266 Substitution and Elimination Reactions

266 Substitution and Elimination Reactions - Substitution...

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Unformatted text preview: Substitution and Elimination Reactions This document has been written to provide you with an overview of substitution and elimination reactions. This is a condensed overview and provides you with the essential knowledge needed to answer substitution and elimination reactions questions involving alkyl halides. For more detailed descriptions consult your text book. Any comments or questions please email Dr. Steven Forsey: spforsey@uwaterloo.ca 1.0 Introduction to Substitution Reactions 1.1 The SN2 mechanism 1.2 The SN1 mechanism 1.3 The effect of the substrate on the rate of SN2 and SN1 reactions. 1.4 Solvent Effects in SN1 and SN2 reactions 1.5 Nucleophilicity 1.6 Steric hindrance of nucleophiles in SN2 reactions 2.0 Elimination Reactions of Haloalkanes: E1 and E2 Mechanisms 2.1 E1 Mechanism 2.2 E2 Mechanism 3.0 How to determining if an SN2, E2, SN1 or E1 reaction will occur with Primary and Secondary Alkyl Halides 4.0 Regiochemistry and Stereochemistry 4.1 E2 Reactions with Unsymmetrical Alkyl Halides 4.2 SN1/E1 reactions and Unsymmetrical Alkyl Halides 5.0 Carbocation Rearrangement and SN1/E1 reactions 6.0 Sterically Hindered nucleophile/base 7.0 Overview 8.0 Vinyl, aryl, allylic and benzylic compounds 2 6 7 8 10 13 17 19 19 21 22 24 24 28 29 30 31 37 Dr. Steven Forsey 1 Substitution and Elimination Reactions 1.0 Introduction to Substitution Reactions Substitution reactions replace one atom with another. For example in the substitution reaction of methyl bromide with sodium hydroxide replaces the Br atom by a hydroxide atom to give methanol. CH3Br + Na+ + OH CH3OH + Cl + Na+ This type of chemical reaction is called a nucleophilic substitution reaction because the hydroxide ion is a nucleophile a reactant that seeks centers of positive charge in a molecule. The nucleophile is electron rich and it is the attacking species in the reaction that donates a pair of electrons to another compound. We learned that the carbontochlorine bond is polar, so the C atom is a region of slight positive charge and is said to be electrophilic (electron attracting). The electrophile, also known as the substrate is the species that is being attacked by the nucleophile and accepts a pair of electrons. Electrophiles are compounds or atoms that are electron deficient. In this substitution reaction, the hydroxide ion attacks at the slightly positive C atom, displacing pushing out the chloride ion, which is called the leaving group When considering the feasibility of a substitution reaction, we must consider not only the nucleophilicity of the nucleophile but also the leaving ability of the leaving group. An important rule in organic chemistry is as follows. The weaker the base, the better the leaving group Nucleophiles, often denoted by the abbreviation Nu, may be negatively charged or neutral, but every nucleophile contains at least one pair of unshared electrons. Nucleophiles are, in fact, Lewis bases. The nucleophilic substitution of a haloalkanes is described by either of two general equations. As an example consider the following reaction. The nucleophile is the methylacetylide anion, the electrophile is bromoethane, and the Br anion is the leaving group. Arrows are used to show bond formation and breaking. The red arrow shows the movement of the lone pair of electrons from the negatively charged carbon to the delta positive carbon bonded to the halogen, labeled carbon. This arrows show that the lone pair of electrons on the carbon Dr. Steven Forsey 2 are used to form a new bond with carbon. Now, since the carbon can not have 5 bonds, a bond must be broken. The blue arrow shows the electrons from the carbonbromine sigma bond moving onto the bromine atom to form the bromide ion. If you use arrows to show bond formation and breaking you know the structure of the products The reaction will proceed to the right because the bromide ion is a much weaker base than the methylacetylide ion. Nomenclature: The carbon to which the function group is bonded to is called the alpha carbon (in this case a halogen). The carbon bonded to the alpha carbon is called the beta carbon, the next is called gama and so forth. This nomenclature is important when discussing reactions. In this reaction the nucleophile attacks the carbon to displace the leaving group. It does not attack the carbon. functional group Br O O Examples: CH 2 CH2 C H3C OH CH2 CH2 C H3C CH 2 CH3 functional functional group group Similarly, in the reaction below between sodium ethoxide and iodomethane the nucleophile is the ethoxide anion, the electrophile is iodomethane, and the leaving group is the I anion, a weak base. Finally, in the following reaction, the nucleophile is the ammonia molecule, the electrophile is bromoethane and the leaving group is the bromide ion (Br). In studying these reactions, you may have noticed the similarity of the reaction; To the BronstedLowry acidbase reaction In the BronstedLowry acidbase reaction, the stronger base and the stronger acid react to form the weaker acidbase pair. Similarly, in a nucleophilic substitution reaction, both the nucleophile and the leaving group are bases. In all of the above reactions the weaker acidbase pairs are formed and the equilibriums are shifted to the right. This concept can be used to determine the likelihood of a reaction occurring. Thus, for example we would not expect the following reactions to occur, HO + H3CH H3COH + H B + H X H B + X Dr. Steven Forsey 3 Br + CH3CH2OH CH3CH2Br + OH In both reactions a stronger base has formed, the hydride ion is a much stronger base than the hydroxide ion and the hydroxide ion is a much stronger base than the bromide ion. These reactions will not take place. It is important to state at this point that there is difference between basicity and nucleophilicity. Nucleophilicity is the rate at which a nucleophile attacks an electrophile and is thus a kinetic property. A better nucleophile will react faster. Basicity is a measure of base strength in an equilibrium reaction and is a thermodynamic property. When measuring the basicity of a compound in an equilibrium reaction the reaction could take seconds or days. It does not matter because we are not concerned about how fast the reaction proceeds but how far the equilibrium is shifted to the reactants or products. In organic chemistry it is important to be able to label the attacking species as a base or as a nucleophile because they are different reactions. What makes it even more confusing is that an attacking species can act as either a base or a nucleophile. For example a methoxide ion is both a strong base and a good nucleophile. In both cases the reaction equilibrium would have a large equilibrium constant and equilibriums would be shifted completely to the right. The methoxide ion is a much stronger base than the bromide and the acetate anions. If the difference in base strength between the two is not very large, the equilibrium constant for the resulting reaction will be close to unity. The substitution reaction will reach equilibrium rather than go to completion, as it would if the equilibrium constant were large. For example, the iodide and chloride ion are both weak bases however the iodide ion is a weaker base because it is larger and more polarizable, thus the equilibrium for the reaction below is shifted to the chloroalkane. Such equilibria can be shifted to the right or left by the removal of one of the products. For example, if the reaction is carried out in acetone, the product is iodopropane because sodium chloride is insoluble in acetone, whereas sodium iodide is soluble. Dr. Steven Forsey 4 Worked example Identifying Electrophiles, Nucleophiles, Acids and Bases. Decide which of the following compounds and ions are electrophiles and nucleophiles or acids and bases? Analyze: First put lone pairs on the atoms and look at the products that are given. Determine which bonds are being formed and which are being broken. This will tell you which molecule is the attacking species that is donating the electrons to form a new bond. Once you have determined the attacking species look at what is being attacked. If it is removing a proton, it is a base. If it is attacking a nucleus, it is a nucleophile. Dr. Steven Forsey 5 1.1 The SN2 mechanism SN2 symbolizes Substitution Nucleophilic Bimolecular. An example of a SN2 reaction is reaction of methyl bromide with sodium hydroxide to produce methanol. If we double the concentration of methyl bromide the reaction rate doubles and if we double the concentration of sodium hydroxide the rate also double. This means that this is a bimolecular reaction and both the nucleophile and the electrophile are involved in the rate determining step. The reaction rate is thus second order. The Free Energy Reaction Coordinate diagram