Part A: Electrophilic Aromatic Substitution (What products are formed when a strong electrophile is added to benzene?)

Transcription

1 hemactivity 29 lectrophilic Aromatic Substitution 1 hemactivity 29 Part A: lectrophilic Aromatic Substitution (What products are formed when a strong electrophile is added to benzene?) Model 1: (review) lectrophilic Addition of l xn 1 l l l cyclohexene carbocation intermediate xn 2 l l x l benzene This product carbocation intermediate DS T form! ritical Thinking Questions 1. For xn 1 (above) draw curved arrows showing the mechanism of electrophilic addition of l. Include an appropriate carbocation intermediate in the box above. Figure 1: eaction Diagrams for lectrophilic Addition of l V.. (Potential nergy) l V.. carbocation intermediate l carbocation intermediate l l eaction Progress (xn 1) eaction Progress (xn 2) 2. xn 1 is slightly down-hill in terms of energy. xn 2 is very up-hill in terms of energy (see Figure 1). onstruct an explanation for the large difference in energy between the reactants and the product in xn 2. In xn 2 the product is not aromatic so it is much higher in energy than the aromatic reactant. 3. Draw the carbocation that would form in xn 2. xplain why this carbocation goes back to the starting material ( l and benzene) instead of forming the product.

2 hemactivity 29 lectrophilic Aromatic Substitution 2 The elimination reaction shown (the reverse of the first step) restores aromaticity, and is therefore favored over step two which results in a non-aromatic product.

3 hemactivity 29 lectrophilic Aromatic Substitution 3 Model 2: lectrophilic Aromatic Substitution ecall that deuterium (D) has nearly identical reactivity as hydrogen (). In the following reaction D l reacts the same way as would l. When benzene is treated with D-l, an is replaced with a D. (With excess D-l this continues until all the s have been replaced and the product is 6 D 6. D l D l D xn 3 l ritical Thinking Questions 4. Use curved arrows to show a reasonable mechanism for the reaction in Model 2. (int: the first step is formation of a carbocation intermediate, just as in xn 2.) 5. The energy diagram for xn 2 (using Dl in place of l) is shown below using a dotted line. n this same set of axes, draw a solid line to show an energy diagram for xn 3 in Model 2. (ote: xns 2 and 3 have the same carbocation intermediate.) D l V.. carbocation intermediate for xns 2 & 3 D D l eaction Progress (xn 2 with Dl) eaction Progress (xn 3) 6. onstruct an explanation for why xn 3 is much more likely to occur than xn 2. (ote: xn 2 does not occur under normal circumstances.) xn 3 yields a much lower energy product than xn 2, so the former is much more likely.

4 hemactivity 29 lectrophilic Aromatic Substitution 4 7. Draw a generalized mechanism for AS (lectrophilic Aromatic Substitution) showing the substitution of an electrophile ( + ) for one of the s of benzene in the presence of a mild base (B). (ote: in xn 3, + = D + ) B B Information: ot All lectrophiles are reated qual A very strong electrophile ( + ) is required to break up an aromatic ring. The of a strong acid will work since it exists essentially as +. Know the following strong acids l S I hydrochloric acid hydrobromic acid hydroiodic acid sulfuric acid nitric acid Suitable electrophiles ( + ) for AS (lectrophilic Aromatic Substitution) include an + or D + donated by a strong acid, or the electrophiles ( + ) listed below. Table 2: ther Suitable lectrophiles ( + ) for AS eactant + = eagents Product/s 2 S 4 2 and 3 anhydrous S 3 sulfuric S acid 2 and Fe 3 or l or or * l 2 and Fel 3 l benzene or other aromatic ring * -X, AlX 3 (X = l or ) AlX 3 X (X = l or ) = alkyl group = alkyl group * *The last three rows of electrophiles are generated only in the presence of a Lewis acid catalyst such as FeX 3 or AlX 3. A supplementary activity on Lewis acid catalysts is assigned for homework.

