EAS+for+Org+210+2008[1] - 25D ChamActlvlty 29 Electrophillc...

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Unformatted text preview: 25D ChamActlvlty 29 Electrophillc Aromatic Substitution ghemAgtivitx 29 Part A: Electrophilic Aromatic Substitution (What products are formed when a strong electrophile is added to benzene?) Model 1: (review) Electrophilic Addition of HG! H (:1 Rm 1 H-‘- C] —.—=-— -" cyclohexene carbocation intermediate H . Cl Rm 2 H— Cl fa" 96 benzene This product ; carbocation intermediate DOES NOT form! : Critical Thinking Questions 1. For Rxn 1 (above) draw curved arrows showing the mechanism of electrophilic addition of HCl. Include an appropriate carbocation intermediate in the box above. Fiure 1: Reaction Diagrams for Electrophilic Addition of HCI V.E. (Potential Energy) Reaction Progress (Rxn i) Reaction Progress (Rain 2) 2. Km 1 is slightly down-hill in terms of energy. Rxn 2 is very up—hill in terms of energy (see Figure 1). Construct an explanation for the large difference in energy between the reactants and the product in Rxnl. 3. Draw the carbocation that would form in Rm 2. Explain why this carbocation goes back to the starting material (H—Cl and benzene) instead of forming the product. ChemActivity 29 Electrophlllc Aromatic Substitution 251 Model 2: Electrophilic Aromatic Substitution I Recall that deuterium (D) has nearly identical reactivity as hydrogen (H). o In the following reaction D—Cl reacts the same way as would H—Cl. - When benzene is treated with D-Cl, an H is replaced with a D. (With excess D-Cl this continues until all the H's have been replaced and the product is C5D6. T o—c1 'l3 H H ' \c/c“c/ “\C/Cs‘c/ H/ \C¢ \H H/C\C¢C\ | I H H Critical Thinking Questions 4. Use curved arrows to show a reasonable mechanism for the reaction in Model 2. (Hint: the first step is formation of a carbocation intermediate, just as in Rxn 2.) 5. The energy diagram for Rxn 2 (using DCl in place of HCl) is shown below using a dotted line. On this same set of axes, draw a solid line to show an energy diagram for Rxn 3 in Model 2. (Note: Rxns 2 and 3 have the same carbocation intermediate.) intermediate for Rxns 2 & 3' ---- - - Reaction Progress (Rxn 2 with DC]) — Reaction Progress (Rxn 3) 6. Construct an explanation for why Rxn 3 is much more likely to occur than Rxn 2. (Note: Rxn 2 does not occur under normal circumstances.) 252 ChemActivity 29 Electrophilic Aromatic Substitution 7. Draw a generalized mechanism for EAS (Electrophilic Aromatic Substitution) showing the substitution of an electrophile (F) for one of the PPS of benzene in the presence of a mild base (B). (Note: in Rxn 3, E+ ='D+) ' ' H E E9 :B —-——n— O H—B‘9 Information: Not All Electrophiles are Created Equal - A very strong electrophile (E) is required to break up an aromatic ring. 0 The H of a strong acid will work since it exists essentially as H+. Know the following strong acids. ' ‘- ‘ 5+ 0 0 5+ 5— 5+ 3- 5+ 5* “\OI_ g... OH [3+ ills) H— C] H—Br H—I 5_ \0/ \ds g ' 8— hydtochloric acid hydrobromic acid hydroiodic acid SUlfuric acid I nitric acid 0 Suitable electrophiles (F) for EAS (Electmphilic Aromatic Substitution) include- an H‘L or D+ donated by a strongxacid, or the electrophiles (E4) listed below. BIZ and FeBl‘z, or Cl; and FeCl; other aromatic 'The last three rows of elecuophiles are generated only in the presence of a Lewis acid catalyst such as Fax; or ADC} A supplementary activity on Lewis acid catalysts is assigned for homework. .‘ mun—‘rm-i-Iiwfidl ,wc;a.am..n_,.l..m____acan. u. ChemActivlty 29 Electrophllic Aromatic Substitution 259 Table 6: Directing Effects of Various R Groups in EAS Reactions R R R R 1-:9 ' E =-'"_"-'-'— ortho para ‘ meta . Base E E Inductive Resonance Product Relative Effects Effects Regio— Reaction chemis Rate strong 6” ortho & very very donation para fast -- to moderate 6' strong e‘ ortho & Phenols fiH’ _2R withdraw donation weak e‘ weak e‘ moderatel Alkyl groups —CH3, 43H2CH3 etc. donation* donation* ara fast ———— H‘hsm strong e' weak e' ortho & _ _'1': _..].3.1~: _Z;.1: _":_ withdraw donation para slow on u 0'. - 00 strong e‘ strong e‘ " withdraw withdraw meta very slow strong e’ meta very slow withdraw very slow —-fl Cyanide group withdraw withdraw meta v slow a strong ei strong e' —NR3 withdraw withdraw very slow AITIITDnium ‘Eleco-on donation by on alkyi group can be considered an inductive or a pseudo-resonance effect (see hyperconjugation). Hyperconjagarion cannot be demonstrated with 2’"f order resonance structures. weak e' withdraw Sulfate > u - Critical Thinking Questions 20. Summarize how Column 3 (Resonance Effects) is related to Column 4 (Product Regiochemistry)? 