Baixe Manual soluções Paula Bruice Química Orgânica Cap08 e outras Notas de estudo em PDF para Química, somente na Docsity! 240 Chapter 8 Solutions to Problems 1. a. 1,5-cyclooctadiene b. 1-hepten-4-yne c. 4-methyl-1,4-hexadiene d. 5-vinyl-5-octen-1-yne a HC / HC e. 1,6-dimethyl-1,3-cyclohexadiene f. 3-butyn-1-ol g. 1,3,5-heptatriene h. 2,4-dimethyl-4-hexen-1-ol a and e have only two stereoisomers because in each case there are two identical substituents bonded to one of the sp? carbons. He H c=c cH, HC” c=0" MON H H (E)-2-methyl-2,4-hexadiene (Z)-2-methyl-2,4-hexadiene b. HC H HC H Ne=c” = H C=€ CHoCHs H C=c N / N CH,CH; H H (2E,4E)-2,4-heptadiene (2E,4Z)-2,4-heptadiene H H H N N / ,E=€, CH,CH, HC HC C=c” N CcH,CH, H “E (2Z,4E)-2,4-heptadiene (22,47)-2,4-heptadiene e H N / H N / E ,C=€ C=C CH; , N Pá H H C=C / N H H (E)-1,3-pentadiene (Z)-1,3-pentadiene Chapter 8 241 CH,=CHCH,CH=CH, < CH;CH=CHCH=CH, < CH;CH=CHCH=CHCH, < 1,4-pentadiene 1,3-pentadiene 2,4-hexadiene CH; CH, I I CH;C=CHCH=CCH, 2,5-dimethyl-2,4-hexadiene The indicated double bond is the most reactive in an electrophilic substitution reaction because addition of an electrophile to this double bond forms the most stable carbocation (a tertiary allylic carbocation). Br Br Br Br Br Br or CH; CH; CH; H——Br H——Br Br——H CH, Ho Ho CH, cH, CH, H——Br Br——H H——Br cH, cH; cH,; cH; = No stereoisomers are CH=CHCH;CH;CCR; possible for this compound. cl 244 10. 11. 12. Chapter 8 a. Addition at C-1 forms the more stable carbocation, because the positive charge is shared by two secondary allylic carbons. b. DCI was used to cause the 1,2- and 1,4-products to be different. If HCI had been used the 1,2- and 1,4-products would have been the same. a. The rate-determining step is formation of the carbocation b. The product-determining step is reaction of the carbocation with the nucleophile. a. Solved in the text. je = b. CH;CH= CHECHs CHsÇHCH=CCH; ci a Kinetic product thermodynamic product ê CHs CH; ci kinetic product thermodynamic product a d. cl CH=CHCH; CHCHCH,; kinetic product thermodynamic product In order for a Diels-Alder reaction to occur, the overlapping orbitals of the reactants must have the same color (the same symmetry). In a [2+2] cycloaddition reaction at room temperature (in the ground state electronic configuration), the HOMO of one of the reactants will be symmetric and the LUMO of the other will be asymmetric (see Figure 7.8 on p. 287 of the text). Thus, they will not have the same symmetry and the reaction will not occur. In contrast, a [2+2] cycloaddition reaction does occur under photochemical conditions. Under photochemical conditions one of the alkenes will be in an excited state. Therefore, its HOMO will be asymmetric and will be able to overlap with the asymmetric LUMO of the other alkene. Chapter 8 245 13. R à: CCH; e EC AX > GCH; O HC 0 ne HC. cHO b. Or a 2 HC cH,0 14. a. Itis not optically active because it is a meso compound. (It has 2 asymmetric carbons and a plane of symmetry.) CX. ci b. Itis not optically active because it is a racemic mixture. (Identical amounts of the enantiomers will be obtained.) ci cl (+ DO 15. First draw the resonance contributors to determine where the charges are on the reactants. The major product is obtained by joining the negatively charged carbon of the diene with the positively charged carbon of the dienophile. + CHO sz CH,0, ” Cm cao diene - N CH | Il mea ho 0 [a dienophile CH Because the reaction creates an asymmetric o or carbon, the product will be a racemic mixture. 246 16. 17. 18. 19. 20. Chapter 8 3 CH às C=N C=N b. C=N HC C=N (NY. o Sr. CY HC CH; a and d will not react, because they are both locked in an s-trans conformation. cand e will react, because they are both locked in an s-cis conformation. b and f will react, because they can rotate into an s-cis conformation. Solved in the text. R GocH; = H az E N e D + ç e (a i + 1 Cc 9 + é x 1 doc H HH OCH; o o -O no dg 0 H dom . £ do 4 Se” dz A Ng + lí + + II e Ka & Aa H” “CH HOC” H Diethyl ether does not have any 1 bonds. Therefore it does not have a 17* molecular orbital. Without a 7* molecular orbital neither an n — 1* nor a 7 — n* electronic transition can occur. 