Lecture Notes-Chapter 6: Alkyl Halides

CHAPTER 6: Alkyl Halides Nucleophilic Substitution and Elimination Learning Outcomes (1) Name alkyl halides, explain their physical properties, and describe their common uses. (2) Predict the products of substitution and elimination reactions, and explain wha...

CHAPTER 6: Alkyl Halides Nucleophilic Substitution and Elimination Learning Outcomes (1) Name alkyl halides, explain their physical properties, and describe their common uses. (2) Predict the products of substitution and elimination reactions, and explain what factors favor each type of reaction. (3) Identify the differences between first-order and second-order substitution and elimination reactions, and explain what factors determine the order of the reaction. (4) Given a set of reaction conditions, identify the possible mechanisms, and predict which mechanism(s) and product(s) are most likely. 1  Introduction There are three major classes of halogenated organic compounds: Alkyl halide An alkyl halide has a halogen atom bonded to one of the sp3 carbon atoms of an alkyl group. Vinyl halide A vinyl halide has a halogen atom bonded to one of the sp2 carbon atoms of an alkene. Aryl halide An aryl halide has a halogen atom bonded to one of the sp2 carbon atoms of an aromatic ring. The chemistry of vinyl halides and aryl halides is different from that of alkyl halides because their bonding and hybridization are different. 2  Alky, Vinyl and Aryl halides PROBLEM 6-1 Classify each compound as an alkyl halide, a vinyl halide, or an aryl halide. (a) CH3CHCFCH3 (b) (CH3)3CBr (c) CH3CCl3 3  Reactivity of Alkyl Halides  The carbon-halogen bond in an alkyl halide is polar because halogen atoms are more electronegative than carbon atoms.  The carbon atom has a partial positive charge, making it electrophilic.  A nucleophile can attack this electrophilic carbon, and the halogen atom can leave as a halide ion, taking the bonding pair of electrons with it.  By serving as a leaving group, the halogen can be eliminated from the alkyl halide, or it can be substituted by a wide variety of functional groups. 4  Nomenclature of Alkyl Halides There are two types of nomenclature: IUPAC nomenclature and common names. IUPAC nomenclature  An alkyl halide is named as an alkane with a halo-substituent: Fluorine is fluoro-, chlorine is chloro-, bromine is bromo-, and iodine is iodo-.  For example: 1-chlorobutane or 2-bromopropane. Common Names  Common names are constructed by naming the alkyl group and then the halide, as in isopropyl bromide.  Common names are useful only for simple alkyl halides, such as the following: Special Common Names Some of the halomethanes have common names that are not related to their structures. 5  CH2X2 is called methylene halide  CHX3 is called a haloform  CHX4 is called a carbon tetrahalide. PROBLEM 6-2 Give the structures of the following compounds. (a) methylene iodide (b) carbon tetrabromide (c) 3-bromo-2-methylpentane (d) iodoform (e) 2-bromo-3-ethyl-2-methylhexane (f) isobutyl bromide (g) cis-1-fluoro-3-(fluoromethyl)cyclohexane (h) tert-butyl chloride Classification of Alkyl Halide Alkyl halides are classified according to the nature of the carbon atom bonded to the halogen.  Primary halide: Halogen-bearing carbon is bonded to one carbon atom.  Secondary halide: Halogen-bearing carbon is bonded to two carbon atoms.  Tertiary halide: Halogen-bearing carbon is bonded to three carbon atoms.  Methyl halide: If the halogen is bonded to a methyl group, the compound is a methyl halide. 6  Dihalides There are two types of dihalides: Geminal dihalides and vicinal dihalides.  Geminal dihalide: Two halogen atoms bonded to the same carbon atom.  Vicinal dihalide: Two halogens bonded to adjacent carbon atoms. PROBLEM 6-3 For each of the following compounds, 1. give the IUPAC name. 2. give the common name (if possible). 3. classify the compound as a methyl, primary, secondary, or tertiary halide. (a) (CH3)2CHCH2Cl (b) (CH3)3CBr 7  Common Uses of Alkyl Halides Solvents  Alkyl halides are used primarily as industrial and household solvents.  Carbon tetrachloride (CCl4) was once used for dry cleaning, spot removing, and other domestic cleaning. It is toxic and carcinogenic.  Dry cleaners now use 1,1,1-trichloroethane and other solvents instead.  Methylene chloride (CH2Cl2) is also a good solvent for cleaning and degreasing work.  It was once used to dissolve the caffeine from coffee beans to produce decaffeinated coffee.  Chloroform (CHCl3) is also a good solvent.  However, it is more toxic and carcinogenic than CH2Cl2. Reagents  Many syntheses use alkyl halides as starting materials for making more complex molecules.  The conversion of alkyl halides to organometallic reagents is a particularly important tool for organic synthesis. 8  Anesthetics  Chloroform (CHCl3) produces general anesthesia making a patient unconscious and relaxed.  It was originally used for surgery.  However, it is toxic and carcinogenic and was abandoned in favor of safer anesthetics, such as diethyl ether.  A less toxic halogenated anesthetic is a mixed alkyl halide, CF3CHClBr which goes by the trade name halothane.  Ethyl chloride is often used as a topical anesthetic for minor procedures.  When sprayed on the skin, its evaporation (b.p. 12o C) cools the area and enhances the numbing effect. 9  Freons: Refrigerants and Foaming Agents  The freons (CFCs) are fluorinated haloalkanes that were developed to replace ammonia as a refrigerant gas.  Ammonia is toxic, and leaking refrigerators often killed people who were working or sleeping nearby.  Freon-12®, CF2Cl2 was at one time the most widely used refrigerant.  Low-boiling freons (Freon-11®, CCl3F) were once used as foaming agents that were added to a plastic to vaporize and form a froth that hardens into a plastic foam.  The release of freons into the atmosphere has raised concerns about their reactions with the earth’s protective ozone layer.  CFCs gradually diffuse up into the stratosphere, where the chlorine atoms catalyze the decomposition of ozone (O3) into oxygen (O2).  Freon-12 has been replaced in aerosol cans by low-boiling hydrocarbons or carbon dioxide.  In refrigerators and automotive air conditioners, Freon-12 has been replaced by Freon-22®, CHClF2. 10  Freons with C–H bonds (such as Freon-22), called HCFCs, are generally destroyed at lower altitudes before they reach the stratosphere.  Propane, CO2 and HCFC-123 (CHCl2CF3) are used as substitutes for Freon-11 in making plastic foams. Pesticides  Alkyl halides have contributed to human health through their use as insecticides.  Since antiquity, people have died from famine and disease caused or carried by mosquitoes, fleas, lice, and other vermin.  The “black death” of the Middle Ages wiped out nearly a third of the population of Europe through infection by the flea-borne bubonic plague.  Whole regions of Africa and tropical America were uninhabited and unexplored because people could not survive insect-borne diseases such as malaria, yellow fever, and sleeping sickness.  Arsenic compounds, nicotine, and other crude insecticides were developed in the 19th century, but these compounds are just as toxic to birds, animals, and people as they are to insects.  Their use is extremely hazardous, but a hazardous insecticide was still preferable to certain death by disease or starvation.  The war against insects changed dramatically in 1939 with the discovery of DDT. DDT is extremely toxic to insects, but its toxicity in mammals is quite low. 11  About an ounce of DDT is required to kill a person, but that same amount of insecticide protects an acre of land against locusts or mosquitoes.  In 1970, the U.S. National Academy of Sciences reported, “in little more than two decades DDT has prevented 500 million deaths due to malaria.”  Similar advances were made against the mosquitoes carrying yellow fever and the tsetse flies carrying sleeping sickness.  Using DDT as a body dust protected people against louse-borne typhus, and dusting rodent burrows controlled the threat of plague.  As with many inventions, DDT showed undesired side effects.  It is a long-lasting insecticide, and its residues accumulate in the environment.  The widespread use of DDT as an agricultural insecticide led to the development of substantial DDT concentrations in wildlife, causing declines in several species.  In 1972, DDT was banned by the U.S. Environmental Protection Agency for use as an agricultural insecticide. 12  It is still used, however, in places where insect-borne diseases threaten human life.  DDT-treated bed netting is still the most cost-effective protection against malaria, and careful spraying of DDT around dwellings and in rodent burrows has helped to control the spread of deadly diseases.  Many other chlorinated insecticides have been developed.  Some of them also accumulate in the environment, gradually producing toxic effects in wildlife.  