Chapter 11: Ionic Substitution Reactions

Transcription

1 hapter 11: onic Substitution eactions Note to students: This is a single chapter from a textbook that is under construction. Therefore you can ignore references to other textbook sections Why Should Study This? n a substitution reaction, one portion of the molecule is replaced by something new. n most organic substitution reactions, the new piece that adds to the molecule, or the old piece that is replaced, or both, are ions. Therefore this important process is frequently referred to as an ionic substitution reaction. onic substitution at sp 3 carbon (the topic of this chapter) is also called nucleophilic aliphatic substitution. onic substitution is among the simplest and most fundamental, as well as most common organic reactions. t usually results in the conversion of one functional group, such as an alkyl halide, into a new functional group. eplacement of one functional group with another is a fundamental operation in organic synthesis, the science (and art!) of converting simple organic molecules into more complex molecules such as pharmaceuticals. (We study organic synthesis in chapters 14 and 33.) eaction 11.1 illustrates an ionic substitution reaction in which an alkyl chloride is converted into a thiol. This reaction is the last step in the manufacture of captopril, an antihypertensive drug that works by inhibiting angiotension converting enzyme. Alkyl chloride Thiol l N NaS DMF, 4 hr, 50 o S N aptopril onic substitution is also important in numerous biochemical processes. For example, S-adenosylmethionine (SAM) can add a methyl group to another molecule to meet a variety of cellular needs. SAM is the focus of this chapter s n the eal World discussion, section This methyl group is transferred N 2 N N S N N N 3 S-adenosylmethionine (SAM) onic Substitution eactions Page 1

2 While exploring ionic substitution reactions we will also become familiar with some basic features that are common to many other organic reactions: nucleophiles, leaving groups, steric effects and solvents The S N 2 eaction eview onjugate base (9.xx), curved arrows (3.xx) and reaction mechanism (3.xx). Let s begin our study of ionic substitution reactions by revisiting a substitution reaction that we are already familiar with: an acid-base reaction. n an acid-base reaction, a base (often an ion) attacks a hydrogen atom of an acid, causing the conjugate base to leave. n other words, the base replaces the conjugate acid. For example, hydroxide ion reacts with hydrogen chloride to produce water and chloride ion (reaction 11.2). Note the curved arrows that show the bond changes and summarize the reaction mechanism. l + l δ + δ You can imagine that reaction 11.2 is initiated by the attraction of opposite charges: the negative formal charge of - and the δ + of l. This electrostatic attraction is a common feature of ionic substitution reactions, and many other organic reactions in general. (Flip through this text and you will find many examples.) n hapter 9 we learned quite a bit about acid/base reactions and what influences them. Now it s time to extend what we have learned to a new and very important class of reactions. Think Ahead Question 11.1 Based on reaction 11.2, write the reaction that might occur between - and l. nclude all curved arrows, formal charges and lone pairs. Will this reaction occur as written? Answer: hanging l to l gives us reaction δ + l δ - + l 11.3 Will this reaction occur as written? Like the - /l reaction (11.2), we can also imagine this reaction is initiated by interaction of opposite charges. n this case the positive charge is the δ + carbon atom of l. So if our charge interaction model is valid, reaction 11.2 will occur as written. t is useful at this point to review some terms we learned previously that also apply to this substitution reaction (as well as many other reactions we will encounter throughout our studies of organic chemistry). onic Substitution eactions Page 2

3 Nucleophile: The reactant that provides the electron pair to form a new covalent bond. Also called a Lewis base. Nucleophiles usually have lone pairs, pi bonds, formal negative charges, or (less commonly) δ - charges. Electrophile: The reactant that accepts the pair of electrons to form a new covalent bond. Also called a Lewis acid. Electrophiles usually have open octets, formal positive charges or δ + charges. Some electrophiles also have a leaving group as part of their structure. Leaving group: An atom or group that breaks away, taking with it the electron pair that used to form the covalent bond between the leaving group and the rest of the molecule. Nucleophile Electrophile Leaving group eview Nucleophile, electrophile and leaving group (8.xx). Despite its name, a leaving group is not always a group of atoms. -, Br -, l - and F - are called leaving groups even though they are single atoms. δ + l δ + l Why is this reaction interesting? We start with methyl chloride (an alkyl halide) and convert it into methanol (an alcohol). We have achieved a functional group conversion. As we learned in section 11.1, functional group conversions are important in a variety of organic reactions, so this ionic substitution is worthy of further scrutiny. n the coming sections we will explore certain features of this simple reaction that apply to other substitution reactions and many other types of more complex organic reactions as well. n general the single step ionic substitution mechanism can be drawn this way, where Nuc = nucleophile and LG = leaving group: 1 1 Nuc LG Nuc LG oncept Focus Question 11.1 Write the products and mechanisms for the following ionic substitution reactions. nclude all formal charges and lone pairs. Label each nucleophile, electrophile and leaving group. Describe the functional group change. (a) l + (c) l + (b) + l (d) + N Br oncept Focus Question 11.2 Write the missing reactants on the left side of the arrow as necessary to complete each reaction. nclude all formal charges, lone pairs and curved arrows. Label onic Substitution eactions Page 3

