Names and Structures of Alkyl Halides Although commonly called alkyl halides , halogen-substituted alkanes are named systematically as haloalkanes ( ), treating the halogen as a substituent on a parent alkane chain. There are three steps: STEP 1 Find the longest chain, and name it as the parent. If a double or triple bond is present, the parent chain must contain it. STEP 2 Number the carbons of the parent chain beginning at the end nearer the first substituent, whether alkyl or halo. Assign each substituent a number according to its position on the chain. If different halogens are present, number each one and list them in alphabetical order when writing the name. STEP 3 If the parent chain can be properly numbered from either end by step 2, begin at the end nearer the substituent that has alphabetical precedence. In addition to their systematic names, many simple alkyl halides are also named by identifying first the alkyl group and then the halogen. For example, CH 3 I can be called either iodomethane or methyl iodide. Such names are well entrenched in the chemical literature and in daily usage, but they won’t be used in this book. Halogens increase in size going down the periodic table, so the lengths of the corresponding carbon–halogen bonds increase accordingly ( ). In addition, C−X bond strengths decrease going down the periodic table. As we’ve been doing thus far, we’ll continue using an X to represent any of the halogens F, Cl, Br, or I. A Comparison of the Halomethanes Halomethane Bond length (pm) Bond strength Dipole moment ( D ) (kJ/mol) (kcal/mol) CH 3 F 139 460 110 1.85 CH 3 Cl 178 350 84 1.87 CH 3 Br 193 294 70 1.81 CH 3 I 214 239 57 1.62 In our discussion of bond polarity in functional groups in , we noted that halogens are more electronegative than carbon. The C−X bond is therefore polar, with the carbon atom bearing a slight positive charge ( δ +) and the halogen a slight negative charge ( δ −). This polarity results in a dipole moment for all the halomethanes ( ) and implies that the alkyl halide C−X carbon atom should behave as an electrophile in polar reactions. We’ll soon see that this is indeed the case. Problems

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Preparing Alkyl Halides from Alkanes: Radical Halogenation As we saw briefly in , simple alkyl halides can sometimes be prepared by the radical reaction of an alkane with Cl 2 or Br 2 in the presence of ultraviolet light. The detailed mechanism is shown in for chlorination. Mechanism of the radical chlorination of methane. Three kinds of steps are required in radical substitution reactions: initiation, propagation, and termination. The propagation steps are a repeating cycle, with Cl · a reactant in step 1 and a product in step 2, and with · CH 3 a product in step 1 and a reactant in step 2. The symbol hν shown in the initiation step is the standard way of indicating irradiation with light ( ν is the lowercase Greek letter nu). Radical substitution reactions require three kinds of steps: initiation, propagation , and termination . Once an initiation step has started the process by producing radicals, the reaction continues in a self-sustaining cycle. The cycle requires two repeating propagation steps in which a radical, the halogen, and the alkane yield alkyl halide product plus more radical to carry on the chain. The chain is occasionally terminated by the combination of two radicals. Unfortunately, alkane halogenation is a poor synthetic method for preparing alkyl halides because mixtures of products invariably result. For example, chlorination of methane does not stop cleanly at the monochlorinated stage but continues to give a mixture of dichloro, trichloro, and even tetrachloro products. The situation is even worse for chlorination of alkanes that have more than one kind of hydrogen. Chlorination of butane, for instance, gives two monochlorinated products in a 30 : 70 ratio in addition to multiple chlorinated products such as dichlorobutane, trichlorobutane, and so on. As another example, 2-methylpropane yields 2-chloro-2-methylpropane and 1-chloro-2-methylpropane in a 35 : 65 ratio, along with more highly chlorinated products. From these and similar reactions, it’s possible to calculate a reactivity order toward chlorination for different kinds of hydrogen atoms in a molecule. Take the butane chlorination, for instance. Butane has six equivalent primary hydrogens (−CH 3 ) and four equivalent secondary hydrogens (−CH 2 −). The fact that butane yields 30% of 1-chlorobutane product means that each one of the six primary hydrogens is responsible for 30% ÷ 6 = 5% of the product. Similarly, the fact that 70% of 2-chlorobutane is formed means that each of the four secondary hydrogens is responsible for 70% ÷ 4 = 17.5% of the product. Thus, a secondary hydrogen reacts 17.5% ÷ 5% = 3.5 times as often as a primary hydrogen. A similar calculation for the chlorination of 2-methylpropane indicates that each of the nine primary hydrogens accounts for 65% ÷ 9 = 7.2% of the product, while the single tertiary hydrogen (R 3 CH) accounts for 35% of the product. Thus, a tertiary hydrogen is 35% ÷ 7.2% = 5 times as reactive as a primary hydrogen toward chlorination. The observed reactivity order of alkane hydrogens toward radical chlorination can be explained by looking at the bond dissociation energies given previously in . The data show that a tertiary C−H bond (400 kJ/mol; 96 kcal/mol) is weaker than a secondary C−H bond (410 kJ/mol; 98 kcal/mol), which is in turn weaker than a primary C−H bond (421 kJ/mol; 101 kcal/mol). Since less energy is needed to break a tertiary C−H bond than to break a primary or secondary C−H bond, the resultant tertiary radical is more stable than a primary or secondary radical. Problems

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Preparing Alkyl Halides from Alkenes: Allylic Bromination We’ve already seen several methods for preparing alkyl halides from alkenes, including the reactions of HX and X 2 with alkenes in electrophilic addition reactions ( and ). The hydrogen halides HCl, HBr, and HI react with alkenes by a polar mechanism to give the product of Markovnikov addition. Bromine and chlorine undergo anti addition through halonium ion intermediates to give 1,2-dihalogenated products. Another laboratory method for preparing alkyl halides from alkenes is by reaction with N -bromosuccinimide (abbreviated NBS), in the presence of ultraviolet light, to give products resulting from substitution of hydrogen by bromine at the position next to the double bond—the allylic position. Cyclohexene, for example, gives 3-bromocyclohexene. This allylic bromination with NBS is analogous to the alkane chlorination reaction discussed in the previous section and occurs by a radical chain-reaction pathway ( ). As in alkane halogenation, a Br · radical abstracts an allylic hydrogen atom, forming an allylic radical plus HBr. The HBr then reacts with NBS to form Br 2 , which in turn reacts with the allylic radical to yield the brominated product and a Br · radical that cycles back into the first step and carries on the chain. Mechanism of allylic bromination of an alkene with NBS. The process is a radical chain reaction in which ( 1 ) a Br · radical abstracts an allylic hydrogen atom of the alkene and gives an allylic radical plus HBr. ( 2 ) The HBr then reacts with NBS to form Br 2 , which ( 3 ) reacts with the allylic radical to yield the bromoalkene product and a Br · radical that continues the chain. Why does bromination with NBS occur exclusively at an allylic position rather than elsewhere in the molecule? The answer, once again, is found by looking at bond dissociation energies to see the relative stabilities of various kinds of radicals. Although a typical secondary alkyl C−H bond has a strength of about 410 kJ/mol (98 kcal/mol) and a typical vinylic C−H bond has a strength of 465 kJ/mol (111 kcal/mol), an allylic C−H bond has a strength of only about 370 kJ/mol (88 kcal/mol). An allylic radical is therefore more stable than a typical alkyl radical with the same substitution by about 40 kJ/mol (9 kcal/mol). We can thus expand the stability ordering to include vinylic and allylic radicals.

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Stability of the Allyl Radical: Resonance Revisited To see why an allylic radical is so stable, look at the orbital picture in . The radical carbon atom with an unpaired electron can adopt sp 2 hybridization, placing the unpaired electron in a p orbital and giving a structure that is electronically symmetrical. The p orbital on the central carbon can therefore overlap equally well with a p orbital on either of the two neighboring carbons. An orbital view of the allyl radical. The p orbital on the central carbon can overlap equally well with a p orbital on either neighboring carbon, giving rise to two equivalent resonance structures. Because the allyl radical is electronically symmetrical, it has two resonance forms—one with the unpaired electron on the left and the double bond on the right and another with the unpaired electron on the right and the double bond on the left. Neither structure is correct by itself; the true structure of the allyl radical is a resonance hybrid of the two. (You might want to review to to brush up on resonance.) As noted in , the greater the number of resonance forms, the greater the stability of a compound, because bonding electrons are attracted to more nuclei. An allyl radical, with two resonance forms, is therefore more stable than a typical alkyl radical, which has only a single structure. In molecular orbital terms, the stability of the allyl radical is due to the fact that the unpaired electron is delocalized , or spread out, over an extended π -orbital network rather than localized at only one site, as shown by the computer-generated MO in . This delocalization is particularly apparent in the so-called spin-density surface in , which shows the calculated location of the unpaired electron. The two terminal carbons share the unpaired electron equally. The spin density surface of the allyl radical locates the position of the unpaired electron and shows that it is equally shared between the two terminal carbons. In addition to its effect on stability, delocalization of the unpaired electron in the allyl radical has other chemical consequences. Because the unpaired electron is delocalized over both ends of the π orbital system, reaction with Br 2 can occur at either end. As a result, allylic bromination of an unsymmetrical alkene often leads to a mixture of products. For example, bromination of 1-octene gives a mixture of 3-bromo-1-octene and 1-bromo-2-octene. The two products are not formed in equal amounts, however, because the intermediate allylic radical is not symmetrical and reaction at the two ends is not equally likely. Reaction at the less hindered, primary end is favored. The products of allylic bromination reactions are useful for conversion into dienes by dehydrohalogenation with base. Cyclohexene can be converted into 1,3-cyclohexadiene, for example. Predicting the Product of an Allylic Bromination Reaction What products would you expect from the reaction of 4,4-dimethylcyclohexene with NBS? Strategy Draw the alkene reactant, and identify the allylic positions. In this case, there are two different allylic positions; we’ll label them A and B . Now abstract an allylic hydrogen from each position to generate the two corresponding allylic radicals. Each of the two allylic radicals can add a Br atom at either end ( A or A′ ; B or B′ ), to give a mixture of up to four products. Draw and name the products. In the present instance, the “two” products from reaction at position B are identical, so only three products are formed in this reaction. Solution Problems

