Organic chemistry 8th edition (2017) part 2

416  Chapter 9: Nucleophilic Substitution and b-Elimination Because the solvent is polar protic, there could be a minor extent of SN1/E1 If the sodium acetate were left out of the reaction and it were heated, the prediction would be SN1/E1 Competition The haloalkane is secondary The Nuc/Base has a pKa of its conjugate acid near or slightly below 11 (HCN, pKa 9.3) and hence is a moderate to weak base However, cyanide anion is an excellent nucleophile Consequently, SN2 will dominate over E2 Furthermore, the solvent DMF (dimethylformamide) is polar and aprotic and supports SN2 or E2, but it does not assist SN1 or E1 Because the reactant is chiral, the SN2 inverts the configuration Competition The haloalkane is tertiary; therefore, SN2 cannot occur The Nuc/Base is a weak base (pKa HN3 4.9); therefore, E2 is not obviously favored However, the solvent is not protic but is simply polar Therefore, SN1 and E1 are not going to be favored This is a case that is difficult to predict using Figure 9.8 or Table 9.11 However, the lack of a polar protic solvent means that E2 is most likely, even with the weak base The E2 occurs with an anti and coplanar arrangement of the Br and H that are eliminated, giving an E alkene Any substitution from an SN1 pathway would lead to racemization of the chiral center that possessed the leaving group 1 Example 9.8  SN1 or SN2, E1 or E2 Predict whether each reaction proceeds predominantly by substitution (SN1 or SN2) or elimination (E1 or E2) or whether the two compete Write structural formulas for the major organic product(s) (a) (b) Solution (a) A 3° haloalkane is heated with a strong base/good nucleophile Elimination by an E2 reaction predominates to give 2-methyl-2-butene as the major product Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 9.9  Analysis of Several Competitions Between Substitutions and Eliminations   417 1 (b) Reaction of a 1° haloalkane with this moderate nucleophile/weak base gives substitution by an SN2 reaction 1 Problem 9.8 Watch a video explanation Predict whether each reaction proceeds predominantly by substitution (SN1 or (a) SN2) or elimination (E1 or E2) or whether the two compete Write structural (a) formulas for the major organic product(s) (b) (a) 1 (c) (b) 1 (c) 1 2 1 1 2 (c) (b)    1 2 MCAT Practice: Passage and Questions Solvents and Solvation Choosing the best solvent for a chemical reaction is an extremely important aspect of organic chemistry When deciding upon a solvent, chemists consider the solubility of the reactants and products, as well as the mechanism of the reaction and the solvation of intermediates Further, for reactions that need heating to proceed in a reasonable amount of time, the choice of solvent is guided by its boiling point because this sets the temperature at which the reaction refluxes Lastly, unless the solvent is intentionally used as a reactant, such as in a solvolysis, it must remain inert Questions A When performing an SN1 solvolysis, which of the following solvents would be a poor choice for tertbutyl iodide (“dried” means that water has been removed)? 80% water, 20% ethanol Pure water Dried acetonitrile Dried acetic acid B When attempting to enhance the extent of SN2 substitution by the nucleophile ethylamine (EtNH2), which of the following solvents would be a poor choice for sec-butyl iodide? Pure water Acetonitrile DMSO tert-Butyl alcohol C When performing an SN2 reaction using NaCN as the nucleophile reacting with n-butyl iodide, which of the following solvents would be the worst choice? DMSO DMF Acetonitrile Toluene D The reaction of diethylamine (Et2NH) and sec-butyl iodide requires heating, but to optimize the extent of SN2 over E2 the reaction cannot be too hot Which of the following solvents would best represent a compromise solvent in which to reflux this reaction? Diphenyl ether Diethyl ether THF DMSO Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 418  Chapter 9: Nucleophilic Substitution and b-Elimination An important take-home lesson from this chapter is that understanding key transition state or reactive intermediate geometries as well as relative transition state energies allows the prediction of product stereochemistry and regiochemistry Backside attack in SN2 reactions, the anti and coplanar geometry of the H atom and leaving group in E2 reactions, and the presence of carbocation intermediates in SN1 reactions are important examples of reaction geometries that dictate stereochemistry ­Understanding the relative energies of alternative p ­ ossible transition states is also important In the case of b-elimination reactions, relative transition state energies provide a rationale for Zaitsev’s rule of regiochemistry As you go through the rest of this book, try to learn key features of reaction mechanisms that dictate the stereochemistry and regiochemistry of reaction products You should think of mechanisms as more than just electron pushing: they involve three-dimensional molecular interactions with associated relative energies that control the formation of products 9.10  Neighboring Group Participation So far, we have considered two limiting mechanisms for nucleophilic substitutions that focus on the degree of covalent bonding between the nucleophile and the substitution center during departure of the leaving group In an SN2 mechanism, the leaving group is assisted in its departure by the nucleophile In an SN1 mechanism, the leaving group is not assisted in this way An essential criterion for distinguishing between these two pathways is the order of reaction Nucleophile-assisted substitutions are second order: first order in RX and first order in nucleophile Nucleophile-unassisted substitutions are first order: first order in RX and zero order in nucleophile Chemists recognize that certain nucleophilic substitutions have the kinetic characteristics of first-order (SN1) substitution but, in fact, involve two successive displacement reactions A characteristic feature of a great many of these reactions is the presence of an internal nucleophile (most commonly sulfur, nitrogen, or oxygen) on the carbon atom beta to the leaving group This neighboring nucleophile participates in the departure of the leaving group to give an intermediate, which then reacts with an external nucleophile to complete the reaction The mustard gases are one group of compounds that react by participation of a neighboring group The characteristic structural feature of a mustard gas is a twocarbon chain, with a halogen on one carbon and a divalent sulfur or trivalent nitrogen on the other carbon (S-C-C-Lv or N-C-C-Lv) An example of a mustard gas is bis(2ClCH2CH2SCH2CH2Cl chloroethyl)sulfide, a poison gas used extensively in World War I and at one time, at bis(2-Chloroethyl)sulfide least, manufactured by Iraq This compound is a deadly vesicant (blistering agent) and quickly causes conjunctivitis and blindness ClCH2CH2SCH2CH2Cl ClCH2CH2SCH2CH2Cl bis(2-Chloroethyl)sulfide bis(2-Chloroethyl)sulfide CH3 ClCH2CH2 N CH2CH2Cl bis(2-Chloroethyl)methylamine Bis(2-chloroethyl)sulfide and bis(2-chloroethyl)methylamine are not gases at all They are oily liquids with a high vapor pressure, hence the designation “gas.” Nitrogen and sulfur mustards react very rapidly with moisture in the air and in the mucous membranes of the eye, nose, and throat to produce HCl, which then burns and blisters CH Copyright 2018 Cengage Learning All Rights Reserved May3 not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 ClCH2CH2 N CH3 CH2CH2Cl 9.10  Neighboring Group Participation   419 these sensitive tissues What is unusual about the reactivity of the mustard gases is that they react very rapidly with water, a very poor nucleophile 1 Mustard gases also react rapidly with other nucleophiles, such as those in biological molecules, which makes them particularly dangerous chemicals Of the two steps in the mechanism of the hydrolysis of a sulfur mustard, the first is the slower and is ratedetermining As a result, the rate of reaction is proportional to the concentration of the sulfur mustard but independent of the concentration of the external nucleophile Thus, although this reaction has the kinetic characteristics of an SN1 reaction, it actually involves two successive SN2 displacement reactions Mechanism 9.9 Hydrolysis of a Sulfur Mustard—Participation by a Neighboring Group Step 1:  Make a new bond between a nucleophile and an electrophile and ­simultaneously break a bond to give stable molecules or ions.  