4 (b)??? + S 2 S (d)??? + ( ) 3 (3)3 each nucleophile, electrophile and leaving group. Describe the functional group change. (a) Br +??? (c) 2 +??? 2 N 3 The reaction of an alkoxide ion ( - ) with an alkyl halide or alkyl sulfonate is called the Williamson ether synthesis, and is discussed in more detail in section 20.xx. oncept Focus Question 11.3 What nucleophile is necessary to convert l into each of the following? (a) (e) S 6 5 (i) N + 3 l - (b) (f) N (j) ( ) 4 N + l - (c) (g) (k) 2+ l - (d) S (h) N 3 (l) A. Mechanism and Kinetics n order to make any practical use of the ionic substitution reaction, we must understand how various factors influence its rate. For example, if we use - instead of Br - will the reaction be faster or slower, or perhaps so slow that the product will not be produced at a measurable rate? (The rate issue is of very broad concern. For example, in the chemical industry, time is money, and a reaction that takes too long is of no value.) eview onnection between reaction rate and mechanism (8.xx). Studying the reaction rate is also a useful window on the reaction mechanism. We can write a mechanism on paper, but how do we know that this is what the reactants are actually doing? We proposed a mechanism in the previous section: the two bond changes occur simultaneously. n other words the reaction is concerted. eview if necessary the relationship between reaction rate and mechanism (section 8.xx). Think Ahead Question 11.2 Write the reaction rate expression for reaction Answer: ecall from section 8.xx (and your studies of reaction kinetics in introductory chemistry) that for a reaction in which all bond changes occur in a single step, the rate expression is the product of the concentration of each reactant. The rate expression is first-order in each reactant and second-order overall: ate = k [ l][ - ] This rate expression tells us that changing the concentration of either l or - changes the rate of the reaction. For example, doubling the concentration of l doubles the rate. educing the concentration of - to 1/10 of the starting value cuts the reaction rate by ten. onic Substitution eactions Page 4

5 The rate expression is meaningful only if the mechanism is accurate. onversely, if the mechanism prediction is inaccurate then the effects on reaction rate caused by changing concentration of reactants will not agree with the rate expression. (We are assuming that the reaction does not have a more complex mechanism whose observed rate expression is much simpler than the mechanism suggests. There are other ways to test for this in the lab.) n the lab we find that the predicted rate expression agrees with experiment, which suggests that the mechanism is reasonable. (A rate study can disprove a predicted mechanism, but it can never prove the mechanism because we can write other mechanisms whose rate expressions will appear to be identical.) n fact, lab studies have shown that many simple ionic substitution reactions follow this same rate law, which we can write in a generic form: ate = k [nucleophile][electrophile] onic substitution reactions that follow this rate law are called S N 2, an abbreviation for substitution nucleophilic bimolecular. (Bimolecular means the rate law is first-order in two reactants.) n modern sense, any ionic substitution process in which the nucleophile bonds to carbon at the same time the leaving group leaves is called S N 2, even if it is not bimolecular. For example, the nucleophile and leaving group might be contained within the same molecule, resulting in an intramolecular substitution reaction and a first-order rate expression (reaction 11.4). We still call this an S N 2 reaction. ate = k 11.4 The term S N 2 was first introduced by Edward ughes and Sir hristopher ngold of the University ollege in London, who carried out very extensive studies on ionic substitution reactions beginning about They preformed early studies in physical organic chemistry, especially in substitution and elimination reactions. They introduced many concepts and terms in wide use today such as nucleophile, electrophile, inductive effect, S N 2 and S N 1. ngold was the co-author of the ahn-ngold-prelog priority rules (section 6.xx). eaction 11.4 may appear incomplete because it does not include the leaving group ( - ) that has departed or the cation (such as Li + or Na + ) that is associated with the oxyanion ( - ). This is normal. These inorganic species are of little interest to us (we want to synthesize more complex organic molecules, not make simple metal salts), so we often leave them out when we write reactions and mechanisms. onic Substitution eactions Page 5

6 aution! n a significant number of organic reaction mechanisms, the leaving group still participates after it has left. t is not necessary to draw it in the mechanism step in which it has left, but do not forget about its presence. f you are the forgetful type (even slightly so), it might be a good habit to draw the departed leaving group just to remind yourself. n the other hand, metal cations rarely play an important role in the reaction mechanisms we will study, so they can usually be ignored without significant consequences. oncept Focus Question 11.4 Each of the following is an S N 2 reaction. Write the product(s) (including departed leaving group), mechanism and rate law. f you are having trouble with the products and mechanism, look for patterns from previous examples, such as the kinds of things that can be leaving groups or nucleophiles. (a) Br Na (c) 2 (b) S Li (d) l B. Transition State eview Transition state and energy of activation (8.xx). When we studied the fundamental concepts of reactions in hapter 8, we learned that the structure of the transition state and the energy of activation influence the reaction rate. Altering any factor in the reaction that changes the stability (energy) of the transition state and/or reactants can change the reaction rate. Therefore if we are interested in understanding how various factors influence the rate of an S N 2 reaction, we must pay attention to the transition state. transition state energy This energy difference controls rate eactants Products reaction coordinate Figure 11.1: Energy diagram for a typical exergonic S N 2 reaction. onic Substitution eactions Page 6