Others can be used with little adverse impact if they are applied properly. Other uses of Chlorinated Insecticides  Because of their persistent toxic effects, chlorinated insecticides are rarely used in agriculture.  They are generally used when a potent insecticide is needed to protect life or property.  For example, lindane is used in shampoos to kill lice, and chlordane is used to protect wooden buildings from termites. PROBLEM 6-4 13 Kepone and chlordane are synthesized from hexachlorocyclopentadiene and other five-membered-ring compounds. Show how these two pesticides are composed of two five-membered rings. Structure of Alkyl Halides  The halogen atom is bonded to an sp3 carbon atom.  The halogen is more electronegative than carbon.  The C-X bond is polarized with a partial positive charge on carbon and a partial negative charge on the halogen.  The electronegativities of the halogens increase in the order I < Br < Cl < F Electronegativity: 2.7 3.0 3.2 4.0  The carbon-halogen bond lengths increase as the halogen atoms become bigger in the order: 14  C-F < C-Cl < C-Br < C-I Bond length: 1.38 Å 1.78 Å 1.94 Å 2.14 Å  These two effects oppose each other, with the larger halogens having longer bonds but weaker electronegativities.  The overall result is that the bond dipole moments increase in the order: C-I < C-Br < C-F < C-Cl Dipole moment, : 1.29 D 1.48 D 1.51 D 1.56 D Molecular Dipole Moment  The dipole moment of a molecule is the vector sum of the individual bond dipole moments.  It depends on the molecular geometry that vary with the specific molecule.  The four symmetrically oriented polar bonds of the carbon tetrachloride cancel to give a molecular dipole moment of zero. Experimental Dipole Moments of Methyl Halides 15 PROBLEM 6-5 For each pair of compounds, predict which one has the higher molecular dipole moment, and explain your reasoning. (a) Ethyl chloride or ethyl iodide (b) 1-Bromopropane or cyclopropane (c) cis-2,3-Dibromobut-2-ene or trans-2,3-dibromobut-2-ene (d) cis-1,2-Dichlorocyclobutane or trans-1,3-dichlorocyclobutane Boiling Points of Alkyl Halides Two types of intermolecular forces influence the boiling points of alkyl halides: (1) The London dispersion force and (2) the dipole- dipole attraction. The London Dispersion Force  The London force is the strongest intermolecular attraction in alkyl halides:  Molecules with larger surface areas have larger London attractions, resulting in higher boiling points. 16  The surface areas of the alkyl halides vary with the surface areas of halogens. Linear vs. Branched Alkyl Halides  Branched compounds are more spherical and have smaller surface areas resulting in lower intermolecular attractions.  As a result, they have lower boiling points.  Linear compounds, on the other hand have larger surface areas, higher intermolecular attractions.  They have higher boiling point 17  For example, n-butyl bromide has a boiling point of 102 °C, while the more spherical tert-butyl bromide has a boiling point of only 73 °C. Dipole-Dipole Attractions Dipole-dipole attractions also affect the boiling points, but to a smaller extent. 18 PROBLEM 6-6 For each pair of compounds, predict which compound has the higher boiling point. Explain why that compound has the higher boiling point. (a) Isopropyl bromide and n-butyl bromide (b) Isopropyl chloride and tert-butyl bromide (c) 1-Bromobutane and 1-chlorobutane 19  Densities  Alkyl fluorides and alkyl chlorides are less dense than water.  Alkyl chlorides with two or more chlorine atoms are denser than water.  Alkyl bromides and alkyl iodides are denser than water. PROBLEM 6-7 (a) When water is shaken with hexane, the two liquids separate into two phases. Which compound is present in the top phase, and which is present in the bottom phase? (b) When water is shaken with chloroform, a similar two-phase system results. Again, which compound is present in each phase? (c) Explain the difference in the two experiments. What do you expect to happen when water is shaken with ethanol (CH3CH2OH)? Preparation of Alkyl Halides Free-Radical Halogenation  Free-radical halogenation is not an effective method for the synthesis of alkyl halides.  It usually produces mixtures of products because there are different kinds of hydrogen atoms that can be abstracted.  Also, more than one halogen atom may react, giving multiple substitutions. 20  Industrial Synthesis In industry, free-radical halogenation is sometimes useful because:  the reagents are cheap,  the mixture of products can be separated by distillation, and  each of the individual products is sold separately. Laboratory Synthesis Free-radical halogenation are generally limited to specialized compounds that give a single major product, such as the following examples. Free-radical Chlorination of Cyclohexane  All the hydrogen atoms in cyclohexane are equivalent.  Free-radical chlorination gives a usable yield of chlorocyclohexane.  Formation of dichlorides and trichlorides is possible.  These side reactions are controlled by using only a small amount of chlorine and an excess of cyclohexane. 21  Free-radical Bromination Free-radical bromination is highly selective, and it gives good yields of products that have one type of hydrogen atom that is more reactive than the others. Bromination of Isobutane Isobutane has only one 3o hydrogen atom, and this atom is preferentially abstracted to give a 3o free radical. Allylic Bromination  Free-radical bromination of alkenes can be carried out in a highly selective manner.  Allylic intermediates are stabilized by resonance with the double bond, allowing the charge or radical to be delocalized. 22  Less energy is required to form a resonance-stabilized 1o allylic radical than a typical 2o radical. Bromination of Cyclohexene  Bromination is highly selective, with only the most stable radical being formed.  If there is an allylic hydrogen, the allylic radical is usually the most stable of the radicals that might be formed.  In the free-radical bromination of cyclohexene, the bromine atom substitutes an allylic hydrogen on the carbon atom next to the double bond. 23  Mechanism of Bromination  A bromine radical abstracts an allylic hydrogen atom to give a resonance-stabilized allylic radical.  This radical reacts with a bromine molecule regenerating a bromine radical that continues the chain reaction.  The general mechanism for allylic bromination shows that either end of the resonance-stabilized allylic radical can react with bromine to give products.  In one of the products, the bromine atom appears in the same position where the hydrogen atom was abstracted.  The other product results from reaction at the carbon atom that bears the radical in the second resonance form of the allylic radical. 24  This second compound is said to be the product of an allylic shift. Bromination with N- bromosuccinamide  For efficient allylic bromination, a large concentration of bromine must be avoided because bromine can also add to the double bond.  N-Bromosuccinimide (NBS) is often used as the bromine source in free-radical brominations.  Because it combines with the HBr side product to regenerate a constant low concentration of bromine.  No additional bromine is needed because most samples of NBS contains traces of Br2 to initiate the reaction.  NBS also works well for brominating benzylic positions, next to an aromatic ring. 25 PROBLEM 6-8 Propose a mechanism for the following reaction: PROBLEM 6-9 The light-initiated reaction of 2,3-dimethylbut-2-ene with N- bromosuccinimide (NBS) gives two products: (a) Give a mechanism for this reaction, showing how the two products arise as a consequence of the resonance-stabilized intermediate. (b) The bromination of cyclohexene using NBS gives only one major product. Explain why there is no second product from an allylic shift. PROBLEM 6-10 Show how free-radical halogenation might be used to synthesize the following compounds. In each case, explain why you expect to get a single major product. 26 (a) 1-chloro-2,2-dimethylpropane (b) 2-bromo-2-methylbutane Summary: Methods of making Alkyl Halides 27 28  Reactions of Alkyl Halides There are two major types of reactions of alkyl halide: substitution and elimination are Substitution  A nucleophile replaces the halogen atom.  The halogen atom leaves with its bonding pair of electrons to form a stable halide ion.  Halide is called a leaving group. 29  Elimination  The halide ion leaves with another atom or ion and forms a new  bond.  A molecule of HX is lost from the alkyl halide to give an alkene.  This reaction is called dehydrohalogenations because a hydrogen halide (HX) has been removed from the alkyl halide. Nucleophilic Substitution A nucleophile replaces a leaving group from a carbon atom, using its lone pair of electrons to form a new bond to the carbon atom. Elimination  In an elimination, both the halide ion and another substituent are lost.  