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Reatividade de Compostos Orgânicos – Aula 2 1 UNIVERSIDADE FEDERAL DA GRANDE DOURADOS FACULDADE DE CIÊNCIAS EXATAS E TECNOLOGIA Curso de Licenciatura e Bacharelado em QUÍMICA Aula 2 – Substituição Nucleofílica a C saturado Profa. ROZANNA MARQUES MUZZI Reatividade de Compostos Orgânicos – Aula 2 Ex.: (CH3)3CBr +CH3OH à (CH3)3COCH3 +HBr EQUAÇÃO GERAL: R-GS + Nu R-Nu + GS t-butyl bromide Nu methyl t-butyl ether 2 Reatividade de Compostos Orgânicos – Aula 2 1. MECANISMO SN1: SN1: substitution, nucleophilic, unimolecular velocidade = k[substrato] - REAÇÃO DE 1ª. ORDEM The substrate forms a carbocation intermediate. The carbocation is planar with an empty 2p orbital available for bonding. 3 Reatividade de Compostos Orgânicos – Aula 2 POSSÍVEIS MECANISMOS: A) EM DUAS ETAPAS - SN1 B) EM UMA ETAPA - SN2 4 MECANISMO SN1 Reatividade de Compostos Orgânicos – Aula 2 rate = k[(CH3)3CBr] 5 Reatividade de Compostos Orgânicos – Aula 2 rate = k [alkyl halide] First-order kinetics implies a unimolecular rate-determining step. 6 Reatividade de Compostos Orgânicos – Aula 2 The carbocation intermediate quickly reacts with the nucleophile. Attack can be from either side. + 7 Reatividade de Compostos Orgânicos – Aula 2 8 Reatividade de Compostos Orgânicos – Aula 2 CH3Br CH3CH2Br (CH3)2CHBr (CH3)3CBr Reatividade decrescente para SN1 Reatividade de Compostos Orgânicos – Aula 2 10 Table 10.1 Relative Rates of SN2 Reactions for Several Alkyl Halides SN2R Br Cl−+ R Cl Br−+ CCH3 CH3 CH3 CH3 Br CH3 Br CH3CH2 Br CH3CH2CH2 Br CH3CH Br Alkyl halide Class of alkyl halide Relative rate methyl 1200 primary 40 primary 16 secondary tertiary too slow to measure 1 Section 10.2 The Mechanism of an Reaction 363SN2 Movie: Bimolecular reaction Edward Davies Hughes (1906–1963) was born in North Wales. He earned two doctoral degrees: a Ph.D. from the University of Wales and a D.Sc. from the University of London, working with Sir Christopher Ingold. He was a professor of chemistry at University College, London. The rate law tells us which molecules are involved in the transition state of the rate- determining step of the reaction. From the rate law for the reaction of methyl bromide with hydroxide ion, we know that both methyl bromide and hydroxide ion are involved in the rate-determining transition state. The reaction of methyl bromide with hydroxide ion is an example of an reac- tion, where “S” stands for substitution, “N” for nucleophilic, and “2” for bimolecu- lar. Bimolecular means that two molecules are involved in the rate-determining step. In 1937, Edward Hughes and Christopher Ingold proposed a mechanism for an reaction. Remember that a mechanism describes the step-by-step process by which reactants are converted into products. It is a theory that fits the experimental evi- dence that has been accumulated concerning the reaction. Hughes and Ingold based their mechanism for an reaction on the following three pieces of experimental evidence: 1. The rate of the reaction depends on the concentration of the alkyl halide and on the concentration of the nucleophile. This means that both reactants are involved in the rate-determining step. 2. When the hydrogens of methyl bromide are successively replaced with methyl groups, the rate of the reaction with a given nucleophile becomes progressively slower (Table 10.1). 3. The reaction of an alkyl halide in which the halogen is bonded to an asymmetric carbon leads to the formation of only one stereoisomer, and the configuration of the asymmetric carbon is inverted relative to its configuration in the reacting alkyl halide. Hughes and Ingold proposed that an reaction is a concerted reaction—it takes place in a single step, so no intermediates are formed. The nucleophile attacks the car- bon bearing the leaving group and displaces the leaving group. Because a productive collision is a collision in which the nucleophile hits the car- bon on the side opposite the side bonded to the leaving group, the carbon is said to un- dergo back-side attack. Why does the nucleophile attack from the back side? The simplest explanation is that the leaving group blocks the approach of the nucleophile to the front side of the molecule. +HO CH3 OHCH3Br + Br − − mechanism of the SN2 reaction SN2 SN2 SN2 SN2 Sir Christopher Ingold (1893–1970) was born in Ilford, England. In addition to determining the mechanism of the reaction, he was a member of the group that developed nomenclature for enantiomers. (See p. 188.) He also participated in developing the theory of resonance. SN2 BRUI10-359_399r2 18-02-2003 2:50 PM Page 363 Reatividade de Compostos Orgânicos – Aula 2 2. ESTEREOQUÍMICA NO MECANISMO SN1 11 RACEMIZACÃO Reatividade de Compostos Orgânicos – Aula 2 12 HEt Me Lv Lv - Me Et H + sp2 hybridized Sn1 reactive chemical intermediate ESTEREOQUÍMICA NO MECANISMO SN1 Reatividade de Compostos Orgânicos – Aula 2 13 ESTEREOQUÍMICA NO MECANISMO SN1 Me Et HHEt Me Lv + nucleophile adds at the top Nu HEt Me Nu nucleophile adds at the bottomNu Nu HMe Et Mirror images Reatividade de Compostos Orgânicos – Aula 2 Stereochemistry of the SN1 Reaction • Because the carbocation is planar, the nucleophilic attack can come from either side. -H+ -H+ The products have more inversion, because the LG partially blocks that side. Reatividade de Compostos Orgânicos – Aula 2 • SN1 reactions are exothermic. • SN1 reactions occur in at least two steps. • The first step is slow. It is the formation of the carbocation intermediate, which is a very strong electrophile. • The second step is very fast. It is the reaction of the carbocation with the nucleophile. • Reaction is not stereospecific: racemization in reactions of optically active alkyl halides 15 Reatividade de Compostos Orgânicos – Aula 2 Factors Affecting SN1 Reactions ² Structure of the substrate ² Can a stable carbocation be formed? ² Strength of the nucleophile ² Nature of the leaving group ² The solvent in which the reaction is run. ² Must be able to stabilize the carbocation and the LG (which is usually an ion). Reatividade de Compostos Orgânicos – Aula 2 Factors Affecting SN1 Reactions – Structure of the Substrate ² The carbocation is stabilized by any alkyl groups that are bonded to the electrophilic C atom. – inductive effect from σ bonds – hyperconjugation with p orbital ² Resonance will also stabilize the carbocation. • Relative rates for SN1: 3°>2°>1° (resonance)>”1°” Reatividade de Compostos Orgânicos – Aula 2 ESTABILIDADE DO CARBOCÁTION This is the most important factor influencing the rate of SN1 reactions. Stability: 3° > 2° >1° > methyl > > > 3° 2° 1° methyl 18 Reatividade de Compostos Orgânicos – Aula 2 19 Reatividade de Compostos Orgânicos – Aula 2 Factors Affecting SN1 Reactions Strength of the Nucleophile • The rate is not much affected by the nucleophile. • It may be weak, moderate, or strong. Reatividade de Compostos Orgânicos – Aula 2 Factors Affecting SN1 Reactions – the Leaving Group • The LG should be excellent. • Review: a good LG must be ² electron withdrawing, to polarize the bond and make the C atom electrophilic, ² polarizable, to stabilize the transition state, and ² stable in the solvent(so it cannot be a strong base). Best LGs are neutral species or anions with a stabilized charge. Reatividade de Compostos Orgânicos – Aula 2 Factors Affecting SN1 Reactions – the Leaving Group • The leaving group (LG) serves two purposes in an SN1 reaction. These are the same as for the SN2 reaction. – It polarizes the bond that makes the C atom electrophilic. – It carries away a pair of electrons from the electrophilic C atom. Reatividade de Compostos Orgânicos – Aula 2 Factors Affecting SN1 Reactions – Solvent Effects • The solvent must be capable of dissolving both the carbocation and the leaving group. • SN1 reactions require highly polar solvents that strongly solvate ions. • Typical solvents: water, an alcohol, acetone (to help the alkyl halide to dissolve), or a mix. Reatividade de Compostos Orgânicos – Aula 2 SN1 Reactions - Summary ² The structure of the carbocation is the most important factor: ² Relative rates for SN1: 3°>2°>”1°”. ² The nucleophile is typically weak or moderate in strength. ² The LG should be excellent. ² The solvent be polar and protic to stabilize the carbocation and LG. ² Products will exhibit racemization and possibly rearrangements. Reatividade de Compostos Orgânicos – Aula 2 • 1 7 2. MECANISMO SN2: SN2: substitution, nucleophilic, bimolecular velocidade = k[substrato].