The reason for the extremely rapid hydrolysis of the sulfur mustards is neighboring group ­participation by sulfur in the ionization of the carbon-chlorine bond to form a cyclic sulfonium ion This is the rate-determining step of the reaction; although it is the slowest step, it is much faster than reaction of a typical primary chloroalkane with water At this point, you should review halogenation of alkenes (Sections 6.3D and 6.3F) and compare the cyclic halonium ions formed there with the cyclic ­sulfonium ion formed here 1 Step 2:  Make a new bond between a nucleophile and an electrophile.  The cyclic sulfonium ion contains a highly strained three-membered ring and reacts rapidly with an external nucleophile to open the ring followed by proton transfer to H2O to give H3O1 In this SN2 reaction, H2O is the nucleophile and sulfur is the leaving group 1 Step 3:  Take a proton away.  Proton transfer to water completes the reaction 1 9 1 The net effect of these reactions is nucleophilic substitution of Cl by OH We continue to use the terms SN2 and SN1 to describe nucleophilic substitution reactions You should realize, however, that these designations not adequately describe all nucleophilic substitution reactions Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 420  Chapter 9: Nucleophilic Substitution and b-Elimination Example 9.9  Hydrolysis of Nitrogen Mustards Write a mechanism for the hydrolysis of the nitrogen mustard bis(2-chloroethyl)methylamine Solution Following is a three-step mechanism Step 1:  Make a new bond between a nucleophile and an electrophile and simultaneously break a bond to give stable molecules or ions.  This is an internal SN2 reaction in which ionization of the C!Cl bond is assisted by the neighboring nitrogen atom to form a highly strained three-membered ring 1 Step 2:  Make a new bond between a nucleophile and an electrophile.  Reaction of the cyclic ammonium ion with water opens the three-membered ring In this SN2 reaction, H2O is the nucleophile and nitrogen is the leaving group 1 Step 3:  Take a proton away.  Proton transfer to the basic nitrogen completes the reaction 1 Problem 9.9 Knowing what you about the stereochemisry of SN2 reactions, predict the product of hydrolysis of this compound Connections to Biological Chemistry Mustard Gases and the Treatment of Neoplastic Diseases Autopsies of soldiers killed by sulfur mustards in World War I revealed, among other things, very low white blood cell counts and defects in bone marrow development From these observations, it was realized that sulfur mustards have profound effects on rapidly dividing cells This became a lead observation in the search for less toxic ­alkylating agents for use in treatment of cancers, which have rapidly dividing cells Attention turned to the less reactive nitrogen mustards One of the first compounds tested was Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 9.10  Neighboring Group Participation   421 Mechlorethamine mechlorethamine As with other mustards, the reaction of mechlorethamine with nucleophiles is rapid because of the formation of an aziridinium ion melphalan is chiral It has been demonstrated that the R and S enantiomers have approximately equal therapeutic potency Mechlorethamine undergoes very rapid reaction with ­water (hydrolysis) and with other nucleophiles, so much so that within minutes after injection into the body, it has completely reacted The problem for the chemist, then, was to find a way to decrease the nucleophilicity of nitrogen while maintaining reasonable water solubility Substitution of ­phenyl for methyl reduced the The clinical value of the nitrogen mustards lies in the fact that they undergo reaction with certain nucleophilic sites on the heterocyclic aromatic amine bases in DNA (see Chapter 28) For DNA, the most reactive nucleophilic site is N-7 of guanine Next in reactivity is N-3 of adenine, followed by N-3 of cytosine Chlorambucil nucleophilicity, but the resulting compound was not sufficiently soluble in water for intravenous injection The solubility problem was solved by adding a carboxyl group When the carboxyl group was added directly to the aromatic ring, however, the resulting compound was too stable and therefore not biologically active Adding a propyl bridge (chlorambucil) or an aminoethyl bridge (melphalan) between the aromatic ring and the carboxyl group solved both the solubility problem and the reactivity problem Note that Melphalan The nitrogen mustards are bifunctional alkylating agents; one molecule of nitrogen mustard undergoes reaction with two molecules of nucleophile Guanine alkylation leaves one free reactive alkylating group, which can react with another base, giving cross links that lead to miscoding during DNA replication The therapeutic value of the nitrogen mustards lies in their ability to disrupt normal base pairing This prevents replication of the cells, and the rapidly dividing cancer cells are more sensitive than normal cells (Continued) Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 422  Chapter 9: Nucleophilic Substitution and b-Elimination Guanine Study Guide 9.1  Nucleophilic Substitution in Haloalkanes ●● Nucleophilic substitution is any reaction in which a nucleophile replaces another electron-rich group called a leaving group (Lv) – A nucleophile (Nu:2) is an electron-rich molecule or ion that donates a pair of electrons to another atom or ion to form a new covalent bond P 9.1 9.2 Mechanisms of Nucleophilic Aliphatic Substitution ●● There are two limiting mechanisms for nucleophilic substitution, namely SN2 and SN1 – In the SN2 reaction mechanism, bond forming and bond breaking occur simultaneously – SN2 reactions are bimolecular because both nucleophile and haloalkane concentrations influence reaction rate – The nucleophile must approach the carbon-leaving group (C!Lv) bond from the backside in order to populate the C!Lv antibonding orbital and allow reaction – In the SN1 mechanism, the leaving group departs first in the rate-determining step, leaving a carbocation intermediate that reacts with the nucleophile in a second step – SN1 reactions are unimolecular because only the haloalkane concentration influences reaction rate 9.3 Experimental Evidence for SN1 and SN2 Mechanisms ●● The SN2 reaction can be identified based on kinetics of the reaction and stereochemistry of the products Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 Study Guide   423 – Because an SN2 reaction is bimolecular, doubling the concentration of either haloalkane or nucleophile will double the rate of the reaction – Because backside attack geometry is required, reaction at a chiral center results in inversion of configuration – The SN1 reaction can also be identified based on kinetics of the reaction and stereochemistry of the products – Because an SN1 reaction is unimolecular, doubling the concentration of only the haloalkane can double the rate