A new bond () is formed. 30  The reagent reacts as a base (B:-), abstracting a proton from the alkyl halide.  Most nucleophiles are also basic and can engage in either substitution or elimination.  This depends on the alkyl halide and the reaction conditions. PROBLEM 6-11 Classify each reaction as a substitution, elimination, or neither. Identify the leaving group in each reaction, and the nucleophile in substitutions. PROBLEM 6-12 Give the structures of the substitution products expected when 1- bromohexane reacts with: 31  (a) NaOCH2CH3 (b) KCN (c) NaOH The SN2 Reaction  The abbreviation SN2 stands for substitution, nucleophilic, bimolecular.  An SN2 reaction has the following general form where Nuc:- is the nucleophile and X- is the leaving group.  Reaction of iodomethane with hydroxide ion. The product is methanol. 32  Hydroxide ion is a strong nucleophile because the oxygen atom has unshared pairs of electrons and a negative charge.  Iodomethane is called the substrate, the compound that is attacked by the reagent.  The carbon atom of iodomethane is electrophilic because it is bonded to an electronegative iodine atom.  Electron density is drawn away from carbon by the halogen atom, giving the carbon atom a partial positive charge.  The negative charge of OH- is attracted to this partial positive charge. SN2 Reaction Mechanism  Hydroxide ion attacks the back side of the electrophilic carbon atom, donating a pair of electrons to form a new bond. 33  Curved arrows show the movement of electron pairs, from the electron-rich nucleophile to the electron-poor carbon atom of the electrophile.  Carbon can accommodate only eight electrons in its valence shell, so the C-I bond must begin to break as the C-O bond begins to form.  Iodide ion is the leaving group; it leaves with the pair of electrons. Kinetic Order of the SN2 Reaction  The rate doubles when the concentration of hydroxide ion is doubled.  The rate doubles when the concentration of iodometahne is doubled.  The reaction is therefore first order in each of the reactants and second order overall.  The rate equation has the following form: rate = kr[CH3I][-OH]  This rate equation is consistent with a mechanism that requires a collision between a molecule of CH3I and a OH– ion.  Both of these species are present in the transition state.  And the collision frequency is proportional to both concentrations.  The rate constant kr depends on several factors, including the energy of the transition state and the temperature. The SN2 Reaction-Energy Diagram 34  The SN2 mechanism is a one-step nucleophilic substitution.  The transition state involves the collision of two molecules.  Bimolecular reactions usually are second order overall.  The SN2 reaction is a concerted reaction taking place in a single step with bonds breaking and bond forming at the same time.  The middle structure is a transition state, a point of maximum energy.  In this transition state, the bond to the nucleophile (OH-) is partially formed, and the bond to the leaving group (I-) is partially broken.  The reactants are shown slightly higher in energy than the products because this reaction is known to be exothermic.  The transition state is much higher in energy because it involves a five-coordinate carbon atom with two partial bonds. The SN2 Reaction Mechanism 35  The reaction takes place in a single (concerted) step.  A strong nucleophile attacks the electrophilic carbon, forcing the leaving group to leave.  The order of reactivity for substrates is CH3X > 1° > 2°. (3° alkyl halides cannot react by this mechanism.)  1-bromobutane reacts with sodium methoxide to give 1- methoxybutane. PROBLEM 6-13 (a) Under certain conditions, the reaction of 0.5 M 1-bromobutane with 1.0 M sodium methoxide forms 1-methoxybutane at a rate of 0.05 mol L per second. What would be the rate if 0.1 M 1- bromobutane and 2.0 M NaOCH3 were used? (b) Consider the reaction of 1-bromobutane with a large excess of ammonia (NH3). Draw the reactants, the transition state, and the products. Note that the initial product is the salt of an amine ( RNH3+Br-), which is deprotonated by the excess ammonia to give the amine. 36  Generality of the SN2 Reaction The following table summarizes some of the types of compounds that can be formed by SN2 displacement of alkyl halides. 37  Halogen Exchange Reactions  Alkyl iodides (R-I) and alkyl fluorides (R-F) are more difficult to make than alkyl Chlorides (R-Cl) and bromides (R-Br).  Halides can be converted to other halides by halogen exchange reactions, in which one halide displaces another. Synthesis of Alkyl Iodides Iodide is a good nucleophile, and many alkyl chlorides react with sodium iodide to give alkyl iodides. CH3-CH2-CH2-CH2-Br + NaI → CH3-CH2-CH2-CH2-I + NaBr 1-bromobutane 1-Iodobutane 38  Synthesis of Alkyl Fluorides  Alkyl fluorides are difficult to synthesize directly.  They are often made by treating alkyl chlorides or bromides with KF under conditions that use a crown ether to dissolve the fluoride salt in an aprotic solvent.  These conditions enhance the normally weak nucleophilicity of the fluoride ion. ethyl chloride ethyl fluroide PROBLEM 6-14 Predict the major products of the following substitutions. (a)CH3CH2Br + (CH3)3CCO–K+ (b) HC≡C: –Na+ + CH3CH2CH2CH2Cl (c) (CH3)2CHCH2Br + excess NH3 (d) CH3CH2CH2I + NaCN (e)1-chloropentane + NaI PROBLEM 6-15 Show how you might use SN2 reactions to convert 1-chlorobutane into the following compounds. (a) butan-1-ol (b) 1-iodobutane (c) CH3–(CH2)3–C≡CH (d) CH3–(CH2)3–NH2 (e) 1-fluorobutane (f) CH3–(CH2)3–CN (g) CH3CH2–O–(CH2)3–CH3 39  Factors Affecting SN2 Reactions Strength of the Nucleophile  A stronger nucleophile reacts faster than a weaker nucleophile under the same conditions.  Methanol (CH3OH) and methoxide ion (CH3O-) have easily shared pairs of nonbonding electrons, but CH3O- reacts with electrophiles in the reaction about 1 million times faster than CH3OH.  A species with a negative charge is a stronger nucleophile than a similar, neutral species.  Methoxide ion has nonbonding electrons that are readily available for bonding.  In the transition state, the negative charge is shared by the oxygen of methoxide ion and by the halide leaving group.  Methanol has no negative charge; the transition state has a partial negative charge on the halide but a partial positive charge on the methanol oxygen atom. A conjugate base is always a stronger nucleophile than its parent acid. 40  Basicity Versus Nucleophilicity  Basicity and nucleophilicity are different properties.  Basicity is defined by the equilibrium constant for abstracting a proton.  Nucleophilicity is defined by the rate of attack on an electrophilic carbon.  In both cases, the nucleophile (or base) forms a new bond.  If the new bond is to a proton, it has reacted as a base.  If the new bond is to carbon, it has reacted as a nucleophile.  Most good nucleophiles are also strong bases, and vice versa. 41  Trends in Nucleophilicity 1. A species with a negative charge is a stronger nucleophile than a similar neutral species. In particular, a base is a stronger nucleophile than its conjugate acid. 2. Nucleophilicity decreases from left to right in the periodic table. The more electronegative elements have more tightly held nonbonding electrons that are less reactive toward forming new bonds. 3. Nucleophilicity increases down the periodic table, following the increase in size and polarizability, and the decrease in electronegativity. 42  Polarizibility Effect  Larger atoms have more electrons at a greater distance from the nucleus.  The electrons are more loosely held, and the atom is more polarizable.  Its electrons can move more freely toward a positive charge, resulting in stronger bonding in the transition state.  The increased mobility of its electrons enhances the atom’s ability to begin to form a bond at a relatively long distance. 43  Fluoride ion (F–) has tightly bound electrons that cannot begin to form a C—F bond until the atoms are close together.  Fluoride ion is a “hard” (low polarizability) nucleophile  Iodide ion (I-) has more loosely bound outer electrons that begin bonding earlier in the reaction.  iodide ion is a “soft” (high polarizability) nucleophile. Steric Effects on Nucleophilicity  To serve as a nucleophile, an ion or molecule must get in close to a carbon atom to attack it.  Bulky groups on the nucleophile hinder this close approach, and they slow the reaction rate.  The tert-butoxide ion has three methyl groups that hinder any close approach to a more crowded carbon atom.  Therefore, ethoxide ion is a stronger nucleophile than tert- butoxide ion. 44  When bulky groups interfere with a reaction by virtue of their size, we call the effect steric hindrance. PROBLEM 6-16 For each pair, predict the stronger nucleophile in the reaction (using an alcohol as the solvent). Explain your prediction. (a) (CH3CH2)3N or (CH3CH2)2NH (b) (CH3)2O or (CH3)2S (c) NH3 or PH3 (d) CH3S– or H2S (e) (CH3)3N or (CH3)2O (f) CH3COO– or CF3COO– (g) (CH3)2CHO– or CH3CH2CH2O– (h) I–or Cl– Solvent Effects on Nucleophilicity Protic Solvents  A protic solvent has acidic protons, usually in the form of O-H or N-H groups.  