Chapter 7 Substitution Reactions

Chapter 7 Substitution Reactions Review of Concepts Fill in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 7. Each of the sentences below appears verbatim in the section entitled Review of Concepts and Vocabulary. Substitution reactions exchange one for another. Evidence for the concerted mechanism, called S N 2, includes the observation of a - order rate equation. The reaction proceeds with of configuration. S N 2 reactions are said to be because the configuration of the product is determined by the configuration of the substrate. Evidence for the stepwise mechanism, called S N 1, includes the observation of a - order rate equation. The step of an S N 1 process is the rate-determining step. There are four factors that impact the competition between the S N 2 mechanism and S N 1: 1) the, 2) the, 3) the, and 4) the. solvents favor S N 2. Review of Skills Follow the instructions below. To verify that your answers are correct, look in your textbook at the end of Chapter 7. The answers appear in the section entitled SkillBuilder Review. SkillBuilder 7.1 Drawing the Curved Arrows of a Substitution Reaction SkillBuilder 7.2 Drawing the Product of an S N 2 Process SkillBuilder 7.3 Drawing the Transition State of an S N 2 Process

CAPTER 7 179 SkillBuilder 7.4 Drawing the Carbocation Intermediate of an S N 1 Process DRAW TE CARBCATIN TAT WULD BE FRMED IF A CLRIDE IN IS EXPELLED FRM TE FLLW ING CMPUND Cl Cl SkillBuilder 7.5 Drawing the Products of an S N 1 Process PREDICT TE PRDUCTS F TE FLLWING S N 1REACTIN Br NaCN + SkillBuilder 7.6 Drawing the Complete Mechanism of an S N 1 Process SkillBuilder 7.7 Drawing the Complete Mechanism of an S N 2 Process SkillBuilder 7.8 Determining Whether a Reaction Proceeds via an S N 1 or S N 2 Mechanism

180 CAPTER 7 SkillBuilder 7.9 Identifying the Reagents Necessary for a Substitution Reaction Review of Reactions Follow the instructions below. To verify that your answers are correct, look in your textbook at the end of Chapter 7. The answers appear in the section entitled Review of Reactions. Useful reagents The following is a list of commonly encountered nucleophilic agents: Nucleophilic agent Name Function NaCl Sodium chloride An ionic salt consisting of Na + and Cl ions. The former (Na + ) can be ignored in most cases, while the latter (chloride) is a strong nucleophile. NaCl is a source of chloride ions. NaBr NaI Na Sodium bromide Sodium iodide Sodium hydroxide An ionic salt consisting of Na + and Br ions. The former (Na + ) can be ignored in most cases, while the latter (bromide) is a strong nucleophile. NaBr is a source of bromide ions. An ionic salt consisting of Na + and I ions. The former (Na + ) can be ignored in most cases, while the latter (iodide) is a strong nucleophile. NaI is a source of iodide ions. ydroxide ( ) is a strong nucleophile, used in S N 2 reactions. In Chapter 8, we will see that hydroxide is also a strong base, used in E2 reactions. In many cases, S N 2 and E2 reactions compete with each other, as will be explored in Chapter 8.

CAPTER 7 181 NaR NaS 2 R X (X = Cl, Br, or I) Sodium alkoxide Sodium hydrosulfide Water An alcohol ydrogen halides R is an alkyl group. Examples include sodium methoxide (NaMe) and sodium ethoxide (NaEt). Alkoxide ions are strong nucleophiles. In Chapter 8, we will see that alkoxide ions are also strong bases, used in E2 reactions. In many cases, S N 2 and E2 reactions compete with each other, as will be explored in Chapter 8. S is a strong nucleophile, used in S N 2 reactions. Water is a weak nucleophile, used in S N 1 reactions. In Chapter 8, we will see that water is also a weak base, used in E1 reactions. In most cases, S N 1 and E1 reactions compete with each other, as will be explored in Chapter 8. Examples include methanol (C 3 ) and ethanol (C 3 C 2 ) A strong acid that serves as both a source of + and nucleophilic X where X = Cl, Br, or I. Common Mistakes to Avoid When drawing the mechanism of a reaction, you must always consider what reagents are being used, and your mechanism must be consistent with the conditions employed. As an example, consider the following S N 1 reaction: The following proposed mechanism is unacceptable, because the reagent employed in the second step is not present: This is a common student error. To see what s wrong, let s look closely at the reagent. Methanol (C 3 ) is not a strong acid. Rather, it is a weak acid, because its conjugate base, methoxide (C 3 ), is a strong base. Therefore, methoxide is not present in substantial quantities, so the mechanism in this case should not employ methoxide. Below is the correct mechanism: Methanol (rather than methoxide) functions as the nucleophile in the second step, because methoxide was not indicated as a reagent, and it is not expected to be present. The result of the nucleophilic attack is an oxonium ion (an intermediate with a positive charge on an oxygen atom), which is then deprotonated by another molecule of methanol. nce again, in this final step of the mechanism, methanol functions as the base, rather than methoxide, because the latter is not present. This example is just one illustration of the importance of analyzing the reagent and considering what entities can be used in your mechanism. This will become increasingly important in upcoming chapters.