of the reaction – Because a planar and achiral carbocation intermediate is formed that can be attacked with roughly equal probability from either face, reaction at a chiral center results in racemization of stereochemistry – The chiral center is often not completely racemized because the leaving group forms an ion pair with the carbocation intermediate, partially blocking one face ●● The structure of the haloalkane influences the reaction rate and mechanism – Haloalkanes that can form more stable carbocations react faster if an SN1 mechanism occurs – Because SN1 reactions involve carbocations, rearrangements (1,2 shifts) can occur if they lead to more stable carbocation intermediates – Steric hindrance on the backside of the C—Lv bond of a haloalkane slows down or possibly prevents an SN2 mechanism ●● The more stable the anion produced upon reaction, the better the leaving group ability ●● Solvent properties can have an important influence on reaction mechanisms – Protic solvents are hydrogen-bond donors The most common protic solvents are those containing —OH groups – Aprotic solvents cannot serve as hydrogen-bond donors Common aprotic solvents are acetone, diethyl ether, dimethyl sulfoxide, and N,N-dimethylformamide – Polar solvents interact strongly with ions and polar molecules – Nonpolar solvents not interact strongly with ions and polar molecules – The dielectric constant is the most commonly used measure of solvent polarity – Solvolysis is a nucleophilic substitution reaction in which the solvent is the nucleophile – Polar protic solvents accelerate SN1 reactions by stabilizing the charged carbocation intermediate – Polar aprotic solvents accelerate SN2 reactions because they not interact strongly with the nucleophile ●● Nucleophiles are categorized as good, moderate, or poor – Good nucleophiles are generally anions Moderate nucleophiles are generally neutral, with one or more available lone pairs Poor nucleophiles are generally polar protic solvents – All things being equal, the stronger the interaction of a nucleophile with solvent, the lower the nucleophilicity – Small nucleophiles with very little steric hindrance are better nucleophiles for SN2 reactions P 9.2–9.4, 9.10–9.13, 9.15–9.22, 9.24–9.36 Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 424  Chapter 9: Nucleophilic Substitution and b-Elimination Stereochemistry of an SN1 Reaction Key Reactions Nucleophilic Aliphatic Substitution: SN2 (Section 9.3) SN2 reactions occur in one step; departure of the leaving group is assisted by the incoming nucleophile, and both nucleophile and leaving group are involved in the transition state The nucleophile may be negatively charged as in the first example or neutral as in the second example 9 9 SN2 reactions result in inversion of configuration at the reaction center They are accelerated more in polar aprotic solvents than in polar protic solvents The relative rates of SN2 reactions are governed by steric factors, namely the degree of crowding around the site of reaction Nucleophilic Aliphatic Substitution: SN1 (Section 9.3) An SN1 reaction occurs in two steps Step is a slow, rate-determining ionization of the C—Lv bond to form a carbocation intermediate followed in Step by rapid reaction of the carbocation intermediate with a nucleophile to complete the substitution Reaction at a chiral center gives l­argely racemization, often accompanied with a slight excess of inversion of configuration Reactions often involve carbocation rearrangements and are accelerated by polar protic solvents SN1 reactions are governed by electronic factors, namely the relative stabilities of carbocation intermediates The following reaction involves an SN1 reaction with a ­hydride shift 1 9.4 Analysis of Several Nucleophilic Substitution Reactions ●● ●● ●● P Methyl or primary haloalkanes react through SN2 mechanisms because of an absence of steric hindrance and lack of carbocation stability Secondary haloalkanes react through an SN2 mechanism in aprotic solvents with good nucleophiles, but through an SN1 mechanism in protic solvents with poor nucleophiles Tertiary haloalkanes react through an SN1 mechanism because the steric hindrance disfavors SN2 backside attack, and the attached alkyl groups stabilize a carbocation 9.5, 9.14, 9.23 Problem 9.5 Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 Study Guide   425 9.5  b-Elimination ●● A b-elimination reaction involves removal of atoms or groups of atoms from adjacent carbon atoms – Dehydrohalogenation is a b-elimination reaction that involves loss of an H and a halogen atom from adjacent carbons to create an alkene from a haloalkane – Zaitzev's rule predicts that b-elimination reactions give primarily the more highly substituted alkene Such reactions are called Zaitsev eliminations P 9.6 9.6  Mechanisms of b-Elimination ●● The two limiting mechanisms for b-elimination reactions are the E1 and E2 mechanisms – In the E1 mechanism, the leaving group departs to give a carbocation; then a proton is taken off an adjacent carbon atom by base to create the product alkene – E1 reactions are unimolecular because only the haloalkane concentration influences the rate of the reaction – In the E2 mechanism, the halogen departs at the same time that an H atom is removed by base from an adjacent carbon atom to create the product alkene – E2 reactions are bimolecular because both the haloalkane and base concentrations influence the rate of the reaction 9.7 Experimental Evidence for E1 and E2 Mechanisms ●● E2 reactions are stereoselective in that the lowest energy transition state is the state in which the leaving group and H atoms that depart are oriented anti and coplanar – This anti and coplanar requirement determines whether E or Z alkenes are produced For cyclohexane derivatives, both the leaving group and departing H atom must be axial ●● Both E1 and E2 reactions are regioselective, favoring formation of the more stable (Zaitsev) product alkene (as long as Lv and H can be oriented anti and coplanar in the case of E2) – The more stable alkene is generally the more highly substituted alkene P 9.7, 9.37–9.42 Problem 9.7 Key Reactions b-Elimination: E1 (Sections 9.6, 9.7) An E1 reaction occurs in two steps: slow, rate-determining breaking of the C—Lv bond to form a carbocation intermediate followed by rapid proton transfer to solvent to form an alkene An E1 reaction is first order in haloalkane and zero order in base Skeletal rearrangements are common (Continued) Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 20.2  Electrophilic Addition to Conjugated Dienes   901 1 With electrophilic addition to standard alkenes such as propene, the product predicted by Markovnikov’s rule is also more stable For reactions under t­ hermodynamic (equilibrium) control, the distribution of products is determined by the relative stability of each Thus, kinetically controlled and thermodynamically controlled electrophilic additions of H—X to standard alkenes results in the same dominant product This is the case with many reactions: the product formed fastest is also most stable Yet, many other reactions not behave this way Below we will see that the addition of HBr to conjugated dienes exemplifies reactions in which kinetic and thermodynamic control produce different ­dominant products Whether a reaction is under kinetic or thermodynamic control can be manipulated by changing the experimental conditions A common approach to switching between kinetic or thermodynamic control is to change the temperature chosen for a reaction In general, at lower temperatures, little to no equilibrium is established between reactants and products, and the reactions must therefore be under kinetic control At higher