有机化学（英文原版）：Chapt05

167 CHAPTER 5 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS A lkenes are hydrocarbons that contain a carbon–carbon double bond. A car- bon–carbon double bond is both an important structural unit and an important functional group in organic chemistry. The shape of an organic molecule is influ- enced by the presence of this bond, and the double bond is the site of most of the chem- ical reactions that alkenes undergo. Some representative alkenes include isobutylene (an industrial chemical), H9251-pinene (a fragrant liquid obtained from pine trees), and farnesene (a naturally occurring alkene with three double bonds). This chapter is the first of two dealing with alkenes; it describes their structure, bonding, and preparation. Chapter 6 discusses their chemical reactions. 5.1 ALKENE NOMENCLATURE We give alkenes IUPAC names by replacing the -ane ending of the corresponding alkane with -ene. The two simplest alkenes are ethene and propene. Both are also well known by their common names ethylene and propylene. Isobutylene (used in the production of synthetic rubber) (CH 3 ) 2 C CH 2 H9251-Pinene (a major constituent of turpentine) CH 3 H CH 3 CH 3 Farnesene (present in the waxy coating found on apple skins) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 168 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions Ethylene is an acceptable synonym for ethene in the IUPAC system. Propylene, isobuty- lene, and other common names ending in -ylene are not acceptable IUPAC names. CH 2 ?CH 2 IUPAC name: ethene Common name: ethylene CH 3 CH?CH 2 IUPAC name: propene Common name: propylene ETHYLENE E thylene was known to chemists in the eigh- teenth century and isolated in pure form in 1795. An early name for ethylene was gaz oléfi- ant (French for “oil-forming gas”), a term suggested to describe the fact that an oily liquid product is formed when two gases—ethylene and chlorine—re- act with each other. The term gaz oléfiant was the forerunner of the gen- eral term olefin, formerly used as the name of the class of compounds we now call alkenes. Ethylene occurs naturally in small amounts as a plant hormone. Hormones are substances that act as messengers and play regulatory roles in biological processes. Ethylene is involved in the ripening of many fruits, in which it is formed in a complex series of steps from a compound containing a cyclopropane ring: Even minute amounts of ethylene can stimulate ripening, and the rate of ripening increases with the concentration of ethylene. This property is used to advantage, for example, in the marketing of ba- nanas. Bananas are picked green in the tropics, kept green by being stored with adequate ventilation to limit the amount of ethylene present, and then in- duced to ripen at their destination by passing ethyl- ene over the fruit.* several steps NH 3 CO 2 H11002 H11001 1-Amino- cyclopropane- carboxylic acid CH 2 CH 2 Ethylene H11001 other products H11001CH 2 ?CH 2 Ethylene (bp: H11002104°C) Cl 2 Chlorine (bp: H1100234°C) ClCH 2 CH 2 Cl 1,2-Dichloroethane (bp: 83°C) Ethylene is the cornerstone of the world’s mam- moth petrochemical industry and is produced in vast quantities. In a typical year the amount of ethylene produced in the United States (5 H11003 10 10 lb) exceeds the combined weight of all of its people. In one process, ethane from natural gas is heated to bring about its dissociation into ethylene and hydrogen: This reaction is known as dehydrogenation and is si- multaneously both a source of ethylene and one of the methods by which hydrogen is prepared on an in- dustrial scale. Most of the hydrogen so generated is subsequently used to reduce nitrogen to ammonia for the preparation of fertilizer. Similarly, dehydrogenation of propane gives propene: Propene is the second most important petrochemical and is produced on a scale about half that of ethylene. Almost any hydrocarbon can serve as a starting material for production of ethylene and propene. Cracking of petroleum (Section 2.13) gives ethylene and propene by processes involving cleavage of carbon–carbon bonds of higher molecular weight hydrocarbons. The major uses of ethylene and propene are as starting materials for the preparation of polyethyl- ene and polypropylene plastics, fibers, and films. These and other applications will be described in Chapter 6. H11001CH 3 CH 2 CH 3 Propane H 2 Hydrogen CH 3 CH?CH 2 Propene 750°C H11001CH 3 CH 3 Ethane H 2 Hydrogen CH 2 ?CH 2 Ethylene 750°C *For a review, see “Ethylene—An Unusual Plant Hormone” in the April 1992 issue of the Journal of Chemical Education (pp. 315–318). Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The longest continuous chain that includes the double bond forms the base name of the alkene, and the chain is numbered in the direction that gives the doubly bonded carbons their lower numbers. The locant (or numerical position) of only one of the dou- bly bonded carbons is specified in the name; it is understood that the other doubly bonded carbon must follow in sequence. Carbon–carbon double bonds take precedence over alkyl groups and halogens in determining the main carbon chain and the direction in which it is numbered. Hydroxyl groups, however, outrank the double bond. Compounds that contain both a double bond and a hydroxyl group use the combined suffix -en H11001 -ol to signify that both functional groups are present. PROBLEM 5.1 Name each of the following using IUPAC nomenclature: (a) (CH 3 ) 2 C?C(CH 3 ) 2 (d) (b) (CH 3 ) 3 CCH?CH 2 (e) (c) (CH 3 ) 2 C?CHCH 2 CH 2 CH 3 SAMPLE SOLUTION (a) The longest continuous chain in this alkene contains four carbon atoms. The double bond is between C-2 and C-3, and so it is named as a derivative of 2-butene. Identifying the alkene as a derivative of 2-butene leaves two methyl groups to be accounted for as substituents attached to the main chain. This alkene is 2,3- dimethyl-2-butene. (It is sometimes called tetramethylethylene, but that is a com- mon name, not an IUPAC name.) We noted in Section 2.10 that the common names of certain frequently encoun- tered alkyl groups, such as isopropyl and tert-butyl, are acceptable in the IUPAC sys- tem. Three alkenyl groups—vinyl, allyl, and isopropenyl—are treated the same way. C CH 3 CH 3 H 3 C H 3 C 1 23 4 C 2,3-Dimethyl-2-butene CH 2 ?CHCH 2 CHCH 3 W OH CH 2 ?CHCH 2 CHCH 3 W Cl C CH 3 CH 3 HOCH 2 CH 2 CH 2 H 123 45 6 C 5-Methyl-4-hexen-1-ol (not 2-methyl-2-hexen-6-ol) 432 1 CH 3 CHCH?CH 2 W CH 3 3-Methyl-1-butene (not 2-methyl-3-butene) 6543 21 W CH?CH 2 BrCH 2 CH 2 CH 2 CHCH 2 CH 2 CH 3 6-Bromo-3-propyl-1-hexene (longest chain that contains double bond is six carbons) 1234 CH 2 ?CHCH 2 CH 3 1-Butene (not 1,2-butene) 6543 21 CH 3 CH 2 CH 2 CH?CHCH 3 2-Hexene (not 4-hexene) 5.1 Alkene Nomenclature 169 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website When a CH 2 group is doubly bonded to a ring, the prefix methylene is added to the name of the ring. Cycloalkenes and their derivatives are named by adapting cycloalkane terminol- ogy to the principles of alkene nomenclature. No locants are needed in the absence of substituents; it is understood that the double bond connects C-1 and C-2. Substituted cycloalkenes are numbered beginning with the double bond, proceeding through it, and continuing in sequence around the ring. The direction of numbering is chosen so as to give the lower of two possible locants to the substituent. PROBLEM 5.2 Write structural formulas or build molecular models and give the IUPAC names of all the monochloro-substituted derivatives of cyclopentene. 5.2 STRUCTURE AND BONDING IN ALKENES The structure of ethylene and the orbital hybridization model for the double bond were presented in Section 1.17. To review, Figure 5.1 depicts the planar structure of ethylene, its bond distances, and its bond angles. Each of the carbon atoms is sp 2 -hybridized, and the double bond possesses a H9268 component and a H9266 component. The H9268 component results when an sp 2 orbital of one carbon, oriented so that its axis lies along the internuclear axis, overlaps with a similarly disposed sp 2 orbital of the other carbon. Each sp 2 orbital contains one electron, and the resulting H9268 bond contains two of the four electrons of the double bond. The H9266 bond contributes the other two electrons and is formed by a “side- by-side” overlap of singly occupied p orbitals of the two carbons. Cyclopentene Cl 1 2 3 4 56 7 3-Chlorocycloheptene (not 1-chloro-2-cycloheptene) CH 3 1 2 3 4 5 6 1-Methylcyclohexene Methylenecyclohexane CH 2 CH 2 ?CH± Vinyl as in CH 2 ?CHCl Vinyl chloride CH 2 ?CHCH 2 ± Allyl as in CH 2 ?CHCH 2 OH Allyl alcohol as inCH 2 ?C± W CH 3 Isopropenyl W CH 3 CH 2 ?CCl Isopropenyl chloride 170 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions Vinyl chloride is an industrial chemical produced in large amounts (10 10 lb/year in the United States) and is used in the preparation of poly(vinyl chloride). Poly(vinyl chlo- ride), often called simply vinyl, has many applications, including siding for houses, wall coverings, and PVC pip- ing. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The double bond in ethylene is stronger than the C±C single bond in ethane, but it is not twice as strong. The C?C bond energy is 605 kJ/mol (144.5 kcal/mol) in eth- ylene versus 368 kJ/mol (88 kcal/mol) for the C±C bond in ethane. Chemists do not agree on exactly how to apportion the total C?C bond energy between its H9268 and H9266 com- ponents, but all agree that the H9266 bond is weaker than the H9268 bond. There are two different types of carbon–carbon bonds in propene, CH 3 CH?CH 2 . The double bond is of the H9268 H11001 H9266 type, and the bond to the methyl group is a H9268 bond formed by sp 3 –sp 2 overlap. PROBLEM 5.3 We can use bond-line formulas to represent alkenes in much the same way that we use them to represent alkanes. Consider the following alkene: (a) What is the molecular formula of this alkene? (b) What is its IUPAC name? (c) How many carbon atoms are sp 2 -hybridized in this alkene? How many are sp 3 - hybridized? (d) How many H9268 bonds are of the sp 2 –sp 3 type? How many are of the sp 3 –sp 3 type? SAMPLE SOLUTION (a) Recall when writing bond-line formulas for hydrocar- bons that a carbon occurs at each end and at each bend in a carbon chain. The appropriate number of hydrogens are attached so that each carbon has four bonds. Thus the compound shown is CH 3 CH 2 CH?C(CH 2 CH 3 ) 2 H HH C H H H C±C bond length H11005 150 pm C?C bond length H11005 134 pm sp 3 hybridized carbon C C sp 2 hybridized carbon 5.2 Structure and Bonding in Alkenes 171 FIGURE 5.1 (a) The framework of H9268 bonds in eth- ylene showing bond dis- tances in picometers and bond angles in degrees. All six atoms are coplanar. The carbon–carbon bond is a double bond made up of the H9268 component shown and the H9266 component illustrated in b. (b) The p orbitals of two sp 2 hybridized carbons overlap to produce a H9266 bond. An electron pair in the H9266 bond is shared by the two carbons. The simplest arithmetic ap- proach subtracts the C±C H9268 bond energy of ethane (368 kJ/mol; 88 kcal/mol) from the C?C bond energy of ethyl- ene (605 kJ/mol; 144.5 kcal/mol). This gives a value of 237 kJ/mol (56.5 kcal/mol) for the H9266 bond energy. 