Chapter 9

Chapter 9 - CHAPTER 9 S UBS TI TU TION REACTIONS In the...

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Unformatted text preview: CHAPTER 9 S UBS TI TU TION REACTIONS In the last chapter we saw the importance of understanding mechanisms. We said that mechanisms are the keys to understanding everything else. In this chapter, we will see a very special case of this. Students often have difficulty with substitution reactions—specifically, being able to predict whether a reaction is an S312 or an SN]. These are different types of substitution reactions and their mechanisms are very dif— ferent from each other. By focusing on the differences in their mechanisms, we can understand why we get 8N2 in some cases and SN] in other cases. Four factors are used to determine which reaction takes place. These four fac~ tors make perfect sense when we understand the mechanisms. So, it makes sense to start ol‘f with the mechanisms. 9.1 THE MECHANISMS Ninety-five percent of the reactions that we see in organic chemistry occur between a nucleophile and an electrophile. A nucleophilc is a compound that either is nega~ tively charged or has a region of high electron density (like a lone pair or a double bond). An electrophile is a compound that either is positively charged or has a region of low electron density. When a nucleophile and an clectrophile find each other space, they can attract each other (opposite charges). If the conditions are right, we can have a reaction between them. In both 8N2 and SNI reactions, a nucleoli/tile is attacking an electrophile, giv- ing us a substitution reaction. That explains the SN part of the name. But what do the “l” and “2” stand for? To see this. we need to look at the mechanisms. Let’s start with 8N2: H H Flu]: (\G 8N2 F R1 F L NUC R R Nug. 2 2 On the left, we see a nucleophile. It is attacking a compound that has an clectrophilic carbon that is attached to a leaving group (LG). A leaving group is any group that is willing to be kicked off (we will see examples of this very soon). This leaving group is generally electronegative (because it needs to be happy leaving with a negative charge), which is why the carbon is elcctrophilic. The leaving group is withdrawing electron density from the carbon, making it electron poor. 211 21 2 CHAPTER 9 SUBSTITUTION REACTIONS Our 8N2 mechanism has two arrows: one going from a lone pair on the nucle— ophile to form a bond between the nucleophile and carbon, and the other going from the bond between the carbon and the LG to form a lone pair on the LG. Notice that the configuration at the carbon atom gets inverted in this reaction. So the stereo— chemistry of this reaction is inversion of configuration. Why does this happen? It is kind of like an umbrella flipping inside out in a strong wind. It takes a good force to do it, but it is possible to flip the umbrella. The satne is true here. If the nucleophile is good enough, and it‘ all of the other conditions are just right. we can invert the stereocenter (by bringing the nucleophile in on one side, and kicking off the LG on the other side). Now we get to the meaning of “2" in 5N2. Remember from the last chapter that nucleophilicity is a measure of kinetics (how fast something happens). Since this is a nucleophilic substitution reaction, then we care about how fast the reaction is happening. In other words. what is the rate of the reaction? This mechanism has only one step, and in that step, two things need to find each other: the nucleophile and the electrophile. So it makes sense that the rate of the reaction will be dependent on how much electrophile is around and how mttch nucleophile is around. In other words, the rate of the reaction is dependent on the concentrations of two compounds. There- fore, we call the reaction bimolecular and we put a in the name of the reaction. Now let’s look at the mechanism for an SNI reaction: R1 R Fl 1 1 1 R1 25133 S” A —> R3 R2 6) R3 R2 R3 j Nuc Nuc: Race-mic In this reaction, there are two steps. The first step has the LG leaving all by itself, without any help from an attacking nucleophile. This leaves behind a carbocation, which then gets attacked by the nucleophile in step 2. This is the major difference between 8N2 and SKI reactions. In 8N2 reactions, everything happens in one step. In Syl reactions, it