displays how the more reactive reactants come together to form a high energy transitions state and then decreases in energy to form the more stable products. The transition state is the rate determining step in a bimolecular reaction. In the transition step bonds are partially formed and broken. This is called a concerted reaction. Also notice that in the transition state there is only one pathway for the nucleophile to react with the electrophile to displace the leaving group. This is commonly called a back side attack. Because the nucleophile can only follow one pathway to displace the leaving group to produce the more stable products, inversion at the carbon occurs. If the carbon being attacked is not chiral inversion does change the stereochemistry, however if the carbon is chiral the stereochemistry can change. Dr. Steven Forsey 6 The arrows drawn above are your thought process and helps you visualize how electrons flow to form and break bonds. Thus, the arrow starts from the electronrich nucleophile (the electrons on the hydrogen sulfide anion) to an electronpoor electrophile (the C atom in the polar carbonbromine bond). The red arrow shows how two electrons from the nucleophile flow to form a new bond between the sulfur and the carbon and the blue arrow illustrates how the electrons that form the CBr bond are broken and flow onto the bromine to form the bromide anion. 1.2 The SN1 mechanism SN1 symbolizes Substitution Nucleophilic Unimolecular. The reaction of tertbutylbromide in water follows first order kinetics. Which means the reaction is only dependent on the concentration of tertbutylchloride. The mechanism is very different than the SN2 mechanism. The first step is a slow endothermic reaction that forms an unstable carbocation intermediate. The first step in an SN1 reaction is the rate determining step. This mechanism is completely different than the fast exothermic SN2 reaction that proceeds directly to the final products. This difference is can be very important in determining if a SN2 or a SN1 reaction dominates. We will discus this later. The intermediate that is formed is an unstable trigonal planar sp2 hybridized carbocation. The carbocation is an electrophile and will then react with a nucleophile. In this reaction the nucleophile is water and a protonated alcohol is formed. The conjugate acid of the alcohol rapidly dissociates in excess water to form tertbutanol and a hydronium ion. Fast Fast H 3C H3C H 2O Slow H3C H 2O H3C C CH3 + Br O + H3O C C Br C O H3C H3C H H3C H3C Intermediate H H H3C H3C H 3C Product In the above the reaction we did not have to worry about stereochemistry but what happens to the stereochemistry of the products if the carbon is chrial. When we react an optically active solution of (R)3bromo3 methylhexane in a solvent mixture of water and acetone we obtain two substitution products, (S)3 methyl3hexanol and (R)3 methyl3hexanol. The resulting solution is optically inactive which means that there is a 50:50 mixture of each enantiomer. If the Dr. Steven Forsey 7 solution was not an equal mixture and an enantiomeric excess (ee) of one stereoisomer was formed the solution would still be optically active. (for further discussion consult your text book) A solution that contains and equal mixture of enantiomers is called a racemic mixture. Enantiomers are also called racemates. The reason why a racemic mixture is formed can be explained by looking at the mechanism. The first step of the mechanism involves the formation of a carbocation. The carbocation is sp2 hybridized and trigonal planar. (Again this is totally different than an SN2 mechanism) The intermediate is now attacked by the nucleophile and since the intermediate is planar the attack can occur from either face of the molecule as shown with the red and blue arrows above. The products formed are the protonated alcohols, but we can go directly to the alcohols and skip this drawing by using H+ on the arrow to indicate that the proton has been removed by the solvent. You may be asking why add acetone. Acetone is a non reactive cosolvent. (R)3bromo3methylhexane is not soluble in water but is in acetone. Water is also miscible with acetone thus by adding acetone you can solubilize the haloalkane and allow it to react with the nucleophile, water. With out the acetone there would be no reaction. 1.3 The effect of the substrate on the rate of SN2 and SN1 reactions. In the SN2 mechanism, the nucleophile attacks the electrophilic center on the side opposite the leaving group. Because the nucleophile attacks the backside of the carbon that is bonded to the halogen, substituents attached to the carbon will make it harder for the nucleophile to the get to the backside as shown in the relative rates of reaction with the following branched bromoalkanes with iodide. H H H3C H C Br C Br C Br C Br Bromoalkane: H H3C H3C H H3C H3C H3C H Relative Rate: 145 1 0.0078 negligible The rate of reaction decreases with increasing branching of the carbon. This is caused by an increase in the free energy of the transition state. Thus branching decreasing the ability of the nucleophile to form a bond with the electrophilic carbon and increase in the energy barrier (activation energy) of the reaction and consequently slows down the reaction. The obstruction of the nucleophile from interacting with the center of partial positive charge is called steric hindrance. The figure below depicts the increasing steric interaction and increasing activation energy associated with greater alkyl substitution on the electrophilic carbon. The backside of bromomethane has little steric interaction created from the hydrogens and reacts readily with a nucleophile. Contrastingly, the backside of 1,1 dimethylbromoethane (tertbutyl bromide) is completely block from a nucleophilic attack and will not undergo an SN2 reaction. bromomethane bromoethane 1methylbromoethane 1,1dimethylbromoethane Dr. Steven Forsey 8 Thus a tertiary haloalkanes will not undergo a SN2 reaction because of steric hindrance. However as seen in section 1.3, tertbutyl bromide does undergo an SN1 reaction with water. What effect does alkyl substitution have on the rate of an SN1 reaction? When we look at a series of bromoalkanes reacting with water a very different order of reactivity is observed. H H H3C H C Br C Br C Br C Br Bromoalkane: H H3C H3C H H3C H3C H3C H Relative Rate: 1 1 12 1.2 X 106 The reactivity is opposite of an SN2 reaction. The reason for this is that the SN1 reaction undergoes a completely different mechanism. In the SN1 reaction an intermediate is formed in an endothermic process. If the intermediate is to form at all it must be some what stable otherwise the energy barrier or the activation energy is far too great to create the unstable intermediate. The intermediate formed in a SN1 reaction is a carbocation. A carbocation is stabilized by electron donating groups such as alkyl groups. Alkyl groups donate electron density inductively through bonds or hyperconjugation (See acid and base notes) Thus increasing the number of alkyl groups attached the carbocation increases the stability of the cation and will increase the likelihood that a SN1 reaction will occur. Under normal SN1 conditions only secondary and tertiary carbocations will form. Dr. Steven Forsey 9 Thus, in conclusion, we expect haloalkanes that are highly substituted at the C atom to which the halogen is attached to undergo nucleophilic substitution by an SN1 mechanism. We would also expect haloalkanes that are less substituted at the C atom to undergo nucleophilic substitution by an SN2 mechanism because there is less steric hindrance for the attacking nucleophile. Relative Stability of Carbocations tertiary > secondary > primary > methyl 1.4 Solvent Effects in SN1 and SN2 reactions Solvent effects have a critical role in helping a reaction follow a SN1 or an SN2 mechanism. First consider the SN1 pathway. The rate determining step involves the formation of a cation and an anion intermediate. If the solvent does not stabilize the salt formed the reaction will not occur. As discussed in the physical Dr. Steven Forsey 10 properties sections of the course, to dissolve a solute the solvent molecules must surround the solute molecules and suspend the solute in the solvent. To dissolve a salt strong iondipole bonds are needed between the solvent and the ions. Thus good solvents for ionic compounds must be highly polar and have the ability to surround the ions and form strong bonds to dissolve the ions. Water is such a solvent. The water molecule has an electronegative oxygen with two small hydrogens attached to it. The electron density is pulled away from the hydrogens nucleus to create a very polar molecule. Water solvates cations through the donation of unshared electron pairs of the oxygen atom to the vacant orbitals of the cation and because the hydrogen atoms are tiny, water molecules are able to surround the cation tightly. Water