Presentation on theme: "Chapter 6 Ionic Reactions--- Nucleophilic substitution and elimination reactions of alkylhalides ( 卤代烃的亲核取代反 应和消除反应） Because halogen atoms are more electronegative."— Presentation transcript:
Chapter 6 Ionic Reactions--- Nucleophilic substitution and elimination reactions of alkylhalides ( 卤代烃的亲核取代反 应和消除反应） Because halogen atoms are more electronegative than carbon, the carbon-halogen bond of alkyl halides is polarized; the carbon atom bears a partial positive charge, the halogen atom a partial negative charge.
Table 6.1 Carbon-halogen bond lengths
Vinyl halides ( 卤代乙烯） or phenyl halides （卤代苯基）
6.2 Physical properties of organic halides Very low solubilities in water They are miscible with each other and with other relatively nonpolar solvents CH 2 Cl, CHCl 3 and CCl 4 are often used as solvents for nonpolar CHCl 3 and CCl 4 have a cumulative toxicity and are carcinogenic. Polyfluoroalkanes have low boiling points, (Hexafluoroethane boils at –79 o ) ( 致癌物 ( 质 ) 的 )
6.3 Reaction Mechanisms Mechanism of the reaction—The events that are postulated to take place at the molecular level as reactants become products If the reaction takes place in more than one step, then what are these steps, and what kinds of intermediates intervene between reactants and products?
6.3A Homolysis and heterolysis of covalent bonds ( 共价键的均裂和异 裂） Covent bond may break in three possible ways:
6.3B Reactive intermediates in organic chemistry Organic reactions that take place in more than one step involve the formation of an intermediate----one that results from either homolysis or heterolysis of a bond. Homolysis of a bond to carbon leads to an intermediate known as a carbon radical (free radical)
Heterolysis of a bond can lead either to a trivalent carbon cation or carbon anion
6.3C Ionic reactions ( 离子反应） and radical reactions （自由基反应 or 游离基反应） In ionic reactions the bonds of the reacting molecules undergo heterolysis; In radical reactions, they undergo homolysis (in detail in chapter 7) In this chapter we concern ourselves only with ionic reactions.
Nucleophilic substitution reactions （亲核取代反应） The carbon-halogen bond of the substrate undergoes heterolysis, and the unshared pair of the nucleophile is used to form a new bond to the carbon atom
6.5 Nucleophiles [nju:kliefail] n.[ 化 ] 亲核试剂 A nucleophile is a reagent that seeks a positive center
6.5 A Leaving groups ( 离去基 团） To be a good leaving group the substituent must be able to leave as a relatively stable, weakly basic molecule or ion
6.7 Kinetics of a nucleophilic substitution reaction: An S N 2 reaction
6.8 A mechanism for the S N 2 reaction （ S N 2 的反应机理）
S N 2 Reaction 伯卤代烃一般按 S N 2 历程进行.
6.9 Transition state theory: Free- energy diagrams (S N 2)
Fig 6.6 A potential energy diagram for the reaction of methyl chloride with hydroxide ion at 60 o C
Fig 6.3 A free-energy diagram for a hypothetical reaction with a positive free- energy change
6.10 The stereochemistry of S N 2 reactions In S N 2 reaction the nucleophile attacks from the backside, that is, from the side directly opposite the leaving group. This mode of attack causes a change in the configuration of the carbon atom that is the object of nucleophilic attack
S N 2 reaction------a configuration inversion
S N 2 reactions always lead to inversion of configuration
6.11 The reaction of tert-butyl chloride with hydroxide ion: An S N 1 reaction
6.12 A mechanism for the S N 1 reaction (multistep reactions)
Fig 6.8 A free-energy diagram for the S N 1 reaction of tert-butyl chloride with water
The important transition state for the S N 1 reaction is the transition state of the rate-determining step. In it the carbon-chlorine bond of tert-butyl chloride is largely broken and ions are beginning to develop. The solvent (water) stabilizes these developing ions by solvation
6.13 Carbocations ( 碳阳离子 ) 6.13A The structure of carbocations The central carbon atom in a carbocation is electron deficient; it has only six electrons in its outside energy level.