temperatures, reactions become increasingly reversible, and equilibrium can be established between reactants and products, leading to thermodynamic control The electrophilic addition of HBr to standard alkenes such as propene would give the same dominant product at all temperatures, although the extent of domination will vary somewhat with temperature We can now return to the addition of HBr to conjugated dienes We saw in Section 20.2A that electrophilic addition to conjugated dienes gives a mixture of 1,2-addition and 1,4-addition products Following are some additional experimental observations about the products of electrophilic additions to 1,3-butadiene For addition of HBr at 278°C and addition of Br2 at 215°C, the 1,2-addition­ products predominate over the 1,4-addition products Generally, at lower temperatures, the 1,2-addition products predominate over 1,4-addition products For addition of HBr and Br2 at higher temperatures (generally, 40–60°C), the 1,4-addition products predominate If the products of low temperature addition are allowed to remain in solution and then are warmed to a higher temperature, the composition of the product changes over time and becomes identical to that obtained when the reaction is carried out at higher temperature Thus, under these higher temperature conditions, an ­equilibrium is established between 1,2- and 1,4-addition products in which 1,4-addition products predominate These experimental observations can be explained by considering kinetic versus thermodynamic products, which dominate at lower and higher temperatures, ­respectively At the lower temperatures, no equilibrium is established between the 1,2- and 1,4-addition products Because the 1,2-addition products dominate u ­ nder these conditions, they must be the products formed by kinetic control (i.e., the 1,2-addition is faster than 1,4-addition) Alternatively, at the higher temperatures, the 1,4-addition products dominate; therefore, we can conclude that an equilibrium is established between the 1,2- and 1,4-addition products and that the 1,4-addition products are thermodynamically more stable than 1,2-addition products Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 902  Chapter 20: Dienes, Conjugated Systems, and Pericyclic Reactions Figure 20.3  Kinetic versus thermodynamic control A plot of Gibbs free energy versus reaction coordinate for Step in the electrophilic addition of HBr to 1,3-butadiene The resonancestabilized allylic carbocation intermediate reacts with bromide ion by way of the transition state on the left to give the 1,2-addition product It reacts with bromide ion by way of the alternative transition state on the right to give the 1,4-addition product DG DG Br2 d+ CH3CH CH d+ CH2 Stabilized by charge delocalization Br CH3CHCH CH2 CH3CH 1,2-Addition product (less stable) Watch a video explanation CHCH2Br 1,4-Addition product (more stable) Relationships between kinetic and thermodynamic control for electrophilic a­ ddition of HBr to 1,3-butadiene are illustrated graphically in Figure 20.3 The ­structure shown in the Gibbs free energy well in the center of Figure 20.3 is the ­resonance-stabilized allylic cation intermediate formed by proton transfer from HBr to C1 of 1,3-butadiene The dashed lines in this intermediate show the partial ­double bond character between C2 and C3 and between C3 and C4 in the resonance ­hybrid To the left of this intermediate is the activation energy for its reaction with bromide ion to form the less stable 1,2-addition product; to the right is the activation e­ nergy for its reaction with bromide ion to form the more stable 1,4-addition product As shown in Figure 20.3, the activation energy for 1,2-addition is less than that for 1,4-addition; therefore, the 1,2-addition product is favored under kinetic control The 1,4-addition product is more stable and is favored when the reaction is under ­thermodynamic control To complete our discussion of electrophilic addition to conjugated dienes and of kinetic versus thermodynamic control, we need to ask the following questions Why is the 1,2-addition product (the less stable product) formed more rapidly at lower temperatures? First, we need to look at the resonance-stabilized allylic ­carbocation intermediate and determine which Lewis structure makes the greater contribution to the hybrid We must consider the degree of substitution of both the positive carbon and the carbon-carbon double bond in each contributing structure " 9 " A secondary carbocation is more stable than a primary carbocation If the degree of substitution of the carbon bearing the positive charge were the more important factor, the Lewis structure on the left would make the greater contribution to the hybrid However, a more substituted double bond is more stable than a less substituted double bond (Section 6.6B) If the degree of s­ ubstitution of the carbon-carbon double bond were the more important f­ actor, the Lewis structure on the right would make the greater contribution to the hybrid We know from other experimental evidence that the location of the positive charge in the allylic carbocation is more important than the location of the double bond Therefore, in the hybrid, the greater fraction of positive charge is on the secondary carbon Reaction with bromide ion occurs more rapidly at this carbon, giving 1,2-addition, simply because it has a greater density of positive charge The electrostatic potential map shows that the positive charge (blue) is more intense on the secondary carbon Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 20.2  Electrophilic Addition to Conjugated Dienes   903 Is the 1,2-addition product also formed more rapidly at higher temperatures even though the 1,4-addition product predominates under these conditions? The ­answer is yes The factors affecting the structure of a resonance-stabilized allylic carbocation intermediate and the reaction of this intermediate with a nucleophile are not greatly affected by changes in temperature Why is the 1,4-addition product the thermodynamically more stable product? The answer to this question has to with the relative degree of substitution of ­double bonds In general, the greater the degree of substitution of a carbon lectrostatic potential map carbon ­double bond, the greater the stability of the compound or ion containing E it Following are pairs of 1,2- and 1,4-addition products In each case, the more of the allylic carbocation formed by protonating 1,3-butadiene stable alkene is the 1,4-addition product " 3-Bromo-1-butene (E)-1-Bromo-2-butene " 3,4-Dibromo-1-butene (E)-1,4-Dibromo-2-butene However, there are cases where the 1,2-addition product is more stable and would be the product of thermodynamic control For example, addition of ­bromine to 1,4-dimethyl-1,3-cyclohexadiene under conditions of thermodynamic control gives 3,4-dibromo-1,4-dimethylcyclohexene because its trisubstituted ­double bond is more stable than the disubstituted double bond of the 1,4-addition product What is the mechanism by which the thermodynamically less stable product is converted to the thermodynamically more stable product at higher temperatures? At higher temperatures used for electrophilic addition of HBr and Br2 to conjugated dienes, collisions between the 1,2- and 1,4-addition products with the s­ olvent are energetic enough to reform the resonance-stabilized allylic carbocation intermediate via ionization of the C—Br bonds When the 1,2-addition product reverts to this allylic carbocation, re-addition of bromide can give the more stable 1,4-addition product At lower temperature, however, the increase in potential energy of the products upon collisions is not sufficient to overcome the activation energy