有机化学（英文原版）：Chapt08

302 CHAPTER 8 NUCLEOPHILIC SUBSTITUTION W hen we discussed elimination reactions in Chapter 5, we learned that a Lewis base can react with an alkyl halide to form an alkene. In the present chapter, you will find that the same kinds of reactants can also undergo a different reaction, one in which the Lewis base acts as a nucleophile to substitute for the halide substituent on carbon. We first encountered nucleophilic substitution in Chapter 4, in the reaction of alcohols with hydrogen halides to form alkyl halides. Now we’ll see how alkyl halides can them- selves be converted to other classes of organic compounds by nucleophilic substitution. This chapter has a mechanistic emphasis designed to achieve a practical result. By understanding the mechanisms by which alkyl halides undergo nucleophilic substitution, we can choose experimental conditions best suited to carrying out a particular functional group transformation. The difference between a successful reaction that leads cleanly to a desired product and one that fails is often a subtle one. Mechanistic analysis helps us to appreciate these subtleties and use them to our advantage. 8.1 FUNCTIONAL GROUP TRANSFORMATION BY NUCLEOPHILIC SUBSTITUTION Nucleophilic substitution reactions of alkyl halides are related to elimination reactions in that the halogen acts as a leaving group on carbon and is lost as an anion. The car- bon–halogen bond of the alkyl halide is broken heterolytically: the pair of electrons in that bond are lost with the leaving group. H11001 Alkyl halide R X Lewis base Y H11002 Product of nucleophilic substitution R Y H11001 Halide anion X H11002 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.1 Functional Group Transformation by Nucleophilic Substitution 303 The most frequently encountered nucleophiles in functional group transformations are anions, which are used as their lithium, sodium, or potassium salts. If we use M to represent lithium, sodium, or potassium, some representative nucleophilic reagents are Table 8.1 illustrates an application of each of these to a functional group transfor- mation. The anionic portion of the salt substitutes for the halogen of an alkyl halide. The metal cation portion becomes a lithium, sodium, or potassium halide. Notice that all the examples in Table 8.1 involve alkyl halides, that is, compounds in which the halogen is attached to an sp 3 -hybridized carbon. Alkenyl halides and aryl halides, compounds in which the halogen is attached to sp 2 -hybridized carbons, are essentially unreactive under these conditions, and the principles to be developed in this chapter do not apply to them. To ensure that reaction occurs in homogeneous solution, solvents are chosen that dis- solve both the alkyl halide and the ionic salt. The alkyl halide substrates are soluble in organic solvents, but the salts often are not. Inorganic salts are soluble in water, but alkyl halides are not. Mixed solvents such as ethanol–water mixtures that can dissolve enough of both the substrate and the nucleophile to give fairly concentrated solutions are fre- quently used. Many salts, as well as most alkyl halides, possess significant solubility in dimethyl sulfoxide (DMSO), which makes this a good medium for carrying out nucle- ophilic substitution reactions. sp 2 -hybridized carbonsp 3 -hybridized carbon Alkyl halide C X Alkenyl halide X CC Aryl halide X Nucleophilic reagent M H11001 H11002 Y H11001 R X Alkyl halide R Y Product of nucleophilic substitution H11001 X H11002 M H11001 Metal halide MOR MOCR O X MSH MCN MN 3 (a metal alkoxide, a source of the nucleophilic anion )RO H11002 (a metal hydrogen sulfide, a source of the nucleophilic anion )HS H11002 (a metal cyanide, a source of the nucleophilic anion )CPN H11002 (a metal azide, a source of the nucleophilic anion N?N?N H11002H11002H11001 ) (a metal carboxylate, a source of the nucleophilic anion RC±O O X ) H11002 H11002 Y R X R Y H11001 X H11002 R XX H11005 I, Br, Cl, F H9254H11001 H9254H11002 The carbon–halogen bond in an alkyl halide is polar and is cleaved on attack by a nucleophile so that the two electrons in the bond are retained by the halogen Alkenyl halides are also re- ferred to as vinylic halides. The use of DMSO as a sol- vent in dehydrohalogenation reactions was mentioned earlier, in Section 5.14. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 304 CHAPTER EIGHT Nucleophilic Substitution TABLE 8.1 Representative Functional Group Transformations by Nucleophilic Substitution Reactions of Alkyl Halides Nucleophile and comments Cyanide ion ( : C H11546 PN:) The negatively charged carbon atom of cyanide ion is usually the site of its nucleophilic character. Use of cyanide ion as a nucleophile permits the extension of a carbon chain by carbon–carbon bond formation. The product is an alkyl cyanide, or nitrile. (Continued) Alkoxide ion (RO : H11546 ) The oxygen atom of a metal alkoxide acts as a nucleophile to replace the halogen of an alkyl halide. The product is an ether. : : Hydrogen sulfide ion (HS : H11546 ) Use of hydrogen sulfide as a nucleophile permits the conversion of alkyl hal- ides to compounds of the type RSH. These compounds are the sulfur ana- logs of alcohols and are known as thiols. : : Azide ion ( : N H11546 ?N H11545 ?N H11546 : ) Sodium azide is a reagent used for carbon–nitro- gen bond formation. The product is an alkyl azide. : : Carboxylate ion (RC±O : H11546 ) An ester is formed when the negatively charged oxygen of a carboxylate replaces the halogen of an alkyl halide. : : : O : X General equation and specific example Sodium isobutoxide (CH 3 ) 2 CHCH 2 ONa H11001 Ethyl bromide CH 3 CH 2 Br Ethyl isobutyl ether (66%) (CH 3 ) 2 CHCH 2 OCH 2 CH 3 H11001 Sodium bromide NaBr isobutyl alcohol Potassium octadecanoate KOC(CH 2 ) 16 CH 3 O X H11001 Ethyl iodide CH 3 CH 2 I Ethyl octadecanoate (95%) CH 3 CH 2 OC(CH 2 ) 16 CH 3 O X H11001 Potassium iodide KI acetone water Pentyl iodide CH 3 (CH 2 ) 4 I Sodium azide NaN 3 H11001 Sodium iodide NaIH11001 Pentyl azide (52%) CH 3 (CH 2 ) 4 N 3 1-propanol- water H11001H11001 Ether RH11032OR Halide ion X H11002 Alkoxide ion RH11032O H11002 Alkyl halide R X H11001H11001 Halide ion X H11002 Alkyl halide R X Carboxylate ion RH11032CO O X H11002 Ester RH11032COR O X H11001H11001 Halide ion X H11002 Alkyl halide R X Hydrogen sulfide ion HS H11002 Thiol RSH Potassium hydrogen sulfide KSH H11001H11001 2-Bromononane CH 3 CH(CH 2 ) 6 CH 3 Br W 2-Nonanethiol (74%) CH 3 CH(CH 2 ) 6 CH 3 SH W Potassium bromide KBr ethanol water H11001H11001 Halide ion X H11002 Alkyl halide R X Cyanide ion NPC H11002 Alkyl cyanide RCPN H11001H11001 Halide ion X H11002 Alkyl halide R X Alkyl azide RN?N?N H11001H11002 Azide ion N?N?N H11001H11002H11002 Sodium cyanide NaCN H11001 Cl Cyclopentyl chloride CN Cyclopentyl cyanide (70%) H11001 Sodium chloride NaCl DMSO Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.2 Relative Reactivity of Halide Leaving Groups 305 TABLE 8.1 Representative Functional Group Transformations by Nucleophilic Substitution Reactions of Alkyl Halides (Continued) Nucleophile and comments Iodide ion ( : I : H11546 ) Alkyl chlorides and bromides are converted to alkyl iodides by treatment with sodium iodide in acetone. NaI is soluble in acetone, but NaCl and NaBr are insoluble and crystallize from the reaction mixture, driving the reac- tion to completion. : : General equation and specific example 2-Bromopropane CH 3 CHCH 3 W Br H11001 Sodium iodide NaI 2-Iodopropane (63%) CH 3 CHCH 3 I W H11001 Sodium bromide NaBr (solid) acetone H11001H11001 Chloride or bromide ion X H11002 Iodide ion I H11002 Alkyl chloride or bromide R X Alkyl iodide R I acetone PROBLEM 8.1 Write a structural formula for the principal organic product formed in the reaction of methyl bromide with each of the following compounds: (a) NaOH (sodium hydroxide) (b) KOCH 2 CH 3 (potassium ethoxide) (c) (d) LiN 3 (lithium azide) (e) KCN (potassium cyanide) (f) NaSH (sodium hydrogen sulfide) (g) NaI (sodium iodide) SAMPLE SOLUTION (a) The nucleophile in sodium hydroxide is the negatively charged hydroxide ion. The reaction that occurs is nucleophilic substitution of bro- mide by hydroxide. The product is methyl alcohol. With this as background, you can begin to see how useful alkyl halides are in syn- thetic organic chemistry. Alkyl halides may be prepared from alcohols by nucleophilic substitution, from alkanes by free-radical halogenation, and from alkenes by addition of hydrogen halides. They then become available as starting materials for the preparation of other functionally substituted organic compounds by replacement of the halide leav- ing group with a nucleophile. The range of compounds that can be prepared by nucle- ophilic substitution reactions of alkyl halides is quite large; the examples shown in Table 8.1 illustrate only a few of them. Numerous other examples will be added to the list in this and subsequent chapters. 8.2 RELATIVE REACTIVITY OF HALIDE LEAVING GROUPS Among alkyl halides, alkyl iodides undergo nucleophilic substitution at the fastest rate, alkyl fluorides the slowest. H11001H11001 Hydroxide ion (nucleophile) HO H11002 Methyl bromide (substrate) CH 3 Br Bromide ion (leaving group) Br H11002 Methyl alcohol (product) CH 3 OH NaOC O (sodium benzoate) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The order of alkyl halide reactivity in nucleophilic substitutions is the same as their order in eliminations. Iodine has the weakest bond to carbon, and iodide is the best leaving group. Alkyl iodides are several times more reactive than alkyl bromides and from 50 to 100 times more reactive than alkyl chlorides. Fluorine has the strongest bond to car- bon, and fluoride is the poorest leaving group. Alkyl fluorides are rarely used as sub- strates in nucleophilic substitution because they are several thousand times less reactive than alkyl chlorides. PROBLEM 8.2 A single organic product was obtained when 1-bromo-3-chloro- propane was allowed to react with one molar equivalent of sodium cyanide in aqueous ethanol. What was this product? Leaving-group ability is also related to basicity. A strongly basic anion is usually a poorer leaving group than a weakly basic one. Fluoride is the most basic and the poor- est leaving group among the halide anions, iodide the least basic and the best leaving group. 