Chapter 1_ THE SKILLS YOU NEED 1 1 The Goal of This book 1 1.2 Mechanisms Are Your Keys to Success 2 Chapter 2_ INTRO TO IONIC MECHANISMS 4 2.1 Curved Arrows 4 2.2 The Basic Moves 8 23 Combining the Basic Moves 18 Chapter 3 ELECTROPHILIC AROMATIC SUBSTITUTION 22 3.1 Halogenation and the role of Lewis Acids 23 3.2 Nivration 28 3.3 Friedel-Crafts Alkylation and Acylation 31 34 Sulfonation 38. 3.5 Modifying the Nucleophilicity of the Nucleophile 43 3.6 Predicting Directing Effects 46 3.7 Identifying Activators and Deactivators 59 3.8 Predicting and Exploiting Steric Effects 68 3.9 Synthesis Strategies 76 Chapter 4_NUCLEOPHILIC AROMATIC SUBSTITUTION 83 4.1 Criteria for Nucleophilic Aromstic Substitution 83 4.2 SwAr Mechanism 86 4.3 Elimination-Addition 90 4.4 Mechanism Strategies 96 Chapter §_ KETONES AND ALDEHYDES 98 5.1 Preparation of Ketones and Aldehydes 5.2 Stability and Reactivity of the Carbonyl 102 5.3 H-Nucleophiles 104 5.4 O-Nucleophiles 109 5.5 S-Nucleophiles 122iv. Tavis oF conrente: 5.6 N-Nucleophiles 131 5.7 C-Nucleophiles 139 5.8 Some Important Exceptions to the Rule 148 5.9 How to Approach Synthesis Problems 182 Chapter CARBOXYLIC ACID DERIVATIVES 189 6.1 General Rules 160 6.2 Acyl Halides 167 6.3 Anbydrides 178 64 Esters 179 6.5 Amides and Nitriles 188 6.6 Synthesis Problems 194 Chepter 7 ENOLS AND ENOLATES 208 Shepter 7 ENOLS AND ENOLATES 208 7.1 Keto-Enol Tautomerism 208 7.2 Reactions Involving Enols 210 773 Making Enolates 213 7.4 Haloform Reactions 216 7.5 Alkylation of Enolates 219 7.6 Aldol Reaction and Aldol Condensation 223 7.7 Claisen Condensation 230 7.8 Decarboxylation Provides Some Useful Synthetic Techniques 238 7.9 Michael Reactions 245 Chapter 8_AMINES 21 Chapter 8 AMINES 26300 8.1 Nucleophilicity and Basicity of Amines and Amides 253, 8.2 Preparation of Amines through Sx2 Reactions 28: 33 Preparation of Amines through Reductive Amination 259 8.4 Preparation of Amines from Amides 263 8.5 Acylation of Amines 267 8.6 Reactions of Amines with Nitrous Acid 271 8.7 Aromatic Diazonium Salis. 274 Answer Key 277 Index 3111.1_THE GOAL OF THIS BOOK ‘There ae a lot of reactions that you will learn this year. Perhaps you could try to memorize them. ‘Some people are good at tat. But rather than memorizing, you would actually be beter off if you tried to focus on building your skills. You will need certain skills in order to do wel in this course In this book, we will focus on those skills. ‘Specifically, you will learn the skills that you need to do three very imporant types of problems: 1. Proposing a mechanism 2. Predicting products 3. Proposing a synthesis 'As you progress through the course, you will soon realize that it is not enough just to learn the reactions. To do well in this course, you MUST learn how to approach and solve these three types ‘of problems. You must become a master of very specific skis. These skills will uide you in solv~ ing problems. These thee types of problems represent the core of an organic chemistry course. If ‘you master the skills you need. you will do very well. Each chapter in this book will focus on the skills that you need in order to master a particlar topic. The chapters inthis book are designed to map out fasly well onto the chapters in your text book. For instance, when you are leaning about carboxylic acid derivatives, there will Bea chap- ter in this book with the same ttle. ‘We will not have enough space to cover every topic in your textbook. This supplement is not designed to replace your textbook or your instrucior. Rather, itis meant to provide you with the core skills that wil allow you to study more efficiently. Although we will focus on three major types of problems, we must place the major emphasis on ‘mechanisms. Mechanisms are your keys to succes in organic chemistry. If you master the mechanisms, ‘you will do very well inthe elas; if you don't master them, you will do poorly. It is hard to talk about synthesis problems if you don’t know the reactions well enough (and the same is true for predicting roduts). That i why Chapter 2 is devoted to laying the foundation you need to master mechanisms. ‘That chapter is important. So, even though it won't corespond to a specific chapter in your textbook, make sure to go through Chapter 2 anyway. ‘In Chapter 2, we will se that mechanisms follow a small number of basic themes and ideas. Ry focusing on these basic themes, you will se the common threads between mechanisms that would oth- cexwise appear tobe very differet. This approach will minimize the need for memonzatin. tn fact, we ‘ill soon argue that students who focus on memorization will miss problems that se trivial when you ‘understand the basic concepts. This book will provide you with the fundamental language and tools that you need in order to master mechanisms. And while we are at it, we will work on the kills you reed to solve synthesis problems and predicting products as well2 CHAPTER 1 THE SKLLS YOU NEED 1.2 MECHANISMS ARE YOUR KEYS TO SUCCESS ‘What are mechanisms, and why are they so important? ‘To understand the important roe that mechanisms play, let's consider an analogy. I recall the time that I had to teach my children how to put on their shoes. It is amazing bow many steps are involved when tying your sheelaces. Next time you tie your shoelaces, think about how many in- dividual steps your hands are doing, and think about how you would explain each of these steps 10 someone who never learned how to tie their own shoes. It is a difficult thin to teach. I am em- barrassed to say that my first response was to take the eas) road—I bought them shoes with Vel- ro straps. I had temporarily solved my problem because the new shoes required so few steps 10 put on. ‘The end resalt (a child wearing shoes) looks the same regardless of the type of shoe used. But the process of getting the shoes on is very much dependent on the type of shoe. In particular, the umber of steps involved to compete the process is vastly different from one type of shoe to the ‘other. The same is true with reactions. All reactions involve steps to get from starting materials to products. Some reactions proceed through a lot of steps, while others go through just a few steps. [A detailed list of steps that a reacton follows is called a mechanism. ‘When two compounds react with each other to form new and different products, we try to un- derstand how the reaction occurred—what are the individual steps in the process? Every step in- volves the flow of electron density; electrons more to break bonds or to form new bonds. Mech nisms illustrate how the electrons move during each step of a reaction. The flow of electrons is shown with curved arrows, for example: det yg Ey ‘The curved arrows in each of the tree steps above show us how the reaction took place. You should think of a mechanism as the “book-keeping of electrons.” Just like an accountant will do the book-keeping of a company’s cash flow (money coming in and money going oUt), 80, t00, a re- ‘action mechanism is the book-keeping of the elecron-flow in each step of a reaction. In my previous book (Orranic Chemistry as a Second Language: Translating the Basic Con- cepts), we saw that bond-line drawings (the way we draw molecules in organic chemistry—see the reaction above) are the “hieroglyphics” of organic chemistry. We saw that these drawings focus on. the electrons (the atoms themselves are generally not drawn, but are implied) Each line inthe draw= ing represents 2 bond, which shows you where the electrons are. Look atthe mechanism above, and. in your mind, think of it asa sentence. The drawings of the molecules (the bondb-ine drawings) rep- resent the nours of the sentence. The curved arrows are the verbs of the sentence. So a mechanism {s essentially a sentence. Iti truly a language, and you need to leam how to combine the nouns and verbs, ustas you do in any other language. In my previous book, we focused on the nouns of the sentences—we focused on molecules. In this book, we focus or the verbs. Imagine leaming French without leaming any verbs. You wouldn't be able to get out a single sentence!!! ‘Sa if vom want tn master mechanicme then vost rely need th master curved arrows. which are 41.2 MECHANISMS ARE YOUR KEYS TO success 3 iknow that, because of gravity, water flows to the lowest spot. That is true on a mountain, on 3 ‘bumpy road, on a roof, and so on. In each situation, you could probably predict how water would flow on the terrain by simply inspecting it and looking for grooves that lead to the lowest point. Simple, right? Well, imagine if you had a friend who did not understand that water always flows to the lowest spot. Your friend would be absolutely amazed at your ability to predict how the wa- ter would flow in every (seemingly different) situation. After all, a roof is certainly a different uation from a bampy roed. But in fact the situations are not all that different when you understand the one simple concept that water flows to the lowest point. ‘Similarly, the flow of electrons can be predicted by understanding just a few simple concepts. If you really want to make your life miserable, then you could memorize every single mechanism {nthe course. But that would be silly—that would be like your friend (who does not know how wa- ter flows) trying to memorize every different situation (the roof, the mountain, the road, ete.). Your friend might have a good memory, but if he does not understand how water flows, he will be com- pletely stumped when he sees a new terrain that he has not memorized. Without knowing the one guiding principle, he would be lost. But when you know the guiding principle it is trivial to make the prediction of how the water will flow on ANY new terrain. ‘When you understand a mechanism, you will understand why the reaction took place, why the