for C—Br bond ionization; therefore, the reactions are not reversible In summary, although the thermodynamically most stable product is often the most rapidly formed product, such is not always the case Whether the thermodynamically more stable product is formed at a greater rate—from a reactant or a common intermediate—very much depends on the particular reactants, the reaction mechanism, and the reaction conditions Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 904  Chapter 20: Dienes, Conjugated Systems, and Pericyclic Reactions 20.3  UV-Visible Spectroscopy An important property of conjugated systems is that they absorb energy in the ­u ltraviolet-visible region of the spectrum as a result of electronic transitions (­Table 12.3) In this section, we study the information this absorption gives us about the conjugation of carbon-carbon and carbon-oxygen double bonds and their substitution A. Introduction The region of the electromagnetic spectrum covered by most ultraviolet spectrophotometers is from 200 to 400 nm, a region commonly referred to as the near ­ultraviolet Wavelengths shorter than 200 nm require special instrumentation and are not used routinely The region covered by most visible spectrophotometers runs from 400 nm (violet) to 700 nm (red), with extensions into the (near) IR region to 800 or 1000 nm available on many instruments Example 20.4  UV-Vis Radiation Calculate the energy of radiation at either end of the near-ultraviolet spectrum [i.e., at 200 nm and 400 nm (review Section 12.1)] Solution Use the relationship E hc/l Make sure you express the dimension of length in consistent units hc kJ s m E5 3.99 10213 3.00 108 s l mol 200 1029 m 598 kJ (143 kcal)/mol By a similar calculation, the energy of radiation of wavelength 400 nm is found to be 299 kJ (71.5 kcal)/mol Problem 20.4 Wavelengths in ultraviolet-visible spectroscopy are commonly expressed in nanometers; wavelengths in infrared spectroscopy are sometimes expressed in micrometers Carry out the following conversions (a)  2.5 mm to nanometers    (b)  200 nm to micrometers Wavelengths and corresponding energies for near-ultraviolet and visible radiation are summarized in Table 20.2 Table 20.2  Wavelengths and Energies of Near Ultraviolet and Visible Radiation Region of Spectrum Energy Wavelength (nm) kJ/mol Near Ultraviolet 200–400 299–598 71.5–143 Visible 400–700 171–299 40.9–71.5 kcal/mol Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 20.3  UV-Visible Spectroscopy   905 Ultraviolet and visible spectral data are recorded as plots of absorbance (A) on the vertical axis versus wavelength on the horizontal axis Absorbance sAd log I0 I where I0 is the intensity of radiation incident on the sample and I is the intensity of the radiation transmitted through the sample The quantity ( I/I0 ) 100 is called ­percent transmittance; many spectrophotometers read in this scale Typically, UV-visible spectra consist of a small number of broad absorption bands, sometimes just one Figure 20.4 is an ultraviolet absorption spectrum of 2,5dimethyl-2,4-hexadiene Absorption of ultraviolet radiation by this conjugated diene begins at wavelengths below 200 nm and continues to almost 270 nm, with maximum ­absorption at 242 nm This spectrum is reported as a single absorption peak using the notation lmax 242 nm Absorbance (A) A quantitative measure of the extent to which a compound absorbs radiation of a particular wavelength A log( I0/I ) where I0 is the incident radiation and I is the transmitted radiation 0.8 0.7 CH3 Absorbance 0.6 CH3C 0.5 CH3 CHCH CCH3 0.4 0.3 0.2 0.1 200 210 220 230 240 250 260 270 Wavelength (nm) 280 290 300 310 320 The extent of absorption of ultraviolet-visible radiation is proportional to the number of molecules capable of undergoing the observed electronic transition; therefore, ultraviolet-visible spectroscopy can be used for quantitative analysis of samples The relationship between absorbance, concentration, and length of the sample cell (cuvette) is known as the Beer-Lambert law The proportionality constant in this equation is given the name molar absorptivity («) or extinction coefficient A5«cl where A is the absorbance (unitless), « is the molar absorptivity (in per moles per liter per centimeter, M21cm21), c is the concentration of solute (in moles per liter, M), and l is the length of the sample cell, or cuvette (in centimeters, cm) The molar absorptivity is a characteristic property of a compound and is not affected by its concentration or the length of the light path Values range from zero to 106 M21cm21 Values above 104 M21cm21 correspond to high-intensity absorptions; values below 104 M21cm21, to low-intensity absorptions The molar absorptivity of 2,5-dimethyl-2,4-hexadiene, for example, is 13,100 M21cm21, a high-intensity absorption An interesting aspect of absorption by molecules in the visible region is that a sample will appear to our eyes as the combination of reflected wavelengths White light is composed of all wavelengths of light in the visible region (400–740 nm), present in approximately equal intensity Individual wavelengths of light have individual colors, as indicated on the spectrum (Figure 20.5a) Light of the given wavelength appears as the color indicated to our eyes For example, monochromatic 400 nm light appears violet and monochromatic 700 nm light appears red Figure 20.4  Ultraviolet spectrum of 2,5-dimethyl2,4-hexadiene (in methanol) Molar absorptivity (´) The absorbance of a M solution of a compound Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 906  Chapter 20: Dienes, Conjugated Systems, and Pericyclic Reactions Watch a video explanation Absorption by a substance removes the absorbed wavelengths from white light, l­eaving the remaining wavelengths to be reflected, the combination of which determines the color our eyes see Figure 20.5b shows the approximate color that a substance would appear if a single wavelength were absorbed For example, if a molecule absorbs strongly only at 500 nm (lighter blue light), it appears red to our eyes because the remaining ­reflected colors combine to appear red Similarly, a molecule that strongly absorbs around 600 nm (orange light) appears blue, because orange is removed from the reflected light, and the remaining reflected wavelengths combine to appear blue 560 nm 575 nm 600 nm 540 nm 650 nm 525 nm 510 nm 200 300 400 500 600 nm 700 800 900 200 300 400 500 600 nm 700 800 900 495 nm 465 nm 405 nm Colored arrows are complementary Figure 20.5  (a) Visible light color-wavelength correlation (b) Approximate color of substance (reflected light) if a single wavelength (i.e., the wavelength listed on the numerical scale of the x-axis) is absorbed (c) Complementary colors on a color wheel The correlation between absorbed wavelength and reflection can be approximated using the concept of complementary colors illustrated by an artist’s color wheel (Figure 20.5c) A molecule that absorbs light in one region of the spectrum will reflect the nonabsorbed wavelengths A good rule of thumb is that the ­reflected wavelengths combine to appear more or less as the complement of the absorbed color To a first approximation then, a molecule that absorbs one color will ­appear to our eyes to be the color on the opposite side of the color wheel The color wheel shown has approximate wavelengths of monochromatic light indicated for reference More complicated absorptions, with two or more strong absorptions by a single molecule, lead to a more complex interpretation of reflected color, but the concept is the same A substance will appear to our eyes as the combination of reflected (not absorbed) wavelengths Example 20.5  Beer-Lambert Law The molar absorptivity of 2,5-dimethyl-2,4-hexadiene