These groups form H-bonds to negatively charged nucleophiles.  Small anions are solvated more strongly than large anions in a protic solvent because the solvent approaches a small anion more closely and forms stronger hydrogen bonds. 45  When an anion reacts as a nucleophile, energy is required to “strip off” some of the solvent molecules, breaking some of the hydrogen bonds that stabilized the solvated anion.  More energy is required to strip off solvent from a small, strongly solvated ion such as fluoride than from a large, diffuse, less strongly solvated ion like iodide.  The enhanced solvation of smaller anions in protic solvents, requiring more energy to strip off their solvent molecules, reduces their nucleophilicity.  The polarizability increases with increasing atomic number, and the solvation energy decreases with increasing atomic number.  Therefore, nucleophilicity generally increases down a column in the periodic table, as long as we compare similar species with similar charges. Aprotic Solvents  Aprotic solvents (solvents without O-H or N-H groups) enhance the nucleophilicity of anions.  An anion is more reactive in an aprotic solvent because it is not so strongly solvated.  There are no hydrogen bonds to be broken when the nucleophile approaches an electrophilic carbon atom. 46  The relatively weak solvating ability of aprotic solvents is also a disadvantage: Most polar, ionic reagents are insoluble in simple aprotic solvents such as alkanes.  Polar aprotic solvents have strong dipole moments to enhance solubility, yet they have no O-H or N-H groups to form hydrogen bonds with anions.  We can add specific solvating reagents to enhance solubility without affecting the reactivity of the nucleophile. Crown Ethers  Crown ethers solvate the cation, so the nucleophilic strength of the anion increases.  Using the potassium salt of a nucleophile and solvating the potassium ions causes the nucleophilic anion to be dragged along into solution.  Fluoride ion, normally a poor nucleophile in protic solvents, can be a good nucleophile in an aprotic solvent.  Although KF is not very soluble in acetonitrile, 18-crown-6 solvates the potassium ions, and the poorly solvated (and therefore nucleophilic) fluoride ion follows. 47  Reactivity of the Substrate in SN2 Reactions To be a good substrate for SN2 attack by a nucleophile:  A molecule must have an electrophilic carbon atom.  The electrophilic carbon atom must not be too sterically hindered.  The molecule must have a good leaving group. Effect of Substituents on the Rates of SN2 Reactions Leaving-Group Effects  A leaving group serves two purposes in the reaction: 48  It polarizes the C-X bond, making the carbon atom electrophilic.  It leaves with the pair of electrons from the electrophilic carbon atom.  To fill these roles, a good leaving group should be:  Electron withdrawing, to polarize the carbon atom.  Stable (not a strong base) once it has left.  Polarizable, to stabilize the transition state. 1. The leaving group must be electron withdrawing to create a partial positive charge on the carbon atom, making the carbon electrophilic.  An electron-withdrawing leaving group also stabilizes the negatively charged transition state.  Halogen atoms are strongly electronegative, so alkyl halides are common substrates for reactions.  Oxygen, nitrogen, and sulfur also form strongly polarized bonds with carbon; given the right substituents, they can form the basis for excellent leaving groups. Strongly polarized 2. The leaving group must be stable once it has left with the pair of electrons that bonded it to carbon.  A stable leaving group is needed for favorable energetics. 49  The leaving group is leaving in the transition state; a reactive leaving group would raise the energy of the transition state, slowing the reaction.  Also, the energy of the leaving group is reflected in the energy of the products.  A reactive leaving group would raise the energy of the products, driving the equilibrium toward the reactants.  Good leaving groups should be weak bases; therefore, they are the conjugate bases of strong acids.  The hydrohalic acids HCl, HBr, and HI are strong, and their conjugates bases (Cl-, Br- and I-) are all weak bases.  Other weak bases, such as sulfate ions, sulfonate ions, and phosphate ions, can also serve as good leaving groups.  Hydroxide ion, alkoxide ions, and other strong bases are poor leaving groups for SN2 reactions.  For example, the O-H group of an alcohol is a poor leaving groubecause it would have to leave as hydroxide ion. These ions are strong bases and poor leaving groups: 50  – – – OH OR NH2 Hydroxide alkoxide amide  Table 6-4 also lists some neutral molecules that can be good leaving groups.  A neutral molecule often serves as the leaving group from a positively charged species.  For example, if an alcohol is placed in an acidic solution, the OH group is protonated.  Water then serves as the leaving group.  Note that the need to protonate the alcohol (requiring acid) limits the choice of nucleophiles to those few that are weak bases, such as bromide and iodide.  A strongly basic nucleophile would become protonated in acid. 51 3. A good leaving group should be polarizable, to maintain partial bonding with the carbon atom in the transition state.  This bonding helps stabilize the transition state and reduce the activation energy.  Polarizable nucleophiles and polarizable leaving groups both stabilize the transition state by engaging in more bonding at a longer distance.  Iodide ion, one of the most polarizable ions, is both a good nucleophile and a good leaving group.  In contrast, fluoride ion is a small, “hard” ion. Fluoride is both a poor nucleophile (in protic solvents) and a poor leaving group in SN2 reactions. PROBLEM 6-17 When diethyl ether (CH3CH2OCH2CH3) is treated with concentrated HBr, the initial products are CH3CH2Br and CH3CH2OH. Propose a mechanism to account for this reaction. Effect of Substituents on the SN2 Reaction Rates  The SN2 reaction goes rapidly with methyl halides and with most 1o substrates.  It is more sluggish with 2o halides.  3o Halides fail to react at all by the SN2 mechanism. Relative rates for SN2: CH3X > 1° > 2° >> 3°  All slow-reacting compounds have the back side of the electrophilic carbon atom crowded by the presence of bulky groups.   3o Halides are more hindered than 2o halides, which are more hindered than 1o halides. 52  Even a bulky 1o halides (neopentyl bromide) undergoes SN2 reaction at a rate similar to that of a 3O halide.  The relative rates show that it is the bulk of the alkyl groups, rather than an electronic effect, that hinders the reactivity of bulky alkyl halides in the SN2 displacement.  When the nucleophile approaches the back side of the electrophilic carbon atom, it must come within bonding distance of the back lobe of the C–X sp3 orbital. 53  If there are two alkyl groups bonded to the carbon atom, this process is difficult.  Three alkyl groups make it impossible.  Just one alkyl group can produce a large amount of steric hindrance if it is unusually bulky, like the tert-butyl group of neopentyl bromide. PROBLEM 6-18 Rank the following compounds in decreasing order of their reactivity toward the reaction with sodium ethoxide (Na+ – OCH2CH3) in ethanol. Methyl chloride Methyl iodide Ethyl chloride Isopropyl bromide tert-Butyl iodide Neopentyl bromide PROBLEM 6-19 For each pair of compounds, state which compound is the better substrate. (a) 2-methyl-1-iodopropane or tert-butyl iodide (b) cyclohexyl bromide or 1-bromo-1-methylcyclohexane (c) 2-bromobutane or isopropyl bromide (d) 1-chloro-2,2-dimethylbutane or 2-chlorobutane (e) 1-iodobutane or 2-iodopropane 54  Stereochemistry of the SN2 Reaction  The SN2 reaction requires attack by a nucleophile on the back side of an electrophilic carbon atom.  A carbon atom can have only four filled bonding orbitals, so the leaving group must leave as the nucleophile bonds to the carbon atom.  The nucleophile’s electrons insert into the back lobe of carbon’s hybrid orbital in its antibonding combination with the orbital of the leaving group.  These electrons in the antibonding MO help to weaken the C–Br bond as bromide leaves.  The transition state shows partial bonding to both the nucleophile and the leaving group. Inversion of Configuration  Back-side attack literally turns the tetrahedron of the carbon atom inside out, like an umbrella caught by the wind.  In the product, the nucleophile assumes a stereochemical position opposite the position the leaving group originally occupied.  