stereocenters turned out the way they did, and the like. If you do not understand the mechanism, then you will find yourself memorizing the exact details of every single reaction. Unless you have 1a photographic memory, that will be a very difficult challenge, and as we have just seen, you will not be able to extrapolate to situations that you have never seen. By understanding mechanisms, ‘you will be able to make more sense of the course content, you will be able to better organize all of the reactions in your mind, and you will be able to propose mechanisms in new situations. ‘At this point, you might be wondering why everyone always says that organic chemistry is all about memorization. Well, the truth is, they are all WRONG. Before you can master organic chem istry, you must let go of the myth that so many former students have engrained in your psyche. ‘There is actually very little memorization in organic chemistry. And if you try to replace true un- derstanding with rote memorization, you will not do as well. Organic chemistry is all about taking the principles you learn and applying them to new situa- tions. This is easy to do if you understand the rules. It is VERY hard to do if you try to memorize ‘200 mechanisms. So, don’t memorize mechanisms. Instead, focus on understanding them with an ‘emphasis on the guiding principles of how electrons flow. That way, you will be able to predict re- ‘actions that you have never seen. And when you can do that, you will feel really good about or- ‘ganic chemistry. It might seem like a lofty goal right now, but be patient; this book will guide you through the process, step by step. Every year, my students ask me how many mechanisms they need to know. I always tell them that it depends how you count the mechanisms. If your strategy is memorization, then you will need to know about 200 mechanisms. However, if you focus on just a few simple principles and rules, ‘you will see that there are only about a dozen unique mechanisms. In fact, those dozen mechanisms are just different combinations of four basic moves. So, we will begin our step-by-step process by going over the basic moves. We will learn them ‘and practice them, Then we will explore the various combinations of these basic moves, once again in a step-by-step fashion. In the end, you will see that proposing a mechanism is just as simple as predicting how water will flow down your roof.fa z TONIC MECHANISMS Tonic reactions (those reactions that involve either full charges or partial charges) represent most (05%) of the mechanisms you will see this semester. The other two major categories, radical mechanisms and pericyclic mechanisms, occupy a much smaller focus in the typical uader- ‘graduate organic chemistry course. Accordingly, we will devote most of our attention to ionic mechanisms. Tn this chapter, we will learn about the basic steps involved in all ionic mechanisms. There are only four basic moves. Let's begin with a quick review (from last semester) of curved arrows, 2.1 CURVED ARROWS —— Before you can understand the four basic moves, you must first become a master of drawing curved arrows. These ace the tools of drawing mechanisms. Inthe previous book, Organic Chemistry as a Second Language: Translating the Basie Concepts, there was an entire chapter devoted to mecha nisms. The frst nine pages ofthat chapter were devoted to building skills for drawing proper curved rows. If you have a copy of that book, I highly recommend that you review those nine puges be- fore continuing in this book. If you do not have a copy of that book, you should lock in your text- ‘book to see if there is any introduction to carved arrows and mechanisms. ‘Even for those students who feel comfortable with curved arrows, a short and quick review can- not hurt. To quickly summarize, every curved arrow has a head and a tail Itis essential that the head and tail of every arrow be drawn in precisely the proper place. The tail shows where the elee- trons are coming from, and the head shows where the electrons are going: Tail, ~~ _-\Head Some students confuse the meaning of an arrow. They think the arow shows where atoms are mov- ing. Bat ths is wrong. Curved arrows actually show the moton of electrons. As an example, con- sider s simple acid-base reaction: 9 AH ‘We see that one curved arrow comes from the base (OH™), showing the electrons ofthe base grab- bing the proton. The next curved arrow shows what happens to the electrons that were originally holding orto the proton. In the end, 2 proton has been transferred from one place to another. So, 421 CURVED ARROWS 5 ‘ve call this a proton transfer. But don't Jet the name fool you into thinking that the mechanism is like this: This is wrong because « mechanism does NOT show how a proton moves. Rather, the mechanism shows how the electrons move. So, for every curved arrow that you draw, you must always make sure that itis going in the right direction. Otherwise, your arrow (and therefore your mechanism) will be wrong ‘There is one more thing to clarify before we can move on. Notice that the tail of the curved ar~ row (coming from the oxygen) is placed on a lone pair: So Aw &. i, “This makes sense because the tail of the artow represents where the electrons are coming from. And electrons can either come from lone pairs or from bonds (we will see many examples very soon). In the example above, lone pair is grabbing the proton. However, itis very common for organic chemists to draw compounds without drawing the lone pairs because itis faster that way. For example: 9 On OH — is fastertodrawthan 0H and here is another, more striking, example: % og” laterto raw inan Lone pairs do not have fo be drawn because they are implied (we saw that inthe first semester). In other words, you are supposed to know that the lone pairs are really there, despite their absence from our drawing. As 4 result, you will often see something like this: [Notice thatthe tail ofthe curved arrow is placed oa the negative charge. This is the way we must ‘draw the curved arrow in situations where we omit the lone pairs from our drawings. But don't be ‘confused; it really isn’t the negative charge that is grabbing the proton. Rather, it i alone pair that 1s grabbing the proton.6 CHAPTER 2 INTRO TO IONIC MECHANISMS There are two reasons why chemists will often omit the lone pairs when drawing mechanisms. First of all, itis faster to draw mechanisms if we omit most of the Tone pairs. But more importantly, itis a lot easier to follow a mechanism when it is tot cluttered with lone pairs, Compare for yourself Som Clearly, the second way of drawing itis less cluttered and easier to follow. ‘Your textbook will most likely éraw all lone pairs (at least the ones participating in a reaction), but your instructor might draw some mechanisms without lone pairs during the lecture (to save time). Both ‘ways are correct. For purposes of clarity and simplicity, most drawings in this book will leave out the Jone pairs, will only show the lone pairs when their presence does not compromise the clarity of the presentation ‘Since we will be leaving out the lone pairs very often, itis important that you get accustomed to “seeing” the lone pairs even though they are not drawn. Oxygen is the most common element that ‘you will see in the course (that possesses lone pairs), so let's start there. Each oxygen atom with no ‘charge will have two lone pairs: By isthesameas ‘An oxygen atom with one bond and a negative charge will always have three lone pairs: ° 2 Be ‘sthesameas °)" ‘And an oxygen atom with thice bonds and a positive charge will always have one lone pair: ae or aA isthe same as A and NgH Be ne ‘Yeu will sce many other elements in this course (nitrogen, sulfur, phosphorus, etc), but oxygen is the most common of them all. So, you would do yourself a big favor if you would remember the rules above.21. CURVED ARROWS 7 One other very common example is a carbon atom with « negative charge: SL wrennem s ° Itis important tbat you can see the lone pairs even when they are not drawn, so let's get some quick practice to make sure that you can see the lone pais: EXERCISE 2.1. In te following intermediate, the lone pairs are not drawn. Draw them. ye me Answer: Oxygen atoms with no charge will have two lone pairs: te, HO3 0°R ‘And oxygen atoms with three bonds and a positive charge, will always have one lone pair: a HO: OF For each of the following, draw in the lone pairs that are not shown: 8 22, MPP ag BPR oo L a Hage 9° o 2s. C3 20. ar 27. Ar ‘When drawing curved arrows, you must focus on two things: 1. The ‘ail of every curved arrow needs to be in the right place, and 2. The head of every curved arrow needs to be in the right place.% cHapTen2 INTRO TO IONIC MECHANIEMS ‘Now let's make sure that you are comfortable with curved arrows. For each of the reactions below, try 10 draw the curved arrows without flipping to the answers in the back of the book. When you are finished, look up the answers in the back of the book to make sure that you drew the arrows correctly. ‘When looking at the answers, make sure to focus on the position of every head and every tail of the arrows in your answers. If any heads or tails were not drawn in exactly the correct place, then you should go back and review the first nine pages of the mechanisms chapter from the previous book. PROBLEMS. For cach transformation below, complete the mechanism by drawing the proper arrows (most of these reactions are from the first semester of organic chemistry): 6 cr 2.11. 8 —— yo + Br Br oP cl 2a, SO — A + Bre DS + B® 1 Se am CO" SE Ot rms H 2.15. 