117.2H11034 134 pm 110 pm 121.4H11034 (b)(a) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The general molecular formula for an alkene is C n H 2n . Ethylene is C 2 H 4 ; propene is C 3 H 6 . Counting the carbons and hydrogens of the compound shown (C 8 H 16 ) reveals that it, too, corresponds to C n H 2n . 5.3 ISOMERISM IN ALKENES Although ethylene is the only two-carbon alkene, and propene the only three-carbon alkene, there are four isomeric alkenes of molecular formula C 4 H 8 : 1-Butene has an unbranched carbon chain with a double bond between C-1 and C-2. It is a constitutional isomer of the other three. Similarly, 2-methylpropene, with a branched carbon chain, is a constitutional isomer of the other three. The pair of isomers designated cis- and trans-2-butene have the same constitution; both have an unbranched carbon chain with a double bond connecting C-2 and C-3. They differ from each other, however, in that the cis isomer has both of its methyl groups on the same side of the double bond, but the methyl groups in the trans isomer are on oppo- site sides of the double bond. Recall from Section 3.12 that isomers that have the same constitution but differ in the arrangement of their atoms in space are classified as stereoisomers. cis-2-Butene and trans-2-butene are stereoisomers, and the terms “cis” and “trans” specify the configuration of the double bond. Cis–trans stereoisomerism in alkenes is not possible when one of the doubly bonded carbons bears two identical substituents. Thus, neither 1-butene nor 2-methyl- propene can have stereoisomers. PROBLEM 5.4 How many alkenes have the molecular formula C 5 H 10 ? Write their structures and give their IUPAC names. Specify the configuration of stereoisomers as cis or trans as appropriate. In principle, cis-2-butene and trans-2-butene may be interconverted by rotation about the C-2?C-3 double bond. However, unlike rotation about the C-2±C-3 single bond in butane, which is quite fast, interconversion of the stereoisomeric 2-butenes does not occur under normal circumstances. It is sometimes said that rotation about a car- bon–carbon double bond is restricted, but this is an understatement. Conventional labo- ratory sources of heat do not provide enough thermal energy for rotation about the dou- ble bond in alkenes to take place. As shown in Figure 5.2, rotation about a double bond requires the p orbitals of C-2 and C-3 to be twisted from their stable parallel alignment— in effect, the H9266 component of the double bond must be broken at the transition state. Identical C H H CH 2 CH 3 H C 1-Butene (no stereoisomers possible) Identical CH 3 CH 3 C H H C 2-Methylpropene (no stereoisomers possible) Identical C H H CH 2 CH 3 H C 1-Butene CH 3 CH 3 C H H C 2-Methylpropene cis-2-Butene CH 3 H CH 3 C H C trans-2-Butene H CH 3 CH 3 C H C 172 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions Stereoisomeric alkenes are sometimes referred to as geometric isomers. The activation energy for ro- tation about a typical carbon–carbon double bond is very high—on the order of 250 kJ/mol (about 60 kcal/mol). This quantity may be taken as a measure of the H9266 bond contribution to the total C?C bond strength of 605 kJ/mol (144.5 kcal/mol) in ethylene and compares closely with the value esti- mated by manipulation of thermochemical data on page 171. Make molecular models of cis-and trans-2-butene to ver- ify that they are different. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 5.4 NAMING STEREOISOMERIC ALKENES BY THE E–Z NOTATIONAL SYSTEM When the groups on either end of a double bond are the same or are structurally simi- lar to each other, it is a simple matter to describe the configuration of the double bond as cis or trans. Oleic acid, for example, a material that can be obtained from olive oil, has a cis double bond. Cinnamaldehyde, responsible for the characteristic odor of cin- namon, has a trans double bond. PROBLEM 5.5 Female houseflies attract males by sending a chemical signal known as a pheromone. The substance emitted by the female housefly that attracts the male has been identified as cis-9-tricosene, C 23 H 46 . Write a structural formula, including stereochemistry, for this compound. The terms “cis” and “trans” are ambiguous, however, when it is not obvious which substituent on one carbon is “similar” or “analogous” to a reference substituent on the other. Fortunately, a completely unambiguous system for specifying double bond stereo- chemistry has been developed based on an atomic number criterion for ranking sub- stituents on the doubly bonded carbons. When atoms of higher atomic number are on the same side of the double bond, we say that the double bond has the Z configuration, where Z stands for the German word zusammen, meaning “together.” When atoms of higher atomic number are on opposite sides of the double bond, we say that the config- uration is E. The symbol E stands for the German word entgegen, meaning “opposite.” C 6 H 5 H CH O C H C CinnamaldehydeOleic acid CH 3 (CH 2 ) 6 CH 2 CH 2 (CH 2 ) 6 CO 2 H C HH C 5.4 Naming Stereoisomeric Alkenes by the E–Z Notational System 173 trans-2-Butene p orbitals aligned: Optimal geometry for π bond formation cis-2-Butene p orbitals aligned: Optimal geometry for π bond formation p orbitals perpendicular: Worst geometry for π bond formation FIGURE 5.2 Interconversion of cis- and trans-2-butene proceeds by cleavage of the H9266 com- ponent of the double bond. The red balls represent the two methyl groups. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The substituent groups on the double bonds of most alkenes are, of course, more com- plicated than in this example. The rules for ranking substituents, especially alkyl groups, are described in Table 5.1. PROBLEM 5.6 Determine the configuration of each of the following alkenes as Z or E as appropriate: (a) (c) (b) (d) SAMPLE SOLUTION (a) One of the doubly bonded carbons bears a methyl group and a hydrogen. According to the rules of Table 5.1, methyl outranks hydro- gen. The other carbon atom of the double bond bears a methyl and a ±CH 2 OH group. The ±CH 2 OH group is of higher priority than methyl. Higher ranked substituents are on the same side of the double bond; the config- uration is Z. A table on the inside back cover (right page) lists some of the more frequently encountered atoms and groups in order of increasing precedence. You should not attempt to memorize this table, but should be able to derive the relative placement of one group versus another. 5.5 PHYSICAL PROPERTIES OF ALKENES Alkenes resemble alkanes in most of their physical properties. The lower molecular weight alkenes through C 4 H 8 are gases at room temperature and atmospheric pressure. The dipole moments of most alkenes are quite small. Among the C 4 H 8 isomers, 1-butene, cis-2-butene, and 2-methylpropene have dipole moments in the 0.3–0.5 D range; trans-2-butene has no dipole moment. Nevertheless, we can learn some things about alkenes by looking at the effect of substituents on dipole moments. Experimental measurements of dipole moments give size, but not direction. We normally deduce the overall direction by examining the directions of individual bond Higher Lower Higher Lower ±C(O,H,H) ±C(H,H,H) (C) (H) H 3 C H CH 2 OH CH 3 CC CH 3 CH 2 H CH 3 C C H 3 C H CH 2 CH 2 F CH 2 CH 2 CH 2 CH 3 CC H 3 C H CH 2 CH 2 OH C(CH 3 ) 3 CC H 3 C H CH 2 OH CH 3 CC Cl Br F Higher Lower Higher Lower C H C Z configuration Higher ranked substituents (Cl and Br) are on same side of double bond Cl F Br Higher Lower Lower Higher C H C E configuration Higher ranked substituents (Cl and Br) are on opposite sides of double bond 174 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions The priority rules were devel- oped by R. S. Cahn and Sir Christopher Ingold (England) and Vladimir Prelog (Switzer- land) in the context of a dif- ferent aspect of organic stereochemistry; they will ap- pear again in Chapter 7. The physical properties of se- lected alkenes are collected in Appendix 1. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 5.5 Physical Properties of Alkenes 175 TABLE 5.1 Cahn–Ingold–Prelog Priority Rules Rule 1. Higher atomic number takes precedence over lower. Bromine (atomic number 35) outranks chlor- ine (atomic number 17). Methyl (C, atomic number 6) outranks hydrogen (atomic number 1). 2. When two atoms directly attached to the double bond are identical, compare the atoms attached with these two on the basis of their atomic numbers. Pre- cedence is determined at the first point of difference: 3. Work outward from the point of attachment, com- paring all the atoms attached to a particular atom before proceeding further along the chain: 4. When working outward from the point of attach- ment, always evaluate substituent atoms one by one, never as a group. Since oxygen has a higher atomic number than carbon, Example The compound has the Z configuration. Higher ranked atoms (Br and C of CH 3 ) are on the same side of the double bond. The compound The compound The compound has the Z configuration. has the E configuration. Similarly, tert-butyl outranks isopropyl, and isopropyl outranks ethyl: Higher Lower Higher Lower Br Cl CH 3 H CC Higher Lower Lower Higher Br Cl CH 3 CH 2 CH 3 CC has the E configuration. Higher Lower Lower Higher Br Cl CH 2 CH 2 OH CH(CH 3 ) 2 CC Higher Lower Higher Lower Br Cl CH 2 OH C(CH 3 ) 3 CC Ethyl [±C(C,H,H)] The group ±CH?O [±C(O,O,H)] outranks ±CH 2 OH [±C(O,H,H)] ±CH(CH 3 ) 2 [±C(C,C,H)] ±CH 2 CH 2 OH [±C(C,H,H)] methyl [±C(H,H,H)]outranks outranks ±CH 2 OH [±C(O,H,H)] ±C(CH 3 ) 3 [±C(C,C,C)] outranks 5. An atom that is multiply bonded to another atom is considered to be replicated as a substituent on that atom: The compound has the E configuration. Higher Lower Lower Higher Br Cl CH 2 OH CH CC O ±C(O,O,H)is treated as if it were±CH X O ±C(CH 3 ) 3 H11022 ±CH(CH 3 ) 2 H11022 ±CH 2 CH 3 ±C(C,C,C) H11022 ±C(C,C,H) H11022 ±C(C,H,H) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website dipoles. With alkenes the basic question concerns the alkyl groups attached to C?C. Does an alkyl group donate electrons to or withdraw electrons from a double bond? This question can be approached by comparing the effect of an alkyl group, methyl for exam- ple, with other substituents. Ethylene, of course, has no dipole moment. Replacing one of its hydrogens by chlorine gives chloroethene, which has a dipole moment of 1.4 D. The effect is much smaller when one of the hydrogens is replaced by methyl; CH 3 CH?CH 2 has a dipole moment of only 0.3 D. Now place CH 3 and Cl trans to each other on the double bond. If methyl releases electrons better than H, then the dipole moment of trans-CH 3 CH?CHCl should be larger than that of CH 2 ?CHCl, because the effects of CH 3 and Cl reinforce each other. If methyl is electron attracting, the opposite should occur, and the dipole moment of trans-CH 3 CH?CHCl will be smaller than 1.4 D. In fact, the dipole moment of trans- CH 3 CH?CHCl is larger than that of CH 2 ?CHCl, indicating that a methyl group is an electron-donating substituent on the double bond. A methyl group releases electrons to a double bond in much the same way that it releases electrons to the positively charged carbon of a carbocation—by an inductive effect and by hyperconjugation (Figure 5.3). Other alkyl groups behave similarly and, as we go along, we’ll see several ways in which the electron-releasing effects of alkyl substituents influence the properties of alkenes. The first is described in the following section. 