8.3 THE S N 2 MECHANISM OF NUCLEOPHILIC SUBSTITUTION The mechanisms by which nucleophilic substitution takes place have been the subject of much study. Extensive research by Sir Christopher Ingold and Edward D. Hughes and their associates at University College, London, during the 1930s emphasized kinetic and stereochemical measurements to probe the mechanisms of these reactions. Recall that the term “kinetics” refers to how the rate of a reaction varies with changes in concentration. Consider the nucleophilic substitution in which sodium hydrox- ide reacts with methyl bromide to form methyl alcohol and sodium bromide: The rate of this reaction is observed to be directly proportional to the concentration of both methyl bromide and sodium hydroxide. It is first-order in each reactant, or second- order overall. Rate H11005 k[CH 3 Br][HO H11002 ] Hughes and Ingold interpreted second-order kinetic behavior to mean that the rate- determining step is bimolecular, that is, that both hydroxide ion and methyl bromide are involved at the transition state. The symbol given to the detailed description of the mech- anism that they developed is S N 2, standing for substitution nucleophilic bimolecular. The Hughes and Ingold S N 2 mechanism is a single-step process in which both the alkyl halide and the nucleophile are involved at the transition state. Cleavage of the bond between carbon and the leaving group is assisted by formation of a bond between car- bon and the nucleophile. In effect, the nucleophile “pushes off” the leaving group from H11001 Methyl bromide CH 3 Br Hydroxide ion HO H11002 Bromide ion Br H11002 Methyl alcohol CH 3 OH H11001 Increasing rate of substitution by nucleophiles RF H11021H11021 RCl H11021 RBr H11021 RI Least reactive Most reactive 306 CHAPTER EIGHT Nucleophilic Substitution The relationship between leaving group ability and ba- sicity is explored in more de- tail in Section 8.14. The S N 2 mechanism was in- troduced earlier in Section 4.13. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.4 Stereochemistry of S N 2 Reactions 307 its point of attachment to carbon. For this reason, the S N 2 mechanism is sometimes referred to as a direct displacement process. The S N 2 mechanism for the hydrolysis of methyl bromide may be represented by a single elementary step: Carbon is partially bonded to both the incoming nucleophile and the departing halide at the transition state. Progress is made toward the transition state as the nucleophile begins to share a pair of its electrons with carbon and the halide ion leaves, taking with it the pair of electrons in its bond to carbon. PROBLEM 8.3 Is the two-step sequence depicted in the following equations con- sistent with the second-order kinetic behavior observed for the hydrolysis of methyl bromide? The S N 2 mechanism is believed to describe most substitutions in which simple pri- mary and secondary alkyl halides react with anionic nucleophiles. All the examples cited in Table 8.1 proceed by the S N 2 mechanism (or a mechanism very much like S N 2— remember, mechanisms can never be established with certainty but represent only our best present explanations of experimental observations). We’ll examine the S N 2 mecha- nism, particularly the structure of the transition state, in more detail in Section 8.5 after first looking at some stereochemical studies carried out by Hughes and Ingold. 8.4 STEREOCHEMISTRY OF S N 2 REACTIONS What is the structure of the transition state in an S N 2 reaction? In particular, what is the spatial arrangement of the nucleophile in relation to the leaving group as reactants pass through the transition state on their way to products? Two stereochemical possibilities present themselves. In the pathway shown in Fig- ure 8.1a, the nucleophile simply assumes the position occupied by the leaving group. It attacks the substrate at the same face from which the leaving group departs. This is called “front-side displacement,” or substitution with retention of configuration. In a second possibility, illustrated in Figure 8.1b, the nucleophile attacks the sub- strate from the side opposite the bond to the leaving group. This is called “back-side dis- placement,” or substitution with inversion of configuration. Which of these two opposite stereochemical possibilities operates was determined in experiments with optically active alkyl halides. In one such experiment, Hughes and Ingold determined that the reaction of 2-bromooctane with hydroxide ion gave 2-octanol having a configuration opposite that of the starting alkyl halide. (S)-(H11001)-2-Bromooctane C H H 3 C CH 3 (CH 2 ) 5 Br (R)-(H11002)-2-Octanol H CH 3 (CH 2 ) 5 CH 3 CHO NaOH ethanol-water CH 3 Br CH 3 H11001 H11001 Br H11002 slow CH 3 H11001 CH 3 OHH11001 HO H11002 fast H11001 Hydroxide ion HO H11002 Methyl bromide CH 3 Br H11001 Transition state HO CH 3 H9254H11002 H9254H11002 Br Bromide ion Br H11002 Methyl alcohol HOCH 3 Although the alkyl halide and alcohol given in this ex- ample have opposite config- urations when they have opposite signs of rotation, it cannot be assumed that this will be true for all alkyl halide/alcohol pairs. (See Sec- tion 7.5) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Nucleophilic substitution had occurred with inversion of configuration, consistent with the following transition state: PROBLEM 8.4 The Fischer projection formula for (H11001)-2-bromooctane is shown. Write the Fischer projection of the (H11002)-2-octanol formed from it by nucleophilic substitution with inversion of configuration. PROBLEM 8.5 Would you expect the 2-octanol formed by S N 2 hydrolysis of (H11002)- 2-bromooctane to be optically active? If so, what will be its absolute configura- tion and sign of rotation? What about the 2-octanol formed by hydrolysis of racemic 2-bromooctane? Numerous similar experiments have demonstrated the generality of this observation. Substitution by the S N 2 mechanism is stereospecific and proceeds with inversion of con- figuration at the carbon that bears the leaving group. There is a stereoelectronic require- ment for the nucleophile to approach carbon from the side opposite the bond to the leav- ing group. Organic chemists often speak of this as a Walden inversion, after the German chemist Paul Walden, who described the earliest experiments in this area in the 1890s. 8.5 HOW S N 2 REACTIONS OCCUR When we consider the overall reaction stereochemistry along with the kinetic data, a fairly complete picture of the bonding changes that take place during S N 2 reactions emerges. The potential energy diagram of Figure 8.2 for the hydrolysis of (S)-(H11001)-2- bromooctane is one that is consistent with the experimental observations. CH 3 BrH CH 2 (CH 2 ) 4 CH 3 C CH 3 (CH 2 ) 5 H HO Br CH 3 H9254H11002 H9254H11002 308 CHAPTER EIGHT Nucleophilic Substitution H11001 H11002 H11002 H11002 H11002 H11001 (a) Nucleophilic substitution with retention of configuration H11001 H11001 (b) Nucleophilic substitution with inversion of configuration H11002 H11002 H11002H11002 H9254 H9254H9254 H9254 The first example of a stereo- electronic effect in this text concerned anti elimination in E2 reactions of alkyl halides (Section 5.16). For a change of pace, try doing Problem 8.4 with molecu- lar models instead of making structural drawings. FIGURE 8.1 Two contrasting stereochemical pathways for substitution of a leaving group (red) by a nucleophile (blue). In (a) the nucleophile attacks carbon at the same side from which the leaving group departs. In (b) nucle- ophilic attack occurs at the side opposite the bond to the leaving group. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.5 How S N 2 Reactions Occur 309 Hydroxide ion acts as a nucleophile, using an unshared electron pair to attack car- bon from the side opposite the bond to the leaving group. The hybridization of the car- bon at which substitution occurs changes from sp 3 in the alkyl halide to sp 2 in the tran- sition state. Both the nucleophile (hydroxide) and the leaving group (bromide) are partially bonded to this carbon in the transition state. We say that the S N 2 transition state is pentacoordinate; carbon is fully bonded to three substituents and partially bonded to both the leaving group and the incoming nucleophile. The bonds to the nucleophile and the leaving group are relatively long and weak at the transition state. Once past the transition state, the leaving group is expelled and carbon becomes tetracoordinate, its hybridization returning to sp 3 . During the passage of starting materials to products, three interdependent and syn- chronous changes take place: 1. Stretching, then breaking, of the bond to the leaving group 2. Formation of a bond to the nucleophile from the opposite side of the bond that is broken 3. Stereochemical inversion of the tetrahedral arrangement of bonds to the carbon at which substitution occurs Although this mechanistic picture developed from experiments involving optically active alkyl halides, chemists speak even of methyl bromide as undergoing nucleophilic substitution with inversion. By this they mean that tetrahedral inversion of the bonds to carbon occurs as the reactant proceeds to the product. H11001 Hydroxide ion HO H11002 Methyl bromide C H H H Br Transition state HO C H9254H11002 H9254H11002 Br HH H H11001 Bromide ion Br H11002 Methyl alcohol H H H CHO Potential energy Pentacoordinate carbon is sp 2 - hybridized Bonding is weak between carbon and bromine and carbon and oxygen in the transition state Reaction coordinate CH 3 (CH 2 ) 5 CH 3 (CH 2 ) 5 CH 3 CH 3 CH 3 (CH 2 ) 5 CH 3 Br Br H H H C C CHO HO H11002 Br H11002 HO H11002 σO bondC(sp 3 ) σBr bondC(sp 3 ) δ δ FIGURE 8.2 Hybrid orbital description of the bonding changes that take place at carbon during nucleophilic substitution by the S N 2 mechanism. For an animation of this S N 2 reaction, see Learning By Modeling. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website We saw in Section 8.2 that the rate of nucleophilic substitution depends strongly on the leaving group—alkyl iodides are the most reactive, alkyl fluorides the least. In the next section, we’ll see that the structure of the alkyl group can have an even greater effect. 8.6 STERIC EFFECTS IN S N 2 REACTIONS There are very large differences in the rates at which the various kinds of alkyl halides— methyl, primary, secondary, or tertiary—undergo nucleophilic substitution. As Table 8.2 shows for the reaction of a series of alkyl bromides: the rates of nucleophilic substitution of a series of alkyl bromides differ by a factor of over 10 6 when comparing the most reactive member of the group (methyl bromide) and the least reactive member (tert-butyl bromide). The large rate difference between methyl, ethyl, isopropyl, and tert-butyl bromides reflects the steric hindrance each offers to nucleophilic attack. The nucleophile must approach the alkyl halide from the side opposite the bond to the leaving group, and, as illustrated in Figure 8.3, this approach is hindered by alkyl substituents on the carbon that is being attacked. The three hydrogens of methyl bromide offer little resistance to approach of the nucleophile, and a rapid reaction occurs. Replacing one of the hydro- gens by a methyl group somewhat shields the carbon from attack by the nucleophile and causes ethyl bromide to be less reactive than methyl bromide. Replacing all three hydro- gen substituents by methyl groups almost completely blocks back-side approach to the tertiary carbon of (CH 3 ) 3 CBr and shuts down bimolecular nucleophilic substitution. In general, S N 2 reactions exhibit the following dependence of rate on substrate structure: Least reactive, most crowded Most reactive, least crowded Tertiary R 3 CX H11021 Secondary R 2 CHX H11021 Primary RCH 2 X H11021 Methyl CH 3 X Increasing rate of substitution by the S N 2 mechanism H11001 Alkyl bromide RBr Lithium iodide LiI Lithium bromide LiBr Alkyl iodide RI H11001 acetone 310 CHAPTER EIGHT Nucleophilic Substitution TABLE 8.2 Reactivity of Some Alkyl Bromides Toward Substitution by the S N 2 Mechanism* Alkyl bromide Methyl bromide Ethyl bromide Isopropyl bromide tert-Butyl bromide CH 3 Br CH 3 CH 2 Br (CH 3 ) 2 CHBr (CH 3 ) 3 CBr Structure Unsubstituted Primary Secondary Tertiary Class 221,000 1,350 1 Too small to measure Relative rate ? *Substitution of bromide by lithium iodide in acetone. ? Ratio of second-order rate constant k for indicated alkyl bromide to k for isopropyl bromide at 25°C. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.6 Steric Effects in S N 2 Reactions 311 PROBLEM 8.6 Identify the compound in each of the following pairs that reacts with sodium iodide in acetone at the faster rate: (a) 1-Chlorohexane or cyclohexyl chloride (b) 1-Bromopentane or 3-bromopentane (c) 2-Chloropentane or 2-fluoropentane (d) 2-Bromo-2-methylhexane or 2-bromo-5-methylhexane (e) 2-Bromopropane or 1-bromodecane SAMPLE SOLUTION (a) Compare the structures of the two chlorides. 1-Chloro- hexane is a primary alkyl chloride; cyclohexyl chloride is secondary. Primary alkyl halides are less crowded at the site of substitution than secondary ones and react faster in substitution by the S N 2 mechanism. 1-Chlorohexane is more reactive. Alkyl groups at the carbon atom adjacent to the point of nucleophilic attack also decrease the rate of the S N 2 reaction. Compare the rates of nucleophilic substitution in the series of primary alkyl bromides shown in Table 8.3. Taking ethyl bromide as the standard and successively replacing its C-2 hydrogens by methyl groups, we see that each additional methyl group decreases the rate of displacement of bromide by iodide. The effect is slightly smaller than for alkyl groups that are attached directly to the car- bon that bears the leaving group, but it is still substantial. When C-2 is completely sub- stituted by methyl groups, as it is in neopentyl bromide [(CH 3 ) 3 CCH 2 Br], we see the unusual case of a primary alkyl halide that is practically inert to substitution by the S N 2 mechanism because of steric hindrance. Cyclohexyl chloride (secondary, less reactive) H Cl 1-Chlorohexane (primary, more reactive) CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 Cl Least crowded– most reactive Most crowded– least reactive CH 3 Br CH 3 CH 2 Br (CH 3 ) 2 CHBr (CH 3 ) 3 CBr FIGURE 8.3 Ball-and-spoke and space-filling models of alkyl bromides, showing how sub- stituents shield the carbon atom that bears the leaving group from attack by a nucleophile. The nucleophile must attack from the side opposite the bond to the leaving group. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.7 NUCLEOPHILES AND NUCLEOPHILICITY The Lewis base that acts as the nucleophile often is, but need not always be, an anion. Neutral Lewis bases can also serve as nucleophiles. Common examples of substitutions involving neutral nucleophiles include solvolysis reactions. Solvolysis reactions are sub- stitutions in which the nucleophile is the solvent in which the reaction is carried out. Solvolysis in water converts an alkyl halide to an alcohol. Solvolysis in methyl alcohol converts an alkyl halide to an alkyl methyl ether. In these and related solvolyses, the first stage is the one in which nucleophilic substitution takes place and is rate-determining. The proton-transfer step that follows it is much faster. Since, as we have seen, the nucleophile attacks the substrate in the rate- determining step of the S N 2 mechanism, it follows that the rate at which substitution occurs may vary from nucleophile to nucleophile. Just as some alkyl halides are more reactive than others, some nucleophiles are more reactive than others. Nucleophilic strength, or nucleophilicity, is a measure of how fast a Lewis base displaces a leaving group from a suitable substrate. By measuring the rate at which various Lewis bases react with methyl iodide in methanol, a list of their nucleophilicities relative to methanol as the standard nucleophile has been compiled. It is presented in Table 8.4. Neutral Lewis bases such as water, alcohols, and carboxylic acids are much weaker nucleophiles than their conjugate bases. When comparing species that have the same nucleophilic atom, a negatively charged nucleophile is more reactive than a neutral one. Methyl alcohol O H 3 C H Alkyl halide R X slow fast Dialkyloxonium halide O H 3 C H H11001 R H11001 X H11002 Alkyl methyl ether ROCH 3 H11001 Hydrogen halide HXH11001 Water O H H Alkyl halide R XH11001 slow fast Alkyloxonium halide O H H H11001 R H11001 X H11002 Alcohol ROH H11001 Hydrogen halide HX 312 CHAPTER EIGHT Nucleophilic Substitution TABLE 8.3 Effect of Chain Branching on Reactivity of Primary Alkyl Bromides Toward Substitution Under S N 2 Conditions* Alkyl bromide Ethyl bromide Propyl bromide Isobutyl bromide Neopentyl bromide CH 3 CH 2 Br CH 3 CH 2 CH 2 Br (CH 3 ) 2 CHCH 2 Br (CH 3 ) 3 CCH 2 Br Structure 1.0 0.8 0.036 0.00002 Relative rate ? *Substitution of bromide by lithium iodide in acetone. ? Ratio of second-order rate constant k for indicated alkyl bromide to k for ethyl bromide at 25°C. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.7 Nucleophiles and Nucleophilicity 313 As long as the nucleophilic atom is the same, the more basic the nucleophile, the more reactive it is. An alkoxide ion (RO H11002 ) is more basic and more nucleophilic than a carboxylate ion (RCO 2 H11002 ). The connection between basicity and nucleophilicity holds when comparing atoms in the same row of the periodic table. Thus, HO H11002 is more basic and more nucleophilic than F H11002 , and H 3 N is more basic and more nucleophilic than H 2 O. It does not hold when proceeding down a column in the periodic table. For example, I H11002 is the least basic of the halide ions but is the most nucleophilic. F H11002 is the most basic halide ion but the least nucleophilic. The factor that seems most responsible for the inverse relationship between basicity and nucleophilicity among the halide ions is the degree to which they are sol- vated by hydrogen bonds of the type illustrated in Figure 8.4. Smaller anions, because of their high charge-to-size ratio, are more strongly solvated than larger ones. In order to act as a nucleophile, the halide must shed some of the solvent molecules that surround it. Among the halide anions, F H11002 forms the strongest hydrogen bonds to water and alco- hols, and I H11002 the weakest. Thus, the nucleophilicity of F H11002 is suppressed more than that of Cl H11002 , Cl H11002 more than Br H11002 , and Br H11002 more than I H11002 . Similarly, HO H11002 is smaller, more sol- vated, and less nucleophilic than HS H11002 . Nucleophilicity is also related to polarizability, or the ease of distortion of the elec- tron “cloud” surrounding the nucleophile. The partial bond between the nucleophile and the alkyl halide that characterizes the S N 2 transition state is more fully developed at a longer distance when the nucleophile is very polarizable than when it is not. An increased degree of bonding to the nucleophile lowers the energy of the transition state and is more nucleophilic thanRO H11002 Stronger base Conjugate acid is ROH: K a H11005 10 H1100216 (pK a H11005 16) RCO H11002 O X Weaker base Conjugate acid is RCO 2 H: K a H11005 10 H110025 (pK a H11005 5) is more nucleophilic thanRO H11002 Alkoxide ion ROH Alcohol is more nucleophilic thanRCO H11002 O X Carboxylate ion RCOH O X Carboxylic acid TABLE 8.4 Nucleophilicity of Some Common Nucleophiles Reactivity class Very good nucleophiles Good nucleophiles Fair nucleophiles Weak nucleophiles Very weak nucleophiles I H11002 , HS H11002 , RS H11002 Br H11002 , HO H11002 , RO H11002 , CN H11002 , N 3 H11002 NH 3 , Cl H11002 , F H11002 , RCO 2 H11002 H 2 O, ROH RCO 2 H Nucleophile H1102210 5 10 4 10 3 1 10 H110022 Relative reactivity* *Relative reactivity is k(nucleophile)/k(methanol) for typical S N 2 reactions and is approximate. Data pertain to methanol as the solvent. A descriptive term applied to a highly polarizable species is soft. Iodide is a very soft nucleophile. Conversely, fluoride ion is not very polar- izable and is said to be a hard nucleophile. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 314 CHAPTER EIGHT Nucleophilic Substitution AN ENZYME-CATALYZED NUCLEOPHILIC SUBSTITUTION OF AN ALKYL HALIDE N ucleophilic substitution is one of a variety of mechanisms by which living systems detoxify halogenated organic compounds introduced into the environment. Enzymes that catalyze these reactions are known as haloalkane dehalogenases. The hydrolysis of 1,2-dichloroethane to 2- chloroethanol, for example, is a biological nucle- ophilic substitution catalyzed by a dehalogenase. The haloalkane dehydrogenase is believed to act by using one of its side-chain carboxylates to dis- place chloride by an S N 2 mechanism. (Recall the reac- tion of carboxylate ions with alkyl halides from Table 8.1.) ±C±O H11001 O X S N 2 Enzyme H11002 CH 2 ±Cl CH 2 Cl W H11001±C±O±CH 2 O X Enzyme CH 2 Cl W Cl H11002 ClCH 2 CH 2 Cl 1,2-Dichloroethane 2H 2 O Water H11001 ClCH 2 CH 2 OH 2-Chloroethanol H 3 O H11001 Hydronium ion Cl H11002 Chloride ion H11001H11001 dehalogenase enzyme The product of this nucleophilic substitution then re- acts with water, restoring the enzyme to its original state and giving the observed products of the reac- tion. This stage of the reaction proceeds by a mechanism that will be discussed in Chapter 20. Both stages are faster than the reaction of 1,2-dichloroethane with water in the absence of the enzyme. Some of the most common biological S N 2 reac- tions involve attack at methyl groups, especially a methyl group of S-adenosylmethionine. Examples of these will be given in Chapter 16. ±C±O H11001 O X Enzyme H11002 HOCH 2 CH 2 Cl W H11001 H 3 O H11001 several steps H11001±C±O±CH 2 O X Enzyme CH 2 Cl W 2H 2 O Cl H11002 H11001 H11001 H11001 H11001 H9254 H9254 H9254 H9254 FIGURE 8.4 Solvation of a chloride by ion–dipole attractive forces with water. The negatively charged chloride ion interacts with the positively polarized hydrogens of water. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.9 Carbocation Stability and S N 1 Reaction Rates 315 increases the rate of substitution. Among related atoms, polarizability increases with increasing size. Thus iodide is the most polarizable and most nucleophilic halide ion, fluoride the least. PROBLEM 8.7 Sodium nitrite (NaNO 2 ) reacted with 2-iodooctane to give a mix- ture of two constitutionally isomeric compounds of molecular formula C 8 H 17 NO 2 in a combined yield of 88%. Suggest reasonable structures for these two isomers. 