2.2 THE BASIC MOVES To truly become the master of arrow pushing, you must master the “basic moves” that are at your disposal when proposing a mechanism. These basic moves are your tools for proposing mechanisms. ‘That's what this chapter is all about: mastering the most basic moves of any ioaie mechanism. Tt is comforting to realize that there are only four basic moves. (That will cover you for any ionic mech- anism that you will se in this course.) Let’s go through them, one by one: di ile. For example: A ap A Remember that a nucleophile is a compound with a region of high electron density, In some cases, the nucleophile will actually have a negative charge (lke the bromide ion in the example above). In‘22 THE Basie Moves 9 many other cases, the nucleophile will not have a charge—you must keep in mind that lone pairs and + bonds can also be nucleoptilic: Here is one example of a Jone pair acting as a nucleophile: ‘The nucleophile here isthe lone pair n the oxygen of the alcohol above (ROH). This oxygen atom 's using one of its lone pairs to attack the C=O bond. Notice tht there is no negative charge on the atacking oxygen atom. ROH does not bear a negative charge. That’s OK, because a nucleophile is jus a region in space of high-electron density. And the lone pairs ofthe oxygen atom are regions in space of high-elecuron density ‘Also notice that there is more than one curved arrow on the step above. The frst curved arrow shows the nucleophile (ROH) attacking the elecrophile. The second cuved arrow (pushing the elec- tron density up onto the oxygen) is part of the same basic move—nucleophile attacking an elec trope. To see how this second arrow is part of the same move, consider the following: what if 1 were to draw a resonance structure of the starting compound BEFORE it gets atacked ‘9 Aa oe Now, I can draw the atiack taking place at the second resonance structure, like this: R-O-H 2 S, — ihe, | a 4 ‘When we think of it like this, we realize that only one of the arrows is actually showing the attack:10 cHarreR? INTRO TO IONIC MECHANISMS ‘The second arrow can be thought of as resonance, or alternatively, you can think of the second arrow as an actual flow of electron density that goes up onto the oxygen when the nucleophile attacks: oO Etectron density | Ag, is pushed up C ‘onto the oxygen ‘There is a subtle difference between these two ways of looking at the second arrow (either think- ing of it as resonance or thinking of it as an actual flow of electron density). It is probably more ac- curate to think of it as an actual flow of electron density up onto the oxygen because that is the way most organic chemisis think about it. However you wish to view this, you should realize that both curved arrows are used to show only one basic move: a nucleophile is attacking an electrophile. In fact, it is common to see even more than two arrows when a nucleophile attacks an elec- trophile. For example, consider the following: 8.0, Su cl Al of these arrows show one thing, happening: a nucleophile (OH) is attacking an electrophile. ‘The first curved arrow (coming from OH) shows this attack. The other curved arrows can be viewed as resonance arrows before the attack: (Or they can be viewed as showing the flow of electron density that takes place when the nucleo- ance %@-0, Electron density 4s pushed up ‘onto the oxygen te OH22THERasiC Moves 11 ‘To be consistent with the way organic chemists speak, you will probably be better off if you think of it as a flow of electron density (rather than just one attacking arrow and the remaiaing arrows being considered as resonance arrows). I only gave the resonance argument so that we can justify in our minds that all of these arrows are really just showing one basic thing happening. In all of the examples we have seen, the nucleophile had a lone pair that was responsible for at- tacking the electrophile, But we said that a m bond can also serve as a nuclecphile. For example: @ 8 o*%o 0=8=0 O Oo Once again, there is more than one curved arrow here. There are two cucved arrows. The frst ar- row (coming from the ring to attack SO,) is the arrow showing the attack. The second arrow can be thought ofin two ways. You can think of it asa resonance arrow: @ ° g o*S0 oto O Or you can think of it as the flow of electron density when the nucleophile attacks: Q Electron density Cg is pushed up conto the oxygen Either way, you should realize that this i just one basic move: a nucleophile attscking an electrophile. ‘As we move through this course, we will see many more examples of nucleo-phils. For now, let's just appreciate the basic feature of a nucleophile—it isa region of high-clectron density (either a lone pair or a bond). And nucleophiles always do what they ae supposed to do—they attack slectophiles. ‘An electrophileis a compound with a region of low-electron density, In some cases, the eletrophile will actually have a postive charge (as in the example we saw earlier where the bromide ion is at- tacking a compound wit a postive charge). In many other cases the clectrophile will not have a charge. For example, a ketone is an excellent electrophile, which we can understand when we draw the reso- nance strictures12 CHAPTER2 INTRO TOIONIC MECHANISMS We can see from the second resonance structure that the carbon of the C—O double bond is electron-poor (a site of Jow-electron density). When you normally draw a Ketone, you don’t see a positive charge on the carbon atom: 9° AA But even though there is no net charge, there is still an electrophilic center (that is waiting to be at- tacked by a nucleophile). Later in this semester, we vill see « lot of reactions that deal with the chemistry of compounds containing C—=0 bonds. Mos of this chemistry revolves around the elec- trophilicity of the C=O group. ‘As we move through the course, we will see more examples of electrophiles. Now that we have identified the first basic move (nucleophiles attacking electrophiles), let’s continue with a summary of the second basic move. 2, Loss ofa leaving group. For example: A —— hee This move can be thought of as the reverse process of the first basic move that we saw. In the reection we are showing here, a leaving group leaves to form a carbocation. If you were to imag: ine taking a video of this process, and then playing the video backwards (rewinding itso that you can see it while it is rewinding), you would see a bromide ion (a nucleophile) attacking a carboca- tion (an electrophile) to form the compound on the left above. So, in fact, the tasic move we are discussing right now (loss of a leaving group) is just the reverse process of the first basic move that ‘we saw (nucleophile attacking an electrophile) ‘Sometimes, you will See more than one arrow being used for the loss of a leaving group. For example, consider the following: RF Ro 0 ° Gon OH ‘There is one curved arrow that actually shows the Cl leaving. The remaining arrows can be thought of in two different ways (just as we saw earlier). You can think of the other arrows as being reso- group leaves: O09 %@,0 i “ Q O @ OH OH OH22 THEBASIC Moves 13 Or you can think of this as a flow of electron density that pushes out the leaving group: %e.9 Electron density is pushed down 0 to kick of @ leaving group Cl OH ‘You are probably better off if you think of this as a flow of electron density that pushes out the teaving group. But however you think of it, you should realize that all ofthese arrows are just show- ing one basic move: a leaving group is leaving In the first semester of organic chemistry, we saw tips for identifying good leaving groups and bad caving groups. If you are rusty on Jeaving groups, I recommend that you go back and review that ‘material. You will need to be able to identify good leaving groups in order to propose mechanisms. So far, we have seen two basic moves: a nucleophile attacking an electrophile and a leaving group leaving. Now let's take a look at our third basic move: 3. Broton Transfers. For example: ee ialer Qu 5 A A, *8 Proton transfers are just acid-base reactions. We talked about acidity (proton transfers) at length in the first semester of organic chemistry. For now, let's just focus on the fact that this basic move always requires at least two curved arrows. As an example, look at the proton transfer we just saw. ‘The ketone is grabbing a proton, and we need two curved arrows to show this. One arrow goes from the ketone to the proton, and the second arrow shows what happens to the electrons that were pre- viously holding the proton. There are always at least two arrows, whether the compound is grabbing a proton (like the case above), or whether a compound is losing a proton, like this: Notice that, once again, we need two arrows to show the proton transfer. When a compound is los- ing a proton, some textbooks and instructors will skip showing what grabs the proton. And they ‘will only use one arrow, like this: QaH foo) oa +H A Other instructors are particular that you show both arrows. In other words, you need to show what grabs the proton. Certainly, it can’t hurt to get into the habit of always showing what grabs the pro- ton, right? So get in the habit of using two arrows.14 CHAPTER2 INTRO TO IONIC MECHANISMS Sometimes, you will see more than two arrows. For example, consider this case which shows three arro. 9 roe § tf A To explain this, we will use the same logic we used when describing the other basic moves. We ‘can understand this in either of two ways, We can say that there are only two arrows needed for the reaction, and the remaining arrows are just resonance arrows: ~ A A * in the equilibrium mixture, and that NO>* functions as the electrophile we need in ‘order to pat a nitro group on a benzene ring.30 CHAPTER 3 ELECTROPHILIC AROMATIC SUBSTITUTION ‘Once again, this mechanism is essentially the same mechanism that we saw in the previous sec. tion: NO;* on and then H* off. There are only two very subile differences, Let's take a look at the ‘mechanism, and then we will focus on the two subile