5.6 RELATIVE STABILITIES OF ALKENES Earlier (Sections 2.15, 3.12) we saw how to use heats of combustion to compare the sta- bilities of isomeric alkanes. We can do the same thing with isomeric alkenes. Consider the heats of combustion of the four isomeric alkenes of molecular formula C 4 H 8 . All undergo combustion according to the equation C 4 H 8 H11001 6O 2 ±￡ 4CO 2 H11001 4H 2 O When the heats of combustion of the isomers are plotted on a common scale as in Fig- ure 5.4, we see that the isomer of highest energy (the least stable one) is 1-butene, CH 2 ?CHCH 2 CH 3 . The isomer of lowest energy (most stable) is 2-methylpropene (CH 3 ) 2 C?CH 2 . C H HH H C Ethylene H9262 H11005 0 D H Cl C H H C Chloroethene H9262 H11005 1.4 D H HH CH 3 C C Propene H9262 H11005 0.3 D Cl H H CH 3 C C trans-1-Chloropropene H9262 H11005 1.7 D 176 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions sp 2 -hybridized carbons of an alkene are more electronegative than sp 3 -hybridized carbon and are stabilized by electron-donating substituents. CC H CH 3 Methyl group is a better electron-donating substituent than hydrogen. FIGURE 5.3 Alkyl groups donate electrons to sp 2 - hybridized carbons of an alkene. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Analogous data for a host of alkenes tell us that the most important factors gov- erning alkene stability are: 1. Degree of substitution (alkyl substituents stabilize a double bond) 2. Van der Waals strain (destabilizing when alkyl groups are cis to each other) Degree of substitution. We classify double bonds as monosubstituted, disubstituted, trisubstituted, or tetrasubstituted according to the number of carbon atoms that are directly attached to the C?C structural unit. Monosubstituted alkenes: Disubstituted alkenes: (R and RH11032 may be the same or different) Trisubstituted alkenes: (R, RH11032, and RH11033 may be the same or different) C RH11033 HRH11032 R C as in (CH 3 ) 2 C CHCH 2 CH 3 (2-methyl-2-pentene) RCH CHRH11032 as in CH 3 CH CHCH 3 (cis- or trans-2-butene) C H HRH11032 R C as in (CH 3 ) 2 C CH 2 (2-methylpropene) RCH?CH 2 CH 3 CH 2 CH?CH 2 (1-butene)as in 5.6 Relative Stabilities of Alkenes 177 cis-2-Butene trans-2-Butene Alkene 4CO 2 H11001 4H 2 O CH 3 H110016O 2 CC H CH 3 H 1-Butene 2-Methylpropene 7 7 2710 3 2700 Energy ?HH11034 ?HH11034 ?HH11034 ?HH11034 CH 2 CHCH 2 CH 3 CH 3 CC H CH 3 H CH 3 C CH 3 CH 2 2717 2707 FIGURE 5.4 Heats of com- bustion of C 4 H 8 alkene iso- mers plotted on a common scale. All energies are in kilo- joules per mole. (An energy difference of 3 kJ/mol is equivalent to 0.7 kcal/mol; 7 kJ/mol is equivalent to 1.7 kcal/mol.) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Tetrasubstituted alkenes: (R, RH11032, RH11033, and RH11630 may be the same or different) In the example shown, each of the highlighted ring carbons counts as a separate sub- stituent on the double bond. PROBLEM 5.7 Write structural formulas or build molecular models and give the IUPAC names for all the alkenes of molecular formula C 6 H 12 that contain a trisub- stituted double bond. (Don’t forget to include stereoisomers.) From the heats of combustion of the C 4 H 8 alkenes in Figure 5.5 we see that each of the disubstituted alkenes is more stable than the monosubstituted alkene In general, alkenes with more highly substituted double bonds are more stable than iso- mers with less substituted double bonds. PROBLEM 5.8 Give the structure or make a molecular model of the most stable C 6 H 12 alkene. Like the sp 2 -hybridized carbons of carbocations and free radicals, the sp 2 - hybridized carbons of double bonds are electron attracting, and alkenes are stabilized by substituents that release electrons to these carbons. As we saw in the preceding section, alkyl groups are better electron-releasing substituents than hydrogen and are, therefore, better able to stabilize an alkene. An effect that results when two or more atoms or groups interact so as to alter the electron distribution in a system is called an electronic effect. The greater stability of more highly substituted alkenes is an example of an electronic effect. van der Waals strain. Alkenes are more stable when large substituents are trans to each other than when they are cis. As was seen in Figure 5.4, trans-2-butene has a lower heat of combustion and is more stable than cis-2-butene. The energy difference between the two is 3 kJ/mol (0.7 kcal/mol). The source of this energy difference is illustrated in C CH 2 CH 3 HH H C 1-Butene C CH 3 HH CH 3 C cis-2-Butenetrans-2-Butene C H CH 3 H CH 3 C 2-Methylpropene C H HCH 3 CH 3 C C RH11033 RH11630RH11032 R C as in CH 3 CH 3 (1,2-dimethylcyclohexene) 178 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Figure 5.5, where it is seen that methyl groups approach each other very closely in cis- 2-butene, but the trans isomer is free of strain. An effect that results when two or more atoms are close enough in space that a repulsion occurs between them is one type of steric effect. The greater stability of trans alkenes compared with their cis counterparts is an example of a steric effect. PROBLEM 5.9 Arrange the following alkenes in order of decreasing stability: 1-pentene; (E)-2-pentene; (Z)-2-pentene; 2-methyl-2-butene. The difference in stability between stereoisomeric alkenes is even more pronounced with larger alkyl groups on the double bond. A particularly striking example compares cis- and trans-2,2,5,5-tetramethyl-3-hexene, in which the heat of combustion of the cis stereoisomer is 44 kJ/mol (10.5 kcal/mol) higher than that of the trans. The cis isomer is destabilized by the large van der Waals strain between the bulky tert-butyl groups on the same side of the double bond. Energy difference H11005 44 kJ/mol (10.5 kcal/mol) trans-2,2,5,5-Tetramethyl-3-hexene More stable H CC C CH 3 H H 3 C H 3 C C CH 3 CH 3 H 3 C CH 3 C CC C CH 3 CH 3 HH CH 3 H 3 C H 3 C cis-2,2,5,5-Tetramethyl-3-hexene Less stable 5.6 Relative Stabilities of Alkenes 179 cis-2 Butene trans-2 Butene FIGURE 5.5 Ball-and- spoke and space-filling mod- els of cis- and trans-2-butene. The space-filling model shows the serious van der Waals strain between two of the hydrogens in cis-2- butene. The molecule adjusts by expanding those bond an- gles that increase the separa- tion between the crowded atoms. The combination of angle strain and van der Waals strain makes cis-2 butene less stable than trans- 2-butene. A similar steric effect was seen in Section 3.12, where van der Waals strain be- tween methyl groups on the same side of the ring made cis-1,2-dimethylcyclopropane less stable than its trans stereoisomer. The common names of these alkenes are cis- and trans- di-tert-butylethylene. In cases such as this the com- mon names are somewhat more convenient than the IUPAC names because they are more readily associated with molecular structure. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEM 5.10 Despite numerous attempts, the alkene 3,4-di-tert-butyl-2,2,5,5- tetramethyl-3-hexene has never been synthesized. Can you explain why? Try mak- ing a space-filling model of this compound. 5.7 CYCLOALKENES Double bonds are accommodated by rings of all sizes. The simplest cycloalkene, cyclo- propene, was first synthesized in 1922. A cyclopropene ring is present in sterculic acid, a substance derived from one of the components of the oil present in the seeds of a tree (Sterculia foelida) that grows in the Philippines and Indonesia. As we saw in Section 3.9, cyclopropane is destabilized by angle strain because its 60° bond angles are much smaller than the normal 109.5° angles associated with sp 3 - hybridized carbon. Cyclopropene is even more strained because the deviation of the bond angles at its doubly bonded carbons from the normal sp 2 hybridization value of 120° is greater still. Cyclobutene has, of course, less angle strain than cyclopropene, and the angle strain of cyclopentene, cyclohexene, and higher cycloalkenes is negligible. So far we have represented cycloalkenes by structural formulas in which the dou- ble bonds are of the cis configuration. If the ring is large enough, however, a trans stereoisomer is also possible. The smallest trans cycloalkene that is stable enough to be isolated and stored in a normal way is trans-cyclooctene. trans-Cycloheptene has been prepared and studied at low temperature (H1100290°C) but is too reactive to be isolated and stored at room temperature. Evidence has also been presented for the fleeting existence of the even more strained trans-cyclohexene as a reactive intermediate in certain reactions. PROBLEM 5.11 Place a double bond in the carbon skeleton shown so as to rep- resent (a) (Z)-1-Methylcyclodecene (d) (E)-3-Methylcyclodecene (b) (E)-1-Methylcyclodecene (e) (Z)-5-Methylcyclodecene (c) (Z)-3-Methylcyclodecene (f) (E)-5-Methylcyclodecene CH 3 HH (Z)-Cyclooctene (cis-cyclooctene) More stable H H (E)-Cyclooctene (trans-cyclooctene) Less stable Energy difference H11005 39 kJ/mol (9.2 kcal/mol) (CH 2 ) 7 CO 2 HCH 3 (CH 2 ) 7 HH Sterculic acid HH HH Cyclopropene 180 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions Sterculic acid and related substances are the subject of an article in the July 1982 is- sue of Journal of Chemical Education (pp. 539–543). Make molecular models of (E) and (Z)-cyclooctene and compare their H±C?C±H di- hedral angles. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website SAMPLE SOLUTION (a) and (b) Since the methyl group must be at C-1, there are only two possible places to put the double bond: In the Z stereoisomer the two lower priority substituents—the methyl group and the hydrogen—are on the same side of the double bond. In the E stereoisomer these substituents are on opposite sides of the double bond. The ring carbons are the higher ranking substituents at each end of the double bond. Because larger rings have more carbons with which to span the ends of a double bond, the strain associated with a trans cycloalkene decreases with increasing ring size. The strain eventually disappears when a 12-membered ring is reached and cis and trans- cyclododecene are of approximately equal stability. When the rings are larger than 12 membered, trans cycloalkenes are more stable than cis. In these cases, the ring is large enough and flexible enough that it is energetically similar to a noncyclic alkene. As in noncyclic cis alkenes, van der Waals strain between carbons on the same side of the dou- ble bond destabilizes a cis cycloalkene. 5.8 PREPARATION OF ALKENES: ELIMINATION REACTIONS The rest of this chapter describes how alkenes are prepared by reactions of the type: Alkene formation requires that X and Y be substituents on adjacent carbon atoms. By making X the reference atom and identifying the carbon attached to it as the H9251 carbon, we see that atom Y is a substituent on the H9252 carbon. Carbons succeedingly more remote from the reference atom are designated H9253, H9254, and so on. Only H9252 elimination reactions will be discussed in this chapter. [Beta (H9252) elimination reactions are also known as 1,2 eliminations.] You are already familiar with one type of H9252 elimination, having seen in Section 5.1 that ethylene and propene are prepared on an industrial scale by the high-temperature dehydrogenation of ethane and propane. Both reactions involve H9252 elimination of H 2 . Many reactions classified as dehydrogenations occur within the cells of living sys- tems at 25°C. H 2 is not one of the products, however. Instead, the hydrogens are lost in separate steps of an enzyme-catalyzed process. The enzyme indicated in the reaction: H11001CH 3 CH 3 Ethane CH 2 ?CH 2 Ethylene H 2 Hydrogen 750°C H11001CH 3 CH 2 CH 3 Propane CH 3 CH?CH 2 Propene H 2 Hydrogen 750°C X CYC H9251H9252 C C H11001 XY CH 3 H 1 2 (E )-1-Methylcyclodecene CH 3 H 1 2 (Z )-1-Methylcyclodecene 5.8 Preparation of Alkenes: Elimination Reactions 181 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website is a special kind, known as a flavoprotein. Dehydrogenation of alkanes is not a practical laboratory synthesis for the vast majority of alkenes. The principal methods by which alkenes are prepared in the labo- ratory are two other H9252 eliminations: the dehydration of alcohols and the dehydrohalo- genation of alkyl halides. A discussion of these two methods makes up the remainder of this chapter. 