5 hemactivity 29 lectrophilic Aromatic Substitution 5 Part B: Substituent ffects (ow do substituents on an aromatic ring direct the placement of + and rate of AS?) Model 3: lectrophilic Aromatic Substitution with Toluene ritical Thinking Questions 8. Add curved arrows showing the movement of electrons in the two reactions above. a) Draw any missing important resonance structures for each of the two carbocation intermediates. (ne from each set is re-drawn for you below.) meta para 2 2 b) In the meta set, none of the three resonance structures stand out as being most important, but in the para set one resonance structure stands out as being most important. xplain this statement and circle the most important res. structure. All three resonance structures in the meta set are secondary. In the para set, the middle resonance structure is tertiary, and therefore a larger contributor to the overall picture of electron distribution. c) f the two products at the top of the page, one forms and the other does not. ircle the product that forms and explain your reasoning. The para product is the only one that forms since the carbocation intermediate on this pathway is lower in potential energy than the carbocation intermediate on the meta pathway. 2

6 hemactivity 29 lectrophilic Aromatic Substitution 6 9. Draw the other two resonance structures of the intermediate that forms if the nitro group adds to the ortho position (as shown in the mechanism below) a) The carbocation above is very close in potential energy to the intermediate in the para set on the previous page. In what way are these two sets of resonance structures similar? Both sets of resonance structures consist of two secondary carbocation resonance structures and one tertiary carbocation resonance structure. b) f the ortho intermediate and the para intermediate, one is slightly lower in potential energy due to steric effects. Which one do you expect to be lower in potential energy? ortho intermediate or para intermediate [circle one]. c) The following energy diagram shows pathways to the ortho, meta, and para products. Label each pathway with the correct name. = meta = ortho = para V (weak base) ortho meta para reaction progress d) When toluene is mixed with nitric acid and sulfuric acid (giving + 2 ) two of the three products in the box above are formed. ircle these two and cross out the other one. The ortho and para products form. The meta product does not form. e) A methyl group, or any other alkyl group () is said to be an ortho & para director. xplain why the group "directs" the nitro to the o or p positions.

7 hemactivity 29 lectrophilic Aromatic Substitution 7 The presence of the alkyl group appears to direct the electrophile to substitute for an either ortho or para to itself (the methyl group in this case). Model 4: Steric vs. lectronic ffects in AS For the reaction below, chemists say that the meta product is T formed because of unfavorable electronic effects. (The electron arrangement in the meta intermediate is not as favorable as in the ortho and para intermediates.) Generally, electronic effects are much more powerful than steric effects, but steric effects can have an impact. With large + like nitro ( + 2 ), the para product is favored over ortho product. 2 2 = ortho (63 %) meta (~3 %) para (34 %) = 3 ortho (12 %) meta (~3 %) para (85 %) 3 ritical Thinking Questions 10. xplain the very strong preference for the para product when = isopropyl (as compared to when = methyl). An isopropyl group is much larger than a methyl group so the former generates much larger steric interactions. It makes sense, therefore, that para would dominate in the case of = isopropyl. 11. "Without steric effects, you would expect the ortho product to be twice as abundant as the para product." xplain this statement (see the int below) ortho (66 %) meta (~0 %) para (33 %) 2 int: What is the difference between this product and the ortho structure in the box at left? = alkyl group xpected atio if there were no Steric ffects There are two different ortho s that can be substituted for, whereas there is only one para. The two ortho products shown above are identical to one another. 12. An alkyl group () is slightly electron donating. This means it donates electron density into an attached aromatic ring. Based on this information, which of the following AS reactions do you expect to be faster? xplain your reasoning. = alkyl group 2 S 4 / 3 (gives + 2 ) 2 2 S 4 / 3 (gives ) int: Which is better for the AS reaction, to have a ring that more is electron rich or more electron poor? 2 +