molecules will also form strong iondipole electrostatic forces between the hydrogen atoms and the anion through hydrogen bonding. Again because the hydrogen atoms are small a number of water molecules can surround the anion without crowding. The figure below illustrates the solvation of a salt with water. The water molecules oxygen and hydrogens create electrostatic bonds with the ions to form a cage around the ions to solubilize them. This is called a solvent shell. The water molecules in the bulk solution are freer to move around while the solvent shell molecules are bound to the ions. H H H O H H O H H H O O H H O H H O H H H O H H H H O H O O H H H O O H H H H H H + O H H O O O - H H H O O H H + H + O + H H O H H + H H H O O O H O O H H O H H H H H H H H O O O H H O H O H H H O H H H O H H H O O H H H Compounds like methanol are also able to solubilize salts because of the OH group. However because of the hydrophobic alkyl group and the steric hindrance created by the methyl group, methanol is less polar than water and is inferior to water in its ability to solvate salts. Solvents like water and methanol are called protic solvents. Protic solvents are solvents whose molecules have hydrogen atoms bonded to electronegative atoms such as oxygen or nitrogen. Solvents like methanol, ethanol, acetic acid, methyl amine are protic solvents that stabilize anions through hydrogen bonding and cations through unshared pairs of electrons. So why is this so important in SN1 reactions? In order for the reaction to occur the solvent must solubilize the carbocation intermediate and the anionic leaving group. Thus we must use polar protic solvents to promote an SN1 reaction. Notice that in all of the above reactions the solvent was also the nucleophile. These reactions are called solvolysis reactions. Dr. Steven Forsey 11 Now consider the SN2 reaction, what type of solvent would promote this type of reaction? This is a concerted reaction and no intermediate is formed. The nucleophile has a lone pair of electrons or is negatively changed which then attacks an electrophile. As mentioned above polar protic solvents form strong dipoleion intermolecular forces with anions and in this case will slow down the rate of reaction. This makes the nucleophile less reactive in two ways; a) the protic solvent stabilizes the nucleophile through hydrogen bonding and decreases its basicity thus decreasing it reactivity b) the nucleophile must shed the solvent shell before it can reach the electrophile as illustrated below. The stronger the intermolecular forces between the solvent and the nucleophile the slower the reaction will proceed. O O C S H 3C CH 3 H 3C CH 3 So how can we change the solvent to encourage a SN2 mechanism? We can use Propanone Dimethylsulfoxide (Acetone) (DMSO) polar aprotic solvents or solvents that do not have a hydrogen atom bonded to an O electronegative atom such as propanone, dimethyl sulfoxide, diethylether, O O Diethylether Dimethoxyethane dimethlyformamaide or dimethylacetamide. These solvents are polar and have (DME) atoms that can donate a pair of electrons to the cation thus they can solvate O O cations well but since they can not hydrogen bond these solvents leave the anions H3C C H3C C N CH 3 H N unsolvated. This makes these solvents excellent solvents for SN2 reactions CH 3 CH 3 because the nucleophile is not hindered by a tightly bound solvent shell. Note Dimethylformamide Dimethylacetamide there is another class of solvents called nonpolar aprotic solvents such as hexane (DMF) (DMA) and benzene. While these solvent will not solvate anions they most likely will not solubilize your starting material because the nucleophile in an SN2 reaction is either polar or has a charge on it. Thus to dissolve the starting materials the solvent will also have to be polar (like dissolves like) Worked Example: For the following reactions a) identify the nucleophile and electrophile, b) is the electrophile accessible to a nucleophilic attack or is it sterically hindered, c) is the solvent polar protic or polar aprotic d) will the reaction follow a SN1 or a SN2 pathway and e) put lone pairs on the appropriate atoms and use arrows to illustrate the mechanism to show bond formation and breaking. Br a) CH 3CH2SNa + CH 3CHCH 3 Br b) + CH3OH DMF Dr. Steven Forsey 12 Analyze: Reaction 1) a) The nucleophile is the ethanethiolate ion since the negatively charged sulfur atom has basic lone pairs of electrons that it can be donated to the electrophile. The electrophile is 2bromopropane, the delta positively charged carbon can accept electrons. b) The electrophile in this case is a secondary alkyl halide and there is some crowding at the electrophilic C atom thus the reaction may go by a SN2 or a SN1 mechanism if we just look at the electrophile. c) If we look at the solvent we see that it is a polar aprotic solvent that can not hydrogen bond. Thus this solvent would not promote an SN1 reaction because the solvent would not help stabilize the formation of a carbocation intermediate. The solvent would however increase the reactivity of the nucleophile. d) The reaction would thus undergo a SN2 mechanism. e) Br CH3 CH 2S nucleophile Na + CH 3CHCH 3 + electrophile DMF S N2 S + Na + Br Reaction 2 a) The nucleophile is methanol because it has lone pairs of electrons to donate to an electrophile and the electrophile is 1methyl1bromocyclohexane. b) The delta positive electrophilic carbon is bonded to 3 other alkyl groups (3o carbon) and is sterically hindered towards a nucleophilic attack. The reaction will not undergo a SN2 reaction. This does not mean the reaction is a SN1 reaction because there is a possibility that no reaction will occur at all. c) The solvent is polar protic and will help stabilize a carbocation intermediate and the anionic leaving group. d) The reaction will proceed through an SN1 mechanism. e) H Br + H3COH + Br O CH 3 -H + O CH 3 Notice in the last step H+ was used in the mechanism instead of showing the solvent removing the proton. 1.5 Nucleophilicity A strong base is a compound that is not stable and is reactive. It wants the proton in an equilibrium reaction. But what makes a good nucleophile? Nucleophilicity is a measured rate of reaction and not an equilibrium reaction like acids and bases. Nucleophilicity is a measure on how fast a pair of electrons are donated to an electrophilic center Dr. Steven Forsey 13 bearing a leaving group. It is difficult to tell sometimes when comparing a series of nucleophiles which nucleophile will react faster. This is because it is not only dependent on the nucleophile but also the electrophile. Thus it is a property that depends on reaction partners. Generally nucleophilicity is dependent on a number of factors, charge, basicity, solvent, polarizability, and the nature of the electrophile. This can be confusing but there are some general trends that can be used. Generally the stronger the base the better the nucleophile, thus nucleophilicity decreases from left to right across the periodic table. H2N > RO > OH > F Also nucleophilicity follows basicity if we look at series of nucleophiles in which the nucleophilic atom is the same. RO > HO > RO2 > ROH > H2O Weak bases can be weak nucleophiles such as ROH, H2O, NH3, RNH2, R2NH, R3N But not all weak bases are poor nucleophiles. As we go down the periodic table we find that nucleophilicity increases. I > Br > Cl > F HSe > HS > HO H2S > H2O The reason for this is that as you go down the periodic table the atoms become larger and more polarizable. The atoms electrons are freer to move towards the delta positive charge as it approaches the electrophile resulting in stronger bonding in the transition state. Consider the reactivity of the halogens in a protic solvents. We see that the trend in reactivity is the same as nucleophilicity. Protic solvent I > Br > Cl > F The iodide ion reacts faster than the fluoride ion. But how can this be. The fluoride ion is a stronger base and is less stable and should react very quickly compared to the more stable iodide ion even if the activation energy of transition state is lowered by the polarizability of the iodide ion. The answer is hydrogen bonding of the protic solvent with the fluoride ion. Fluoride will form stronger hydrogen bonds with the solvent because it is a stronger base. Thus a fluoride ion is hindered by the solvent and must shed or displace the solvent as the ion is attacking the electrophilic carbon. Iodide forms weaker hydrogen Dr. Steven Forsey 14 bonds and more readily displaces the solvent shell in its nucleophilic attack. Thus less energy is needed to break the hydrogen bonds that are solvating the iodide anion. When we do the reactions again but in an aprotic solvent, we find the reactivity is reversed. Aprotic solvent F > Cl > Br > I In this case reactivity follows basicity because now the halogen is unhindered by the solvent. Thus to increase the nucleophilicity of a compound we should use