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Preparing Alkyl Halides from Alcohols The most generally useful method for preparing alkyl halides is to make them from alcohols, which themselves can be obtained from carbonyl compounds as we’ll see in Sections 17.4 and 17.5 . Because of the importance of this process, many different methods have been developed to transform alcohols into alkyl halides. The simplest method is to treat the alcohol with HCl, HBr, or HI. For reasons that will be discussed in Section 11.5 , this reaction works best with tertiary alcohols, R 3 COH. Primary and secondary alcohols react much more slowly and at higher temperatures. The reaction of HX with a tertiary alcohol is so rapid that it’s often carried out simply by bubbling pure HCl or HBr gas into a cold ether solution of the alcohol. 1-Methylcyclohexanol, for example, is converted into 1-chloro-1-methylcyclohexane by treatment with HCl. Primary and secondary alcohols are best converted into alkyl halides by treatment with either thionyl chloride (SOCl 2 ) or phosphorus tribromide (PBr 3 ). These reactions, which normally take place readily under mild conditions, are less acidic and less likely to cause acid-catalyzed rearrangements than the HX method. As the preceding examples indicate, the yields of these SOCl 2 and PBr 3 reactions are generally high and other functional groups such as ethers, carbonyls, and aromatic rings don’t usually interfere. We’ll look at the mechanisms of these and other related substitution reactions in Section 11.3 . Alkyl fluorides can also be prepared from alcohols. Numerous alternative reagents are used for such reactions, including diethylaminosulfur trifluoride [(CH 3 CH 2 ) 2 NSF 3 ] and HF in pyridine solvent. Problems

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Reactions of Alkyl Halides: Grignard Reagents Alkyl halides, RX, react with magnesium metal in ether or tetrahydrofuran (THF) solvent to yield alkylmagnesium halides, RMgX. The products, called Grignard reagents (RMgX) after their discoverer, Francois Auguste Victor Grignard, who received the 1912 Nobel Prize in Chemistry, are examples of organometallic compounds because they contain a carbon–metal bond. In addition to alkyl halides, Grignard reagents can also be made from alkenyl (vinylic) and aryl (aromatic) halides. The halogen can be Cl, Br, or I, although chlorides are less reactive than bromides and iodides. Organofluorides rarely react with magnesium. As you might expect from the discussion of electronegativity and bond polarity in , the carbon–magnesium bond is polarized, making the carbon atom of Grignard reagents both nucleophilic and basic. An electrostatic potential map of methylmagnesium iodide, for instance, indicates the electron-rich (red) character of the carbon bonded to magnesium. A Grignard reagent is formally the magnesium salt, R 3 C –+ MgX, of a carbon acid, R 3 C–H, and is thus a carbon anion, or carbanion . But because hydrocarbons are such weak acids, with p K a ’s in the range 44 to 60 ( ), carbon anions are very strong bases. Grignard reagents must therefore be protected from atmospheric moisture to prevent their being protonated and destroyed in acid–base reactions: R–Mg–X + H 2 O → R–H + HO–Mg–X. Grignard reagents themselves don’t occur in living organisms, but they serve as useful carbon-based nucleophiles in several important laboratory reactions, which we’ll look at in detail in Section 17.5. In addition, they act as a simple model for other, more complex carbon-based nucleophiles that are important in biological chemistry. We’ll see many examples of these in Chapter 29. Problems

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Organometallic Coupling Reactions Many other kinds of organometallic compounds can be prepared in a manner similar to that of Grignard reagents. For instance, alkyllithium reagents, RLi, can be prepared by the reaction of an alkyl halide with lithium metal. Alkyllithiums are both nucleophiles and strong bases, and their chemistry is similar in many respects to that of alkylmagnesium halides. One particularly valuable reaction of alkyllithiums occurs when making lithium diorganocopper compounds, R 2 CuLi, by reaction with copper(I) iodide in diethyl ether as solvent. Often called Gilman reagents (LiR 2 Cu) , lithium diorganocopper compounds are useful because they undergo a coupling reaction with organochlorides, bromides, and iodides (but not fluorides). One of the alkyl groups from the lithium diorganocopper reagent replaces the halogen of the organohalide, forming a new carbon–carbon bond and yielding a hydrocarbon product. Lithium dimethylcopper, for instance, reacts with 1-iododecane to give undecane in a 90% yield. This organometallic coupling reaction is useful in organic synthesis because it forms carbon–carbon bonds, thereby allowing the preparation of larger molecules from smaller ones. As the following examples indicate, the coupling reaction can be carried out on aryl and vinylic halides as well as on alkyl halides. An organocopper coupling reaction is carried out commercially to synthesize muscalure, (9 Z )-tricosene, the sex attractant secreted by the common housefly. Minute amounts of muscalure greatly increase the lure of insecticide-treated fly bait and provide an effective and species-specific means of insect control. The mechanism of the coupling reaction involves initial formation of a triorganocopper intermediate, followed by coupling and loss of a mono-organocopper, RCu. The coupling is not a typical polar nucleophilic substitution reaction of the sort considered in the next chapter. In addition to the coupling reaction of diorganocopper reagents with organohalides, related processes also occur with other organometallic reagents, particularly organopalladium compounds. One of the most commonly used procedures is the coupling reaction of an aromatic or vinyl substituted boronic acid [R—B(OH) 2 ] with an aromatic or vinyl substituted organohalide in the presence of a base and a palladium catalyst. This reaction is less general than the diorganocopper reaction because it doesn’t work with alkyl substrates, but it is preferred when possible because it uses only a catalytic amount of metal rather than a full equivalent and because palladium compounds are less toxic than copper compounds. For example: Called the Suzuki–Miyaura reaction, this process is particularly useful for preparing so-called biaryl compounds, which have two linked aromatic rings. A large number of commonly used drugs fit this description, so the Suzuki–Miyaura reaction is much-used in the pharmaceutical industry. As an example, valsartan, marketed as Diovan, is widely prescribed to treat high blood pressure, heart failure, and diabetic kidney disease. Its synthesis begins with a Suzuki–Miyaura coupling of ortho -chlorobenzonitrile with para -methylbenzeneboronic acid. Shown in a simplified form in , the mechanism of the Suzuki–Miyaura reaction involves initial reaction of the aromatic halide with the palladium catalyst to form an organopalladium intermediate, followed by reaction of that intermediate with the aromatic boronic acid. The resultant diorganopalladium complex then decomposes to the coupled biaryl product plus regenerated catalyst. Mechanism of the Suzuki–Miyaura coupling reaction of an aromatic boronic acid with an aromatic halide to give a biaryl. The reaction takes place by ( 1 ) reaction of the aromatic halide, ArX, with the catalyst to form an organopalladium intermediate, followed by ( 2 ) reaction with the aromatic boronic acid. ( 3 ) Subsequent decomposition of the diarylpalladium intermediate gives the biaryl product. Problems