We call this result an inversion of configuration at the carbon atom. 55 SN2 Reaction Mechanism: Inversion of Configuration Back-side attack inverts the configuration of the carbon atom. In the case of an asymmetric carbon atom, back-side attack gives the opposite configuration of the carbon atom. Back-Side Attack in the SN2 Reaction  The SN2 reaction takes place through nucleophilic attack on the back lobe of carbon’s sp3 hybrid orbital.  This back-side attack inverts the tetrahedral carbon atom, like a strong wind inverts an umbrella.  Such inversion of configuration of an asymmetric carbon atom is called a Walden inversion. 56  Proof of Inversion of Configuration  When cis-1-bromo-3-methylcyclopentane undergoes SN2 displacement by hydroxide ion, inversion of configuration gives trans-3-methylcyclopentanol. Stereospecific Reaction The SN2 displacement is a stereospecific reaction: one in which different stereoisomers react to give different stereoisomers of the product. PROBLEM 6-20 Draw a perspective structure or a Fischer projection for the products of the following reactions. (a) trans-1-bromo-3-methylcyclopentane + KOH (b) (R)-2-Bromopentane + KCN 57 PROBLEM 6-21 Under appropriate conditions, (S)-1-bromo-1-fluoroethane reacts with sodium methoxide to give pure (S)-1-fluoro-1- methoxyethane. CH3CHBrF + NaOCH3 CH3CHFOCH3 + NaBr (S) (S) (a) Why is bromide rather than fluoride replaced? (b) Draw perspective structures for the starting material, the transition state, and the product. (c) Does the product show retention or inversion of configuration? (d) Is this result consistent with reaction by the mechanism? 58  The SN1 Reaction  When tert-butyl bromide is placed in boiling methanol, methyl tert-butyl ether is produced.  In this reaction the solvent acts as the nucleophile.  Therefore, it is called a solvolysis reaction.  It is a substitution reaction because methoxide has replaced bromide on the tert-butyl group.  This reaction does not go through the SN2 mechanism.  The SN2 requires a strong nucleophile and a substrate that is not too hindered.  CH3OH is a weak nucleophile, and tert-butyl bromide is a hindered 3O alkyl halide—a poor SN2 substrate. 59  Its rate does not depend on the concentration of the nucleophile, methanol.  The rate depends only on the concentration of the substrate, tert-butyl bromide. rate = kr[(CH3)3C–Br]  The reaction is first order in the concentration of the alkyl halide.  The reaction is zeroth order in the concentration of the nucleophile.  The rate of the reaction is first order overall.  Because the rate does not depend on the concentration of the nucleophile, the nucleophile is not present in the transition state of the rate-limiting step.  The nucleophile must react after the slow step.  This type of substitution is called an SN1 reaction, for substitution, nucleophilic, unimolecular.  The term unimolecular means there is only one molecule involved in the transition state of the rate-limiting step.  The mechanism of the SN1 reaction of tert-butyl bromide with methanol is shown here.  Ionization of the alkyl halide (first step) is the rate-limiting step. Mechanism of the SN1 Reaction 60  The SN1 mechanism is a multistep process.  The first step is a slow ionization to form a carbocation.  The second step is a fast attack on the carbocation by a nucleophile.  The carbocation is a strong electrophile, and it reacts very fast with nucleophiles.  If the nucleophile is an uncharged molecule like water or an alcohol, the positively charged product must lose a proton to give the final uncharged product. The SN1 Reaction Mechanism  The SN1 reaction involves a two-step mechanism.  A slow ionization gives a carbocation that reacts quickly with a weak nucleophile.  Order of Reactivity: 3o > 2o > 1o Step 1. Formation of the carbocation (rate-limiting). 61 Step 2. Nucleophilic attack on the carbocation (fast) If the nucleophile is water or an alcohol, a third step is needed to deprotonate the product. Solvolysis of 1-Iodo-1-methylcyclohexane in Methanol Step 1: Formation of a carbocation (slow rate-limiting). Step 2: Nucleophilic attack by the solvent (methanol). 62 Step 3: Deprotonation to form the product. PROBLEM 6-22 Propose an SN1 mechanism for the solvolysis of 3-bromo-2,3- dimethylpentane in ethanol. Energy Diagram for SN1 Reaction  The energy diagram shows why the rate does not depend on the strength or concentration of the nucleophile.  The ionization (first step) is highly endothermic, and its large activation energy determines the overall reaction rate.  The nucleophilic attack (second step) is strongly exothermic, with a lower-energy transition state. 63  In effect, a nucleophile reacts with the carbocation almost as soon as it forms.  The intermediate appears as a relative minimum in the reaction-energy diagram.  Reagents and conditions that favor formation of the carbocation (the slow step) accelerate the SN1 reaction.  Reagents and conditions that hinder its formation retard the reaction.  The SN1 is a two-step mechanism with two transition states (‡1 and ‡2) and a carbocation intermediate. Factors Affecting SN1 Reactions Substituent Effects  The rate-limiting step of the SN1 reaction is ionization to form a carbocation, a strongly endothermic process.  Rates of SN1 reactions depend strongly on carbocation stability. 64  Alkyl groups stabilize carbocations by donating electrons through sigma bonds (inductive effect) and through overlap of filled orbitals with the empty p orbital of the carbocation (hyperconjugation).  Highly substituted carbocations are therefore more stable. Order of Reactivity  Reactivity toward SN1 substitution reaction follows the stability of carbocations: SN1 reactivity: 3° > 2° > 1° > CH3X  This order is opposite that of the SN2 reaction.  Alkyl groups hinder the SN2 by blocking attack of the strong nucleophile, but alkyl groups enhance the SN1 by stabilizing the carbocation intermediate. SN1 Reactions of Allyl Halide  Resonance stabilization of the carbocation can also promote the SN1 reaction.  Allyl bromide is a 1o halide, but it undergoes the SN1 reaction about as fast as a 2o halide. 65  The carbocation formed by ionization is resonance stabilized, with the positive charge spread equally over two carbon atoms. SN1 Reactions of Vinyl and Aryl Halides  Vinyl and aryl halides do not undergo SN1 or SN2 reactions.  An SN1 reaction would require ionization to form a vinyl or aryl cation, either of which is less stable than most alkyl carbocations.  An SN2 reaction would require back-side attack by the nucleophile, which is made impossible by the repulsion of the electrons in the double bond or aromatic ring. Leaving-Group Effects  The leaving group is breaking its bond to carbon in the rate- limiting ionization step of the SN1 mechanism. 66  A highly polarizable leaving group helps stabilize the rate- limiting transition state through partial bonding as it leaves.  The leaving group should be a weak base, very stable after it leaves with the pair of electrons that bonded it to carbon. Transition State of the Ionization Step  The leaving group is taking on a negative charge while it stabilizes the new carbocation through partial bonding.  The leaving group should be stable as it takes on this negative charge, and it should be polarizable to engage in effective partial bonding as it leaves.  A good leaving group is just as necessary in the SN1 reaction as it is in the SN2, and similar leaving groups are effective for either reaction.  In the transition state of the SN1 ionization, the leaving group is taking on a negative charge.  The C—X bond is breaking, and a polarizable leaving group can still maintain substantial overlap. PROBLEM 6-23 Choose the member of each pair that will react faster by the mechanism. (a) 1-bromopropane or 2-bromopropane (b) 2-bromo-2-methylbutane or 2-bromo-3-methylbutane 67 (c) n-propyl bromide or allyl bromide (d) 1-bromo-2,2-dimethylpropane or 2-bromopropane PROBLEM 6-24* 3-Bromocyclohexene is a secondary halide, and benzyl bromide is a primary halide. Both halides undergo substitution about as fast as most tertiary halides. Use resonance structures to explain this enhanced reactivity. Solvent Effects  The SN1 reaction goes much more readily in polar solvents that stabilize ions.  The rate-limiting step forms two ions, and ionization is taking place in the transition state.  Polar solvents solvate these ions by an interaction of the solvent’s dipole moment with the charge of the ion.  Protic solvents such as alcohols and water are even more effective solvents because:  Anions form hydrogen bonds with the –OH hydrogen atom, and cations complex with the nonbonding electrons of the –OH oxygen atom.  Ionization of an alkyl halide requires formation and separation of positive and negative charges. 68  Therefore, SN1 reactions require highly polar solvents that strongly solvate ions.  Measure of a solvent’s ability to solvate ions is its dielectric constant (ε) a measure of the solvent’s polarity.  