differences: os 4 1 whe ONE Be 6 SIGMA COMPLEX Inthe first step, when we attack the electrophile, we need to use two arrows (whereas we only used one arrow to attack E* in the previous reactions). We have the second arrow here so that we won't give five bonds to nitrogen. (This is similar to the argument we gave when we looked at the structure of nitric acid NEVER give nitrogen five bonds because it only has four orbitals with which to form bonds.) ‘The only other subtle difference is in the last step of the mechanism. In this case, we are using HSO,~ to pull off the proton (instead of AIBr«~), which should make sense because we don't have any AIB1a~ in this reaction Other than those two subtle differences, the mechanism is identical to what we have already seen in the previous reactions. ‘So far, we have seen how to put a halogen on the ring (Cl, Br, oF 1) and we have seen how to put « nitro group onto the ring. Before we move on, let's just make sure that you are familiar with the reagents necessary to do these reactions. Far each of the following problems, fill in the reagents that you would need in order to carry out the transformation: Br 3.5. Co —_—> NO. aC cl ‘3.8. Without looking back at the previous section, try to draw the mechanism for the nitra- tion of benzene. You will need a separate piece of paper to record your answer. Make sure to start by drawing the mechanism for formation of NO2*, and then show the reaction of benzene with NOs*,3.3 FRIEDEL-CRAFTS ALKYLATION AND ACYLATION 31 3.3 FRIEDEL-CRAFTS ALKYLATION AND ACYLATION In the previous sections, we saw how to put several different groups (Br, Cl, 1, ot NO2) onto a benzene ring, using an electrophilic aromatic substitution. In each case, the mechanism was the same: E* om the ring. and then H* off. In this section, we will learn how to put an alkyl group oon the ring. Let's start with the simplest of all alkyl groups: a methyl group. So, the question is: what reagents would we need to do the following transformation: CHs o—o Using the logic that we have developed in this chapter, we would want to use CH3* as our electrophile. But you should probably cringe when you see CH". After all, you probably re- member what you leamed about the stability of carbocations—that tertiary carbocations are more stable than secondary carbocations, and so on. Ceriainly a methyl carbocation would not be very stable at all, In fact, we deliberately avoid using methyl or primary carbocations in our mechanisms. But here we are, trying to make a methyl carbocation. Is it even possible? The answer is: yes. In fact, we will make it using the same method we have used in the previous sections, If We take methy! chloride and we mix it with a pinch of AICIs, we will have a source of CH3*: oes oS a1 ics —~ cI Does not really exist as free CHs* ‘The truth is that we are NOT really forming a free methyl carbocation that can float off into solu- tion. CH;* would be too unstable by itself (think back to carbocation stability). So, instead, we ‘must view this as a complex that can serve as a “source” of CH”. Gl Source of CHs* This gives us a way to methylate a benzene ring: CHSC AICI32 CHAPTERS ELECTROPHILIC AROMATIC SUBSTITUTION ‘And the mechanism is, once again, the same mechanism that we have seen over and over again. It is an electrophilic aromatic substitution: CHs* om the ring and then HY off: eRe a oeARa ” H pHs H He Hows] _ o a _ <4 eo 6 SIGMA COMPLEX We can use the exact same process to put an ethyl group on a ring: Oo eo AICls “This process (putting an alkyl group onto a ring) is called a Friedel-Crafts Alkylation. It works very ‘well for putting a methyl group or an ethy! group on the ring. BUT we run into problems when we try to put a propyl group onto the ring. We actually get a mixture of products when we try to put «a propyl group on the rings ove 6-6 ‘The reason for this is simple. Since we are forming a carbocation, it is possible to get a carboca- tion rearrangement. It was not possible for a methyl carbocation to rearrange. Similarly, an ethyl carbocation cannot rearrange to become any more stable, But a propyl carbocation CAN rearrange (via a hydride shift): Sts H Xs 2 rm ‘And since we are forming a propyl carbocation, we can expect that ‘sometimes it will rearrange be- fore reacting with benzene (while other times, it will not get a chance to rearrange before it reacts fr aig wha roe le ae atta ca yu eed be ced ‘Now, if we wanted to make isopropyl benzene, we could avoid this whole issue by just using a AN oO AICI,33 FRIEDEL-CRAFTS ALKYLATION AND ACYLATION 33. But what if we want to make propylbenzene? How would we do that? If we use propyl chloride, we have already scen that we will get some re- serangement, and we will not get a good yield of the desired product. In fact, we can generalize the question like this: how do we attach any alky! group and avoid a potential carbocation rearrangement. For example, how could we do the following transformation, without a carbocation rearrangement? If we just use chlorohexane (and AICI3), we will likely get a mixture of products Clearly, we need a trick. And there isa trick. To see how it works, we need to take a close look at a similar reaction that also bears the name Freidel-Crafts. But this reaction is not an alkylation. Rather, itis called an acylation. To see the difference, let’s quickly compare an alkyl group with an acyl group. < Alkyl group Z 9 Acyl group ( SS A ‘We can put an aeyl group onto a benzene ring in exactly the same way that we put an alky! group ‘on the ring. We just need to use the following reagents: A CO AICI, ‘The first reagent is called an acyl chloride (or acid chloride), and we are slready familiar with the role of AICI; (the Lewis acid). The Lewis acid is used to pull a Cl atom off of the acyl chloride, like this: cl g a oO Bags — a] of A ‘The end result is that we have a new kind of electrophile that we can use to react with benzene. This lectrophile is called an acylium ion (that should make sense—“acyl” because we can see that this34 CHAPTERS ELECTROPHILIC AROMATIC SUBSTITUTION clectrophile has an acyl group; and “dum” because there is a positive charge). This electrophile ac- tually has an important resonance structure: & Ve LF ACYLIUM ION ‘These resonance structures are important because they tell us that an acylium ion is stabilized by resonance, and therefore, it will NOT undergo a carbocation rearrangement. (If it did, it would lose this resonance stabilization.) Compare the following two cases: \~@ CAN REARRANGE ~ J CANNOT REARRANGE ‘So, we see that if we do a Friedel-Crafts Aeylation, we can cleanly hook an acyl group onto a ben- zene ring (without any rearrangements): ° °. Aa CO ACI, ‘And we don’t get any side products that would result from a carbocation rearrangement, Once again, ‘compare a Friedel-Crafts Allylation with a Friedel-Crafts Acylation: aucemoe & oS aren . 7 Mixture of products ° 0 ACYLATION oO eS AICIg, (One product Now take a close look at the acylation above, and let's point out a very important feature, Notice that we have hooked on a three-carbon chain onto the ring, but this chain is attached by the first ‘carbon of the chain, as opposed to the middle carton of the chain: s 8 We get this We don’t get this3.3 FRIEDEL-CRAFTS ALKYLATION AND ACYLATION 35 Once agai, it hooks on by the first carbon because we don’t get a rearrangement (the acylium fon is resonance stabilized, and does not rearrange). Now, all we need is a way to remove the oxygen, and then we will have a two-step synthesis for ataching a propyl group to a benzene ring: oO A, o 2 [And luckily, there isa simple way to remove the oxygen; in fact, there are three very common ways to remove the oxygen. We will just focus on one method right now (Which uses acidic conditions), but Keep in mind that we will sce two other ways of doing this in the upcoming chapters (one way ruses basic conditions and the other way uses neutral conditions). When we reduce a ketone under acidic conditions, we call the reaction a Clemmensen reduction: Zn (Hg) HCI heat Let's quickly look at the reagents we used here, Inthe first reagent, Zn{Hg], the brackets around the mercury indicate that we are using an amalgam of zine and mercury. An amalgam is what you get when you take two metals, heat them up until they are liquids, mix those liquids, and then let them cool back down to give you one solid that is composed of both metals. In addition, we also need HCI and heat to have the conditions for a Clemmensen reduction ‘Chemists have not reached a clear consensus abost the mechanism of a Clemmensen reduction, so most texthooks do not go into the mechanism of this reaction. The important thing to know here js that you can use a Clemmensen reduction as the second step of a two-step synthesis that places an alkyl group onto a ring without any rearrangement taking place: oO oO 1» WAG Ace 2) Zn [Hg], HCI, heat Refore we move on, there is one subtle point to mention about a Friedel-Crafts Acylation (step 1 ‘of the synthesis above). Remember that this reaction takes place in the presence of a Lewis acid (AICIs), which is constantly on the lookout for electrons that it can Jatch onto. Well, the product of the acylation step is a ketone. And a ketone has electrons that the Lewis acid can latch onto: ‘Therefore, whenever we do an acylation reaction, we need to pull the Lewis acid off our product in the end. And there is a simple way to do this. We just give the Lewis acid some other source of flectrons 10 latch onto instead. The easiest thing to use is water (HO). So, whenever you use a36 CHAPTER 3 ELECTROPHILIC AROMATIC SUBSTITUTION Friedel-Crafts Acylation in a synthesis problem, you should make sure to include water in your list of reagents (immediately after the acylation): 9 Oo 1) AQ » AlCls 2H] 8) Zn [Hg] , HCI, heat ‘To summarize, we have seen that a Friedel-Crafts Acylation can be followed up by a Clem mensen reduction, as a clever way of putting an alkyl group on a ring (without rearrangements). But there are actually times when you will want to put an acyl group on the ring, and you won't want to do a Clemmensen reduction afterward. For example: 0. To do this transformation, you would just use a Friedel-Crafts Acylation (including the step ‘here you use water), and that's it. There is no need for a Clemmensen reduction, because we don't ‘want to pull off the oxygen in the end EXERCISE 3.9. Show the reagents you would use to cary out the following synthesis C Answer: In this problem, we need to attach an alkyl group to a benzene ring. So, we first look to see if we could do this one step, using a Friedel-Crafts Alkylation, In this case, we cannot do it in one step because we have to worry about a carbocation rearrangement. If we think about the elec- trophile that we would need to make, we will see that it could rearrange: oy Hydride shift Av And this would give us a mixture of products ok &. &|3.3 FRIEDEL-CRAFTS ALKYLATION AND ACYLATION 37 So, instead we will have to use a Friedel-Crafts Acylation (don't forget the water) followed by a ‘Clemmensen reduction: at 1) AlCl ° OS re 3) Zn [Hg] , HCI, heat For each of the following problems, show what reagents you would use to accomplish the trans- formation. In some situations, you will want to use a Friedel-Crafts Alkylation, while in other situations, you will want to use a Friedel-Crafts Acylation, 3:10. Co — “~o—O6 w~o—é no ~G— CO 3.15, Predict the products that would form from the following reaction. awk AICI (Hint: There should be a mixture of three produets in this case. Be sure to consider all of the possible rearrangements that can take place. If you are rusty on carbocation re- arrangements, then you should take a break to go back and review them now.)3B owwens execmoriac anomaric susstTuTiION BE Ow a separate piece of paper, draw the mechanism of formation for each one of the ‘are prodects from the previous problem. 3.17. On a separate piece of paper, draw the mechanism of the following reaction. Make sure ‘0 show the mechanism of formation of the seylium ion before you use itt react with the ing cl Whe my » AIC, ° CO 2) HO 3) Zn [Hg], HCI, heat Using Friedel-Crafts reactions involves a few limitations, You should take a momentto read about them. {n your texibook. The two mast imporiant limitations are as follows: 1, When doing a Friedel-Crafts Alkylation, itis often difficult to put on just one alkyl group. Each alkyl group makes the ring more reactive toward a subsequent attack on the same ring. ‘When doing @ Friedel-Crafts Acylation, it is generally not possible to place more than one seyl group onto a ring, The presence of one acyl group makes the ring less reactive toward « second acylation, We need to understand WHY an alkyl group makes the ring more reactive, and WHY an acyl ‘group makes the ring less reactive. We will explain this in greater detail when we turn our atten- tion to the nucleophile of an electrophilic aromatic substitution reaction, (Remember that the ben Zene ring is the nucleophile of the reaction, and it attacks some electrophile, E*.) Until now, we have focused on the electrophile of the reaction. So far, we have seen how to use Br*, CI*.T*, NO2", alkyl", and acyl. Before we can shift our attention to the nucleophile of the reaction, we have one more electrophile to discuss. 3.4 SULFONATION a The reaction we will discuss now is probably one of the most important reactions for you to have at your fingertips. This reaction will be used extensively in synthesis problems later in this chap- ‘er. If you do not keep this reaction in mind while you are solving synthesis problems, then you ‘Will be at a severe loss. We will explain why this reaction is so important inthe upcoming sections, For now, just take my word for it, and let's just master the reaction. Al of the elestrophiles we have seen so far have always been positively charged. But now we ‘will deal with one clectrophile that is not positively charged. The electrophile is $03, Let's take a close look at the structure: Notice that there are three SO double bonds here. But these double bonds are not such great dou- ble bonds. Recall that a double bond is formed from the overlap of two p orbitals:34 SULFONATION 39 3 When we are talking about a carbon-carbon double bond, the overlap of the p orbitals is pretty ‘good because the p orbitals are the same size. But what happens when you try to overlap the p or bital of an oxygen atom with the p orbital ofa sulfur atom? The p orbitals are different sizes (oxy zen isin the second row of the periodic table, which means that its using a p orbital from the sec- ‘ond energy level: but sulfur is in the third row of the periodic table, soit is using a p orbital from the third energy level): ‘Therefore, the overlap is not so good, and itis misleading to think of S=O as being a double bond. Its probably much closer in nature to being like ths: @o S-O ratherthan S=0 When we do this analysis for each of the three double bonds in SOs, we begin to see that the sul- fur atom is VERY electron-poor: Q g o> ~ 7 SoG? In fact, the sulfur atom is so electron-poor that itis an excellent eleetropkile, even though the com- pound is overall neutral (no net charge). Now we are going to see a reaction that uses SOs as an electrophile. But first, let's see where SO3 comes from. Sulfuric acid is constantly in equilibrium with $O3 and water: H2SO4 SOs; + HO ‘That means that any botlle of sulfuric acid will have some SOs in it. At room temperature, SOs is a gas, and itis possible to add extra SO3 gas to sulfuric acid (which shifts the equilibrium a bit)40 cHaPTER's ELECTROPHILIC AROMATIC SUBSTITUTION ‘When we do this, we call the mixture fuming sulfuric acid. So, from now on, whenever you see concentrated, fuming sulfuric acid, you should realize that we are talking ubout SOs as the reagent, ‘And here is the reaction: Co cone, fuming HoSO4 Notice that, in the end, we have placed the SOsH group on the ring. The obvious question is: why is the H attached to the SO3 in the end? To see why, let's take a closer look at the mech- anism, Remember our two steps for any electrophilic aromatic substitution: E* goes on the ting, and then H* comes off. But wait a second... . In this ease, we are not using an electrophile with a net positive charge. The electrophile in this case has no net charge. In all of the reactions ‘we have seen so far, we put something positively charged onto the ring, and then we took some- thing positively charged off of the ring. So in the end, our ring never gained or lost any charges. But in this case, we are putting something neutral (S03) onto the ring, and then we are remov- ing something positively charged (H). That should leave our product with an overall negative charge, which is exactly what happens: SO3H SIGMA COMPLEX So, we need to add one more step to our mechanism. The negative charge on our product grabs a proton from sulfuric acid (remember that there is plenty of sulfuric acid around because we used fuming sulfuric acid as our source of SOs): :O Although this reaction has this one extra step at the end of the meckanism, Keep in mind that this extra step is just a proton transfer. The core reaction is still the same as we have seen in all of the previous reactions: the electrophile comes on the ring, and then H* comes off of the ring. An important feature of this reaction (and this is the feature that will make this reaction so important for synthesis problems) is how easily the reaction can be reversed. The amount of product that you get is equilibrium controlled, and it is very sensitive to the conditions. So, if24suLronaTION 41 you use dilute sulfuric acid instead, the equibrium Keans the other way (look closely at the equilibrium arrows below): SO3H cone. fuming H2SOx ean es SO3H dilute H2SO4 SS We can use this to our advantage because this gives us a way to remove the SO3H group whenever we want, We would just use dilute sulfuric acid to pull it off: cone. fuming H2S04 Sy ee SO3H lu So we now have the ability to put the SOsH group onto the ring whenever we want, AND we can take it off whenever we want as well. You might wonder why you woukd want to put a group on in order just to take it off later. On the surface, that would seem like a waste of time. But in the upcoming chapters, we will see that this will become very important in synthesis problems, For now, let's make sure that we are comfortable with the reagents. EXERCISE 3.18. Ldentify the reagents that you would use to cary out the transformation shown below: Br Br ‘SO3H Answer: We know that we use fuming sulfuric acid to put an SO3H group onto the ring, and ‘we use dilute sulfuric acid to take the group off. In this case, we are taking the group off, so we need to use dilute sulfuric acid Br Br dilute HySO, Np |42 CHAPTERS ELECTROPHILIC AROMATIC SUBSTITUTION. For each of the following problems, identity the reagents you would use to carry out the transfor- Sa A gl” SOsH Br Br SO3H °° ie ( S 3.21. 2 7 ‘SO3H Z =O) CO t SO3H ‘And to make sure that you are not getting rusty on the other reactions we have learned in this chap- ter so far, fill in the reageats you would use for the following transformations: Br S 3.23. | ———— NOz S 3.24, a cl cs sas. ( ) ———___-35 MODIFYING THE NUCLEOPHILICITY OF THE NUCLEOPHILE 43 3.28. Now, let's just make sure that you can draw the mechanism of a salfonation reaction, ‘Thats just the fancy name we give to the reaction where we put an SOsH group onto the ring.) On a separate piece of paper, take a moment and try to draw the mechanism for the sulfonation of benzene, Remember that there will be three steps: (1) electrophile comes on the ring, (2) H* comes off, and then (3) proton transfer gots rid of the negative charge. Don't forget to ‘draw the resonance structures of the intermediate sigma complex. Try to draw the mechanism without looking back to where itis shown earlier inthis section. 