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Oxidation and Reduction in Organic Chemistry We’ve pointed out on several occasions that some of the reactions discussed in this and earlier chapters are either oxidations or reductions. As noted in , an organic oxidation results in a loss of electron density by carbon, caused either by bond formation between carbon and a more electronegative atom (usually O, N, or a halogen) or by bond-breaking between carbon and a less electronegative atom (usually H). Conversely, an organic reduction results in a gain of electron density by carbon, caused either by bond formation between carbon and a less electronegative atom or by bond-breaking between carbon and a more electronegative atom ( ). Oxidation Decreases electron density on carbon by: – forming one of these: C− O C− N C− X – or breaking this: C– H Reduction Increases electron density on carbon by: – forming this: C– H – or breaking one of these: C– O C– N C– X Based on these definitions, the chlorination reaction of methane to yield chloromethane is an oxidation because a C−H bond is broken and a C−Cl bond is formed. The conversion of an alkyl chloride to an alkane via a Grignard reagent followed by protonation is a reduction, however, because a C−Cl bond is broken and a C−H bond is formed. As other examples, the reaction of an alkene with Br 2 to yield a 1,2-dibromide is an oxidation because two C−Br bonds are formed, but the reaction of an alkene with HBr to yield an alkyl bromide is neither an oxidation nor a reduction because both a C−H and a C−Br bond are formed. A list of compounds of increasing oxidation level is shown in . Alkanes are at the lowest oxidation level because they have the maximum possible number of C−H bonds per carbon, and CO 2 is at the highest level because it has the maximum possible number of C−O bonds per carbon. Any reaction that converts a compound from a lower level to a higher level is an oxidation, any reaction that converts a compound from a higher level to a lower level is a reduction, and any reaction that doesn’t change the level is neither an oxidation nor a reduction. Oxidation levels of some common compounds. shows how to compare the oxidation levels of different compounds with the same number of carbon atoms. Comparing Oxidation Levels Rank the following compounds in order of increasing oxidation level: Strategy Compounds that have the same number of carbon atoms can be compared by adding the number of C−O, C−N, and C−X bonds in each and then subtracting the number of C−H bonds. The larger the resultant value, the higher the oxidation level. Solution The first compound (propene) has six C−H bonds, giving an oxidation level of −6; the second (2-propanol) has one C−O bond and seven C−H bonds, giving an oxidation level of −6; the third (acetone) has two C−O bonds and six C−H bonds, giving an oxidation level of −4; and the fourth (propane) has eight C−H bonds, giving an oxidation level of −8. Thus, the order of increasing oxidation level is Problems Naturally Occurring Organohalides Just forty years ago in 1980, only about 30 naturally occurring organohalides were known. It was simply assumed that chloroform, halogenated phenols, chlorinated aromatic compounds called PCBs, and other such substances found in the environment were industrial pollutants. Now, less than half a century later, the situation is quite different. More than 5000 organohalides have been found to occur naturally, and tens of thousands more surely exist. From a simple compound like chloromethane to an extremely complex one like the antibiotic vancomycin, a remarkably diverse range of organohalides exists in plants, bacteria, and animals. Many even have valuable physiological activity. The pentahalogenated alkene halomon, for instance, has been isolated from the red alga Portieria hornemannii and found to have anticancer activity against several human tumor cell lines. Some naturally occurring organohalides are produced in massive quantities. Forest fires, volcanic eruptions, and marine kelp release up to 5 million tons of CH 3 Cl per year, for example, while annual industrial emissions total about 26,000 tons. Termites are thought to release as much as 10 8 kg of chloroform per year. A detailed examination of the Okinawan acorn worm Ptychodera flava found that the 64 million worms living in a 1 km 2 study area excreted nearly 8000 pounds per year of bromophenols and bromoindoles, compounds previously thought to be non-natural pollutants. Why do organisms produce organohalides, many of which are undoubtedly toxic? The answer seems to be that many organisms use organohalogen compounds for self-defense, either as feeding deterrents, irritants to predators, or natural pesticides. Marine sponges, coral, and sea hares, for example, release foul-tasting organohalides that deter fish, starfish, and other predators. Even humans appear to produce halogenated compounds as part of their defense against infection. The human immune system contains a peroxidase enzyme capable of carrying out halogenation reactions on fungi and bacteria, thereby killing the pathogen. And most remarkable of all, even free chlorine—Cl 2 —has been found to be present in humans. Marine corals secrete organohalogen compounds that act as a feeding deterrent to fish. (credit: “Coral reef” by Qui Nguyen, United Nations Environment Programme/Flickr, Public Domain) Much remains to be learned—only a few hundred of the more than 500,000 known species of marine organisms have been examined—but it’s clear that organohalides are an integral part of the world around us. alkyl halide allylic carbanion delocalized Gilman reagent (LiR 2 Cu) Grignard reagent (RMgX) organohalide organometallic Summary Alkyl halides are not often found in terrestrial organisms, but the kinds of reactions they undergo are among the most important and well-studied reaction types in organic chemistry. In this chapter, we saw how to name and prepare alkyl halides, and we’ll soon make a detailed study of their substitution and elimination reactions. Simple alkyl halides can be prepared by radical halogenation of alkanes, but mixtures of products usually result. The reactivity order of alkanes toward halogenation is identical to the stability order of radicals: R 3 C · > R 2 CH · > RCH 2 · . Alkyl halides can also be prepared from alkenes by reaction with N -bromosuccinimide (NBS) to give the product of allylic bromination. The NBS bromination of alkenes takes place through an intermediate allylic radical, which is stabilized by resonance. Alcohols react with HX to form alkyl halides, but the reaction works well only for tertiary alcohols, R 3 COH. Primary and secondary alkyl halides are normally prepared from alcohols using either SOCl 2 , PBr 3 , or HF in pyridine. Alkyl halides react with magnesium in ether solution to form organomagnesium halides, called Grignard reagents (RMgX) , which are both nucleophilic and strongly basic. Alkyl halides also react with lithium metal to form organolithium reagents, RLi. In the presence of CuI, these form diorganocoppers, or Gilman reagents (LiR 2 Cu) . Gilman reagents react with organohalides to yield coupled hydrocarbon products. Summary of Reactions No stereochemistry is implied unless specifically indicated with wedged, solid, and dashed lines. Preparation of alkyl halides From alkenes by allylic bromination ( ) From alcohols ( ) (1) Reaction with HX (2) Reaction of 1° and 2° alcohols with SOCl 2 (3) Reaction of 1° and 2° alcohols with PBr 3 (4) Reaction of 1° and 2° alcohols with HF–pyridine Reactions of alkyl halides Formation of Grignard (organomagnesium) reagents ( ) Formation of Gilman (diorganocopper) reagents ( ) Organometallic coupling ( ) (1) Diorganocopper reaction (2) Palladium-catalyzed Suzuki–Miyaura reaction Additional Problems Visualizing Chemistry Mechanism Problems Naming Alkyl Halides Synthesizing Alkyl Halides Oxidation and Reduction General Problems

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Why This Chapter? Competition occurs throughout nature. In chemistry, competition often occurs between alternative reaction pathways, such as in the substitution and elimination reactions of alkyl halides. (credit: modification of work “Bull moose fight” by Grand Teton, National Parks Service/Flickr, Public Domain) Nucleophilic substitution and base-induced elimination are two of the most widely occurring and versatile reactions in organic chemistry, both in the laboratory and in biological pathways. We’ll look at them closely in this chapter to see how they occur, what their characteristics are, and how they can be used. We’ll begin with substitution reactions. We saw in the preceding chapter that the carbon–halogen bond in an alkyl halide is polar and that the carbon atom is electron-poor. Thus, alkyl halides are electrophiles, and much of their chemistry involves polar reactions with nucleophiles and bases. Alkyl halides do one of two things when they react with a nucleophile/base such as hydroxide ion: they either undergo substitution of the X group by the nucleophile, or they undergo elimination of HX to yield an alkene.

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The Discovery of Nucleophilic Substitution Reactions Discovery of the nucleophilic substitution reaction of alkyl halides dates back to work carried out by the German chemist Paul Walden in 1896. Walden found that the pure enantiomeric (+)- and (–)-malic acids could be interconverted through a series of simple substitution reactions. When Walden treated (–)-malic acid with PCl 5 , he isolated (+)-chlorosuccinic acid. This, on treatment with wet Ag 2 O, gave (+)-malic acid. Similarly, reaction of (+)-malic acid with PCl 5 gave (–)-chlorosuccinic acid, which was converted into (–)-malic acid when treated with wet Ag 2 O. The full cycle of reactions is shown in . Walden’s cycle of reactions interconverting (+)- and (–)-malic acids. At the time, the results were astonishing. The eminent chemist Emil Fischer called Walden’s discovery “the most remarkable observation made in the field of optical activity since the fundamental observations of Pasteur.” Because (–)-malic acid was converted into (+)-malic acid, some reactions in the cycle must have occurred with a change, or inversion, of configuration at the chirality center. But which ones, and how? (Remember from that the direction of light rotation and the configuration of a chirality center aren’t directly related. You can’t tell by looking at the sign of rotation whether a change in configuration has occurred during a reaction.) Today, we refer to the transformations taking place in Walden’s cycle as nucleophilic substitution reactions because each step involves the substitution of one nucleophile (chloride ion, Cl – , or hydroxide ion, HO – ) by another. Nucleophilic substitution reactions are one of the most common and versatile reaction types in organic chemistry. Following the work of Walden, further investigations were undertaken during the 1920s and 1930s to clarify the mechanism of nucleophilic substitution reactions and to find out how inversions of configuration occur. Among the first series studied was one that interconverted the two enantiomers of 1-phenyl-2-propanol ( ). Although this particular series of reactions involves nucleophilic substitution of an alkyl para -toluenesulfonate (called a tosylate ) rather than an alkyl halide, exactly the same type of reaction is involved as that studied by Walden. For all practical purposes, the entire tosylate group acts as if it were simply a halogen substituent. (In fact, when you see a tosylate substituent in a molecule, do a mental substitution and tell yourself that you’re dealing with an alkyl halide.) A Walden cycle interconverting (+) and (–) enantiomers of 1-phenyl-2-propanol. Chirality centers are marked by asterisks, and the bonds broken in each reaction are indicated by red wavy lines. The inversion of chirality occurs in step 2 , where acetate ion substitutes for tosylate ion. In the three-step reaction sequence shown in , (+)-1-phenyl-2-propanol is interconverted with its (–) enantiomer, so at least one of the three steps must involve an inversion of configuration at the chirality center. Step 1 , formation of a tosylate, occurs by breaking the O–H bond of the alcohol rather than the C–O bond to the chiral carbon, so the configuration around the carbon is unchanged. Similarly, step 3 , hydroxide-ion cleavage of the acetate, takes place without breaking the C–O bond at the chirality center. Thus, the inversion of stereochemical configuration must take place in step 2 , the nucleophilic substitution of tosylate ion by acetate ion. From this and nearly a dozen other series of similar reactions, researchers concluded that the nucleophilic substitution reaction of a primary or secondary alkyl halide or tosylate always proceeds with inversion of configuration. (Tertiary alkyl halides and tosylates, as we’ll see shortly, give different stereochemical results and react by a different mechanism than the primary and secondary ones.) Predicting the Stereochemistry of a Nucleophilic Substitution Reaction What product would you expect from a nucleophilic substitution reaction of ( R )-1-bromo-1-phenylethane with cyanide ion, − C≡N , as nucleophile? Show the stereochemistry of both reactant and product, assuming that inversion of configuration occurs. Strategy Draw the R enantiomer of the reactant, and then change the configuration of the chirality center while replacing the – Br with a – CN. Solution Problems

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The S N 2 Reaction In almost all chemical reactions, there is a direct relationship between the rate at which the reaction occurs and the concentrations of the reactants. When we measure this relationship, we measure the kinetics of the reaction. For example, let’s look at the kinetics of a simple nucleophilic substitution—the reaction of CH 3 Br with OH – to yield CH 3 OH plus Br – . With a given temperature, solvent, and concentration of reactants, the substitution occurs at a certain rate. If we double the concentration of OH – , the frequency of encounters between reaction partners doubles and we find that the reaction rate also doubles. Similarly, if we double the concentration of CH 3 Br, the reaction rate again doubles. We call such a reaction, in which the rate is linearly dependent on the concentrations of two species, a second-order reaction . Mathematically, we can express this second-order dependence of the nucleophilic substitution reaction by setting up a rate equation . As either [RX] or [ – OH] changes, the rate of the reaction changes proportionately. Reaction rate = Rate of disappearance of reactant = k × [ RX ] × [ − OH ] where [ RX ] = CH 3 Br concentration in molarity [ − OH] = − OH concentration in molarity k = a constant value (the rate constant) A mechanism that accounts for both the inversion of configuration and the second-order kinetics that are observed with nucleophilic substitution reactions was suggested in 1937 by the British chemists E. D. Hughes and Christopher Ingold, who formulated what they called the S N 2 reaction —short for substitution, nucleophilic, bimolecular. ( Bimolecular means that two molecules, nucleophile and alkyl halide, take part in the step whose kinetics are measured.) The essential feature of the S N 2 mechanism is that it takes place in a single step, without intermediates, when the incoming nucleophile reacts with the alkyl halide or tosylate (the substrate ) from a direction opposite the group that is displaced (the leaving group ). As the nucleophile comes in on one side of the substrate and bonds to the carbon, the halide or tosylate departs from the other side, thereby inverting the stereochemical configuration. The process is shown in for the reaction of ( S )-2-bromobutane with HO – to give ( R )-2-butanol. MECHANISM The mechanism of the S N 2 reaction. The reaction takes place in a single step when the incoming nucleophile approaches from a direction 180° away from the leaving halide ion, thereby inverting the stereochemistry at carbon. As shown in , the S N 2 reaction occurs when an electron pair on the nucleophile Nu : – forces out the group X : – , which takes with it the electron pair from the former C–X bond. This occurs through a transition state in which the new Nu–C bond is partially formed at the same time that the old C–X bond is partially broken and in which the negative charge is shared by both the incoming nucleophile and the outgoing halide ion. The transition state for this inversion has the remaining three bonds to carbon in a planar arrangement ( ). The transition state of an S N 2 reaction has a planar arrangement of the carbon atom and the remaining three groups. Electrostatic potential maps show that negative charge is delocalized in the transition state. The mechanism proposed by Hughes and Ingold is fully consistent with experimental results, explaining both stereochemical and kinetic data. Thus, the requirement for a backside approach of the entering nucleophile (180° away from the departing X group) causes the stereochemistry of the substrate to invert, much like an umbrella turning inside-out in the wind. The Hughes–Ingold mechanism also explains why second-order kinetics are observed: the S N 2 reaction occurs in a single step that involves both alkyl halide and nucleophile. Two molecules are involved in the step whose rate is measured. Problems