8 hemactivity 29 lectrophilic Aromatic Substitution 8 In an AS reaction, the ring acts as a nucleophile so the more electron rich the ring, the faster the reaction. For this reason an electron donating group such as an alkyl is expected to speed the rate of AS. Part : esonance Donating and Withdrawing Groups (ow do π donating and withdrawing groups affect rate and placement in AS?) Model 5: Second rder esonance Structures Second order resonance structures are like "assistant resonance structures." They are not as important as regular (first order) resonance structures, but they can tell us some information about the arrangement of electrons in a complicated molecule. itrobenzene Second rder esonance Structures of itrobenzene ritical Thinking Questions 13. xplain why the three second order resonance structures shown in Model 5 are not full fledged resonance structures. That is, why are they "not as important" as a regular, first order resonance structure such as the one at the far left, above? The first order resonance structure above, left, has only two non-zero formal charges. ach of the three second order resonance structures has four non-zero formal charges. In general, the fewer the number of non-zero formal charges the more important the resonance structure. 14. This set of second order resonance structures tells us that the nitro group withdraws electron density into its p orbitals very strongly from certain carbons on the aromatic ring. omplete the composite drawing of nitrobenzene below by placing a + on appropriate carbons in the ring. omposite Drawing of itrobenzene (showing a combination of all first and second order resonance structures) 15. Add curved arrows to nitrobenzene at the top of the page showing how you would change it into one of the second order resonance structures. Then use curved arrows to generate each of the subsequent second order resonance structures.

9 hemactivity 29 lectrophilic Aromatic Substitution In the first step of an electrophilic aromatic substitution (AS) reaction the aromatic ring is acting as a nucleophile and reacting with the electrophile ( + ). Z Step ne Z Base Step Two a) Which is more likely to react with an electrophile, a carbon that is electron rich, or a carbon that is electron poor and holds a d+ charge [circle one]? b) Assume that Z = nitro group and add + to appropriate ring carbons on the structure of the starting material above. (int: see Model 5.) c) onstruct an explanation for why, when "Z" is an electron withdrawing group such as nitro, the meta product is the major product. Z Base major product The electrophile is less attracted to the positions on the ring with a partial positive charge, and more attracted to the other two positions (meta to a resonance electron withdrawing group such as nitro). 17. Any molecule with a lone pair next to the aromatic ring will have second order resonance structures. The first in a set of three for aniline is shown below. a) Add curved arrows to aniline showing how you would change it into the second order resonance structure shown. Then use curved arrows to generate the two missing second order resonance structures. minor products aniline Second rder esonance Structures for Aniline b) In terms of electrons, does a 2 group donate into or withdraw from [circle one] the aromatic ring? c) omplete the composite drawing of aniline below by placing a on appropriate carbons in the ring.

10 hemactivity 29 lectrophilic Aromatic Substitution 10 omposite Drawing of Aniline (showing a combination of all first and second order resonance structures) 18. ecall that, in the first step of an AS reaction the aromatic ring is acting as a nucleophile and reacting with the + charged electrophile ( + ). Z Z Z Base Step ne Step Two a) Assume that Z = an amino group ( 2 ) and add to appropriate ring carbons on the structure of the starting material above. b) onstruct an explanation for why, when Z = a strong electron donating group such as 2, the major products are ortho and para Base major products minor product The electrophile is attracted to positions on the ring with more negative charge. With a resonance donating group such as an amino group, the ortho and para positions have the largest amount of negative charge. 19. To say a group on the ring is a meta director means it directs an + (in the next AS reaction) to add the to the meta position on the ring. omplete the following: a) A strong electron withdrawing group is an ortho/para director or a meta director [circle one]. b) A strong electron donating group (with a lone pair to donate into the ring) is an ortho/para director or a meta director [circle one]. Part D: Summary of Directing ffects (Why are halogens deactivators but ortho/para directors in an AS reaction?) Model 6: Inductive ffects vs. esonance ffects An inductive effect is the donation or withdrawal of electron density through sigma bonds. (As shown in the diagram below, left.) A resonance effect is the donation or withdrawal of electron density as demonstrated by 1 st or 2 nd order resonance structures (as shown below, right).

11 hemactivity 29 lectrophilic Aromatic Substitution 11 xample of an Inductive ffect xample of a esonance ffect l l is more electronegative than, so it steals electron density from the ring. l l esonance donating effects places extra electron density at the ortho and para positions on the ring. ote: A halogen is an inductive withdrawing group and a resonance donating group.