7 Think Ahead Question 11.3 Draw the structure of the transition state for reaction Answer: Even though a transition state has a very brief existence no more than the time it takes for one bond vibration to occur it still obeys the same rules and forces as any other molecular structure. The most broadly influential of these forces is electron repulsion, which we explored in section 2.xx. Just to review, the attachments (atoms, groups of atoms and lone pairs) arrange themselves around a central atom in a way that minimizes electron repulsion. Size matters: larger groups cause more electron repulsion, causing smaller groups to move away a little bit. The position of the attachments around the central atom controls its hybridization. f the central atom has four attachments, a tetrahedral arrangement gives the least electron repulsion and the lowest energy, and the central atom uses sp 3 hybrid orbitals to bond with the attachments. eview Electron repulsion effects on molecular structure (2.xx). The central atom of our transition state (a carbon) has five attachments. These are best accommodated using a trigonal bipyramid: two pyramids with triangular faces and a common base. So far our transition state looks like this: The nucleophile ( - ) is approaching the carbon in order form a bond, but that bond is not yet complete. We indicate this by using a dashed line that is longer than a normal (full) bond. The nucleophile is using a lone pair to make its new bond to carbon, so its electron density (as measured by formal charge) is decreasing. n - the oxygen bears a formal negative charge and in the product it is neutral. n the transition state its charge is not quite -1 but not yet zero, so we indicate this partial negative charge using our familiar δ - symbol. The leaving group (l - ) is on its way out (along with the electron pair that was the -l bond) but not yet completely gone. Like the partial carbon-nucleophile bond, we use an elongated dashed line to depict the partial carbon-leaving group bond. Like the nucleophile, its charge is changing (neutral in l to negative in l - ) so in this transitory structure its charge is also transitory (δ - ). Where should we place the nucleophile and leaving group? They both have a δ - charge. To maximum stability (and minimize electron repulsion), we place the nucleophile and leaving group as far apart as possible. This arrangement in which the nucleophile approaches the carbon-leaving group bond from the carbon end is called backside attack. δ - δ - l The hydrogen atoms of the methyl group move around as the reaction occurs, but the - bonds do not break. They lie in the space somewhere between the onic Substitution eactions Page 7

8 leaving group and nucleophile. We cannot predict exactly where they are (as discussed in section 8.xx), so we draw them about halfway between the nucleophile and leaving group. Don t forget the wedge and dashed line, to show the hydrogens are coming out of and going into the plane of the paper. This reminds us of the three-dimensional shape of the structure. Lastly, we enclose the structure in brackets and add the little double dagger symbol ( ) indicate this is a transition state. The completed transition state drawing is shown in Figure δ - δ - l Simplified drawing alculated Figure 11.2: Transition state representations for the S N 2 reaction of - with l. This drawing is just a crude sketch to give us an approximation of the relative position of each atom or group in the transition state, but it is good enough for our simple purposes. The actual transition state for this reaction, using molecular modeling software, is also shown in Figure Build a model of the transition state to get a better feeling for the three-dimensional relationship of the atoms. oncept Focus Question 11.5 What is the formal charge or δ charge on the carbon atom in the transition state of Figure 11.2? oncept Focus Question 11.6 arbon is limited to eight valence shell electrons (so it can form four covalent bonds at most) yet the carbon of the transition state in Figure 11.2 has five bonds. Explain. oncept Focus Question 11.7 Students often ask if it is necessary to draw a transition state in a certain way. For example, does the nucleophile have to be on the right, or does one - bond always have to be vertical? The transition state can be drawn from any viewpoint or orientation, as long as it accurately depicts the relative positions of the atoms involved. To explore this point, build a model of transition state A, then use it decide if structures B-D depict the same transition state. l δ - δ - 2 δ - δ - δ - δ - l δ - l l δ A B D onic Substitution eactions Page 8

9 oncept Focus Question 11.8 (a) According to the ammond postulate (section 8.xx), the structure of the transition state is influenced by the energy of activation. Draw the transition state for an exergonic S N 2 reaction with a generic nucleophile (nuc) at a methyl group bearing a generic leaving group ( -LG). (b) Draw the transition state for an endergonic S N 2 reaction with a generic nucleophile with -LG. (c) What is necessary for the three hydrogens of these generic transition states to be exactly coplanar? oncept Focus Question 11.9 Draw the transition states for the reactions in oncept Focus Question Assume each reaction is exergonic.. nversion of onfiguration n the previous section we developed a transition state structure for the S N 2 reaction of - and l. We assumed the nucleophile would attack the backside of the carbon-leaving group bond, because the nucleophile and leaving group δ - charges repel. an we verify this assumption experimentally? What is the experimentally observable difference (if any) between backside attack and front side attack (attack on the same face as the leaving group)? Think Ahead Question 11.4 onsider the S N 2 reaction of (S)-2-iodobutane with cyanide ion (reaction 11.5). The product has a stereocenter so two enantiomers are possible (review enantiomers in section 6.xx). Based on the mechanism and transition state, predict which enantiomer(s) are produced. Models of the starting iodide and the transition state are very useful here. eview Enantiomers (6.xx). 2 N 2 N + N Answer: As cyanide approaches the backside of the carbon-iodine bond, the ethyl group, methyl group and hydrogen atom begin to move away. (When someone crowds you on a bench, don t you move away, too?) As the carbon bearing the leaving group changes from tetrahedral (sp 3 ) in 2-iodobutane to trigonal planar (sp 2 ) in the transition state, the attachments move from a tetrahedral to a planar arrangement. (The partial N--- and --- bonds lie along the sp 2 carbon s unhybridized p orbital.) As the leaving group departs, more space is made available. To reduce repulsion the attachments move away from the incoming nucleophile toward the newly vacated space, and the carbon returns to sp 3. onic Substitution eactions Page 9