[Nu] REAÇÃO DE 2ª. ORDEM Reatividade de Compostos Orgânicos – Aula 2 26 Reaction coordinate diagrams for (a) the SN2 reaction of methyl bromide and (b) an SN2 reaction of a sterically hindered alkyl bromide Reatividade de Compostos Orgânicos – Aula 2 27 CH3Br CH3CH2Br (CH3)2CHBr (CH3)3CBr Decreasing SN2 Reactivity Reatividade de Compostos Orgânicos – Aula 2 28 Stereochemistry of SN2 Reactions Inversion of Configuration nucleophile attacks carbon from side opposite bond to the leaving group three-dimensional arrangement of bonds in product is opposite to that of reactant Reatividade de Compostos Orgânicos – Aula 2 29 Inversion of configuration (Walden inversion) in an SN2 reaction is due to back side attack Reatividade de Compostos Orgânicos – Aula 2 30 A stereospecific reaction is one in which stereoisomeric starting materials give stereoisomeric products. The reaction of 2-bromoctane with NaOH (in ethanol-water) is stereospecific. (+)-2-Bromooctane (–)-2-Octanol (–)-2-Bromooctane (+)-2-Octanol Stereospecific Reaction Reatividade de Compostos Orgânicos – Aula 2 31 Q U ÍM IC A O R G Â N IC A 2 - P ro fa . R o z a n n a M a rq u e s M u z z i SUBSTITUIÇÃO NUCLEOFÍLICA A CARBONO SATURADO CC HH CHCH33 BrBr CHCH33(CH(CH22))55 CC HH CHCH33 HOHO NaOHNaOH ((SS)-(+)-2-Bromooctane)-(+)-2-Bromooctane Stereospecific Reaction Stereospecific Reaction Reatividade de Compostos Orgânicos – Aula 2 32 A bulky substituent in the alkyl halide reduces the reactivity of the alkyl halide: steric hindrance Q U ÍM IC A O R G Â N IC A 2 - P ro fa . R o z a n n a M a rq u e s M u z z i SUBSTITUIÇÃO NUCLEOFÍLICA A CARBONO SATURADO A bulky substituent in the alkyl halide reduces the reactivity of the alkyl halide: steric hindrance Q U ÍM IC A O R G Â N IC A 2 - P ro fa . R o z a n n a M a rq u e s M u z z i SUBSTITUIÇÃO NUCLEOFÍLICA A CARBONO SATURADO A bulky substituent in the alkyl halide reduces the reactivity of the alkyl halide: steric hindrance A velocidade da SN2 é governada por fatores estéricos Reatividade de Compostos Orgânicos – Aula 2 33 RBr + LiI RI + LiBr Effect of chain branching on rate of SN2 substitution Alkyl Structure Relative bromide rate Ethyl CH3CH2Br 1.0 Propyl CH3CH2CH2Br 0.8 Isobutyl (CH3)2CHCH2Br 0.036 Neopentyl (CH3)3CCH2Br 0.00002 Reatividade de Compostos Orgânicos – Aula 2 34 Factors Affecting SN2 Reactions The Leaving Group Section 10.3 Factors Affecting Reactions 367SN2 The weaker the base, the better it is as a leaving group. 10.3 Factors Affecting Reactions The Leaving Group If an alkyl iodide, an alkyl bromide, an alkyl chloride, and an alkyl fluoride (all with the same alkyl group) were allowed to react with the same nucleophile under the same conditions, we would find that the alkyl iodide is the most reactive and the alkyl fluo- ride is the least reactive. The only difference among these four reactions is the nature of the leaving group. From the relative reaction rates, we can see that the iodide ion is the best leaving group and the fluoride ion is the worst. This brings us to an important rule in organic chemistry—one that you will see frequently: The weaker the basicity of a group, the better is its leaving ability. The reason leaving ability depends on basicity is because weak bases are stable bases—they readily bear the electrons they formerly shared with a proton (Section 1.18). Because weak bases don’t share their electrons well, a weak base is not bonded as strongly to the carbon as a strong base would be and a weaker bond is more easily broken (Section 1.13). We have seen that the halide ions have the following relative basicities (or relative stabilities) because larger atoms are better able to stabilize their negative charge (Section 1.18): Because stable (weak) bases are better leaving groups, the halides have the following relative leaving abilities: Therefore, alkyl iodides are the most reactive of the alkyl halides, and alkyl fluorides are the least reactive. In fact, the fluoride ion is such a strong base that alkyl fluorides essentially do not undergo reactions. In Section 10.1, we saw that it is the polar carbon–halogen bond that causes alkyl halides to undergo substitution reactions. Carbon and iodine, however, have the same electronegativity. (See Table 1.3 on p. 10.) Why, then, does an alkyl halide undergo a substitution reaction? We know that larger atoms are more polarizable than smaller atoms. (Recall from Section 2.9 that polarizability is a measure of how easily an atom’s electron cloud can be distorted.) The high polarizability of the large iodine atom causes it to react as if it were polar even though, on the basis of the electronega- tivity of the atoms, the bond is nonpolar. relative reactivities of alkyl halides in an SN2 reaction RI > RBr > RCl > RF least reactivemost reactive SN2 relative leaving abilities of the halide ions I− > Br− > Cl− > F− worst leaving group best leaving group relative basicities of the halide ions I− < Br− < Cl− < F− strongest base, least stable base weakest base, most stable base RCH2I RCH2OH+ +HO− I− relative rates of reaction 30,000 RCH2Br RCH2OH+ +HO− Br− 10,000 RCH2Cl RCH2OH+ +HO− Cl− 200 RCH2F RCH2OH+ +HO− F− 1 SN2 Stable bases are weak bases. BRUI10-359_399r2 18-02-2003 2:50 PM Page 367 Reatividade de Compostos Orgânicos – Aula 2 35 Section 10.3 Factors Affecting Reactions 367SN2 The weaker the base, the better it is as a leaving group. 10.3 Factors Affecting Reactions The Leaving Group If an alkyl iodide, an alkyl bromide, an alkyl chloride, and an alkyl fluoride (all with the same alkyl group) were allowed to react with the same nucleophile under the same conditions, we would find that the alkyl iodide is the most reactive and the alkyl fluo- ride is the least reactive. The onlydifference among these four reactions is the nature of the leaving group. From the relative reaction rates, we can see that the iodide ion is the best leaving group and the fluoride ion is the worst. This brings us to an important rule in organic chemistry—one that you will see frequently: The weaker the basicity of a group, the better is its leaving ability. The reason leaving ability depends on basicity is because weak bases are stable bases—they readily bear the electrons they formerly shared with a proton (Section 1.18). Because weak bases don’t share their electrons well, a weak base is not bonded as strongly to the carbon as a strong base would be and a weaker bond is more easily broken (Section 1.13). We have seen that the halide ions have the following relative basicities (or relative stabilities) because larger atoms are better able to stabilize their negative charge (Section 1.18): Because stable (weak) bases are better leaving groups, the halides have the following relative leaving abilities: Therefore, alkyl iodides are the most reactive of the alkyl halides, and alkyl fluorides are the least reactive. In fact, the fluoride ion is such a strong base that alkyl fluorides essentially do not undergo reactions. In Section 10.1, we saw that it is the polar carbon–halogen bond that causes alkyl halides to undergo substitution reactions. Carbon and iodine, however, have the same electronegativity. (See Table 1.3 on p. 10.) Why, then, does an alkyl halide undergo a substitution reaction? We know that larger atoms are more polarizable than smaller atoms. (Recall from Section 2.9 that polarizability is a measure of how easily an atom’s electron cloud can be distorted.) The high polarizability of the large iodine atom causes it to react as if it were polar even though, on the basis of the electronega- tivity of the atoms, the bond is nonpolar. relative reactivities of alkyl halides in an SN2 reaction RI > RBr > RCl > RF least reactivemost reactive SN2 relative