happens in two steps, and we are forming a carbocation in the pro— cess. The existence of the carbocation as an intermediate in only the SNl mechanism is the key. By understanding this, we can understand everything else. For example, let’s look at the stereochemistry of the SN 1. We already saw that the 8N2 reaction went through inversion of configuration. But the SN] reaction is very different. Recall that a carbocation is sp2 hybridized, so its geometry is trigonal planar. When the nucleophile attacks, there is no preference as to which side it can attack. and we get both possible configurations in equal amounts. Half of the mole- cules would have one configuration and the other half would have the other config- uration. We learned before that this is called a racemic mixture. Notice that we can explain the stereochemical outcome of this reaction by understanding the nature of the carbocation intermediate that is formed. This also allows us to understand why we have the in SNl. There are two steps in this reaction. The first step is very slow (the LG just leaves on its own to form C+ and LG‘. which is pretty strange when you think about it), and the second 9.1 THE MECHANISMS 2 1 3 step is very fast. Therefore. the rate of the second step is irrelevant. Let‘s use an anal- ogy to understand this. Imagine that you have an hourglass with two openings that the sand had to pass by: First opening Second opening The first opening is much smaller. and the sand can travel through this open- ing only at a certain speed. The size of the second opening doesn‘t really matter. If you made the second opening a little bit wider. it would not help the sand get to the bottom any faster. As long as the top opening is smaller, the rate of the falling sand will depend only on the size of the top opening. The same is true in a two-step reaction. If the first step is slow and the second step is fast, then the speed of the second step is irrelevant. The rate at which you get product will depend only on the rate of the first step (the slow step). So in our SKI reaction, the first step is the slow stop (loss of the LG to form the carbocation) and the second step is fast (nucleophile attacking the carbocation). Just as we saw in the hourglass, the second step of our mechanism will not affect the rate of the reaction. Notice that the nucleophile does not appear in the mechanism until the second step. If we added more nucleophile, it would not affect the rate of the first step. Adding more nucleophile would only speed up the second step. But we already saw that the rate of the second step does not matter for the overall reaction rate. Speeding up the second step will not change anything. So the concentration of nucleophile does not affect the rate of the reaction. Of coursc. it is important that we have a nucleophile present, but how much we have doesn’t matter. So now we can understand the “l” in SNl. The rate of the reaction is dependent only on the concentration of the electrophile. and not that of the nucleophile. Since the rate is dependent on the concentration of only one thing, we call the reaction unimolecular, and we put a l in the name. Of course. this does not mean that you only need the electrophile. You still need the nucleophile for the reaction to happen. You still need two different things (nucleophile and electrophile). The simply means that the rate is not dependent on the concentration of both of them. The rate is dependent on the concentration of only one of them. The mechanisms of the SNl and 8N2 reaction helped us understand the stereo- chemistry of each reaction, and we were also able to see why we call them SKI and 8N2 reactions (based on reaction rates that are justified by looking at the mecha- nisms). So, the mechanisms really do explain a lot. 214 CHAPTER 9 SUBSTITUTION REACTIONS We mentioned before that we need to consider four factors when choosing whether a reaction will go by an SNI or 5N2 mechanism. These four factors are: elec- trophile. nucleophile, leaving group, and solvent. We will go through each factor one at a time. and we will see that the difference between the two mechanisms is the key to understanding each of these four factors. Before we move on, it is very important that you understand the two mechanisms. For practice? try to draw them in the space below