in methanol is 13,100 M21cm21 What concentration of this diene in methanol is required to give an absorbance of 1.6? Assume a light path of 1.00 cm Calculate concentration in these units (a) Moles per liter (b) Milligrams per milliliter Solution Solve the Beer-Lambert equation for concentration and substitute appropriate ­values for length, absorbance, and molar absorptivity (a) c A 1.6 5 1.22 1024 molyL l « 1.00 cm 13,100 L ? mol21 ? cm21 (b) The molecular weight of 2,5-dimethyl-2,4-hexadiene is 110 g/mol The concentration of the sample in milligrams per milliliter is 1000 mg mol 110 g lL 1.22 1024 3 1.34 1022 mgymL g L mol 1000 mL Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 20.3  UV-Visible Spectroscopy   907 Problem 20.5 The visible spectrum of b-carotene (C40H56, MW 536.89, the orange pigment in carrots) dissolved in hexane shows intense absorption maxima at 463 nm and 494 nm, both in the blue-green region Because light of these wavelengths is ­absorbed by b-carotene, we perceive the color of this compound as that of the complement to blue-green, namely red-orange -Carotene « « Calculate the concentration in milligrams per milliliter of b-carotene that gives an absorbance of 1.8 at 463 nm B.  The Origin of Transitions Between Electronic Energy Levels Absorption of electromagnetic radiation in the ultraviolet-visible region results in promotion of electrons from a lower energy, occupied MO to a higher energy, unoccupied MO The energy of this radiation is generally insufficient to affect electrons in the much lower energy, s-bonding molecular orbitals It is, however, sufficient to cause an electron in a nonbonding (lone pair) or p orbital to be promoted to an ­antibonding p* orbital (called an n S p* and p S p* transition, respectively) ­Conjugated p systems have particularly noteworthy p S p* transitions Three examples of ­conjugated systems follow " " 1,3-Butadiene " p p h p p 9 3-Buten-2-one Benzaldehyde As an example of a p S p* transition, consider ethylene The double bond in ­ethylene consists of one s bond formed by combination of sp orbitals and one p bond formed by combination of 2p orbitals The relative energies of the p-bonding and p-­antibonding molecular orbitals are shown schematically in Figure 20.6 The p S p* transitions for simple, unconjugated alkenes occur below 200 nm (at 165 nm for ­ethylene) Because these transitions occur at extremely short wavelengths, they are not observed in conventional ultraviolet spectroscopy and therefore are not useful to us for determining molecular structure For 1,3-butadiene, the difference in energy between the highest occupied p molecular orbital and the lowest unoccupied p-antibonding molecular orbital is less than it is for ethylene with the result that a p S p* transition for 1,3-butadiene (Figure 20.7) takes less energy (occurs at longer wavelength) than that for ethylene This transition for 1,3-butadiene occurs at 217 nm Electronic excitation in molecules is accompanied by changes in vibrational or ­rotational energy levels The energy levels for these excitations are considerably smaller than the energy differences between electronic excitations These transitions are superposed on the electronic excitations, which results in a large number of ­absorption peaks so closely spaced that the spectrophotometer cannot resolve them For this reason, UV-visible absorption peaks usually are much broader than IR ­absorption peaks Figure 20.6  A p S p* transition in excitation of ethylene Absorption of ultraviolet radiation causes a transition of an electron from a p-bonding MO in the ground state to a p-antibonding MO in the excited state There is no change in electron spin Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 908  Chapter 20: Dienes, Conjugated Systems, and Pericyclic Reactions p p p p p p p p h Figure 20.7  Electronic excitation of 1,3-butadiene; a p S p* transition Simple aldehydes and ketones show only weak absorption in the ultraviolet ­region of the spectrum owing to an n to p* electronic transition of the carbonyl group If, however, the carbonyl group is conjugated with one or more carbon-­carbon double bonds, intense absorption (« 8,000 20,000 M21cm21) occurs as a result of a p to p* transition; as with polyenes, the position of absorption is shifted to l­onger wavelengths and the molar absorptivity, «, of the absorption maximum ­increases sharply For the a,b-unsaturated ketone 3-penten-2-one, for example, lmax is 224 nm (log « 4.10) 2-Pentanone « 3-Penten-2-one « Acetophenone « The greater the extent of conjugation of unsaturated systems with the carbonyl group, the more the absorption maximum is shifted toward the visible region of the spectrum Like the carbonyl groups of simple aldehydes and ketones, the carboxyl group shows only weak absorption in the ultraviolet spectrum unless it is conjugated with a carbon-carbon double bond or an aromatic ring The important point is that conjugation decreases the energy gap between filled and unfilled p orbitals Therefore, in general, the greater the number of double bonds in conjugation, the longer the wavelength of ultraviolet radiation absorbed Shown in Table 20.3 are wavelengths and energies required for p S p* transitions in several conjugated alkenes Table 20.3  Wavelengths and Energies Required for p S p* Transitions of Ethylene and Three Conjugated Polyenes Name Structural Formula lmax (nm) Energy [kj (kcal) / mol] Ethylene CH2"CH2 165 724 (173) 1,3-Butadiene CH2"CHCH"CH2 217 552 (132) (3E)-1,3,5-Hexatriene CH2"CHCH"CHCH"CH2 268 448 (107) (3E,5E)-1,3, 5,7-Octatetraene CH2"CH(CH"CH)2CH"CH2 290 385 (92) Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 20.4  Pericyclic Reaction Theory   909 20.4  Pericyclic Reaction Theory Pericyclic reaction Chemical Connections Although we cover only one approach to the understanding of pericyclic reactions, the various approaches developed over the years have given important different contri­ butions to our understanding R B Woodward (Harvard University), Roald Hoffmann (then at Harvard, now at Cornell University), Kenichi Fukui (Kyoto University), and Howard Zimmerman (University of Wisconsin) provided the key insights into pericyclic reaction mechanisms Hoffman and Fukui were awarded the Nobel Prize for this work in 1981 (after the death of Woodward) ▲ To this point in this book, there have been only a few cases for which we analyzed the orbitals of reactants in order to understand the reaction mechanisms, as, for example, SN2 (Section 9.2A) and E2 (Section 9.7C) Even though we have drawn on orbital analysis infrequently, the interactions of orbitals actually dictate all chemical reactions In fact, there is a class of reactions called pericyclic for which an analysis of orbitals is critical for even a rudimentary understanding of the mechanisms Pericyclic reactions occur in a single step with a closed loop of orbitals; that is, we can draw orbitals interacting at the transition states of the reactions in a cyclic ring Further, because the reactions occur in a single step, there are no radical or ionic i­ntermediates The examples given below will make this characteristic clear Because many of these reactions involve conjugated and nonconjugated dienes, we discuss them in this chapter One of the hallmarks of pericyclic reactions is precise control of the stereochemistry of the reactions, and the examples given below will highlight this feature Pericyclic reactions are routinely classified as “  allowed” or “  forbidden” with a particular structure for the transition state In practice, this classification means that one geometry for the reaction has a low energy transition state (allowed) or that a different geometry has a very high energy transition state (forbidden) To determine whether a reaction is allowed or forbidden, a handful of approaches exist We will examine one approach: frontier molecular orbital theory A reaction that takes place in a single step without intermediates and that involves a cyclic distribution of orbitals Curry and Cancer Curcumin is a natural dye from the root of Curcuma longa L In pure form, it is an orange-yellow crystalline powder that is isolated from the spice turmeric, one of the major ingredients of curry Its color is a result of the highly conjugated system in curcumin (it is probable that the molecule is actually enolized as shown) It has been known for some time that curcumin retards the growth of new cancers by inhibiting the formation of blood ­vessels that are necessary for the cancers to grow (angiogenesis) Recently, Korean biochemists have shown that curcumin acts by inhibiting an enzyme that is important to angiogenesis So curry may be good for you Curcumin A.  