Baixe Manual soluções Paula Bruice Química Orgânica Cap24 e outras Notas de estudo em PDF para Química, somente na Docsity! Chapter 24 707 Solutions to Problems 1. aHt, Ea, Ast, AG, krate (These are the parameters that measure the difference in energy between the reactant and the transition state.) 2. Note that (1) and (2) have only the first phase (because the final product of the reaction is a tetrahedral intermediate), (3) has only the second phase (because the initial reactant is a tetrahedral intermediate) and (4) has both the first and second phases. First Phase 7 NA, ein qn ça R—C-ocH, CÊ rC-ocH, > R—Ç—0CHs om R=G—OCH; Dão do acid-catalyzed ester hydrolysis OH OH+ HB 0: A e QH OH OH +p+ (1) R-C—R EB pgC-R Ho A E 2 R—C—R 50: E . ” +0H OH + HB No QE) 6: (A OH on om Il H-B* | (2) R-C—-H R—C-H — > R—>C-H —» R-C-H ROH ! B I i * +OR OR + HB H a E O: 0H oH 0H l dos Sh Í (4) R—C—NHCH; —>» R—C—NHCH, R—C—NHCH; —> R—C—NHCH, l O a. similarities: the first step is protonation of the carbonyl compound, the second step is attack of a nucleophile on the protonated carbonyl compound, and the third step is loss of a proton. b. differences: the carbonyl compound that is used as the starting material, and the nucleophile that is used in (2) is an alcohol rather than water. 708 4. Chapter 24 Second Phase SH sã ON dA (50H +0H Ls lH N : R=C—ÓCH; É R-CÓCH; — R—C—ocH, + R-É-ocH, OH 0H + HB* acid-catalyzed ester hydrolysis io H de q OH +OCH OcH. E dd 4 Soo Taio 9H N (3) R=C-H HE REICH —»> R—C-H dn (E e R—C-H OCH; GocH, RN +OCH; OCH; CN + HB E ac SE Es (4) R=C—NHCE; > R=CTNECH, —» R-C-oH “B» p-C-oH oH o + HB' a. similarities: the first step is protonation of the tetrahedral intermediate, and the second step is elimination of a group from the tetrahedral intermediate. In two of the three reactions, the third step is loss of a proton. b. differences: the nature of the group that is eliminated from the tetrahedral intermediate. In acetal formation, the third step is not loss of a proton because the intermediate does not have a proton to lose. Instead, the third step is attack of a nucleophile, and the fourth step is loss of a proton. HB* CH, Solved in the text. 13. 14. Chapter 24 qm The tetrahedral intermediate has two leaving groups, a carboxylate ion and a phenolate ion. The carboxylate ion is a weaker base (a better leaving group) than the phenolate ion, so the tetrahedral intermediate re-forms A. The 2,4-dinitrophenoxide ion is a weaker base (better leaving group) than the carboxylate ion, so the tetrahedral intermediate forms B. Tf the ortho- carboxyl substituent acts as an intramolecular general-base catalyst, 180 would not be incorporated into salicylic acid. q á cp cH;CO ——» cHe ——» CHÇTO 180H 180H c Cc HO” N -07W So “o AE [od Sa -0N salicylate If the ortho-carboxyl substituent acts as an intramolecular nucleophilic catalyst, 180 would be incorporated into salicylic acid. es CH;Co 8 ——» + Cc H Ao mo % ' era a is | H HO: HO Ho 8 E c = crio o -” So ur “o St est 1 + cHço- 712 Chapter 24 Not all the salicylic acid would contain 180, because the anhydride intermediate can be hydrolyzed in two different ways. “42 o vo ) :O— Õ 18 “QrgH CHÇOH H Õ 15. —Ser-Ala-Phe would be more readily cleaved by carboxypeptidase A because the phenyl substituent of phenylalanine would be more attracted to the hydrophobic pocket of the enzyme than would the negatively charged substituent of aspartate. 