without looking back to see them again. Remember, an 3N2 mechanism has one step: the nucleophile attacks the elec— trophile. kicking off the leaving group. An SN] mechanism has two steps: first the leaving groups leaves to form a carbocation, and then the nucleophile attacks that carbocation. Also remember that 3N2 involves inversion of configuration, while SN] involves racemization. Now, try to draw them. 5N2: 5N1: 9.2 FACTOR 1—THE ELECTROPHILE (SUBSTRATE) The electrophile is the compottnd being attacked by the nttcleophile. ln substitution and elimination reactions (which we will see in the next chapter), we generally refer to the electrophile as the substrate. Remember that carbon has four bonds. So. other than the bond to the leaving group, the carbon atom that we are attacking has three other bonds: 6) @%LG @ The question is, how many of these groups are alkyl groups (methyl, ethyl, propyl, etc.)? We represent alkyl groups with the letter “R”. If there is one alkyl group, we call the substrate “primary” (1°). If there are two alkyl groups, we call the substrate “sec- ondary“ [2"). And if there are three alkyl groups, we call the substrate “tertiary” (3"): Ft Fl R Hfi—LG Rj—LG RR9—K; Primary Secondary Tcrt iary 9.2 FACTOR 1—THE ELECTROPHILE [SUBSTRATE] 2 1 5 In an 8N2 reaction, alkyl groups make it very crowded at the electrophilic cen- ter where the nucleophile needs to attack. If there are three alkyl groups. then it is virtually impossible for the nucleophile to get in and attack (this is an argument based on sterics): ). LG Nuc: So. for 3512 reactions. 1" is the best, 2° is OK. and 3” rarely happens. But SN] reactions are totally different. The first step is not attack 01‘ the nucleo— phile. The first step is loss of the leaving group to form the carbocation. Then the nu- cleophile attacks the carbocation. Remember that carbocations are trigonal planar, so it doesn't matter how big the groups are. The groups go out into the plane, so it is easy for the nucleophile to attack. Sterics is not a problem: Nuczj [n SNl reactions, the stability of the carbocation is the paramount issue. Recall that alkyl groups are electron donating. Therefore, 3” is best because the three alkyl groups stabilize the carbocation. l0 is the worst because there is only one alkyl group to stabilize the carbocation. This has nothing to do with sterics; this is an argument of electronics (stability of charge). So we have two opposite trends. for completely different reasons: 8N2 3N1 Rate Rate 1u 2: 30 10 2o 39 These charts show the rate of reaction, If you have a l" substrate, then the reaction will go Via an SN? mechanism, with inversion of configuration. If you have a 3” sub strate. then the reaction will go via an SN] mechanism, with raccmization. What do you do if the substrate is 23‘! You move on to factor 2. 21 6 CHAPTER 9 suesm’unow REACTIONS EXERCISE 9.1 For the following compound, determine whether a nucleophile is more likely to attack it in an 8N2 or an SNl mechanism: ON... Answer The substrate is primary, so we predict an 8N2 reaction. PROBLEMS For each compound below, determine whether a nucleophile is more likely to attack it in an SNZ or an 5N1 mechanism or whether both mechanisms are possible. Br 9.2 0/ 9.3 CIM There is one other way to stabilize a carbocation (other than alkyl groups)— resonance. If a carbocation is resonance stabilized, then it will be easier to form that carbocation: CI) 9 9 M—’ MHM The carbocation above is stabilized by resonance. Therefore, the LG is willing to leave, and we can have an SNl reaction. There are two kinds of systems that you should learn to recognize: an LG in a benzylic position and an LG in an allylic position. Compounds like this will be res— onance stabilized when the LG leaves: @X AX Benzylic Allylic If you see a double bond near the LG and you are not sure if it is a benzylic or allylic system, just draw the carbocation you would get and see if there are any resonance structures. 