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Characteristics of the S N 2 Reaction Now that we know how S N 2 reactions occur, we need to see how they can be used and what variables affect them. Some S N 2 reactions are fast, and some are slow; some take place in high yield and others in low yield. Understanding the factors involved can be of tremendous value. Let’s begin by recalling a few things about reaction rates in general. The rate of a chemical reaction is determined by the activation energy ∆ G ‡ , the energy difference between reactant ground state and transition state. A change in reaction conditions can affect ∆ G ‡ either by changing the reactant energy level or by changing the transition-state energy level. Lowering the reactant energy or raising the transition-state energy increases ∆ G ‡ and decreases the reaction rate; raising the reactant energy or decreasing the transition-state energy decreases ∆ G ‡ and increases the reaction rate ( ). We’ll see examples of all these effects as we look at S N 2 reaction variables. The effects of changes in reactant and transition-state energy levels on reaction rate. (a) A higher reactant energy level (red curve) corresponds to a faster reaction (smaller ∆ G ‡ ). (b) A higher transition-state energy level (red curve) corresponds to a slower reaction (larger ∆ G ‡ ). Steric Effects in the S N 2 Reaction The first S N 2 reaction variable to look at is the structure of the substrate. Because the S N 2 transition state involves partial bond formation between the incoming nucleophile and the alkyl halide carbon atom, it seems reasonable that a hindered, bulky substrate should prevent easy approach of the nucleophile, making bond formation difficult. In other words, the transition state for reaction of a sterically hindered substrate, whose carbon atom is “shielded” from the approach of the incoming nucleophile, is higher in energy and forms more slowly than the corresponding transition state for a less hindered substrate ( ). Steric hindrance to the S N 2 reaction. As the models indicate, the carbon atom in (a) bromomethane is readily accessible, resulting in a fast S N 2 reaction. The carbon atoms in (b) bromoethane (primary), (c) 2-bromopropane (secondary), and (d) 2-bromo-2-methylpropane (tertiary) are successively more hindered, resulting in successively slower S N 2 reactions. As shows, the difficulty of nucleophile approach increases as the three substituents bonded to the halo-substituted carbon atom increase in size. Methyl halides are by far the most reactive substrates in S N 2 reactions, followed by primary alkyl halides such as ethyl and propyl. Alkyl branching at the reacting center, as in isopropyl halides (2°), slows the reaction greatly, and further branching, as in tert -butyl halides (3°), effectively halts the reaction. Even branching one carbon away from the reacting center, as in 2,2-dimethylpropyl (neopentyl) halides, greatly hinders nucleophilic displacement. As a result, S N 2 reactions occur only at relatively unhindered sites and are normally useful only with methyl halides, primary halides, and a few simple secondary halides. Relative reactivities for some different substrates are as follows: Vinylic halides ( R 2 C═CRX ) and aryl halides are not shown on this reactivity list because they are unreactive toward S N 2 displacement. This lack of reactivity is due to steric factors: the incoming nucleophile would have to approach in the plane of the carbon–carbon double bond and burrow through part of the molecule to carry out a backside displacement. The Nucleophile Another variable that has a major effect on the S N 2 reaction is the nature of the nucleophile. Any species, either neutral or negatively charged, can act as a nucleophile as long as it has an unshared pair of electrons; that is, as long as it is a Lewis base. If the nucleophile is negatively charged, the product is neutral; if the nucleophile is neutral, the product is positively charged. A wide array of substances can be prepared using nucleophilic substitution reactions. In fact, we’ve already seen examples in previous chapters. For instance, the reaction of an acetylide anion with an alkyl halide, discussed in , is an S N 2 reaction in which the acetylide nucleophile displaces a halide leaving group. lists some nucleophiles in the order of their reactivity, shows the products of their reactions with bromomethane, and gives the relative rates of their reactions. There are large differences in the rates at which various nucleophiles react. Some S N 2 Reactions with Bromomethane Nu: – + CH 3 Br → CH 3 Nu + Br – Nucleophile Product Relative rate of reaction Formula Name Formula Name H 2 O Water CH 3 OH 2 + Methylhydronium ion 1 CH 3 CO 2 – Acetate CH 3 CO 2 CH 3 Methyl acetate 500 NH 3 Ammonia CH 3 NH 3 + Methylammonium ion 700 Cl – Chloride CH 3 Cl Chloromethane 1,000 HO – Hydroxide CH 3 OH Methanol 10,000 CH 3 O – Methoxide CH 3 OCH 3 Dimethyl ether 25,000 I – Iodide CH 3 I Iodomethane 100,000 – CN Cyanide CH 3 CN Acetonitrile 125,000 HS – Hydrosulfide CH 3 SH Methanethiol 125,000 What are the reasons for the reactivity differences observed in ? Why do some reactants appear to be much more “nucleophilic” than others? The answers to these questions aren’t straightforward. Part of the problem is that the term nucleophilicity is imprecise. The term is usually taken to be a measure of the affinity of a nucleophile for a carbon atom in the S N 2 reaction, but the reactivity of a given nucleophile can change from one reaction to the next. The exact nucleophilicity of a species in a given reaction depends on the substrate, the solvent, and even the reactant concentrations. Detailed explanations for the observed nucleophilicities aren’t always simple, but some trends can be detected from the data of . Nucleophilicity roughly parallels basicity when comparing nucleophiles that have the same reacting atom. Thus, OH – is both more basic and more nucleophilic than acetate ion, CH 3 CO 2 – , which in turn is more basic and more nucleophilic than H 2 O. Since “nucleophilicity” is usually taken as the affinity of a Lewis base for a carbon atom in the S N 2 reaction and “basicity” is the affinity of a base for a proton, it’s easy to see why there might be a correlation between the two kinds of behavior. Nucleophilicity usually increases going down a column of the periodic table. Thus, HS – is more nucleophilic than HO – , and the halide reactivity order is I – > Br – > Cl – . Going down the periodic table, elements have their valence electrons in successively larger shells where they are successively farther from the nucleus, less tightly held, and consequently more reactive. This matter is complex, though, and the nucleophilicity order can change depending on the solvent. Negatively charged nucleophiles are usually more reactive than neutral ones. As a result, S N 2 reactions are often carried out under basic conditions rather than neutral or acidic conditions. Problems The Leaving Group Still another variable that can affect the S N 2 reaction is the nature of the group displaced by the incoming nucleophile, the leaving group . Because the leaving group is expelled with a negative charge in most S N 2 reactions, the best leaving groups are those that best stabilize the negative charge in the transition state. The greater the extent of charge stabilization by the leaving group, the lower the energy of the transition state and the more rapid the reaction. But as we saw in , the groups that best stabilize a negative charge are also the weakest bases. Thus, weak bases such as Cl – , Br – , and tosylate ion make good leaving groups, while strong bases such as OH – and NH 2 – make poor leaving groups. It’s just as important to know which are poor leaving groups as to know which are good, and the preceding data clearly indicate that F – , HO – , RO – , and H 2 N – are not displaced by nucleophiles. In other words, alkyl fluorides, alcohols, ethers, and amines do not typically undergo S N 2 reactions. To carry out an S N 2 reaction with an alcohol, it’s necessary to convert the – OH into a better leaving group. This, in fact, is just what happens when a primary or secondary alcohol is converted into either an alkyl chloride by reaction with SOCl 2 or an alkyl bromide by reaction with PBr 3 ( ). Alternatively, an alcohol can be made more reactive toward nucleophilic substitution by treating it with para -toluenesulfonyl chloride to form a tosylate. As noted previously, tosylates are even more reactive than halides in nucleophilic substitutions. Note that tosylate formation does not change the configuration of the oxygen-bearing carbon because the C–O bond is not broken. The one general exception to the rule that ethers don’t typically undergo S N 2 reactions pertains to epoxides, the three-membered cyclic ethers that we saw in . Because of the angle strain in their three-membered ring, epoxides are much more reactive than other ethers. They react with aqueous acid to give 1,2-diols, as we saw in , and they react readily with many other nucleophiles as well. Propene oxide, for instance, reacts with HCl to give 1-chloro-2-propanol by an S N 2 backside attack on the less hindered primary carbon atom. We’ll look at the process in more detail in Section 18.5 . Problems The Solvent The rates of S N 2 reactions are strongly affected by the solvent. Protic solvents—those that contain an –OH or –NH group—are generally the worst for S N 2 reactions, while polar aprotic solvents, which are polar but don’t have an –OH or –NH group, are the best. Protic solvents, such as methanol and ethanol, slow down S N 2 reactions by solvation of the reactant nucleophile. The solvent molecules hydrogen-bond to the nucleophile and form a cage around it, thereby lowering its energy and reactivity. In contrast with protic solvents—which decrease the rates of S N 2 reactions by lowering the ground-state energy of the nucleophile—polar aprotic solvents increase the rates of S N 2 reactions by raising the ground-state energy of the nucleophile. Acetonitrile (CH 3 CN), dimethylformamide [(CH 3 ) 2 NCHO, abbreviated DMF], and dimethyl sulfoxide [(CH 3 ) 2 SO, abbreviated DMSO] are particularly useful. A solvent known as hexamethylphosphoramide {[(CH 3 ) 2 N] 3 PO, abbreviated HMPA} can also be useful but it should only be handled with great care and not be allowed to touch the eyes or skin. These solvents can dissolve many salts because of their high polarity, but they tend to solvate metal cations rather than nucleophilic anions. As a result, the bare, unsolvated anions have a greater nucleophilicity and S N 2 reactions take place at correspondingly increased rates. For instance, a rate increase of 200,000 has been observed on changing from methanol to HMPA for the reaction of azide ion with 1-bromobutane. Problems A Summary of S N 2 Reaction Characteristics The effects on S N 2 reactions of the four variables—substrate structure, nucleophile, leaving group, and solvent—are summarized in the following statements and in the energy diagrams of : Variable Effects Substrate Steric hindrance raises the energy of the S N 2 transition state, increasing ∆ G ‡ and decreasing the reaction rate ( a). As a result, S N 2 reactions are best for methyl and primary substrates. Secondary substrates react slowly, and tertiary substrates do not react by an S N 2 mechanism. Nucleophile Basic, negatively charged nucleophiles are less stable and have a higher ground-state energy than neutral ones, decreasing ∆ G ‡ and increasing the S N 2 reaction rate ( b). Leaving group Good leaving groups (more stable anions) lower the energy of the transition state, decreasing ∆ G ‡ and increasing the S N 2 reaction rate ( c). Solvent Protic solvents solvate the nucleophile, thereby lowering its ground-state energy, increasing ∆ G ‡ , and decreasing the S N 2 reaction rate. Polar aprotic solvents surround the accompanying cation but not the nucleophilic anion, thereby raising the ground-state energy of the nucleophile, decreasing ∆ G ‡ , and increasing the reaction rate ( d). Energy diagrams showing the effects of (a) substrate, (b) nucleophile, (c) leaving group, and (d) solvent on S N 2 reaction rates. Substrate and leaving group effects are felt primarily in the transition state. Nucleophile and solvent effects are felt primarily in the reactant ground state.