Note that ionization occurs much faster in highly polar solvents such as water and alcohols.  Although most alkyl halides are not soluble in water, they often dissolve in highly polar mixtures of acetone and alcohols with water. Stereochemistry of SN1 Reaction 69  In contrast to the SN2 reaction, the SN1 reaction is not stereospecific.  The carbocation intermediate is sp2 hybridized and planar and achiral.  A nucleophile can attack the carbocation from either face giving both enantiomers of the product.  Such a process, giving both enantiomers of the product, is called racemization.  If a nucleophile attacks the carbocation from the front side, the product molecule shows retention of configuration.  Attack from the back side gives a product molecule showing inversion of configuration. Racemization in SN1 Reaction 70  Racemization is simply a combination of retention and inversion.  When racemization occurs, the product is rarely completely racemic.  There is often more inversion than retention of configuration.  As the leaving group leaves, it partially blocks the front side of the carbocation.  The back side is unhindered, so attack is more likely there. 71 Proof of Racemization in SN1 Reaction  In the figure below, the bromine atom is cis to the deuterium in the reactant, so the nucleophile is cis to the deuterium in the retention product.  The nucleophile is trans to the deuterium in the inversion product.  The product mixture contains both cis and trans isomers.  The trans isomer slightly favored because the leaving group hinders approach of the nucleophilic solvent from the front side. 72  Mechanism of Racemization The SN1 reaction involves ionization to a flat carbocation, which can be attacked from either side. Step 1: Ionization of a tetrahedral carbon gives a flat carbocation. Step 2: A nucleophile may attack either side of the carbocation. 73 Rearrangements in the SN1 Reactions  Carbocations frequently undergo structural changes, called rearrangements, to form more stable ions.  A rearrangement may occur after a carbocation has formed or it may occur as the leaving group is leaving.  An example of a reaction with rearrangement is the SN1 reaction of 2-bromo-3-methylbutane in boiling ethanol.  The product is a mixture of 2-ethoxy-3-methylbutane (not rearranged) and 2-ethoxy-2-methylbutane (rearranged).  The rearranged product, 2-ethoxy-2-methylbutane, results from a hydride shift, the movement of a hydrogen atom with its bonding pair of electrons.  A hydride shift is represented by the symbol ~H.  In this case, the hydride shift converts the initially formed 2o carbocation to a more stable 3o carbocation.  Attack by the solvent gives the rearranged product. PROBLEM 6-25 Give the mechanism for the formation of 2-ethoxy-3- methylbutane, the unrearranged product in this reaction. 74  Mechanism of Hydride Shift  Carbocations often rearrange to form more stable carbocations.  This may occur when a hydrogen atom moves with its bonding pair of electrons.  Formally, this is the movement of a hydride ion although no actual free hydride ion is involved. Step 1: Unimolecular ionization gives a carbocation. Step 2: A hydride shift forms a more stable carbocation.  This rearrangement involves movement of a H atom with its bonding pair of electrons over to the empty p orbital of the carbocation.  In three dimensions, the rearrangement looks like this: 75 Step 3: Solvent (a weak nucleophile) attacks the rearranged carbocation. Step 4: Deprotonation gives the rearranged product.  When neopentyl bromide is boiled in ethanol, it gives only a rearranged substitution product.  This product results from a methyl shift (~CH3), the migration of a CH3 group together with its pair of electrons.  Without rearrangement, ionization of neopentyl bromide would give a very unstable primary carbocation.  The methyl shift occurs while bromide ion is leaving, so that only the more stable tertiary carbocation is formed. 76  Mechanism of Methyl Shift An alkyl group can rearrange to make a carbocation more stable. Step 1: Ionization occurs with a methyl shift. In three dimensions, Step 2: Attack by ethanol gives a protonated version of the rearranged product. Step 3: Deprotonation gives the rearranged product.  Because rearrangement is required for ionization, only rearranged products are observed. 77  In general, we should expect rearrangements in reactions involving carbocations whenever a hydride shift or an alkyl shift can form a more stable carbocation.  Most rearrangements convert 2° (or incipient 1°) carbocations to 3° or resonance-stabilized carbocations. Problem-solving Hint Primary halides and methyl halides rarely ionize to carbocations in solution. If a primary halide ionizes, it will likely ionize with rearrangement. PROBLEM 6-26 Propose a mechanism involving a hydride shift or an alkyl shift for each solvolysis reaction. Explain how each rearrangement forms a more stable intermediate. 78 Comparison of SN1 and SN2 Reactions Effect of the Nucleophile  The nucleophile takes part in the slow step (the only step) of the SN2 reaction but not in the slow step of the SN1.  Therefore, a strong nucleophile promotes the SN2 but not the SN1.  Weak nucleophiles fail to promote the SN2 reaction; therefore, reactions with weak nucleophiles often go by the SN1 mechanism if the substrate is secondary or tertiary. SN1: Nucleophile strength is unimportant (usually weak). SN2: Strong nucleophiles are required. Effect of the Substrate Structure  The structure of the substrate (R-X) is an important factor in determining which of these substitution mechanisms might operate.  Most methyl halides and 1o halides are poor substrates for SN1 substitutions because they cannot easily ionize to high-energy methyl and 1o carbocations.  They are relatively unhindered, however, so they make good SN2 substrates.  Tertiary halides are too hindered to undergo SN2 displacement, but they can ionize to form 3o carbocations.  Tertiary halides undergo substitution exclusively through the SN1 mechanism.  Secondary halides can undergo substitution by either mechanism, depending on the conditions. SN1 substrates: 3° > 2° (1° and CH3X are unlikely) SN2 substrates: CH3X > 1° > 2° (3° is unsuitable) 79  If silver nitrate (AgNO3) is added to an alkyl halide in a good ionizing solvent, the silver ion removes the halide ion to give a carbocation. Effect of the Solvent  The slow step of the SN1 reaction involves formation of two ions.  Solvation of these ions is crucial to stabilizing them and lowering the activation energy for their formation.  Very polar ionizing solvents such as water and alcohols are needed for the SN1.  The solvent may be heated to reflux (boiling) to provide the energy needed for ionization.  Less charge separation is generated in the transition state of the SN2 reaction.  Strong solvation may weaken the strength of the nucleophile because of the energy needed to strip off the solvent molecules.  Thus, the SN2 reaction often goes faster in less polar solvents if the nucleophile will dissolve.  Polar aprotic solvents may enhance the strength of weak nucleophiles. SN1: Good ionizing solvent required. SN2: Good ionizing solvent required. 80  Kinetics  The rate of the SN1 reaction is proportional to the concentration of the alkyl halide but not the concentration of the nucleophile.  It follows a first-order rate equation.  The rate of the SN2 reaction is proportional to the concentrations of both the alkyl halide [R–X] and the nucleophile [Nuc:–].  It follows a second-order rate equation. SN1 rate = kr[R–X] SN2 rate = kr[R–X][Nuc:–] Stereochemistry  The SN1 reaction involves a flat carbocation intermediate that can be attacked from either face.  Therefore, the SN1 usually gives a mixture of inversion and retention of configuration.  The SN2 reaction takes place through a back-side attack, which inverts the stereochemistry of the carbon atom.  Complete inversion of configuration is the result. SN1: Mixture of retention and inversion; racemization. SN2: Complete inversion. 81  Rearrangements  The SN1 reaction involves a carbocation intermediate.  This intermediate can rearrange, usually by a hydride shift or an alkyl shift, to give a more stable carbocation.  The SN2 reaction takes place in one step with no intermediates.  No rearrangement is possible in the SN2 reaction. SN1: Rearrangements are common. SN2: Rearrangements are impossible. SUMMARY Nucleophilic Substitutions PROBLEM 6-27 For each reaction, give the expected substitution product, and predict whether the mechanism will be predominantly first order (SN1) or second order (SN2). (a) 2-chloro-2-methylbutane + CH3COOH (b) Isobutyl bromide + sodium methoxide (c) 1-Iodo-1-methylcyclohexane + ethanol (d) Cyclohexyl bromide + methanol (e) Cyclohexyl bromide + sodium ethoxide 82  Elimination Reactions  An elimination reaction involves the loss of two atoms or groups from the substrate with formation of a  bond.  