16. Glu 270 attacks the carbonyl group of the ester, forming a tetrahedral intermediate. Collapse of the tetrahedral intermediate is most likely catalyzed by a general-acid group of the enzyme in order to increase the leaving ability of the RO group. (Perhaps the HO substituent of tyrosine is close enough in the esterase to act as the catalyst.) The group that donates the proton can then act as a general-base catalyst to remove a proton from water as it hydrolyzes the anhydride. a RO c Glu270-C—O” ce Glu 270— Glu 270-C—0" [4 R no, Chapter 24 713 17. Because arginine extends farther into the binding pocket, it must be the one that forms direct hydrogen bonds. Lysine, which is shorter, needs the mediation of a water molecule in order to engage in bond formation with aspartate. arginine lysine GHo cH, cH, cH, CH, Go NH CH, de No | > H71>H ses H. Hg É + - É N N HOCH y nº Og NE 9 HOCH, 8d , Seo! 4 - - x q O-ô q D-8 18. | The side chains of D-Arg and D-Lys are not positioned to bind correctly at the active site. 19. NAM would contain “O because it is the ring that would undergo nucleophilic attack by H"o. à» o É o NAM ( NAG NAM HH HO E ;-º RR NAG 20. Lemon juice contains citric acid. Some of the side chains of an enzyme will become protonated in an acidic solution. This will change the charge of the group (for example a negatively charged aspartate, when protonated, becomes neutral), and because the shape of an enzyme is determined by the interaction of the side chains, changing the charges of the side chains will cause the enzyme to undergo a conformational change that leads to denaturation. When the enzyme is denatured, it loses its ability to catalyze the reaction that causes apples to tum brown. 21. In the absence of imine formation, D-fructose isomerizes to D-glucose and D-mannose (an equilibrium is set up between D-fructose, D-glucose, and D-mannose) because a new asymmetric carbon is formed at C-2 and it can have either the R or the S configuration. Enzyme catalyzed reactions are enantioselective-—the enzyme catalyzes the formation of a single enantiomer of a pair. Thus D-fructose isomerizes only to D-glucose. 716 29. 30. 31. Chapter 24 In order to hydrolyze an amide, the NH2 group in the tetrahedral intermediate has to leave in preference to the less basic OH group. This can happen if the NH2 group is protonated because +NHg3 is a weaker base and, therefore, easier to eliminate than OH. Of the four compounds, two have substituents that can protonate the NH by acting as general-acid catalysts, ortho-carboxy- benzamide and ortho-hydroxybenzamide. R R R R 4 prêntm ad ad (prt coH cH oH ! y ortho-hydroxybenzamide ortho-carboxybenzamide Because the carboxy group is electron-withdrawing and the phenolic OH group is electron- donating, formation of the tetrahedral intermediate will be faster for o-carboxybenzamide. Because the carboxy group is a stronger acid than the phenolic OH group, the tetrahedral intermediate of o-carboxybenzamide will collapse to products faster. Therefore, the o-carboxybenzamide has the faster rate of hydrolysis. :0H :0H as a C—0oH NH, O NH, o LAN a. CH;CH,SCH,CH, CI intramolecular nucleophilic catalysis Hop oH b. A intramolecular general-acid catalysis c ço The OH substituent is protonating the 0H leaving group as it departs, causing it to be a weaker base and, thus, a better leaving group. If the ortho-carboxy substituent is acting as a general-base catalyst, the kinetic isotope effect will be greater than 1.0 because an OH (or OD) bond is broken in the slow step of the reaction. 7 cH,Co Chapter 24 717 If the ortho-carboxyl substituent is acting as a nucleophilic catalyst, the kinetic isotope effect will be about 1.0 because an OH (or OD) bond is not broken in the slow step of the reaction. f cH;C 32. Because the catalytic group is a general-acid catalyst, it will be active in its acidic form and inactive in its basic form. The pH at the midpoint of the curve corresponds to the pKa of the group responsible for the catalysis. = o Activity 5.6 pH 33. Co?