9.3 FACTOR Z—THE NUCLEOPHILE 2 1 7 EXERCISE 9.6 In the compound below, circle the LGs that are benzylic or allylic: Br \ CI Cl Br Answer \ O 0‘ l Br Br PROBLEMS For each compound below. determine whether the LG leaving would form a resonance-stab”izecl carbocation. If you are not sure, try to draw resonance structures of the carbocation you would get if you pulled off the leaving group. Cl CV B 9.10 A r 9.3 FACTOR 2—THE NUCLEOPHILE # In many cases we can determine from the substrate alone whether we will get SN] or $512. If we have 21 1° substrate. then the reaction will go Via an 8N2 mechanism, with inversion of configuration. If you have a 3“ substrate, then the reaction will go 21 8 CHAPTER 9 sussrnunow REACTIONS via an SN] mechanism. with racemization. What do we do if the substrate is 2°? We move on to factor 2—thc nucleophile. There are three categories of nucleophiles: very strong, moderate, and weak. Let’s go through each of these categories, one at a time, and then we will see how the strength of the nucleophile helps us determine whether the reaction will be SNl or 8N2. Weak nucleophiles are those that do not have a negative charge at all. They have lone pairs, and they use these lone pairs to attack the electrophilic site of the substrate. Examples are 0.0.. .6. N. .. .- / \ / \ H7 \H / H H R H H RH R Each of these compounds has no charge. Students often forget that compounds like this can be nucleophiles. But since there is no charge. they are certainly very weak nucleophiles. Then we have the moderate nucleophiles. These are nucleophiles with a nega- tive charge that is very stable. The halogens (F, Cl, Br, I) are excellent examples. When they have a negative charge, they are fairly stable. Other examples include resonance-stabilized ions. where the charge is spread out over more than one electro- negative atom: O O u @ Q§_O RACE) 0 Finally, there are strong nucleophiles to consider. These are nucleophiles with a negative charge that is not stabilized. The charge is not on a halogen, and there is no resonance that spreads the charge out. Examples include 9 e R O R’N‘R To put everything together, weak nucleophiles have no charge at all, moderate nu- cleophiles have a stabilized negative charge, and strong nucleophiles have an unsta~ bilized negative charge. Now let’s see what effect this has. Once again, we turn to the mechanisms to understand the nucleophile’s effect on the rates of reaction. We have actually already discussed this, when we explained the meaning of the and the “2” in the names SN] and 8N2. Remember that the “2" in 8N2 meant that the rate of the reaction is dependent on the concentrations of the substrate and the nucleophile. If we increased the concentration of the nuclei ophile. we Would speed up the reaction. But in an SNl reaction, the “l” meant that the rate of reaction is dependent only on the concentration of the substrate (remem~ her the hourglass). The concentration of the nucleophile was not relevant in deter— mining the rate of reaction. Based on this, it makes sense to say that the strength of the nucleophile will make a difference 01in for 5N2. If we use a strong nucleophile, then the 3N2 rate will 9.3 FACTOR Z—THE NUCLEOPHILE 2 1 9 be fast. If we use a weak nucleophile. then the rate will be slow. But in an SNl reac— tion, it does not matter at all how strong or weak our nucleophile is. It only matters that a nucleophile is present (once the carbocation forms. it is not very picky—it will take any nucleophile it finds. whether the nucleophile has a negative charge or not). SO the trends are as follows: 3N2 8N1 Rate Rate Strong Moderate Weak Strong Moderate Weak We see from these trends that a strong nucleophile favors 3N2. We often say that a weak nucleophile favors SN 1 . but we really mean that a weak nucleophile disfitvors 8N2. Therefore. if there is a competition between 8N2 and SNI, then SNI will be comparatively faster than an 5512 if we use a weak nucleophile, because the rate of an 3512 will be very slow. Here is the bottom line: a strong nucleophile suggests an 8N2 mechanism, and a weak nucleophile suggests an SN] mechanism. What do we do if the nucleophile is moderate? Move on to factor 3. EXERCISE 9.1 1 Predict whether the nucleophile below will favor 8N2 or 5le 6 NH2 Answer There is a charge. so we know right away that this is not a weak nucleo— phile. Now we need to ask whether the charge is stabilized. We said that “stabilized” meant that the charge either was on a halogen or was resonance stabilized on more than one electronegative atom. In this case. the charge is not on a halogen, and the charge is not resonance stabilized. Therefore, this is a strong nucleophile. Strong nu- cleophiles favor 8N2 reactions. PROBLEMS For each nucleophile below. predict whether the nucleophile will favor SN] or 5N2. Identify cases where you cannot tell (and you would need to move on to factor 3). O u @ Hac—s—o ii 9.1 2 0 Answer: 9.1 3 Answer: 220 CHAPTERS suasrnunow REACTIONS e 0' e 9. 