leaving abilities of the halide ions I− > Br− > Cl− > F− worst leaving group best leaving group relative basicities of the halide ions I− < Br− < Cl− < F− strongest base, least stable base weakest base, most stable base RCH2I RCH2OH+ +HO− I− relative rates of reaction 30,000 RCH2Br RCH2OH+ +HO− Br− 10,000 RCH2Cl RCH2OH+ +HO− Cl− 200 RCH2F RCH2OH+ +HO− F− 1 SN2 Stable bases are weak bases. BRUI10-359_399r2 18-02-2003 2:50 PM Page 367 Reatividade de Compostos Orgânicos – Aula 2 36 368 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides The Nucleophile When we talk about atoms or molecules that have lone-pair electrons, sometimes we call them bases and sometimes we call them nucleophiles. What is the difference be- tween a base and a nucleophile? Basicity is a measure of how well a compound (a base) shares its lone pair with a proton. The stronger the base, the better it shares its electrons. Basicity is measured by an equilibrium constant (the acid dissociation constant, ) that indicates the tenden- cy of the conjugate acid of the base to lose a proton (Section 1.17). Nucleophilicity is a measure of how readily a compound (a nucleophile) is able to attack an electron-deficient atom. Nucleophilicity is measured by a rate constant (k). In the case of an reaction, nucleophilicity is a measure of how readily the nucle- ophile attacks an hybridized carbon bonded to a leaving group. When comparing molecules with the same attacking atom, there is a direct relation- ship between basicity and nucleophilicity: Stronger bases are better nucleophiles. For example, a species with a negative charge is a stronger base and a better nucleophile than a species with the same attacking atom that is neutral. Thus is a stronger base and a better nucleophile than stronger base, weaker base, better nucleophile poorer nucleophile When comparing molecules with attacking atoms of approximately the same size, the stronger bases are again the better nucleophiles. The atoms across the second row of the periodic table have approximately the same size. If hydrogens are attached to the second-row elements, the resulting compounds have the following relative acidities (Section 1.18): Consequently, the conjugate bases have the following relative base strengths and nucleophilicities: Note that the amide anion is the strongest base, as well as the best nucleophile. When comparing molecules with attacking atoms that are very different in size, an- other factor comes into play—the polarizability of the atom. Because the electrons are farther away in the larger atom, they are not held as tightly and can, therefore, move more freely toward a positive charge. As a result, the electrons are able to overlap from farther away with the orbital of carbon, as shown in Figure 10.5. This results in a greater degree of bonding in the transition state, making it more stable. Whether the greater polarizability of the larger atoms makes up for their decreased basicity depends on the conditions under which the reaction is carried out. relative base strengths and relative nucleophilicities −NH2 > HO− > F−strongest base best nucleophile relative acid strengths NH3 < H2O < HFweakest acid HO- CH3O- -NH2 CH3CH2NH- 7 7 7 7 H2O CH3OH NH3 CH3CH2NH2 H2O. HO- sp3 SN2 Ka BRUI10-359_399r2 18-02-2003 2:50 PM Page 368 368 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides The Nucleophile When we talk about atoms or molecules that have lone-pair electrons, sometimes we call them bases and sometimes we call them nucleophiles. What is the difference be- tween a base and a nucleophile? Basicity is a measure of how well a compound (a base) shares its lone pair with a proton. The stronger the base, the better it shares its electrons. Basicity is measured by an equilibrium constant (the acid dissociation constant, ) that indicates the tenden- cy of the conjugate acid of the base to lose a proton (Section 1.17). Nucleophilicity is a measure of how readily a compound (a nucleophile) is able to attack an electron-deficient atom. Nucleophilicity is measured by a rate constant (k). In the case of an reaction, nucleophilicity is a measure of how readily the nucle- ophile attacks an hybridized carbon bonded to a leaving group. When comparing molecules with the same attacking atom, there is a direct relation- ship between basicity and nucleophilicity: Stronger bases are better nucleophiles. For example, a species with a negative charge is a stronger base and a better nucleophile than a species with the same attacking atom that is neutral. Thus is a stronger base and a better nucleophile than stronger base, weaker base, better nucleophile poorer nucleophile When comparing molecules with attacking atoms of approximately the same size, the stronger bases are again the better nucleophiles. The atoms across the second row of the periodic table have approximately the same size. If hydrogens are attached to the second-row elements, the resulting compounds have the following relative acidities (Section 1.18): Consequently, the conjugate bases have the following relative base strengths and nucleophilicities: Note that the amide anion is the strongest base, as well as the best nucleophile. When comparing molecules with attacking atoms that are very different in size, an- other factor comes into play—the polarizability of the atom. Because the electrons are farther away in the larger atom, they are not held as tightly and can, therefore, move more freely toward a positive charge. As a result, the electrons are able to overlap from farther away with the orbital of carbon, as shown in Figure 10.5. This results in a greater degree of bonding in the transition state, making it morestable. Whether the greater polarizability of the larger atoms makes up for their decreased basicity depends on the conditions under which the reaction is carried out. relative base strengths and relative nucleophilicities −NH2 > HO− > F−strongest base best nucleophile relative acid strengths NH3 < H2O < HFweakest acid HO- CH3O- -NH2 CH3CH2NH- 7 7 7 7 H2O CH3OH NH3 CH3CH2NH2 H2O. HO- sp3 SN2 Ka BRUI10-359_399r2 18-02-2003 2:50 PM Page 368 368 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides The Nucleophile When we talk about atoms or molecules that have lone-pair electrons, sometimes we call them bases and sometimes we call them nucleophiles. What is the difference be- tween a base and a nucleophile? Basicity is a measure of how well a compound (a base) shares its lone pair with a proton. The stronger the base, the better it shares its electrons. Basicity is measured by an equilibrium constant (the acid dissociation constant, ) that indicates the tenden- cy of the conjugate acid of the base to lose a proton (Section 1.17). Nucleophilicity is a measure of how readily a compound (a nucleophile) is able to attack an electron-deficient atom. Nucleophilicity is measured by a rate constant (k). In the case of an reaction, nucleophilicity is a measure of how readily the nucle- ophile attacks an hybridized carbon bonded to a leaving group. When comparing molecules with the same attacking atom, there is a direct relation- ship between basicity and nucleophilicity: Stronger bases are better nucleophiles. For example, a species with a negative charge is a stronger base and a better nucleophile than a species with the same attacking atom that is neutral. Thus is a stronger base and a better nucleophile than stronger base, weaker base, better nucleophile poorer nucleophile When comparing molecules with attacking atoms of approximately the same size, the stronger bases are again the better nucleophiles. The atoms across the second row of the periodic table have approximately the same size. If hydrogens are attached to the second-row elements, the resulting compounds have the following relative acidities (Section 1.18): Consequently, the conjugate bases have the following relative base strengths and nucleophilicities: Note that the amide anion is the strongest base, as well as the best nucleophile. When comparing molecules with attacking atoms that are very different in size, an- other factor comes into