Elimination reactions frequently accompany and compete with substitutions.  By varying the reagents and conditions, we can often modify a reaction to favor substitution or to favor elimination.  Depending on the reagents and conditions involved, an elimination might be a first-order (E1) or second-order (E2) process. 83  The E1 Reaction: First-Order Elimination Mechanism and Kinetics  The abbreviation E1 stands for elimination, unimolecular.  Unimolecular means the rate-limiting transition state involves a single molecule.  The slow step of an E1 reaction is the same as in the SN1 reaction: unimolecular ionization to form a carbocation.  In a fast second step, a base abstracts a proton from the carbon atom adjacent to the C+.  The electrons that once formed the C–H bond now form a  bond between two carbon atoms. The E1 Reaction Mechanism  The E1 reaction requires ionization to a carbocation intermediate like the SN1, so it follows the same order of reactivity: 3° > 2° >> 1°.  A base (usually weak) abstracts a proton from the carbocation to give an alkene. Step 1: Unimolecular ionization to give a carbocation (rate- limiting). 84 Step 2: Deprotonation by a weak base (often the solvent) gives the alkene (fast). E1 elimination of bromocyclohexane in methanol Step 1: Ionization gives a carbocation and bromide ion in a slow step. Step 2: Methanol abstracts a proton to give cyclohexene in a fast step. E1 Rate Equation  Because the rate-limiting step involves unimolecular ionization of the alkyl halide, the rate equation is first-order.  The rate depends only on the concentration of the alkyl halide, and not on the strength or concentration of the base. E1 rate = kr[RX] The weak base (often the solvent) takes part in the fast second step of the reaction. 85  Competition with the SN1 Reaction  The E1 reaction almost always competes with the SN1 reaction.  Whenever a carbocation is formed, it can undergo either substitution or elimination, and mixtures of products often result.  The 2-methylpropene results from dehydrohalogenation, an elimination of a H and a X atom.  In the absence of a strong base, dehydrohalogenation takes place by the E1 mechanism:  Ionization of the alkyl halide gives a carbocation intermediate, which loses a proton to give the alkene.  Ethanol serves as a base in the elimination and as a nucleophile in the substitution. Step 1: Ionization to form a carbocation. 86 Step 2 (E1): Basic attack by the solvent abstracts a proton to give an alkene. Step 2 (SN1): Nucleophilic attack by the solvent on the carbocation.  Often the carbocation intermediates react in two or more ways to give mixtures of products.  For this reason, SN1 and E1 reactions of alkyl halides are not often used for organic synthesis. PROBLEM 6-30 SN1 substitution and E1 elimination frequently compete in the same reaction. (a) Propose a mechanism and predict the products for the solvolysis of 1-bromo-1-methylcyclopentane in ethanol. (b) Compare the function of the solvent (ethanol) in the E1 and reactions. Orbital Changes in the E1 Reaction  In the second step, the carbon atom next to the C+ rehybridize to sp2 as the base attacks the proton and electrons flow into the new  bond. 87  Energy Diagram for E1 Reaction  The energy diagram for the E1 reaction is similar to that for the SN1 reaction.  The ionization step is strongly endothermic, with a rate-limiting transition state.  The second step is a fast exothermic deprotonation by a base.  The base is not involved in the reaction until after the rate- limiting step, so the rate depends only on the concentration of the R-X.  Weak bases are common in E1 reactions. 88  Rearrangements in E1 Reactions  Like other reactions involving carbocations, the E1 may be accompanied by rearrangement.  Compare the following E1 reaction with the SN1 reaction of the same substrate.  Solvent acts as a base in the E1 reaction and a nucleophile in the SN1 reaction. Mechanism of Rearrangement Step 1: Ionization to form a carbocation (slow) 89 Step 2: A hydride shift forms a more stable carbocation (fast). Step 3: The weakly basic solvent removes either adjacent proton (fast). PROBLEM 6-1 (Partially solved) When the following compound is heated in methanol, several different products are formed. Propose mechanisms to account for the four products shown. SOLUTION With no strong base and a good ionizing solvent, we would expect a first-order reaction. But this is a 1o alkyl halide, so ionization is 90 difficult unless it rearranges. It might rearrange as it forms, but we’ll imagine the cation forming then rearranging. From these rearranged intermediates, either loss of a proton (E1) or attack by the solvent (SN1) gives the observed products. SUMMARY Carbocation Reactions 1. A carbocation can react with its own leaving group to return to the reactant: R+ + :X– R–X 2. It can react with a nucleophile to form a substitution product (SN1): R+ + :Nuc– R–Nuc 91 3. It can lose a proton to form an elimination product (an alkene) (E1): 4. It can react rearrange to a more stable carbocation, then react further. The order of stability of carbocations is: resonance-stabilized, 3° > 2° > 1°. PROBLEM 6-33 Give the substitution and elimination products for the following reactions. (a) 3-bromo-3-ethylpentane heated in methanol (b) 1-iodo-1-methylcyclopentane heated in ethanol (c) 3-bromo-2,2-dimethylbutane heated in ethanol (d) 1-iodo-2-methylcyclohexane + silver nitrate in water (see Problem 6-29) Positional Orientation of Elimination  Many compounds can eliminate in more than one way, to give mixtures of alkenes.  In many cases, we can predict which elimination product will predominate. 92  In the example below, the carbocation can lose a proton on either of two adjacent carbon atoms.  The first product has a trisubstituted double bond.  The second product has a disubstituted double bond.  Where there are two or more possible elimination products, the product with the most substituted double bond will predominate.  This general principle is called Zaitsev’s rule, and reactions that give the most substituted alkene are said to follow Zaitsev orientation. ZAITSEV’S RULE: In elimination reactions, the most substituted alkene usually predominates. R2C=CR2 > R2C=CHR > RHC=CHR and R2C=CH2 > RHC= CH2 tetrasubstituted trisubstituted disubstituted monosubstituted This order of preference is the same as the order of stability of alkenes. PROBLEM 6-34 When 1-bromo-1-methylcyclohexane is heated in ethanol for an extended period of time, three products result: one ether and two 93 alkenes. Predict the products of this reaction, and propose a mechanism for their formation. Predict which of the two alkenes is the major elimination product. SOLVED PROBLEM When 3-iodo-2,2-dimethylbutane is treated with silver nitrate in ethanol, three elimination products are formed. Give their structures, and predict which ones are formed in larger amounts. SOLUTION Silver nitrate reacts with the alkyl iodide to give solid silver iodide and a cation. This secondary carbocation can lose a proton to give an unrearranged alkene (A), or it can rearrange to a more stable tertiary cation. The tertiary cation can lose a proton in either of two positions. One of the products (B) is a tetra-substituted alkene, and the other (C) is disubstituted. 94 Product B predominates over product C because the double bond in B is more substituted. Whether product A is a major product will depend on the specific reaction conditions and whether proton loss or rearrangement occurs faster. 95  The E2 Reaction: Second-Order Elimination Mechanism and Kinetics  E2 stands for elimination, bimolecular.  E2 take place under second-order conditions with a strong base present.  tert-Butyl bromide undergoes E2 reaction with methoxide ion in methanol.  This is a second order reaction because CH3O- is a strong base as well as a strong nucleophile.  It attacks the alkyl halide faster than the halide can ionize to give a first-order reaction.  No substitution product (methyl tert-butyl ether) is observed, however.  The SN2 mechanism is blocked because the 3o alkyl halide is too hindered.  The observed product is 2-methylpropene, resulting from elimination of HBr and formation of a double bond. 96  Rate Equation  The rate of E2 elimination is proportional to the concentrations of both the alkyl halide and the base, giving a second-order rate equation.  This is a bimolecular process, with both the base and the alkyl halide participating in the transition state.  So, this mechanism is abbreviated as E2 for elimination, bimolecular. E2 rate = kr[RX][B:–]  In the E2 reaction just shown, methoxide reacts as a base rather than as a nucleophile.  Most strong nucleophiles are also strong bases, and elimination commonly results when a strong base / nucleophile is used with a poor SN2 substrate such as a 3° or hindered 2° alkyl halide.  