aprotic solvents like acetone, dimethylformamide or DMSO. An elegant example of basicity versus nucleophilicity is the synthesis of sulfides from thiols using sodium hydroxide. RSH + RBr + NaOH RSR + NaBr (Where RBr is primary) In this reaction, HO is a stronger base than RS and RS is a better nucleophile than HO. The proton transfer between the acid (RSH) and base (HO), is faster than the SN2 reaction between the nucleophile (HO) and the electrophile (RBr). Once the thiolate anion is form the weak base, good nucleophile readily reaction the primary alkylhalide to form the SN2 product. The table below shows some nucleophiles segregated into categories. It is not possible to do much better than this as nucleophilicity is a property that depends on the reaction partner. If we change the reaction partner and thus the energy of the orbitals involved, we will change the stabilization involved as well. Excellent Relative Good Relative Fair Nucleophile Relative Nucleophile Rate Nucleophile Rate Rate NC Cyanide 126,000 HO Hydroxide 16,000 Cl Chloride 1,000 HS Thiolate 126,000 Br Bromide 10,000 CH3COO Acetate 630 I Iodide 80,000 N3 Azide 8,000 F Fluoride 80 NH3 Ammonia 8,000 CH3OH Methanol 1 Basicity is a thermodynamic property measure by an equilibrium K = equilibrium NO2 Nitrite 5,000 H2O Water 1 Nucleophilicity is a kinetic phenomenon, quantified by comparing rate of reactions k = rate constant Dr. Steven Forsey 15 Worked example: Which of the following reactions will react faster in methanol? O O a) + Cl S O O + Br b) S CH 3 O Analyze: First determine the nucleophile and the electrophile. In both reactions the halogens are the nucleophiles, the sulfonates are the electrophiles and methanesulfonate is the leaving group for both reactions. What type of substitution reactions are they, SN1 or SN2. We have a protic polar solvent that will stabilize the formation of a carbocation. But carbocation formation is very unlikely because the carbocation intermediate formed would be primary and very unstable. The leaving group is bonded to a primary carbon thus the reaction will follow a SN2 pathway. Now look at the nucleophiles. The chloride ion is a stronger base but the bromide ion is a better nucleophile because it is larger and more polarizable. Which factor would have a greater influence on the reactivity, basicity or nucleophilicity? Well both do. Since the chloride ion is a stronger base it will hydrogen bond stronger to the solvent than the bromide ion. This makes the chloride ion it less nucleophilic. Bromine is a good nucleophile because of its polarizablility thus reaction b) will be faster than reaction a). O a) H O H H O H H O H H H O H CH3 O O S CH 3 O Cl + O S CH3 O + H O O Cl O b) O H H O H H O S CH 3 O Br + O S O CH 3 + Br O H O H Worked example: Would you expect the reaction between nbutylbromide and sodium cyanide to react faster in ethanol of DMSO? Analyze: This is an SN2 reaction. The cyanide ion is an excellent nucleophile and the alkyl halide is primary. The reaction will be faster in the solvent that does not hinder and slow down the nucleophilic attack. Thus the reaction will proceed faster in DMSO than in ethanol. Ethanol is a protic solvent that can form strong iondipole hydrogen bonds whereas DMSO is an aprotic solvent that will not stabilize the cyanide ion. CN + Na Br Br + Na N C Br + NaC N Dr. Steven Forsey 16 1.6 Steric hindrance of nucleophiles in SN2 reactions For an SN2 reaction to advance to the transition state the nucleophile must get close to the electrophilic carbon before bonding can occur. Bulky groups adjacent to the nucleophilic atom will hinder this approach and slow down the rate of reaction. For example consider two oxygen nucleophiles, sodium tertbutoxide and sodium methoxide. The nucleophilic oxygen of the methoxide ion is unhindered by adjacent groups. However, the oxygen of the tertbutoxide ion is hindered by three methyl groups. methoxide ion H 3C O H3C C O ter t-butoxide ion H 3C H 3C Sodium tertbutoxide is a stronger base than sodium methoxide because of the electron donating alkyl groups adjacent to the negatively charged oxygen. However when these nucleophiles are reacted with the same electrophile we find that the rate of reaction is faster for the methoxide anion. + Br + H3C Fast H3C O H 3C O CH3 + Br H 3C H3C + Br CH 3 + C O + Slow H 3C Br C O H3C H 3C H3C H 3C Thus just like a solvent that impedes the approach of nucleophile and the formation of the transition state, bulky group adjacent to the nucleophilic atom will slow down the rate of reaction. The methoxide ion is therefore a stronger nucleophile than the bulky tert butoxide ion even though the tertbutoxide is a stronger base. You may ask wouldn't the bulkiness of the base effect it's base strength as well. In fact it might but in most cases we find that steric hindrance has little effect on basicity because basicity generally involves an attack on an unhindered proton. Dr. Steven Forsey 17 Summary All unhindered strong bases are good nucleophiles (i.e. CH3O)and weak polarizable bases are good nucleophiles (i.e. Br), however small none polarizable weak bases (i.e. CH3OH) are poor nucleophiles and bulky hindered strong bases (i.e. (CH3)3C O) are poor nucleophiles. Unhindered electrophiles favor SN2 reactions and hindered electrophiles favor SN1 reactions. Polar protic solvents promote SN1 reactions by stabilizing the formation of the carbocation intermediate. Polar protic solvent decrease the reactivity of nucleophiles in SN2 reactions because of hydrogen bonding with the nucleophile. The stronger the base the stronger the hydrogen bonding. Polar aprotic solvents form weak intermolecular forces with the nucleophile thus promote SN2 reactions by increasing the reactivity of the nucleophile. Polar aprotic solvents do not solvate anions well because they can not hydrogen bond thus polar aprotic solvents do not aid in the formation of a carbocation intermediate in a SN1 reaction. Thus SN1 reactions are unlikely to occur in polar aprotic solvents Notice for secondary alkyl halides that either SN2 or SN1 reactions can occur. Thus by choosing the appropriate solvent and nucleophile we can promote either substitution mechanism. Worked example: Do the reactions given below, follow a SN2 or a SN1 reaction? Br a) + CH3OH heat (R)-2-bromo-4-methylpentane Br b) + CH 3SNa acetone (R)-2-bromo-4-methylpentane Analyze: In reaction a) methanol is the nucleophile and the alkyl halide is the electrophile. The alkyl halide is secondary and sterically inhibits an SN2 reaction. Methanol is a small weak base and thus a poor nucleophile. A SN2 reaction will not occur. Is a SN1 reaction possible? Yes, the solvent is polar protic and will help stabilize the formation of a carbocation and anion. Once the carbocation has formed the solvent becomes the nucleophile (solvolysis). In this case two products are formed because the attack of the nucleophile can occur on either face of the carbocation to produce a pair of enantiomers. Br H heat Br + CH 3OH + CH 3OH -H + -H + H3C H3 C O O H H3 C H C 3 H (R)-2-methoxy-4-methylpentane (S)-2-methoxy-4-methylpentane Dr. Steven Forsey 18 In reaction b) the methanethiolate anion is the nucleophile and the alkyl halide is the electrophile. The alkyl halide is secondary and sterically inhibits the attack by a nucleophile. Will a SN1 reaction occur? Not likely because the solvent does help stabilize the formation of a salt. Actually the solvent decreases the stability of an anion and thus increases its nucleophilicity. Is the negatively charged sulfur atom a good nucleophile? Yes it is. The thiolate anion is a polarizable weak base and is thus a good nucleophile. This will be a SN2 reaction. Since the attack occurs on a stereogenic carbon inversion will occur to produce a chiral product that is optically active. Br + + H3C S Na acetone H 3C S + Na + Br (S)-methyl(4-methylpentan-2-yl)sulfane 2.0 Elimination Reactions of Haloalkanes: E1 and E2 Mechanisms Haloalkanes can undergo an elimination reaction, in which a hydrogen atom and a halogen atom are removed from the molecule. As is the case for substitution reactions of haloalkanes, elimination reactions of haloalkanes can occur by different mechanisms. If the ratedetermining step is unimolecular, then the mechanism is called E1. If the ratedetermining step is bimolecular, the mechanism is called E2. In this section, we shall explore these mechanisms in some detail and discuss the competition that occurs between elimination and substitution reactions. Elimination General Reaction H Base - B Br + B H + Br 2.1 E1 Mechanism We now know that an attacking species can either act as a base or a nucleophile. The attacking species in all of the reactions we examined so far either attacked exclusively as a base or a nucleophile. However life is hardly ever so black and white. For H3C H 3C example when 2bromo2methyl CH3 + Br C Br C propane is dissolved in methanol the H 3C H 3C expected