Frontier Molecular Orbital Theory (FMOT) There are five classes of pericyclic reactions, only two of which are covered in this book, cycloadditions and sigmatropic shifts The most common, cycloaddition ­involves the reaction of a conjugated diene with an alkene, although we also ­examine the reaction of an alkene with an alkene Hence, we are examining what are commonly called and 2 cycloadditions, respectively, to keep track of the number of p electrons involved in the reaction 1 Cycloaddition reaction A reaction in which two reactants add together in a single step to form a cyclic product 1 Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 910  Chapter 20: Dienes, Conjugated Systems, and Pericyclic Reactions HOMO Highest occupied molecular orbital LUMO Lowest unoccupied molecular orbital Suprafacial When the same face of a p bond performs a reaction Antarafacial When the opposite faces of a p bond perform a reaction To analyze whether these reactions are allowed or forbidden, chemists focus on the frontier molecular orbitals of the reactants The frontier molecular orbitals consist of the highest occupied molecular orbitals (HOMOs) and the lowest unoccupied molecular orbitals (LUMOs) The terms highest and lowest refer to the energy of the orbitals, and the terms occupied and unoccupied refer to whether the orbital is populated with two electrons or is empty For example, the HOMO and LUMO of butadiene would be orbitals and in Figure 20.2, respectively The HOMO and LUMO of ethylene are simply the p and p* orbitals given in Figure 1.21 After identifying the frontier molecular orbitals of the reactants, chemists predict a reaction geometry The goal is to decide if the predicted geometry for a reaction is allowed or forbidden In the analysis, we examine the HOMO of one reactant and the LUMO of the other It does not matter which reactant is assigned the HOMO or LUMO in the analysis because the answer will be the same What is important is how one reactant’s HOMO interacts (contacts) with the other reactant’s LUMO in the proposed reaction geometry When the phasing of the orbitals that are undergoing contact matches (zero phase changes), the reaction is allowed In fact, if there are any even number of phase changes, the reaction is allowed However, if there is one (or any odd number of ) contacts between the orbitals in which the phasing does not match, the reaction is forbidden To demonstrate the frontier molecular orbital analysis, let’s analyze a collision geometry for butadiene and ethylene that is called suprafacial for each reactant A suprafacial interaction occurs when the same face (or side) of the p system of an individual reactant undergoes collision (an alternative interaction is referred to as ­antarafacial, shown in Example 20.6) In Figure 20.8(a), the bottoms of the p orbitals on carbons and are undergoing reaction (hence the same face) and the tops of the p orbitals of ethylene are reacting (hence again the same face) Therefore, each reactant is interacting in a suprafacial manner Suprafacial collision geometries are shown for both the and 2 reactions in Figure 20.8 Note that the reaction is allowed because when the HOMO and LUMO on the reactants contact each other, the phasing matches; red on red and blue on blue The 2 reaction is forbidden because there is one red on blue contact The frontier molecular orbital analysis led to the conclusion that butadiene will react with ethylene to give cyclohexene if both butadiene and ethylene collide in a suprafacial manner However, ethylene will not react with ethylene in an analogous manner We can extend these conclusions to the reaction of any conjugated diene with any alkene and to any alkene with another alkene Presented below are several examples and practical considerations of the allowed reaction (the Diels-Alder reaction) The suprafacial approach of both reactants has important consequences on the stereochemistry of the Diels-Alder reaction Figure 20.8  (a) A frontier molecular orbital analysis for the cycloaddition of butadiene with ethylene The phasing is found to match between the HOMO of butadiene and the LUMO of ethylene when they collide face to face, and hence, the reaction is allowed (b) A frontier molecular orbital analysis for the 2 cycloaddition of two ethylene molecules Because the phasing does not match in one spot when the HOMO and LUMO are combined, the reaction is termed forbidden Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 20.5  The Diels-Alder Reaction   911 Example 20.6  FMOT Let’s examine an alternative collision geometry for the reaction In an antarafacial interaction, one reactant collides in such a manner that the opposite faces of the p orbitals in the p bonds interact Show that the cycloaddition reaction is forbidden when butadiene interacts in an antarafacial manner Solution Shown below is the reaction of butadiene with ethylene, except that we have the ethylene collide with the butadiene at a skewed angle Recall that molecules in a reaction flask are constantly colliding with the solvent and each other in all possible orientations, but only certain orientations lead to chemical reactions In the antarafacial geometry for the butadiene, there is one contact between the butadiene and ethylene in which the phasing does not match Hence, this geometry for a collision between the two reactants will not lead to product and is considered forbidden Problem 20.6 The 2 cycloaddition with one suprafacial and one antarafacial interaction is allowed Show this conclusion via a frontier molecular orbital analysis Although the reaction is allowed, it is seldom seen Can you think of a reason not based on an orbital analysis of why this reaction is difficult? 