182 CAPTER 7 Solutions 7.1. (a) The parent is the longest chain, which is seven carbon atoms in this case (heptane). There are two substituents (ethyl and chloro), both of which are located at the C4 position. This will be the case whether we number the parent from left to right or from right to left. Substituents are alphabetized in the name (chloro precedes ethyl). 7.2. (a) The substrate is 2-bromopropane, and the nucleophile is S. Bromide functions as the leaving group. In a concerted process, nucleophilic attack and loss of the leaving group occur in a simultaneous fashion (in one step). Two curved arrows are required. ne curved arrow shows the nucleophilic attack, and the other curved arrow shows loss of the leaving group. (b) The parent is a six-membered ring (cyclohexane). There are two substituents (methyl and bromo), both of which are located at the C1 position. Substituents are alphabetized in the name (bromo precedes methyl). (b) The substrate is 1-iodopropane, and the nucleophile is methoxide (Me ). Iodide functions as the leaving group. In a concerted process, nucleophilic attack and loss of the leaving group occur in a simultaneous fashion (in one step). Two curved arrows are required. ne curved arrow shows the nucleophilic attack, and the other curved arrow shows loss of the leaving group. (c) The parent is the longest chain, which is five carbon atoms in this case (pentane). There are three substituents (bromo, bromo, and chloro), and their locants are assigned as 4, 4, and 1, respectively. In this case, the parent was numbered from right to left, so as to give the lowest number to the first substituent (1,4,4 rather than 2,2,5). Notice that two locants are necessary (rather than one) to indicate the locations of the two bromine atoms, even though they are connected to the same position (4,4-dibromo rather than 4-dibromo). Br 5 4 Br 3 4,4-dibromo-1-chloropentane (d) The parent is the longest chain, which is six carbon atoms in this case (hexane). There are three substituents (fluoro, methyl, and methyl), and their locants are assigned as 5, 2, and 2, respectively. In this case, the parent was numbered from right to left, so as to give the lowest number to the second substituent (2,2,5 rather than 2,5,5). The substituents are arranged alphabetically in the name, so fluoro precedes dimethyl (the former is f and the latter is m ). In this case, there is also a chirality center, so we must assign the configuration (S), which must be indicated at the beginning of the name. 2 1 Cl 7.3. (a) The substrate is 1-bromo-1-methylcyclohexane, and the nucleophile is an acetate ion (C 3 C 2 ). Bromide functions as the leaving group. In a stepwise process, there are two separate steps. The first step is loss of a leaving group to generate a carbocation intermediate. Then, the second step is a nucleophilic attack, in which an acetate ion attacks the carbocation intermediate. The first step requires one curved arrow (showing loss of the leaving group), and the second step requires one curved arrow (showing the nucleophilic attack). Br Br (b) The substrate is 2-iodo-2-methylbutane, and the nucleophile is a chloride ion (Cl ). Iodide functions as the leaving group. In a stepwise process, there are two separate steps. The first step is loss of the leaving group to generate a carbocation intermediate. Then, the second step is a nucleophilic attack, in which a chloride ion attacks the carbocation intermediate. The first step requires one curved arrow (showing loss of the leaving group), and the second step requires one curved arrow (showing the nucleophilic attack).

CAPTER 7 183 7.4. In a concerted process, nucleophilic attack and loss of the leaving group occur in a simultaneous fashion (in one step). Since the nucleophilic center and the electrophilic center are tethered to each other, the reaction occurs in an intramolecular fashion, as shown below. Two curved arrows are required. ne curved arrow shows the nucleophilic attack, and the other curved arrow shows loss of the leaving group (bromide). (b) The substrate is (R)-3-iodohexane, and the nucleophile is chloride (Cl ). Iodide is ejected as a leaving group, with inversion of configuration. (c) The substrate is (R)-2-bromohexane, and the nucleophile is hydroxide ( ). Bromide is ejected as a leaving group, with inversion of configuration. 7.5. In a stepwise process, the first step is loss of a leaving group to generate a carbocation intermediate. In this case, the carbocation intermediate is resonancestabilized, so we draw both resonance structures. Nucleophilic attack (at the electrophilic position indicated in the second resonance structure) affords the observed product. Br (R)-2-bromohexane + 7.8. The reaction does proceed with inversion of configuration. owever, the Cahn-Ingold-Prelog system for assigning a stereodescriptor (R or S) is based on a prioritization scheme. Specifically, the four groups connected to a chirality center are ranked (one through four). In the reactant, the highest priority group is the leaving group (bromide) which is then replaced by a group that does not receive the highest priority. In the product, the fluorine atom has been promoted to the highest priority as a result of the reaction, and as such, the prioritization scheme has changed. In this way, the stereodescriptor (S) remains unchanged, despite the fact that the chirality center undergoes inversion. Br 7.6. (a) The reaction has a second-order rate equation, which means that the rate should be linearly dependent on the concentrations of two compounds (the nucleophile AND the substrate). If the concentration of the substrate is tripled, the rate should also be tripled. (b) As described above, the rate is linearly dependent on the concentrations of both the nucleophile and the substrate. If the concentration of the nucleophile is doubled, the rate of the reaction is doubled. (c) As described above, the rate is linearly dependent on the concentrations of both the nucleophile and the substrate. If the concentration of the substrate is doubled and the concentration of the nucleophile is tripled, then the rate of the reaction will be six times faster ( 2 3). 7.9. (a) The leaving group is a bromide ion (Br ) and the nucleophile is a hydroxide ion ( ). In the transition state, each of these groups is drawn as being connected to the position with a dotted line (indicating these bonds are in the process of forming or breaking), and a is placed on each group to indicate that the charge is spread over both locations. Don t forget the brackets and the symbol that indicate the drawing is a transition state. 7.7. (a) The substrate is (S)-2-chloropentane, and the nucleophile is S. Chloride is ejected as a leaving group, with inversion of configuration. (b) The leaving group is an iodide ion (I ) and the nucleophile is an acetate ion (C 3 C 2 ). In the transition state, each of these groups is drawn as being connected to the position with a dotted line (indicating