5.9 DEHYDRATION OF ALCOHOLS In the dehydration of alcohols, the H and OH are lost from adjacent carbons. An acid catalyst is necessary. Before dehydrogenation of ethane became the dominant method, ethylene was pre- pared by heating ethyl alcohol with sulfuric acid. Other alcohols behave similarly. Secondary alcohols undergo elimination at lower tem- peratures than primary alcohols, and tertiary alcohols at lower temperatures than secondary alcohols. Sulfuric acid (H 2 SO 4 ) and phosphoric acid (H 3 PO 4 ) are the acids most frequently used in alcohol dehydrations. Potassium hydrogen sulfate (KHSO 4 ) is also often used. H 2 SO 4 heat C CH 3 OH CH 3 CH 3 2-Methyl-2-propanol C H 3 C H 3 C CH 2 2-Methylpropene (82%) H 2 O Water H11001 OH Cyclohexanol Cyclohexene (79–87%) H11001 H 2 O Water H 2 SO 4 140°C H11001CH 3 CH 2 OH Ethyl alcohol CH 2 ?CH 2 Ethylene H 2 O Water H 2 SO 4 160°C H11001H COHC Alcohol C C Alkene H 2 O Water H H11001 HOCCH 2 CH 2 COH O X O X Succinic acid O X X O C?C HOC H H COH ± ± ± ± Fumaric acid succinate dehydrogenase 182 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions A quote from a biochemistry text is instructive here. “This is not an easy reaction in or- ganic chemistry. It is, how- ever, a very important type of reaction in metabolic chemistry and is an integral step in the oxidation of car- bohydrates, fats, and several amino acids.” G. L. Zubay, Biochemistry, 4th ed., William C. Brown Publishers, 1996, p. 333. HSO 4 H11002 and H 3 PO 4 are very similar in acid strength. Both are much weaker than H 2 SO 4 , which is a strong acid. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEM 5.12 Identify the alkene obtained on dehydration of each of the fol- lowing alcohols: (a) 3-Ethyl-3-pentanol (c) 2-Propanol (b) 1-Propanol (d) 2,3,3-Trimethyl-2-butanol SAMPLE SOLUTION (a) The hydrogen and the hydroxyl are lost from adjacent carbons in the dehydration of 3-ethyl-3-pentanol. The hydroxyl group is lost from a carbon that bears three equivalent ethyl sub- stituents. Beta elimination can occur in any one of three equivalent directions to give the same alkene, 3-ethyl-2-pentene. 5.10 REGIOSELECTIVITY IN ALCOHOL DEHYDRATION: THE ZAITSEV RULE In the preceding examples, including those of Problem 5.12, only a single alkene could be formed from each alcohol by H9252 elimination. What about elimination in alcohols such as 2-methyl-2-butanol, in which dehydration can occur in two different directions to give alkenes that are constitutional isomers? Here, a double bond can be generated between C-1 and C-2 or between C-2 and C-3. Both processes occur but not nearly to the same extent. Under the usual reaction conditions 2-methyl-2-butene is the major product, and 2-methyl-1-butene the minor one. Dehydration of this alcohol is selective in respect to its direction. Elimination occurs in the direction that leads to the double bond between C-2 and C-3 more than between C-2 and C-1. Reactions that can proceed in more than one direction, but in which one direction is preferred, are said to be regioselective. As a second example, consider the regioselective dehydration of 2-methylcyclo- hexanol to yield a mixture of 1-methylcyclohexene (major) and 3-methylcyclohexene (minor). CH 3 3-Methylcyclohexene (16%) H11001 CH 3 1-Methylcyclohexene (84%) CH 3 OH 2-Methylcyclohexanol H 3 PO 4 heat H11001 C H 3 C H 3 C CHCH 3 2-Methyl-2-butene (90%) CH 2 CH 2 CH 3 CH 3 C 2-Methyl-1-butene (10%) CH 3 OH CH 3 1342 1 C CH 2 CH 3 2-Methyl-2-butanol H 2 SO 4 80°C H H11001 C CH 2 CH 3 OH H9252 H9252 CH 3 CH 2 H9252 CH 2 CH 3 H9251 3-Ethyl-3-pentanol C CH 3 CH 2 CH 3 CH 2 CHCH 3 3-Ethyl-2-pentene H11001 H 2 O Water 5.10 Regioselectivity in Alcohol Dehydration: The Zaitsev Rule 183 The term “regioselective” was coined by Alfred Hass- ner, then at the University of Colorado, in a paper pub- lished in the Journal of Or- ganic Chemistry in 1968. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website In 1875, Alexander M. Zaitsev of the University of Kazan (Russia) set forth a gen- eralization describing the regioselectivity H9252-eliminations. Zaitsev’s rule summarizes the results of numerous experiments in which alkene mixtures were produced by H9252 elimi- nation. In its original form, Zaitsev’s rule stated that the alkene formed in greatest amount is the one that corresponds to removal of the hydrogen from the H9252 carbon having the fewest hydrogens. Zaitsev’s rule as applied to the acid-catalyzed dehydration of alcohols is now more often expressed in a different way: H9252 elimination reactions of alcohols yield the most highly substituted alkene as the major product. Since, as was discussed in Section 5.6, the most highly substituted alkene is also normally the most stable one, Zaitsev’s rule is sometimes expressed as a preference for predominant formation of the most stable alkene that could arise by H9252 elimination. PROBLEM 5.13 Each of the following alcohols has been subjected to acid- catalyzed dehydration and yields a mixture of two isomeric alkenes. Identify the two alkenes in each case, and predict which one is the major product on the basis of the Zaitsev rule. (a) (b) (c) SAMPLE SOLUTION (a) Dehydration of 2,3-dimethyl-2-butanol can lead to either 2,3-dimethyl-1-butene by removal of a C-1 hydrogen or to 2,3-dimethyl-2- butene by removal of a C-3 hydrogen. The major product is 2,3-dimethyl-2-butene. It has a tetrasubstituted double bond and is more stable than 2,3-dimethyl-1-butene, which has a disubstituted double bond. The major alkene arises by loss of a hydrogen from the H9252 carbon that has fewer attached hydrogens (C-3) rather than from the H9252 carbon that has the greater number of hydrogens (C-1). 5.11 STEREOSELECTIVITY IN ALCOHOL DEHYDRATION In addition to being regioselective, alcohol dehydrations are stereoselective. A stereo- selective reaction is one in which a single starting material can yield two or more stereoisomeric products, but gives one of them in greater amounts than any other. H11001 H 3 C H 3 C CH 3 CH 3 C C 2,3-Dimethyl-2-butene (major product) CH 2 CH 3 CH(CH 3 ) 2 C 2,3-Dimethyl-1-butene (minor product) C CH 3 OH 123 CH 3 CH(CH 3 ) 2 2,3-Dimethyl-2-butanol H11002H 2 O OH H OHH 3 C(CH 3 ) 2 CCH(CH 3 ) 2 OH R 2 C CH 3 CH 2 R C Alkene present in greatest amount in product OH CH 3 C CH 2 RR 2 CH H9252H9252 H9252 H9251 Hydrogen is lost from H9252-carbon having the fewest attached hydrogens H11002H 2 O 184 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions Although Russian, Zaitsev published most of his work in German scientific journals, where his name was translit- erated as Saytzeff. The spelling used here (Zaitsev) corresponds to the currently preferred style. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Alcohol dehydrations tend to produce the more stable stereoisomer of an alkene. Dehy- dration of 3-pentanol, for example, yields a mixture of trans-2-pentene and cis-2-pen- tene in which the more stable trans stereoisomer predominates. PROBLEM 5.14 What three alkenes are formed in the acid-catalyzed dehydra- tion of 2-pentanol? The biological dehydrogenation of succinic acid described in Section 5.8 is 100% stereoselective. Only fumaric acid, which has a trans double bond, is formed. High lev- els of stereoselectivity are characteristic of enzyme-catalyzed reactions. 5.12 THE MECHANISM OF ACID-CATALYZED DEHYDRATION OF ALCOHOLS The dehydration of alcohols and the conversion of alcohols to alkyl halides by treatment with hydrogen halides (Section 4.8) are similar in two important ways: 1. Both reactions are promoted by acids. 2. The relative reactivity of alcohols decreases in the order tertiary H11022 secondary H11022 primary. These common features suggest that carbocations are key intermediates in alcohol dehy- dration, just as they are in the conversion of alcohols to alkyl halides. Figure 5.6 por- trays a three-step mechanism for the sulfuric acid-catalyzed dehydration of tert-butyl alcohol. Steps 1 and 2 describe the generation of tert-butyl cation by a process similar to that which led to its formation as an intermediate in the reaction of tert-butyl alcohol with hydrogen chloride. Step 3 in Figure 5.6, however, is new to us and is the step in which the double bond is formed. Step 3 is an acid-base reaction in which the carbocation acts as a Br?nsted acid, transferring a proton to a Br?nsted base (water). This is the property of carbocations that is of the most significance to elimination reactions. Carbocations are strong acids; they are the conjugate acids of alkenes and readily lose a proton to form alkenes. Even weak bases such as water are sufficiently basic to abstract a proton from a carbocation. PROBLEM 5.15 Write a structural formula for the carbocation intermediate formed in the dehydration of each of the alcohols in Problem 5.13 (Section 5.10). Using curved arrows, show how each carbocation is deprotonated by water to give a mixture of alkenes. SAMPLE SOLUTION (a) The carbon that bears the hydroxyl group in the start- ing alcohol is the one that becomes positively charged in the carbocation. H H11001 H11002H 2 O OH (CH 3 ) 2 CCH(CH 3 ) 2 (CH 3 ) 2 CCH(CH 3 ) 2 H11001 C H 3 C HCH 2 CH 3 H C trans-2-Pentene (75%) (major product) H11001C H 3 C H CH 2 CH 3 H C cis-2-Pentene (25%) (minor product) CH 3 CH 2 CHCH 2 CH 3 OH 3-Pentanol H 2 SO 4 heat 5.12 The Mechanism of Acid-Catalyzed Dehydration of Alcohols 185 Step 3 in Figure 5.6 shows water as the base which ab- stracts a proton from the car- bocation. Other Br?nsted bases present in the reaction mixture that can function in the same way include tert- butyl alcohol and hydrogen sulfate ion. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Water may remove a proton from either C-1 or C-3 of this carbocation. Loss of a proton from C-1 yields the minor product 2,3-dimethyl-1-butene. (This alkene has a disubstituted double bond.) Loss of a proton from C-3 yields the major product 2,3-dimethyl-2-butene. (This alkene has a tetrasubstituted double bond.) As noted earlier (Section 4.13) primary carbocations are too high in energy to be intermediates in most chemical reactions. If primary alcohols don’t form primary car- H OH 2 H11001 H11001 2,3-Dimethyl-2-butene H 3 C H 3 C CH 3 CH 3 CC H 3 C H 3 C 2 3 1 H C C(CH 3 ) 2 H11001 OH 2 CH 2 CH(CH 3 ) 2 CH 3 C 2,3-Dimethyl-1-butene H11001H 2 O H11001 HHCH 2 CH 3 CH(CH 3 ) 2 12 3 H11001 CH 2 O 186 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions The overall reaction: (CH 3 ) 3 COH ±±￡ (CH 3 ) 2 C CH 2 H11001 H 2 O tert-Butyl alcohol H 2 SO 4 heat 2-Methylpropene Water Step (1): Protonation of tert-butyl alcohol. tert-Butyl alcohol Hydronium ion tert-Butyloxonium ion Water Step (2): Dissociation of tert-butyloxonium ion. tert-Butyloxonium ion tert-Butyl cation Water Step (3): Deprotonation of tert-butyl cation Water 2-Methylpropene Hydronium iontert-Butyl cation H H11001H11001 H H fast H H H H (CH 3 ) 3 C±O H11001 H±O ±￡ (CH 3 ) 3 C±O H11001 O H11001 H H slow H H (CH 3 ) 3 C±O ±￡ (CH 3 ) 3 C H11001 H11001 O CH 3 H11001 CH 3 H H fast CH 3 CH 3 H H H11001 C±CH 2 ±H H11001 O ±￡ C CH 2 H11001 H±O FIGURE 5.6 The mechanism for the acid-catalyzed dehy- dration of tert-butyl alcohol. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website bocations, then how do they undergo elimination? A modification of our general mech- anism for alcohol dehydration offers a reasonable explanation. For primary alcohols it is believed that a proton is lost from the alkyloxonium ion in the same step in which car- bon–oxygen bond cleavage takes place. For example, the rate-determining step in the sulfuric acid-catalyzed dehydration of ethanol may be represented as: Like tertiary alcohols, secondary alcohols normally undergo dehydration by way of carbocation intermediates. In Chapter 4 you learned that carbocations could be captured by halide anions to give alkyl halides. In the present chapter, a second type of carbocation reaction has been introduced—a carbocation can lose a proton to form an alkene. In the next section a third aspect of carbocation behavior will be described, the rearrangement of one carbocation to another. 5.13 REARRANGEMENTS IN ALCOHOL DEHYDRATION Some alcohols undergo dehydration to yield alkenes having carbon skeletons different from the starting alcohols. Not only has elimination taken place, but the arrangement of atoms in the alkene is different from that in the alcohol. A rearrangement is said to have occurred. An example of an alcohol dehydration that is accompanied by rearrange- ment is the case of 3,3-dimethyl-2-butanol. This is one of many such experiments car- ried out by F. C. Whitmore and his students at Pennsylvania State University in the 1930s as part of a general study of rearrangement reactions. A mixture of three alkenes was obtained in 80% yield, having the composition shown. The alkene having the same carbon skeleton as the starting alcohol, 3,3-dimethyl-1- butene, constituted only 3% of the alkene mixture. The two alkenes present in greatest amount, 2,3-dimethyl-2-butene and 2,3-dimethyl-1-butene, both have carbon skeletons different from that of the starting alcohol. Whitmore proposed that the carbon skeleton rearrangement occurred in a separate step following carbocation formation. Once the alcohol was converted to the corre- sponding carbocation, that carbocation could either lose a proton to give an alkene hav- ing the same carbon skeleton or rearrange to a different carbocation, as shown in Fig- ure 5.7. The rearranged alkenes arise by loss of a proton from the rearranged carbocation. Why do carbocations rearrange? The answer is straightforward once we recall that tertiary carbocations are more stable than secondary carbocations (Section 4.10). Thus, rearrangement of a secondary to a tertiary carbocation is energetically favorable. As shown in Figure 5.7, the carbocation that is formed first in the dehydration of H11001 H H O Water H11001 Ethylene CH 2 CH 2 H11001H 2 O H11001 H Hydronium ion H 2 O Water Ethyloxonium ion H CH 2 CH 2 H11001 H H O 5.13 Rearrangements in Alcohol Dehydration 187 H 3 C H 3 C CH 3 CH 3 H11001 2,3-Dimethyl- 2-butene (64%) CCCH 3 CH 3 CH 3 OH CCHCH 3 3,3-Dimethyl- 2-butanol H 3 PO 4 heat CH 3 CH 3 CH 3 CH 2 C CH 3,3-Dimethyl- 1-butene (3%) H11001 C H 2 C H 3 C CH 3 CH CH 3 2,3-Dimethyl- 1-butene (33%) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 3,3-dimethyl-2-butanol is secondary; the rearranged carbocation is tertiary. Rearrange- ment occurs, and almost all of the alkene products come from the tertiary carbocation. How do carbocations rearrange? To understand this we need to examine the struc- tural change that takes place at the transition state. Again referring to the initial (sec- ondary) carbocation intermediate in the dehydration of 3,3-dimethyl-2-butanol, rearrangement occurs when a methyl group shifts from C-3 to the positively charged car- bon. The methyl group migrates with the pair of electrons that made up its original H9268 bond to C-3. In the curved arrow notation for this methyl migration, the arrow shows the movement of both the methyl group and the electrons in the H9268 bond. At the transition state for rearrangement, the methyl group is partially bonded both to its point of origin and to the carbon that will be its destination. This rearrangement is shown in orbital terms in Figure 5.8. The relevant orbitals of the secondary carbocation are shown in structure (a), those of the transition state for rearrangement in (b), and those of the tertiary carbocation in (c). Delocalization of the electrons of the C±CH 3 H9268 bond into the vacant p orbital of the positively charged car- bon by hyperconjugation is present in both (a) and (c), requires no activation energy, and stabilizes each carbocation. Migration of the atoms of the methyl group, however, occurs only when sufficient energy is absorbed by (a) to achieve the transition state (b). The activation energy is modest, and carbocation rearrangements are normally quite fast. CH 3 CH 3 CH 3 H11001 C CHCH 3 1,1,2-Trimethylpropyl cation (tertiary, more stable) CH 3 CH 3 CH 3 C CHCH 3 H9254H11001 H9254H11001 Transition state for methyl migration (dashed lines indicate partial bonds) 1,2,2-Trimethylpropyl cation (secondary, less stable) CH 3 CH 3 CH 3 1234 C CHCH 3 H11001 188 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions H11001 OH 3,3-Dimethyl-2-butanol H H11001 H11002H 2 O 3,3-Dimethyl-1-butene (3%) (CH 3 ) 3 CCHCH 3 CH 3 C CH 3 CH 3 CHCH 3 methyl shift from C-3 to C-2 1,1,2-Trimethylpropyl cation (a tertiary carbocation) CH 3 C CH 3 CH 3 CHCH 3 CH 3 C CH 3 CH 3 CH 1,2,2-Trimethylpropyl cation (a secondary carbocation) CH 2 CC CH 3 CH 3 CH 3 CH 3 2,3-Dimethyl-2-butene (64%) H11001 CH 2 C CH 3 CH(CH 3 ) 2 2,3-Dimethyl-1-butene (33%) 32 1 H11002H H11001 H11002H H11001 H11001 To simplify the accompany- ing discussion, the carbons of the carbocation are num- bered so as to correspond to their positions in the starting alcohol 3,3-dimethyl-2-bu- tanol. These numbers are dif- ferent from the locants in the IUPAC cation names, which are given under the structural formulas. Once a carbocation is formed, anything that hap- pens afterward happens rapidly. FIGURE 5.7 The first formed carbocation from 3,3-dimethyl-2-butanol is secondary and re- arranges to a more stable tertiary carbocation by a methyl migration. The major portion of the alkene products is formed by way of the tertiary carbocation. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEM 5.16 The alkene mixture obtained on dehydration of 2,2-dimethylcy- clohexanol contains appreciable amounts of 1,2-dimethylcyclohexene. Give a mechanistic explanation for the formation of this product. Alkyl groups other than methyl can also migrate to a positively charged carbon. Many carbocation rearrangements involve migration of a hydrogen. These are called hydride shifts. The same requirements apply to hydride shifts as to alkyl group migrations; they proceed in the direction that leads to a more stable carbocation; the ori- gin and destination of the migrating hydrogen are adjacent carbons, one of which must be positively charged; and the hydrogen migrates with a pair of electrons. Hydride shifts often occur during the dehydration of primary alcohols. Thus, although 1- butene would be expected to be the only alkene formed on dehydration of 1-butanol, it is in fact only a minor product. The major product is a mixture of cis- and trans-2-butene. H11001CH 3 CH 2 CH 2 CH 2 OH CH 3 CH 2 CH?CH 2 1-Butene (12%) CH 3 CH?CHCH 3 Mixture of cis-2-butene (32%) and trans-2-butene (56%) H 2 SO 4 140–170°C A H BY CCX H11001 H YB ACCX H11001 Hydride shift 5.13 Rearrangements in Alcohol Dehydration 189 CC C H 3 C H 3 C H 3 C CH 3 CH 3 H H H 3 C Hybridization changing from sp 3 to sp 2 Hybridization changing from sp 2 to sp 3 sp 3 sp 2 H 3 C CH 3 H H 3 C sp 3 sp 2 p orbital p orbital bond bond (a) (b) (c) 1,2,2-Trimethylpropyl cation (secondary) 1,1,2-Trimethylpropyl cation (tertiary) C C H 3 C H 3 C C H 3 C σ σ FIGURE 5.8 An orbital repre- sentation of methyl migra- tion in 1,2,2-trimethylpropyl cation. Structure (a) is the ini- tial secondary carbocation; structure (b) is the transition state for methyl migration, and structure (c) is the final tertiary carbocation. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A mechanism for the formation of these three alkenes is shown in Figure 5.9. Dis- sociation of the primary alkyloxonium ion is accompanied by a shift of hydride from C- 2 to C-1. This avoids the formation of a primary carbocation, leading instead to a sec- ondary carbocation in which the positive charge is at C-2. Deprotonation of this carbocation yields the observed products. (Some 1-butene may also arise directly from the primary alkyloxonium ion.) This concludes discussion of our second functional group transformation involv- ing alcohols: the first was the conversion of alcohols to alkyl halides (Chapter 4), and the second the conversion of alcohols to alkenes. In the remaining sections of the chap- ter the conversion of alkyl halides to alkenes by dehydrohalogenation is described. 5.14 DEHYDROHALOGENATION OF ALKYL HALIDES Dehydrohalogenation is the loss of a hydrogen and a halogen from an alkyl halide. It is one of the most useful methods for preparing alkenes by H9252 elimination. When applied to the preparation of alkenes, the reaction is carried out in the presence of a strong base, such as sodium ethoxide (NaOCH 2 CH 3 ) in ethyl alcohol as solvent. CXH C Alkyl halide CC Alkene CH 3 CH 2 OH Ethyl alcohol NaOCH 2 CH 3 Sodium ethoxide H11001 NaX Sodium halide H11001H11001 H H H Cl Cyclohexyl chloride H H Cyclohexene (100%) NaOCH 2 CH 3 CH 3 CH 2 OH, 55°C CXH C Alkyl halide CC Alkene HX Hydrogen halide H11001 190 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions 1-Butanol CH 3 CH 2 CH 2 CH 2 OH H H11001 O H11001 H H CH 2 H CH 3 CH 2 CH H11002H 2 O H11001 CH 3 CH 2 CHCH 3 CH 3 CH 2 CH CH 2 1-Butene (12%) H11001 CH 3 CH CHCH 3 cis and trans-2-Butene (32% cis; 56% trans) 321 32 1 hydride shift concerted with dissociation H11002H H11001 Sodium ethoxide is prepared by the reaction of sodium metal with ethanol. FIGURE 5.9 Dehydration of 1-butanol is accompanied by a hydride shift from C-2 to C-1. