Baixe Manual soluções Paula Bruice Química Orgânica Cap23 e outras Notas de estudo em PDF para Química, somente na Docsity! Chapter 23 683 Solutions to Problems 1. 2. a. Protonation of the doubled-bonded nitrogen forms a conjugate acid that is stabilized by electron delocalization. Protonation of the other nitrogen forms a conjugate acid that is not stabilized by electron delocalization. The resonance-stabilized conjugate acid is the one that is more readily N/NH ——> HN NH <—> HNk NH = Ht = E NA NH Ê ví Tv b. Protonation of the doubled-bonded nitrogen forms a conjugate acid that is stabilized by electron delocalization. Protonation of either of the other nitrogen atoms forms a conjgate acid that is not stabilized by electron delocalization. The resonance-stabilized conjugate acid is the one that is more readily formed. JH H,N—C-NH— guanidino group I H H H H H H Ne Nes NR N N o . e o + + o. H;)N—-C—-NH— <—>» HN-C=NH— <€—>» HN=C-NH— a. (R)-alanine b. (R)-aspartate e. The o-carbons of all the D-amino acids except cysteine have the R-configuration. Similarly, the o-carbon of all the L-amino acids except cysteine has the S-configuration. (In all the amino acids except cysteine, the amino group has the highest priority and the carboxyl group has the second-highest priority. In cysteine, the thiomethyl group has a higher priority than the carboxyl group because sulfur has a greater atomic number than oxygen.) isoleucine and threonine naturally occurring L-isoleucine is (28,38)-isoleucine naturally occurring L-threonine is (28,3R)-threonine The electron-withdrawing +NH3 substituent on the a-carbon increases the acidity of the carboxyl group. 684 Chapter 23 a. Solved in the text. f b. CHÇHCO” = +NH; NH l N e. H;NCCH,CH,CHCO” +NH; o ll ll a. HOCCH,CH,CHCOH +NH; l l b. HOCCH,CHÇHCO” +NH, l d. HyNCH,CH,CH,CHaCHCO” e. f. d. +NH; +NH, Il l NH;CNHCH;CH;CHoCHOO” +NH; no S-cngudo- +NH; = ID OCCH;CH;CHCO +NH, - ll OCCH,CH;ÇCHCO” NH, a. The carboxyl group of the aspartic acid side chain is a stronger acid than the carboxyl group of the glutamic acid side chain because the aspartic acid side chain is closer to the electron- withdrawing protonated amino group. b. The Iysine side chain is a stronger acid that the arginine side chain. The arginine side chain has less of a tendency to lose a proton because the positive charge is delocalized over three nitrogen atoms. In order for the amino acid to have no net charge, the two amino groups must have a +1 charge between them in order to cancel out the -1 charge of the carboxylate group. Because they are positively charged in their acidic forms and neutral in their basic forms, the sum of their charges will be +1 at the midpoint of their pKa values. Chapter 23 687 19. o o é H é H ” No Ns “en” Dn <> cn” SN | pt fe 20. a. 208 = 25,600,000,000 b. 20100 21. The bonds on either side of the a-carbon can freely rotate. In other words, the bond between the a-carbon and the carbonyl carbon and the bond between the o-carbon and the nitrogen (the bonds indicated by arrows) can freely rotate. H O R H o NS E DD Cc Sr a f q fe > H o R Q 22. — In forming the amide linkage, the amino group of cysteine reacts with the y-carboxyl group rather than with the a-carboxyl group of glutamic acid. y-carboxyl group c-carboxyl group y-carboxyl group | l CH, “OCCHCH;CHCO! (OLE HEHSCH CNH — +NH; +NH; glutamate a segment of glutathione 688 23. 24. Chapter 23 Leu-Val and Val-Val will be formed because the amino group of leucine is not reactive (so leucine could not be the C-terminal amino acid) but the amino group of valine would react equally easily with the carboxyl group of leucine and the carboxyl group of valine. Hs ? 