184 CAPTER 7 these bonds are in the process of forming or breaking), and a is placed on each group to indicate that the charge is spread over both locations. Don t forget the brackets and the symbol that indicate the drawing is a transition state. 7.11. This step is favorable (downhill in energy) because ring strain is alleviated when the three-membered ring is opened. (c) The leaving group is a chloride ion (Cl ) and the nucleophile is a hydroxide ion ( ). In the transition state, each of these groups is drawn as being connected to the position with a dotted line (indicating these bonds are in the process of forming or breaking), and a is placed on each group to indicate that the charge is spread over both locations. Don t forget the brackets and the symbol that indicate the drawing is a transition state. 7.12. (a) The lone pair on the sp 3 hybridized nitrogen atom functions as a nucleophilic center and attacks the electrophilic methyl group in SAM, forming an ammonium ion which loses a proton to give the product. (d) The leaving group is a bromide ion (Br ) and the nucleophile is S. In the transition state, each of these groups is drawn as being connected to the position with a dotted line (indicating these bonds are in the process of forming or breaking), and a is placed on each group to indicate that the charge is spread over both locations. Don t forget the brackets and the symbol that indicate the drawing is a transition state. (b) The nitrogen atom functions as a nucleophilic center and attacks the electrophilic methyl group in SAM, forming an ammonium ion. 7.10. The leaving group is a bromide ion (Br ) and the nucleophilic center is the oxygen atom bearing the negative charge. In the transition state, each of these groups is drawn as being connected to the position with a dotted line (indicating these bonds are in the process of forming or breaking), and a is placed on each group to indicate that the charge is spread over both locations. Don t forget the brackets and the symbol that indicate the drawing is a transition state. 7.13. (a) The reaction occurs via an S N 1 pathway, which means that the rate is not dependent on the concentration of the nucleophile. The rate is only dependent on the concentration of the substrate. If the concentration of substrate is doubled, the rate will be doubled (the change in concentration of the nucleophile will not affect the rate). (b) The reaction occurs via an S N 1 pathway, which means that the rate is not dependent on the concentration of the nucleophile. If that is the only factor that is changed, then the rate will remain the same.

CAPTER 7 185 7.14. In each case, the bond between the position and the leaving group is broken, and the carbon atom obtains a positive charge. chirality center remains unchanged. This gives the following two stereoisomers. (a) (c) (b) (d) Notice that in each stereoisomer, the chirality center on the right side has the S configuration, as it did in the starting material. The difference between these two stereoisomers is the configuration of the other chirality center (where the reaction took place). These compounds are stereoisomers that are not mirror images of each other, so they are diastereomers. 7.15. The first compound will generate a tertiary carbocation, while the second compound will generate a tertiary benzylic carbocation that is resonance stabilized. The second compound leads to a more stable carbocation, so the second compound will lose its leaving group more rapidly than the first compound. 7.16. (a) The leaving group is iodide, and the nucleophile is chloride. The former is replaced by the latter, giving the following products. (b) The leaving group is bromide, and the nucleophile is S. The former is replaced by the latter. In this case, the reaction is taking place at a chirality center, so we expect that both enantiomers will be produced (as expected for an S N 1 process). (c) The leaving group is chloride, and the nucleophile is an acetate ion (C 3 C 2 ). The former is replaced by the latter. In this case, the reaction is taking place at a chirality center, so we expect that both enantiomers will be produced (as expected for an S N 1 process). 7.17. The leaving group is bromide, and the nucleophile is S. The former is replaced by the latter. In this case, the reaction is taking place at one of the two chirality centers that are present in the compound. The other 7.18. (a) The substrate is an alkyl iodide, which has an excellent leaving group (iodide), so an S N 1 process will not require a proton transfer at the beginning of the mechanism. (b) The substrate is an alcohol (R), which does not have a good leaving group. The group must be protonated to function as a leaving group. So, an S N 1 process will require a proton transfer at the beginning of the mechanism. (c) The substrate is an alkyl bromide, which has an excellent leaving group (bromide), so an S N 1 process will not require a proton transfer at the beginning of the mechanism. (d) The substrate is an alcohol (R), which does not have a good leaving group. The group must be protonated to function as a leaving group. So, an S N 1 process will require a proton transfer at the beginning of the mechanism. (e) The substrate is an alcohol (R), which does not have a good leaving group. The group must be protonated to function as a leaving group. So, an S N 1 process will require a proton transfer at the beginning of the mechanism. (f) The substrate is an alkyl chloride, which has an excellent leaving group (chloride), so an S N 1 process will not require a proton transfer at the beginning of the mechanism. 7.19. (a) The nucleophile (S ) is negatively charged, so a proton transfer will not be necessary at the end of the mechanism. (b) The nucleophile ( 2 S) is neutral, so nucleophilic attack will produce a positively charged species. Removal of the positive charge will require a proton transfer at the end of the mechanism. (c) The nucleophile ( 2 ) is neutral, so nucleophilic attack will produce a positively charged species. Removal of the positive charge will require a proton transfer at the end of the mechanism. (d) The nucleophile (Et) is neutral, so nucleophilic attack will produce a positively charged species. Removal of the positive charge will require a proton transfer at the end of the mechanism.