play—the polarizability of the atom. Because the electrons are farther away in the larger atom, they are not held as tightly and can, therefore, move more freely toward a positive charge. As a result, the electrons are able to overlap from farther away with the orbital of carbon, as shown in Figure 10.5. This results in a greater degree of bonding in the transition state, making it more stable. Whether the greater polarizability of the larger atoms makes up for their decreased basicity depends on the conditions under which the reaction is carried out. relative base strengths and relative nucleophilicities −NH2 > HO− > F−strongest base best nucleophile relative acid strengths NH3 < H2O < HFweakest acid HO- CH3O- -NH2 CH3CH2NH- 7 7 7 7 H2O CH3OH NH3 CH3CH2NH2 H2O. HO- sp3 SN2 Ka BRUI10-359_399r2 18-02-2003 2:50 PM Page 368 Reatividade de Compostos Orgânicos – Aula 2 37 Section 10.3 Factors Affecting Reactions 369SN2 H XC H H C H XC H H I− transition state H XC H H C H XC H H F− transition state more bonding little bonding ++ ++ δ− δ− > Figure 10.5 An iodide ion is larger and more polarizable than a fluoride ion. Therefore, the relatively loosely held electrons of the iodide ion can overlap from farther away with the orbital of carbon undergoing nucleophilic attack. The tightly bound electrons of the fluoride ion cannot start to overlap until the atoms are closer together. If the reaction is carried out in the gas phase, the direct relationship between basici- ty and nucleophilicity is maintained—the stronger bases are still the best nucleophiles. If, however, the reaction is carried out in a protic solvent—meaning the solvent mole- cules have a hydrogen bonded to an oxygen or to a nitrogen—the relationship between basicity and nucleophilicity becomes inverted. The largest atom is the best nucleophile even though it is the weakest base. Therefore, iodide ion is the best nucleophile of the halide ions in a protic solvent. PROBLEM 4! a. Which is a stronger base, or b. Which is a better nucleophile in an aqueous solution? The Effect of the Solvent on Nucleophilicity Why, in a protic solvent, is the smallest atom the poorest nucleophile even though it is the strongest base? How does a protic solvent make strong bases less nucleophilic? When a negatively charged species is placed in a protic solvent, the ion becomes sol- vated (Section 2.9). Protic solvents are hydrogen bond donors. The solvent molecules arrange themselves so that their partially positively charged hydrogens point toward the negatively charged species. The interaction between the ion and the dipole of the protic solvent is called an ion–dipole interaction. Because the solvent shields the nu- cleophile, at least one of the ion–dipole interactions must be broken before the nucle- ophile can participate in an reaction. Weak bases interact weakly with protic solvents, whereas strong bases interact more strongly because they are better at shar- ing their electrons. It is easier, therefore, to break the ion–dipole interactions between SN2 RS-?RO- increasing basicity increasing nucleophilicity in the gas phase increasing size increasing nucleophilicity in a protic solvent F− Cl− Br− I− A protic solvent contains a hydrogen bonded to an oxygen or a nitrogen; it is a hydrogen bond donor. BRUI10-359_399r2 18-02-2003 2:50 PM Page 369 Reatividade de Compostos Orgânicos – Aula 2 38 Section 10.3 Factors Affecting Reactions 369SN2 H XC H H C H XC H H I− transition state H XC H H C H XC H H F− transition state more bonding little bonding ++ ++ δ− δ− > Figure 10.5 An iodide ion is larger and more polarizable than a fluoride ion. Therefore, the relatively loosely held electrons of the iodide ion can overlap from farther away with the orbital of carbon undergoing nucleophilic attack. The tightly bound electrons of the fluoride ion cannot start to overlap until the atoms are closer together. If the reaction is carried out in the gas phase, the direct relationship between basici- ty and nucleophilicity is maintained—the stronger bases are still the best nucleophiles. If, however, the reaction is carried out in a protic solvent—meaning the solvent mole- cules have a hydrogen bonded to an oxygen or to a nitrogen—the relationship between basicity and nucleophilicity becomes inverted. The largest atom is the best nucleophile even though it is the weakest base. Therefore, iodide ion is the best nucleophile of the halide ions in a protic solvent. PROBLEM 4! a. Which is a stronger base, or b. Which is a better nucleophile in an aqueous solution? The Effect of the Solvent on Nucleophilicity Why, in a protic solvent, is the smallest atom the poorest nucleophile even though it is the strongest base? How does a protic solvent make strong bases less nucleophilic? When a negatively charged species is placed in a protic solvent, the ion becomes sol- vated (Section 2.9). Protic solvents are hydrogen bond donors. The solvent molecules arrange themselves so that their partially positively charged hydrogens point toward the negatively charged species. The interaction betweenthe ion and the dipole of the protic solvent is called an ion–dipole interaction. Because the solvent shields the nu- cleophile, at least one of the ion–dipole interactions must be broken before the nucle- ophile can participate in an reaction. Weak bases interact weakly with protic solvents, whereas strong bases interact more strongly because they are better at shar- ing their electrons. It is easier, therefore, to break the ion–dipole interactions between SN2 RS-?RO- increasing basicity increasing nucleophilicity in the gas phase increasing size increasing nucleophilicity in a protic solvent F− Cl− Br− I− A protic solvent contains a hydrogen bonded to an oxygen or a nitrogen; it is a hydrogen bond donor. BRUI10-359_399r2 18-02-2003 2:50 PM Page 369 370 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides An aprotic solvent does not contain a hydrogen bonded to either an oxygen or a nitrogen; it is not a hydrogen bond donor. an iodide ion (a weak base) and the solvent than between a fluoride ion (a stronger base) and the solvent. As a result, iodide ion is a better nucleophile than fluoride ion in a protic solvent (Table 10.2). Fluoride ion would be a better nucleophile in a nonpolar solvent than in a polar sol- vent because there would be no ion–dipole interactions between the ion and the nonpo- lar solvent. However, ionic compounds are insoluble in most nonpolar solvents, but they can dissolve in aprotic polar solvents, such as dimethylformamide (DMF) or dimethyl- sulfoxide (DMSO). An aprotic polar solvent is not a hydrogen bond donor because it does not have a hydrogen attached to an oxygen or to a nitrogen, so there are no posi- tively charged hydrogens to form ion–dipole interactions. The molecules of an aprotic polar solvent have a partial negative charge on their surface that can solvate cations, but the partial positive charge is on the inside of the molecule, which makes it less accessi- ble. The relatively “naked” anion can be a powerful nucleophile in an aprotic polar sol- vent. Fluoride ion, therefore, is a better nucleophile in DMSO than it is in water. PROBLEM 5! Indicate whether each of the following solvents is protic or aprotic: a. chloroform c. acetic acid b. diethyl ether d. hexane [CH3(CH2)4CH3](CH3CH2OCH2CH3) (CH3COOH)(CHCl3) δ−δ− δ− δ− δ+ δ+ δ+ δ+ δ− δ+ δ+ δ+ δ+ H H OY− H H O H H O H HO ion–dipole interactions between a nucleophile and water 3-D Molecules: N,N-Dimethylformamide; Dimethyl sulfoxide N,N-dimethylformamide DMF C H CH3 Oδ− CH3 N dimethyl sulfoxide DMSO H3C CH3 the δ− is on the surface of the molecule S Oδ− δ+ δ+ the δ+ is not very accessible δ+ δ+ δ+ δ− δ−δ+ δ− CH3 CH3 CH3 CH3 CH3 CH3 CH3 CH3 S S O δ− O OK+ O S S DMSO can solvate a cation better than it can solvate an anion increasing nucleophilicity Table 10.2 Relative Nucleophilicity Toward CH3l in Methanol RS− > I− > −C N > CH3O− > Br− > NH3 > Cl− > F− > CH3OH BRUI10-359_399r2 18-02-2003 2:50 PM Page 370370 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides An aprotic solvent does not contain a hydrogen