8.8 THE S N 1 MECHANISM OF NUCLEOPHILIC SUBSTITUTION Having just learned that tertiary alkyl halides are practically inert to substitution by the S N 2 mechanism because of steric hindrance, we might wonder whether they undergo nucleophilic substitution at all. We’ll see in this section that they do, but by a mecha- nism different from S N 2. Hughes and Ingold observed that the hydrolysis of tert-butyl bromide, which occurs readily, is characterized by a first-order rate law: Rate H11005 k[(CH 3 ) 3 CBr] They found that the rate of hydrolysis depends only on the concentration of tert-butyl bromide. Adding the stronger nucleophile hydroxide ion, moreover, causes no change in the rate of substitution, nor does this rate depend on the concentration of hydroxide. Just as second-order kinetics was interpreted as indicating a bimolecular rate-determining step, first-order kinetics was interpreted as evidence for a unimolecular rate-determining step—a step that involves only the alkyl halide. The proposed mechanism is outlined in Figure 8.5 and is called S N 1, standing for substitution nucleophilic unimolecular. The first step, a unimolecular dissociation of the alkyl halide to form a carbocation as the key intermediate, is rate-determining. An energy diagram for the process is shown in Figure 8.6. PROBLEM 8.8 Suggest a structure for the product of nucleophilic substitution obtained on solvolysis of tert-butyl bromide in methanol, and outline a reason- able mechanism for its formation. The S N 1 mechanism is an ionization mechanism. The nucleophile does not participate until after the rate-determining step has taken place. Thus, the effects of nucleophile and alkyl halide structure are expected to be different from those observed for reactions pro- ceeding by the S N 2 pathway. How the structure of the alkyl halide affects the rate of S N 1 reactions is the topic of the next section. 8.9 CARBOCATION STABILITY AND S N 1 REACTION RATES In order to compare S N 1 substitution rates in a range of alkyl halides, experimental con- ditions are chosen in which competing substitution by the S N 2 route is very slow. One such set of conditions is solvolysis in aqueous formic acid (HCO 2 H): H11001 Alkyl halide RX Water H 2 O Hydrogen halide HX Alcohol ROH H11001 formic acid H11001 tert-Butyl bromide (CH 3 ) 3 CBr Water H 2 O Hydrogen bromide HBr tert-Butyl alcohol (CH 3 ) 3 COH H11001 The S N 1 mechanism was ear- lier introduced in Section 4.11. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 316 CHAPTER EIGHT Nucleophilic Substitution (CH 3 ) 3 C±Br ±￡ (CH 3 ) 3 C H11001 H11001 Br H11002 The Overall Reaction: Step 1: The alkyl halide dissociates to a carbocation and a halide ion. Step 2: The carbocation formed in step 1 reacts rapidly with a water molecule. Water is a nucleophile. This step completes the nucleophilic substitution stage of the mechanism and yields an alkyloxonium ion. slow (CH 3 ) 3 C H11001 H11001 O ±￡ (CH 3 ) 3 C±O tert-Butyl cation fast H H Water tert-Butyloxonium ion H H H H H11001 Step 3: This step is a fast acid-base reaction that follows the nucleophilic substitution. Water acts as a base to remove a proton from the alkyloxonium ion to give the observed product of the reaction, tert-butyl alcohol. (CH 3 ) 3 C±O H11001 O ±￡ (CH 3 ) 3 C±O H11001 H±O tert-Butyloxonium ion H11001 fast Water tert-Butyl alcohol H Hydronium ion tert-Butyl bromide tert-Butyl bromide tert-Butyl cation Bromide ion Water (CH 3 ) 3 CBr H11001 2H 2 O ±￡ (CH 3 ) 3 COH H11001 H 3 O H11001 H11001 Br H11002 Hydronium iontert-Butyl alcohol Bromide ion H H H H H11001 2H 2 O Potential energy Reaction coordinate (CH 3 ) 3 C - - - Br H11002H11001 (CH 3 ) 3 C - - - OH 2 , Br H11002 , H 2 O (CH 3 ) 3 C H11001 Br H11002 , 2H 2 O (CH 3 ) 3 CBr, 2H 2 O (CH 3 ) 3 CO - - - H - - - OH 2 , Br H11002 H (CH 3 ) 3 COH 2 , Br H11002 , H 2 O H11001 (CH 3 ) 3 COH, Br H11002 , H 3 O H11001 E act δ H11001δ H11001δ H11001δ H11001δ δ FIGURE 8.6 Energy diagram illustrating the S N 1 mecha- nism for hydrolysis of tert- butyl bromide. FIGURE 8.5 The S N 1 mecha- nism for hydrolysis of tert- butyl bromide. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.9 Carbocation Stability and S N 1 Reaction Rates 317 Neither formic acid nor water is very nucleophilic, and so S N 2 substitution is suppressed. The relative rates of hydrolysis of a group of alkyl bromides under these conditions are presented in Table 8.5. The relative rate order in S N 1 reactions is exactly the opposite of that seen in S N 2 reactions: S N 1 reactivity: methyl H11021 primary H11021 secondary H11021 tertiary S N 2 reactivity: tertiary H11021 secondary H11021 primary H11021 methyl Clearly, the steric crowding that influences reaction rates in S N 2 processes plays no role in S N 1 reactions. The order of alkyl halide reactivity in S N 1 reactions is the same as the order of carbocation stability: the more stable the carbocation, the more reactive the alkyl halide. We have seen this situation before in the reaction of alcohols with hydrogen halides (Section 4.12), in the acid-catalyzed dehydration of alcohols (Section 5.9), and in the conversion of alkyl halides to alkenes by the E1 mechanism (Section 5.17). As in these other reactions, an electronic effect, specifically, the stabilization of the carboca- tion intermediate by alkyl substituents, is the decisive factor. PROBLEM 8.9 Identify the compound in each of the following pairs that reacts at the faster rate in an S N 1 reaction: (a) Isopropyl bromide or isobutyl bromide (b) Cyclopentyl iodide or 1-methylcyclopentyl iodide (c) Cyclopentyl bromide or 1-bromo-2,2-dimethylpropane (d) tert-Butyl chloride or tert-butyl iodide SAMPLE SOLUTION (a) Isopropyl bromide, (CH 3 ) 2 CHBr, is a secondary alkyl halide, whereas isobutyl bromide, (CH 3 ) 2 CHCH 2 Br, is primary. Since the rate- determining step in an S N 1 reaction is carbocation formation and since secondary carbocations are more stable than primary carbocations, isopropyl bromide is more reactive than isobutyl bromide in nucleophilic substitution by the S N 1 mechanism. Primary carbocations are so high in energy that their intermediacy in nucleophilic substitution reactions is unlikely. When ethyl bromide undergoes hydrolysis in aqueous formic acid, substitution probably takes place by a direct displacement of bromide by water in an S N 2-like process. TABLE 8.5 Reactivity of Some Alkyl Bromides Toward Substitution by the S N 1 Mechanism* Alkyl bromide Methyl bromide Ethyl bromide Isopropyl bromide tert-Butyl bromide CH 3 Br CH 3 CH 2 Br (CH 3 ) 2 CHBr (CH 3 ) 3 CBr Structure Unsubstituted Primary Secondary Tertiary Class 1 2 43 100,000,000 Relative rate ? *Solvolysis in aqueous formic acid. ? Ratio of rate constant k for indicated alkyl bromide to k for methyl bromide at 25°C. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.10 STEREOCHEMISTRY OF S N 1 REACTIONS Although S N 2 reactions are stereospecific and proceed with inversion of configuration at carbon, the situation is not as clear-cut for S N 1 reactions. When the leaving group is attached to the stereogenic center of an optically active halide, ionization gives a carbo- cation intermediate that is achiral. It is achiral because the three bonds to the positively charged carbon lie in the same plane, and this plane is a plane of symmetry for the car- bocation. As shown in Figure 8.7, such a carbocation should react with a nucleophile at the same rate at either of its two faces. We expect the product of substitution by the S N 1 mechanism to be racemic and optically inactive. This outcome is rarely observed in prac- tice, however. Normally, the product is formed with predominant, but not complete, inversion of configuration. For example, the hydrolysis of optically active 2-bromooctane in the absence of added base follows a first-order rate law, but the resulting 2-octanol is formed with 66% inversion of configuration. (R)-(H11002)-2-Bromooctane C H CH 3 CH 3 (CH 2 ) 5 Br (R)-(H11002)-2-Octanol C H CH 3 CH 3 (CH 2 ) 5 OHH11001 (S)-(H11001)-2-Octanol H CH 3 (CH 2 ) 5 CH 3 CHO 66% net inversion corresponds to 83% S, 17% R H 2 O ethanol Bimolecular transition state for hydrolysis of ethyl bromide C Br CH 3 O H H H9254H11001 H9254H11002 HH 318 CHAPTER EIGHT Nucleophilic Substitution 50% 50% H11001 H11001 H11002 H11002 FIGURE 8.7 Forma- tion of a racemic product by nucleophilic substitution via a carbocation intermediate. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.11 Carbocation Rearrangements in S N 1 Reactions 319 Partial but not complete loss of optical activity in S N 1 reactions probably results from the carbocation not being completely “free” when it is attacked by the nucleophile. Ionization of the alkyl halide gives a carbocation–halide ion pair, as depicted in Figure 8.8. The halide ion shields one side of the carbocation, and the nucleophile captures the carbocation faster from the opposite side. More product of inverted configuration is formed than product of retained configuration. In spite of the observation that the prod- ucts of S N 1 reactions are only partially racemic, the fact that these reactions are not ste- reospecific is more consistent with a carbocation intermediate than a concerted bimolec- ular mechanism. PROBLEM 8.10 What two stereoisomeric substitution products would you expect to isolate from the hydrolysis of cis-1,4-dimethylcyclohexyl bromide? From hydrolysis of trans-1,4-dimethylcyclohexyl bromide? 8.11 CARBOCATION REARRANGEMENTS IN S N 1 REACTIONS Additional evidence for carbocation intermediates in certain nucleophilic substitutions comes from observing rearrangements of the kind normally associated with such species. For example, hydrolysis of the secondary alkyl bromide 2-bromo-3-methylbutane yields the rearranged tertiary alcohol 2-methyl-2-butanol as the only substitution product. 2-Bromo-3-methylbutane Br CH 3 CH 3 CHCHCH 3 H 2 O 2-Methyl-2-butanol (93%) CH 3 CH 3 CCH 2 CH 3 OH More than 50% Less than 50% H11001 H11001 H11002 H11002 FIGURE 8.8 Inversion of configuration predomi- nates in S N 1 reactions be- cause one face of the carbocation is shielded by the leaving group (red). Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A reasonable mechanism for this observation assumes rate-determining ionization of the substrate as the first step followed by a hydride shift that converts the secondary carbocation to a more stable tertiary one. The tertiary carbocation then reacts with water to yield the observed product. PROBLEM 8.11 Why does the carbocation intermediate in the hydrolysis of 2- bromo-3-methylbutane rearrange by way of a hydride shift rather than a methyl shift? Rearrangements, when they do occur, are taken as evidence for carbocation inter- mediates and point to the S N 1 mechanism as the reaction pathway. Rearrangements are never observed in S N 2 reactions. 