1 4 Answer: __ 9. 1 5 N Answer: _— /\ /H e 9.16 '.S.‘ Answer: __ 9.1 7 Br Answer: 9.4 FACTOR 3—THE LEAVING GROUP In many cases we can determine from the substrate and the nucleophile whether we will get SN] or 8N2. If we have a 1° substrate and a strong nucleophile, then the re— action will go via an 5N2 mechanism, with inversion of configuration. If we have a 3° substrate and a weak nucleophile, then the reaction will go via an SN] mechanism, with racemization. What do we do if the substrate is 2‘3 and the nucleophile is mod— erate? We move on to factor 3—the leaving group. There are three categories of leaving group: excellent. good, and bad. These three categories are exactly the same as the three categories of nucleophile that we just learned above. This makes sense if we think about the reactions as movies that can be played in reverse. [f you play an 8N2 in reverse, the roles of nucleophile and leaving group are reversed. The same thing happens when you play an SN] in re- verse. Go back to the two mechanisms and try to play both reactions forward and backward in your mind. You should be able to see that for Sci and for 8N2, the for— ward and reverse reactions simply exchange the roles of the nucleophile and the leaving group. Let’s go through all three categories once more. Excellent leaving groups are those that do not have a negative charge at all. Examples are :0: :0: .n. /N. ,N. H H R H HH H RH H Fifi H Then we have the good leaving groups. These are leaving groups with a negative charge that is very stable. The halogens (F. Cl, Br, I) are examples. When they have a negative charge, they are fairly stable. Other examples include resonance stabilized ions, where the charge is spread out over more than one electronegative atom: o o n @ Gigi—O RJkOe O 9.4 FACTOR 3—THE LEAVING GROUP 221 Finally. there are bad leaving groups. These are nucleophiles with a negative charge that is not stabilized. The charge is not on a halogen. and there is no resonance that spreads the charge out. Examples include 9 G _ ’N\ R 0 Fl H To summarize, excellent leaving groups have no charge at all. good leaving groups have a stabilized negative charge, and bad leaving groups have an unstabilized neg- ative charge. Now let‘s see what effect the leaving group has on the rates of SN! and 8N2. Once again. we turn to the mechanisms to understand the leaving group’s effect on the rates of reaction. The 5512 reaction certainly needs at least a good leaving group to work. but the SNl reaction is much more sensitive to the stability of the leaving group. The first and only step of an 8512 reaction is the nucleophile coming in and displacing the leaving group. As long as the leaving group is more stable than the nucleophile, then the reaction can go. But in an SNl reaction, the first (and rate-determining) step is loss of the leaving group. The more stable the leaving group is, the faster the reac— tion will happen. So, for an SNl reaction, the leaving group should really be excel- lent (although it can still happen for good leaving groups). Let‘s compare the trends: 5N2 3N1 Rate Rate Excellent Good Bad Excellent Good Bad Both reactions are faster when the leaving group is better. The trends are the same. but SN] is more sensitive to this factor. At this point. it is worth noting the differences in the trends for 8N2 and SNl when it comes to each of the factors we have seen. For the substrate (factor 1), we saw opposite trends for SN] and 3N2. For the nucleophile (factor 2), we saw that SNl did not have a trend but 5N2 did. In that case we could either favor 8N2 or disfavor 5N2. For the leaving group (factor 3), the trends for 5N2 and SNI are similar, but the SN] trend is stronger. If this sounds like jibberish to you. perhaps you should look at the charts showing these trends on the previous few pages. Compare each of the fac— tors to each other. Bottom line for factor 3: if you want to favor an SNl mechanism over an 8512, you should really use an excellent leaving group. If you use a good leaving