20.5  The Diels-Alder Reaction In 1928, Otto Diels and Kurt Alder in Germany discovered a unique reaction of conjugated dienes: they undergo cycloaddition reactions with certain types of carbon-carbon double and triple bonds For their discovery and subsequent studies of this reaction, Diels and Alder were jointly awarded the 1950 Nobel Prize in Chemistry The compound with the double or triple bond that reacts with the diene in a Diels-Alder reaction is given the special name dienophile (diene-loving), and the product of a Diels-Alder reaction is given the special name Diels-Alder adduct The designation cycloaddition refers to the fact that two reactants add together to give a cyclic product Following are two examples of Diels-Alder reactions: one with a compound containing a carbon-carbon double bond and the other containing a carboncarbon triple bond Dienophile A compound containing a double or triple bond (consisting of one or two C, N, or O atoms) that can react with a conjugated diene to give a Diels-Alder adduct Diels-Alder adduct A cyclohexene resulting from the cycloaddition reaction of a diene and a dienophile Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 912  Chapter 20: Dienes, Conjugated Systems, and Pericyclic Reactions 1,3-Butadiene 3-Buten-2-one 3-Cyclohexenyl methyl ketone 1,3-Butadiene Diethyl 2-butynedioate Diethyl 1,4-cyclohexadiene1,2-dicarboxylate Note that the four carbon atoms of the diene and two carbon atoms of the dienophile combine to form a six-membered ring Note further that there are two more s bonds and two fewer p bonds in the product than in the reactants This exchange of two (weaker) p bonds for two (stronger) s bonds is a major driving force in Diels-Alder reactions We can write a Diels-Alder reaction in the following way, showing only the carbon skeletons of the diene and dienophile In this representation, curved arrows are used to show that two new s bonds are formed, three p bonds are broken, and one new p bond is formed It must be emphasized here that in this particular case, the curved arrows in this diagram are not meant to show a mechanism Rather, they are intended to show which bonds are broken, which new bonds are formed, and how many electrons are involved (six in this case) The real mechanism is pericyclic and is a cycloaddition, as was presented in Section 20.4 The special values of the reaction discovered by Diels and Alder are that (1) it is one of the simplest reactions that can be used to form six-membered rings; (2) it is one of few reactions that can be used to form two new carbon-carbon bonds at the same time; and, as we will see later in this section, (3) it is completely stereospecific and quite ­regioselective For these reasons, the Diels-Alder reaction has proved to be enormously valuable in synthetic organic chemistry Example 20.7  The Diels-Alder Reaction I Draw a structural formula for the Diels-Alder adduct formed by reaction of each diene and dienophile pair (a) 1,3-Butadiene and propenal (b) 2,3-Dimethyl-1,3-butadiene and 3-buten-2-one Solution First, draw the diene and dienophile so that each molecule is properly aligned to form a six-membered ring Then complete the reaction to form the sixmembered ring Diels-Alder adduct Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 20.5  The Diels-Alder Reaction   913 (a) 1,3-Butadiene (b) Propenal (Acrolein) 2,3-Dimethyl1,3-butadiene 3-Buten-2-one (Methyl vinyl ketone) Problem 20.7 What combination of diene and dienophile undergoes Diels-Alder reaction to give each adduct? (a) (b) Watch a video explanation (c) Now let us look more closely at the scope and limitations, stereochemistry, and mechanism of Diels-Alder reactions A.  Diene Must Be Able to Assume an s-Cis Conformation We can illustrate the significance of conformation of the diene by reference to 1,3-­butadiene For maximum stability of a conjugated diene, overlap of the four ­unhybridized 2p orbitals making up the p system must be complete, a condition that occurs only when all four carbon atoms of the diene lie in the same plane If the carbon skeleton of 1,3-butadiene is planar, the six atoms bonded to the skeleton of the diene are also contained in the same plane Bond rotation is somewhat restricted around the central single bond due to conjugation; if the atoms are not coplanar, conjugation is imperfect or broken completely There are two planar conformations of 1,3-butadiene, called the s-trans conformation and the s-cis conformation where the designation s refers to the carbon-carbon single bond of the diene Of these, the s-trans conformation is slightly lower in energy and therefore is slightly more stable Although s-trans-1,3-butadiene is the more stable conformation, s-cis1,3-­butadiene is the reactive conformation in Diels-Alder reactions In the s-cis conformation, carbon atoms and of the conjugated system are close enough to react with the carbon-carbon double or triple bond of the dienophile and to form a six-membered ring In the s-trans conformation, they are too far apart for this to happen The energy barrier for interconversion of the s-trans and s-cis conformations for 1,3-butadiene is low, approximately 11.7 kJ (2.8 kcal)/mol; consequently, 1,3-butadiene can still be a reactive diene in Diels-Alder reactions Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 914  Chapter 20: Dienes, Conjugated Systems, and Pericyclic Reactions H H C H H C C H H C H H C C C C H H H H s trans s cis (2Z,4Z)-2,4-Hexadiene is unreactive in Diels-Alder reactions because steric hindrance prevents it from assuming the required s-cis conformation s trans s cis (2Z,4Z)-2,4-Hexadiene Example 20.8  The Diels-Alder Reaction II Which molecules can function as dienes in Diels-Alder reactions? (a) (c) (b) Solution The dienes in both (a) and (b) are fixed in the s-trans conformation and therefore are not capable of participating in Diels-Alder reactions The diene in (c) is fixed in the s-cis conformation and therefore has the proper orientation to participate in Diels-Alder reactions Problem 20.8 Which molecules can function as dienes in Diels-Alder reactions? (a) (b) (c) B.  The Effect of Substituents on Rate The simplest example of a Diels-Alder reaction is that between 1,3-butadiene and ethylene, both gases at room temperature Although this reaction does occur, Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 20.5  The Diels-Alder Reaction   915 it is very slow and takes place only when the reactants are heated at 200°C under pressure 1,3-Butadiene Ethylene Cyclohexene Diels-Alder reactions are facilitated by a combination of electron-withdrawing substituents on one of the reactants and electron-releasing substituents on the other Most commonly, the dienophile is electron deficient and the diene is electron rich For example, placing a carbonyl group (electron withdrawing because of the partial positive charge on its carbon) on the dienophile facilitates the reaction To illustrate, 1,3-butadiene and 3-buten-2-one form a Diels-Alder adduct when heated at 140°C 1,3-Butadiene 3-Buten-2-one Placing electron-releasing methyl groups on the diene further facilitates reaction; 2,3-dimethyl-1,3-butadiene and 3-buten-2-one form a Diels-Alder adduct at 30°C 2,3-Dimethyl1,3-butadiene 3-Buten-2-one Several of the electron-releasing and electron-withdrawing groups most commonly encountered in Diels-Alder reactions are given in Table 20.4 Note that the ester group can be either electron donating or electron withdrawing depending on whether the oxygen or the carbonyl is attached to the double bond Table 20.4  Electron-Releasing and Electron-Withdrawing Groups Electron-Releasing Electron-Withdrawing Electron-Releasing Groups Electron-Withdrawing Groups Groups Groups 9 9 9 9 9 9 # C.  Diels-Alder Reactions Can Be Used to Form Bicyclic Systems Conjugated cyclic dienes, in which the double bonds are of necessity held in an ­s-cis conformation, are highly reactive in Diels-Alder reactions Two particularly useful Copyright 2018 Cengage Learning All Rights Reserved May not be copied, scanned, or duplicated, in whole or in part WCN 02-200-203 ... CH3CH2CH2CH2OH 1-Butanol 74 117 g/100 g CH3CH2CH2CH2CH3 Pentane 72 36 Insoluble HOCH2CH2CH2CH2OH 1,4-Butanediol 90 23 0 Infinite CH3CH2CH2CH2CH2OH 1-Pentanol 88 138 2. 3 g/100 g CH3CH2CH2CH2CH2CH3... blindness ClCH2CH2SCH2CH2Cl ClCH2CH2SCH2CH2Cl bis ( 2- Chloroethyl)sulfide bis ( 2- Chloroethyl)sulfide CH3 ClCH2CH2 N CH2CH2Cl bis ( 2- Chloroethyl)methylamine Bis ( 2- chloroethyl)sulfide and bis ( 2- chloroethyl)methylamine... whole or in part WCN 0 2- 2 0 0 -2 03 10 .2 Physical Properties of Alcohols   441 Solution (a) 2- Propen-1-ol Its common name is allyl alcohol (b) (E ) -2 -Hexen-1-ol (trans -2 - Hexen-1-ol) d– Problem 10.3