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The S N 1 Reaction Most nucleophilic substitutions take place by the S N 2 pathway just discussed. The reaction is favored when carried out with an unhindered substrate and a negatively charged nucleophile in a polar aprotic solvent, but is disfavored when carried out with a hindered substrate and a neutral nucleophile in a protic solvent. You might therefore expect the reaction of a tertiary substrate (hindered) with water (neutral, protic) to be among the slowest of substitution reactions. Remarkably, however, the opposite is true. The reaction of the tertiary halide 2-bromo-2-methylpropane (CH 3 ) 3 CBr with H 2 O to give the alcohol 2-methyl-2-propanol is more than 1 million times faster than the corresponding reaction of CH 3 Br to give methanol. What’s going on here? A nucleophilic substitution reaction is occurring—a hydroxyl group is replacing a halogen—yet the reactivity order seems backward. These reactions can’t be taking place by the S N 2 mechanism we’ve been discussing, so we must therefore conclude that they are occurring by an alternative substitution mechanism. This alternative mechanism is called the S N 1 reaction , for substitution, nucleophilic, unimolecular . In contrast to the S N 2 reaction of CH 3 Br with OH – , the S N 1 reaction of (CH 3 ) 3 CBr with H 2 O has a rate that depends only on the alkyl halide concentration and is independent of the H 2 O concentration. In other words, the process is a first-order reaction ; the concentration of the nucleophile does not appear in the rate equation. Reaction rate = Rate of disappearance of alkyl halide = k × [ RX ] To explain this result, we need to know more about kinetics measurements. Many organic reactions occur in several steps, one of which usually has a higher-energy transition state than the others and is therefore slower. We call this step with the highest transition-state energy the rate-limiting step , or rate-determining step . No reaction can proceed faster than its rate-limiting step, which acts as a kind of traffic jam, or bottleneck. In the S N 1 reaction of (CH 3 ) 3 CBr with H 2 O, the fact that the nucleophile concentration does not appear in the first-order rate equation means that it is not involved in the rate-limiting step and must therefore be involved in some other, non-rate-limiting step. The mechanism shown in accounts for these observations. MECHANISM The mechanism of the S N 1 reaction of 2-bromo-2-methylpropane with H 2 O involves three steps. Step 1 —the spontaneous, unimolecular dissociation of the alkyl bromide to yield a carbocation—is rate-limiting. Unlike what occurs in an S N 2 reaction, where the leaving group is displaced while the incoming nucleophile approaches, an S N 1 reaction takes place by loss of the leaving group before the nucleophile approaches. 2-Bromo-2-methylpropane spontaneously dissociates to the tert -butyl carbocation (CH 3 ) 3 C + , plus Br – in a slow, rate-limiting step, and the intermediate carbocation is then immediately trapped by the nucleophile water in a faster second step. Thus, water is not a reactant in the step whose rate is measured. The energy diagram is shown in . An energy diagram for an S N 1 reaction. The rate-limiting step is the spontaneous dissociation of the alkyl halide to give a carbocation intermediate. Reaction of the carbocation with a nucleophile then occurs in a second, faster step. Because an S N 1 reaction occurs through a carbocation intermediate, its stereochemical outcome is different from that of an S N 2 reaction. Carbocations, as we’ve seen, are planar, sp 2 -hybridized, and achiral. Thus, if we carry out an S N 1 reaction on one enantiomer of a chiral reactant and go through an achiral carbocation intermediate, the product loses its optical activity ( ). That is, the symmetrical intermediate carbocation can react with a nucleophile equally well from either side, leading to a racemic, 50 : 50 mixture of enantiomers ( ). Stereochemistry of the S N 1 reaction. Because the reaction goes through an achiral intermediate, an enantiomerically pure reactant gives an optically inactive racemic product. The conclusion that S N 1 reactions on enantiomerically pure substrates should give racemic products is nearly, but not exactly, what is found. In fact, few S N 1 displacements occur with complete racemization. Most give a minor (0–20%) excess of inversion. The reaction of ( R )-6-chloro-2,6-dimethyloctane with H 2 O, for example, leads to an alcohol product that is approximately 80% racemized and 20% inverted (80% R,S + 20% S is equivalent to 40% R + 60% S ). This lack of complete racemization in S N 1 reactions is due to the fact that ion pairs are involved. According to this explanation, first proposed by Saul Winstein at UCLA, dissociation of the substrate occurs to give a structure in which the two ions are still loosely associated and in which the carbocation is effectively shielded from reaction on one side by the departing anion. If a certain amount of substitution occurs before the two ions fully diffuse apart, then a net inversion of configuration will be observed ( ). Ion pairs in an S N 1 reaction. The leaving group shields one side of the carbocation intermediate from reaction with the nucleophile, thereby leading to some inversion of configuration rather than complete racemization. Problems