8.12 EFFECT OF SOLVENT ON THE RATE OF NUCLEOPHILIC SUBSTITUTION The major effect of the solvent is on the rate of nucleophilic substitution, not on what the products are. Thus we need to consider two related questions: 1. What properties of the solvent influence the rate most? 2. How does the rate-determining step of the mechanism respond to this property of the solvent? Because the S N 1 and S N 2 mechanisms are so different from each other, let’s examine each one separately. Solvent Effects on the Rate of Substitution by the S N 1 Mechanism. Table 8.6 lists the relative rate of solvolysis of tert-butyl chloride in several media in order of increasing dielectric constant (H9280). Dielectric constant is a measure of the ability of a material, in this case the solvent, to moderate the force of attraction between oppositely charged par- ticles compared with that of a standard. The standard dielectric is a vacuum, which is assigned a value H9280 of exactly 1. The higher the dielectric constant H9280, the better the medium is able to support separated positively and negatively charged species. Solvents with high dielectric constants are classified as polar solvents. As Table 8.6 illustrates, the rate of solvolysis of tert-butyl chloride (which is equal to its rate of ionization) increases dramatically as the dielectric constant of the solvent increases. H 2 O fast fast 2-Methyl-2-butanol CH 3 CH 3 CCH 2 CH 3 OH 1,1-Dimethylpropyl cation CH 3 CH 3 CCH 2 CH 3 H11001 O HH CH 3 CH 3 CCH 2 CH 3 H11001 slow H11002Br H11002 fast 1,1-Dimethylpropyl cation (a tertiary carbocation) CH 3 CH 3 CCHCH 3 H11001 H 2-Bromo-3-methylbutane H CH 3 CH 3 C Br CHCH 3 1,2-Dimethylpropyl cation (a secondary carbocation) H CH 3 CH 3 C CHCH 3 H11001 320 CHAPTER EIGHT Nucleophilic Substitution Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.12 Effect of Solvent on the Rate of Nucleophilic Substitution 321 According to the S N 1 mechanism, a molecule of an alkyl halide ionizes to a pos- itively charged carbocation and a negatively charged halide ion in the rate-determining step. As the alkyl halide approaches the transition state for this step, a partial positive charge develops on carbon and a partial negative charge on the halogen. Figure 8.9 con- trasts the behavior of a nonpolar and a polar solvent on the energy of the transition state. Polar and nonpolar solvents are similar in their interaction with the starting alkyl halide, but differ markedly in how they affect the transition state. A solvent with a low dielec- tric constant has little effect on the energy of the transition state, whereas one with a high dielectric constant stabilizes the charge-separated transition state, lowers the acti- vation energy, and increases the rate of reaction. TABLE 8.6 Relative Rate of S N 1 Solvolysis of tert-Butyl Chloride as a Function of Solvent Polarity* Solvent Acetic acid Methanol Formic acid Water 6 33 58 78 Dielectric constant H9280 1 4 5,000 150,000 Relative rate *Ratio of first-order rate constant for solvolysis in indicated solvent to that for solvolysis in acetic acid at 25°C. H11006H11006 H11006 H11006 H11006 H11006 H11001 R H11002 X ---- R±X Nonpolar solvent Polar solvent Transition state is more polar than starting state; polar solvent can cluster about transition state so as to reduce electrostatic energy associated with separation of opposite charges. Energy of alkyl halide is approximately the same in either a nonpolar or a polar solvent. E act E act H11006H11006 H11006 H11006 H11006 H11006 H11002H11002 H11002 H11002 H11002 H11002 H11001H11001 H11001 H11001 H11001 H11001 H11002H11002 H11001 H11001 H11001 H11001 H11001H11001 H11002 H11002 H11002 H11002 R±X H11001 R H11002 X ---- δ δ δ δ FIGURE 8.9 A polar solvent stabilizes the transition state of an S N 1 reaction and increases its rate. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Solvent Effects on the Rate of Substitution by the S N 2 Mechanism. Polar solvents are required in typical bimolecular substitutions because ionic substances, such as the sodium and potassium salts cited earlier in Table 8.1, are not sufficiently soluble in nonpolar sol- vents to give a high enough concentration of the nucleophile to allow the reaction to occur at a rapid rate. Other than the requirement that the solvent be polar enough to dis- solve ionic compounds, however, the effect of solvent polarity on the rate of S N 2 reac- tions is small. What is most important is whether or not the polar solvent is protic or aprotic. Water (HOH), alcohols (ROH), and carboxylic acids (RCO 2 H) are classified as polar protic solvents; they all have OH groups that allow them to form hydrogen bonds to anionic nucleophiles as shown in Figure 8.10. Solvation forces such as these stabilize the anion and suppress its nucleophilicity. Aprotic solvents, on the other hand, lack OH groups and do not solvate anions very strongly, leaving them much more able to express their nucleophilic character. Table 8.7 compares the second-order rate constants k for S N 2 substitution of 1-bromobutane by azide ion (a good nucleophile) in some common polar aprotic solvents with the corresponding k’s for the much slower reactions observed in the polar protic solvents methanol and water. H11001 1-Bromobutane CH 3 CH 2 CH 2 CH 2 Br Azide ion N 3 H11002 Bromide ion Br H11002 1-Azidobutane CH 3 CH 2 CH 2 CH 2 N 3 H11001 322 CHAPTER EIGHT Nucleophilic Substitution TABLE 8.7 Relative Rate of S N 2 Displacement of 1-Bromobutane by Azide in Various Solvents* Solvent Methanol Water Dimethyl sulfoxide N,N-Dimethylformamide Acetonitrile 32.6 78.5 48.9 36.7 37.5 Dielectric constant H9280 Polar protic Polar protic Polar aprotic Polar aprotic Polar aprotic Type of solvent CH 3 OH H 2 O (CH 3 ) 2 S?O (CH 3 ) 2 NCH?O CH 3 CPN Structural formula 1 7 1300 2800 5000 Relative rate *Ratio of second-order rate constant for substitution in indicated solvent to that for substitution in methanol at 25°C. FIGURE 8.10 Hydrogen bonding of the solvent to the nucleophile stabilizes the nu- cleophile and makes it less reactive. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.13 Substitution and Elimination as Competing Reactions 323 The large rate enhancements observed for bimolecular nucleophilic substitutions in polar aprotic solvents are used to advantage in synthetic applications. An example can be seen in the preparation of alkyl cyanides (nitriles) by the reaction of sodium cyanide with alkyl halides: When the reaction was carried out in aqueous methanol as the solvent, hexyl bromide was converted to hexyl cyanide in 71% yield by heating with sodium cyanide. Although this is a perfectly acceptable synthetic reaction, a period of over 20 hours was required. Changing the solvent to dimethyl sulfoxide brought about an increase in the reaction rate sufficient to allow the less reactive substrate hexyl chloride to be used instead, and the reaction was complete (91% yield) in only 20 minutes. The rate at which reactions occur can be important in the laboratory, and under- standing how solvents affect rate is of practical value. As we proceed through the text, however, and see how nucleophilic substitution is applied to a variety of functional group transformations, be aware that it is the nature of the substrate and the nucleophile that, more than anything else, determines what product is formed. 8.13 SUBSTITUTION AND ELIMINATION AS COMPETING REACTIONS We have seen that an alkyl halide and a Lewis base can react together in either a sub- stitution or an elimination reaction. Substitution can take place by the S N 1 or the S N 2 mechanism, elimination by E1 or E2. How can we predict whether substitution or elimination will be the principal reac- tion observed with a particular combination of reactants? The two most important fac- tors are the structure of the alkyl halide and the basicity of the anion. It is useful to approach the question from the premise that the characteristic reaction of alkyl halides with Lewis bases is elimination, and that substitution predominates only under certain special circumstances. In a typical reaction, a typical secondary alkyl halide such as iso- propyl bromide reacts with a typical nucleophile such as sodium ethoxide mainly by elimination: Figure 8.11 illustrates the close relationship between the E2 and S N 2 pathways for this case, and the results cited in the preceding equation clearly show that E2 is faster than S N 2 when the alkyl halide is secondary and the nucleophile is a strong base. Isopropyl bromide CH 3 CHCH 3 Br Ethyl isopropyl ether (13%) CH 3 CHCH 3 OCH 2 CH 3 NaOCH 2 CH 3 CH 3 CH 2 OH, 55°C CH 3 CH CH 2 Propene (87%) H11001 H11001 C C H X H11001 Y H11002 H9252 elimination nucleophilic substitution CCH11001 HY X H11002 C C H Y H11001 X H11002 H11001 Hexyl halide CH 3 (CH 2 ) 4 CH 2 X Sodium cyanide NaCN Sodium halide NaX Hexyl cyanide CH 3 (CH 2 ) 4 CH 2 CN H11001 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website As crowding at the carbon that bears the leaving group decreases, the rate of nucle- ophilic attack by the Lewis base increases. A low level of steric hindrance to approach of the nucleophile is one of the special circumstances that permit substitution to pre- dominate, and primary alkyl halides react with alkoxide bases by an S N 2 mechanism in preference to E2: If, however, the base itself is a crowded one, such as potassium tert-butoxide, even pri- mary alkyl halides undergo elimination rather than substitution: A second factor that can tip the balance in favor of substitution is weak basicity of the nucleophile. Nucleophiles that are less basic than hydroxide react with both pri- mary and secondary alkyl halides to give the product of nucleophilic substitution in high yield. To illustrate, cyanide ion is much less basic than hydroxide and reacts with 2- chlorooctane to give the corresponding alkyl cyanide as the major product. Azide ion ( ) is a good nucleophile and an even weaker base than cyanide. It reacts with secondary alkyl halides mainly by substitution: Hydrogen sulfide ion HS H11002 , and anions of the type RS H11002 , are substantially less basic than hydroxide ion and react with both primary and secondary alkyl halides to give mainly substitution products. Cyclohexyl iodide I NaN 3 Cyclohexyl azide (75%) N N H11001H11002 N N?N?N H11002H11002H11001 2-Chlorooctane CH 3 CH(CH 2 ) 5 CH 3 Cl 2-Cyanooctane (70%) CH 3 CH(CH 2 ) 5 CH 3 CN KCN DMSO Propyl bromide CH 3 CH 2 CH 2 Br Ethyl propyl ether (91%) CH 3 CH 2 CH 2 OCH 2 CH 3 NaOCH 2 CH 3 CH 3 CH 2 OH, 55°C CH 3 CH CH 2 Propene (9%) H11001 324 CHAPTER EIGHT Nucleophilic Substitution E2 Br H C CH 3 CH 2 O S N 2 FIGURE 8.11 When a Lewis base reacts with an alkyl halide, either substitution or elimination can occur. Sub- stitution (S N 2) occurs when the nucleophile attacks car- bon to displace bromide. Elimination occurs when the Lewis base abstracts a pro- ton from the H9252 carbon. The alkyl halide shown is iso- propyl bromide. The carbon atom that bears the leaving group is somewhat sterically hindered, and elimination (E2) predominates over sub- stitution with alkoxide bases. 