3.29. And now, for a challenging problem, ty to draw the mectuauisi of a desulfonation reaction (a reaction where we take the SOsH group off of the ring). The process will be exactly the reverse of what you just drew in the previous problem. There will be three steps: (1) remove the proton from the SOH group, (2) H* comes on the ring, and then (3) SOs comes off of the ring. The truth is that there are only two core steps here: H* comes on the ring, and then SOs comes off of the ring. You can actually pull the proton off of the SO3H group at the same time that SOs comes off of the ring. Try to do it yourself, and if you get stuck, you can look at the “answer in the back of the book. But try to do the problem without looking at the answer. 3.5 MODIFYING THE NUCLEOPHILICITY OF THE NUCLEOPHILE ‘When we began this chapter, we saw that benzene and bromine do not react with each other (with- ‘out a Lewis acid): Bro No reaction So, we began to investigate how we could force the reaction to take place. We said that the ben ‘ene ting is the nucleophile and Br, is the electrophile. And if we want to force the reaction to 0, then we have two choices: we can either use a better electrophile (something more electrophilic than Bra) or we can use @ better nucleophile (something more nucleophilic than benzenc). We then focused all of our aitention on making the electrophile more electrophilic. We saw how to use a Lewis acid to make Br*. We then saw many other positively charged electrophiles (CI*,1*, NOa*, alkyl*, and acyl*), and we even saw one electrophile that is an excellent electophile, even though it does not have a net positive charge (SO;). Now, we will turn our attention to the nucleophile— the aromatic ring. How éo we make an aromatic ring more nucleophilic?44 CHAPTER 3 ELECTROPHILIC AROMATIC SUBSTITUTION ‘To anvwer this question, We need to focus on the groups that are already connected to the ring. Benzene itself (CaHa) has no groups on it. But consider the siucture of methyl benvene (commonly called toluene): Methyibenzene (Common name = Toluene) Here, we have an aromatic ring with a methyl group on it, and the question is: what effect does {hat methyl group have on the nucleophiliity ofthe aromatic ring? Is this compound bence oe When Bromine does go on the ring, why is it going in specific locations? Why don’t we get 4 Penta-brominated product? To better understand this question, let's make sure we are se.3.5 MODIFYING THE NUCLEOPHILICITY OF THE NUCLEOPHILE 45 ing the correct terminology. When you have an aromatic ring with one group on it, we de- scribe the other positions with the following terms: R ortho Z~ ortho meta / meta para “The two positions that are closest to the R group are called ortho positions. Then we have the ‘meta positions. And finally the farthest position from the group is called the para position. Us- ing this terminology, we see in our reactions above thatthe substitution reactions always seem to take place atthe ortho and para positions. Why? Why don’t we get a substitution reaction at the meta position” ‘order to answer these questions, we will need to take a close look atthe factors that determine + electronics of any compound. If you ever want to know whether a compound is nucleophilic + how nucleophilic itis), you are really asking about the electronics of that compound, And itis ‘mforting to know that there are only two factors that you ever have to consider: induction and sonance. ‘Let's use phenol as our example, and let’s explore the electronics of that compound. Let's be- 1 by looking at the first factor: induction, Recall from the first semester that induction is fairly inple to assess, You just look at the relative electronegativity of the atoms. If you want to know hat inductive effect the OH group will have on the aromatic ring, you look at the bond connect- & the OH group to the ring: hoa -xygen is more electronegative than carbon, so there is an inductive effect (shown by the arrow sove). The oxygen is withdrawing electron density from the ring. Remember that the ring is only ‘nucleophile because itis electron-rich (from all of those.pi electrons), s0 if we take away elec ‘on density from the ring, then we are decreasing the nucleophilicity of the ring. So the inductive Ffect of the OH group is to make the ring less nucleophilic. But we're not done yet. We need to onsider the other factor that determines the electronics of a compound: resonance. In order to see how resonance affects the electronics of our compound, we need to draw the res- nance structures (if you did not master the steps for drawing resonance structures in the first se- rester, then you should go back and do so now—resonance Will reappear in almost every topic in re second semester): @ Q OH & SH4G CHAPTER 3. ELECTROPHILIC AROMATIC SUBSTITUTION Notice that there is a negative charge spread throughout the ring. When we meld all of these reso- nance stuctures together in our minds, we get the following picture for the molecule: OH ‘The &- shows that there is a partial negative charge spread throughout the ring. Therefore, the ef- fect of resonance is to give electron density to the ring. So, now we have « competition. By induc~ tion, the OH group is electron-withdrawing, which makes the ting less nucleophilic. But by reso- nance, the OH group is electron-donating, which makes the ring more nucleophilic. So the question is: which effect is stronger? Resonance or induction? This is a common scenario in organic chem- istry (where induction and resonance are in competition), and the general rule is: resonance usually ins. There are some important exceptions. We will soon see one of these exceptions, but in gen- ‘eral, resonance is much stronger than induction, Now let's apply this general rule to our case of phenol. If we say that resonance wins, that means that the net effect of the OH group is to donate electron density to the ring. Therefore, the net ef- fect of the OH group is to make the ring more nucleophilic (as compared with benzene). {A simple analogy sums this up. Imagine that you have $100 in your hand. You come to me, and I do two things. Fist, [take some money away from you, Then, I give you some money back. And the ‘question is: do you have more money than you came with, or less? The only way to answer the ques- tionis to know whether I took more than I gave you back, or whether I gave back more than I took. The cease of phen is like « case where I take $10 away from you (which brings you down to $90), but then T give you back $200, So, in the end, you end up with $290, which is a lot more than you had origi- nally. Is it true that I frst took some money away from you? Yes, it i. But it is almost insignificant Compared with the major gift of $200 that I gave you. ‘This analogy is overly simplistic, and it is probably completely unnecessary in this case because it is easy enough to understand without the analogy. I am using the analogy, however, because we will soon see arguments that are somewhat difficult to understand. Use of this money analogy throughout ‘our discussion will help you gresp the more difficult concepts that we will see in the upcoming section. We took a very close look at phenol, and we were able to see that the effect of the OH group. is to make the ring more nucleophilic. We call this “activation.” In other words, the OH group is activating the ring (making it more nucleophilic). So, the OH group is called an activator. The methyl group is also an activator (eecall from the first semester that alkyl groups are electron do- nating). Some groups, however, actually withdraw electron density from the ring; we call those ‘groups deactivators because they deactivate the ring (make the ring less nucleophilic). In the next section, we will see an example of a deactivator. But we have not answered one of the questions we asked earlier inthis section. We saw two re- actions (bromination of toluene and bromination of phenol) where the substitution took place at the cotho and para positions only. We did mor get any meta substitution in those cases, Why n0t? Ate there cases where you do get meta substitution? The answers to these questions are critical for solv- ing synthesis problems. Let's explore these questions in more detail now. 3.6 PREDICTING DIRECTING EFFECTS ‘When we talk about the preference for ortho and para substitution, we are talking about an issue of regiochemistry. In other words, in what region of the aromatic ring does the reaction take place? Let's review whst we saw about the case of phenol248 PREDICTING DIRECTING EFFECTS. 