7 + 7 - + DRA CcH;C —0O—C—NHCHCO + H;NCHCO = H;NCHC —NHCHCO em, Cm, CHsÇH Ho CHsÇH CHÇH CH; CHÇH CH, CH, CH; N-protected leucine valine Leu-Val In + IL HNcHdo” + H;NÇHCOT ——> — H;NCHO—NHCHCO CH;CH CHsÇH CHGH CHsÇH ém, cH; CH, CH; valine valine Val-Val If valine's carboxyl group is activated with thionyl chloride, the OH group of serine, as well as the NHh group of serine, would react readily with the very reactive acyl halide, forming both an ester and an amide. Da Tso cd fmom Rf q OH u - u HNCueO” — EoNçHCCI — RCOCH,ÇHCO + RENHÇHCO CH(CH5), CH(CH5), NH, CcH,0H ester amide If valine's carboxyl group is activated with DCC, an imidate will be formed. Because an imidate is less reactive than an acyl chloride, the imidate will react with the more reactive NH) group in preference to the OH group. 7 o HaNÇHCO o o NH HI DCC Wo CcH,0H l Io I HoNÇHCO- e H;NÇHCO—Ç — Ei H;NCHC—NHÇHCO: + 0=€ (CH;)CH (CH)CH NH (CH;)CH CH,0H NH an imidate Chapter 23 689 25. GH 0 0 CH 9 SH O 5 CH;C—OCOCO-CCH, + H;NÇHCO” — CHsÇ—OCNHÇECO” CH, cm, GH CH; Ho 1. DCC q (CH;)CH Leu (CH)CH 2. H;NÇHCO” CH,CçHs cH O Phe CH, O o o o 3 [od N A. DCC Cê n N. CHs—OCNHÇHENHCHCNHÇHCO = CH — OCNHÇHENHÇHEO l cH CH CH CH; GB cm GH 2. EyNCHCO” 3 o] > (CH)CH CH; NH, CHyNH, (CHs) CoHs n Lys 1. DCC |2. H;NÇHCO” CH(CH;), Val Es q N N l CH;C— OCNHCHCNHCHCNHCHCNHCHCO Í dm I I CF;COOH CH GH CH (GH CHCHG) Nem (CH3)pCH CH NH, Hs s Moo To cH;C + CO, + HaNCHCNHCHENHOHCNHCHCO dg, GEL CH, (CH, CH(CH; (CH5)CH CH NH; Leu Phe Lys Val 26. a 58% 2 3 4 5 6 7 8 9 70% 49% 34% 24% 17% 12% 82% 58% b. 4.4% 2 3 4. 5 6 7/89/10 w 12 13 14 15 80% 64% 51% 41% 33% 26% 21% 17% 13% 11% 86% 69% 55% 44% 692 Chapter 23 28. Ss: I CH,COOH + H* 29. — Knowing that the N-terminal amino acid is Gly, look for a peptide fragment that contains Gly. "f" tells you that the 2nd amino acid is Arg "e" tells you the next two are Ala-Trp or Trp-Ala. “d?” tells you that Glu is next to Ala, so 3 and 4 must be Trp-Ala and the 5th is Glu "g” tells you the 6th amino acid is Leu "kh" tells you the next two are Met-Pro or Pro-Met. “e” tells you that Pro is next to Val, so 7 and 8 must be Met-Pro and the 9th is Val "b" tells you the last amino acid is Asp Gly-Arg-Trp-Ala-Glu-Leu-Met-Pro-Val-Asp 30. — Cysteine can react with cyanogen bromide, but the lactone will not be formed, because it would be a strained four-membered-ring lactone and the sulfur would not be positively charged, causing it to be a poor leaving group. Without lactone formation, cleavage cannot occur. 31. Treatment with Edman's reagent would release two PTH-amino acids in approximately equal amounts. 32. Solved in the text. 33. 34. 35. 36. 37. Chapter 23 693 The data from treatment with Edman's reagent and carboxypeptidase A identify the first and last amino acids. Leu Ser The data from cleavage with cyanogen bromide identify the position of Met and identify the other amino acids in the pentapeptide and tripeptide but not their order. O cleavage with cyanogen bromide Arg, Lys, Tyr Leu Met Arg, Phe Ser The data from treatment with trypsin put the remaining amino acids in the correct position. Leu Tyr Lys Ser Arg | Met Phe Arg 74 amino acids/ 3.6 amino acids per tum = 20.6 turns of the helix 20.6x 5.4Á = 110 À in an o-helix 74 amino acids x 3.5 À = 260 À in a straight chain It would fold so that its nonpolar residues are on the outside of the protein in contact with the nonpolar membrane and its polar residues are on the inside of the protein. a. A cigar-shaped protein has the greatest surface area to volume ratio, so it has the highest percentage of polar amino acids. b. A subunit of a hexamer would have the smallest percentage of polar amino acids because part of the surface of the subunit can be on the inside of the hexamer and, therefore, have nonpolar amino acids on its surface. An amino acid is insoluble in diethyl ether (a relatively nonpolar solvent) because an amino acid exits as a zwitterion at neutral pH. In contrast, carboxylic acids and amines have a single charge at neutral pH and, therefore, are less polar. 694 38. 39. 40. 41. 