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Characteristics of the S N 1 Reaction Just as the S N 2 reaction is strongly influenced by the structure of the substrate, the leaving group, the nucleophile, and the solvent, the S N 1 reaction is similarly influenced. Factors that lower ∆ G ‡ , either by lowering the energy level of the transition state or by raising the energy level of the ground state, favor faster S N 1 reactions. Conversely, factors that raise ∆ G ‡ , either by raising the energy level of the transition state or by lowering the energy level of the reactant, slow down the S N 1 reaction. The Substrate According to the Hammond postulate ( ), any factor that stabilizes a high-energy intermediate also stabilizes the transition state leading to that intermediate. Because the rate-limiting step in an S N 1 reaction is the spontaneous, unimolecular dissociation of the substrate to yield a carbocation, the reaction is favored whenever a stabilized carbocation intermediate is formed. The more stable the carbocation intermediate, the faster the S N 1 reaction. We saw in that the stability order of alkyl carbocations is 3° > 2° > 1° > methyl. To this list we should also add the resonance-stabilized allyl and benzyl cations. Just as allylic radicals are unusually stable because the unpaired electron can be delocalized over an extended π orbital system ( ), so allylic and benzylic carbocations are unusually stable. (The word benzylic means “next to an aromatic ring.”) As indicates, an allylic cation has two resonance forms. In one form, the double bond is on the “left”; in the other form it’s on the “right.” A benzylic cation has five resonance forms, all of which contribute to the overall resonance hybrid. Resonance forms of allylic and benzylic carbocations. The positive charge is delocalized over the π system in both. Electron-poor atoms are indicated by blue arrows. Because of resonance stabilization, a primary allylic or benzylic carbocation is about as stable as a secondary alkyl carbocation, and a secondary allylic or benzylic carbocation is about as stable as a tertiary alkyl carbocation. This stability order of carbocations is the same as the order of S N 1 reactivity for alkyl halides and tosylates. We should also note parenthetically that primary allylic and benzylic substrates are particularly reactive in S N 2 reactions as well as in S N 1 reactions. Allylic and benzylic C–X bonds are about 50 kJ/mol (12 kcal/mol) weaker than the corresponding saturated bonds and are therefore more easily broken. Problems The Leaving Group We said during the discussion of S N 2 reactivity that the best leaving groups are those that are most stable; that is, those that are the conjugate bases of strong acids. An identical reactivity order is found for the S N 1 reaction because the leaving group is directly involved in the rate-limiting step. Thus, the S N 1 reactivity order is Note that in the S N 1 reaction, which is often carried out under acidic conditions, neutral water is sometimes the leaving group. This occurs, for example, when an alkyl halide is prepared from a tertiary alcohol by reaction with HBr or HCl ( ). As shown in , the alcohol is first protonated and then spontaneously loses H 2 O to generate a carbocation, which reacts with halide ion to give the alkyl halide. Knowing that an S N 1 reaction is involved in the conversion of alcohols to alkyl halides explains why the reaction works well only for tertiary alcohols. Tertiary alcohols react fastest because they give the most stable carbocation intermediates. MECHANISM The mechanism of the S N 1 reaction of a tertiary alcohol with HBr to yield an alkyl halide. Neutral water is the leaving group (step 2 ). The Nucleophile The nature of the nucleophile plays a major role in the S N 2 reaction but does not affect an S N 1 reaction. Because the S N 1 reaction occurs through a rate-limiting step in which the added nucleophile has no part, the nucleophile can’t affect the reaction rate. The reaction of 2-methyl-2-propanol with HX, for instance, occurs at the same rate regardless of whether X is Cl, Br, or I. Furthermore, neutral nucleophiles are just as effective as negatively charged ones, so S N 1 reactions frequently occur under neutral or acidic conditions. The Solvent What about the solvent? Do solvents have the same effect in S N 1 reactions that they have in S N 2 reactions? The answer is both yes and no. Yes, solvents have a large effect on S N 1 reactions, but no, the reasons for the effects on S N 1 and S N 2 reactions are not the same. Solvent effects in the S N 2 reaction are due largely to stabilization or destabilization of the nucleophile reactant, while solvent effects in the S N 1 reaction are due largely to stabilization or destabilization of the transition state . The Hammond postulate says that any factor stabilizing the intermediate carbocation should increase the rate of an S N 1 reaction. Solvation of the carbocation—the interaction of the ion with solvent molecules—has such an effect. Solvent molecules orient around the carbocation so that the electron-rich ends of the solvent dipoles face the positive charge ( ), thereby lowering the energy of the ion and favoring its formation. Solvation of a carbocation by water. The electron-rich oxygen atoms of solvent molecules orient around the positively charged carbocation and thereby stabilize it. The properties of a solvent that contribute to its ability to stabilize ions by solvation are related to the solvent’s polarity. S N 1 reactions take place much more rapidly in strongly polar solvents, such as water and methanol, than in less polar solvents, such as ether and chloroform. In the reaction of 2-chloro-2-methylpropane, for example, a rate increase of 100,000 is observed upon going from ethanol (less polar) to water (more polar). The rate increases when going from a hydrocarbon solvent to water are so large they can’t be measured accurately. It should be emphasized again that both the S N 1 and the S N 2 reaction show solvent effects, but that they do so for different reasons. S N 2 reactions are disfavored in protic solvents because the ground-state energy of the nucleophile is lowered by solvation. S N 1 reactions are favored in protic solvents because the transition-state energy leading to carbocation intermediate is lowered by solvation. A Summary of S N 1 Reaction Characteristics The effects on S N 1 reactions of the four variables—substrate, leaving group, nucleophile, and solvent—are summarized in the following statements: Variable Effects Substrate The best substrates yield the most stable carbocations. As a result, S N 1 reactions are best for tertiary, allylic, and benzylic halides. Leaving group Good leaving groups increase the reaction rate by lowering the energy level of the transition state for carbocation formation. Nucleophile The nucleophile must be nonbasic to prevent a competitive elimination of HX ( ), but otherwise does not affect the reaction rate. Neutral nucleophiles work well. Solvent Polar solvents stabilize the carbocation intermediate by solvation, thereby increasing the reaction rate. Predicting the Mechanism of a Nucleophilic Substitution Reaction Predict whether each of the following substitution reactions is likely to be S N 1 or S N 2: Strategy Look at the substrate, leaving group, nucleophile, and solvent. Then decide from the summaries at the ends of and whether an S N 1 or an S N 2 reaction is favored. S N 1 reactions are favored by tertiary, allylic, or benzylic substrates, by good leaving groups, by nonbasic nucleophiles, and by protic solvents. S N 2 reactions are favored by primary substrates, by good leaving groups, by good nucleophiles, and by polar aprotic solvents. Solution (a) This is likely to be an S N 1 reaction because the substrate is secondary and benzylic, the nucleophile is weakly basic, and the solvent is protic. (b) This is likely to be an S N 2 reaction because the substrate is primary, the nucleophile is a good one, and the solvent is polar aprotic. Problems

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Biological Substitution Reactions Both S N 1 and S N 2 reactions are common in biological chemistry, particularly in the pathways for biosynthesis of the many thousands of plant-derived substances called terpenoids, which we saw briefly in the Chapter 8 Chemistry Matters and will discuss in Section 27.5 . Unlike what typically happens in the laboratory, however, the substrate in a biological substitution reaction is usually an organodiphosphate rather than an alkyl halide. Thus, the leaving group is the diphosphate ion, abbreviated PP i , rather than a halide ion. In fact, it’s useful to think of the diphosphate group as the “biological equivalent” of a halide. The dissociation of an organodiphosphate in a biological reaction is typically assisted by complexation to a divalent metal cation such as Mg 2+ to help neutralize charge and make the diphosphate a better leaving group. As an example, two S N 1 reactions occur during the biosynthesis of geraniol, a fragrant alcohol found in roses and used in perfumery. Geraniol biosynthesis begins with dissociation of dimethylallyl diphosphate to give an allylic carbocation, which reacts with isopentenyl diphosphate ( ). From the viewpoint of isopentenyl diphosphate, the reaction is an electrophilic alkene addition, but from the viewpoint of dimethylallyl diphosphate, the process is an S N 1 reaction in which the carbocation intermediate reacts with a double bond as the nucleophile. Biosynthesis of geraniol from dimethylallyl diphosphate. Two S N 1 reactions occur, both with diphosphate ion as the leaving group. Following this initial S N 1 reaction, loss of the pro- R hydrogen gives geranyl diphosphate, itself an allylic diphosphate that dissociates a second time. Reaction of the geranyl carbocation with water in a second S N 1 reaction, followed by loss of a proton, then yields geraniol. As another example, S N 2 reactions are involved in almost all biological methylations, which transfer a –CH 3 group from an electrophilic donor to a nucleophile. The donor is S -adenosylmethionine (abbreviated SAM), which contains a positively charged sulfur (a sulfonium ion, ), and the leaving group is the neutral S -adenosylhomocysteine molecule. In the biosynthesis of epinephrine (adrenaline) from norepinephrine, for instance, the nucleophilic nitrogen atom of norepinephrine attacks the electrophilic methyl carbon atom of S -adenosylmethionine in an S N 2 reaction, displacing S -adenosylhomocysteine ( ). In effect, S -adenosylmethionine is simply a biological equivalent of CH 3 Cl. The biosynthesis of epinephrine from norepinephrine occurs by an S N 2 reaction with S -adenosylmethionine. Problems

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Elimination Reactions: Zaitsev’s Rule We said at the beginning of this chapter that two kinds of reactions can take place when a nucleophile/Lewis base reacts with an alkyl halide. The nucleophile can either substitute for the halide by reaction at carbon or can cause elimination of HX by reaction at a neighboring hydrogen: Elimination reactions are more complex than substitution reactions for several reasons. One is the problem of regiochemistry. What product results by loss of HX from an unsymmetrical halide? In fact, elimination reactions almost always give mixtures of alkene products, and the best we can usually do is to predict which will be the major product. According to Zaitsev’s rule , formulated in 1875 by the Russian chemist Alexander Zaitsev, base-induced elimination reactions generally (although not always) give the more stable alkene product—that is, the alkene with more alkyl substituents on the double-bond carbons. In the following two cases, for example, the more highly substituted alkene product predominates. ZAITSEV’S RULE In the elimination of HX from an alkyl halide, the more highly substituted alkene product predominates. Another factor that complicates a study of elimination reactions is that they can take place by different mechanisms, just as substitutions can. We’ll consider three of the most common mechanisms—the E1, E2, and E1cB reactions—which differ in the timing of C–H and C–X bond-breaking. In the E1 reaction, the C–X bond breaks first to give a carbocation intermediate, which undergoes subsequent base abstraction of H + to yield the alkene. In the E2 reaction, base-induced C–H bond cleavage is simultaneous with C–X bond cleavage, giving the alkene in a single step. In the E1cB reaction (cB for “ conjugate base ”), base abstraction of the proton occurs first, giving a carbanion (R : – ) intermediate. This anion, the conjugate base of the reactant “acid,” then undergoes loss of X – in a subsequent step to give the alkene. All three mechanisms occur frequently in the laboratory, but the E1cB mechanism predominates in biological pathways. Predicting the Product of an Elimination Reaction What product would you expect from reaction of 1-chloro-1-methylcyclohexane with KOH in ethanol? Strategy Treatment of an alkyl halide with a strong base such as KOH yields an alkene. To find the products in a specific case, locate the hydrogen atoms on each carbon next to the leaving group, and then generate the potential alkene products by removing HX in as many ways as possible. The major product will be the one that has the most highly substituted double bond—in this case, 1-methylcyclohexene. Solution Problems