solvolysis reaction occurs, H3C CH 3OH 2-bromo-2-methylpropane however a second product is also OCH3 formed, 2methylpropene (19%). Kinetic H H 3C studies show that the formation of the C + H st H3C alkene follows 1 order kinetics and the + H C C CH3 rate is dependent only on the alkane. H 3C H H 3C The second product is formed by the 2-methoxy-2-methylpropane 2-methylpropene elimination of HBr from the substrate (81%) (19%) and since the reaction follows first order kinetics, this is called an E1 reaction. SN1 E1 Substituting -OCH 3 Elimination of HBr and removing -Br Dr. Steven Forsey 19 H 3C H3C H3C But how and why is the reaction occurring? Let's look at the mechanism. After the formation of the carbocation, methanol can act like a nucleophile and produce the SN1 product. But methanol is also a weak base, thus after the formation of the intermediate methanol can also abstract a proton to produce the alkene. CH 3 H CH3 O O H 3C -H SN1 mechanism CH3OH C CH3 + Br C C H 3C H3C H3C CH 3 CH3 H 3C H3C C Br H H H3C H E1 mechanism H3C CH3OH C C + H3C O C CH 2 + Br H H3C H H3C Thus when the carbocation is formed a mixture of SN1 and E1 products are produced. It is difficult to influence the ratio of SN1 and E1 products because they both proceed through the same intermediate. However, generally tertiary alkyl halides produce a mixture of products with the major product being the substitution product. This is because it takes less energy to form a sigma bond in a SN1 reaction than it does to form a new basehydrogen bond, break the carbonhydrogen bond and form a new bond i.e. the more bond changes in the E1 reaction increases the barrier to the reaction compared to a SN1 reaction. To increase the amount of the elimination product your could increase the temperature of the reaction which would increase the number of molecules with enough energy to surmount the energy barrier of the E1 reaction but this would not stop the SN1 reaction from occurring and thus a mixture of products would still form. A stronger base could also be used but a mixture of products is still produced because a strong base is also generally a good nucleophile. Thus the ratio of SN1 and E1 products remains approximately the same. Synthetically this method is not widely used because mixtures of products are produced and other methods can be employed to obtain a major product. Dr. Steven Forsey 20 2.2 E2 Mechanism What reactions conditions do we need to produce the elimination product as the major product? As stated, increasing the base strength may increase the yield of the E1 product but if a carbocation is formed a substitution product will also be produced. What would happen if we increased the base strength to such an extent that the rate of abstracting a proton was greater than the rate of formation of the carbocation intermediate? What type of reaction mechanism would this follow? Let's do an experiment. Since we do not want a substitution reaction to occur we should choose a tertiary substrate such as 2bromo2methyl propane. If we then combine 2bromo2methyl propane with a strong base, what product would be formed and what does the kinetics tell us about the mechanism? When we perform the reaction, 2methylpropene is produced and the kinetics follows a bimolecular reaction. The rate of the reaction is dependent on both the substrate and the base. H3C H3C H3C C Br + CH 3CH2O Na CH 3CH2OH H3C C H3C C H H Rate = k[(CH3)3CBr][CH3CH2O] This means that the transition state involves both species and a carbocation intermediate does not form. Since this is a bimolecular elimination reaction that occurs in one step it is abbreviated as E2, for Elimination, bimolecular. Notice that the reaction condition uses a polar protic solvent that can stabilize the formation of a carbocation. Why does a carbocation not form and produce SN1/E1 products? Well, it might if you use a low concentration of the strong base or perform the reaction at a reduced temperature. H3C H 3C H3C C Br + CH3CH2O Na CH3CH 2OH 25oC H3C Mainly E2 (91%) H 3C C C H H + H 3C H3C H3C C OCH2CH3 SN1 (9%) But if the reaction conditions are done correctly with a slight excess of the strong base and at higher temperatures the E2 reaction will dominate. This is because the strong base is very reactive and wants to abstract a proton or react with an electrophilic carbon to form a weaker acidbase pair. Since the SN2 reaction can not occur with the tertiary alkyl halide, the strong base will abstract a proton and follow the Dr. Steven Forsey 21 fast exothermic E2 reaction pathway before a carbocation can form. Remember the first step in SN1/E1 reactions is an endothermic process that produces an unstable intermediate that is slow to form. Thus if two reaction pathways are possible the fast exothermic reaction will dominate the slow endothermic reaction. 3.0 How to determining if an SN2, E2, SN1 or E1 reaction will occur with Primary and Secondary Alkyl Halides. So far we have discussed what happens when a strong or weak base is added to a tertiary alkyl halide to form E2 and SN1/E1 products respectively. What is the ratio of products when a strong base or a weak base is reacted with a primary or a secondary alkyl halide? Which reaction dominates SN2, E2 or SN1/E1? Let's consider a series of experiments and find out what happens and then rationalize the results by investigating the attacking species (nucleophile/base), the species being attacked (electrophile/acid) and the solvents. 1) CH3CH 2CH 2Br + CH3OH Primary Poor nucleophile Weak base No Reaction In the first experiment no reaction is observed because methanol is a weak base and a poor nucleophile. Methanol is not reactive and the only possible reaction that methanol may enhance because it is a polar protic solvent, is the formation of a carbocation. However primary carbocations are very unstable and unlikely to form in these conditions. Thus no reaction is observed. In the second reaction the attacking species is a good nucleophile and is a weak base. The electrophile is a primary alkyl halide, thus SN1/E1 reactions will not occur since primary carbocations are unstable and will not form especially in an aprotic solvent. What about E2 and SN2 reactions. Since the attacking species is a weak base E2 products are not formed but because it is a polarizable good nucleophile a SN2 reaction readily occurs on the unhindered primary electrophilic carbon. In reaction 3 reaction the substrate is the same but now the attacking species is both a good nucleophile and a strong base thus both SN2 and E2 reactions are possible. In this reaction we find that the SN2 reaction dominates because there is little steric interaction around the electrophilic carbon, however because the ethoxide ion is a strong base a minor E2 product is formed. Dr. Steven Forsey 22 In the fourth experiment the substrate is primary however, there is branching close to the electrophilic carbon. Branching this close the carbon sterically interferes with the attack of the nucleophile and impedes the reaction. Since the nucleophile is also a strong base the elimination reaction becomes dominates. In the experiment 5 the attacking species is sterically hindered and even though the electrophile is primary the steric interaction caused by the branching of the nucleophile greatly increases the energy barrier or activation energy of the SN2 reaction. Since it is also a strong base the E2 reaction becomes the major pathway. Notice, in reactions 3, 4 and 5 polar protic solvents were used which promotes SN1/E1 reactions. Will an SN1/E1 reaction occur? No. As seen in reaction 1 in which no reaction occurs with the polar protic solvents and a primary alkyl halide. The reaction of secondary alkyl halides with nucleophiles and bases cause students the most problems because they can undergo both types of substitution and elimination reactions. Let us examine some experiments and discuss the results. In experiment 6 the attacking species is a weak base and a poor nucleophile thus we would not expect a SN2 or E2 reaction. However, water will help stabilize the formation of a secondary carbocation intermediate. Secondary carbocations are not very stable thus heat is needed to make this reaction proceed to the SN1/E1 products. In this reaction we are using a better nucleophile than water that is still a weak base. A weak base does not favor elimination reactions. Also, the solvent is acetone which is a polar aprotic solvent that enhanced the nucleophilicity of the attacking species and promotes SN2 reactions over SN1 reactions. Thus we observe approximately 100% of the SN2 product. Dr. Steven Forsey 23 Br 8) CH 3CHCH 3 Secondary + CH 3CH2O Na Good nucleophile Strong base CH 3CH2OH H3C OCH 2CH 3 CH3CHCH3 SN2 (13%) + H C C H H E2 (87%) In the final reaction we are using a good nucleophile that is also a strong base. The secondary alkyl halide is sterically hindered towards a nucleophilic attack. Thus the E2 reaction dominates. Shouldn't we also get SN1/E1 products like what was observed in reaction 6 since we are using a polar protic solvent that enhances SN1/E1 reactions? The answer is not likely. The fast exothermic SN2 and E2 reactions will occur before the formation of the unstable secondary carbocation which is a slow endothermic process. 