group, it 222 CHAPTERS SUBSTITUTION REACTIONS is hard to say from this factor alone which mechanism would predominate. If you use a bad leaving group. neither reaction will occur at all. EXERCISE 9.18 Consider the compound shown below, and predict whether the leaving group will favor 8N2 or SNl: Answer Whenever you are assessing a leaving group or a nucleophile to deter— mine how stable it is, you need to look at what it looks like when it is not attached to the molecule. ln other words. how stable is it when it is by itself. Let‘s draw what we get if the leaving group leaves: H to We get H20, which is neutral. It does not have a negative charge, so this is an ex— cellent leaving group. Fora leaving group to be neutral when it leaves, it must have a positive charge while it is still attached to the molecule. But remember that you need to look at it alier it leaves. An excellent leaving group favors both 5N2 and SNI, but we said that S511 re— actions are more sensitive to this factor. So, an excellent leaving group suggests an SNl reaction. [t is possible. to convert a bad leaving group into an excellent leaving group. For example, if we protonate an OH group, we convert it into a good leaving group: AXC Bad leaving group Excellent leaving group PROBLEMS For each compound below, determine whether the leaving group is excellent, good, or bad. 0\ p H I 6’8 €09 9.19 Answer: 9.20 R Answer: 9.5 FACTOR 4—THE SOLVENT 223 S\ N H2 i/n/ 9.21 Answer: 9.22 Answer: 9.23 Answer: 9.24 W Answer: 9.25 Using your answers to the last six problems. which of the six compounds above would you expect to be most likely to undergo an SN] reaction? 9.26 If you wanted to get the compound in Problem 9.24 to undergo an SN] re‘ action with chloride as the nucleophile. what would you need to do to the leaving group. What reagent would you use (to change the leaving group and to provide Cl‘ at the same time)? 9.5 FACTOR 4—THE SOLVENT So far, we have looked at the substrate. the nucleophile, and the leaving group. This takes care of all of the parts of the compounds that are reacting with each other. Let’s summarize substitution reactions in a way that allows us to see this: Not: Nuc Substrate ———> Substrate \ LG LG So, by talking about the substrate, the nucleophile, and the leaving group, we have covered almost everything. Btlt there is one more thing to take into account. What solvent are these compounds dissolved in? It can make a difference. Let’s see how. There is a really strong solvent effect that greatly affects the competition be— tween SNl and 8N2, and here it is: polar aprotic solvents favor 8N2 reactions. So, what are polar aprotic solvents. and why do they favor 8N2 reactions? Let’s break it down into two parts: polar and apmric. Hopefully, you remem- ber from general chemistry what the term “polar” means, and you should also re- member that “like dissolves like” (so polar solvents dissolve polar compounds, and nonpolar solvents dissolve nonpolar compounds). Therefore, we really need a polar solvent to run substitution reaCtions. SN] desperately needs the polar solvent to sta- bilize the carbocation, and 8N2 needs a polar solvent to dissolve the nucleophile. Syl certainly needs the polar solvent more than 5N2 does, but you will rarely see a sub— stitution reaction in a nonpolar solvent. So, let’s focus on the term aprotic. Let‘s begin by defining a protic solvent. We will need to jog our memories about acid—base chemistry. Recall that in Chapter 3 we talked about the acidities of protons (these are hydrogen atoms without the electrons. symbolized by H+), and we 224 CHAPTERS SUBSTITUTION REACTIONS saw that protons can be pulled off of a compound if the compound can stabilize the negative charge that develops when H+ goes away. A protic solvent is a solvent that has a proton connected to an electronegative atom (for example. H20 or EtOH). It is called protic because the solvent can serve as a source of protons. In other words, the solvent can give a proton because the solvent can stabilize a negative charge (at least a little bit). So what is an aprotic solvent? Aprotic means that the solvent does not have a proton on an electronegative atom. The solvent can still have hydrogen atoms. but none of them are connected to electronegative atoms. The most common examples of polar aprotic solvents are