10 2 2 δ - δ - N N N 2 + eactants Transition state Products Figure 11.3: Atomic motions in the S N 2 reaction leading to inversion of configuration. The arrows indicate direction of motion. eview Stereochemical configuration (6.xx). This model predicts that the reaction produces ()-2-methybutyronitrile, a prediction that is consistent with experimental results. n fact studies of many S N 2 reactions have shown that backside attack always occurs. Even if the nucleophile and leaving group have opposite charges in the transition state, backside attack still occurs, even though we might expect front side attack because opposite charges attract. (Explore this more in oncept Focus Question ) This backside attack leads to inversion of configuration at the asymmetric carbon. (eview the concept of stereochemical configuration in section 6.xx.) f the configuration does not change we call it retention of configuration.) aution! S N 2 inversion of configuration is also called Walden inversion, after Paul Walden ( ), a pioneer in the study of substitution reactions. nversion of configuration requires making and breaking of bonds, and therefore occurs only at the carbon bearing the leaving group. n the following example, inversion occurs at the carbon bearing the chlorine atom (the leaving group) but not the methyl group (not the leaving group). n this case inversion converts an stereocenter into an S stereocenter, as well as trans into cis. Not inverted Na l nverted ()-2-hloro-1-methyl- (S)-2-odo-1-methylcyclohexane (trans) cyclohexane (cis) n an S N 2 reaction, why does backside attack occur? epulsion of δ - charges on the nucleophile and leaving group is a simple explanation, but it fails to explain the cases where the nucleophile and leaving group have opposite charges in the transition state. An explanation that covers all S N 2 reactions and is independent of the charges is this: the nucleophile attacks the σ* (sigma antibonding) orbital associated with the carbon-leaving group sigma bond. (eview bonding and antibonding orbitals from section 2.xx if necessary. Why these particular orbitals are involved is beyond the scope of our introductory exploration of ionic substitutions.) The nucleophile approaches from the backside, an orientation that leads to the onic Substitution eactions Page 10

11 greatest overlap between its lone pair orbital and the σ* orbital. (ecall that greater overlap gives stronger bonds, more stability, and in the case of a transition state, lower energy of activation.) lone pair σ* σ bond 1 nuc LG nuc LG nuc LG eview Bonding and antibonding orbitals (2.xx). Figure 11.4: Simplified scheme showing orbital involvement in the S N 2 reaction. oncept Focus Question Provide the product, mechanism and transition state for each of these S N 2 reactions. Pay carefully attention to the stereochemistry. Label the stereocenter(s) of each reactant and product as or S. (a) l (c) Br Br + (excess) (b) LiS (d) D Br Na (D = deuterium = 2 ) oncept Focus Question Provide the missing electrophile for each S N 2 reaction. Pay careful attention to stereochemistry. (a)??? (c)??? SPh (b)??? PhS (excess) (d) 6 12 S 3 S SPh oncept Focus Question Does inversion occur in an S N 2 reaction even if the carbon bearing the leaving group is not a stereocenter? As an example, consider the S N 2 reaction of iodomethane with cyanide ion. onic Substitution eactions Page 11

12 oncept Focus Question (a) From the view point of the ahn-ngold-prelog priority rules (section 6.xx), explain why inversion of configuration converts almost any stereocenter into S and S into. (b) Every S N 2 reaction occurs with inversion of configuration but does every S N 2 reaction convert an stereocenter to S (or S to )? Design an example S N 2 reaction to illustrate your discussion. oncept Focus Question n order to explore the effect of leaving group and nucleophile charge on the S N 2 transition state the following reaction was performed. NaN 3 S( ) 2 N 3 (a) Draw the best Lewis structure for azide ion (N 3- ). (b) Draw the transition states and products expected from both backside and front side attack. Pay careful attention to the stereochemistry. (The product has a stereocenter.) (c) Which enantiomer of the product was formed? oncept Focus Question Students often have trouble seeing inversion of configuration when an S N 2 reaction occurs at the carbon of a cyclohexane ring. as inversion occurred in the reaction shown below? By adding, subtracting or otherwise changing just one atom of the electrophile, rewrite the reaction so that inversion is now clear and obvious. N N 11.3 Factors nfluencing the S N 2 eaction You probably noticed in the previous sections and oncept Focus Questions that S N 2 reactions include a wide variety of nucleophiles, electrophiles and leaving groups. an we predict how these factors might influence an S N 2 reaction, to make it faster, slower or perhaps not even occur at all? (Not every S N 2 reaction that we might write is viable.) n this section we explore some important factors (or variables) and learn how to predict their effects on S N 2 reaction rate and products. We will look at the nucleophile, the leaving group, the affect of groups attached to the carbon bearing the leaving group and the reaction solvent. onic Substitution eactions Page 12