bonded to either an oxygen or a nitrogen; it is not a hydrogen bond donor. an iodide ion (a weak base) and the solvent than between a fluoride ion (a stronger base) and the solvent. As a result, iodide ion is a better nucleophile than fluoride ion in a protic solvent (Table 10.2). Fluoride ion would be a better nucleophile in a nonpolar solvent than in a polar sol- vent because there would be no ion–dipole interactions between the ion and the nonpo- lar solvent. However, ionic compounds are insoluble in most nonpolar solvents, but they can dissolve in aprotic polar solvents, such as dimethylformamide (DMF) or dimethyl- sulfoxide (DMSO). An aprotic polar solvent is not a hydrogen bond donor because it does not have a hydrogen attached to an oxygen or to a nitrogen, so there are no posi- tively charged hydrogens to form ion–dipole interactions. The molecules of an aprotic polar solvent have a partial negative charge on their surface that can solvate cations, but the partial positive charge is on the inside of the molecule, which makes it less accessi- ble. The relatively “naked” anion can be a powerful nucleophile in an aprotic polar sol- vent. Fluoride ion, therefore, is a better nucleophile in DMSO than it is in water. PROBLEM 5! Indicate whether each of the following solvents is protic or aprotic: a. chloroform c. acetic acid b. diethyl ether d. hexane [CH3(CH2)4CH3](CH3CH2OCH2CH3) (CH3COOH)(CHCl3) δ−δ− δ− δ− δ+ δ+ δ+ δ+ δ− δ+ δ+ δ+ δ+ H H OY− H H O H H O H HO ion–dipole interactions between a nucleophile and water 3-D Molecules: N,N-Dimethylformamide; Dimethyl sulfoxide N,N-dimethylformamide DMF C H CH3 Oδ− CH3 N dimethyl sulfoxide DMSO H3C CH3 the δ− is on the surface of the molecule S Oδ− δ+ δ+ the δ+ is not very accessible δ+ δ+ δ+ δ− δ−δ+ δ− CH3 CH3 CH3 CH3 CH3 CH3 CH3 CH3 S S O δ− O OK+ O S S DMSO can solvate a cation better than it can solvate an anion increasing nucleophilicity Table 10.2 Relative Nucleophilicity Toward CH3l in Methanol RS− > I− > −C N > CH3O− > Br− > NH3 > Cl− > F− > CH3OH BRUI10-359_399r2 18-02-2003 2:50 PM Page 370 Reatividade de Compostos Orgânicos – Aula 2 39 390 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides SOLVATION EFFECTS The tremendous amount of energy that is provid- ed by solvation can be appreciated by considering the energy required to break the crystal lattice of sodium chlo- ride (table salt) (Figure 1.1). In the absence of a solvent, sodium chloride must be heated to more than 800 °C to overcome the forces that hold the oppositely charged ions together. However, sodium chloride readily dissolves in water at room temperature because solvation of the and ions by water provides the energy necessary to separate the ions. Cl-Na+ Stabilization of charges by solvent interaction plays an important role in organic re- actions. For example, when an alkyl halide undergoes an reaction, the first step is dissociation of the carbon–halogen bond to form a carbocation and a halide ion. Ener- gy is required to break the bond, but with no bonds being formed, where does the en- ergy come from? If the reaction is carried out in a polar solvent, the ions that are produced are solvated. The energy associated with a single ion–dipole interaction is small, but the additive effect of all the ion–dipole interactions involved in stabilizing a charged species by the solvent represents a great deal of energy. These ion–dipole in- teractions provide much of the energy necessary for dissociation of the carbon–halo- gen bond. So in an reaction, the alkyl halide does not fall apart spontaneously, but rather, polar solvent molecules pull it apart. An reaction, therefore, cannot take place in a nonpolar solvent. It also cannot take place in the gas phase, where there are no solvent molecules and, consequently, no solvation effects. SN1 SN1 SN1 Table 10.7 The Dielectric Constants of Some Common Solvents Solvent Structure Dielectric constantAbbreviation (ε , at 25 °C) Boiling point (°C) Water H2O 79 Formic acid HCOOH 59 Methanol CH3OH 33 Ethanol CH3CH2OH 25 tert-Butyl alcohol (CH3)3COH 11 Acetic acid CH3COOH 6 Dimethyl sulfoxide (CH3)2SO 47 Acetonitrile CH3CN 38 Dimethylformamide (CH3)2NCHO 37 Hexamethylphosphoric acid triamide [(CH3)2N]3PO 30 Acetone (CH3)2CO 21 Eth Dichloromethane Tetrahydrofuran yl acetateCH CH2Cl2 3COOCH2CH3 6 9.1 Diethyl ether CH3CH2OCH2CH3 4.3 Benzene 2.3 Hexane CH3(CH2)4CH3 1.9 100 100.6 64.7 78.3 82.3 117.9 189 81.6 153 233 56.3 77.1 40 7.6 66 34.6 80.1 68.7 — — MeOH EtOH tert-BuOH HOAc DMSO MeCN DMF HMPA Me2CO THF EtOAc — Et2O O — — Protic solvents Aprotic solvents BRUI10-359_399r2 18-02-2003 2:50 PM Page 390 Reatividade de Compostos Orgânicos – Aula 2 40 370 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides An aprotic solvent does not contain a hydrogen bonded to either an oxygen or a nitrogen; it is not a hydrogen bond donor. an iodide ion (a weak base) and the solvent than between a fluoride ion (a stronger base) and the solvent. As a result, iodide ion is a better nucleophile than fluoride ion in a protic solvent (Table 10.2). Fluoride ion would be a better nucleophile in a nonpolar solvent than in a polar sol- vent because there would be no ion–dipole interactions between the ion and the nonpo- lar solvent. However, ionic compounds are insoluble in most nonpolar solvents, but they can dissolve in aprotic polar solvents, such as dimethylformamide (DMF) or dimethyl- sulfoxide (DMSO). An aprotic polar solvent is not a hydrogen bond donor because it does not have a hydrogen attached to an oxygen or to a nitrogen, so there are no posi- tively charged hydrogens to form ion–dipole interactions. The molecules of an aprotic polar solvent have a partial negative charge on their surface that can solvate cations, but the partial positive charge is on the inside of the molecule, which makes it less accessi- ble. The relatively “naked” anion can be a powerful nucleophile in an aprotic polar sol- vent. Fluoride ion, therefore, is a better nucleophile in DMSO than it is in water. PROBLEM 5! Indicate whether each of the following solvents is protic or aprotic: a. chloroform c. acetic acid b. diethyl ether d. hexane [CH3(CH2)4CH3](CH3CH2OCH2CH3) (CH3COOH)(CHCl3) δ−δ− δ− δ− δ+ δ+ δ+ δ+ δ− δ+ δ+ δ+ δ+ H H OY− H H O H H O H HO ion–dipole interactions between a nucleophile and water 3-D Molecules: N,N-Dimethylformamide; Dimethyl sulfoxide N,N-dimethylformamide DMF C H CH3 Oδ− CH3 N dimethyl sulfoxide DMSO H3C CH3 the δ− is on the surface of the molecule S Oδ− δ+ δ+ the δ+ is not very accessible δ+ δ+ δ+ δ− δ−δ+ δ− CH3 CH3 CH3 CH3 CH3 CH3 CH3 CH3 S S O δ− O OK+ O S S DMSO can solvate a cation better than it can solvate an anion increasing nucleophilicity Table 10.2 Relative Nucleophilicity Toward CH3l in Methanol RS− > I− > −C N > CH3O− > Br− > NH3 > Cl− > F− > CH3OH BRUI10-359_399r2 18-02-2003 2:50 PM Page 370 370 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides An aprotic solvent does not contain a hydrogen bonded to either an oxygen or a nitrogen; it is not a hydrogen bond donor. an iodide ion (a weak base) and the solvent than between a fluoride ion (a stronger base) and the solvent. As a result, iodide ion is a better nucleophile than fluoride ion in a protic solvent (Table 10.2). Fluoride ion would be a better nucleophile in a nonpolar solvent than in a polar sol- vent because there would be no ion–dipole interactions between the ion and the nonpo- lar solvent. However, ionic compounds are insoluble in most nonpolar solvents, but they can dissolve in aprotic polar solvents, such as dimethylformamide (DMF) or dimethyl- sulfoxide (DMSO). An aprotic polar solvent is not a hydrogen bond donor because it does not have a hydrogen attached to an oxygen or to a nitrogen, so there are no posi- tively charged hydrogens to form ion–dipole interactions. The molecules of an aprotic polar solvent have a partial negative charge on their surface that can solvate cations, but the partial positive charge is on the inside of the molecule, which makes it less accessi- ble. The relatively “naked” anion can be a powerful nucleophile in an aprotic polar sol- vent. Fluoride