12 hemactivity 29 lectrophilic Aromatic Substitution 12 Table 6: Directing ffects of Various Groups in AS eactions Base Identity of ortho Inductive ffects weak e 2,, 2 withdraw Amines moderate e, Phenols withdraw weak e Alkyl groups 3, 2 3 etc. donation* esonance ffects strong e donation strong e donation weak e donation* para Product egiochemistry ortho & para ortho & para ortho & para meta elative eaction ate very very fast very fast moderatel y fast Benzene 1 alogens I l F strong e withdraw weak e donation ortho & para slow Acyl groups Sulfate group itro group yanide group Ammonium + - strong e ' S - withdraw strong e withdraw strong e withdraw + - strong e withdraw strong e 3 withdraw strong e withdraw meta very slow strong e withdraw meta very slow strong e withdraw meta very slow strong e withdraw meta very slow strong e withdraw meta very slow *lectron donation by an alkyl group can be considered an inductive or a pseudo-resonance effect (see hyperconjugation). yperconjugation cannot be demonstrated with 2 nd order resonance structures. ritical Thinking Questions 20. Summarize how olumn 3 (esonance ffects) is related to olumn 4 (Product egiochemistry)? Those groups with resonance donating effects are o,p directors. Those with resonance withdrawing effects are meta directors. 21. Summarize how olumn 5 (eaction ate elative to Benzene) is related to the overall level of electron withdrawal or donation? (consider both inductive and resonance effects) If a ring is more electron rich than benzene, it undergoes AS faster than benzene and vice versa.

13 hemactivity 29 lectrophilic Aromatic Substitution According to Model 6, are halogens o/p directors or m directors? o/p directors a) alogens are powerful inductive withdrawing groups. This means that a halogen removes electron density from all carbons of the ring. Put a + next to each ring carbon in the structure of bromobenzene drawn above. b) But, a halogen has a lone pair to donate into the ring. This means they are resonance donating groups. n the same drawing of bromobenzene, put a small next to the three carbons that receive electron donation from. c) These two effects (inductive withdrawal and resonance donation) compete with one another, but the inductive withdrawing effects win. Is this consistent with the reaction rate for bromobenzene (on Table 6) relative to plain benzene? xplain. alobenzenes undergo AS more slowly than benzene. This is consistent with the hypothesis that they are less electron rich than benzene, and that the overall affect of the halogen is to withdraw electron density from the ring. d) If you were an electophile ( + ), which carbons on bromobenzene would you be most attracted to? xplain your reasoning. The ortho an para positions are most electron rich and are most likely to react with an electrophile. e) Is your answer in part d) consistent with the data in Table 6? Yes. The halogens are ortho/para directors, but overall deactivators with regard to AS. Information: activating group = group that makes the rate of AS faster than with benzene deactivating group = group that makes the rate of AS slower than with benzene examples of activating groups examples of deactivating groups 2 X S very strong activators FAST weak activator weak deactivator SPD F AS ATI very strong deactivators SLW

14 hemactivity 29 lectrophilic Aromatic Substitution 14 Model 7: AS eactions with Di-Substituted ings /Fe 3 2 /Fe 3 xn A xn B /Fe 3 xn /Fe 3 xn D only trace amounts of products ritical Thinking Questions 23. onsider the starting material in xn A, in Model 7. a) Mark the position/s on the ring where the 3 group would direct an +. b) Mark the position/s on the ring where the 2 group would direct an +. c) Are these two directing effects opposed to one another or in agreement [circle one]? 24. onsider the starting material in xn B, in Model 7. a) Mark the position/s on the ring where the 3 group would direct an +. b) Mark the position/s on the ring where the group would direct an +. c) Are these two directing effects opposed to one another or in agreement [circle one]? d) Label each group on the ring ( and 3 ) as a strong activator, weak activator, weak deactivator or strong deactivator. e) onstruct an explanation for why the product shown is the major product. In this case, the group is a strong activator and the methyl group is a weak activator. As expected, the strong activator wins in terms of directing effects yielding the product shown. 25. onsider the starting material in xn, in Model 7. a) Mark the position/s on the ring where each 3 group would direct an +. b) onstruct an explanation for why the product below is formed in very small amounts compared to the product shown above. 3 Minor Product 3 Sterics direct the bromine to add to an activated position that is T between the two methyl groups. 26. Which reaction/s in Model 7 demonstrate the general rule that Friedel-rafts reactions are extremely slow with a deactivated aromatic ring.