13 nuc + 3 LG solvent nuc 3 + LG Figure 11.5: The fundamental features that influence the rate and product of every S N 2 reaction are the nucleophile (nuc), the leaving group (LG), the attachments to the carbon that bears the leaving group ( 3 ) and the solvent. A. The Nucleophile Figure 11.5 suggests two ways in which the nucleophile influences an S N 2 reaction: product and rate. ow does the nucleophile influence the product of an S N 2 reaction? The nucleophile provides the new functional group. For example, if the nucleophile is cyanide ion ( - N), then the product is a nitrile ( N). (eview oncept Focus Question 11.3 for more examples.) ow does the nucleophile influence the rate of an S N 2 reaction? A useful approach to this issue starts with the question what is the role of a nucleophile in an S N 2 reaction, and how does its structure influence its ability to carry out this role? The role of the nucleophile is simple: to share an electron pair with the electrophile, resulting in a new covalent bond. igher electron density makes the nucleophile less stable (more electron repulsion) therefore more strongly driven to share some of that electron density. (Don t you have a greater desire to find a new place to live when your dorm room or apartment becomes too crowded?) Therefore any structural feature that influences electron density will also influence nucleophilicity. These structural features might include resonance, atomic radius, electronegativity, inductive effects and formal charge. We measure nucleophilicity by the relative rate of S N 2 reactions with various nucleophiles, and the same electrophile, solvent, concentration and temperature. Perhaps you are saying to yourself ey, this sounds familiar! as you recall our previous studies on the relationship between basicity and electron density (sections 9.xx-9.xx). t is a good parallel: a nucleophile and a base both provide an electron pair for the new bond. They differ in the context of the bond formation: a nucleophile shares its electron pair with an electrophile, whereas a base shares its electron pair with the hydrogen of an acid. n other words, in a reaction where X shares its electron pair with Y to form a new X-Y bond, if Y is a hydrogen atom we call X a base. f Y is a carbon atom we call X a nucleophile. f this basicity/nucleophilicity parallel is true, then do the same features that influence basicity also influence nucleophilicity? The S N 2 rate data of Table 11.1 will be useful to guide us to a conclusion. onic Substitution eactions Page 13

14 Table 11.1: elative rate for the S N 2 reaction of selected nucleophiles with Br in ethanol. 3 Br Nuc 2 3 Nuc + Br Nucleophile S 2 elative rate 57,000, ,000 60,000 12,000 -Nuc pk a Nucleophile Br ( 2 ) 3 N l elative rate 5,000 2,000 1,400 1,100 -Nuc pk a Nucleophile 2 F l 4 elative rate Nuc pk a Despite our prediction, however, the relative rate data in Table 11.1 make it obvious that the relationship between nucleophilicity and pk a is not universal. The problem here is that nucleophilicity and pk a relate to similar, but not identical reactions. A nucleophile bonds to carbon atom, whereas a base bonds to a hydrogen atom. Are there any trends at all in this data? an we make any useful predictions about structure and nucleophilicity? 1. esonance eview Structural effects on basicity (9.xx-9.xx). Think Ahead Question 11.5 Table 11.1 reveals that acetate ion ( 2 - ) is a weaker nucleophile and weaker base than ethoxide ion ( 2 - ). What structural difference makes acetate ion a poorer base and a poorer nucleophile than ethoxide ion? t may be useful to review how structure influences basicity from sections 9.xx-9.xx. Answer: As we learned earlier, the mechanistic role of a nucleophile and a base is similar: to provide an electron pair to form a new covalent bond. We also predicted that the same structural features that influence basicity might also influence nucleophilicity. Table 11.1 tells us this parallel does not cover all onic Substitution eactions Page 14

15 nucleophiles, but it seems to apply in the case of acetate ion versus ethoxide ion. What structural difference makes acetate ion a poorer base and a poorer nucleophile than ethoxide ion? ne obvious difference is resonance. Acetate ion has significant resonance stabilization and delocalization of its electron density. Ethoxide ion does not have this resonance delocalization. ts resonance hybrid is identical to its only significant resonance contributor. Acetate ion has two significant resonance contributors. δ - δ - esonance hybrid eminder: When making a comparison between two or more things, focus on differences instead of similarities. eview elative significance of resonance contributors (3.xx). X no additional resonance contributors Ethoxide ion has a single significant resonance contributor. Because of its resonance delocalization, each oxygen atom of acetate ion has less electron density than the oxygen of ethoxide ion. (Each oxygen atom of acetate ion bears a δ - charge whereas the oxygen of ethoxide ion has a full 1 formal charge.) Lower electron density reduces the drive to share electron density, so acetate ion is a poorer (less aggressive) nucleophile than ethoxide ion. This is exactly the same reason that acetate ion is a poorer base than ethoxide ion, as we learned in section 9.xx. ere is an alternate and equally valid explanation. We know that resonance is a stabilizing feature. A molecule that has resonance stabilization wants to keep it. A molecule that does not have resonance stabilization wants to get it. Upon forming a new covalent bond with the carbon atom of an electrophile such as methyl iodide, acetate ion suffers a reduction in the number of significant resonance contributors. This resonance loss destabilizes the transition state, and slows the reaction. Ethoxide does not suffer this resonance loss. Therefore acetate ion is a poorer nucleophile than ethoxide ion. X no additional significant resonance contributors Before reaction: acetate ion has two significant resonance contributors After reaction: the ester product has just one significant resonance contributor. A study of many other resonance-stabilized nucleophiles leads to this general rule: onic Substitution eactions Page 15