186 CAPTER 7 (e) The nucleophile (N C ) is negatively charged, so a proton transfer will not be necessary at the end of the mechanism. (f) The nucleophile (Cl ) is negatively charged, so a proton transfer will not be necessary at the end of the mechanism. (g) The nucleophile ( 2 N ) is negatively charged, so a proton transfer will not be necessary at the end of the mechanism. (h) The nucleophile (N 3 ) is neutral, so nucleophilic attack will produce a positively charged species. Removal of the positive charge will require a proton transfer at the end of the mechanism. (i) The nucleophile (Me ) is negatively charged, so a proton transfer will not be necessary at the end of the mechanism. (j) The nucleophile (Et ) is negatively charged, so a proton transfer will not be necessary at the end of the mechanism. (k) The nucleophile (Me) is neutral, so nucleophilic attack will produce a positively charged species. Removal of the positive charge will require a proton transfer at the end of the mechanism. (l) The nucleophile (Br ) is negatively charged, so a proton transfer will not be necessary at the end of the mechanism. 7.20. (a) Loss of a leaving group results in a tertiary carbocation, which cannot rearrange to become more stable. So a carbocation rearrangement will not occur in this case. (b) In order for this compound to function as a substrate in an S N 1 reaction, acidic conditions will be necessary (to protonate the group, rendering it a better leaving group). Then, loss of the leaving group will result in a secondary carbocation. This carbocation can rearrange via a methyl shift to produce a more stable, tertiary carbocation. So we do expect that a rearrangement can and will occur. (c) In order for this compound to function as a substrate in an S N 1 reaction, acidic conditions will be necessary (to protonate the group, rendering it a better leaving group). Then, loss of the leaving group will result in a secondary carbocation. This carbocation can rearrange via a hydride shift to produce a more stable, tertiary carbocation. So we do expect that a rearrangement can and will occur. (d) In order for this compound to function as a substrate in an S N 1 reaction, acidic conditions will be necessary (to protonate the group, rendering it a better leaving group). Then, loss of the leaving group will result in a secondary carbocation. In this case, there is no possible rearrangement that will lead to a more stable, tertiary carbocation. As such, we expect no rearrangement will occur. (e) Loss of a leaving group results in a tertiary carbocation, which cannot rearrange to become more stable. So a carbocation rearrangement will not occur in this case. (f) Loss of a leaving group results in a secondary carbocation. In this case, there is no possible rearrangement that will lead to a more stable, tertiary carbocation. As such, we expect no rearrangement will occur. 7.21. (a) The substrate is an alcohol, so acidic conditions are employed so that the group can be protonated, rendering it a better leaving group. Then, loss of a leaving group generates a carbocation, which is then captured by a bromide ion to give the product. Notice that the mechanism is comprised of a proton transfer, followed by the two core steps of an S N 1 process (loss of a leaving group and nucleophilic attack). (b) The substrate is an alcohol, so acidic conditions are employed so that the group can be protonated, rendering it a better leaving group. Then, loss of a leaving group generates a secondary carbocation, which undergoes a hydride shift to give a more stable, tertiary carbocation. This carbocation is then captured by a bromide ion to give the product.

CAPTER 7 187 (c) The leaving group is bromide. Loss of the leaving group generates a tertiary carbocation, which is captured by a water molecule to generate an oxonium ion. Deprotonation of the oxonium ion gives the product. Notice that the mechanism is comprised of the two core steps of an S N 1 process (loss of a leaving group and nucleophilic attack), followed by a proton transfer step. (d) The leaving group is iodide. Loss of the leaving group generates a secondary carbocation, which then undergoes a methyl shift to give a more stable, tertiary carbocation. This carbocation is then captured by a molecule of ethanol to generate an oxonium ion. Deprotonation of the oxonium ion gives the product. (e) The substrate is an alcohol, so acidic conditions are employed so that the group can be protonated, rendering it a better leaving group. The likely source of the proton is Me 2 +, which received its proton from sulfuric acid (a comparison of the pk a values for Me 2 + and 2 S 4 indicates that there is likely very little 2 S 4 present at equilibrium). Loss of a leaving group generates a tertiary carbocation, which is captured by a molecule of methanol to give an oxonium ion. Deprotonation of the oxonium ion gives the product. (f) The substrate is an alcohol, so acidic conditions are employed so that the group can be protonated, rendering it a better leaving group (as in the previous problem, the likely source of the proton is Me 2 +, which received its proton from sulfuric acid). Then, loss of a leaving group generates a secondary carbocation, which undergoes a methyl shift to give a more stable, tertiary carbocation. This carbocation is then captured by a molecule of methanol to give an oxonium ion. Deprotonation of the oxonium ion gives the product.

188 CAPTER 7 (g) The leaving group is bromide. Loss of the leaving group generates a tertiary carbocation, which is then captured by the nucleophile (S ) to give the product. (h) The leaving group is iodide. Loss of the leaving group generates a tertiary carbocation, which is then captured by a molecule of ethanol to generate an oxonium ion. Deprotonation of the oxonium ion gives the product. 7.22. (c) In problem 7.21c, we saw that the two core steps (loss of leaving group and nucleophilic attack) are accompanied by a proton transfer at the end of the mechanism. Therefore, the mechanism has a total of three steps: 1) loss of a leaving group, 2) nucleophilic attack, and 3) proton transfer. (f) In problem 7.21f, we saw that the two core steps (loss of leaving group and nucleophilic attack) are accompanied by three additional steps (a proton transfer at the beginning of the mechanism, a carbocation rearrangement between the two core steps, and finally, a proton transfer at the end of the mechanism). Therefore, the mechanism has a total of five steps: 1) proton transfer, 2) loss of a leaving group, 3) rearrangement, 4) nucleophilic attack, and 5) proton transfer. (d) In problem 7.21d, we saw that the two core steps (loss of leaving group and nucleophilic attack) are accompanied by a carbocation rearrangement (in between the two core steps) and a proton transfer at the end of the mechanism. Therefore, the mechanism has a total of four steps: 1) loss of a leaving group, 2) carbocation rearrangement, 3) nucleophilic attack, and 4) proton transfer. (g) In problem 7.21g, we saw that the two core steps (loss of leaving group and nucleophilic attack) are not accompanied by any additional steps. Therefore, the mechanism has only two steps: 1) loss of a leaving group and 2) nucleophilic attack. (e) In problem 7.21e, we saw that the two core steps (loss of leaving group and nucleophilic attack) are accompanied by a proton transfer at the beginning of the mechanism and another proton transfer at the end of the mechanism. Therefore, the mechanism has a total of four steps: 1) proton transfer, 2) loss of a leaving group, 3) nucleophilic attack, and 4) proton transfer. (h) In problem 7.21h, we saw that the two core steps (loss of leaving group and nucleophilic attack) are accompanied by a proton transfer at the end of the mechanism. Therefore, the mechanism has a total of three steps: 1) loss of a leaving group, 2) nucleophilic attack, and 3) proton transfer. Problem 7.21c and 7.21h exhibit the same pattern. Both problems are characterized by three mechanistic steps: 1) loss of a leaving group, 2) nucleophilic attack, and 3) proton transfer.