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Similarly, sodium methoxide (NaOCH 3 ) is a suitable base and is used in methyl alco- hol. Potassium hydroxide in ethyl alcohol is another base–solvent combination often employed in the dehydrohalogenation of alkyl halides. Potassium tert-butoxide [KOC(CH 3 ) 3 ] is the preferred base when the alkyl halide is primary; it is used in either tert-butyl alcohol or dimethyl sulfoxide as solvent. The regioselectivity of dehydrohalogenation of alkyl halides follows the Zaitsev rule; H9252 elimination predominates in the direction that leads to the more highly substi- tuted alkene. PROBLEM 5.17 Write the structures of all the alkenes that can be formed by dehydrohalogenation of each of the following alkyl halides. Apply the Zaitsev rule to predict the alkene formed in greatest amount in each case. (a) 2-Bromo-2,3-dimethylbutane (d) 2-Bromo-3-methylbutane (b) tert-Butyl chloride (e) 1-Bromo-3-methylbutane (c) 3-Bromo-3-ethylpentane (f) 1-Iodo-1-methylcyclohexane SAMPLE SOLUTION (a) First analyze the structure of 2-bromo-2,3-dimethylbu- tane with respect to the number of possible H9252 elimination pathways. The two possible alkenes are The major product, predicted on the basis of Zaitsev’s rule, is 2,3-dimethyl-2- butene. It has a tetrasubstituted double bond. The minor alkene has a disubsti- tuted double bond. In addition to being regioselective, dehydrohalogenation of alkyl halides is stereo- selective and favors formation of the more stable stereoisomer. Usually, as in the case of 5-bromononane, the trans (or E) alkene is formed in greater amounts than its cis (or Z) stereoisomer. H 3 C C H 3 C CH 3 CH 3 C 2,3-Dimethyl-2-butene (major product) andC CH 3 CH(CH 3 ) 2 CH 2 2,3-Dimethyl-1-butene (minor product) Bromine must be lost from C-2; hydrogen may be lost from C-1 or from C-3 CH 3 ±C±CHCH 3 CH 3 W W Br W CH 3 12 4 CH 3 CCH 2 CH 3 CH 3 Br 2-Bromo-2-methylbutane KOCH 2 CH 3 CH 3 CH 2 OH, 70°C CH 2 CH 3 CH 3 CH 2 C 2-Methyl-1-butene (29%) H11001 CHCH 3 H 3 C H 3 C C 2-Methyl-2-butene (71%) 1-Chlorooctadecane CH 3 (CH 2 ) 15 CH 2 CH 2 Cl CH 3 (CH 2 ) 15 CH?CH 2 1-Octadecene (86%) KOC(CH 3 ) 3 DMSO, 25°C 5.14 Dehydrohalogenation of Alkyl Halides 191 Dimethyl sulfoxide has the structure and is commonly referred to as DMSO. It is a relatively inex- pensive solvent, obtained as a byproduct in paper manu- facture. (CH 3 ) 2 S±O H11001H11002 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEM 5.18 Write structural formulas for all the alkenes that can be formed in the reaction of 2-bromobutane with potassium ethoxide. Dehydrohalogenation of cycloalkyl halides lead exclusively to cis cycloalkenes when the ring has fewer than ten carbons. As the ring becomes larger, it can accommo- date either a cis or a trans double bond, and large-ring cycloalkyl halides give mixtures of cis and trans cycloalkenes. 5.15 MECHANISM OF THE DEHYDROHALOGENATION OF ALKYL HALIDES: THE E2 MECHANISM In the 1920s, Sir Christopher Ingold proposed a mechanism for dehydrohalogenation that is still accepted as a valid description of how these reactions occur. Some of the infor- mation on which Ingold based his mechanism included these facts: 1. The reaction exhibits second-order kinetics; it is first-order in alkyl halide and first- order in base. Rate H11005 k[alkyl halide][base] Doubling the concentration of either the alkyl halide or the base doubles the reaction rate. Doubling the concentration of both reactants increases the rate by a factor of 4. 2. The rate of elimination depends on the halogen, the reactivity of alkyl halides increasing with decreasing strength of the carbon–halogen bond. Increasing rate of dehydrohalogenation Alkyl fluoride (slowest rate of elimination; strongest carbon–halogen bond) RF RCl RBr H11021H11021H11021 RI Alkyl iodide (fastest rate of elimination; weakest carbon–halogen bond) H H trans-Cyclodecene [(E)-cyclodecene] (15%) H11001 H H cis-Cyclodecene [(Z)-cyclodecene] (85%) HBr Bromocyclodecane KOCH 2 CH 3 CH 3 CH 2 OH CH 2 CH 2 CH 2 CH 3 HH CH 3 CH 2 CH 2 CC cis-4-Nonene (23%) H11001 CH 2 CH 2 CH 2 CH 3 H H CH 3 CH 2 CH 2 CC trans-4-Nonene (77%) CH 3 CH 2 CH 2 CH 2 CHCH 2 CH 2 CH 2 CH 3 Br 5-Bromononane KOCH 2 CH 3 , CH 3 CH 2 OH 192 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Cyclohexyl bromide, for example, is converted to cyclohexene by sodium ethoxide in ethanol over 60 times faster than cyclohexyl chloride. Iodide is the best leaving group in a dehydrohalogenation reaction, fluoride the poorest leaving group. Fluoride is such a poor leaving group that alkyl fluorides are rarely used as starting materials in the prepa- ration of alkenes. What are the implications of second-order kinetics? Ingold reasoned that second- order kinetics suggest a bimolecular rate-determining step involving both a molecule of the alkyl halide and a molecule of base. He concluded that proton removal from the H9252 carbon by the base occurs during the rate-determining step rather than in a separate step following the rate-determining step. What are the implications of the effects of the various halide leaving groups? Since it is the halogen with the weakest bond to carbon that reacts fastest, Ingold concluded that the carbon–halogen bond breaks in the rate-determining step. The weaker the car- bon–halogen bond, the easier it breaks. On the basis of these observations, Ingold proposed a concerted (one-step) mech- anism for dehydrohalogenation and gave it the mechanistic symbol E2, standing for elimination bimolecular. In the E2 mechanism the three key elements 1. C±H bond breaking 2. C?C H9266 bond formation 3. C±X bond breaking are all taking place at the same transition state. The carbon–hydrogen and carbon–halo- gen bonds are in the process of being broken, the base is becoming bonded to the hydro- gen, a H9266 bond is being formed, and the hybridization of carbon is changing from sp 3 to sp 2 . An energy diagram for the E2 mechanism is shown in Figure 5.10. PROBLEM 5.19 Use curved arrows to track electron movement in the dehydro- halogenation of tert-butyl chloride by sodium methoxide by the E2 mechanism. The regioselectivity of elimination is accommodated in the E2 mechanism by not- ing that a partial double bond develops at the transition state. Since alkyl groups stabi- lize double bonds, they also stabilize a partially formed H9266 bond in the transition state. The more stable alkene therefore requires a lower energy of activation for its formation and predominates in the product mixture because it is formed faster than a less stable one. Ingold was a pioneer in applying quantitative measurements of reaction rates to the understanding of organic reaction mechanisms. Many of the reactions to be described in this text were studied by him and his students during the period of about 1920 to 1950. The facts disclosed by Ingold’s experiments have been verified many times. His interpretations, although considerably refined during the decades that followed his H11001 X H11002 CCH11001BHC C H H9254H11002 B X H9254H11002 B H11002 C H X C Transition state for bimolecular elimination 5.15 Mechanism of the Dehydrohalogenation of Alkyl Halides: The E2 Mechanism 193 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website original reports, still serve us well as a starting point for understanding how the funda- mental processes of organic chemistry take place. Beta-elimination of alkyl halides by the E2 mechanism is one of those fundamental processes. 5.16 ANTI ELIMINATION IN E2 REACTIONS: STEREOELECTRONIC EFFECTS Further insight into the E2 mechanism comes from stereochemical studies. One such experiment compares the rates of elimination of the cis and trans isomers of 4-tert-butyl- cyclohexyl bromide. 194 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions CC CC C B X H C BB H X H Carbon–hydrogen bond is breaking B—H bond is forming Carbon–carbon bond is forming Carbon–halogen bond is breaking Transition state Potential energy Reaction coordinate Reactants Products Lone pair of base Lone pair of halide H 1s —base bond C 2p —C 2p bond C sp3 —Halogen bond C sp3 —H 1s bond X π π σ σ σ α β FIGURE 5.10 Potential energy diagram for con- certed E2 elimination of an alkyl halide. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Although both stereoisomers yield 4-tert-butylcyclohexene as the only alkene, they do so at quite different rates. The cis isomer reacts over 500 times faster than the trans. The difference in reaction rate results from different degrees of H9266 bond develop- ment in the E2 transition state. Since H9266 overlap of p orbitals requires their axes to be parallel, H9266 bond formation is best achieved when the four atoms of the H±C±C±X unit lie in the same plane at the transition state. The two conformations that permit this relationship are termed syn periplanar and anti periplanar. Because adjacent bonds are eclipsed when the H±C±C±X unit is syn periplanar, a transition state having this geometry is less stable than one that has an anti periplanar relationship between the proton and the leaving group. As Figure 5.11 shows, bromine is axial in the most stable conformation of cis-4- tert-butylcyclohexyl bromide, but it is equatorial in the trans stereoisomer. An axial bromine is anti periplanar with respect to the axial hydrogens at C-2 and C-6, and so X H Syn periplanar; orbitals aligned but bonds are eclipsed X H Gauche; orbitals not aligned for double bond formation X H Anti periplanar; orbitals aligned and bonds are staggered KOC(CH 3 ) 3 (CH 3 ) 3 COH KOC(CH 3 ) 3 (CH 3 ) 3 COH 4-tert-Butylcyclohexene Br trans-4-tert-Butylcyclohexyl bromide Br cis-4-tert-Butylcyclohexyl bromide 5.16 Anti Elimination in E2 Reactions: Stereoelectronic Effects 195 The peri- in periplanar means “almost” or “nearly.” Although coplanarity of the p orbitals is the best geome- try for the E2 process, mod- est deviations from this ideal can be tolerated. cis-4-tert-Butylcyclohexyl bromide Axial halide is in proper orientation for anti elimination with respect to axial hydrogens on adjacent carbon atoms. Dehydrobromination is rapid. trans-4-tert-Butylcyclohexyl bromide Equatorial halide is gauche to axial and equatorial hydrogens on adjacent carbon; cannot undergo anti elimination in this conformation. Dehydrobromination is slow. FIGURE 5.11 Confor- mations of cis- and trans-4- tert-butylcyclohexyl bromide and their relationship to the preference for an anti peri- planar arrangement of pro- ton and leaving group. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website the proper geometry between the proton and the leaving group is already present in the cis bromide, which undergoes E2 elimination rapidly. The less reactive stereoisomer, the trans bromide, has an equatorial bromine in its most stable conformation. An equatorial bromine is not anti periplanar with respect to any of the hydrogens that are H9252 to it. The relationship between an equatorial leaving group and all the C-2 and C-6 hydrogens is gauche. In order to undergo E2 elimination, the trans bromide must adopt a geometry in which the ring is strained. The transition state for its elimination is therefore higher in energy, and reaction is slower. PROBLEM 5.20 Use curved arrow notation to show the bonding changes in the reaction of cis-4-tert-butylcyclohexyl bromide with potassium tert-butoxide. Be sure your drawing correctly represents the spatial relationship between the leav- ing group and the proton that is lost. Effects that arise because one spatial arrangement of electrons (or orbitals or bonds) is more stable than another are called stereoelectronic effects. There is a stereoelec- tronic preference for the anti periplanar arrangement of proton and leaving group in E2 reactions. 