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The E2 Reaction and the Deuterium Isotope Effect The E2 reaction (for elimination, bimolecular ) occurs when an alkyl halide is treated with a strong base, such as hydroxide ion or alkoxide ion (RO – ). It is the most commonly occurring pathway for elimination and can be formulated as shown in . MECHANISM Mechanism of the E2 reaction of an alkyl halide. The reaction takes place in a single step through a transition state in which the double bond begins to form at the same time the H and X groups are leaving. Like the S N 2 reaction, the E2 reaction takes place in one step without intermediates. As the base begins to abstract H + from a carbon next to the leaving group, the C–H bond begins to break, a C═C bond begins to form, and the leaving group begins to depart, taking with it the electron pair from the C–X bond. Among the pieces of evidence supporting this mechanism is the fact that E2 reactions show second-order kinetics and follow the rate law: rate = k × [RX] × [Base]. That is, both the base and alkyl halide take part in the rate-limiting step. A second piece of evidence in support of the E2 mechanism is provided by a phenomenon known as the deuterium isotope effect . For reasons that we won’t go into, a carbon–hydrogen bond is weaker by about 5 kJ/mol (1.2 kcal/mol) than the corresponding carbon–deuterium bond. Thus, a C–H bond is more easily broken than an equivalent C–D bond, and the rate of C–H bond cleavage is faster. For instance, the base-induced elimination of HBr from 1-bromo-2-phenylethane proceeds 7.11 times faster than the corresponding elimination of DBr from 1-bromo-2, 2-dideuterio-2-phenylethane. This result tells us that the C–H (or C–D) bond is broken in the rate-limiting step, consistent with our picture of the E2 reaction as a one-step process. If it were otherwise, we wouldn’t observe a rate difference. Yet a third piece of mechanistic evidence involves the stereochemistry of E2 eliminations. As shown by a large number of experiments, E2 reactions occur with periplanar geometry, meaning that all four reacting atoms—the hydrogen, the two carbons, and the leaving group—lie in the same plane. Two such geometries are possible: syn periplanar geometry, in which the H and the X are on the same side of the molecule, and anti periplanar geometry, in which the H and the X are on opposite sides of the molecule. Of the two, anti periplanar geometry is energetically preferred because it allows the substituents on the two carbons to adopt a staggered relationship, whereas syn geometry requires that the substituents be eclipsed. What’s so special about periplanar geometry? Because the sp 3 σ orbitals in the reactant C–H and C–X bonds must overlap and become p π orbitals in the alkene product, there must also be some overlap in the transition state. This can occur most easily if all the orbitals are in the same plane to begin with—that is, if they’re periplanar ( ). The transition state for the E2 reaction of an alkyl halide with base. Overlap of the developing p orbitals in the transition state requires periplanar geometry of the reactant. You can think of E2 elimination reactions with periplanar geometry as being similar to S N 2 reactions with 180° geometry. In an S N 2 reaction, an electron pair from the incoming nucleophile pushes out the leaving group on the opposite side of the molecule. In an E2 reaction, an electron pair from a neighboring C–H bond also pushes out the leaving group on the opposite side of the molecule. Anti periplanar geometry for E2 eliminations has specific stereochemical consequences that provide strong evidence for the proposed mechanism. To take just one example, meso -1,2-dibromo-1,2-diphenylethane undergoes E2 elimination on treatment with base to give only the E alkene. None of the isomeric Z alkene is formed because the transition state leading to the Z alkene would have to have syn periplanar geometry and would thus be higher in energy. Predicting the Double-Bond Stereochemistry of the Product in an E2 Reaction What stereochemistry do you expect for the alkene obtained by E2 elimination of (1 S ,2 S )-1,2-dibromo-1,2-diphenylethane? Strategy Draw (1 S ,2 S )-1,2-dibromo-1,2-diphenylethane so that you can see its stereochemistry and so that the –H and –Br groups to be eliminated are anti periplanar. Then carry out the elimination while keeping all substituents in approximately the same positions, and see what alkene results. Solution Anti periplanar elimination of HBr gives ( Z )-1-bromo-1,2-diphenylethylene. Problems

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The E2 Reaction and Cyclohexane Conformation Anti periplanar geometry for E2 reactions is particularly important in cyclohexane rings, where chair geometry forces a rigid relationship between the substituents on neighboring carbon atoms ( ). The anti periplanar requirement for E2 reactions overrides Zaitsev’s rule and can be met in cyclohexanes only if the hydrogen and the leaving group are trans diaxial ( ). If either the leaving group or the hydrogen is equatorial, E2 elimination can’t occur. The geometric requirement for an E2 reaction in a substituted cyclohexane. The leaving group and the hydrogen must both be axial for anti periplanar elimination to occur. The elimination of HCl from the isomeric menthyl and neomenthyl chlorides shown in gives a good illustration of this trans-diaxial requirement. Neomenthyl chloride undergoes elimination of HCl on reaction with ethoxide ion 200 times faster than menthyl chloride. Furthermore, neomenthyl chloride yields 1-menthene as the major alkene product, whereas menthyl chloride yields 2-menthene. Dehydrochlorination of menthyl and neomenthyl chlorides. (a) Neomenthyl chloride loses HCl directly from its more stable conformation, but (b) menthyl chloride must first ring-flip to a higher energy conformation before HCl loss can occur. The abbreviation “Et” represents an ethyl group. The difference in reactivity between the isomeric menthyl chlorides is due to the difference in their conformations. Neomenthyl chloride has the conformation shown in a, with the methyl and isopropyl groups equatorial and the chlorine axial—a perfect geometry for E2 elimination. Loss of the hydrogen atom at C1 occurs easily to yield the more substituted alkene product, 1-menthene, as predicted by Zaitsev’s rule. Menthyl chloride, by contrast, has a conformation in which all three substituents are equatorial ( b). To achieve the necessary geometry for elimination, menthyl chloride must first ring-flip to a higher-energy chair conformation, in which all three substituents are axial. E2 elimination then occurs with loss of the only trans-diaxial hydrogen available, leading to the non-Zaitsev product 2-menthene. The net effect of the simple change in chlorine stereochemistry is a 200-fold change in reaction rate and a complete change of product. The chemistry of the molecule is controlled by its conformation. Problems

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The E1 and E1cB Reactions The E1 Reaction Just as the E2 reaction is analogous to the S N 2 reaction, the S N 1 reaction has a close analog called the E1 reaction (for elimination, unimolecular). The E1 reaction can be formulated as shown in , with the elimination of HCl from 2-chloro-2-methylpropane. MECHANISM Mechanism of the E1 reaction. Two steps are involved, the first of which is rate-limiting, and a carbocation intermediate is present. E1 eliminations begin with the same unimolecular dissociation to give a carbocation that we saw in the S N 1 reaction, but the dissociation is followed by loss of H + from the adjacent carbon rather than by substitution. In fact, the E1 and S N 1 reactions normally occur together whenever an alkyl halide is treated in a protic solvent with a nonbasic nucleophile. Thus, the best E1 substrates are also the best S N 1 substrates, and mixtures of substitution and elimination products are usually obtained. For example, when 2-chloro-2-methylpropane is warmed to 65 °C in 80% aqueous ethanol, a 64 : 36 mixture of 2-methyl-2-propanol (S N 1) and 2-methylpropene (E1) results. Much evidence has been obtained in support of the E1 mechanism. For example, E1 reactions show first-order kinetics, consistent with a rate-limiting, unimolecular dissociation process. Furthermore, E1 reactions show no deuterium isotope effect because rupture of the C–H (or C–D) bond occurs after the rate-limiting step rather than during it. Thus, we can’t measure a rate difference between a deuterated and nondeuterated substrate. A final piece of evidence involves the stereochemistry of elimination. Unlike the E2 reaction, where anti periplanar geometry is required, there is no geometric requirement on the E1 reaction because the halide and the hydrogen are lost in separate steps. We might therefore expect to obtain the more stable (Zaitsev’s rule) product from E1 reaction, which is just what we find. To return to a familiar example, menthyl chloride loses HCl under E1 conditions in a polar solvent to give a mixture of alkenes in which the Zaitsev product, 3-menthene, predominates ( ). Elimination reactions of menthyl chloride. E2 conditions ( 1 , strong base in 100% ethanol) lead to 2-menthene through an anti periplanar elimination, whereas E1 conditions ( 2 , dilute base in 80% aqueous ethanol) lead to a mixture of 2-menthene and 3-menthene. The E1cB Reaction In contrast to the E1 reaction, which involves a carbocation intermediate, the E1cB reaction takes place through a carbanion intermediate. Base-induced abstraction of a proton in a slow, rate-limiting step gives an anion, which expels a leaving group on the adjacent carbon. The reaction is particularly common in substrates that have a poor leaving group, such as –OH, two carbons removed from a carbonyl group, as in HOC–CH–C═O . The poor leaving group disfavors the alternative E1 and E2 possibilities, and the carbonyl group makes the adjacent hydrogen unusually acidic by resonance stabilization of the anion intermediate. We’ll look at this acidifying effect of a carbonyl group in Section 22.5 .

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Biological Elimination Reactions All three elimination reactions—E2, E1, and E1cB—occur in biological pathways, but the E1cB mechanism is particularly common. The substrate is usually an alcohol rather than an alkyl halide, and the H atom removed is usually adjacent to a carbonyl group, just as in laboratory reactions. Thus, 3-hydroxy carbonyl compounds are frequently converted to unsaturated carbonyl compounds by elimination reactions. A typical example occurs during the biosynthesis of fats and oils when a 3-hydroxybutyryl thioester is dehydrated to the corresponding unsaturated (crotonyl) thioester. The base in this reaction is a histidine amino acid in the enzyme, and the loss of the –OH group is assisted by simultaneous protonation.