21. Summarize how Column 5 (Reaction Rate Relative to Benzene) is related to the overall level of electron withdrawal or donation? (consider both inductive and resonance effects) 280 ChernAetivity 29 Electrophilic Aromatic Substitution 22. According to Model 6, are halogens o/p directors or m directors? .0 :Br 3 a) Halogens are powerful inductive withdrawing groups. This means that a halogen removes electron density from all carbons of the ring. Put a 6+ next to each ring carbon in the structure of bromobenzene drawn above. in) But, a halogen has a lone pair to donate into the ring. This means they are resonance donating groups. On the same drawing of bromobenzene, put a small 6-— next to the three carbons that receive electron donation from Br. c) These two effects (inductive withdrawal and resonance donation) compete with one another, but the inductive withdran effects win. Is this consistent with the reaction rate for bromobenzene (on Table 6) relative to plain benzene? Explain. d) lfyou were an electophile (E), which carbon; on bromobenzene would you be most attracted to? Explain your reasoning. o) Is your answer in part d) consistent with the data in Table 6‘?’ Information: . . r activating group = group that makes the rate of EAS faster than with benzene deactivating group = group that makes the rate of EAS sloWer than with benzene examples of deactivating groups 0 n ,c‘ . t ‘ NH2 OH on HN O‘C’OR o‘c’o” 0c“ $03H on N02 neoehoeho e honor?“ v stron activators weak weak v stron deactivators my 8 activator deactivate-r m g FAST W SLOW SPEED OF EAS REACTION examples of activating groups I -unlit”hwyihmiMWJhMIQfi-irweihdfl4' " " manhole-ii Loki-4W 44.4,- u , ChomActlvity 29 Eiectrophilic Aromatic Substitution 261 Model 7: EAS Reactions with Di-Substituted Rings CH3 ., . CH3 Critical Thinking Questions 23. Consider the Starting material in Rxn A, in Model 7. a) Mark the position/s on the ring where the CH3 group would direct an 13". 13) Mark the positionfs on the ring where the N02 group would direct an 13+. c) Are these two directing effects opposed to one another Q1; in agreement [circle one]? 24. Consider the starting material in km B, in Model 7. a) Mark the position/s on the ring where the CH3 group would direct an E”. b) Mark the position/s on the ring where the OH group would direct an E". o) Are these two directing effects opposed to one another or in agreement [circle one]? ' d) Label each group on the ring (OH and CH3) as a strong activator, weak activator, weak deactivator or strong deactivator. e) Construct an explanation for why the product shown is the major product. 25. Consider the starting material in Rm C, in Model 7. a) Mark the positionfs on the ring where each CH3 group would direct an 15+. b) Construct an explanation for why the product below is formed in very small amounts compared to the product shown above. CH3 Minor Product Br CH3 26. Which reaction/s in Model 7 demonstrate the general rule that Friedel-Crafts reactions are extremely slow with a deactivated aromatic ring. 262 ChemActivity 29 Electrophllic Aromatic Substitution Exercises for Part A 1. Show the mechanism and most likely products that result from the following reactants. (Note: two weak bases, water and bisulfate ion are also in solution.) as e u 0 N 0 . so?J H20 2. Sulfuric acid with absolutely no water in it is called fuming sulfuric acid and contains small amounts of the powerful electrophile 803 (one resonance structure is shown below). Construct a mechanism for the following reaction. Hint: the final step is an intramolecular H atom transtter. ' O\\ /0H 0 i s 6,n \\ @o/S‘b 0 3. Draw all possible resonance structures for the carbocation intermediate in Model 2. 