6.13B The relative stabilities of carbocations Tertiary carbocations are the most stable, secondary carbocations are the stable, and the primary carbocations are not stable.
Fig 6.10 How a methyl group helps stabilize the positive charge of a carbocation.
Because of sigama-p conjugating. As a result, the delocalization of charge and the order of stability of the carbocations as follows
6.14 The stereochemistry of S N 1 reactions
6.14A Reactions that involve racemization ( 涉及外消旋化反应）
6.14B Solvolysis ( 溶剂化作 用） The S N 1 reaction of an alkyl halide with water is an example of solvolysis Examples of Solvolysis
In the last example the solvent is formic acid ( 甲酸，蚁酸 HCOOH) and the following take place
6.15 Factors affecting the rates of S N 1 and S N 2 reactions( 影响 S N 1 and S N 2 的反应速率因素） 1. The structure of the substrate 2. The concentration and reactivity of the nucleophile (for bimolecular reactions only) The effect of the solvent. The nature of the leaving group.
6.15A The effect of the structure of the substarate ( 底物 结构的影响） General order of reactivity in S N 2 reaction
The important factor behind this order of reactivity is a steric effect ( 立体效应对 S N 2 反应影响大）
S N 1 reaction. The primary factor that determines the reactivity of organic substrates in an S N 1 reaction is the relative stability of the carbocation that is formed
S N 1 reaction
An S N 1 Mechanism For a methyl, primary, or secondary halide to react by an S N 1 mechanism it would have to ionize to form a methyl, primary, or secondary carbocation. These carbocations, however, are much higher in energy than a tertiary carbocation, and the transition states leading to these carbocations are higher in energy. The activation energy for an S N 1 reaction of a simple methyl, primary or secondary halide, consequently, is so large (the reaction is so slow).
6.15B The effect of the concentration and strength of the nucleophile ( 亲核试剂浓度和强度效应） Since the nucleophile does not participate in the rate-determining step of an S N 1 reaction, the rates of S N 1 reactions are unaffected by either the concentration or the identity of the nucleophile. The rates of S N 2 reactions, however, depend on both the concentration and the identity of the attacking nucleophile. We saw that how increasing the concentration of the nucleophile increases the rate of an S N 2 reaction.
We describe nucleophiles as being strong or weak. When we do this we are really describing their relative reactives in S N 2 reactions. A strong nucleophile is one that reacts rapidly with a given substrate. A weak nucleophile is one that reacts slowly with the same substrate under the same reaction conditions.
The relative strengths of nucleophiles can be correlated with two structural features; 1. A negatively charged nucleophile is always a stronger nucleophile than its conjugate acid in a S N 2 reaction.
2. In a group of nucleophiles in which the nucleophilic atom is the same, nucleophilicities parallel basicities This is also their order of basicity. An alkoxide ion (RO - ) is a slightly stronger base than a hydroxide ion (HO - ), a hydroxide ion is a much stronger base than a carboxylate ion (RCOO - ), and so on
6.15C Solvent effects on S N 2 reactions: Polar protic and aprotic solvents ( 极性质子溶剂和非质子溶 剂 ） Protic solvent-----has a hydrogen atom attached to an atom of a strongly electronegative element (oxygen). ( water, alcohol etc.) Aprotic solvent------Aprotic solvents are those solvents whose molecules do not have a hydrogen atom that is attached to an atom of a strongly electronegative element.(Benzene, the alkanes, etc.)
Relative Nucleophilicity in protic Solvents Why?
Because molecules of protic solvents can form hydrogen bonds to nucleophiles in the following way:
Polar Aprotic Solvents （极性 非质子溶剂） Aprotic solvents are those solvents whose molecules do not have a hydrogen atom that is attached to an atom of a strongly electronegative element.
Problem 6.6 Classify the following solvents as being protic or aprotic: Formic acid HCOOH Acetone CH 3 COCH 3 Acetonitrile CH 3 CN Formamide HCONH 2 Sulfur dioxide SO 2 Ammonia NH 3 Trimethylamine N(CH 3 ) 2 Ethylene glycol HOCH 2 CH 2 OH
Problem 6.7 Would you expect the reaction of propyl bromide with sodium cyanide (NaCN), that is, to occur faster in DMF or in ethanol? Explain your answer.