42. Chapter 23 a. HisLys Leu-Val-Glu-Pro-Arg Ala-Gly-Ala b. Leu-Gly-Ser-Met-Phe-Pro-Tyr Gly-Val c. Val-Arg-Gly-Met-Arg-Ala Ser d. Ser-Phe-Lys-Met Pro-Ser-Ala-Asp e. Arg Ser-Pro-Lys Lys Ser-Glu-Gly ú ú HNCHC—NHÇHCOCH; GH CH coo Tl N T Mo Ji No “1 N a. HOCCH;ÇHOOH | b, HOCCH;ÇHCO e DCCHÇHCO” d. DECHCHCO” +NH; +NH; +NH; NH, The student is correct. At the pl the total of the positive charges on the tripeptide"s amino groups must be 1 to balance the one negative charge of the carboxyl group. When the pH of the solution is equal to the pKa of a lysine residue, the three lysine groups each have one-half a positive charge for a total of one and one-half positive charges. Thus, the solution must be more basic than this in order to have just one positive charge. Since the mixture of amino acids is in a solution of pH = 5, His will have an overall positive charge and Glu will have an overall negative charge. His, therefore, will migrate to the cathode, and Glu will migrate to the anode. Ser is more polar than Thr (both have OH groups, but Thr has an additional carbon); Thr is more polar than Met (Met has S instead of O and an additional carbon); and Met is more polar than Leu. o a n Hi Ea O is 9 5 Sa E £ z o Thr O Leu o s O Glu Met 2 chromatography mi Chapter 23 697 Then, from the pieces given, you can determine where the disulfide brides are. Val- Mais Eyre las fogo SeraPho=AlacfGlusder s I s Ser-Cys-Phe-Ly: s-Cys-Trp-Lys-Tyr-Cys-Phe-Arg-Cys-Ser S—s 48. The methyl ester of phenylalanine rather than phenylalanine itself should be added in the peptide bond-forming step because if esterification is done after amide bond formation, both carboxyl groups could be esterified. Some product will be obtained in which the amide bond is formed with the y-carboxyl group of aspartate rather than with the o-carboxyl group. ND E cho HaNÇHCO- + CHÇ—-0C—0-Co-CCH; —» CH;C—OCNHÇHCO Ho CH; CH; CH; qro coo- o coo- EoNCHÉOCH, CH, pec, E Pcs Rancid EP adod CHÇ—OCNHÇHENHÇHCOCH; «SÃO cg —oCNHCHCO—C CH CH GH CH; GH NH coo- coo” cnscoontenes CH; CH + CO, + MyniçHê—nHeHÉOC, cH, CH, CH, coo 698 49. 50. Chapter 23 Ser-Glu-Leu-Trp-Lys-Ser-Val-Glu-His-Gly-Ala-Met From the experiment with carboxypeptidase, we know the C-terminal amino acid is Met. “4º” tells us the amino acid adjacent to Met is Ala “e” tells us the next amino acid is Gly “sp” tells us the next amino acid is His “9” tells us the next amino acid is Glu “sj” tells us the next amino acid is Val “e” tells us the next amino acid is Ser “s” tells us the next amino acid is Lys “a?” tells us the next amino acid is Trp “sh? tells us the next amino acid is Leu “sk” tells us the next amino acid is Glu “Kk” tells us the next (first) amino acid is Ser The pKa of the carboxylic acid of the dipeptide is higher than the pKa of the carboxylic acid of the amino acid because the positively charged amino group of the amino acid is more strongly electron withdrawing than the amide group of the peptide. This causes the amino acid to be a stronger acid and have a lower pKa. o +. +. Il H;NCH,COH H;NCH,CNHCH,COH Ro higher The pKa of the amino group of the peptide is lower than the pKa of the amino group of the amino acid because the amide group of the peptide is more strongly electron withdrawing than the carboxylate group of the amino acid. + 1 + 1 Io H;NCILCO- H;NCH,CNHCH,CO lower 51. 52. Chapter 23 699 Finding that there is one less spot than the number of amino acids tells you that the spots for two of the amino acids superimpose. Since leucine and isoleucine have identical polarities, they are good candidates for being the amino acids that migrate to the same location. Val (pl = 5.97), Trp (pI = 5.89), and Met (pl = 5.75) can be ordered based on their pI values because the one with the greatest pJ will be the one with the greatest amount of positive charge at pH=5. o CDArg Q Zola tn y Ala En OS Thr OS Val E SS Oo O e) EE & S O OT Ss Le g S Po Tyr > Pe Te % Glu Met SDAsp 9 chromatography ——» Oxidation of dithiothreitol is an intramolecular reaction so it occurs with a larger rate constant than the oxidation of 2-mercaptoethanol, which is an intermolecular reaction. The reverse reduction reaction should occur with about the same rate constant in both cases. Increasing the rate of the oxidation reaction while keeping the rate of the reduction reaction constant is responsible for the greater equilibrium constant, since Keg= kk. HO HO. Ta + mm H —<— HO k1 HO k 2 HOCH,CH;SH == HOCH,CH;S—SCH,CH,0H -1