4. +NO2 is formed when nitric acid (HN03) and sulfuric acid (H2804) are mixed. Draw the Lewis structure of each and construct a mechanism that explains formation of +N02. Hint: water and HSO4‘ are the other products formed in this reaction. - 5. When toluene is treated with sulfuric and nitric acids under special conditions, three nitro (N02) groups are substituted for hydrogens (at the 2, 4 and'é 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. st04 HNO3 toluene common name = chemical name = 6. Construct a reasonable mechanism for the following reaction called a Friedel-Crafis alkylation. 9 Note: when R-X and A1X3 are mixed, you can assume the result is R9 and A1X4 ‘ assume this to be... . CHcha AlCl3(cat.) " - . ‘ Cl-CHZCH3 ""“ ' H—CI i ChemActivity 29 Electrophlllc Aromatle Substitution 263 7. Draw the mechanism (use curved arrows) and most likely product/s that would result from the following EAS reaction called a Fn'edel-Crafts acylation. assume these 5 ; ect'es are present ll 8. Draw a mechanism to explain the formation of each of the two Friedel-Crafis I products. Hint: think of (and draw) the R6+ group in the R—X—AlClg complex as a carbocation (R+), then think about possible carbocation rearrangements. ' CH3 - I . CHZCHZCHa CH\ O Br—CHZCHchs A1C13 (cat.) 0 0/ CH3 MIXTURE of above two products H_Br 9. Give an example (not appearing in this ChemActivity) of. . . a) an alkyl halide (R—X) that will likely undergo rearranganent during a Friedel-Crafis alkylation. b) an alkyl halide (R—X) that will m undergo rearrangement during a Friedel—Crafls 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. Explain why. Br\)\ 3 MCla Scat! Br HCl heated in 0 Mg amalgum b ._.___... AICIS (you are NOT responsible O for this mechanism) 11. Read the assigned pages in your text and do the assigned problems. 12. Complete the mini-activity on Lewis acid catalysts found at the end of this ‘ ChemActivity. ChemActivity 29 Electrophlllc Aromatic Substitution 267 22. Aniline reacts with a given electrophile 100 times faster than toluene and 1000 faster than benzene. a) Explain why an EAS intermediate for aniline such as the one in part a) above is lower in potential energy than the intermediate you drew for pam- substitution of toluene in Model 3. b) Construct an explanation for why aniline undergoes EAS much faster than toluene. 23. Read the assigned pages in your text and do the assigned problems. Exercises for Part D 24. Consider the following reactions: N02 N02 N02 FeC13 Ci2 c; I '.- ‘ a . C‘ b c and d NOT formed Cl ' 0, c1 Br n FeBr3 Brz - e Br , fNOT formed a) Construct an explanation for why Product I: is not formed in Rxn I. b) Construct an explanation for why Product (I is not formed in Rim 1. c) Construct an explanation for why Product 1‘ is not formed in Rxn ll 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. in) Which of the two major products do you expect to dominate and why? on3 ClleeCl3 O¢C\OH 26. 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 over-powers the directing effects of a deactivator. c) T or F: All activators are 0/}: directors. d) T or P: All deactivators are m directors. 268 ChemAetivity 29 Electrophlllc Aromatic Substitution 27. Circle m of the following that help explain why EAS with fluorobenzene is slower than EAS with phenol (hydroxybenzene)? - I. F is more electronegative than 0, generating stronger inductive effects. II. F holds it lone pairs very tightly, making it a Weaker pi donator than 0. III. F has more lone pairs than 0. IV. F is smaller in size than 0. 28. Consider the following reactions: I (:12 __F§_CL Cl only product OCH; OCH3 OCH3 F603 11 C12 CI Cl _ 6§ % 35 % m U U- 0: Cl Cl 37 % 63 % .1) Construct an explanation for why o_nly the para product is observed in reaction I. b) Fact: Neglecting steric effects, the expected ratio of paramrtho is 1:2 for reactions 1, II and III. Construct an explanation for this fact. c) Why is Reaction III closest to the expected 1:2 ratio? . . 29. According to the note at the end of Part D of this ChemActivity, which of the following starting materials will yield detectable products in a Friedel—Craf‘ts alkylation or acylation. benzene bromobenzene phenol nitrobenzene onho-cltloroaniline 2-bromo-6-chlorophenol Note: for the last two compounds the ring is net activated since NH; and OH are strong enough activators to overpower even two weak deactivators. 