16 General ule ompared to similar nucleophiles without resonance, most nucleophiles with resonance delocalization of the electron pair used to form the new covalent bond with the electrophile are less nucleophilic. There are some exceptions to watch out for. aution! esonance does not always reduce electron density. A molecule may have resonance that does not influence the electron density at the atom that attacks the electrophile. n some rare circumstances resonance can even increase the electron density. (See oncept Focus Question for more on this.) oncept Focus Question Select the weaker nucleophile in each pair. ffer a brief explanation in each case. (a) and (b) N 2 and N 2 (c) and (d) N 2 and N 2 oncept Focus Question By adding, subtracting or otherwise changing at most three atoms, redraw each structure so that it has resonance (or even more resonance) that makes it a poorer nucleophile. Draw the resonance contributor(s) that account for the reduced nucleophilicity. (a) (b) ( ) 3 2 N 2 (c) oncept Focus Question t was mentioned previously that resonance does not always decrease nucleophilicity. With this thought in mind, select the faster reaction in each pair, and briefly explain why it is faster. onic Substitution eactions Page 16

17 (a) N 2 3 N 2 versus N 3 N 2 N N 2 2 N N 2 2 N N 2 2 N N 2 (b) 2 2 l 2 2 versus 2 2 l 2 2 oncept Focus Question Write two S N 2 products for this reaction. int: think about resonance. A nucleophile that can react at more than one site on its structure is called an ambident nucleophile. 2. Electronegativity n the previous section we learned that resonance influences nucleophilicity in the same way it influences basicity. s this also true for electronegativity? Think Ahead Question 11.6 Based on what you know about electronegativity effects on basicity (section 9.xx), rank these ions in order of increasing nucleophilicity: F -, - and 2 N -. Answer: We have already established that basicity and nucleophilicity both involve sharing of an electron pair. Therefore we predicted that any structural feature that increases or decreases basicity has a parallel effect on nucleophilicity. eview Electronegativity effects on basicity (9.xx). Back in section 9.xx we deduced that electronegativity also influences basicity. f the three ions in question fluoride ion is the weakest base because fluorine is more electronegative (i.e., less willing to share electron density) than oxygen or nitrogen. xygen is more electronegative than nitrogen, so the order of basicity is F - < - < 2 N -. Therefore the order of nucleophilicity is F - < - < 2 N -. Experiments conducted under conditions where no other factors interfere verify this prediction, and further show that the effect is general. General ule The strength of a nucleophile is influenced by the electronegativity of the atom(s) that share an electron pair to form a covalent bond with the electrophile: higher electronegativity causes lower nucleophilicity. onic Substitution eactions Page 17

18 oncept Focus Question For each reaction pair, select the faster reaction, and write its product. (a) hloride ion and methanethiolate ion ( S - ) reacting with 2-iodopropane. (b) Water and methylamine reacting with ethyl chloride. (c) Triphenylphosphine (Ph 3 P) and diphenylsulfide (Ph 2 S) reacting with ()-1- bromo-1-phenylethane. (d) N and reacting with methyl iodide. 3. Atomic adius n section 9.xx we learned that basicity is influenced by atomic radius: larger atomic radius results in lower electron density and reduced basicity. Does this effect apply to nucleophilicity as well? Think Ahead Question 11.7 Methyl iodide was reacted with halide ions in DMF (the solvent) at 25 o. The relative S N 2 reaction rates (k rel ) were found to be: F - k rel > 7.5 (fastest), l - k rel = 6.3, Br - k rel = 3.3 and - k rel = 1.0 (slowest). What is the relationship between atomic radius and reaction rate? Suggest an origin for this effect. eview Effect of atomic radius on basicity (9.xx). Answer: These ions are all in the same family (column) of the periodic table. ecall from general chemistry that atomic radius increases as you move down the family. Fluoride ion is the smallest and most nucleophilic, whereas iodide is the largest and least nucleophilic. This atomic radius effect on nucleophilicity is parallel to what we learned previously (section 9.xx) in relationship to basicity: smaller atomic radius results in more concentrated electron density, and greater basicity (or nucleophilicity). Atomic radius also changes somewhat as you move across the row (period) of the periodic table. owever, atomic radius differences within a period are much less than atomic radius differences within a family. When comparing atoms in the same column of the periodic table, atomic radius has more influence on nucleophilicity (and basicity) than electronegativity. When comparing atoms in the same row of the periodic table, electronegativity has more influence on nucleophilicity (and basicity) than atomic radius. (A more precise analysis uses ionic radius instead of atomic radius, but the qualitative results are the same.) onic Substitution eactions Page 18

19 ncreasing electronegativity Same period: electronegativity dominates 2.5 N F 4.0 P 2.5 S 2.5 Se 2.4 l 3.0 Br 2.8 Same family: atomic radius dominates ncreasing atomic radius 2.5 Figure 11.6: The northeast corner of the periodic table showing trends in atomic radius (size of circles; not to scale) and electronegativity (number in the circle). A more rigorous analysis reveals that the origin of the effect is subtler and more complex than just electron density. (For example, Table 11.1 reveals that when the solvent is ethanol, iodide ion is a much more aggressive nucleophile than fluoride ion.) The atomic radius effect is also strongly influenced by how the solvent interacts with the reactants and transition state. Therefore we must reserve making a General ule about atomic radius effects until we learn how polarizability (section 11.3) and solvent (section 11.3E) influence an S N 2 reaction. aution! The atomic radius effect is based on the atomic radius of the atom that is sharing an electron pair with the electrophile, and not on the size of the entire nucleophile. oncept Focus Question Select the strongest nucleophile in each set. Write its S N 2 reaction with ethyl iodide. (a) - and S - (c) F - and S - (b) ( ) 3 N and ( ) 3 P (d) - and ( ) 3 - onic Substitution eactions Page 19