15 hemactivity 29 lectrophilic Aromatic Substitution 15 xn D. ven though the two nitro group direct to the same position, the effect of two strong deactivating groups is to slow the rate of AS to the point where no product is observed.

16 hemactivity 29 lectrophilic Aromatic Substitution 16 xercises for Part A 1. Show the mechanism and most likely products that result from the following reactants. (ote: two weak bases, water and bisulfate ion are also in solution.) S Sulfuric acid with absolutely no water in it is called fuming sulfuric acid and contains small amounts of the powerful electrophile S 3 (one resonance structure is shown below). onstruct a mechanism for the following reaction. int: the final step is an intramolecular atom transfter. S S 3. Draw all possible resonance structures for the carbocation intermediate in Model is formed when nitric acid ( 3 ) and sulfuric acid ( 2 S 4 ) are mixed. Draw the Lewis structure of each and construct a mechanism that explains formation of + 2. int: water and S 4 are the other products formed in this reaction. 5. When toluene is treated with sulfuric and nitric acids under special conditions, three nitro ( 2 ) groups are substituted for hydrogens (at the 2, 4 and 6 positions on the ring). The product is a highly explosive substance commonly known by a three letter name. Draw the structure and write the common name and the chemical name for this explosive substance. 2 S 4 3 toluene common name = chemical name = 6. onstruct a reasonable mechanism for the following reaction called a Friedel-rafts alkylation. ote: when -X and AlX 3 are mixed, you can assume the result is and AlX 4 assume this to be... All 3 (cat.) 2 3 l l All 4

17 hemactivity 29 lectrophilic Aromatic Substitution Draw the mechanism (use curved arrows) and most likely product/s that would result from the following AS reaction called a Friedel-rafts acylation. assume these species are present Al 3 Al 4 8. Draw a mechanism to explain the formation of each of the two Friedel-rafts products. int: think of (and draw) the + group in the X All 3 complex as a carbocation ( + ), then think about possible carbocation rearrangements All 3 3 (cat.) 3 MIXTU of above two products 9. Give an example (not appearing in this hemactivity) of a) an alkyl halide ( X) that will likely undergo rearrangement during a Friedel-rafts alkylation. b) an alkyl halide ( X) that will T undergo rearrangement during a Friedel-rafts alkylation. 10. Shown below are two ways of making the same target product starting from benzene. Synthetic pathway b gives a higher % yield of the desired product. xplain why. a All 3 (cat.) b All 3 l heated in Zn/g amalgum (you are T responsible for this mechanism) 11. ead the assigned pages in your text and do the assigned problems. 12. omplete the mini-activity on Lewis acid catalysts found at the end of this hemactivity.

18 hemactivity 29 lectrophilic Aromatic Substitution 18 xercises for Part B 13. f the choices in brackets, circle the word or phrase that makes the sentence true. a) The pi system of the ring acts as [a nucleophile or an electrophile] in an AS reaction. b) The [more or less] electron rich the pi system of the aromatic ring, the faster the rate of AS. 14. A nitro group is a very powerful electron withdrawing group. Which do you expect will undergo AS reaction faster: benzene or nitrobenzene [circle one], and explain your reasoning. 15. onstruct an explanation for the following finding: ven with the best electrophile ( + ), di-nitro benzene undergoes AS extremely slowly. So slowly that not even trace amounts of product are observed. 2 2 Base very very very slow PDUTS BSVD 16. When a flask containing a mixture of 1 mole of nitrobenzene and 1 mole of toluene is treated with one mole of D-l, deuterium is incorporated into the toluene ring, but not the nitrobenzene ring. xplain. That is, why does the toluene hog all the D-l, while the nitrobenzene does not get any? This notation says that D is on the ring, but does not specify which position on the ring. 1 mole D l 3 2 D D mole toluene 1 mole nitrobenzene observed not observed 17. ead the assigned pages in your text and do the assigned problems. xercises for Part 18. Draw three 2 nd order resonance structures for phenol. phenol a) xplain why the 2 nd order resonance contributors are less important than 1 st order resonance structures, and contribute only a small amount to our