30. Read the assigned pages in your text and do the assigned problems. womanw A,..,..‘.l :_._.... .... g .. on“. . . ChemAetlvity 29 Electrophllic Aromatic Substitution 269 Mini-Activity on Lewis Acid Catalysis (What catalyst is needed to generate 1=+ (:1+ BF r and C+ electrophiles?) Information: Lewis Acid" Catalysts - 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 6‘) is a Lewis base. - A catalyst helps increase the reaction rate, but is not consumed in the reaction. Figure A: Common Lewis Acids c1 5‘ 13:5" c15' Bra— 5— l | a | 3. I Cl——-ll=e 5+ Bit—735+ (El—Tl 5+ Br T15+ c1 5‘ ' 13:5" c1 3“ Bra— iron chloride iron bromide aluminum chlon'dc aluminum bromide In each molecule above, the electronegative halogens steal electron density from the central metal, leaving the Fe or A1 with a lack of electrons and almost a full + charge. Critical Thinking Questions 31. Circle the most Lewis acidic atom in each molecule in Figure A. Information 0 The electron cloud around bromine is normally syrmnetxical (below, lefi). :- A passing molecule with a dipole moment will polarize this cloud (below, middle) 0 An even stronger effect is generated when a Lewis acid catalyst such as FeBr; binds to one of the Br atoms of Brz (below, right) Outline of Electron Cloud of Br; Br; with partial bond to FeBr3 Outline (It-undisturbed near transient + charge disturbance (outline of electron cloud not shown) Electron Cloud of Br2 Br 9 l @ Err—"'“Br' ' ‘ 'Fle—Br 3} Br passing molecule with a small dipole 32. Consider the picture above, right, showing Br; bound to FeBrg. a) Add to this drawing a depiction of the shape of the electron cloud of BIZ when it is bound to FeBr; (above right). Don’t include the electron cloud of FeBr3 in your drawing. b) Add 5+ and 8— where appropriate to the Brz portion of this complex (above right). 270 ChemActivity 29 Electrophlllc Aromatic Substitution 33. Consider the reactions below: g: :' Br I Br—Br .._u... + 3' II. 31—13: —-..—-=== N0 REACTION H - Br FeBr3 (cat) III- fir-"Br ——-—-—... H—Br . 3) Why does Br; react with cyclohexene but not with benzene? (see Rxns I & II, above) _ b) Constructs. reasonable mechanism for Reaction III that shows the role of the catalyst. For simplicity, in your mechanism show... _ ar- - - -Br- - - FeBr3 as (‘9 Br and BrFegra Note: in the last step FeBrr acts as a base, generating FeBmH, which decomposes into HBr and FeBrg, regenerating the Lewis acid catalyst (as shown below). “2.4.4 5‘ ;‘.;_-.-...-.~ mm. '..;..'s..,.. : lu.‘ G) E; Fgra —"""—'—” H FeBra 0U o ChemActivity 29 Electrophilic Aromatic Substitution 271 Information: Friedel-Crafts Alkylation and Acyiation 0 One of the most difficult and important objectives in organic synthesis is the formation of new carbon-carbon sigma bonds. 0 Electrophilic aromatic substitution (EAS) is one of the few ways to do this. 0 As with the EAS bromination in Model 4, a Lewis acid catalyst (usually AlC13) is required to make a sufficiently strong electrophile (see ch“ below left). 0 For simplicity, you can think of this methyl electrophile as a methyl carbocation (see below right), though 1° and methyl caz’oocations do not exist. H 3131': l Note: Al should é; u have a formal - H—— C—‘—Br' ' ‘ ' N—'—Br: N ' charge but this I 5"” " ‘ " N representation is H '3‘: more accurate. . This technique for making carbon—carbon sigma bonds was discovered accidentally by Charles Friedel (1832-1899) and James Crafis (1839—1917). The two were carrying out reactions involving A1C13 using benzene as a solvent (which they erroneously thought be totally inert). Fn'edel—Crafts Reactions work with two different types of carbon groups, below designated as R. ‘i‘ X=C101Br .OI—N-_ —-fl, .3 I g Rialkylorocylgroup so - 0 0 II H m. §’C‘*C:H3 §ZCNCHZCH3 ethauoyl (acetyl) propanoyl 34. What reagents would you use to carry out the following reactions? 0 ...
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EAS+for+Org+210+2008[1] - 25D ChamActlvlty 29 Electrophillc...

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