CAPTER 7 189 7.23. The mechanism has a total of five steps (two core steps, and three additional steps). In the first step, the group is protonated to give a better leaving group. The chirality center at C2 is lost when the leaving group leaves to form a carbocation with trigonal planar geometry. This carbocation undergoes rearrangement via a hydride shift, during which the chirality center at C3 is also lost (once again via the trigonal planar sp 2 hybridized center). The tertiary carbocation is then captured by water to give an oxonium ion, which undergoes deprotonation to generate the product. 7.24. (a) An S N 2 process is a concerted process in which the nucleophile attacks and the leaving group leaves simultaneously. The nucleophile (methanol) is uncharged, so a proton transfer is required at the end of the mechanism in order to remove the positive charge. A molecule of methanol serves as the base for this deprotonation step. (b) An S N 2 process is a concerted process in which the nucleophile attacks and the leaving group leaves simultaneously. The nucleophile (ethanol) is uncharged, so a proton transfer is required at the end of the mechanism in order to remove the positive charge. A molecule of ethanol serves as the base for this deprotonation step. (c) An S N 2 process is a concerted process in which the nucleophile attacks and the leaving group leaves simultaneously. The nucleophile (water) is uncharged, so a proton transfer is required at the end of the mechanism in order to remove the positive charge. A molecule of water serves as the base for this deprotonation step.

190 CAPTER 7 (d) An S N 2 process is a concerted process in which the nucleophile attacks and the leaving group leaves simultaneously. The nucleophile (1-propanol) is uncharged, so a proton transfer is required at the end of the mechanism in order to remove the positive charge. A molecule of 1-propanol serves as the base for this deprotonation step. Cl Cl 7.25. Ammonia (N 3 ) functions as a nucleophile and attacks methyl iodide in an S N 2 reaction, generating an ammonium ion. The positive charge is removed upon deprotonation (a molecule of N 3 serves as the base for this deprotonation step). This sequence of steps (S N 2 followed by deprotonation) is repeated two more times, followed by one final S N 2 reaction to give the quaternary ammonium ion, as shown here. 7.26. (a) The substrate is tertiary, which favors S N 1. (b) The substrate is primary, which favors S N 2. (c) The substrate is an aryl halide, which is expected to be unreactive in substitution reactions. As such, neither S N 1 nor S N 2 is favored. (d) The substrate is tertiary, which favors S N 1. (e) The substrate is benzylic, which favors both S N 1 and S N 2. (f) The substrate is a vinyl halide, which is expected to be unreactive in substitution reactions. As such, neither S N 1 nor S N 2 is favored. (g) The substrate is allylic, which favors both S N 1 and S N 2. 7.27. (a) Ethanol is a weak nucleophile, which disfavors S N 2 (and thereby allows S N 1 to compete successfully). (b) Ethanethiol is a strong nucleophile, which favors S N 2. (c) Ethoxide is a strong nucleophile, which favors S N 2. (d) ydroxide is a strong nucleophile, which favors S N 2. (e) Cyanide is a strong nucleophile, which favors S N 2. not tertiary), AND there must be a suitable leaving group. These criteria are satisfied in the following, highlighted positions. While iodide and bromide are both excellent leaving groups, they occupy tertiary positions, which will not undergo S N 2 reactions. The other two groups are not good leaving groups (Me and N 2 ). (b) In order for an S N 1 reaction to occur, the position must be capable of stabilizing a carbocation (i.e. tertiary rather than primary), AND there must be a suitable leaving group. These criteria are satisfied in the following, highlighted positions. 7.28. (a) In order for an S N 2 reaction to occur, the position must be unhindered (primary, or perhaps secondary, but

CAPTER 7 191 7.29. (a) Ethanol is a protic solvent, which favors S N 1 (see Table 7.2). (b) DMS is a polar aprotic solvent, which favors S N 2 (see Table 7.2). (c) Acetic acid is a protic solvent, which favors S N 1 (see Table 7.2). (d) DMF is a polar aprotic solvent, which favors S N 2 (see Table 7.2). (e) Methanol is a protic solvent, which favors S N 1 (see Table 7.2). (f) Acetonitrile is a polar aprotic solvent, which favors S N 2 (see Table 7.2). (g) MPA is a polar aprotic solvent, which favors S N 2 (see Table 7.2). (h) Ammonia is a protic solvent, which favors S N 1 (see Table 7.2). 7.30. Acetone is a polar aprotic solvent and will favor S N 2 by raising the energy of the nucleophile, giving a smaller E a. 7.31. (a) The substrate is tertiary, and the nucleophile (Me) is weak, both of which favor an S N 1 process, giving the following product. (b) The substrate is secondary, which is not very helpful in the determination between S N 1 and S N 2. But the nucleophile is strong, and the solvent is polar aprotic, and these two factors favor an S N 2 process. The product is shown below. (e) The substrate is tertiary, and the nucleophile ( 2 ) is weak, both of which favor an S N 1 process, giving a pair of enantiomers in this case. (f) The substrate is secondary, which is not very helpful in the determination between S N 1 and S N 2. But the nucleophile is strong, and the solvent is polar aprotic, and these two factors favor an S N 2 process. We therefore expect inversion of configuration, as shown below. 7.32. No. Preparation of this amine via the Gabriel synthesis would require the use of a tertiary alkyl halide, which will not undergo an S N 2 process. 7.33. (a) The substrate is primary, so we will need to use an S N 2 process. We must therefore use a strong nucleophile (hydroxide), rather than a weak nucleophile (water). I Na (b) The substrate is tertiary, so we will need to use an S N 1 process. The group cannot function as a leaving group without protonation, so we must use strongly acidic conditions. Br supplies both the proton and the nucleophile: (c) The substrate is tertiary, which favors S N 1. The group is protonated under these conditions, giving an excellent leaving group (water). The nucleophile (bromide) is strong, but an S N 2 reaction will not occur at a tertiary substrate, so we expect an S N 1 process will occur, giving a pair of enantiomers in this case. (c) The substrate is primary, so we will need to use an S N 2 process. The group cannot function as a leaving group, so we must convert it to a better leaving group, which can be accomplished by first converting the alcohol to a tosylate, followed by an S N 2 reaction with iodide. (d) The substrate is primary, which favors S N 2. The nucleophile is strong, and the solvent is polar aprotic, and these two factors also favor an S N 2 process. The product is shown below. (d) The desired transformation involves inversion of configuration, so we will need to use an S N 2 process. We use a strong nucleophile (S ) and a polar aprotic solvent.