ion, therefore, is a better nucleophile in DMSO than it is in water. PROBLEM 5! Indicate whether each of the following solvents is protic or aprotic: a. chloroform c. acetic acid b. diethyl ether d. hexane [CH3(CH2)4CH3](CH3CH2OCH2CH3) (CH3COOH)(CHCl3) δ−δ− δ− δ− δ+ δ+ δ+ δ+ δ− δ+ δ+ δ+ δ+ H H OY− H H O H H O H HO ion–dipole interactions between a nucleophile and water 3-D Molecules: N,N-Dimethylformamide; Dimethyl sulfoxide N,N-dimethylformamide DMF C H CH3 Oδ− CH3 N dimethyl sulfoxide DMSO H3C CH3 the δ− is on the surface of the molecule S Oδ− δ+ δ+ the δ+ is not very accessible δ+ δ+ δ+ δ− δ−δ+ δ− CH3 CH3 CH3 CH3 CH3 CH3 CH3 CH3 S S O δ− O OK+ O S S DMSO can solvate a cation better than it can solvate an anion increasing nucleophilicity Table 10.2 Relative Nucleophilicity Toward CH3l in Methanol RS− > I− > −C N > CH3O− > Br− > NH3 > Cl− > F− > CH3OH BRUI10-359_399r2 18-02-2003 2:50 PM Page 370 Reatividade de Compostos Orgânicos – Aula 2 41 Section 10.3 Factors Affecting Reactions 371SN2 ethoxide ion tert-butoxide ion Nucleophilicity Is Affected by Steric Effects Base strength is relatively unaffected by steric effects because a base removes a relatively unhindered proton. The strength of a base depends only on how well the base shares its electrons with a proton. Thus, tert-butoxide ion is a stronger base than ethoxide ion since tert-butanol is a weaker acid than ethanol Steric effects, on the other hand, do affect nucleophilicity. A bulky nucleophile cannot approach the back side of a carbon as easily as a less sterically hindered nucle- ophile can. Thus, the bulky tert-butoxide ion, with its three methyl groups, is a poorer nucleophile than ethoxide ion even though tert-butoxide ion is a stronger base. PROBLEM 6 SOLVED List the following species in order of decreasing nucleophilicity in an aqueous solution: SOLUTION Let’s first divide the nucleophiles into groups. We have one nucleophile with a negatively charged sulfur, three with negatively charged oxygens, and one with a neutral oxygen. We know that in the polar aqueous solvent, the one with the negatively charged sulfur is the most nucleophilic because sulfur is larger than oxygen. We also know that the poorest nucleophile is the one with the neutral oxygen atom. So now, to complete the problem, we need to rank the three nucleophiles with negatively charged oxygens in order of the ’s of their conjugate acids. A carboxylic acid is a stronger acid than phenol, which is a stronger acid than water (Section 7.10). Because water is the weakest acid, its conjugate base is the strongest base and the best nucleophile. Thus, the relative nucleophilicities are PROBLEM 7! For each of the following pairs of reactions, indicate which reaction occurs faster: a. b. c. d. CH3CH2Cl + I− CH3CH2Br + I−or (in ethanol) CH3CH2Cl + CH3O− CH3CH2Cl + CH3S−or CH3CHCH2Br CH3CH2CHBr+ HO− CH3 + HO−or CH3 CH3CH2Br CH3CH2Br+ H2O + HO−or SN2 CH3CO−>CH3S− > > >HO− CH3OHO− O pKa CH3CO−CH3OH HO−O− CH3S− O ethoxide ion better nucleophile tert-butoxide ion stronger base CH3CO−CH3CH2O− CH3 CH3 1pKa = 15.92. 1pKa = 182 BRUI10-359_399r2 18-02-2003 2:50 PM Page 371 Section 10.3 Factors Affecting Reactions 371SN2 ethoxide ion tert-butoxide ion Nucleophilicity Is Affected by Steric Effects Base strength is relatively unaffected by steric effects because a base removes a relatively unhindered proton. The strength of a base dependsonly on how well the base shares its electrons with a proton. Thus, tert-butoxide ion is a stronger base than ethoxide ion since tert-butanol is a weaker acid than ethanol Steric effects, on the other hand, do affect nucleophilicity. A bulky nucleophile cannot approach the back side of a carbon as easily as a less sterically hindered nucle- ophile can. Thus, the bulky tert-butoxide ion, with its three methyl groups, is a poorer nucleophile than ethoxide ion even though tert-butoxide ion is a stronger base. PROBLEM 6 SOLVED List the following species in order of decreasing nucleophilicity in an aqueous solution: SOLUTION Let’s first divide the nucleophiles into groups. We have one nucleophile with a negatively charged sulfur, three with negatively charged oxygens, and one with a neutral oxygen. We know that in the polar aqueous solvent, the one with the negatively charged sulfur is the most nucleophilic because sulfur is larger than oxygen. We also know that the poorest nucleophile is the one with the neutral oxygen atom. So now, to complete the problem, we need to rank the three nucleophiles with negatively charged oxygens in order of the ’s of their conjugate acids. A carboxylic acid is a stronger acid than phenol, which is a stronger acid than water (Section 7.10). Because water is the weakest acid, its conjugate base is the strongest base and the best nucleophile. Thus, the relative nucleophilicities are PROBLEM 7! For each of the following pairs of reactions, indicate which reaction occurs faster: a. b. c. d. CH3CH2Cl + I− CH3CH2Br + I−or (in ethanol) CH3CH2Cl + CH3O− CH3CH2Cl + CH3S−or CH3CHCH2Br CH3CH2CHBr+ HO− CH3 + HO−or CH3 CH3CH2Br CH3CH2Br+ H2O + HO−or SN2 CH3CO−>CH3S− > > >HO− CH3OHO− O pKa CH3CO−CH3OH HO−O− CH3S− O ethoxide ion better nucleophile tert-butoxide ion stronger base CH3CO−CH3CH2O− CH3 CH3 1pKa = 15.92. 1pKa = 182 BRUI10-359_399r2 18-02-2003 2:50 PM Page 371 Reatividade de Compostos Orgânicos – Aula 2 42 372 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides 10.4 The Reversibility of an Reaction Many different kinds of nucleophiles can react with alkyl halides. Therefore, a wide variety of organic compounds can be synthesized by means of reactions. It may seem that the reverse of each of these reactions can also occur via nucleophilic substitution. In the first reaction, for example, ethyl chloride reacts with hydroxide ion to form ethyl alcohol and a chloride ion. The reverse reaction would appear to satisfy the requirements for a nucleophilic substitution reaction, given that chloride ion is a nucle- ophile and ethyl alcohol has an leaving group. But ethyl alcohol and chloride ion do not react. Why does a nucleophilic substitution reaction take place in one direction, but not in the other? We can answer this question by comparing the leaving tendency of in the forward direction and the leaving tendency of in the reverse direction. Com- paring leaving tendencies involves comparing basicities. Most people find it easier to compare the acid strengths of the conjugate acids (Table 10.3), so we will compare the acid strengths of HCl and HCl is a much stronger acid than which means that is a much weaker base than (Remember, the stronger the acid, the weaker is its conjugate base.) Because it is a weaker base, is a better leaving group. Consequently, can displace in the forward reaction, but cannot displace in the reverse reaction. The reaction proceeds in the direction that al- lows the stronger base to displace the weaker base (the best leaving group). If the difference between the basicities of the nucleophile and the leaving group is not very large, the reaction will be reversible. For example, in the reaction of ethyl bro- mide with iodide ion, is the leaving group in one direction and is the leaving group in the other direction. The reaction is reversible because the values of the conjugate acids of the two leaving groups are similar ( of of see Table 10.3). You can drive a reversible reaction toward the desired products by removing one of the products as it is formed. Recall that the concentrations of the reactants and products at equilibrium are governed by the equilibrium constant of the reaction + +CH3CH2Br Br−CH3CH2II− an SN2 reaction is reversible when the basicities of the leaving groups are similar HI = -10; HBr = -9; pKapKa pKa I-Br- HO- Cl-Cl-HO- Cl- HO-.Cl- H2O,H2O. HO- Cl- HO- NN CR an alcohol a thiol CH3CH2Cl CH3CH2OH Cl−+ +HO− CH3CH2Br CH3CH2SH Br−+ +HS− an ether a thioether CH3CH2I CH3CH2OR I−+ +RO− CH3CH2Br CH3CH2SR Br−+ +RS− a primary amine CH3CH2Cl CH3CH2NH2 Cl−+ +−NH2 an alkyne a nitrile CH3CH2Br CH3CH2C Br−CR+ +−C CH3CH2I CH3CH2C I−+ +−C SN2 SN2 An reaction proceeds in the direction that allows