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A Summary of Reactivity: S N 1, S N 2, E1, E1cB, and E2 S N 1, S N 2, E1, E1cB, E2—how can you keep it all straight and predict what will happen in any given case? Will substitution or elimination occur? Will the reaction be bimolecular or unimolecular? There are no rigid answers to these questions, but it’s possible to recognize some trends and make some generalizations. Primary alkyl halides S N 2 substitution occurs if a good nucleophile is used, E2 elimination occurs if a strong, sterically hindered base is used, and E1cB elimination occurs if the leaving group is two carbons away from a carbonyl group. Secondary alkyl halides S N 2 substitution occurs if a weakly basic nucleophile is used in a polar aprotic solvent, E2 elimination predominates if a strong base is used, and E1cB elimination takes place if the leaving group is two carbons away from a carbonyl group. Secondary allylic and benzylic alkyl halides can also undergo S N 1 and E1 reactions if a weakly basic nucleophile is used in a protic solvent. Tertiary alkyl halides E2 elimination occurs when a base is used, but S N 1 substitution and E1 elimination occur together under neutral conditions, such as in pure ethanol or water. E1cB elimination takes place if the leaving group is two carbons away from a carbonyl group. Predicting the Product and Mechanism of Reactions Tell whether each of the following reactions is likely to be S N 1, S N 2, E1, E1cB, or E2, and predict the product of each: Strategy Look carefully in each reaction at the structure of the substrate, the leaving group, the nucleophile, and the solvent. Then decide from the preceding summary which kind of reaction is likely to be favored. Solution (a) A secondary, nonallylic substrate can undergo an S N 2 reaction with a good nucleophile in a polar aprotic solvent but will undergo an E2 reaction on treatment with a strong base in a protic solvent. In this case, E2 reaction is likely to predominate. (b) A secondary benzylic substrate can undergo an S N 2 reaction on treatment with a nonbasic nucleophile in a polar aprotic solvent and will undergo an E2 reaction on treatment with a base. Under protic conditions, such as aqueous formic acid (HCO 2 H), an S N 1 reaction is likely, along with some E1 reaction. Problems Green Chemistry Organic chemistry in the 20th century changed the world, giving us new medicines, food preservatives, insecticides, adhesives, textiles, dyes, building materials, composites, and all manner of polymers. But these advances did not come without a cost: Almost every chemical process produces waste that must be dealt with, including reaction solvents and toxic by-products that might evaporate into the air or be leached into groundwater if not disposed of properly. Even apparently harmless by-products must be safely buried or otherwise sequestered. As always, there’s no such thing as a free lunch. With the good also comes the bad. It may never be possible to make organic chemistry completely benign, but awareness of the environmental problems caused by many chemical processes has grown dramatically in recent years, giving rise to a movement called green chemistry. Green chemistry is the design and implementation of chemical products and processes that reduce waste and attempt to eliminate the generation of hazardous substances. There are 12 principles of green chemistry: Prevent waste – Waste should be prevented rather than treated or cleaned up after it has been created. Maximize atom economy – Synthetic methods should maximize the incorporation of all materials used in a process into the final product so that waste is minimized. Use less hazardous processes – Synthetic methods should use reactants and generate wastes with minimal toxicity to health and the environment. Design safer chemicals – Chemical products should be designed to have minimal toxicity. Use safer solvents – Minimal use should be made of solvents, separation agents, and other auxiliary substances in a reaction. Design for energy efficiency – Energy requirements for chemical processes should be minimized, with reactions carried out at room temperature if possible. Use renewable feedstocks – Raw materials should come from renewable sources when feasible. Minimize derivatives – Syntheses should be designed with minimal use of protecting groups to avoid extra steps and reduce waste. Use catalysis – Reactions should be catalytic rather than stoichiometric. Design for degradation – Products should be designed to be biodegradable at the end of their useful lifetimes. Monitor pollution in real time – Processes should be monitored in real time for the formation of hazardous substances. Prevent accidents – Chemical substances and processes should minimize the potential for fires, explosions, or other accidents. The foregoing 12 principles may not all be met in most real-world applications, but they provide a worthy goal and they can make chemists think more carefully about the environmental implications of their work. Real success stories have occurred, and more are in progress. Approximately 7 million pounds per year of ibuprofen (6 billion tablets!) are now made by a “green” process that produces approximately 99% less waste than the process it replaces. Only three steps are needed, the anhydrous HF solvent used in the first step is recovered and reused, and the second and third steps are catalytic. anti periplanar benzylic bimolecular deuterium isotope effect E1 reaction E1cB reaction E2 reaction first-order reaction ion pair kinetics leaving group nucleophilic substitution reaction periplanar rate-determining step rate-limiting step second-order reaction S N 1 reaction S N 2 reaction solvation syn periplanar tosylate unimolecular Zaitsev’s rule Summary The reaction of an alkyl halide or tosylate with a nucleophile/base results either in substitution or in elimination. The resultant nucleophilic substitution and base-induced elimination reactions are two of the most widely occurring and versatile reaction types in organic chemistry, both in the laboratory and in biological pathways. Nucleophilic substitutions are of two types: S N 2 reactions and S N 1 reactions . In the S N 2 reaction, the entering nucleophile approaches the halide from a direction 180° away from the leaving group, resulting in an umbrella-like inversion of configuration at the carbon atom. The reaction is kinetically second-order and is strongly inhibited by increasing steric bulk of the reactants. Thus, S N 2 reactions are favored for primary and secondary substrates. In the S N 1 reaction, the substrate spontaneously dissociates to a carbocation in a slow rate-limiting step , followed by a rapid reaction with the nucleophile. As a result, S N 1 reactions are kinetically first-order and take place with substantial racemization of configuration at the carbon atom. They are most favored for tertiary substrates. Both S N 1 and S N 2 reactions occur in biological pathways, although the leaving group is typically a diphosphate ion rather than a halide. Eliminations of alkyl halides to yield alkenes occur by three mechanisms: E2 reactions , E1 reactions , and E1cB reactions , which differ in the timing of C–H and C–X bond-breaking. In the E2 reaction, C–H and C–X bond-breaking occur simultaneously when a base abstracts H + from one carbon while the leaving group departs from the neighboring carbon. The reaction takes place preferentially through an anti periplanar transition state in which the four reacting atoms—hydrogen, two carbons, and leaving group—are in the same plane. The reaction shows second-order kinetics and a deuterium isotope effect , and occurs when a secondary or tertiary substrate is treated with a strong base. These elimination reactions usually give a mixture of alkene products in which the more highly substituted alkene predominates ( Zaitsev’s rule ). In the E1 reaction, C–X bond-breaking occurs first. The substrate dissociates to yield a carbocation in the slow rate-limiting step before losing H + from an adjacent carbon in a second step. The reaction shows first-order kinetics and no deuterium isotope effect and occurs when a tertiary substrate reacts in polar, nonbasic solution. In the E1cB reaction, C–H bond-breaking occurs first. A base abstracts a proton to give a carbanion, followed by loss of the leaving group from the adjacent carbon in a second step. The reaction is favored when the leaving group is two carbons removed from a carbonyl, which stabilizes the intermediate anion by resonance. Biological elimination reactions typically occur by this E1cB mechanism. In general, substrates react in the following way: Summary of Reactions Nucleophilic substitutions S N 1 reaction of 3°, allylic, and benzylic halides ( and ) S N 2 reaction of 1° and simple 2° halides ( and ) Eliminations E1 reaction ( ) E1cB reaction ( ) E2 reaction ( ) Additional Problems Visualizing Chemistry Mechanism Problems Nucleophilic Substitution Reactions Elimination Reactions General Problems

A Summary of Reactivity: SN1, SN2, E1, E1cB, and E2

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Why This Chapter? More than a thousand different chemical compounds have been isolated from coffee. Their structures were determined using various spectroscopic techniques. (credit: modification of work "Coffee" by Rafael Saldaña/Flickr, CC BY 2.0) Finding the structures of new molecules, whether small ones synthesized in the laboratory or large proteins and nucleic acids found in living organisms, is central to progress in chemistry and biochemistry. We can only scratch the surface of structure determination in this book, but after reading this and the following two chapters, you should have a good idea of the range of structural techniques available and of how and when each is used. Every time a reaction is run, the products must be identified, and every time a new compound is found in nature, its structure must be determined. Determining the structure of an organic compound was a difficult and time-consuming process until the mid-20th century, but powerful techniques and specialized instruments are now routinely used to simplify the problem. In this and the next two chapters, we’ll look at four such techniques—mass spectrometry (MS), infrared (IR) spectroscopy, ultraviolet spectroscopy (UV), and nuclear magnetic resonance spectroscopy (NMR)—and we’ll see the kind of information that can be obtained from each. Technique Information Mass spectrometry What is the size and formula? Infrared spectroscopy What functional groups are present? Ultraviolet spectroscopy Is a conjugated π electron system present? Nuclear magnetic resonance spectroscopy What is the carbon–hydrogen framework?

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Mass Spectrometry of Small Molecules: Magnetic-Sector Instruments At its simplest, mass spectrometry (MS) is a technique for measuring the mass, and therefore the molecular weight (MW), of a molecule. In addition, it’s often possible to gain structural information about a molecule by measuring the masses of the fragments produced when molecules are broken apart. More than 20 different kinds of commercial mass spectrometers are available depending on the intended application, but all have three basic parts: an ionization source in which sample molecules are given an electrical charge, a mass analyzer in which ions are separated by their mass-to-charge ratio, and a detector in which the separated ions are observed and counted. Among the most common mass spectrometers used for routine purposes in the laboratory is the electron-impact, magnetic-sector instrument shown schematically in . A small amount of sample is vaporized into the ionization source, where it is bombarded by a stream of high-energy electrons. The energy of the electron beam can be varied but is commonly around 70 electron volts (eV), or 6700 kJ/mol. When a high-energy electron strikes an organic molecule, it dislodges a valence electron from the molecule, producing a cation radical — cation because the molecule has lost an electron and now has a positive charge; radical because the molecule now has an odd number of electrons. Electron bombardment transfers so much energy that most of the cation radicals fragment after formation. They break apart into smaller pieces, some of which retain the positive charge and some of which are neutral. The fragments then flow through a curved pipe in a strong magnetic field, which deflects them into different paths according to their mass-to-charge ratio ( m / z ). Neutral fragments are not deflected by the magnetic field and are lost on the walls of the pipe, but positively charged fragments are sorted by the mass spectrometer onto a detector, which records them as peaks at the various m / z ratios. Since the number of charges z on each ion is usually 1, the value of m / z for each ion is simply its mass m. Masses up to approximately 2500 atomic mass units (amu) can be analyzed by this type of instrument. Representation of an electron-ionization, magnetic-sector mass spectrometer. Molecules are ionized by collision with high-energy electrons, causing some of the molecules to fragment. Passage of the charged fragments through a magnetic field then sorts them according to their mass. Another common type of mass spectrometer uses what is called a quadrupole mass analyzer , which has a set of four solid rods is arranged parallel to the direction of the ion beam, with an oscillating electrostatic field is generated in the space between the rods. For a given field, only one m/z value will make it through the quadrupole region. The others will crash into the rods or the walls of the instrument and never reach the detector . Representation of a quadrupole mass analyzer. Only ions of a certain m/z will reach the detector; other ions will collide with the rods. The mass spectrum of a compound is typically presented as a bar graph, with masses ( m / z values) on the x axis and intensity, or relative abundance of ions of a given m / z striking the detector, on the y axis. The tallest peak, assigned an intensity of 100%, is called the base peak , and the peak that corresponds to the unfragmented cation radical is called the precursor peak , or the molecular ion ( M + , or simply M ). shows the mass spectrum of propane. Mass spectrum of propane (C 3 H 8 ; MW = 44). Mass spectral fragmentation patterns are usually complex, and the molecular ion is often not the base peak. The mass spectrum of propane in , for instance, shows a molecular ion at m / z = 44 that is only about 30% as high as the base peak at m/z = 29. In addition, many other fragment ions are present.

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