1-Bromooctadecane CH 3 (CH 2 ) 15 CH 2 CH 2 Br tert-Butyl octadecyl ether (13%) CH 3 (CH 2 ) 15 CH 2 CH 2 OC(CH 3 ) 3 KOC(CH 3 ) 3 (CH 3 ) 3 COH, 40°C CH 3 (CH 2 ) 15 CH CH 2 1-Octadecene (87%) H11001 Cyanide is a weaker base than hydroxide because its conjugate acid HCN (pK a 9.1) is a stronger acid than water (pK a 15.7). The conjugate acid of azide ion is called hydrazoic acid (HN 3 ). It has a pK a of 4.6, and so is similar to acetic acid in its acidity. Hydrogen sulfide (pK a 7.0) is a stronger acid than water (pK a 15.7). Therefore HS H11002 is a much weaker base than HO H11002 . Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.13 Substitution and Elimination as Competing Reactions 325 Tertiary alkyl halides are so sterically hindered to nucleophilic attack that the pres- ence of any anionic Lewis base favors elimination. Usually substitution predominates over elimination in tertiary alkyl halides only when anionic Lewis bases are absent. In the solvolysis of the tertiary bromide 2-bromo-2-methylbutane, for example, the ratio of substitution to elimination is 64:36 in pure ethanol but falls to 1:99 in the presence of 2 M sodium ethoxide. PROBLEM 8.12 Predict the major organic product of each of the following reac- tions: (a) Cyclohexyl bromide and potassium ethoxide (b) Ethyl bromide and potassium cyclohexanolate (c) sec-Butyl bromide solvolysis in methanol (d) sec-Butyl bromide solvolysis in methanol containing 2 M sodium methoxide SAMPLE SOLUTION (a) Cyclohexyl bromide is a secondary halide and reacts with alkoxide bases by elimination rather than substitution. The major organic prod- ucts are cyclohexene and ethanol. Regardless of the alkyl halide, raising the temperature causes both the rate of sub- stitution and the rate of elimination to increase. The rate of elimination, however, usu- ally increases faster than the rate of substitution, so that at higher temperatures the pro- portion of elimination products increases at the expense of substitution products. As a practical matter, elimination can always be made to occur quantitatively. Strong bases, especially bulky ones such as tert-butoxide ion, react even with primary alkyl halides by an E2 process at elevated temperatures. The more difficult task is to find the set of conditions that promote substitution. In general, the best approach is to choose conditions that favor the S N 2 mechanism—an unhindered substrate, a good nucleophile that is not strongly basic, and the lowest practical temperature consistent with reason- able reaction rates. Functional group transformations that rely on substitution by the S N 1 mechanism are not as generally applicable as those of the S N 2 type. Hindered substrates are prone to elimination, and there is the possibility of rearrangement when carbocation interme- diates are involved. Only in cases in which elimination is impossible are S N 1 reactions used for functional group transformations. Cyclohexyl bromide Br H11001 Potassium ethoxide KOCH 2 CH 3 Cyclohexene H11001 Ethanol CH 3 CH 2 OH ethanol 25°C 2-Bromo-2-methyl- butane Br CH 3 CH 3 CCH 2 CH 3 2-Ethoxy-2- methylbutane (Major product in absence of sodium ethoxide) OCH 2 CH 3 CH 3 CH 3 CCH 2 CH 3 H11001H11001(CH 3 ) 2 C CHCH 3 2-Methyl-2-butene CH 2 CCH 2 CH 3 CH 3 2-Methyl-1-butene (Alkene mixture is major product in presence of sodium ethoxide) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.14 SULFONATE ESTERS AS SUBSTRATES IN NUCLEOPHILIC SUBSTITUTION Two kinds of starting materials have been examined in nucleophilic substitution reac- tions to this point. In Chapter 4 we saw alcohols can be converted to alkyl halides by reaction with hydrogen halides and pointed out that this process is a nucleophilic sub- stitution taking place on the protonated form of the alcohol, with water serving as the leaving group. In the present chapter the substrates have been alkyl halides, and halide ions have been the leaving groups. A few other classes of organic compounds undergo nucleophilic substitution reactions analogous to those of alkyl halides, the most impor- tant of these being alkyl esters of sulfonic acids. Sulfonic acids such as methanesulfonic acid and p-toluenesulfonic acid are strong acids, comparable in acidity with sulfuric acid. Alkyl sulfonates are derivatives of sulfonic acids in which the proton of the hydroxyl group is replaced by an alkyl group. They are prepared by treating an alcohol with the appropriate sulfonyl chloride. These reactions are usually carried out in the presence of pyridine. Alkyl sulfonate esters resemble alkyl halides in their ability to undergo elimina- tion and nucleophilic substitution. Nucleophile Y H11002 H11001 p-Toluenesulfonate ester O O CH 3 OSR Product of nucleophilic substitution R Y H11001 p-Toluenesulfonate anion O O CH 3 H11002 OS CH 3 CH 2 OH Ethanol H11001 CH 3 S O O Cl p-Toluenesulfonyl chloride pyridine CH 3 CH 2 OS O O CH 3 Ethyl p-toluenesulfonate (72%) H11001RO H Alcohol O O RH11032S Cl Sulfonyl chloride RO SRH11032 O O Sulfonate ester H11001 HCl Hydrogen chloride S O O OHCH 3 Methanesulfonic acid CH 3 S O O OH p-Toluenesulfonic acid 326 CHAPTER EIGHT Nucleophilic Substitution Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.14 Sulfonate Esters as Substrates in Nucleophilic Substitution 327 The sulfonate esters used most frequently are the p-toluenesulfonates. They are com- monly known as tosylates and given the abbreviated formula ROTs. p-Toluenesulfonate (TsO H11002 ) is a very good leaving group. As Table 8.8 reveals, alkyl p-toluenesulfonates undergo nucleophilic substitution at rates that are even faster than those of alkyl iodides. A correlation of leaving-group abilities with carbon–halogen bond strengths was noted earlier, in Section 8.2. Note also the correlation with the basic- ity of the leaving group. Iodide is the weakest base among the halide anions and is the best leaving group, fluoride the strongest base and the poorest leaving group. A similar correlation with basicity is seen among oxygen-containing leaving groups. The weaker the base, the better the leaving group. Trifluoromethanesulfonic acid (CF 3 SO 2 OH) is a much stronger acid than p-toluenesulfonic acid, and therefore trifluoromethanesulfonate is a much weaker base than p-toluenesulfonate and a much better leaving group. Notice too that strongly basic leaving groups are absent from Table 8.8. In gen- eral, any species that has a K a less than 1 for its conjugate acid cannot be a leaving group in a nucleophilic substitution. Thus, hydroxide (HO H11002 ) is far too strong a base to be dis- placed from an alcohol (ROH), and alcohols do not undergo nucleophilic substitution. In strongly acidic media, alcohols are protonated to give alkyloxonium ions, and these do undergo nucleophilic substitution, because the leaving group is a weakly basic water molecule. Since halides are poorer leaving groups than p-toluenesulfonate, alkyl p-toluene- sulfonates can be converted to alkyl halides by S N 2 reactions involving chloride, bro- mide, or iodide as the nucleophile. sec-Butyl p-toluenesulfonate CH 3 CHCH 2 CH 3 OTs H11001 Sodium bromide NaBr sec-Butyl bromide (82%) CH 3 CHCH 2 CH 3 Br H11001 Sodium p-toluenesulfonate NaOTs DMSO (3-Cyclopentenyl)methyl p-toluenesulfonate H CH 2 OTs 4-(Cyanomethyl)cyclo- pentene (86%) H CH 2 CN KCN ethanol-water TABLE 8.8 Approximate Relative Leaving-Group Abilities* Leaving group F H11002 Cl H11002 Br H11002 I H11002 H 2 O TsO H11002 CF 3 SO 2 O H11002 HF HCl HBr HI H 3 O H11001 TsOH CF 3 SO 2 OH Conjugate acid of leaving group 3.5 H11003 10 H110024 10 7 10 9 10 10 55 6 H11003 10 2 10 6 K a of conjugate acid 10 H110025 10 0 10 1 10 2 10 1 10 5 10 8 Relative rate 3.5 H110027 H110029 H1100210 H110021.7 H110022.8 H110026 pK a *Values are approximate and vary according to substrate. Trifluoromethanesulfonate esters are called triflates. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEM 8.13 Write a chemical equation showing the preparation of octade- cyl p-toluenesulfonate. PROBLEM 8.14 Write equations showing the reaction of octadecyl p-toluene- sulfonate with each of the following reagents: (a) Potassium acetate (b) Potassium iodide (KI) (c) Potassium cyanide (KCN) (d) Potassium hydrogen sulfide (KSH) (e) Sodium butanethiolate (NaSCH 2 CH 2 CH 2 CH 3 ) SAMPLE SOLUTION All these reactions of octadecyl p-toluenesulfonate have been reported in the chemical literature, and all proceed in synthetically useful yield. You should begin by identifying the nucleophile in each of the parts to this problem. The nucleophile replaces the p-toluenesulfonate leaving group in an S N 2 reaction. In part (a) the nucleophile is acetate ion, and the product of nucleophilic substitution is octadecyl acetate. Sulfonate esters are subject to the same limitations as alkyl halides. Competition from elimination needs to be considered when planning a functional group transforma- tion that requires an anionic nucleophile, because tosylates undergo elimination reactions, just as alkyl halides do. An advantage that sulfonate esters have over alkyl halides is that their preparation from alcohols does not involve any of the bonds to carbon. The alcohol oxygen becomes the oxygen that connects the alkyl group to the sulfonyl group. Thus, the configuration of a sulfonate ester is exactly the same as that of the alcohol from which it was pre- pared. If we wish to study the stereochemistry of nucleophilic substitution in an opti- cally active substrate, for example, we know that a tosylate ester will have the same con- figuration and the same optical purity as the alcohol from which it was prepared. The same cannot be said about reactions with alkyl halides as substrates. The conver- sion of optically active 2-octanol to the corresponding halide does involve a bond to the stereogenic center, and so the optical purity and absolute configuration of the alkyl halide need to be independently established. (S)-(H11001)-2-Octanol [H9251] D 25 H110019.9° (optically pure) C H H 3 C CH 3 (CH 2 ) 5 OH p-Toluenesulfonyl chloride pyridine (S)-(H11001)-1-Methylheptyl p-toluenesulfonate [H9251] D 25 H110017.9° (optically pure) C H H 3 C CH 3 (CH 2 ) 5 OS CH 3 O O Acetate ion O CH 3 CO H11002 H11001 Octadecyl tosylate CH 2 (CH 2 ) 16 CH 3 OTs Octadecyl acetate O CH 3 COCH 2 (CH 2 ) 16 CH 3 (KOCCH 3 ) O 328 CHAPTER EIGHT Nucleophilic Substitution Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.15 Looking Back: Reactions of Alcohols with Hydrogen Halides 329 The mechanisms by which sulfonate esters undergo nucleophilic substitution are the same as those of alkyl halides. Inversion of configuration is observed in S N 2 reactions of alkyl sulfonates and predominant inversion accompanied by racemization in S N 1 processes. PROBLEM 8.15 The hydrolysis of sulfonate esters of 2-octanol is a stereospecific reaction and proceeds with complete inversion of configuration. Write a structural formula that shows the stereochemistry of the 2-octanol formed by hydrolysis of an optically pure sample of (S)-(H11001)-1-methylheptyl p-toluenesulfonate, identify the product as R or S, and deduce its specific rotation. 