29. 30. 31. Chapter 8 249 (3E,6E)-3,7,11-trimethyl-1,3,6,10-dodecatetraene The configuration of the double bond at the 1-position and at the 10-position is not specified because isomers are not possible at those positions since there are 2 hydrogens on C-1 and two methyl groups on C-11. The reaction of 1,3-cyclohexadiene with Brz forms 3,4-dibromocyclohexene as the 1,2-addition product and 3,6-dibromocyclohexene as the 1,4-addition product. (Recall that in naming the compounds, the double bond is at the 1,2-position.) The reaction of 1,3-cyclohexadiene with HBr forms 3-bromocyclohexenc as both the 1,2-addition product and the 1,4-addition product. Br Br o -Br C ) Br, 4 3,4-dibromocyclohexene Br É 1,2-addition product | 3,6-dibromocyclohexene 1,4-addition product 1,3-cyclohexadiene ss HBr, o 3-bromocyclohexene 1,2-addition product 1,4-addition product Br CH=CH, CH=CH, Br a. b. cH & d. Br cH; Br CH, CH; 250 32. 33. 34. Chapter 8 a. cm=cHcH=cHcH=CH, By CHsCHCH=CHCH=CH, + A 1,3,5-hexatriene Br CH;CH=CHÇHCH=CH, + CH;CH=CHCH=CHCH,Br B Br c b. A will predominate if the reaction is under kinetic control because it is the 1,2-product and therefore will be the product formed most rapdily as a result of the proximity effect. Tn addition, A will be the 1,2-product regardless of which end of the conjugate system reacts with the electrophile. e. C will predominate if the reaction is under thermodynamic control because it is the most stable diene. (It is the most substituted conjugated diene.) a. The absorption at 236 nm is due to a 7 n* transition, and the absorption at 314 nm is due to an n> T* transition. s The one at 236 nm (the 1 — n* transition) shows the greater absorbance. O 1 o GHCCHs Il CcH. o (IT Co,CcH. GHCO,CH; ara CH, —"— 2 a CHE: O 0 8.6 ' o — (UM Õ O a. e. c. RO 35. 36. oH a. 1. HO + Br, ——» Br CH; Br. ,CH; 2 À | + HBr —» or Br. CH; + HBr —>» to , MN CH, H,SO, + HO ——» or CH; HO CH; CH; cH; s H,SO, | + HO ——» CHa HO cH; b. CH, 3 HO Br et AD » The diene is the nucleophile, and the dienophile is the electrophile in a Diels-Alder reaction. Chapter 8 HO CH; CH; 251 a. An electron-donating substituent in the diene would increase the rate of the reaction, because electron donation would increase its nucleophilicity. b. An electron-donating substituent in the dienophile would decrease the rate of the reaction, because electron donation would decrease its electrophilicity. e. An electron-withdrawing substituent in the diene would decrease the rate of the reaction, because electron withdrawal would decrease its nucleophilicity. 254 Chapter 8 o ll b. CH,=CHCH=CHCH=CH, and CH,=CHCH=CHCCH; It will show 1 absorption band. It will show 2 absorption bands. an—> nº transition a n—> 1º transition and * a an n>> 7º transition o e. and o The Amax Will be at a longer wavelength because the carbonyl group is conjugated with the benzene ring. 0H OCH; and The Amax Of the phenolate ion will be at a longer The Amax is PH independent. wavelength than the Amax Of phenol. Since the pK, of phenol is -10, the Amax Will be at a longer wavelength at pH =11 than at pH = 7. 40. — The first pair is the preferred set of reagents because it has the more nucleophilic diene and the more electrophilic dienophile. ? o CCH; Im Í H CcH; A c e > + or No + " c e Z H H Chapter 8 255 41. A Diels-Alder reaction is a reaction between a nucleophilic diene and an electrophilic dienophile. a. The compound shown below is more reactive in both 1 and 2, because electron delocalization increases the electrophilicity of the dienophile. f 2. + CH,=CHCH =» CH,CH=CH b. The compound shown below is more reactive, because electron delocalization increases the nucleophilicity of the diene. - + CH,=CHCH=CHOCH;, <—>» CH,CH=CHCH=0CH,; 42. H ) ) H H — H exo endo 43. en HH a. nas . Cc o: O HÉcem, dh Ii €= O Cla , b. É . Ed ã A + | o HH 256 Chapter 8 44. B a Br r. A OrS Br OS . OO e, (Oem Br Br A B c b. A has two asymmetric carbons but only two stereoisomers are obtained because addition of Bry can occur only in an anti fashion. Br Br N CIO + " Br “Br B has four stereoisomers because it has an asymmetric carbon and a double bond that can be in either the E or Z configuration. ? "CH,Br 2 CH,Br Br de | iBT + “g q C has two stereoisomers because it has one asymmetric carbon. CH,Br CH,Br H Br CH;Br 48. 