20 4. nductive Effects So far in this section we have seen that there is a good (although imperfect) parallel between the structural features that influence basicity and nucleophilicity. We have focused on the atom of the nucleophile that is sharing electron density with the electrophile. What happens if we make changes elsewhere in the nucleophile, away from the business end? Think Ahead Question 11.8 The data in Table 11.1 suggests that compared to other nucleophiles, acetate ion ( 2- ) is rather sluggish. s trifluoroacetate ion (F 3 2- ) a stronger or weaker nucleophile than acetate ion? Answer: Acetate ion is a sluggish nucleophile because its electron density is delocalized by resonance, and because of the high electronegativity of oxygen. Trifluoroacetate ion has the same influences because it also a carboxylate ion (i.e., it has the same functional group). eview nductive effect influence on basicity (9.xx). An obvious difference between the nucleophiles in question is the presence of a methyl group in 2- versus a trifluoromethyl group in F ow might the presence of fluorine influence nucleophilicity? When we think of fluorine one thing that comes to mind is its high electronegativity. f all the atoms in the periodic table, fluorine has the greatest greed for electron density, and will take it from its neighboring atoms. (ydrogen s low electronegativity gives it no cause to steal or donate electron density to other parts of the molecule.) This theft of electron density from elsewhere in the molecule is called an inductive effect, and we have encountered it before when we studied basicity (section 9.xx). An electron-withdrawing inductive effect reduces basicity, and extending our parallel between basicity and nucleophilicity, we predict it also decreases nucleophilicity. Therefore F is a poorer nucleophile than 2 -. Studies of many nucleophiles show that the inductive effect is general, so we can formulate a General ule. General ule Nucleophilicity is decreased by electron-withdrawing inductive effects, and increased by electron-donating inductive effects. aution! Most, but not all, inductive effects that we will encounter are electron withdrawing in nature. A few are electron donating. Therefore do not assume that inductive effects always decrease nucleophilicity. onic Substitution eactions Page 20

21 oncept Focus Question Select the weakest nucleophile in each set. (a) F versus 2 - (c) F versus F (b) Br versus Br (d) l versus F oncept Focus Question Select the major product of this reaction. F F F F or F F oncept Focus Question The data in Table 11.1 reveals that hydroxide ion is a substantially better nucleophile than perchlorate ion (l 4 - ). What structural feature(s) account for this difference? oncept Focus Question Measurement of S N 2 reaction rates for some thiolate ions gives the following order of nucleophilicity: S < 2 S < 2 S What does this suggest about the inductive effect of a benzene ring? 5. Formal harge n our studies of organic molecules as acids and bases we deduced that bases with formal negative charges are stronger than uncharged bases (section 9.xx). Does formal negative charge also enhance nucleophilicity? Think Ahead Question 11.9 The data in Table 11.1 reveals that hydroxide ion is a substantially better nucleophile than water. What structural feature(s) account for this difference? Answer: Water and hydroxide ion do not have resonance to delocalize their electron density. n both cases an oxygen atom provides the electron pair that becomes the new covalent bond, so there is no difference in the atomic radius or electronegativity effects. Neither oxygen atom bears electron-withdrawing or electron-donating groups, so the nucleophilicity difference cannot be due to inductive effects. All that remains is the formal charge. ow does formal charge influence electron density, our root cause of basicity and nucleophilicity? Back in section 9.xx we learned that a base with a formal negative charge is stronger than a neutral base, because a negative formal charge indicates the atom has one more electron than the corresponding uncharged atom of the same element. Keeping with our eview Formal charge effect on basicity (9.xx). eview Meaning of formal charge (general chemistry and 1.xx). onic Substitution eactions Page 21

22 parallel between basicity and nucleophilicity, we therefore predict that a nucleophile with a formal negative charge is more nucleophilic than a nucleophile that is neutral. This explains why - is a stronger nucleophile (and stronger base) than 2. General ule Everything being equal, a nucleophile with a formal negative charge is more nucleophilic than a neutral (uncharged) nucleophile. f we write S N 2 reactions for these nucleophiles with (for example) methyl iodide, we can draw a useful corollary of this formal charge effect xygen formal charge 1 0 Faster reaction xygen formal charge 0 +1 Slower reaction n these reactions, the oxygen atom of hydroxide ion (the better nucleophile) undergoes a formal charge change of 1 to 0 upon S N 2 reaction. n the reaction of water (the weaker nucleophile) the formal charge of oxygen changes from neutral to +1 (it becomes an oxonium ion). ecall (section 1.xx) that atoms prefer to remain neutral. Therefore a formal charge change of negative to neutral enhances nucleophilicity whereas a formal charge change from neutral to positive decreases nucleophilicity. oncept Focus Question For each pair of molecules, draw the conjugate base (review section 9.xx if necessary) and select the strongest nucleophile in each pair of conjugate bases. (a) N 3 versus + N 4 (b) Water versus methyl ether (c) 2 S versus S - (d) xalic acid ( 2-2 ) versus hydronium ion ( 3 + ) 6. elative nfluence of Factors You may have noticed that in several of the oncept Focus Questions leading up to this point, different solutions were possible depending upon what assumptions you made about the relative influence or priority of the various factors that control nucleophilicity. t would be useful, therefore, to be able to rank the relative influence of these factors to help us understand and predict those cases where the factors are not synergistic. To assemble this ranking, we compare reaction rates for nucleophiles that have two factors in conflict, such as atomic radius and electronegativity, but all other factors equal. Think Ahead Question gives you a taste of how this might be done. onic Substitution eactions Page 22