19 hemactivity 29 lectrophilic Aromatic Substitution 19 overall understanding of phenol. b) xplain the following statement: The 2 nd order resonance structures help explain why groups with a lone pair (such as ) activate the ortho and para positions toward electrophilic aromatic substitution (AS). 19. Draw three 2 nd order resonance structures for benzoic acid. benzoic acid a) Based on these 2 nd order resonance structures, do you expect the carboxylic acid group () to be a resonance donating group or a resonance withdrawing group? b) xplain the following statement: The 2 nd order resonance structures in part c) help explain why a carboxylic acid group deactivates the ortho and para positions toward electrophilic aromatic substitution (AS). Information So far we have argued that 2 nd order resonance structures can be used to predict the regiochemistry of an AS reaction. The following questions ask you to consider the potential energies of intermediates on the AS reaction pathways. As with all reactions with a small change in energy between reactant and product, it is the height of the activation barrier that determines which product will form. The potential energy of the intermediate is an excellent approximation (according to the ammond Postulate) of the height of the activation barrier. Simply put The pathway with the most favorable carbocation intermediate will likely dominate. 20. onsider electrophilic aromatic substitution (AS) performed on nitrobenzene Base major product minor products a) ow does the placement of + on the nitrobenzene ring differ from an AS reaction starting with toluene or aminobenzene (aniline)?

20 hemactivity 29 lectrophilic Aromatic Substitution 20 b) Draw the intermediate on the reaction pathway to the major (meta) product. (Be sure to include all important resonance structures.) Base carbocation intermediate meta LY product c) Draw the intermediate on the reaction pathway to the para product. (Be sure to include all important resonance structures.) Base carbocation intermediate para (and ortho) T FMD d) Any resonance structure in which two + charges are next to each other is very unfavorable. ircle all unfavorable resonance structures above. e) onstruct an explanation for why the intermediate on the reaction pathway to the meta product is lowest in potential energy. f) Draw an energy diagram showing all three pathways (ortho, meta and para). g) onstruct an explanation for why the meta product is strongly favored in this reaction over the para (and ortho) product. 21. Aniline gives only ortho and para products in an AS reaction. a) Draw the intermediate on the reaction pathway to the para product. Be sure to draw all FU important resonance structures. 2 2 Base aniline b) Draw the intermediate on the pathway to the meta product. c) onstruct an explanation for why the intermediate on the pathway to the meta product is higher in potential energy than the intermediate on the pathway to the para product.

21 hemactivity 29 lectrophilic Aromatic Substitution 21 d) Draw an energy diagram showing all three pathways (ortho, meta and para). 22. Aniline reacts with a given electrophile 100 times faster than toluene and 1000 faster than benzene. a) xplain why an AS intermediate for aniline such as the one in part a) above is lower in potential energy than the intermediate you drew for parasubstitution of toluene in Model 3. b) onstruct an explanation for why aniline undergoes AS much faster than toluene. 23. ead the assigned pages in your text and do the assigned problems. xercises for Part D 24. onsider the following reactions: I II 2 2 Fel 3 l 2 a l l l Fe 3 2 l 2 b l 2 2 l l c d c and d T formed e f f T formed a) onstruct an explanation for why Product c is not formed in xn I. b) onstruct an explanation for why Product d is not formed in xn I. c) onstruct an explanation for why Product f is not formed in xn II. 25. In the reaction below, two major products are observed. a) Draw them and construct an explanation for why they are formed instead of the other two possibilities. b) Which of the two major products do you expect to dominate and why? 3 l 2 /Fel Mark each of the following statements True or False based on your current understanding. (If false, cite an example of a substituent for which it is false.) a) T or F: A strong activator overpowers the directing effects of a weak activator. b) T or F: A weak activator overpowers the directing effects of a deactivator. c) T or F: All activators are o/p directors.