192 CAPTER 7 (e) The desired transformation involves inversion of configuration, so we will need to use an S N 2 process. The use of a polar aprotic solvent will help overcome the fact that the substrate is secondary. (h) The substrate is tertiary, so we will need to use an S N 1 process. We therefore use a weak nucleophile ( 2 ) rather than a strong nucleophile ( ). (f) The desired transformation involves inversion of configuration, so we will need to use an S N 2 process. The group cannot function as a leaving group, so we must convert it to a better leaving group, which can be accomplished by converting the alcohol to a tosylate. Then, we perform an S N 2 reaction with bromide as the nucleophile. The use of a polar aprotic solvent will help overcome the fact that the substrate is secondary. (g) The substrate is primary, so we will need to use an S N 2 process. We must therefore use a strong nucleophile (ethoxide), rather than a weak nucleophile (ethanol). 7.34. We have not learned a direct way to perform a substitution reaction with retention of configuration. owever, if we perform two successive S N 2 reactions (each of which proceeds via inversion of configuration), the net result of the overall process will be retention of configuration. The first S N 2 reaction cannot be performed with hydroxide as a leaving group, so the alcohol must first be converted into a tosylate (by treatment with TsCl and pyridine). After converting the group into a better leaving group, we are now ready for the first S N 2 reaction. Iodide makes an excellent choice for the nucleophile, because it is both a strong nucleophile AND an excellent leaving group (the former is important for the first S N 2 reaction, while the latter will be important for the second S N 2 reaction). Since the substrate is secondary, polar aprotic solvents are used to enhance the rate of each S N 2 process. 7.35. The lone pair of the nitrogen atom (connected to the aromatic ring) provides anchimeric assistance, ejecting the chloride ion in an intramolecular S N 2-type reaction, generating a high-energy intermediate that exhibits a threemembered ring. The ring is opened upon attack of a nucleophile in an S N 2 process. These two steps are then repeated, as shown here.

CAPTER 7 193 7.36. (a) The parent is the longest chain, which is three carbon atoms in this case (propane). There is only one substituent (chloro), and its locant is assigned as 2 (as shown below), so the systematic name for this compound is 2-chloropropane. The common name is isopropyl chloride. 7.37. The constitutional isomers of C 4 9 I are shown below, arranged in order of increasing reactivity toward S N 2. Notice that the tertiary substrate is the least reactive because it is the most hindered. Among the two primary substrates, butyl iodide is the least sterically hindered (and therefore the most reactive toward S N 2), because it does not contain a substituent at the beta position. Increasing reactivity (S N 2) (b) The parent is the longest chain, which is three carbon atoms in this case (propane). There are two substituents (bromo and methyl), and their locants are assigned as 2 and 2, as shown below. Substituents are alphabetized in the name (bromo precedes methyl), so the systematic name is 2-bromo-2-methylpropane. The common name is tert-butyl bromide. I I 7.38. (a) The secondary substrate is more hindered than the primary substrate, and therefore, the latter reacts more rapidly in an S N 2 reaction. I I (c) The parent is the longest chain, which is three carbon atoms in this case (propane). There is only one substituent (iodo), and its locant is assigned as 1 (as shown below), so the systematic name for this compound is 1-iodopropane. The common name is propyl iodide. (b) Both substrates are primary, but one is more hindered than the other. The less hindered substrate reacts more rapidly in an S N 2 reaction. (d) The parent is the longest chain, which is four carbon atoms in this case (butane). There is only one substituent (bromo), and its locant is assigned as 2 (as shown below). The compound has a chirality center, so the configuration must be indicated at the beginning of the name: (R)-2-bromobutane. The common name is (R)- sec-butyl bromide. (c) The secondary substrate is less hindered than the tertiary substrate, and therefore, the former reacts more rapidly in an S N 2 reaction. (e) The parent is the longest chain, which is three carbon atoms in this case (propane). There are three substituents (chloro, methyl, and methyl), and their locants are assigned as 1, 2, and 2, respectively, as shown below. Substituents are alphabetized in the name (chloro precedes methyl). Make sure that each methyl group receives a locant (2,2-dimethyl rather than 2-dimethyl). The systematic name is therefore 1-chloro-2,2- dimethylpropane. The common name is neopentyl chloride. (d) Both substrates are primary, but they have different leaving groups. The compound with the better leaving group (iodide) will react more rapidly in an S N 2 reaction. 7.39. No. Preparation of this compound via an acetylide ion would require the use of the following tertiary alkyl halide, which will not participate in an S N 2 reaction because of steric crowding.