4.0 Regiochemistry and Stereochemistry 4.1 E2 Reactions with Unsymmetrical Alkyl Halides In the elimination reactions that we have studies so far only one elimination product was possible because the alkyl groups bonded to the carbon of the leaving group were the same. The molecules were symmetrical and it did not matter which hydrogen was removed because the same elimination product would be formed. For example abstraction of any of the 9 hydrogens of 2bromopropane will produce only one product, propene. Br H H + HOCH3 + OCH3 + CH3O H H H H symmetrical However, in many cases more than one elimination product is possible because the molecule is not symmetrical. For example when 2bromobutane is reacted with sodium methoxide in methanol three E2 products are possible, one alkene that has the double bond at the end of the carbon chain, 1butene and two with internal double bonds, cis2butene and trans2butene. 1Butene is a constitutional isomer of the two internally double bonded alkenes and the cis/trans isomers are stereoisomers, more specifically diastereomers. To obtain the constitutional isomers abstraction of the hydrogens must have occurred on different carbons as shown below. Br H CH3 O + H H Br CH 3O H + H H tr ans-2-butene 67% H H cis-2-butene 15% H H 1-butene H H 18% Thus two different regions of the molecule where involved in the reaction mechanism to create the constitutional isomers. This is called regiochemistry and this reaction is also regioselective, meaning that the reaction leads to the preferential formation of one constitutional isomer over another. For this example the internal double bounded alkenes are formed preferentially over the terminally double bonded 1butene. total internal double bond = 82% Dr. Steven Forsey 24 To determine if constitutional isomers will form, label the carbons and look at the groups. If they are different then constitutional isomers are possible. If they are the same then the same product will form. Another possibility is that the carbon has no hydrogens on it. In this case an elimination reaction can not occur. Consider the following reaction between 2bromo2methylbutane and OH-. H H3CH 2C Less highly C C substituted H alkene H 3CH2C H3C H2O + Br- + C Br + OH - 2-methyl-1-butene H 3C H 3C More highly CH 3 H 3C substituted C C alkene H CH 3 2-methyl-2-butene There are two groups of hydrogen atoms that can be eliminated. If a hydrogen from one of the methyl groups (shown in red) is removed in the elimination reaction, then the product is 2 methyl1butene. If a hydrogen from the ethyl group (shown in blue) is eliminated, then the product is 2methyl2butene. Which product is the major product? More often than not, the major product will be the one that is most stable In general, more highly substituted alkenes are more stable than less highly substituted alkenes. The 2 more highly substituted alkenes have fewer hydrogens bonded to the sp carbons, so another statement of Zaitsev's rule is: The smaller the number of hydrogens bonded to the sp2 carbons of an alkene, the greater is its stability. According to Zaitsev's rule, the major product obtained from the reaction of 2 bromo2methylpropane and OH- will be 2methyl2butene (with only one hydrogen atom bonded to the sp2 carbons). Increasing stability of the alkene due to increasing substitution is attributed the electron donating ability of the alkyl groups through hyperconjugation in which adjacent sigma bonds contribute electron density to the bond. This is similar to the trend seen in carbocation stability. Also bulky alkyl groups like to be as far as apart as possible. This is better accomplished with a sp2 hybridized carbon than a sp3 hybridized carbon that has bond angles of 120o and 109.5o respectively. H H H R H R R R < < < CH2 CH2 < RCH CH 2 < C C C C C C C C R cis R tr ans H R R R R R Relative Stability of Alkenes The reaction between 2bromo2methylbutane and OH- to form 2 methyl1 butene and 2methyl2butene is an example of kinetic control of a reaction. The trisubstituted alkenelike transition state is more stable than the disubstituted transition state and thus has a lower free energy of activation. Since more molecules in the system will have enough kinetic energy to surmount the lower energy barrier than the higher energy barrier the more stable product should form at a faster rate. This relates a thermodynamic Dr. Steven Forsey 25 property to a kinetic rate but in most cases this is what is observed. The more stable product is formed faster and is the major product formed. From the chart of relative stabilities of alkenes given above we see that the trans isomer is more stable than the cis isomer, this is because the alkyl groups of the cis isomer are sterically close to each other and is energetically unfavorable. This observation brings us back to our reaction with 2bromobutane and sodium methoxide in which three products were formed. We now understand that the compounds with the internal double bond will be favored because they are more substituted and the trans stereoisomer will be the most abundant because it is the most stable of the three isomers. But we should ask the question, how are the two stereoisomers formed. First we must look at the E2 mechanism more closely. For an E2 reaction to occur the hydrogen and the leaving group must be in the same plane. This is because in the transition state, as the hydrogen and bromine atoms are leaving the carbons are changing hybridization 3 2 from sp to sp and for a new bond to form the developing p orbitals must be in the same plane. Remember bonds are created from the sideways overlap of two p orbitals. Dr. Steven Forsey 26 Now consider the elimination reaction of 2bromobutane that produces the two stereoisomer, cis and trans2butene. There are two different hydrogens that can be removed to produce the internal double bond and depending on the molecules conformation, different alkenes will be produced. Therefore this reaction is stereospecific. In the diagram given below if the blue hydrogen is removed the trans isomer is produced. If we change the conformation so that, the pink hydrogen is in the same plane as the bromine the cis isomer is produced. To determine the stereochemistry of the reaction it is best to redraw the structures as Newman projections. In the Newman projections given below we see that when the methyls are anti the trans isomer is produced and when the methyls are gauche the less stable cis isomer is produced. Worked example: Question: Predict the elimination products of the following reaction. Which alkene would be the major product? Br CH 3ONa / CH 3OH Analyze: The elimination reaction is performed with a strong base, so we know this is an E2 reaction. Next identify the carbons. There are two carbons with different groups attached to them, thus the Dr. Steven Forsey 27 regiochemistry of the reaction will produce two different constitutional isomers. But we must also be concerned with the stereochemistry of the internal double bond because (E) (Z) isomers are possible. However, notice that there is only one hydrogen, thus only one stereoisomer will be produced. Which one is it? It is best to draw the Newman projection to determine the stereochemistry. First determine the terminal alkene Br H CH3O H CH3 H H (R)-3-methyl-1-pentene Next redraw the diagram as a Newman projection. To do this, in your minds eye look down the - carboncarbon bond. The pink methyl is pointing down and the ethyl group is up. The red hydrogen and the methyl group are on the same side and the bromine and the blue hydrogen are on the same side. Make the Newman projection. Next rotate the diagram so that bromine and the hydrogen are anti to each other. This is the conformation that produces the double bond. Notice that the ethyl and the pink methyl are on the same side and the red hydrogen and the black methyl are on the same side. When you draw the alkene keep this orientation. looking down H3C CH2CH 3 Br H carbon-carbon bond CH2CH 3 Br H3C rotate -HBr H Br CH 3 H CH 3 H CH3 CH 3 H H H CH 3 CH 3 CH 3O (Z)-3-methyl-2-pentene The product formed from this reaction will be (Z)3methyl2pentene. This is the major product because it is more substituted than (R)3methyl1pentene. H3C 4.2 SN1/E1 reactions and Unsymmetrical Alkyl Halides The regiochemistry and stereochemistry is very important in the reaction of unsymmetrical alkyl halides under SN1/E1 conditions. For example the reaction of (2 bromobutan2yl)benzene produces 5 different products. Two stereoisomers, (R) and (S) enantiomers, and three alkenes in which two are cis and trans diastereomers. Because of the many