acetone, DMSO. DME, and DMF: O /U\ Acetone 0 g Dimethylsulfgxide (DMSO) / \ /0\/\0/ Dimethoxyethane(DME) \ )L Dimethylformamide (DMF) There are, of course, other polar aprotic solvents. You should look through your textbook and your class notes to determine if there are any other polar aprotic solvents that you will be expected to know. If there are any more, you can add them to the drawing above. You should learn to recognize these solvents when you see them. So why do these solvents speed up the rate of 8N2 reactions? To answer this ques— tion, we need to talk about a solvent effect that is usually present when we dissolve a nucleophile in a solvent. A nucleophile with a negative charge, when dissolved in a polar solvent, will get sunounded by solvent molecules in what is called a solvent shell: Solvent Shell we \ 9.5 FACTOR 4—THE SOLVENT 225 This solvent shell is in the way, holding back the nucleopltile from doing what is supposed to do (go attack something). For the nucleophile to do its job, the nucle— ophile must first shed this solvent shell. This is always the case when you dissolve a nucleophile in a polar solvent, except when you use a polar aprotic solvent. Polar aprotic solvents are not very good at forming solvent shells around neg- ative charges. So if you dissolve a nucleophile in a polar aprotic solvent. the nucle— ophile is said to be a “naked” nucleophile, because it does not have a solvent shell. Therefore, it does not need to first shed a solvent shell before it can react with some— thing. It never had a solvent shell to begin with. This effect is drastic. As you can imagine, a nucleophile with a solvent shell is going to spend most of its existence with the solvent shell, and there will be only brief moments every now and then when it is free to react. By allowing the nucleophile to react all of the time, we are greatly speeding up the reaction. 8N2 reactions performed with nucleophiles in polar aprotic solvents occur about 1000 times faster than those in regular protic solvents. Bottom line: Whenever a solvent is indicated, you should look to see if it is one of the polar aprotic solvents listed above. if it is. it is a safe bet that the reaction is going to be 8N2. EXERCISE 9.27 Predict whether the reaction below will occur via an 8N2 or an SN] mechanism: Br Ole DMSO Answer We look at the substrate and we see that it is secondary. That doesn‘t tell us much, so we move on to the nucleophile and we see that it is a moderate nucle» ophile (not strong and not weak); that doesn’t tell us much either. We then look at the LG and see that it is a good LG (not excellent and not bad). which that doesn’t tell us anything either. Finally, we look at the solvent. This is a polar aprotie solvent, and that serves as the tie breaker. 5N2 wins. PROBLEM 9.28 Go back to the list of polar aprotic solvents, study the list, and then try to copy the list here without looking back. 226 CHAPTERS SUBSTITUTION REACTIONS 9.6 USING ALL FOUR FACTORS Now that we have seen all four factors individually. we need to see how to put them all together. When analyzing a reaction, we need to look at all four factors and make a determination of which mechanism, $311 or 8N2, is predominating. It may not be just one mechanism in every problem. Sometimes both mechanisms occur and it is difficult to predict which one predominates. Nevertheless, it is a lot more common to see situations that are obviously leaning toward one mechanism over the other. For example, it is clear that a reaction will be 5N2 if we have a primary substrate with a strong nucleophile in a polar aprotic solvent. On the fiipside, a reaction will clearly be SN] if we have a tertiary substrate with a weak nucleophile and an excellent leav— ing group. Your job is to look at all of the factors and make an informed decision. Let’s put everything we saw into one chart. Review the chart. If there are any parts that do not make sense, you should return to the section on that factor and review the concepts. lC—Only 3N2, Strong—5542 Bad—Neither Polar uprotic—SNQ N0 Moderate—Both Good—Both (but more 3N2) 33—0111)r SKI Excellent—SKI N0 5N2 EXERCISE 9.28 For the reaction below. look at all of the reagents and conditions, and determine if the reaction will proceed via an 3N2 or an SNI, or both or neither. Cl OHe Answer The substrate is primary, which immediately tells us that it needs to be 8N2. On top of that, we see that we have a strong nucleophile. which also favors $32. The LG is good, which doesn’t tell us much. The solvent is not indicated. So. taking everything into account, we predict that the reaction follows an 8N2 mechanism. PROBLEMS For each reaction below, look at all of the reagents and conditions. and determine if the reaction will proceed via an 5N2 or an SN 1, or both or neither. 