49. so. 51. 52. 53. Chapter 8 259 His recrystallization was not successful. Because maleic anhydride is a dienophile, it reacts with cyclopentadiene in a Diels-Alder reaction. O O O Op — Seo dr o endo O exo o 0 Since only benzene absorbs light of 260 nm, the concentration of benzene can be determined by measuring the amount of absorbance at 260 nm, using the Beer-Lambert law, since the length of the light path of the cell is known and the molar absorptivity of benzene at 260 nm can be looked up (or can be determined from the absorbance at 260 nm given by a solution of benzene with a known concentration). the Beer-Lambertlaw: A =cle High temperatures are required in order to break the bonds formed by the overlapping in-phase orbitals. The Dicls-Alder product will not reform at high temperatures, because a [4+2] cycloaddition will not occur unless both of the reactants are in their ground states. Maleic anhydride reacts with cyclopentadiene as in the above problem. The function of maleic acid in this reaction is to remove the cyclopentadiene, since removal of a product drives the equilibrium toward products. (See Le Chãtelier's principle on page 373 of the text.) The bridgehead carbon cannot have the 120º bond angle required for the sp? hybridized carbon of a double bond. With a 120º bond angle, the compound would be too strained to exist. a. Unless the reaction is being carried out under kinetic control, the amount of product obtained is not dependent on the rate at which the product is formed, so the relative amounts of products obtained will not tell you which product was formed faster. b. In a thermodynamically controlled reaction, the product distribution depends on the realative stabilites of the products since the products come to equilibrium. Thus if the distribution of products that is obtained does not reflect the relative stabilities of the products, the reaction must have been kinetically controlled. 260 Chapter 8 54. He should follow his friend”s advice. If he uses 2-methyl-1-3,cyclohexadiene, the product that is formed faster will be 3-chloro-3-methylcyclohexene both if the proximity effect controls which product is formed faster and if the more stable transition state controls which product is formed faster. Thus the experiment will not be able to differentiate between the two. cl CH; GO am Of 3-chloro-3-methylcyclohexene If he follows his friend's advice and uses I-methyl-1-3,cyclohexadiene, the product that is formed faster will be 3-chloro-1-methylcyclohexene only if the proximity effect controls which product is formed faster. The product will be 3-chloro-3-methylcyclohexene if the more stable transition state controls which product is formed faster. CHs ai CH; CH; O HCl + cl 3-chloro-1-methylcyclohexene | 3-chloro-3-methylcyclohexene 55. a. AtpH<4, the lone pair of the nitrogen of the N(CH3)2 group is protonated, so it cannot interfere with the fully conjugated system. CH; + H5C—N OL N= pH< 4 Pp fo Nat o Chapter 8 261 At pH >4, the nitrogen of the N(CH3), group is not protonated and the lone pair can be delocalized into the benzene ring. This decreases the conjugation and, therefore, light of shorter wavelengths will be absorbed; hence, the change in color from red to yellow. EE H;C-N q SN—N. o” pH> 4 / 4 —o” Nat o b. In acidic solutions, the three benzene rings are isolated from one another. In basic solutions, as a result of loss of the proton from one of the OH groups, there is a greater degree of conjugation. oH