products that can form with SN1/E1 reactions, SN1/E1 reactions are not used frequently in synthetic organic chemistry. O Br CH 3OH heat -Br no hydrogens H C H3C C H H C C H H CH 3OH SN 1 -H + H3C O H 3C C 6H 5 H 3CH 2C C C6H5 C H C H3CH 2C (R) CH3 C 6H5 SN1 -H+ CH2CH 3 (S) H H C H 3C (Z) C CH3 H CH 3OH E1 H H 3C H C C C H H HOCH3 E1 C 6H5 H 3C C H (E) C CH 3 CH 3OH E1 H 3C H C H C CH 3 C 6H 5 Dr. Steven Forsey 28 5.0 Carbocation Rearrangement and SN1/E1 reactions Another potential complication with SN1 and E1 reactions is the rearrangement of the carbocation intermediate to form more stable intermediates before the SN1 or E1 product is formed. For example when 2bromo3methylbutane is heated with water the major SN1 product is not 3methyl2butanol but 2methyl2butanol. This product was produced by the rearrangement of the carbocation through the movement of a hydrogen and its two electrons. This is called a hydride shift. CH3 H3C C H CH Br CH3 H 2O SN1 (-HBr) H 3C CH3 C OH 2-methyl-2-butanol 2-bromo-3-methylbutane major CH2 CH3 + H 3C CH3 C H CH OH CH3 3-methyl-2-butanol minor How did the hydride shift occur? Let us look at the mechanism. The first step is the formation of the secondary carbocation. The carbocation is stabilized by hyperconjugation or the donation of electron density through adjacent sigma bonds. hyperconjugation CH3 H H3C C H CH Br CH 3 H 3C sp3 C C H sp2 C H H H sp3 + Br H3C The next step in the mechanism is the transfer of a hydrogen and its two electrons to the electron deficient carbocation. There are two possible hydride shifts that could occur. The first conceivable hydride shift is the shift of the blue hydrogen in the diagram below to produce a more unstable primary carbocation. This energetically not favored and will not occur under normal conditions. The shift of the red tertiary hydrogen to the electron deficient carbon would produce the more stable tertiary carbocation, and this is what is observed. The hydride shift is very fast and is generally much faster than a SN1 or E1 reaction because this is an intramolecular rearrangement that occurs within the molecule to create a more stable compound, unlike a SN1/E1 reaction which involves the intermediate colliding with another molecule. After the rearrangement to a more stable intermediate an SN1 reaction occurs to form the major product. H H H 3C C C hyperconjugation CH 2 CH 3 H b H H H CH3 b 1 o carbocation a H less stable than 2 o H C C C H 3C CH H CH3 C H C H H H 3C H (H3 C) 2HC a o H Br H3 C 1 carbocation H H 3C C C CH 2 + Br CH3 H H2 O o 3 carbocation hyperconjugation more stable than 2 o H H C CH3 H H H OH H H O H H H C C H 2O H3 C H3 O + H3 C C C CH2 C H 3C C C CH 2 H 3C H CH 3 H CH3 H 3 carbocation o H Dr. Steven Forsey 29 Anytime a carbocation forms there is the possibility of a rearrangement. The rearrangement occurs to form the more stable carbocation and does not only involve the movement of a hydrogen but also methyl groups or whole alkyl groups. Students often ask how do you know what shifts the hydrogen, a methyl or a whole alkyl group. The answer is that the molecule will always rearrange to form the most stable carbocation. What ever you move the carbocation must become more stable. If it does not then the shift does not occur. Given below is an example of a methyl group shift. CH3 H 3C C CH CH3 H 2O SN1 CH3 H 3C CH3 C CH + CH3 methyl shift Br H3C C CH 3 CH CH 3 CH3 Br 3-bromo-2,2-dimethylbutane 2o carbocation H2O H OH H2O O CH CH 3 H3C C CH 3 3o carbocation H H 3O + H3C C CH CH3 CH3 CH3 2,3-dimethyl-2-butanol CH 3 CH 3 6.0 Sterically Hindered nucleophile/base Zaitsev's rule is a useful guideline but, as is often the case, there are important exceptions. In the examples above, the formation of the most highlysubstituted alkene requires that the base attack a hydrogen atom bonded to a secondary or tertiary carbon. Access to the hydrogen atoms on these carbons is hindered by the adjacent groups. Small nucleophiles, such as OH- or CH3O-, can get past the groups but larger nucleophiles, cannot. Consequently, larger nucleophiles such as (CH3)3CO-, which are also poor nucleophiles will preferentially attack the more exposed hydrogens yielding the less stable and least highly substituted alkene. This is called the Hofmann rule. The transition state leading to the more stable product is increased in energy by steric interference with the bulky base and the reaction produces the thermodynamically less favored isomer CH 3 H3C CH3 H3CH2C CH3CH2ONa, CH3CH2OH, 70oC C C + C CH 2 CH 3CH 2CCH 3 -HBr H CH3 H3C Br 30% 70% 2-methyl-2-butene 2-methyl-1-butene 2-bromo-2-methylbutane CH 3 H3C CH3 H3CH2C (CH 3) 3COK, (CH 3)3COH C C + C CH 2 CH 3CH 2CCH 3 -HBr H CH3 H3C Br 27% 73% 2-methyl-1-butene 2-methyl-2-butene 2-bromo-2-methylbutane Dr. Steven Forsey 30 7.0 Overview When determining if a reaction will proceed through a SN2, SN1, E2 or E1 mechanism you must consider both the nucleophile and the electrophile. You must consider the nucleophiles nucleophilicity, basicity and if it is unhindered or hindered. You must determine if the electrophile is sterically hindered or unhindered towards a nucleophilic attack. Will the electrophile undergo cleavage to form a carbocation intermediate? What effect does the solvent have on the reaction? All of these factors must be considered and this is what makes this section of organic chemistry difficult for many students. The follow is a stepwise thought process that can be used to determine if the reaction will proceed through a SN2, SN1, E2 or E1 mechanism. I do not necessarily follow this order but I do think about all of the concepts given below when answering a question or making up a question. Step 1 Put delta and full charges on all atoms. Put lone pairs on the atoms and identify chiral atoms. Step 2 Determine the attacking species that donates electrons (nucleophile or base) and the species that can accept the electrons (electrophile or acid) Step 3 Classify the electrophile as primary, secondary or tertiary Step 4 Classify the attacking species as a strong or weak nucleophile Step 5 Classify the attacking species as a strong or weak base Step 6 Classify the attacking species as unhindered or hindered Step 7 Classify the solvent as protic or aprotic Step 8 At this point you will be able to say if the reaction will preferentially undergo a SN2, E2 or SN1/E1 mechanism. Now think about how the mechanism will proceed. Identify the alpha and beta hydrogens and use arrows to determine the substitution and elimination products that will form depending on what you have decided. Always think about the regeochemistry and the stereochemistry of the reactions. Also think about the relative stability of the alkene products, if they form. You must also always consider the products that are formed because no reaction is also an option. For the reaction to proceed to form products the leaving group must be a weak base and the products formed must be a weaker acid/base pair than the reactants. Dr. Steven Forsey 31 Substrate CH 3X RCH2X o Primary (1 ) R R CH X Methyl Possible Reaction Explanation SN2 SN2: Sterically unhindered electrophile easy access for nucleophilic attack An elimination reaction cannot occur SN2 E2 SN2: Main reaction with strong nucleophiles (e.g., RS, I, CN) or strong bases (e.g., RO). The alkyl halide is sterically unhindered and will readily undergo an SN2 E2: An E2 reaction will mainly occur with a hindered strong base (e.g., (R3CO) SN1/E1 does not occur because a primary carbocation will not form Tertiary (3o) SN1/E1 SN2, E2, SN1/E1 E2 SN2: main reaction when SN1: main reaction if a weak using good base/poor nucleophile (e.g., nucleophiles/weak bases ROH) is used in a polar (e.g., RS, I, CN, RCO2). An protic solvent (e.g., ROH). aprotic solvent. (e.g., Solvent is most likely the acetone, DMF,THF) enhances nucleophile. E1 minor the reactivity of a nucleophile products will also form. Remember to think about E2 internal double bond: regiochemistry and Main reaction when an stereochemistry. Anytime a strong base (e.g., RO) is carbocation forms used. Do not forget (E) and rearrangement can occur (Z) isomers may form E2: Main reaction with a SN1: main reaction if a weak strong base (e.g., RO, base/poor nucleophile (e.g., R3CO) Remember to think ROH) is used in a polar protic about regiochemistry and solvent (e.g., ROH). Solvent is stereochemistry most likely the nucleophile. An SN2 reaction can not E1 products will also form. occur because of steric Remember to think about interaction. The nucleophile regiochemistry and is completely blocked from stereochemistry. Anytime a a nucleophilic attack with a carbocation forms tertiary alkyl halide rearrangement can occur Secondary (2o) R R C R X Dr. Steven Forsey 32 Worked examples: For each of the following reactions determine if the reactions will follow a SN2, SN1, E2 or E1 reaction mechanism. Give all possible products and indicate the major product. NaSCH3 Br