47 In the previous section, we saw that the OH has two effects on the aromatic ring. Iti electron- withdrawing by induction, and it is eleciron-donating by resonance. We saw that resonance was stronger, and therefore, the net effect of the OH group was to donate electron density to the ring (which activates the ring). When we drew the resonance structures for phenol and then melded them. together in our minds, we got the following picture: OH é & a When we draw it like this, we clearly see that the OH group is donating electron density to the ring. But take a close look at where the electron density is being donated. It is not in all positions on the sing. It is only on the ortho and para positioas. So itis true that the OH group makes the ring more nucleophilic, but it is only more nucleophilic at the ortho and para posi- tions, Thus, when this ring reacts with an electrophile, the reaction takes place at the ortho and. OH OH Br, Br. Br Br ‘Notice thatthe ring is SO incredibly activated that the reaction takes place at all three spots (the two ortho positions AND the para position). And we don’t even need a Lewis acid to form Br. ‘The ring is so activated by the OH group that even a poor electrophile (Brz) will do the job (three times). ‘We have just given a specific explanation for the preference for ortho-para direction inthe case ‘of phenol. Our explanation was based on the electronics of the starting material. But another ex- planation can be given and is based on the stability of the intermediate. Luckily, this second explanation gives the same result (that there should be a preference for ortho-para direction). Most ‘textbooks give this second explanation, probably because itis the only way to truly explain why alkyl groups are ortho-para directing (the first explanation. which we just gave, cannot be used to ‘explain the directing effects of alkyl groups). You should look in your textbook to review that sec~ ond explanation for ortho-para direction, but for now, we will just give a quick summary of the explanation. ‘You start by drawing out three mechanisms: You draw one mechanism for a substitution taking place at the ortho position. Then you draw another mechanism for a substitution at the mera posi- tion, Next you draw one last mechanism for a substitution at the para position. Finally, you com- pare the sigma complexes in each of the three mechanisms. You will find thatthe sigma complexes for ortho-substitution and para-substitution are more stable (than the sigma complex of a meta- substitution) because each has an extra resonance structure that is not present in the sigma complex. of meta-substitution. ‘Whether or not you fully understand that second explanation, the take-home message is that the (1 group and the methyl group are both ortho-para dircetors. Infact, this is true not only for these two groups, but for ALL activators. All activators are ortho-para directors. In the next section, we ‘ill learn how to predict whether a group is an activator. But before we are ready for that, we need to explore what happens when we have a group on the ring that is a deactivator.48 CHAPTER 3 ELECTROPHILIC AROMATIC SUBSTITUTION The best example is the nitro group. Consider the structure of nitrobenzene: ogee If we want to understand the electronics of this compound, we will need to look at the two fac- tors that determine electronics: induction and resonance. Induction is simple. Nitrogen is more electronegative than carbon, so the nitro group withdraws electron density from the ring by in- duction. The real question is: what is the effect of resonance? To see that, we will have to draw ‘out the resonance structures: ogo Notice that we now have a positive charge spread throughout the ring. In the case of phenol, we had a negative charge spread throughout the ring. In that case, we argued that the OH group gives electron density to the ring, making the ring more nucleophilic. But here, we have a pos- itive charge on the ring. So, the nitro group is withdrawing electron density from the ring, mak- ing the ring less nucleophi Noe 3+ ZN at 8+ So, the nitro group is electron-withdrawing by resonance and by induction. Therefore, in this cease, there is no competition between resonance and induction. Both factors tell us that a nitro group should deactivate the ring. In other words, it should be difficult to do an electrophilic aromatic sub- stitution on nitrobenzene. So, let's say we take nitrobenzene and try to do an electrophilic aromatic substitution. For ex- ample, let's say we try to brominate it. We know that benzene will not react with Brp. So it goes without saying that nitrobenzene, which is fess nucleophilic than benzene, will definitely not react with Bro. But what happens if we try to force the reaction to go? Recall how we forced benzene to react—we used Bry together with a Lewis acid to form Br*. So, what happens if we try the same trick here? We do, in fact, get a reaction: No. NOp Bry AIBr33.6 PREDICTING DIRECTING EFFECTS 49. But you should immediately notice that the bromine went into the meta position only. To under- stand this, we need to take a close look at the electronics of nitrobenzene. We just saw that reso- nance allows us to predict that the nitro group is withdrawing a significant amount of electron den- sity from the ring: NOz a ot ot But notice that the electron density has not been withdrawn from the entire ring. Rather, itis the or- tho and para positions that have been the most affected. So what happens when we force this reac- tion to go by using Br*? It certainly cannot react at the ortho and para positions. Remember that the ring is acting as our nucleophile. And the ortho and para positions just don’t have the electron den- sity that it would take to attack Br. So, if we force the reaction to go, it will have to go in the meta position BY DEFAULT. The meta position has NOT been activated. But rather, the ortho and para positions have been deactivated. So, by default, the reaction must go meta. The explanation we just gave can be used to show why a nitro group (which is a deactivator) is ‘a meta-director. Our explanation was based on the electronics of the starting material. But once again, a second explanation can be given, which is based on the stability of the intermediate. Lixck- ily, this second explanation gives the same result (that there should be a preference for meta rection). Most textbooks give this second explanation. You should look in your textbook to review that second explanation for meta direction. Most importantly, you should know the bottom line: deactivators are meta-directors. ‘So, we can now summarize the two important concepts that we have leamed so far: * Activators are ortho-para directors. *+ Deactivators are meta-ditectors. And now for the cbvious question: are there any exceptions to these general statements? The an- sswer is: yes. There is one very important exception. Halogens (F, Br, Cl, or I) are deactivators, 50 ‘we might expect them to be meta-directors. But instead, they are actually ortho-para directors, Let's tty to understand why halogens are the exception. Before we explain this, let me just say upfront that this explanation is perhaps one of the most difficult concepts in all of orgenic chemistry. If you get lost and you have trouble understanding WHY halogens are the exception, then don’t feel bad. Most students have a hard time with this. It takes time and patience—you should know that before we start. But if you find that you completely understand this explanation, and it makes perfect sense to you, then you should feel very good about yourself because organic chemistry doesn’t get much tougher then this. And now, with that dis- Claimer in mind, let's jump into it. If we want to understand why halogens are the exception, we need to remember a rule that we saw earlier in this chapter. When we were analyzing the effect of the OH group on an aromatic ring, we saw that there were two competing effects: induction and resonance. We saw that induction withdraws electron density from the ring but that resonance donates electron deasity to the ring. In order to know which factor dominates, we gave a general rule: resonance usually beats induction. We also seid that there was an important exception to this general nule that We would see later. Well, it is now later. Halogens are the exception. We will look at this very closely over the next couple of pages, but for starters, here is the one-paragraph summary:50 CHAPTER3 ELECTROPHILIC AROMATIC SUBSTITUTION ‘We have a general rule that resonance usually beats induction. But halogens are the exception to this rule. So, in the case of halogens, induction actually beats resonance. We will then use this ‘concept to explain why halogens violate our rule for directing effects (that all activators are ortho- para directors and all deactivators are meta directors). We will explain why halogens are deactiva- tocs but ortho-para directors. And we will see that the answer comes directly from the fact that, in the case of halogens, induction actually beats resonance. Now that we have seen the summaty, let’s take a closer look at the explanation. Let's use the ‘example of chlorobenzene: ql Tf.we want to understand the electronics of this compound, we will need to look at two factors: in- duction and resonance. Let's start with induction. The inductive effect is similar to the effect we ‘saw with an OH group on the ring, Just like an OH group, the Cl group is also electron-withdrawing by induction: te However, we also need to look at resonance effects. So we draw the resonance structures: ‘And once again, we see that the Cl group is very similar to the OH group. It is donating electron density by resonance: cl 6 So we have a similar analysis that we had with the case of the OH group. Once again, we have ‘¢ competition between resonance and induction. The Cl group is withdrawing electron density by induction, and donating electron density by resonance. But in the case of the OH group, we used the argument that resonance beats induction (we said that was a general rule that holds true most of the time). Therefore, the net effect of the OH group was to donate electron density to the ring (thus, the OH group was an activator). But with the CI group, resonance does NOT beat induc- tion, This is one of the rare cates where induction actually beats resonance.2.6 PREDICTING DIRECTING EFFECTS 51 If we want to understand why, we must take another look at the resonance structures for chlorobenzene (see above). Notice that these resonance structures show a positive charge on Cl. ‘That is very bad. Halogens do not like to bear positive charges (even more so than oxygen). So these resonance structures do not contribute very much