Instead of attacking the back side of the hindered electrophilic carbon, methoxide abstracts a proton from one of the methyl groups.  This reaction takes place in one-step, with bromide leaving as the base abstracts a proton.  In the general mechanism of the E2 reaction, a strong base abstracts a proton on a carbon atom adjacent to the one with the leaving group.  As the base abstracts a proton, a double bond forms and the leaving group leaves. 97  E2 is a concerted reaction in which bonds break and new bonds form at the same time, in a single step. Substrate Reactivity  The order of reactivity of alkyl halides toward E2 dehydrohalogenation is: 3° > 2° > 1°  This reactivity order reflects the greater stability of highly substituted double bonds.  Elimination of a 3° halide gives a more substituted alkene than elimination of a 2° halide, which gives a more substituted alkene than a 1° halide. The E2 Reaction Mechanism  The concerted E2 reaction takes place in a single step.  A strong base abstracts a proton on a carbon next to the leaving group, and the leaving group leaves.  The product is an alkene. E2 Elimination of 3-bromopentane with sodium ethoxide 98  Mixtures of Products  The E2 reaction requires abstraction of a proton on a carbon atom next to the carbon bearing the halogen.  If there are two or more possibilities, mixtures of products may result.  Zaitsev’s rule predicts that the most substituted alkene will be the major product.  Reaction of 1-bromo-1-methylcyclohexane with sodium ethoxide gives a mixture of a disubstituted alkene and a trisubstituted alkene.  The trisubstituted alkene is the major product. 99 PROBLEM 6-37 1. Predict the elimination products of the following reactions. When two alkenes are possible, predict which one will be the major product. Explain your answers, showing the degree of substitution of each double bond in the products. 2. Which of these reactions are likely to produce both elimination and substitution products? (a) 2-bromopentane + NaOCH3 (b) 3-bromo-3-methylpentane + NaOCH3 (c) 2-bromo-3-ethylpentane + NaOH (d) cis-1-bromo-2-methylcyclohexane + NaOEt (Et = ethyl, CH2CH3) Stereochemistry of the E2 Reaction  The E2 follows a concerted mechanism: Bond breaking, and bond formation take place at the same time.  Concerted mechanisms require specific geometric arrangements so that the orbitals of the bonds being broken can overlap with those being formed and the electrons can flow smoothly from one bond to another.  A coplanar geometric arrangement is required for E2 reaction.  E2 elimination requires partial formation of a new  bond, with its parallel p orbitals, in the transition state.  The electrons that once formed a C–H bond must begin to overlap with the orbital that the leaving group is vacating.  Formation of this new  bond implies that these two sp3 orbitals must be parallel so that  overlap is possible as the hydrogen and halogen leave and the orbitals rehybridize to the p orbitals of the new  bond. 100  Coplanarity in the E2 Reaction The figure shows two conformations that provide the necessary coplanar alignment of the leaving group, the departing hydrogen, and the two carbon atoms.  When the H and the X are anti to each other ( = 180o) their orbitals are aligned.  This is called the anti-coplanar conformation.  When the H and the X eclipse each other ( = 0o) their orbitals are once again aligned.  This is called the syn-coplanar conformation.  Of these possible conformations, the anti-coplanar arrangement is most commonly seen in E2 reactions. 101  The transition state for the anti-coplanar arrangement is a staggered conformation, with the base far away from the leaving group.  In most cases, this transition state is lower in energy than that for the syn-coplanar elimination.  The transition state for syn-coplanar elimination is an eclipsed conformation.  In addition to the higher energy resulting from eclipsing interactions, the transition state suffers from interference between the attacking base and the leaving group.  To abstract the proton, the base must approach quite close to the leaving group.  In most cases, the leaving group is bulky and negatively charged, and the repulsion between the base and the leaving group raises the energy of the syn-coplanar transition state. Rigid Molecules  Some molecules are rigidly held in eclipsed conformations, with a hydrogen atom and a leaving group in a syn-coplanar arrangement.  Such compounds are likely to undergo E2 elimination by a concerted syn-coplanar mechanism.  Deuterium labeling is used in the following reaction to show which atom is abstracted by the base.  Only the hydrogen atom is abstracted, because it is held in a syn-coplanar position with the bromine atom. 102  Remember that syn-coplanar eliminations are unusual, however, and anti-coplanar eliminations are more common.  The E2 is a stereospecific reaction, because different stereoisomers of the starting material react to give different stereoisomers of the product.  This stereospecificity results from the anti-coplanar transition state that is usually involved in the E2. PROBLEM 6-38 When the first compound shown here is treated with sodium methoxide, the only elimination product is the trans isomer. The second diastereomer (blue) gives only the cis product. Use your models and careful drawings of the transition states to explain these results. 103 Comparison of E1 and E2 Mechanisms Effect of the Base 104  The nature of the base is the single most important factor in determining whether an elimination will go by the E1 or E2 mechanism.  If a strong base is present, the rate of the bimolecular reaction will be greater than the rate of ionization, and the E2 reaction will predominate (accompanied by the SN2).  If no strong base is present, then a good solvent makes a unimolecular ionization likely.  Subsequent loss of a proton to a weak base (such as the solvent) leads to elimination.  Under these conditions, the E1 reaction usually predominates, usually accompanied by the SN1. E1: Base strength is unimportant (usually weak). E2: Requires strong bases. Effect of the Solvent  The slow step of the E1 reaction is the formation of two ions.  Like the SN1, the E1 reaction critically depends on polar ionizing solvents such as water and the alcohols.  In the E2 reaction, the transition state spreads out the negative charge of the base over the entire molecule.  There is no more need for solvation in the E2 transition state than in the reactants.  The E2 is therefore less sensitive to the solvent; in fact, some reagents are stronger bases in less polar solvents. E1: Requires a good ionizing solvent. E2: Solvent polarity is not so important. Effect of the Substrate 105  In the E1 reaction, the rate-limiting step is formation of a carbocation, and the reactivity order reflects the stability of carbocations.  In the E2 reaction, the more substituted halides generally form more substituted, more stable alkenes.  For both the E1 and the E2 reactions, the order of reactivity is E1: 3o > 2o > 1o (1° usually will not go E1) E2: 3o > 2o > 1o (1° usually will not go E2) Kinetics  The rate of the E1 reaction is proportional to the concentration of the alkyl halide [RX] but not to the concentration of the base.  It follows a first-order rate equation.  The rate of the E2 reaction is proportional to the concentrations of both the alkyl halide [RX] and the base [B:–].  It follows a second-order rate equation. E1 rate = kr[RX] E2 rate = kr[RX][B:–] Orientation of Elimination 106  In most E1 and E2 eliminations with two or more possible products, the product with the most substituted double bond predominates.  This principle is called Zaitsev’s rule, and the most highly substituted product is called the Zaitsev product. E1: Usually Zaitsev orientation. E2: Usually Zaitsev orientation. Stereochemistry  The E1 reaction begins with an ionization to give a flat carbocation.  No particular geometry is required for ionization.  The E2 reaction takes place through a concerted mechanism that requires a coplanar arrangement of the bonds to the atoms being eliminated.  The transition state is usually anti-coplanar, although it may be syn-coplanar in rigid systems. E1: No particular geometry required for the slow step. E2: Coplanar arrangement (usually anti) required for the transition state. Rearrangements 107  The E1 reaction involves a carbocation intermediate.  This intermediate can rearrange, usually by the shift of a hydride or an alkyl group, to give a more stable carbocation.  The E2 reaction takes place in one-step with no intermediates.  No rearrangement is possible in the E2 reaction. E1: Rearrangements are common. E2: No rearrangements. SUMMARY Elimination Reactions Problem-Solving Strategy 108  Predicting Substitutions and Eliminations Homework Problems 6-41 to 6-46, 6-48, 6-49, 6-51 to 6-57, 6-60 to 6-64, 6-66, 6-68, 6-72 to 6-74 109