5.17 A DIFFERENT MECHANISM FOR ALKYL HALIDE ELIMINATION: THE E1 MECHANISM The E2 mechanism is a concerted process in which the carbon–hydrogen and car- bon–halogen bonds both break in the same elementary step. What if these bonds break in separate steps? One possibility is the two-step mechanism of Figure 5.12, in which the carbon–halogen bond breaks first to give a carbocation intermediate, followed by depro- tonation of the carbocation in a second step. The alkyl halide, in this case 2-bromo-2-methylbutane, ionizes to a carbocation and a halide anion by a heterolytic cleavage of the carbon–halogen bond. Like the dissocia- tion of an alkyloxonium ion to a carbocation, this step is rate-determining. Because the rate-determining step is unimolecular—it involves only the alkyl halide and not the base—this mechanism is known by the symbol E1, standing for elimination unimolec- ular. It exhibits first-order kinetics. Rate H11005 k[alkyl halide] Typically, elimination by the E1 mechanism is observed only for tertiary and some secondary alkyl halides, and then only when the base is weak or in low concentration. The reactivity order parallels the ease of carbocation formation. Increasing rate of elimination by the E1 mechanism RCH 2 X Primary alkyl halide slowest rate of E1 elimination R 3 CX Tertiary alkyl halide fastest rate of E1 elimination H11021 R 2 CHX H11021 196 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Because the carbon–halogen bond breaks in the slow step, the rate of the reaction depends on the leaving group. Alkyl iodides have the weakest carbon–halogen bond and are the most reactive; alkyl fluorides have the strongest carbon–halogen bond and are the least reactive. The best examples of E1 eliminations are those carried out in the absence of added base. In the example cited in Figure 5.12, the base that abstracts the proton from the car- bocation intermediate is a very weak one; it is a molecule of the solvent, ethyl alcohol. At even modest concentrations of strong base, elimination by the E2 mechanism is much faster than E1 elimination. There is a strong similarity between the mechanism shown in Figure 5.12 and the one shown for alcohol dehydration in Figure 5.6. Indeed, we can describe the acid- catalyzed dehydration of alcohols as an E1 elimination of their conjugate acids. The main difference between the dehydration of 2-methyl-2-butanol and the dehydrohalogenation of 2-bromo-2-methylbutane is the source of the carbocation. When the alcohol is the sub- strate, it is the corresponding alkyloxonium ion that dissociates to form the carbocation. The alkyl halide ionizes directly to the carbocation. 5.17 A Different Mechanism for Alkyl Halide Elimination: The E1 Mechanism 197 The reaction: 2-Bromo-2-methylbutane 2-Methyl-1-butene (25%) 2-Methyl-2-butene (75%) The mechanism: Step (1): Alkyl halide dissociates by heterolytic cleavage of carbon–halogen bond. (Ionization step) 2-Bromo-2-methylbutane 1,1-Dimethylpropyl cation Bromide ion Step (2): Ethanol acts as a base to remove a proton from the carbocation to give the alkene products. (Deprotonation step) 1,1-Dimethylpropyl cation Ethanol Ethyloxonium ion 2-Methyl-1-butene H CH 3 CH 3 1,1-Dimethylpropyl cationEthanol CH 3 fast H Ethyloxonium ion H CH 3 CH 3 2-Methyl-2-butene CH 3 CH 3 CH 2 OH heat Br (CH 3 ) 2 CCH 2 CH 3 ±±±±￡ CH 2 CCH 2 CH 3 H11001 (CH 3 ) 2 C CHCH 3 H CH 2 CH 3 CH 3 fast H H CH 2 CH 3 CH 3 CH 3 CH 2 O H±CH 2 ±C ±￡ CH 3 CH 2 O H11001 CH 2 C CH 3 CH 2 O H±CH±C ±￡ CH 3 CH 2 O H11001 CH 3 CH C Br CH 3 slow H11001 H11001H11001 H11001H11001 CH 2 CH 3 CH 3 CH 3 H11002 CH 3 CCH 2 CH 3 ±￡ C H11001 Br FIGURE 5.12 The E1 mech- anism for the dehydro- halogenation of 2-bromo-2- methylbutane in ethanol. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Like alcohol dehydrations, E1 reactions of alkyl halides can be accompanied by carbocation rearrangements. Eliminations by the E2 mechanism, on the other hand, nor- mally proceed without rearrangement. Consequently, if one wishes to prepare an alkene from an alkyl halide, conditions favorable to E2 elimination should be chosen. In prac- tice this simply means carrying out the reaction in the presence of a strong base. 5.18 SUMMARY Section 5.1 Alkenes and cycloalkenes contain carbon–carbon double bonds. Accord- ing to IUPAC nomenclature, alkenes are named by substituting -ene for the -ane suffix of the alkane that has the same number of carbon atoms as the longest continuous chain that includes the double bond. The chain is numbered in the direction that gives the lower number to the first- appearing carbon of the double bond. The double bond takes precedence over alkyl groups and halogens in dictating the direction of numbering, but is outranked by the hydroxyl group. Section 5.2 Bonding in alkenes is described according to an sp 2 orbital hybridization model. The double bond unites two sp 2 -hybridized carbon atoms and is made of a H9268 component and a H9266 component. The H9268 bond arises by over- lap of an sp 2 hybrid orbital on each carbon. The H9266 bond is weaker than the H9268 bond and results from a side-by-side overlap of p orbitals. CH 2 4 CHCH 2 CH 2 OH 32 1 3-Buten-1-ol 2 3 4 5 1 H Br 3-Bromocyclopentene H CH 3 CH 2 CH 3 CH 2 CH 3 23 145 CC 3-Ethyl-2-pentene H11002H 2 O CH 3 CCH 2 CH 3 CH 3 Br Alkyl halide C CH 3 CH 2 CH 3 H 3 C H11001 Carbocation CH 3 CCH 2 CH 3 CH 3 O H11001 HH Alkyloxonium ion BrH11002 H11002 198 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions Sections Isomeric alkenes may be either constitutional isomers or stereoisomers. 5.3–5.4 There is a sizable barrier to rotation about a carbon–carbon double bond, which corresponds to the energy required to break the H9266 component of the double bond. Stereoisomeric alkenes are configurationally stable under normal conditions. The configurations of stereoisomeric alkenes Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website are described according to two notational systems. One system adds the prefix cis- to the name of the alkene when similar substituents are on the same side of the double bond and the prefix trans- when they are on opposite sides. The other ranks substituents according to a system of rules based on atomic number. The prefix Z is used for alkenes that have higher ranked substituents on the same side of the double bond; the prefix E is used when higher ranked substituents are on opposite sides. Section 5.5 Alkenes are relatively nonpolar. Alkyl substituents donate electrons to an sp 2 -hybridized carbon to which they are attached slightly better than hydrogen does. Section 5.6 Electron release from alkyl substituents stabilizes a double bond. In gen- eral, the order of alkene stability is: 1. Tetrasubstituted alkenes (R 2 C?CR 2 ) are the most stable. 2. Trisubstituted alkenes (R 2 C?CHR) are next. 3. Among disubstituted alkenes, trans-RCH?CHR is normally more stable than cis-RCH?CHR. Exceptions are cycloalkenes, cis cycloalkenes being more stable than trans when the ring contains fewer than 11 carbons. Terminally disubstituted alkenes (R 2 C?CH 2 ) may be slightly more or less stable than RCH?CHR, depending on their substituents. 4. Monosubstituted alkenes (RCH?CH 2 ) have a more stabilized dou- ble bond than ethylene (unsubstituted) but are less stable than disubstituted alkenes. The greater stability of more highly substituted double bonds is an exam- ple of an electronic effect. The decreased stability that results from van der Waals strain between cis substituents is an example of a steric effect. Section 5.7 Cycloalkenes that have trans double bonds in rings smaller than 12 mem- bers are less stable than their cis stereoisomers. trans-Cyclooctene can be isolated and stored at room temperature, but trans-cycloheptene is not stable above H1100230°C. Section 5.8 Alkenes are prepared by H9252 elimination of alcohols and alkyl halides. These reactions are summarized with examples in Table 5.2. In both cases, H9252 elimination proceeds in the direction that yields the more highly substituted double bond (Zaitsev’s rule). HH Cyclopropene H H cis-Cyclooctene HH Cyclobutene H H trans-Cyclooctene H H 3 CCH 2 CH 3 H CC cis-2-Pentene [(Z)-2-pentene] H H 3 C CH 2 CH 3 H CC trans-2-Pentene [(E)-2-pentene] 5.18 Summary 199 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Sections See Table 5.2. 5.9–5.11 Section 5.12 Secondary and tertiary alcohols undergo dehydration by way of carbo- cation intermediates. Step 1 Step 2 O H11001 HH R 2 CH CR 2 H11032 Alkyloxonium ion R 2 CH H11001 CR 2 H11032 Carbocation H11002H 2 O O H11001 HH R 2 CH CR 2 H11032 Alkyloxonium ion R 2 CH CR 2 H11032 OH Alcohol H H11001 200 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions TABLE 5.2 Preparation of Alkenes by Elimination Reactions of Alcohols and Alkyl Halides Reaction (section) and comments Dehydration of alcohols (Sections 5.9-5.13) Dehydra- tion requires an acid catalyst; the order of reactivity of alcohols is tertiary H11022 secondary H11022 primary. Elimi- nation is regioselective and proceeds in the direction that produces the most highly substituted double bond. When stereoisomeric alkenes are possible, the more stable one is formed in greater amounts. A car- bocation intermediate is involved, and sometimes rearrangements take place during elimination. Dehydrohalogenation of alkyl halides (Sections 5.14-5.16) Strong bases cause a proton and a halide to be lost from adjacent carbons of an alkyl halide to yield an alkene. Regioselectivity is in accord with the Zaitsev rule. The order of halide reactivity is I H11022 Br H11022 Cl H11022 F. A concerted E2 reaction pathway is followed, carbocations are not involved, and rearrangements do not normally occur. An anti periplanar arrange- ment of the proton being removed and the halide being lost characterizes the transition state. General equation and specific example Alcohol R 2 CHCR 2 H11032 OH W Alkene R 2 C?CR 2 H11032 H11001 Water H 2 O H H11001 Alkyl halide R 2 CHCR 2 H11032 X W Alkene R 2 C?CR 2 H11032 H11001H11001 Conjugate acid of base H±B H11001 Halide X H11002 Base B H11002 H11001 2-Methyl-1-hexene (19%) 2-Methyl-2-hexene (81%) 2-Methyl-2-hexanol HO H 2 SO 4 , 80°C CH 3 Cl 1-Chloro-1-methylcyclohexane CH 2 Methylenecyclohexane (6%) CH 3 1-Methylcyclohexene (94%) H11001 KOCH 2 CH 3 , CH 3 CH 2 OH, 100°C Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Step 3 Primary alcohols do not dehydrate as readily as secondary or tertiary alcohols, and their dehydration does not involve a primary carbocation. A proton is lost from the H9252 carbon in the same step in which carbon– oxygen bond cleavage occurs. Section 5.13 Alkene synthesis via alcohol dehydration is complicated by carbocation rearrangements. A less stable carbocation can rearrange to a more sta- ble one by an alkyl group migration or by a hydride shift, opening the possibility for alkene formation from two different carbocations. Section 5.14 See Table 5.2. Section 5.15 Dehydrohalogenation of alkyl halides by alkoxide bases is not compli- cated by rearrangements, because carbocations are not intermediates. The bimolecular (E2) mechanism is a concerted process in which the base abstracts a proton from the H9252 carbon while the bond between the halo- gen and the H9251 carbon undergoes heterolytic cleavage. Section 5.16 The preceding equation shows the proton H and the halogen X in the anti periplanar relationship that is required for elimination by the E2 mechanism. Section 5.17 In the absence of a strong base, alkyl halides eliminate by the unimo- lecular (E1) mechanism. The E1 mechanism involves rate-determining ionization of the alkyl halide to a carbocation, followed by deprotonation of the carbocation. Step 1 Step 2 H R 2 C H11001 CR 2 H11032 Carbocation H11002H H11001 Alkene R 2 CCR 2 H11032 X R 2 CH CR 2 H11032 Alkyl halide R 2 CH H11001 CR 2 H11032 Carbocation X H11002 H11002 X H11002 Halide ion H11001CC Alkene H11001BH Conjugate acid of base C H X C H9252H9251 Alkyl halide B H11002 Base (G is a migrating group; it may be either a hydrogen or an alkyl group) H11001 RCCR R H G Tertiary carbocation RCCR H11001 G R H Secondary carbocation H R 2 C H11001 CR 2 H11032 Carbocation H11002H H11001 Alkene R 2 CCR 2 H11032 5.18 Summary 201 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEMS 5.21 Write structural formulas for each of the following: (a) 1-Heptene (g) 1-Bromo-3-methylcyclohexene (b) 3-Ethyl-2-pentene (h) 1-Bromo-6-methylcyclohexene (c) cis-3-Octene (i) 4-Methyl-4-penten-2-ol (d) trans-1,4-Dichloro-2-butene (j) Vinylcycloheptane (e) (Z)-3-Methyl-2-hexene (k) 1,1-Diallylcyclopropane (f) (E)-3-Chloro-2-hexene (l) trans-1-Isopropenyl-3-methylcyclohexane 5.22 Write a structural formula or build a molecular model and give a correct IUPAC name for each alkene of molecular formula C 7 H 14 that has a tetrasubstituted double bond. 5.23 Give the IUPAC names for each of the following compounds: (a) (CH 3 CH 2 ) 2 C?CHCH 3 (e) (b) (CH 3 CH 2 ) 2 C?C(CH 2 CH 3 ) 2 (f) (c) (CH 3 ) 3 CCH?CCl 2 (g) (d) 5.24 (a) A hydrocarbon isolated from fish oil and from plankton was identified as 2,6,10,14-tetra- methyl-2-pentadecene. Write its structure. (b) Alkyl isothiocyanates are compounds of the type RN?C?S. Write a structural for- mula for allyl isothiocyanate, a pungent-smelling compound isolated from mustard. 