character to the overall electronics of the ‘molecule. Therefore, very litte electron density is given to the ortho and para positions: al 3 3- Only a'small amount of electron density iven to these three positions. The 5~ ¥ in each position is therefore very small. ‘The effect of resonance in this case is very small, and as a reslt, induction actually beats resonance in this case, Therefore, the net effect of the Cl group is to withdraw electron density from the ring. This explains why the Cl group is a deactivator. This also explains how the Cl group can be an cr. tho-para director (even though it is a deactivator) It is true that resonance is a weak effect in this case, BUT the effect of resonance is not completely negligible. Granted, resonance is s0 weak that induction actually wins the competition here, but resonance does still give a “tiny” amount of elec- tron density back to the ortho and para position To see this more clearly, let's use our money analogy again. Imagine that five people surround ‘you, each carrying $100: You ‘You begin by taking $30 from every person (which leaves each person with $70). Then, you ‘back $20 to each of three people (the ortho and para people): You oi ortho [s90) ortho meta ($70) meta [ee] para ‘The two meta people each have $70. The ortho and para people each have $90. Now think about ‘what you have done. Overall, you have taken more money than you have given back, But when ‘you compare how much money eech person has, you find thatthe ortho and para people have more ‘money than the meta people. Similarly, the net effect of the Cl group is to tate away eleciron density from the entire ring. ‘Therefore, it is a deactivator. However, the Cl group does give a tiny bit of electron density(52 CHAPTER ELECTROPHILIC AROMATIC SUBSTITUTION back to the ortho and para positions. If we force the reaction to go, then, it ortho or para: have to go cl ba od Now we are ready to modify the rules we gave earlier when we said that all activators are ortho- para directors and all dezctivators are meta-directors. Here is our new-and-improved formula: + All activators are ortho-para directors. + All deactivators are meta-directors, except for halogens (which are deactivators, but never- theless, they are ortho-para directors) With that in mind, let's try to predict some directing effects. EXERCISE 3.30. Look closely at the following monosubstituted benzene ring. Br If we tried to do an electrophilic aromatic substitution on this compound, identify where the sub- stitution reaction would take place. Answer: Bris ahalogen (remember thatthe halogens are F, Cl, Br, and I), We have seen that hhalogens are the onc exception to the general rules (halogens are deactivators, but nevertheless they are ortho-para directors). Therefore, if we use this compound in an electrophilic aromatic substitu- tion, we expect the substitution to take place at the ortho and para pcsitions: NI t For each of the eae a predict the directing effects. NO, a8 a This group is a deactivator.26 PREDICTING DIRECTING EFFECTS 53. A 3.35. O This group is an activator. Chk 3.36. ‘group is a deactivator. 3.37. ‘This group is an activator. Clearly, you can predict where the substitution will take place only if you know whether the group js an activator or a deactivator. In the next section, we will learn how to predict whether a group is an activator or a deactivator. But for now, let's get some practice on some real reactions. EXERCISE 3.38. Predict the products of the following reaction: HNOs H,SO, Answer: We begin by looking at the reagents, so that we can determine what kind of reaction is taking place. The reagents are nitric acid and sulfuric acid. We have cen that these reagents give usNO," as an electrophile, which can react with an aromatic ring in an electrophilic aromatic sub- stitution reaction. The end result is to place a nitro group into the ring. So, now the question is: ‘where do we put the nitro group? ‘To answer this question, we must predict the directing effects of the group already on the ring (be- fore the reaction takes place). There is a methyl group on the ring, and we have seen that the methyl ‘group is an activator. Therefore, we predict thatthe reaction will take place at the ortho and para po- sitions, relative fo the methyl group: OG HNO ao” + “HO. NOz Notice that in this case, we don’t draw a substitution at both ortho positions because we get the same product either way: O2N. Np is the same as‘54 CHAPTER 3. ELECTROPHILIC AROMATIC SUBSTITUTION Predict the products of the following reactions: Br aos HNO, “ HeSO4 3.40. Br 341. OO. a2 cone. fuming az sullurie acid Hint. The group on the ring is deactivator. ) os cone. fuming mae sulfuric acid ‘Hint: The group on the ring is an activator. On Bre 3.44, — - AIBrs Hint: The group on the ring is a deactivator. Et Ch 2.48, AICis Hint: The group on the ring is an activator, So far, we have focused on the directing effects when you have only one group on & ring. And we have seen that activators direct toward the ortho and para position, whereas desctivators direct to ward the meta positions:t and we saw only one exception to these general rules (the halogens). ‘But how do you predict the directing effects when you have more than one group on a ring? For example, consider the following compound: NO> ‘What if we used this compound in an electrophilic aromatic substitution reaction. For example, let's say we try to brominate this compound. Where would the bromine go? Let's first consider the effect of the methyl group. We mentioned before that a methyl group is an activator, so We predict that it wil direct toward the ortho ard para positions: ™ a” 4 Notice that we do not point to the ortho position that already bears the nitro group (we only look at postions where there are currently no groups-—remember that in an clectrophilic aromatic sub- stitution, E+ comes on the ring and H-+ comes off) So, the methyl group is directing toward 0 spots, as shown above. Now let's consider the effect of the nitto group. We mentioned before thatthe nitro group is a powerful deactivator. Therefore, we predict that it should direct to the positions that are meta to the nitro group: Meta to the nitro group No» Meta tothe t nitro group So, we see thatthe nitro group and the methyl group are directing toward the same two spots. So, in this case there is no conflict between the directing effects of the nitro group and the methy! group. But consider this case: NOzSG cHarTen 3 ELECTROPHILIC AROMATIC SUBSTITUTION ‘The methyl group and the nitro group are now directing toward different positions: ™~ on 1 NO2 we ‘NO> Directing effects Directing effects of the methyl group of the nitro group (ortho-para director) (meta director) So, the big question is: which group wins? It turns out that the directing effects of the methyl group are stronger than the directing effects of the nitro group. So, if we brominate this ring, we will get the following procucts (Where the Br goes ortho or para to the methyl group): Bre “A Br No, Bra NO2 Nop NO, br It is common to see a situation where the directing effects of two groups are competing with each other (like the methyl and the nitro group in the above example). So we clearly need rules for determining which group wins. It tums out that you need to know just two simple rules in order to determine which group will dominate the directing effects: 1. Ortho-para directors always beat meta-directors. The example we just saw is a perfec illus- tration of this rule. The methyl group is an activator (an ortho-para director), and the nitro ‘group is a deactivator (a meta-director),so the methyl group wins. Thisrrle should make sense, when we consider how directing effects work. Recall that meta directors do aot actually do anything good for the meta positions. Instead, they simply deactivate the ortho and para po- sitions, so if we force a reaction to go. it mast go meta BY DEFAULT. Bat ortho-para direc- tors are actually doing something good for the ortho and para positions. They activate the or- tho and para positions. Thus, ortho-para directors will always beat meta-directors. 2, Strong activators always beat weak activators. For example, consider the following case: OH ‘The OH group is a strong activator, and the methyl group is a wea activator. (We will learn in the next section how to predict which groups are strong and which are weak—for now, just take my word for it) So, the OH group will win, and the directing effects are ortho and para tothe OH group OH OT‘3.6 PREDICTING DIRECTING EFFECTS 57 So, we have seen two rules + Ortho-para directors always beat meta-directors. + Strong activators always beat weak activators. Keep in mind that the first rule always trumps the second rule. So if you have a weak activator against a strong deactivator, the weak activator wins. Even though the activator is weak, it stil beats fa strong deactivator because activators (ortho-para directors) always beat desctivators (meta- directors). To demonstrate this, let's consider the following example: NO, ‘The methyl group is a weak activator, and the nitro group is a strong deactivator. So, in this case, the methyl group wins (and the directing effects are ortho and para to the methyl group; and not meta t the nitwo group): Sy a NO2 4 EXERCISE 3.46. Predict the directing effects in the following scenario, Strong activator Strong deactivator For this problem, you should assume that the deactivator is nor a halogen, Answer: We have two groups. The activator will direct toward the positions that are ortho or ‘para to itself, and the deactivator will direct toward the positions that are meza to itself: Strong Strong activator activator i, a Strong Strong 4 deactivator wr deactivator Directing effect: Directing effects of the activator of the deactivator