+ can catalyze the hydrolysis reaction by complexing with three nitrogen atoms in the substrate as well as with water. Complexation increases the acidity of water, thereby providing a stronger nucleophile for the hydrolysis reaction. Complexation with the substrate locks the nucleophile into the correct position for attack on the carbonyl carbon. ate s, =, “CNH SC—NH A, N 5+ + ô+ + E + R$N----Ço ---"NG] H,O Pos pi + 718 Chapter 24 34. me GHOPOs CH,OPOS = VS STO, 2n?* Es clix Zn? HOLC—H º doll H-cZodu 8 Nem H-é-oH i E —c= : be,oro? H-ç=0 no Sen, Tyr H-C—0H CH,OPOZ 35. ca PN oH A . ar cm-NiÃ oO —> cm- GS CH5—N: Chapter 24 mm OG OM CH, g CH :0 2 à Ho | «Ox | CH, Fo HO em, + ad HC—N É | CH HC” “cH HC A 3 B 3 39. a. The cis reactants each undergo a direct SN2 reaction. Because the acetate displaces the tosyl group by backside attack, each cis reactant forms a trans product. b. The acetate group in a trans reactant is positioned to be able to displace the tosyl leaving group by an intramolccular Sw2 reaction. Acetate ion then attacks in a second SN2 reaction from the backside of the group it displaces, so trans products are formed. Because both trans reactants form the same intermediate, they both will form the same product. Because the acetate ion can attack either of the carbons in the intermediate equally as easily, a racemic mixture will be formed. o Ii..- CH;CO: CH; ÇHs C=0 = ELO s SOw H —» —» <H o + o H Na H (6d ô+0 ,D8+ 9 H H 9 Go 0=ç Y c=0 0=C I i CH; CH; 1 cH cH CH; 3 3 e. The trans reactant is more reactive because the tosyl leaving group is displaced in an intramolecular reaction, forming a positively charged cis intermediate that is considerably more reactive than the neutral cis isomer. 722 40. 41. Chapter 24 A possible mechanism for hydrolysis of the phosphodiester is for Arg to neutralize the negative charge on the phosphate group so a negatively charged nuclcophile can approach it. (See Figure 27.5 on page 1117 of the text.) Ca?+ increases the acidity of water, forming metal-bound hydroxide ion that is a stronger nucleophile than water. Glu can function as a general-acid catalyst, donating a proton to the leaving group, thereby decreasing its basicity and making it a better leaving group. [o CaZ+ 1 1 E o + + a HO! o HO Go => G-dH+H! —> p/o = Psp) o RO7| or RO? | “OR ARS o) o ROTOR q "0 H-0—C—CH,CH,-Glu “MHz H,NCNHCH,CH,CH,-Arg Reduction of the imine linkage with sodium borohydride causes fructose to become permanently attached to the enzyme because the hydrolyzable imine bond has been lost. Acid-catalyzed hydrolysis removes the phosphate groups and hydrolyzes the peptide bonds, so the radioactive fragment that is isolated after hydrolysis is the lysine residue (covalently attached to fructose) of the enzyme that originally formed the imine. CH,OPOS. NH | Hsoros dn RS NaBH, Ri HO-Ç—H ç=o 5 moça G=0 H-Ç—-0-H NH H-C—0-H NH H-C—0H H-C—0H | moro” CH,OPO?” fio CH,0H +NH; Hé-nH—(cH—CH Ho-C-H G=0 H—C—O—H oH H-é-0H í cH,0H Chapter 24 723 42. a. 3-Amino-2-oxindole catalyzes the decarboxylation of an a-keto acid by first forming an iminc that is followed by a prototropic shift. H NH, o + RE-dow mine formatiog O fe = N or-keto acid 2 H H N=CHR Va AN lecarboxylation Eb ma, (OITO H +co fertasn H N=CHR 0=CHR O imine hydrol Lygis OS N H b. 3-Aminoindole would not be as effective a catalyst, because the electrons left behind when COy is eliminated cannot be delocalized onto an electronegative atom. L NH, N=€, Not imi j ço N + RC-cog mine formation N É » N a-keto acid N | H H H 3-aminoindole