Explain; the reaction is an S N 2 reaction. the nucleophile (CN - ) will be relatively unencumbered ( 不受妨碍的 ) by solvent molecules, therefore, it will be more reactive than in ethanol.
6.15D Solvent effects on S N 1 reactions. The ionizing ability of the solvent Because of its ability to solvate cations and anions so effectively, the use of a polar protic solvent will greatly increase the rate of ionization of an alkyl halide in any S N 1 reaction.
S N 1 reaction in polar protic solvent Water is the most effective solvent for promoting ionization, but most organic compounds do not dissolve appreciably in water. They usually dissolve, however, in alcohols, and quite often mixed solvents (methanol-water) are used.
Table 6.5 Dielectric constants of common solvents ( 普通溶剂的介电常数）
6.15E The nature of the leaving group ( 离去基团的本性） The Best leaving groups are those that become the most stable ions after they depart. Since most leaving groups leave as a negative ion, the best leaving groups are those ions that stabilize a negative charge most effectively. Because weak bases do this best, the best leaving groups are weak bases.
In either an S N 1 or S N 2 reaction the leaving group begins to acquire a negative charge as the transition state is reached
An iodide ion is the best leaving group and a fluoride ion is the poorest:
Other weak bases that are good leaving groups
The trifluoromethanesulfonate ion (CF 3 SO 3 -, commonyl called the triflate ion ( 三氟甲磺酸离子） CF 3 SO 3 - Trifate ion (a ‘ super’ leaving group) ROH (OH - rarely act as leaving groups).
Very powerful bases such as hydride ions (H - ) and alkanide ions (R - ) never act as leaving groups.
6.15F Summary: S N 1 versus S N 2
6.16 Organic synthesis 有机合 成 : Functional group transformations using S N 2 reactions The process of making one compound from another is called synthesis. S N 2 reactions are highly useful in organic synthesis because they enable us to convert one functional group into another-- - a process that is called a functionalgroup transformation or a functional group interconversion （官能团转换）
S N 2 reactions in organic synthesis
Alkyl chlorides and bromides are also easily converted to alkyl iodides by nucleophilic substitution reactions.
If we have available ®-2- bromobutane, we can carry out the following synthesis;
Problem 6.11 Starting with ( S)- 2-brombutane, outline syntheses of each of the following compounds:
6.16A The unreactivity of vinylic ( 乙烯式） and phenyl halides （卤 代苯） A halogen atom attached to one carbon atom of a double bond are called vinylic halides; those that have a halogen atom attached to a benzene ring are called phenyl halides
6.17 Elimination reactions of alkylhalides ( 卤代烃的消除反应）） Another characteristic of alkyl halides is that they undergo elimination reactions. In an elimination reaction the fragments of some molecule (YZ) are removed (eliminated) from adjacent atoms of the reactant. This elimination leads to the introduction of a multiple bond ( 重键） :
6.17A Dehydrohalogenation ( 脱卤化 氢） A widely used method for synthesizing alkenes is the elimination of HX from adjacent atoms of an alkyl halide. Heating the alkyl halide with a strong base causes the reaction to take place.
Dehydrohalogenation ( 脱卤化 氢）
6.17B Bases used in dehydrohalogenation Various strong bases have been used for dehydrohalogenations.
6.17C Mechanisms of Dehydrohalogenations ( 脱卤化氢机理） One mechanism is a bimolecular mechanism called the E2 reaction; the other is a unimolecular mechanism called the E1 reaction
6.18 The E2 reaction
The E2 reaction
6.19 The E1 reaction
The E1 reaction mechanism
6.20 Substitution (S N 1 and S N 2) versus Elimination (E2 and E1) Since eliminations occur best by an E2 path when carried out with a high concentration of a strong base ( and thus a high concentration of a strong nucleophile), substitution reactions by an S N 2 path often compete with the elimination reaction. When the nucleophile attacks the carbon atom bearing the leaving group, substitution results. S N 2 compete with E2, S N 1 compete with E1.
The S N 2 compete with E2
Sample problem; In each case give the mechanism (S N 1, S N 2, E1, or E2)