5.25 (a) The sex attractant of the Mediterranean fruit fly is (E)-6-nonen-1-ol. Write a structural formula or build a molecular model for this compound, showing the stereochemistry of the double bond. (b) Geraniol is a naturally occurring substance present in the fragrant oil of many plants. It has a pleasing, roselike odor. Geraniol is the E isomer of Write a structural formula or build a molecular model for geraniol, showing its ste- reochemistry. (c) Nerol is a naturally occurring substance that is a stereoisomer of geraniol. Write its structure or build a molecular model. (d) The sex attractant of the codling moth is the 2Z, 6E stereoisomer of CH 3 CH 2 CH 2 C?CHCH 2 CH 2 C?CHCH 2 OH CH 3 W CH 2 CH 3 W (CH 3 ) 2 C?CHCH 2 CH 2 C?CHCH 2 OH CH 3 W H H H 3 C H 3 C H 3 C 202 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Write the structure of this substance or build a molecular model in a way that clearly shows its stereochemistry. (e) The sex pheromone of the honeybee is the E stereoisomer of the compound shown. Write a structural formula or build a molecular model for this compound. (f) A growth hormone from the cecropia moth has the structure shown. Express the stereochemistry of the double bonds according to the E–Z system. 5.26 Which one of the following has the largest dipole moment (is the most polar)? Compare your answer with the calculated dipole moments on the Learning By Modeling CD. 5.27 Match each alkene with the appropriate heat of combustion: Heats of combustion (kJ/mol): 5293; 4658; 4650; 4638; 4632 Heats of combustion (kcal/mol): 1264.9; 1113.4; 1111.4; 1108.6; 1107.1 (a) 1-Heptene (d) (Z)-4,4-Dimethyl-2-pentene (b) 2,4-Dimethyl-1-pentene (e) 2,4,4-Trimethyl-2-pentene (c) 2,4-Dimethyl-2-pentene 5.28 Choose the more stable alkene in each of the following pairs. Explain your reasoning. (a) 1-Methylcyclohexene or 3-methylcyclohexene (b) Isopropenylcyclopentane or allylcyclopentane (c) (d) (Z)-Cyclononene or (E)-cyclononene (e) (Z)-Cyclooctadecene or (E)-cyclooctadecene 5.29 (a) Suggest an explanation for the fact that 1-methylcyclopropene is some 42 kJ/mol (10 kcal/mol) less stable than methylenecyclopropane. (b) On the basis of your answer to part (a), compare the expected stability of 3-methylcy- clopropene with that of 1-methylcyclopropene and that of methylenecyclopropane. CH 3 1-Methylcyclopropene is less stable than CH 2 Methylenecyclopropane Bicyclo[4.2.0]oct-7-ene or Bicyclo[4.2.0]oct-3-ene H 3 C H 3 C CH 3 CH 3 CC A H 3 C Cl Cl CH 3 CC B H 3 C Cl Cl CH 3 CC C Cl Cl Cl Cl CC D O CH 3 CH 2 H 3 C CH 3 CH 2 CH 3 COCH 3 O 26 37 CH 3 C(CH 2 ) 4 CH 2 CH?CHCO 2 H O X Problems 203 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 5.30 How many alkenes would you expect to be formed from each of the following alkyl bro- mides under conditions of E2 elimination? Identify the alkenes in each case. (a) 1-Bromohexane (e) 2-Bromo-3-methylpentane (b) 2-Bromohexane (f) 3-Bromo-2-methylpentane (c) 3-Bromohexane (g) 3-Bromo-3-methylpentane (d) 2-Bromo-2-methylpentane (h) 3-Bromo-2,2-dimethylbutane 5.31 Write structural formulas for all the alkene products that could reasonably be formed from each of the following compounds under the indicated reaction conditions. Where more than one alkene is produced, specify the one that is the major product. (a) 1-Bromo-3,3-dimethylbutane (potassium tert-butoxide, tert-butyl alcohol, 100°C) (b) 1-Methylcyclopentyl chloride (sodium ethoxide, ethanol, 70°C) (c) 3-Methyl-3-pentanol (sulfuric acid, 80°C) (d) 2,3-Dimethyl-2-butanol (phosphoric acid, 120°C) (e) 3-Iodo-2,4-dimethylpentane (sodium ethoxide, ethanol, 70°C) (f) 2,4-Dimethyl-3-pentanol (sulfuric acid, 120°C) 5.32 Choose the compound of molecular formula C 7 H 13 Br that gives each alkene shown as the exclusive product of E2 elimination. (a) (e) (b) (f) (c) (g) (d) 5.33 Give the structures of two different alkyl bromides both of which yield the indicated alkene as the exclusive product of E2 elimination. (a) CH 3 CH?CH 2 (c) BrCH?CBr 2 (b) (CH 3 ) 2 C?CH 2 (d) 5.34 (a) Write the structures or build molecular models of all the isomeric alkyl bromides hav- ing the molecular formula C 5 H 11 Br. (b) Which one undergoes E1 elimination at the fastest rate? (c) Which one is incapable of reacting by the E2 mechanism? (d) Which ones can yield only a single alkene on E2 elimination? (e) For which isomer does E2 elimination give two alkenes which are not constitutional isomers? (f) Which one yields the most complex mixture of alkenes on E2 elimination? CH 3 CH 3 CH 3 CH 3 C(CH 3 ) 3 CH 3 CH(CH 3 ) 2 CH 2 CH CH 2 204 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 5.35 (a) Write the structures or build molecular models of all the isomeric alcohols having the molecular formula C 5 H 12 O. (b) Which one will undergo acid-catalyzed dehydration most readily? (c) Write the structure of the most stable C 5 H 11 carbocation. (d) Which alkenes may be derived from the carbocation in part (c)? (e) Which alcohols can yield the carbocation in part (c) by a process involving a hydride shift? (f) Which alcohols can yield the carbocation in part (c) by a process involving a methyl shift? 5.36 Predict the major organic product of each of the following reactions. In spite of the struc- tural complexity of some of the starting materials, the functional group transformations are all of the type described in this chapter. (a) (b) (c) (d) (e) (f) (g) KOC(CH 3 ) 3 DMSO, 70°C CH 3 CH 3 H 3 C Cl Cl (C 10 H 14 ) H 2 SO 4 140–145°C HOC(CH 2 CO 2 H) 2 CO 2 H Citric acid (C 6 H 6 O 6 ) KHSO 4 130–150°C HO CN CH 3 O (C 12 H 11 NO) KOC(CH 3 ) 3 (CH 3 ) 3 COH, heat (CH 3 ) 2 CCl NaOCH 2 CH 3 CH 3 CH 2 OH, heat CH 3 Br H H 3 C KOC(CH 3 ) 3 (CH 3 ) 3 COH, heat ICH 2 CH(OCH 2 CH 3 ) 2 KHSO 4 heat Br CHCH 2 CH 3 OH Problems 205 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (h) (i) (j) 5.37 Evidence has been reported in the chemical literature that the reaction (CH 3 CH 2 ) 2 CHCH 2 Br H11001 KNH 2 ±￡ (CH 3 CH 2 ) 2 C?CH 2 H11001 NH 3 H11001 KBr proceeds by the E2 mechanism. Use curved arrow notation to represent the flow of electrons for this process. 5.38 The rate of the reaction (CH 3 ) 3 CCl H11001 NaSCH 2 CH 3 ±￡ (CH 3 ) 2 C?CH 2 H11001 CH 3 CH 2 SH H11001 NaCl is first-order in (CH 3 ) 3 CCl and first-order in NaSCH 2 CH 3 . Give the symbol (E1 or E2) for the most reasonable mechanism, and use curved arrow notation to represent the flow of electrons. 5.39 Menthyl chloride and neomenthyl chloride have the structures shown. One of these stereoisomers undergoes elimination on treatment with sodium ethoxide in ethanol much more readily than the other. Which reacts faster, menthyl chloride or neomenthyl chloride? Why? (Mo- lecular models will help here.) 5.40 The stereoselectivity of elimination of 5-bromononane on treatment with potassium ethox- ide was described in Section 5.14. Draw Newman projections or make molecular models of 5-bromononane showing the conformations that lead to cis-4-nonene and trans-4-nonene, respec- tively. Identify the proton that is lost in each case, and suggest a mechanistic explanation for the observed stereoselectivity. 5.41 In the acid-catalyzed dehydration of 2-methyl-1-propanol, what carbocation would be formed if a hydride shift accompanied cleavage of the carbon–oxygen bond in the alkyloxonium ion? What ion would be formed as a result of a methyl shift? Which pathway do you think will predominate, a hydride shift or a methyl shift? 5.42 Each of the following carbocations has the potential to rearrange to a more stable one. Write the structure of the rearranged carbocation. (a) CH 3 CH 2 CH 2 H11001 (b) (CH 3 ) 2 CHC H11001 HCH 3 CH(CH 3 ) 2 Cl H 3 C Menthyl chloride CH(CH 3 ) 2 Cl H 3 C Neomenthyl chloride NaOCH 3 CH 3 OH, heat (CH 3 ) 3 C CH 3 Cl H KOH heat O Br CH 3 O CH 3 O CH 3 O CH 3 OCH 2 (C 10 H 18 O 5 ) KOC(CH 3 ) 3 DMSO BrBr O O O O (C 14 H 16 O 4 ) 206 CHAPTER FIVE Structure and Preparation of Alkenes: Elimination Reactions Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) (CH 3 ) 3 CC H11001 HCH 3 (e) (d) (CH 3 CH 2 ) 3 CCH 2 H11001 5.43 Write a sequence of steps depicting the mechanisms of each of the following reactions: (a) (b) (c) 5.44 In Problem 5.16 (Section 5.13) we saw that acid-catalyzed dehydration of 2,2-dimethylcy- clohexanol afforded 1,2-dimethylcyclohexene. To explain this product we must write a mechanism for the reaction in which a methyl shift transforms a secondary carbocation to a tertiary one. Another product of the dehydration of 2,2-dimethylcyclohexanol is isopropylidenecyclopentane. Write a mechanism to rationalize its formation. 5.45 Acid-catalyzed dehydration of 2,2-dimethyl-1-hexanol gave a number of isomeric alkenes including 2-methyl-2-heptene as shown in the following formula. (a) Write a stepwise mechanism for the formation of 2-methyl-2-heptene. (b) What other alkenes do you think are formed in this reaction? 5.46 Compound A (C 4 H 10 ) gives two different monochlorides on photochemical chlorination. Treatment of either of these monochlorides with potassium tert-butoxide in dimethyl sulfoxide gives the same alkene B (C 4 H 8 ) as the only product. What are the structures of compound A, the two monochlorides, and alkene B? 5.47 Compound A (C 6 H 14 ) gives three different monochlorides on photochemical chlorination. One of these monochlorides is inert to E2 elimination. The other two monochlorides yield the same alkene B (C 6 H 12 ) on being heated with potassium tert-butoxide in tert-butyl alcohol. Iden- tify compound A, the three monochlorides, and alkene B. H 2 SO 4 heat CH 3 CH 2 CH 2 CH 2 CCH 2 OH CH 3 CH 3 CH 3 CH 2 CH 2 CH 2 CH CH 3 CH 3 C H11001 H H11001 heat C(CH 3 ) 2 Isopropylidenecyclopentane CH 3 CH 3 1,2-Dimethylcyclohexene H OH CH 3 CH 3 2,2-Dimethylcyclohexanol KHSO 4 170°C CH 3 CH 3 CH 3 H OH CH 3 CH 3 CH 3 OH H 2 SO 4 heat H H11001 C(CH 3 ) 3 OH CH 3 CH 2 C CH 3 CH 3 H11001 Problems 207 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website