9.7 SUBSTITUTION REACTIONS TEACH US SOME IMPORTANT LESSONS 227 Cl 0H6 9.29 DME CI H20 9.30 AX: H_Br 9.31 H20 9.32 OR CI H20 9.33 A/ ROH so 0\ ,/O S 9.34 0” \CH3 9.7 SUBSTITUTION REACTIONS TEACH US SOME IMPORTANT LESSONS SNI and 8N2 reactions produce almost the same products. In both reactions, 21 leaving group is replaced by a nucleophile. In fact. the only difference in products between SN] and 5N2 reactions arises when you have a stereocenter where the leaving group is. In this situation. the 8N2 mechanism will invert the stereocenter. while the SNl mechanism will produce a racemic mixture. That's the only difference—the configu— ration of one stereocenter. And if there is no stereocenter. then there is no difference in outcome at all between SKI or 8N2. It seems like a lot ol~ work to go through to de— termine the configuration of one stereocenter (which matters only some of the time). So the obvious question is. why did we go through all of that trouble to learn how to determine whether a reaction is SN] or 8N2? There are many answers to this CHAPTER9 SUBSTITUTION REACTIONS question, and it is important to spend some time on this, because it will help frame the rest of the course for you. Let’s go through some answers one at a time. First we learned the important concept that everything is located in the mech- anisms. By understanding the mechanisms completely, everything else can be justi- fied based on the mechanisms, All of the factors that influence the reaction can be understood by carefully examining the mechanism. This is true for every reaction you will see from now on. Now you have had some practice thinking this way. Next we learned that there are multiple factors at play when analyzing a reac- tion. Sometimes the factors can all be pointing in the same direction, while at other times the factors can be in conflict. When they are in conflict, we need to weigh them and decide which factors win out in determining the path of the reaction. This con— cept of competing factors is a theme in organic chemistry. The experience of going through SNl and $512 mechanisms has prepared you for thinking this way for all re- actions from now on. Finally we learned that if we analyze the first factor (substrate), we will find two effects at play: electronics and sterics. We saw that $512 reactions require pri- mary or secondary substrates because of sterics—it is too crowded for the nucleo— phile to attack a tertiary substrate. On the other hand, SN] reactions did not have a problem with sterics, but electronics was a bigger issue. Tertiary was the best. be- cause the alkyl groups were needed to Stabilize the carbocation. These two effects (sterics and electronics) are major themes in organic chem- istry. Much of what you learn in the rest of the course can be explained with either an electronic or a sterics argument. The sooner you learn to consider these two ef- fects in every problem you encounter. the better off you will be. Electronics is usu— ally the more complicated effect. In fact, the other three factors that we saw (nucleophile, leaving group, and solvent effects) were all electronic arguments. Once you get the hang of the kinds of electronic arguments that are generally made, you will begin to see common threads in all of the reactions that you will encounter in this course. Don’t get me wrong—it is very important to be able to predict whether a stereocenter gets inverted or not when a substitution reaction takes place. That alone would have been enough of a reason to learn all of the factors in this chapter. But I also want you to keep your eye on some of the “bigger picture" issues. They will help you as you move through the course. ...
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Chapter 9 - CHAPTER 9 S UBS TI TU TION REACTIONS In the...

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