194 CAPTER 7 7.40. (a) S bears a negative charge, so it is a stronger nucleophile than 2 S. (b) bears a negative charge, so it is a stronger nucleophile than 2. (c) In a polar aprotic solvent (such as DMS), cations are solvated and surrounded by a solvent shell, but anions are not (see Table 7.2). This greatly enhances the nucleophilicity of the methoxide ions because they are not strongly interacting with the solvent molecules, and are instead more available to function as nucleophiles. In contrast, when dissolved in a protic solvent (such as methanol), methoxide ions interact strongly with the solvent molecules and are therefore less nucleophilic. 7.41. (a) The tertiary substrate will react more rapidly in an S N 1 reaction, because upon loss of a leaving group, the tertiary substrate gives a tertiary carbocation, while the primary substrate gives a primary carbocation. A tertiary carbocation is more stable than a primary carbocation because of hyperconjugation, so the energy of activation (E a ) for formation of the tertiary carbocation is expected to be lower than the E a for formation of the primary carbocation. Therefore, the tertiary carbocation is expected to form more rapidly, which explains why the tertiary substrate is more reactive towards S N 1 than the primary substrate. (b) The tertiary substrate will react more rapidly in an S N 1 reaction (see explanation in Problem 7.41a). (c) The allylic substrate will react more rapidly in an S N 1 reaction, because upon loss of a leaving group, the allylic substrate gives a resonance-stabilized allylic carbocation. This carbocation is more stable than the carbocation that would form from chlorocyclohexane (which would be secondary and not resonancestabilized). The energy of activation (E a ) for formation of the allylic carbocation is expected to be lower than the E a for formation of the secondary carbocation. Therefore, the allylic carbocation is expected to form more rapidly. of the leaving group. Ts is a better leaving group than chloride (Cl ), so we expect the tosylate to undergo S N 1 at a faster rate. Cl Ts better leaving group 7.42. (a) There are many indications that this reaction proceeds via an S N 2 pathway (a strong nucleophile in a polar aprotic solvent, and inversion of configuration). As such, the reaction has a second-order rate equation, which means that the rate should be linearly dependent on the concentrations of two compounds (the nucleophile AND the substrate). If the concentration of the substrate is doubled, the rate should also be doubled. (b) The reaction proceeds via an S N 2 pathway (see solution to Problem 7.42a). As such, the reaction has a second-order rate equation, which means that the rate should be linearly dependent on the concentrations of two compounds (the nucleophile AND the substrate). If the concentration of the nucleophile is doubled, the rate should also be doubled. 7.43. (a) The substrate is a tertiary alcohol, and the reaction proceeds via an S N 1 process (under acidic conditions, the group is protonated to give an excellent leaving group). As such, the reaction has a first-order rate equation, which means that the rate should be linearly dependent only on the concentration of the substrate (not the nucleophile). If the concentration of the substrate is doubled, the rate should also be doubled. (b) The reaction proceeds via an S N 1 pathway (see solution to Problem 7.43a). As such, the reaction has a first-order rate equation, which means that the rate should be linearly dependent only on the concentration of the substrate (not the nucleophile). If the concentration of the nucleophile is doubled, the rate should not be affected. 7.44. (a) DMF is a polar aprotic solvent, because it does not contain any hydrogen atoms that are connected directly to an electronegative atom (see Table 7.2). (d) In this case, both substrates are tertiary, and both substrates lead to the same carbocation intermediate. The difference between these compounds is the identity (b) Ethanol (C 3 C 2 ) is a protic solvent because it contains a hydrogen atom connected to an electronegative atom (oxygen).

CAPTER 7 195 (c) DMS is a polar aprotic solvent, because it does not contain any hydrogen atoms that are connected directly to an electronegative atom (see Table 7.2). (d) Water ( 2 ) is a protic solvent because it contains hydrogen atoms connected to an electronegative atom (oxygen). (e) Ammonia (N 3 ) is a protic solvent because it has hydrogen atoms that are connected to an electronegative atom (nitrogen). 7.45. (a) The chirality center in the substrate has the R configuration, as shown below. (c) The reaction is an S N 2 process, and it does proceed with inversion of configuration. owever, the prioritization scheme changes when the bromo group (#1) is replaced with a cyano group (#2). As a result, the Cahn-Ingold-Prelog system assigns the same configuration to the reactant and the product. 7.46. The leaving group is an iodide ion (I ) and the nucleophile is an acetate ion (C 3 C 2 ). In the transition state, each of these groups is drawn as being connected to the position with a dotted line (indicating these bonds are in the process of forming or breaking), and a is placed on each group to indicate that the charge is spread over both locations. Don t forget the brackets and the symbol that indicate the drawing is a transition state. Me (b) The chirality center in the product has the R configuration, as shown below. 7.47. Iodide functions as a nucleophile and attacks (S)- 2-iodopentane, displacing iodide as a leaving group. The reaction is an S N 2 process, and therefore proceeds via inversion of configuration. The product is (R)-2- iodopentane. The reaction continues repeatedly until a racemic mixture is eventually obtained. 7.48. The substrate is an alcohol, so acidic conditions are employed so that the group can be protonated, rendering it a better leaving group. Then, loss of a leaving group generates a carbocation, which is then captured by a bromide ion to give the product. Notice that the mechanism is comprised of a proton transfer, followed by the two core steps of an S N 1 process (loss of a leaving group and nucleophilic attack). The chirality center is lost when the leaving group leaves to form a carbocation with trigonal planar geometry. The nucleophile can then attack either face of the planar carbocation, leading to a racemic mixture.