a) acetone CH 3ONa/CH 3OH b) Br c) NaCN acetone Cl Br CH 3 CH 3OH d) O Analyze: Reaction a) Step 1 Na SCH 3 Br a) acetone + Step 2 The negatively charged sulfur compound (an alkanethiolate) can either act as a nucleophile in a substitution reaction or a base in an elimination reaction. Step 3 The electrophile is primary and will readily undergo a SN2 reaction and will not form a primary carbocation. Step 4 The negatively charged alkanethiolate is a good nucleophile, polarizable. Step 5 The alkanethiolate is a weak base. Step 6 The alkanethiolate is unhindered. Step 7 The solvent is aprotic and promotes SN2 reactions. Will not stabilize the formation of a carbocation Step 8 The alkanethiolate ion is a strong nucleophile that is attacking an unhindered primary alkyl halide. The dominate reaction will be a SN2 reaction. The E2 product will be very minor because the alkanethiolate is a weak base. However, since the question asked for all possible products the mechanism for the SN2 and E2 reactions are given. Both reactions could proceed as written. The bromide ion is a good leaving group and is a weaker base than the alkanethiolate. S Br Br + Na + + a) acetone S 2 major product SCH 3 H Br acetone + H SCH3 + E2 minor product Br + Na N H Dr. Steven Forsey 33 Reaction b) Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 + b) Br CH 3O Na CH3OH Step 8 The negatively charged methoxide ion can either act as a nucleophile in a substitution reaction or a base in an elimination reaction. The electrophile as secondary. The reaction may undergo SN2, E2, or SN1/E1 reactions depending on the conditions. The methoxide ion is a good nucleophile. The methoxide ion is a strong base The methoxide ion is unhindered The solvent is protic which promotes the formation of a carbocation; however the formation of a carbocation is not likely since a strong base is present and a fast SN2 or E2 exothermic reaction will occur dominate over a slow endothermic process. The attacking species is both a strong base and a good nucleophile but because it is attacking a sterically hindered secondary alkyl halide an E2 reaction will dominate. Determine the products being careful of the stereochemistry and regeochemistry of the reactions. Since the question asked for all possible products the mechanism for both the SN2 and E2 reactions are given. + Br OCH 3 H H CH 3OH + H OCH 3 + Br + Na E2 product CH 3O Na OCH3 (R) H + Br + Na SN2 minor product b) (S) - CH3OH Br CH 3 C H CH3O C Br CH 3 CH 3O H H 3C H CH3OH cis + H OCH3 + Br + Na E2 product H H H 3C C C H CH3OH tr ans + H OCH 3 + Br + Na E2 product Br The mechanism has shown that three E2 products are possible. Since the base is unhindered the base will have not suffer significant steric interaction in reaching the hydrogens on the internal beta carbon. Thus the internally double bonded compounds will form more readily since they are more stable. The trans product is the most stable alkene and will be the major product formed. Both substitution and elimination reactions could proceed as written. The bromide ion is a good leaving group and is a weaker base than the methoxide ion. Dr. Steven Forsey 34 Reaction c) Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 8 c) (S) (R) + - Cl Na CN acetone The negatively charged cyanide ion can either act as a nucleophile in a substitution reaction or a base in an elimination reaction. The electrophile is secondary. The reaction may undergo SN2, E2, or SN1/E1 reactions depending on the conditions. The attacking species is a strong nucleophile The attacking species is weak base (pka of HCN = 9.1, pKb of CN = 4.9) The attacking species is unhindered The solvent is aprotic and promotes SN2 reactions and will not stabilize the formation of a carbocation. The major reaction pathway will be SN2. The alkyl halide is secondary and sterically hindered, however, the attacking species is a very good nucleophile and a weak base and will preferentially attack a carbon center. The SN2 reaction will dominate and the E2 products will be minor. Determine the products being careful of the stereochemistry and regeochemistry of the reactions. Since the question asked for all possible products the mechanism for the SN2 and E2 reactions are given. (R) Na CN acetone H CN H H acetone c) (S) + Cl (S) (S) + CN Cl + Na S N2 major product (S) + H CN + Cl + Na E2 product Cl NC H C H H C Cl CH3 H H acetone (S) (E) H CH 3 + H CN + Cl + Na E2 product NC H H C H C Cl CH3 H H H (S) acetone (Z) CH 3 + H CN + Cl + Na E2 product Both substitution and elimination reactions could proceed as written. The bromide ion is a good leaving group and is a weaker base than the cyanide ion. Dr. Steven Forsey 35 Reaction d) Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 8 - d) Br + (R) O CH 3 + - + CH3OH Methanol potentially can act either as a nucleophile or a base The electrophile is tertiary. A SN2 reaction can not occur The attacking species is a weak nucleophile The attacking species is a weak base. An E2 reaction will not occur The attacking species is unhindered The solvent is protic and promotes SN1 reactions through the stabilization of the carbocation intermediate. The solvent is also a potential nucleophile. This can only be a SN1/E1 reaction. Determine the products being careful of the stereochemistry and regeochemistry of the reactions. For tertiary alkyl halides under standard conditions the SN1 product is always major. The ratio of SN1/E1 products will depend on the temperature. CH3 H3C Br + O (R) CH3 + - + CH 3OH SN1/E1 O H O CH3 H O H C O H H3 C O H H 3C H3C O H H3C H3 C H H O (E) O CH3 O + H3C O H H H H (Z) O O CH3 + H 3C O H E1 products minor H H H O H CH3 H + H 3C O H H O H3 C H 3C (S) CH3 O H + Br -H+ O -H+ (R) CH3 O d) O SN1 products major H CH3 O O H + H3C O H Both substitution and elimination reactions will proceed as written. The bromide ion is a good leaving group and is a weaker base than methanol. Dr. Steven Forsey 36 8.0 Vinyl, aryl, allylic and benzylic compounds benzylic H allylic Allylic and benzylic compounds are structurally Br C H Br C C different than vinyl and aryl compounds and react 3-bromoprop-1-ene H very differently. The halogen bonded to aryl and vinyl H or allyl bromide (bromomethyl)benzene compounds is bonded directly to the sp2 hybridized or benzyl bromide carbon. Where as the halogen bonded in a benzylic or 3 allylic compound is bonded to a sp hybridized carbon aryl vinyl H Br which is then bonded to the sp2 hybridized carbon. C Br H Br C Allylic and benzylic alkyl halides are very reactive in bromoethene H both SN1 and SN2 reactions and vinyl and aryl or vinyl bromide sp2 bromobenzene compounds are not reactive. For SN1 reactions the primary factor that determines the rate of the reaction is the relative stability of the carbocation. Allylic and benzylic carbocations are more stable than tertiary carbocations because they are resonance stabilized where as tertiary carbocations are stabilized by hyperconjugation. Vinyl and benzyl carbocations are very unstable because they can not be resonance stabilized. The empty sp2 orbital is 90o with respect to the bond and can not interact with the bond. H H allylic Br Br -Br C H C H H C + C + C C H resonance stabilized 3-bromoprop-1-ene H H H charge is delocalized onto two carbons H sp2 or allyl bromide sp3 p orbital H H H 3C C stabilized by hyperconjugation - inductive electron donation -Br H C C Br C through bonds H H H3 C not as stabilizing as resonance stabilization H C H H3 C H H 2-bromo-2-methylpropane H vinyl H -Br C C sp2 orbital H Br H Br C C very unstable bromoethene sp2 can not be resonance stabilized H H or vinyl bromide sp2 Thus the order of reactivity of alkyl halides is: Br H sp3 Dr. Steven Forsey 37 Benzylic and allylic compounds are very reactive in SN2 reactions because the adjacent bond helps to stabilize the transition state. Previously we discussed the mechanism of SN2 reactions and how the high energy transition state involves both the nucleophile and the electrophile. In the transition state there are 5 groups surrounding the central carbon atom and the carbon atom will take on a trigonalbipyramidal shape with three coplanar groups. The hybridization of the carbon atom must be sp2 with bond angles of approximately 120o between the 3 coplanar groups. 2 sp 2 In the transition state of the allylic and benzylic compounds, the adjacent bond interacts with the sp hybridized carbon and delocalizes the negative charge. The delocalization of the negative charge helps to stabilize the transition state and reduces the energy barrier to the reaction. Thus benzylic and allylic compounds react very quickly in SN2 reactions Br H Br H allylic C C H H H Br Nu H C C C C H H H H Nu 2 Nu sp transition state sp2 hybridized transition state stabilized by adjacent bond through delocalization of electrons The relative rates of reaction for SN2 reactions are: Dr. Steven Forsey 38 ...
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