cl W W Cl Alkyne RCPCRH11032H11001 H 2 SO 4 Hg 2H11001 Water H 2 O Ketone RCH 2 CRH11032 O X 1-Hexyne HCPCCH 2 CH 2 CH 2 CH 3 H11001 H 2 SO 4 HgSO 4 Water H 2 O 2-Hexanone (80%) CH 3 CCH 2 CH 2 CH 2 CH 3 O X Alkyne RCPCRH11032 Dihaloalkene C?C X RH11032 R X ± ± ± ± X 2 X 2 Tetrahaloalkane RC±CRH11032 X W W X X W W X Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 9.23 Show by writing appropriate chemical equations how each of the following compounds could be converted to 1-hexyne: (a) 1,1-Dichlorohexane (c) Acetylene (b) 1-Hexene (d) 1-Iodohexane 9.24 Show by writing appropriate chemical equations how each of the following compounds could be converted to 3-hexyne: (a) 1-Butene (b) 1,1-Dichlorobutane (c) Acetylene 9.25 When 1,2-dibromodecane was treated with potassium hydroxide in aqueous ethanol, it yielded a mixture of three isomeric compounds of molecular formula C 10 H 19 Br. Each of these compounds was converted to 1-decyne on reaction with sodium amide in dimethyl sulfoxide. Iden- tify these three compounds. 9.26 Write the structure of the major organic product isolated from the reaction of 1-hexyne with (a) Hydrogen (2 mol), platinum (b) Hydrogen (1 mol), Lindlar palladium (c) Lithium in liquid ammonia (d) Sodium amide in liquid ammonia (e) Product in part (d) treated with 1-bromobutane (f) Product in part (d) treated with tert-butyl bromide (g) Hydrogen chloride (1 mol) (h) Hydrogen chloride (2 mol) (i) Chlorine (1 mol) (j) Chlorine (2 mol) (k) Aqueous sulfuric acid, mercury(II) sulfate (l) Ozone followed by hydrolysis 9.27 Write the structure of the major organic product isolated from the reaction of 3-hexyne with (a) Hydrogen (2 mol), platinum (b) Hydrogen (1 mol), Lindlar palladium (c) Lithium in liquid ammonia (d) Hydrogen chloride (1 mol) (e) Hydrogen chloride (2 mol) (f) Chlorine (1 mol) (g) Chlorine (2 mol) (h) Aqueous sulfuric acid, mercury(II) sulfate (i) Ozone followed by hydrolysis 9.28 When 2-heptyne was treated with aqueous sulfuric acid containing mercury(II) sulfate, two products, each having the molecular formula C 7 H 14 O, were obtained in approximately equal amounts. What are these two compounds? 9.29 The alkane formed by hydrogenation of (S)-4-methyl-1-hexyne is optically active, but the one formed by hydrogenation of (S)-3-methyl-1-pentyne is not. Explain. Would you expect the products of hydrogenation of these two compounds in the presence of Lindlar palladium to be optically active? 362 CHAPTER NINE Alkynes Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 9.30 All the following reactions have been described in the chemical literature and proceed in good yield. In some cases the reactants are more complicated than those we have so far encoun- tered. Nevertheless, on the basis of what you have already learned, you should be able to predict the principal product in each case. (a) NaCPCH H11001 ClCH 2 CH 2 CH 2 CH 2 CH 2 CH 2 I ±￡ (b) (c) (d) (e) (f) (g) (h) (i) (j) 9.31 The ketone 2-heptanone has been identified as contributing to the odor of a number of dairy products, including condensed milk and cheddar cheese. Describe a synthesis of 2-heptanone from acetylene and any necessary organic or inorganic reagents. 9.32 (Z)-9-Tricosene [(Z)-CH 3 (CH 2 ) 7 CH?CH(CH 2 ) 12 CH 3 ] is the sex pheromone of the female housefly. Synthetic (Z)-9-tricosene is used as bait to lure male flies to traps that contain insecti- cide. Using acetylene and alcohols of your choice as starting materials, along with any necessary inorganic reagents, show how you could prepare (Z )-9-tricosene. 2-Heptanone CH 3 CCH 2 CH 2 CH 2 CH 2 CH 3 O Product of part (i) H 2 Lindlar Pd H11001 O O(CH2)8Cl NaC CCH 2 CH 2 CH 2 CH 3 1. Na, NH 3 2. H 2 O (Z)-CH 3 CH 2 CH 2 CH 2 CH CHCH 2 (CH 2 ) 7 C CCH 2 CH 2 OH H 2 O, H 2 SO 4 HgO CH 3 CHCH 2 CC CH 3 OH CH 3 CH 1. O 3 2. H 2 O OH C CH Cyclodecyne 1. O 3 2. H 2 O C CNa H11001 CH 3 O O CH 3 CH 2 OS CCH 3 Cl Cl KOC(CH 3 ) 3 , DMSO heat 1. excess NaNH 2 , NH 3 2. H 2 O BrCH 2 CHCH 2 CH 2 CHCH 2 Br Br Br Problems 363 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 9.33 Show by writing a suitable series of equations how you could prepare each of the follow- ing compounds from the designated starting materials and any necessary organic or inorganic reagents: (a) 2,2-Dibromopropane from 1,1-dibromopropane (b) 2,2-Dibromopropane from 1,2-dibromopropane (c) 1,1,2,2-Tetrachloropropane from 1,2-dichloropropane (d) 2,2-Diiodobutane from acetylene and ethyl bromide (e) 1-Hexene from 1-butene and acetylene (f) Decane from 1-butene and acetylene (g) Cyclopentadecyne from cyclopentadecene (h) (i) meso-2,3-Dibromobutane from 2-butyne 9.34 Assume that you need to prepare 4-methyl-2-pentyne and discover that the only alkynes on hand are acetylene and propyne. You also have available methyl iodide, isopropyl bromide, and 1,1-dichloro-3-methylbutane. Which of these compounds would you choose in order to perform your synthesis, and how would you carry it out? 9.35 Compound A has the molecular formula C 14 H 25 Br and was obtained by reaction of sodium acetylide with 1,12-dibromododecane. On treatment of compound A with sodium amide, it was converted to compound B (C 14 H 24 ). Ozonolysis of compound B gave the diacid HO 2 C(CH 2 ) 12 CO 2 H. Catalytic hydrogenation of compound B over Lindlar palladium gave compound C (C 14 H 26 ), and hydrogenation over platinum gave compound D (C 14 H 28 ). Sodium–ammonia reduction of compound B gave compound E (C 14 H 26 ). Both C and E yielded O?CH(CH 2 ) 12 CH?O on ozonolysis. Assign structures to compounds A through E so as to be consistent with the observed transformations. 9.36 Use molecular models to compare ±CPCH, ±CH?CH 2 , and ±CH 2 CH 3 with respect to their preference for an equatorial orientation when attached to a cyclohexane ring. One of these groups is very much different from the other two. Which one? Why? 9.37 Try making a model of a hydrocarbon that contains three carbons, only one of which is sp- hybridized. What is its molecular formula? Is it an alkyne? What must be the hybridization state of the other two carbons? (You will learn more about compounds of this type in Chapter 10.) fromC H C H 3 C H C CH and methyl bromide 364 CHAPTER NINE Alkynes Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website