196 CAPTER 7 7.49. The substrate is an alcohol, so acidic conditions are employed so that the group can be protonated, rendering it a better leaving group. In dilute sulfuric acid, the source of the proton is 3 +, which received its proton from sulfuric acid (a comparison of the pk a values for 3 + and 2 S 4 indicates that there is likely very little 2 S 4 present at equilibrium). Loss of a leaving group generates a secondary carbocation, which is captured by a molecule of water to give an oxonium ion. Deprotonation of the oxonium ion gives the product. The reaction has four steps, so the energy diagram should have four steps, as shown below. Notice that there are three intermediates (valleys on the diagram). The first and last represent oxonium ions, which are lower in energy than the carbocation intermediate. f the three intermediates, the carbocation intermediate is the highest in energy, as can be seen in the middle of the energy diagram. Also note that the starting material and product are represented at the same energy level, because in this case, they have the same identity (2-pentanol). 7.50. The tertiary allylic carbocation is the most stable, because it is stabilized by hyperconjugation as well as resonance. The next most stable is the tertiary carbocation, followed by the secondary carbocation. The primary carbocation is the least stable. (b) The following tertiary carbocation will be formed upon loss of the leaving group. (c) The following primary carbocation would be formed upon loss of the leaving group. 7.51. (a) The following secondary carbocation will be formed upon loss of the leaving group. (d) The following secondary carbocation will be formed upon loss of the leaving group.

CAPTER 7 197 7.52. The substrate is an alcohol, which does not possess a good leaving group. owever, acidic conditions are employed, and under these conditions, the group is protonated, thereby forming an excellent leaving group. Loss of the leaving group ( 2 ) gives a secondary carbocation, which can undergo a hydride shift to form a tertiary carbocation. This carbocation is then captured by chloride to give the product. 7.53. This is a substitution reaction in which an acetate ion (C 3 C 2 ) functions as a nucleophile. The reaction occurs with inversion of configuration, which indicates that an S N 2 process must be operating. As such, we draw a concerted process in which nucleophilic attack and loss of the leaving group occur simultaneously, as shown below. 7.54. (a) In an S N 1 process, there must be at least two core steps: loss of the leaving group, and nucleophilic attack. In addition, there can be other steps that accompany the S N 1 process, including a proton transfer step at the beginning of the mechanism, a carbocation rearrangement in between the two core steps, and a proton transfer step at the end of the mechanism. In this case, the first proton transfer step does not occur because chloride is an excellent leaving group (and acidic conditions are not employed). The intermediate carbocation is tertiary and cannot rearrange to become more stable, so the mechanism will not involve a carbocation rearrangement. A proton transfer step at the end of the mechanism will be required because the attacking nucleophile is uncharged. As such, our mechanism will have three steps: 1) loss of a leaving group, 2) nucleophilic attack, and 3) proton transfer. (b) As in the solution to 7.54a, we consider the three possible steps that can accompany the two core steps of an S N 1 process, shown in circles below. In this case, the first proton transfer step does not occur because chloride is an excellent leaving group (and acidic conditions are not employed). The intermediate carbocation is tertiary and cannot rearrange to become more stable, so the mechanism will not involve a carbocation rearrangement. A proton transfer step at the end of the mechanism will not be required because the attacking nucleophile is an anion. As such, our mechanism will have only the two core steps: 1) loss of a leaving group, and 2) nucleophilic attack. (c) As in the solution to 7.54a, we consider the three possible steps that can accompany the two core steps of an S N 1 process. In this case, a proton transfer step must occur at the beginning of the mechanism, because the group is not

198 CAPTER 7 a good leaving group, and acidic conditions are employed (which cause protonation of the group, thereby converting it into a good leaving group). The intermediate carbocation is tertiary and cannot rearrange to become more stable, so the mechanism will not involve a carbocation rearrangement. A proton transfer step at the end of the mechanism will not be required because the attacking nucleophile is an anion. As such, our mechanism will have three steps: 1) proton transfer to convert the group into a better leaving group, 2) loss of the leaving group, and 3) nucleophilic attack. (d) As in the solution to 7.54a, we consider the three possible steps that can accompany the two core steps of an S N 1 process. In this case, a proton transfer step does not occur at the beginning of the mechanism, because the Ts group is an excellent leaving group (and acidic conditions are not employed). The intermediate carbocation is secondary and can rearrange (via a hydride shift) to give a more stable, tertiary carbocation. A proton transfer step at the end of the mechanism will be required because the attacking nucleophile is uncharged. As such, our mechanism will have four steps: 1) loss of the leaving group, 2) carbocation rearrangement, 3) nucleophilic attack, and 4) proton transfer. 7.55. (a) The substrate is tertiary, and a weak nucleophile is employed, both of which indicate an S N 1 process. As such, we must consider the possibility of a carbocation rearrangement. In this case, a tertiary carbocation will be formed, which cannot rearrange to form a more stable carbocation. Also, we must consider the stereochemical outcome whenever the reaction occurs at a chirality center, but in this case, there is no chirality center, so we don t need to consider the stereochemical outcome. The following substitution product is expected. chloride ion to give the product. Notice the starting material exhibits two chirality centers, while the product has none. (d) Iodide is the leaving group, and the substrate is secondary. The reaction involves a strong nucleophile (N C ) in a polar aprotic solvent, indicating an S N 2 reaction. As such, we expect inversion of configuration, as shown below. (b) The substrate is primary, indicating an S N 2 process. In this case, there is no chirality center, so we don t need to consider the stereochemical outcome. The following substitution product is expected. (c) The substrate is a secondary alcohol, and the group cannot function as a leaving group without being converted into a good leaving group. 7.56. The dianion has two nucleophilic centers, and the electrophile has two electrophilic centers. As such, these compounds can react with each other via two successive S N 2 reactions, as shown below, giving a six-membered ring with molecular formula C 4 8 2. That is, in fact, the purpose of the acidic conditions. Protonation of the group gives an excellent leaving group, which leaves to give a secondary carbocation. This carbocation can rearrange via a hydride shift to give a tertiary carbocation, which is then captured by the