the stronger base to displace the weaker base. SN2 BRUI10-359_399r2 18-02-2003 2:50 PM Page 372 Reatividade de Compostos Orgânicos – Aula 2 43 PROBLEM 22! Which alkyl halide would you expect to be more reactive in an reaction with a given nucleophile? In each case, you can assume that both alkyl halides have the same stability. a. or b. or c. d. e. f. g. PROBLEM 23! For each of the pairs in Problem 22, which compound would be more reactive in an reaction? PROBLEM 24! Show the configuration of the products obtained from the following reaction: a. under conditions that favor an reaction b. under conditions that favor an reaction 10.9 Competition Between and Reactions The characteristics of and reactions are summarized in Table 10.5. Remem- ber that the “2” in and the “1” in refer to the molecularity—how many molecules are involved in the rate-determining step. Thus, the rate-determining step of an reaction is bimolecular, whereas the rate-determining step of an reaction isSN1SN2 “SN1”“SN2” SN1SN2 SN1SN2 CH3OHCH3O−+CHCH2BrCH3CH SN1 SN2 SN1 CH3CH CCH3 or Br CH3CH CHCHCH3 Br CH2Br or Br CH2CH2Br or CH2CHCH3 Br CH3CH2CH2CHBr CH3 CH3CH2CHCH2Br CH3 or CH3CH2CHBr or CH3 CH3CH2CHBr CH2CH3 CH3OCH2ClCH3CH2CH2Cl CH3CH2CH2ICH3CH2CH2Br SN2 Section 10.9 Competition Between and Reactions 385SN1SN2 Table 10.5 Comparison of and ReactionsSN1SN2 A one-step mechanism A stepwise mechanism that forms a carbocation intermediate A bimolecular rate-determining A unimolecular rate-determining step step No carbocation rearrangements Carbocation rearrangements Product has inverted configuration Products have both retained and inverted relative to the reactant configurations relative to the reactant Reactivity order: Reactivity order: methyl 3° 7 2° 7 1° 7 methyl7 1° 7 2° 7 3° SN1SN2 BRUI10-359_399r2 18-02-2003 2:50 PM Page 385 PROBLEM 22! Which alkyl halide would you expect to be more reactive in an reaction with a given nucleophile? In each case, you can assume that both alkyl halides have the same stability. a. or b. or c. d. e. f. g. PROBLEM 23! For each of the pairs in Problem 22, which compound would be more reactive in an reaction? PROBLEM 24! Show the configuration of the products obtained from the following reaction: a. under conditions that favor an reaction b. under conditions that favor an reaction 10.9 Competition Between and Reactions The characteristics of and reactions are summarized in Table 10.5. Remem- ber that the “2” in and the “1” in refer to the molecularity—how many molecules are involved in the rate-determining step. Thus, the rate-determining step of an reaction is bimolecular, whereas the rate-determining step of an reaction isSN1SN2 “SN1”“SN2” SN1SN2 SN1SN2 CH3OHCH3O−+CHCH2BrCH3CH SN1 SN2 SN1 CH3CH CCH3 or Br CH3CH CHCHCH3 Br CH2Br or Br CH2CH2Br or CH2CHCH3Br CH3CH2CH2CHBr CH3 CH3CH2CHCH2Br CH3 or CH3CH2CHBr or CH3 CH3CH2CHBr CH2CH3 CH3OCH2ClCH3CH2CH2Cl CH3CH2CH2ICH3CH2CH2Br SN2 Section 10.9 Competition Between and Reactions 385SN1SN2 Table 10.5 Comparison of and ReactionsSN1SN2 A one-step mechanism A stepwise mechanism that forms a carbocation intermediate A bimolecular rate-determining A unimolecular rate-determining step step No carbocation rearrangements Carbocation rearrangements Product has inverted configuration Products have both retained and inverted relative to the reactant configurations relative to the reactant Reactivity order: Reactivity order: methyl 3° 7 2° 7 1° 7 methyl7 1° 7 2° 7 3° SN1SN2 BRUI10-359_399r2 18-02-2003 2:50 PM Page 385 386 C H A P T E R 1 0 Substitution Reactions of Alkyl Halides Table 10.6 Summary of the Reactivity of Alkyl Halides in Nucleophilic Substitution Reactions Methyl and 1° alkyl halides only Vinylic and aryl halides Neither nor 2° alkyl halides and 1° and 2° benzylic and 1° and 2° allylic halides and 3° alkyl halides only 3° benzylic and 3° allylic halides onlySN1 SN1 SN2SN1 SN2SN1 SN2SN1 SN2 When an alkyl halide can undergo both an reaction and an reaction, both reactions take place simultaneously. The conditions under which the reaction is carried out determine which of the reactions predominates. Therefore, we have some experi- mental control over which reaction takes place. When an alkyl halide can undergo substitution by both mechanisms, what condi- tions favor an reaction? What conditions favor an reaction? These are important questions to synthetic chemists because an reaction forms a single sub- stitution product, whereas an reaction can form two substitution products if the leaving group is bonded to an asymmetric carbon. An reaction is further compli- cated by carbocation rearrangements. In other words, an reaction is a synthetic chemist’s friend, but an reaction can be a synthetic chemist’s nightmare. When the structure of the alkyl halide allows it to undergo both and reactions, three conditions determine which reaction will predominate: (1) the concentration of the nucleophile, (2) the reactivity of the nucleophile, and (3) the solvent in which the reaction is carried out. To understand how the concentration and the reactivity of the nucleophile affect whether an or an reaction predominates, we must examine the rate laws for the two reactions. The rate constants have been given subscripts that indicate the reaction order. The rate law for the reaction of an alkyl halide that can undergo both and re- actions simultaneously is the sum of the individual rate laws. From the rate law, you can see that increasing the concentration of the nucleophile increases the rate of an reaction but has no effect on the rate of an reaction.SN1SN2 contribution to the rate by an SN2 reaction rate = +k2[alkyl halide][nucleophile] k1[alkyl halide] contribution to the rate by an SN1 reaction SN1SN2 Rate law for an SN2 reaction = k2 [alkyl halide] [nucleophile] Rate law for SN1 reaction = k1 [alkyl halide] SN1SN2 SN1SN2 SN1 SN2 SN1 SN1 SN2 SN2SN1 SN2SN1 unimolecular. These numbers do not refer to the number of steps in the mechanism. In fact, just the opposite is true: An reaction proceeds by a one-step concerted mech- anism, and an reaction proceeds by a two-step mechanism with a carbocation intermediate. We have seen that methyl halides and primary alkyl halides undergo only reac- tions because methyl cations and primary carbocations, which would be formed in an reaction, are too unstable to be formed in an reaction. Tertiary alkyl halides undergo only reactions because steric hindrance makes them unreactive in an reaction. Secondary alkyl halides as well as benzylic and allylic halides (unless they are tertiary) can undergo both and reactions because they form relatively stable carbocations and the steric hindrance associated with these alkyl halides is generally not very great. Vinylic and aryl halides do not undergo either or reactions. These results are summarized in Table 10.6. SN2SN1 SN2SN1 SN2SN1 SN2SN1 SN2 SN1 SN2 BRUI10-359_399r2 18-02-2003 2:50 PM Page 386 Reatividade de Compostos Orgânicos – Aula 2 44 Section 10.8 Benzylic Halides, Allylic Halides, Vinylic Halides, and Aryl Halides 383 10.8 Benzylic Halides, Allylic Halides, Vinylic Halides, and Aryl Halides Our discussion of substitution reactions, to this point, has been limited to methyl halides and primary, secondary, and tertiary alkyl halides. But what about benzylic, allylic, vinylic, and aryl halides? Let’s first consider benzylic and allylic halides. Ben- zylic and allylic halides readily undergo reactions unless they are tertiary. Tertiary benzylic and tertiary allylic halides, like other tertiary halides, are unreactive in reactions because of steric hindrance. Benzylic and allylic halides readily undergo reactions because they form relatively stable carbocations. While primary alkyl halides (such as and ) cannot undergo reactions because their carbocations are too unstable, primary benzylic and primary allylic halides readily undergo reactions because their carbocations are stabilized by electron delocalization (Section 7.7). If the resonance contributors of the allylic carbocation intermediate have different groups bonded to their carbons, two substitution products will be obtained. Vinylic