8.15 LOOKING BACK: REACTIONS OF ALCOHOLS WITH HYDROGEN HALIDES The principles developed in this chapter can be applied to a more detailed examination of the reaction of alcohols with hydrogen halides than was possible when this reaction was first introduced in Chapter 4. As pointed out in Chapter 4, the first step in the reaction is proton transfer to the alcohol from the hydrogen halide to yield an alkyloxonium ion. This is an acid-base reaction. With primary alcohols, the next stage is an S N 2 reaction in which the halide ion, bro- mide, for example, displaces a molecule of water from the alkyloxonium ion. With secondary and tertiary alcohols, this stage is an S N 1 reaction in which the alkyl- oxonium ion dissociates to a carbocation and water. Following its formation, the carbocation is captured by halide. fast H11001 Secondary carbocation R 2 CH H11001 Bromide ion Br H11002 Secondary alkyl bromide R 2 CH Br Secondary alkyloxonium ion R 2 CH OH 2 H11001 S N 1 transition state R 2 CH OH 2 H9254H11001 H9254H11001 H11001 Water H 2 O Secondary carbocation R 2 CH H11001 Bromide ion Br H11002 H11001 Primary alkyl- oxonium ion RCH 2 OH 2 H11001 S N 2 transition state C HH OH 2 R Br H9254H11002 H9254H11001 Primary alkyl bromide CH 2 RBr Water H 2 OH11001 H11001H11001 Halide ion (conjugate base) Alcohol (base) R H O Hydrogen halide (acid) HX Alkyloxonium ion (conjugate acid) O R H H H11001 X H11002 H11001 Alcohol ROH Hydrogen halide HX Water H 2 O Alkyl halide RX H11001 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website With optically active secondary alcohols the reaction proceeds with predominant, but incomplete, inversion of configuration. The few studies that have been carried out with optically active tertiary alcohols indicate that almost complete racemization attends the preparation of tertiary alkyl halides by this method. Rearrangement can occur, and the desired alkyl halide is sometimes accompanied by an isomeric halide. An example is seen in the case of the secondary alcohol 2-octanol, which yields a mixture of 2- and 3-bromooctane: PROBLEM 8.16 Treatment of 3-methyl-2-butanol with hydrogen chloride yielded only a trace of 2-chloro-3-methylbutane. An isomeric chloride was isolated in 97% yield. Suggest a reasonable structure for this product. Unbranched primary alcohols and tertiary alcohols tend to react with hydrogen halides without rearrangement. The alkyloxonium ions from primary alcohols react rapidly with bromide ion, for example, in an S N 2 process without significant develop- ment of positive charge at carbon. Tertiary alcohols give tertiary alkyl halides because tertiary carbocations are stable and show little tendency to rearrange. When it is necessary to prepare secondary alkyl halides with assurance that no trace of rearrangement accompanies their formation, the corresponding alcohol is first converted to its p-toluenesulfonate ester and this ester is then allowed to react with sodium chloride, bromide, or iodide, as described in Section 8.14. 8.16 SUMMARY Section 8.1 Nucleophilic substitution is an important reaction type in synthetic organic chemistry because it is one of the main methods for functional group transformations. Examples of synthetically useful nucleophilic sub- stitutions were given in Table 8.1. It is a good idea to return to that table and review its entries now that the details of nucleophilic substitution have been covered. Sections These sections show how a variety of experimental observations led to 8.2–8.12 the proposal of the S N 1 and the S N 2 mechanisms for nucleophilic sub- stitution. Summary Table 8.9 integrates the material in these sections. 2-Octanol CH 3 CHCH 2 (CH 2 ) 4 CH 3 OH 2-Bromooctane (93%) CH 3 CHCH 2 (CH 2 ) 4 CH 3 Br 3-Bromooctane (7%) CH 3 CH 2 CH(CH 2 ) 4 CH 3 Br 1-Methylheptyl cation CH 3 CHCH 2 (CH 2 ) 4 CH 3 H11001 1-Ethylhexyl cation CH 3 CH 2 CH(CH 2 ) 4 CH 3 H11001 HBr Br H11002 Br H11002 (R)-(H11002)-2-Butanol C H CH 3 CH 2 CH 3 OH (R)-(H11002)-2-Bromobutane (13%) C H CH 3 CH 2 CH 3 Br HBr (S)-(H11001)-2-Bromobutane (87%) H CH 2 CH 3 CH 3 CBr H11001 330 CHAPTER EIGHT Nucleophilic Substitution Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.16 Summary 331 TABLE 8.9 Comparison of S N 1 and S N 2 Mechanisms of Nucleophilic Substitution in Alkyl Halides Characteristics of mechanism Rate-determining transition state Molecularity Kinetics and rate law Relative reactivity of halide leaving groups Effect of structure on rate Effect of nucleophile on rate S N 1S N 2 Two elementary steps: Ionization of alkyl halide (step 1) is rate-determining. (Section 8.8) Single step: (Section 8.8) (Sections 8.3 and 8.5) Unimolecular (Section 8.8) Bimolecular (Section 8.3) First order: Rate H11005 k[alkyl halide] (Section 8.8) Second order: Rate H11005 k[alkyl halide][nucleophile] (Section 8.3) RI H11022 RBr H11022 RCl H11022H11022 RF (Section 8.2) RI H11022 RBr H11022 RCl H11022H11022 RF (Section 8.2) R 3 CX H11022 R 2 CHX H11022 RCH 2 X H11022 CH 3 X Rate is governed by stability of car- bocation that is formed in ioniza- tion step. Tertiary alkyl halides can react only by the S N 1 mechanism; they never react by the S N 2 mecha- nism. (Section 8.9) CH 3 X H11022 RCH 2 X H11022 R 2 CHX H11022 R 3 CX Rate is governed by steric effects (crowding in transition state). Methyl and primary alkyl halides can react only by the S N 2 mecha- nism; they never react by the S N 1 mechanism. (Section 8.6) Rate of substitution is independent of both concentration and nature of nucleophile. Nucleophile does not participate until after rate- determining step. (Section 8.8) Rate depends on both nature of nucleophile and its concentration. (Sections 8.3 and 8.7) Effect of solvent on rate Rate increases with increasing polarity of solvent as measured by its dielectric constant H9280. (Section 8.12) Polar aprotic solvents give fastest rates of substitution; solvation of Nu : H11002 is minimal and nucleophilicity is greatest. (Section 8.12) Stereochemistry Potential for rearrangements Not stereospecific: racemization accompanies inversion when leav- ing group is located at a stereogen- ic center. (Section 8.10) Stereospecific: 100% inversion of configuration at reaction site. Nucleophile attacks carbon from side opposite bond to leaving group. (Section 8.4) Carbocation intermediate capable of rearrangement. (Section 8.11) No carbocation intermediate; no rearrangement. Step 1: R X R H11001 H11001 X H11002 Step 2: R H11001 H11001 Nu H11002 R Nu Nucleophile displaces leaving group; bonding to the incoming nucleophile accompanies cleavage of the bond to the leaving group. (Sections 8.3 and 8.5) R X Nu R H11001 X H11002 H11002 Nu H9254H11001 R X H9254H11002 H9254H11002 Nu R X H9254H11002 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Section 8.13 When nucleophilic substitution is used for synthesis, the competition between substitution and elimination must be favorable. However, the normal reaction of a secondary alkyl halide with a base as strong or stronger than hydroxide is elimination (E2). Substitution by the S N 2 mechanism predominates only when the base is weaker than hydroxide or the alkyl halide is primary. Elimination predominates when tertiary alkyl halides react with any anion. Section 8.14 Nucleophilic substitution can occur with leaving groups other than halide. Alkyl p-toluenesulfonates (tosylates), which are prepared from alcohols by reaction with p-toulenesulfonyl chloride, are often used. Section 8.15 In its ability to act as a leaving group, p-toluenesulfonate is comparable to iodide. The reactions of alcohols with hydrogen halides to give alkyl halides (Chapter 4) are nucleophilic substitution reactions of alkyloxonium ions in which water is the leaving group. Primary alcohols react by an S N 2- like displacement of water from the alkyloxonium ion by halide. Sec- ondary and tertiary alcohols give alkyloxonium ions which form carbo- cations in an S N 1-like process. Rearrangements are possible with secondary alcohols, and substitution takes place with predominant, but not complete, inversion of configuration. PROBLEMS 8.17 Write the structure of the principal organic product to be expected from the reaction of 1-bromopropane with each of the following: (a) Sodium iodide in acetone (b) Sodium acetate in acetic acid (c) Sodium ethoxide in ethanol (d) Sodium cyanide in dimethyl sulfoxide (e) Sodium azide in aqueous ethanol (f) Sodium hydrogen sulfide in ethanol (g) Sodium methanethiolate (NaSCH 3 ) in ethanol (CH 3 CONa) O H11001Nu H11002 Nucleophile R OTs Alkyl p-toluenesulfonate Nu R Substitution product H11002 OTs p-Toluenesulfonate ion ROH Alcohol H11001 CH 3 SO 2 Cl p-Toluenesulfonyl chloride pyridine ROS O O CH 3 (ROTs) Alkyl p-toluenesulfonate (alkyl tosylate) 332 CHAPTER EIGHT Nucleophilic Substitution Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Problems 333 8.18 All the reactions of 1-bromopropane in the preceding problem give the product of nucle- ophilic substitution in high yield. High yields of substitution products are also obtained in all but one of the analogous reactions using 2-bromopropane as the substrate. In one case, however, 2- bromopropane is converted to propene, especially when the reaction is carried out at elevated tem- perature (about 55°C). Which reactant is most effective in converting 2-bromopropane to propene? 