23 Think Ahead Question Using the relative rate data given in Table 11.1, decide which of the structural factors that influence nucleophilicity is dominant: (a) 2 - versus l -, and (b) 2 - versus 2. Answer: Let s start by laying out our tool kit (so to speak), with a list of the structural features that we have deduced so far: esonance Electronegativity Atomic radius nductive effect Formal charge n continuing our parallel between nucleophilicity and basicity, note that this is identical to the list of features that influence basicity (section 9.xx). Now let s apply these to the nucleophile pairs in question. 2 - is more nucleophilic than l -. The factors at odds in this pair are electronegativity and atomic radius. The fact that ethoxide ion is a stronger nucleophile than chloride ion suggests that the smaller atomic radius of oxygen carries more weight than the lower electronegativity of chlorine. Therefore in terms of relative influence we can say atomic radius > electronegativity. - 2 is more nucleophilic than 2. The factors at odds in this pair are resonance and formal charge. The fact that acetate ion is a stronger nucleophile than water suggests that the formal negative charge of acetate has a greater effect on its nucleophilicity than its resonance. Therefore in terms of relative influence we can say formal charge > resonance. onsidering many nucleophile pairs like this reveals a general trend of influences, which we use as the basis for a General ule. General ule The relative importance of structural features that influence nucleophilicity is: esonance > atomic radius > electronegativity > inductive effects The influence of formal charge varies greatly, sometimes carrying more weight than resonance and in other cases, less. Therefore we cannot rank it with any certainty. There are exceptions to this ranking, but it makes accurate predictions often enough to be useful. (We will explore some of these exceptions later in this chapter.) ur theory of nucleophilicity would need to be more complex to cover every case and every nuance. A simple, easily understood theory that gets it right most of the time is very useful. (A good professional baseball player may only onic Substitution eactions Page 23

24 get on base 35% of the time, and look how much he gets paid! magine how much more he would get if he got it right 80% of the time.) Since we are discussing nucleophilicity factors as a group, this is a good time to mention a pitfall experienced by some students. aution! When applying the nucleophilicity factors, all but inductive effects apply to the atom(s) that form the new covalent bond with the electrophile. (You might think of these atoms as the business end or warhead functional group of the nucleophile.) f the portion of the molecule under question is not part of the functional group that becomes bonded to the electrophile, then it can only influence nucleophilicity through an inductive effect. esonance Atomic radius operate Electronegativity here Formal charge nductive effect operates here S N 2 X Y 3 LG X Y 3 + LG oncept Focus Question Select the strongest nucleophile in each pair. Briefly explain your reasoning. (a) F versus 2 - (b) The conjugate bases of l and Br (c) Aniline and cyclohexanol oncept Focus Question By adding, subtracting or otherwise changing no more than three atoms, modify each of the strongest nucleophiles from oncept Focus Question so that it is even stronger. B. The Leaving Group Now we turn our attention to the leaving group, the portion of the electrophile that breaks away along with the electron pair that used to be the carbon-leaving group bond. What constitutes a good leaving group? The role of the leaving group appears to be exactly opposite that of the nucleophile. The leaving group accepts an electron pair as the carbon-leaving group bond breaks whereas a nucleophile supplies an electron pair as the carbonnucleophile bond forms. The more effectively the leaving group can accommodate its new electron pair the more readily it will leave. (n more rigorous terms, a more stable leaving group reduces the energy of activation and makes the reaction faster.) Whether we are concerned about nucleophiles or onic Substitution eactions Page 24

25 leaving groups, it appears that ability to stabilize electron density is the central issue. Do the same features that influence electron density for a nucleophile also operate for a leaving group? Think Ahead Question onsider these relative rates of the following reaction pairs. Does this data support the idea that the factors that control nucleophilicity also operate for leaving groups? methanesulfonate (a) 3 S LiS 3 S + S Faster 3 LiS 3 S + No reaction (very slow) (b) 3 LiS 3 S + Faster 3 l LiS 3 S + l Slower Answer: The role of the leaving group is to depart along with an electron pair. The more effectively the leaving group can accommodate or stabilize the electron pair it gains, the more readily it leaves. Methanesulfonate ion is a better leaving group than hydroxide ion: The electron pair ends up on an oxygen atom. xygen is highly electronegative so this assists the leaving group to accommodate the new electron pair. Each leaving group departs with a negative formal charge. ecall that atoms prefer to have zero formal charge, so this factor slows the leaving group s departure. Both leaving groups are influenced equally by these factors, so they are not the factors that account for the difference in leaving group ability. So what structural difference operates here? When it leaves, methanesulfonate ion gains significant resonance to delocalize the pair of electrons it gains upon departure (Figure 11.7), whereas hydroxide ion does not. As we have seen many times before, resonance is a stabilizing feature. Methanesulfonate can accommodate and disperse the new electron pair through resonance more effectively than hydroxide ion, so methanesulfonate ion is a better leaving group. (This is very much like sulfuric acid, which is very strongly acidic due to the resonance it gains upon deprotonation. eview section 9.xx.) onic Substitution eactions Page 25