22 hemactivity 29 lectrophilic Aromatic Substitution 22 d) T or F: All deactivators are m directors. 27. ircle each of the following that help explain why AS with fluorobenzene is slower than AS with phenol (hydroxybenzene)? I. F is more electronegative than, generating stronger inductive effects. II. F holds it lone pairs very tightly, making it a weaker pi donator than. III. F has more lone pairs than. IV. F is smaller in size than. 28. onsider the following reactions: I l 2 Fel 3 l only product II 3 3 l 2 Fel 3 l 65 % 35 % 3 l III l 2 Fel 3 l 37 % 63 % l a) onstruct an explanation for why only the para product is observed in reaction I. b) Fact: eglecting steric effects, the expected ratio of para:ortho is 1:2 for reactions I, II and III. onstruct an explanation for this fact. c) Why is eaction III closest to the expected 1:2 ratio? 29. According to the note at the end of Part D of this hemactivity, which of the following starting materials will yield detectable products in a Friedel-rafts alkylation or acylation. benzene bromobenzene phenol nitrobenzene ortho-chloroaniline 2-bromo-6-chlorophenol ote: for the last two compounds the ring is net activated since 2 and are strong enough activators to overpower even two weak deactivators. 30. ead the assigned pages in your text and do the assigned problems.

23 hemactivity 29 lectrophilic Aromatic Substitution 23 Mini-Activity on Lewis Acid atalysis (What catalyst is needed to generate F + l + + I + and + electrophiles?) Information: Lewis Acid atalysts Any molecule or atom that wants more electrons (lacks electrons) is a Lewis acid. Any molecule or atom that wants to share its electrons (excess e - ) is a Lewis base. A catalyst helps increase the reaction rate, but is not consumed in the reaction. Figure A: ommon Lewis Acids l d+ l d+ d+ l Fe Fe l Al Al d+ l l iron chloride iron bromide aluminum chloride aluminum bromide In each molecule above, the electronegative halogens steal electron density from the central metal, leaving the Fe or Al with a lack of electrons and almost a full + charge. ritical Thinking Questions 31. ircle the most Lewis acidic atom in each molecule in Figure A. Information The electron cloud around bromine is normally symmetrical (below, left). A passing molecule with a dipole moment will polarize this cloud (below, middle) An even stronger effect is generated when a Lewis acid catalyst such as Fe 3 binds to one of the atoms of 2 (below, right) utline of Undisturbed lectron loud of 2 utline of lectron loud of 2 near transient + charge disturbance 2 with partial bond to Fe 3 (outline of electron cloud not shown) + + Fe passing molecule with a small dipole 32. onsider the picture above, right, showing 2 bound to Fe 3. a) Add to this drawing a depiction of the shape of the electron cloud of 2 when it is bound to Fe 3 (above right). Don t include the electron cloud of Fe 3 in your drawing. b) Add + and where appropriate to the 2 portion of this complex (above right).

24 hemactivity 29 lectrophilic Aromatic Substitution onsider the reactions below: I. + II. ATI III. Fe 3 (cat.) a) Why does 2 react with cyclohexene but not with benzene? (see xns I & II, above) b) onstruct a reasonable mechanism for eaction III that shows the role of the catalyst. For simplicity, in your mechanism show Fe 3 as and Fe 3 ote: in the last step Fe 4 _ acts as a base, generating Fe 4, which decomposes into and Fe 3, regenerating the Lewis acid catalyst (as shown below). Fe 3 Fe 3

25 hemactivity 29 lectrophilic Aromatic Substitution 25 Information: Friedel-rafts Alkylation and Acylation ne of the most difficult and important objectives in organic synthesis is the formation of new carbon-carbon sigma bonds. lectrophilic aromatic substitution (AS) is one of the few ways to do this. As with the AS bromination in Model 4, a Lewis acid catalyst (usually All 3 ) is required to make a sufficiently strong electrophile (see 3 + below left). For simplicity, you can think of this methyl electrophile as a methyl carbocation (see below right), though 1 o and methyl carbocations do not exist. + Al Al ote: Al should have a formal - charge but this representation is more accurate. This technique for making carbon-carbon sigma bonds was discovered accidentally by harles Friedel ( ) and James rafts ( ). The two were carrying out reactions involving All 3 using benzene as a solvent (which they erroneously thought be totally inert). Friedel-rafts eactions work with two different types of carbon groups, below designated as. X X Al X = l or X + = alkyl or acyl group X = alkyl group = acyl group 3 methyl 2 3 ethyl 3 isopropyl 3 etc. 3 ethanoyl (acetyl) 2 3 propanoyl etc. 34. What reagents would you use to carry out the following reactions?

26 hemactivity 29 lectrophilic Aromatic Substitution 26