halides and aryl halides do not undergo or reactions. They do not undergo reactions because, as the nucleophile approaches the back side of the carbon, it is repelled by the electron cloud of the double bond or the aromatic ring. a vinylic halide an aryl halide nucleophile C C Cl HH R Br nucleophile a nucleophile is repelled by the pi electron cloud p sp2SN2 SN1SN2 sp2 SN1 SN1CH3CH2CH2Br CH3CH2Br SN1 benzyl chloride 1-bromo-2-butene an allylic halide 2-buten-1-ol benzyl methyl ether CH2Cl CH3O−+ CH3CH CHCH2Br HO−+ CH3CH CHCH2OH Br−+ CH2OCH3 Cl−+ SN2 conditions SN2 conditions SN2 SN2 CH3OHSN1CH2Cl CH2 Cl−+ CH2OCH3 H++ SN1 CH2CH2CH Br−+ H2O CHCH2OHCH2 H++CHCH2BrCH2 CHCH2CH2 + + + + + CH3CH CHCH2Br CH3CH CHCH2 + CH3CHCH CH2 CH3CHCH CH2CH3CH CHCH2OH H+ + Br− H2O + H+OH H2O SN1 Benzylic and allylic halides undergo and reactions.SN2SN1 Vinylic and aryl halides undergo neither nor reactions.SN2SN1 3-D Molecule: Benzyl cation 3-D Molecule: Allyl cation BRUI10-359_399r2 18-02-2003 2:50 PM Page 383 Section 10.8 Benzylic Halides, Allylic Halides, Vinylic Halides, and Aryl Halides 383 10.8 Benzylic Halides, Allylic Halides, Vinylic Halides, and Aryl Halides Our discussion of substitution reactions, to this point, has been limited to methyl halides and primary, secondary, and tertiary alkyl halides. But what about benzylic, allylic, vinylic, and aryl halides? Let’s first consider benzylic and allylic halides. Ben- zylic and allylic halides readily undergo reactions unless they are tertiary. Tertiary benzylic and tertiary allylic halides, like other tertiary halides, are unreactive in reactions because of steric hindrance. Benzylic and allylic halides readily undergo reactions because they form relatively stable carbocations. While primary alkyl halides (such as and ) cannot undergo reactions because their carbocations are too unstable, primary benzylic and primary allylic halides readily undergo reactions because their carbocations are stabilized by electron delocalization (Section 7.7). If the resonance contributors of the allylic carbocation intermediate have different groups bonded to their carbons, two substitution products will be obtained. Vinylic halides and aryl halides do not undergo or reactions. They do not undergo reactions because,as the nucleophile approaches the back side of the carbon, it is repelled by the electron cloud of the double bond or the aromatic ring. a vinylic halide an aryl halide nucleophile C C Cl HH R Br nucleophile a nucleophile is repelled by the pi electron cloud p sp2SN2 SN1SN2 sp2 SN1 SN1CH3CH2CH2Br CH3CH2Br SN1 benzyl chloride 1-bromo-2-butene an allylic halide 2-buten-1-ol benzyl methyl ether CH2Cl CH3O−+ CH3CH CHCH2Br HO−+ CH3CH CHCH2OH Br−+ CH2OCH3 Cl−+ SN2 conditions SN2 conditions SN2 SN2 CH3OHSN1CH2Cl CH2 Cl−+ CH2OCH3 H++ SN1 CH2CH2CH Br−+ H2O CHCH2OHCH2 H++CHCH2BrCH2 CHCH2CH2 + + + + + CH3CH CHCH2Br CH3CH CHCH2 + CH3CHCH CH2 CH3CHCH CH2CH3CH CHCH2OH H+ + Br− H2O + H+OH H2O SN1 Benzylic and allylic halides undergo and reactions.SN2SN1 Vinylic and aryl halides undergo neither nor reactions.SN2SN1 3-D Molecule: Benzyl cation 3-D Molecule: Allyl cation BRUI10-359_399r2 18-02-2003 2:50 PM Page 383 Section 10.8 Benzylic Halides, Allylic Halides, Vinylic Halides, and Aryl Halides 383 10.8 Benzylic Halides, Allylic Halides, Vinylic Halides, and Aryl Halides Our discussion of substitution reactions, to this point, has been limited to methyl halides and primary, secondary, and tertiary alkyl halides. But what about benzylic, allylic, vinylic, and aryl halides? Let’s first consider benzylic and allylic halides. Ben- zylic and allylic halides readily undergo reactions unless they are tertiary. Tertiary benzylic and tertiary allylic halides, like other tertiary halides, are unreactive in reactions because of steric hindrance. Benzylic and allylic halides readily undergo reactions because they form relatively stable carbocations. While primary alkyl halides (such as and ) cannot undergo reactions because their carbocations are too unstable, primary benzylic and primary allylic halides readily undergo reactions because their carbocations are stabilized by electron delocalization (Section 7.7). If the resonance contributors of the allylic carbocation intermediate have different groups bonded to their carbons, two substitution products will be obtained. Vinylic halides and aryl halides do not undergo or reactions. They do not undergo reactions because, as the nucleophile approaches the back side of the carbon, it is repelled by the electron cloud of the double bond or the aromatic ring. a vinylic halide an aryl halide nucleophile C C Cl HH R Br nucleophile a nucleophile is repelled by the pi electron cloud p sp2SN2 SN1SN2 sp2 SN1 SN1CH3CH2CH2Br CH3CH2Br SN1 benzyl chloride 1-bromo-2-butene an allylic halide 2-buten-1-ol benzyl methyl ether CH2Cl CH3O−+ CH3CH CHCH2Br HO−+ CH3CH CHCH2OH Br−+ CH2OCH3 Cl−+ SN2 conditions SN2 conditions SN2 SN2 CH3OHSN1CH2Cl CH2 Cl−+ CH2OCH3 H++ SN1 CH2CH2CH Br−+ H2O CHCH2OHCH2 H++CHCH2BrCH2 CHCH2CH2 + + + + + CH3CH CHCH2Br CH3CH CHCH2 + CH3CHCH CH2 CH3CHCH CH2CH3CH CHCH2OH H+ + Br− H2O + H+OH H2O SN1 Benzylic and allylic halides undergo and reactions.SN2SN1 Vinylic and aryl halides undergo neither nor reactions.SN2SN1 3-D Molecule: Benzyl cation 3-D Molecule: Allyl cation BRUI10-359_399r2 18-02-2003 2:50 PM Page 383 Reatividade de Compostos Orgânicos – Aula 2 45 Recall that protic solvents contain a hydrogen bonded to an oxygen or a nitrogen, so protic solvents are hydrogen bond donors. Aprotic solvents, on the other hand, do not have a hydrogen bonded to an oxygen or a nitrogen, so they are not hydrogen bond donors. Section 10.10 The Role of the Solvent in and Reactions 389SN1SN2 What percentage of the reaction takes place by the mechanism when these conditions are met? a. b. SOLUTION TO 26a 10.10 The Role of the Solvent in and Reactions The solvent in which a nucleophilic substitution reaction is carried out also influences whether an or an reaction will predominate. Before we can understand how a particular solvent favors one reaction over another, however, we must understand how solvents stabilize organic molecules. The dielectric constant of a solvent is a measure of how well the solvent can insu- late opposite charges from one another. Solvent molecules insulate charges by cluster- ing around a charge, so that the positive poles of the solvent molecules surround negative charges while the negative poles of the solvent molecules surround positive charges. Recall that the interaction between a solvent and an ion or a molecule dis- solved in that solvent is called solvation (Section 2.9). When an ion interacts with a polar solvent, the charge is no longer localized solely on the ion, but is spread out to the surrounding solvent molecules. Spreading out the charge stabilizes the charged species. Polar solvents have high dielectric constants and thus are very good at insulating (solvating) charges. Nonpolar solvents have low dielectric constants and are poor insulators. The dielectric constants of some common solvents are listed in Table 10.7. In this table, solvents are divided into two groups: protic solvents and aprotic solvents. Recall that protic solvents contain a hydrogen bonded to an oxygen or a nitrogen, so protic solvents are hydrogen bond donors. Aprotic solvents, on the other hand, do not have a hydrogen bonded to an oxygen or a nitrogen, so they are not hydrogen bond donors. δ−δ− δ− δ− δ+ δ+ δ+ δ+ δ− δ+ δ+ δ+ δ+ H H OY− Y+ H H O δ− δ− δ− δ+ δ+ δ+ δ+ H H HH δ+ δ+HH O O O H H O δ− δ+ δ+ H H O ion–dipole interactions between a negatively charged species and water ion–dipole interactions between a positively charged species and water H HO SN1SN2 SN1 SN2 = 96% = 3.20 * 10 -5 3.20 * 10-5 + 0.15 * 10-5 * 100 = 3.20 * 10 -5 3.35 * 10-5 * 100 = 3.20 * 10-5[2-bromobutane]11 # 002 * 100 3.20 * 10-5[2-bromobutane]11 # 002 + 1.5 * 10-6[2-bromobutane] percentage by SN2 = SN2 SN2 + SN1 * 100 [HO-] = 0.001 M[HO-] = 1.00 M SN2 AU: OK as changed? BRUI10-359_399r2 18-02-2003 2:50 PM Page 389
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