8.19 Each of the following nucleophilic substitution reactions has been reported in the chemical literature. Many of them involve reactants that are somewhat more complex than those we have dealt with to this point. Nevertheless, you should be able to predict the product by analogy to what you know about nucleophilic substitution in simple systems. (a) (b) (c) (d) (e) (f) (g) (h) 8.20 Each of the reactions shown involves nucleophilic substitution. The product of reaction (a) is an isomer of the product of reaction (b). What kind of isomer? By what mechanism does nucle- ophilic substitution occur? Write the structural formula of the product of each reaction. (a) (b) 8.21 Arrange the isomers of molecular formula C 4 H 9 Cl in order of decreasing rate of reaction with sodium iodide in acetone. C(CH 3 ) 3 H11001 SNa Cl Cl C(CH 3 ) 3 H11001 SNa CH 3 O CH 3 O CH 2 CH 2 CH 2 CH 2 OH OCH 3 1. TsCl, pyridine 2. LiI, acetone CH 3 CH 2 Br O CH 2 SNa H11001 NaI acetone TsOCH 2 O O CH 3 CH 3 NaN 3 acetone–water ClCH 2 COC(CH 3 ) 3 O H 2 O, HO H11002 NC CH 2 Cl NaCN ethanol–water CH 3 CH 2 OCH 2 CH 2 Br CH 3 CONa acetic acid O O 2 NCH 2 Cl NaI acetone BrCH 2 COCH 2 CH 3 O Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.22 There is an overall 29-fold difference in reactivity of 1-chlorohexane, 2-chlorohexane, and 3-chlorohexane toward potassium iodide in acetone. (a) Which one is the most reactive? Why? (b) Two of the isomers differ by only a factor of 2 in reactivity. Which two are these? Which one is the more reactive? Why? 8.23 In each of the following indicate which reaction will occur faster. Explain your reasoning. (a) CH 3 CH 2 CH 2 CH 2 Br or CH 3 CH 2 CH 2 CH 2 I with sodium cyanide in dimethyl sulfoxide (b) 1-Chloro-2-methylbutane or 1-chloropentane with sodium iodide in acetone (c) Hexyl chloride or cyclohexyl chloride with sodium azide in aqueous ethanol (d) Solvolysis of 1-bromo-2,2-dimethylpropane or tert-butyl bromide in ethanol (e) Solvolysis of isobutyl bromide or sec-butyl bromide in aqueous formic acid (f) Reaction of 1-chlorobutane with sodium acetate in acetic acid or with sodium methoxide in methanol (g) Reaction of 1-chlorobutane with sodium azide or sodium p-toluenesulfonate in aque- ous ethanol 8.24 Under conditions of photochemical chlorination, (CH 3 ) 3 CCH 2 C(CH 3 ) 3 gave a mixture of two monochlorides in a 4:1 ratio. The structures of these two products were assigned on the basis of their S N 1 hydrolysis rates in aqueous ethanol. The major product (compound A) underwent hydrolysis much more slowly than the minor one (compound B). Deduce the structures of com- pounds A and B. 8.25 The compound KSCN is a source of thiocyanate ion. (a) Write the two most stable Lewis structures for thiocyanate ion and identify the atom in each that bears a formal charge of H110021. (b) Two constitutionally isomeric products of molecular formula C 5 H 9 NS were isolated in a combined yield of 87% in the reaction shown. (DMF stands for N,N-dimethylfor- mamide, a polar aprotic solvent.) Suggest reasonable structures for these two com- pounds. (c) The major product of the reaction cited in (b) constituted 99% of the mixture of isomers. Its structure corresponds to attack by the most polarizable atom of thiocyanate ion on 1-bromobutane. What is this product? 8.26 Reaction of ethyl iodide with triethylamine yields a crystalline compound C 8 H 20 NI in high yield. This compound is soluble in polar solvents such as water but insoluble in nonpolar ones such as diethyl ether. It does not melt below about 200°C. Suggest a reasonable structure for this product. 8.27 Write an equation, clearly showing the stereochemistry of the starting material and the prod- uct, for the reaction of (S)-1-bromo-2-methylbutane with sodium iodide in acetone. What is the configuration (R or S) of the product? 8.28 Identify the product in each of the following reactions: (a) (b) BrCH 2 CH 2 Br H11001 NaSCH 2 CH 2 SNa ±￡ C 4 H 8 S 2 (c) ClCH 2 CH 2 CH 2 CH 2 Cl H11001 Na 2 S ±￡ C 4 H 8 S NaI (1.0 equiv) acetone C 5 H 10 ClIClCH 2 CH 2 CHCH 2 CH 3 Cl [(CH 3 CH 2 ) 3 N] KSCN DMF CH 3 CH 2 CH 2 CH 2 Br 334 CHAPTER EIGHT Nucleophilic Substitution Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Problems 335 8.29 Give the mechanistic symbols (S N 1, S N 2, E1, E2) that are most consistent with each of the following statements: (a) Methyl halides react with sodium ethoxide in ethanol only by this mechanism. (b) Unhindered primary halides react with sodium ethoxide in ethanol mainly by this mechanism. (c) When cyclohexyl bromide is treated with sodium ethoxide in ethanol, the major prod- uct is formed by this mechanism. (d) The substitution product obtained by solvolysis of tert-butyl bromide in ethanol arises by this mechanism. (e) In ethanol that contains sodium ethoxide, tert-butyl bromide reacts mainly by this mechanism. (f) These reaction mechanisms represent concerted processes. (g) Reactions proceeding by these mechanisms are stereospecific. (h) These reaction mechanisms involve carbocation intermediates. (i) These reaction mechanisms are the ones most likely to have been involved when the products are found to have a different carbon skeleton from the substrate. (j) Alkyl iodides react faster than alkyl bromides in reactions that proceed by these mechanisms. 8.30 Outline an efficient synthesis of each of the following compounds from the indicated start- ing material and any necessary organic or inorganic reagents: (a) Cyclopentyl cyanide from cyclopentane (b) Cyclopentyl cyanide from cyclopentene (c) Cyclopentyl cyanide from cyclopentanol (d) NCCH 2 CH 2 CN from ethyl alcohol (e) Isobutyl iodide from isobutyl chloride (f) Isobutyl iodide from tert-butyl chloride (g) Isopropyl azide from isopropyl alcohol (h) Isopropyl azide from 1-propanol (i) (S)-sec-Butyl azide from (R)-sec-butyl alcohol (j) 8.31 Select the combination of alkyl bromide and potassium alkoxide that would be the most effective in the syntheses of the following ethers: (a) CH 3 OC(CH 3 ) 3 (b) (c) (CH 3 ) 3 CCH 2 OCH 2 CH 3 8.32 (Note to the student: This problem previews an important aspect of Chapter 9 and is well worth attempting in order to get a head start on the material presented there.) Alkynes of the type RCPCH may be prepared by nucleophilic substitution reactions in which one of the starting materials is sodium acetylide .(Na H11001 H11002 CPCH) OCH 3 (S)-CH 3 CH 2 CHCH 3 from (R)-sec-butyl alcohol SH Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (a) Devise a method for the preparation of CH 3 CH 2 CH11013CH from sodium acetylide and any necessary organic or inorganic reagents. (b) Given the information that K a for acetylene (HCPCH) is 10 H1100226 (pK a 26), comment on the scope of this preparative procedure with respect to R in RCPCH. Could you prepare (CH 3 ) 2 CHCPCH or (CH 3 ) 3 CCPCH in good yield by this method? 8.33 Give the structures, including stereochemistry, of compounds A and B in the following sequence of reactions: 8.34 (a) Suggest a reasonable series of synthetic transformations for converting trans-2-methyl- cyclopentanol to cis-2-methylcyclopentyl acetate. (b) How could you prepare cis-2-methylcyclopentyl acetate from 1-methylcyclopentanol? 8.35 Optically pure (S)-(H11001)-2-butanol was converted to its methanesulfonate ester according to the reaction shown. (a) Write the Fischer projection of the sec-butyl methanesulfonate formed in this reaction. (b) The sec-butyl methanesulfonate in part (a) was treated with NaSCH 2 CH 3 to give a product having an optical rotation H9251 D of H1100225°. Write the Fischer projection of this product. By what mechanism is it formed? What is its absolute configuration (R or S)? (c) When treated with PBr 3 , optically pure (S)-(H11001)-2-butanol gave 2-bromobutane hav- ing an optical rotation H9251 D H11005H1100238°. This bromide was then allowed to react with NaSCH 2 CH 3 to give a product having an optical rotation H9251 D of H1100123°. Write the Fischer projection for (H11002)-2-bromobutane and specify its configuration as R or S. Does the reaction of 2-butanol with PBr 3 proceed with predominant inversion or retention of configuration? (d) What is the optical rotation of optically pure 2-bromobutane? 8.36 In a classic experiment, Edward Hughes (a colleague of Ingold’s at University College, Lon- don) studied the rate of racemization of 2-iodooctane by sodium iodide in acetone and compared it with the rate of incorporation of radioactive iodine into 2-iodooctane. How will the rate of racemization compare with the rate of incorporation of radioactivity if (a) Each act of exchange proceeds stereospecifically with retention of configuration? (b) Each act of exchange proceeds stereospecifically with inversion of configuration? (c) Each act of exchange proceeds in a stereorandom manner, in which retention and inversion of configuration are equally likely? (I* H11005 radioactive iodine) RI [I*] H11002 H11001 RI* I H11002 H11001 H CH 3 CH 2 CH 3 OH CH 3 SO 2 Cl pyridine sec-butyl methanesulfonate OCCH 3 H 3 C O cis-2-Methylcyclopentyl acetate 336 CHAPTER EIGHT Nucleophilic Substitution pyridine LiBr acetone(CH 3 ) 3 C OH H11001 O 2 N SO 2 Cl compound A compound B Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Problems 337 8.37 The ratio of elimination to substitution is exactly the same (26% elimination) for 2-bromo- 2-methylbutane and 2-iodo-2-methylbutane in 80% ethanol/20% water at 25°C. (a) By what mechanism does substitution most likely occur in these compounds under these conditions? (b) By what mechanism does elimination most likely occur in these compounds under these conditions? (c) Which substrate undergoes substitution faster? (d) Which substrate undergoes elimination faster? (e) What two substitution products are formed from each substrate? (f) What two elimination products are formed from each substrate? (g) Why do you suppose the ratio of elimination to substitution is the same for the two substrates? 8.38 The reaction of 2,2-dimethyl-1-propanol with HBr is very slow and gives 2-bromo-2-methyl- propane as the major product. Give a mechanistic explanation for these observations. 8.39 Solvolysis of 2-bromo-2-methylbutane in acetic acid containing potassium acetate gave three products. Identify them. 8.40 Solvolysis of 1,2-dimethylpropyl p-toluenesulfonate in acetic acid (75°C) yields five differ- ent products: three are alkenes and two are substitution products. Suggest reasonable structures for these five products. 8.41 Solution A was prepared by dissolving potassium acetate in methanol. Solution B was pre- pared by adding potassium methoxide to acetic acid. Reaction of methyl iodide either with solu- tion A or with solution B gave the same major product. Why? What was this product? 8.42 If the temperature is not kept below 25°C during the reaction of primary alcohols with p- toluenesulfonyl chloride in pyridine, it is sometimes observed that the isolated product is not the desired alkyl p-toluenesulfonate but is instead the corresponding alkyl chloride. Suggest a mech- anistic explanation for this observation. 8.43 The reaction of cyclopentyl bromide with sodium cyanide to give cyclopentyl cyanide proceeds faster if a small amount of sodium iodide is added to the reaction mixture. Can you suggest a reasonable mechanism to explain the catalytic function of sodium iodide? 8.44 Illustrate the stereochemistry associated with unimolecular nucleophilic substitution by con- structing molecular models of cis-4-tert-butylcyclohexyl bromide, its derived carbocation, and the alcohols formed from it by hydrolysis under S N 1 conditions. 8.45 Given the molecular formula C 6 H 11 Br, construct a molecular model of the isomer that is a primary alkyl bromide yet relatively unreactive toward bimolecular nucleophilic substitution. NaCN ethanol–waterBr H Cyclopentyl bromide CN H Cyclopentyl cyanide HBr 65°C CH 3 CCH 2 OH CH 3 CH 3 CH 3 CCH 2 CH 3 Br CH 3 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.46 Cyclohexyl bromide is less reactive than noncyclic secondary alkyl halides toward S N 2 sub- stitution. Construct a molecular model of cyclohexyl bromide and suggest a reason for its low reactivity. 8.47 1-Bromobicyclo[2.2.1]heptane (the structure of which is shown) is exceedingly unreactive toward nucleophilic substitution by either the S N 1 or S N 2 mechanism. Use molecular models to help you understand why. Br 1-Bromobicyclo[2.2.1]heptane 338 CHAPTER EIGHT Nucleophilic Substitution Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website