Organic Chemistry 3540

Transcription

1 rganic hemistry 3540 December 10, 2003 (7 Pages, 12 Parts) ame 1. (15%) Many organic compounds found in living systems are complex molecules which can be characterized, in part, by simply listing the chemical class or the specific products of the complete hydrolysis of the complex molecule. Such products might include:. amino acids;. deoxyribose;. glucose; D. glycerol; E. long chain saturated acids; F. long chain unsaturated acids; G. nitrogen bases;. phosphate ion; and I. ribose.. Some products J. might not be in the previous list. dditionally, some organic compounds found in living systems are complex molecules but which cannot be hydrolyzed at all L. ydrolysis involves the breaking of specific bond types which might include: M. acetal;. amide;. anhydride; P. ester; and R. hemiacetal.. Some bonds S might not be in this list.."haracterize" the following groups of naturally occurring molecules by placing in front of the compound the appropriate letter (or letters) above of the product(s) of complete hydrolysis of the naturally occurring molecule..give the bonding which is broken during such hydrolysis by placing after the compound the appropriate letter (or letters) of the bond(s) broken during complete hydrolysis. a. cellulose b. starch c. fat d. oil (as in vegetable oil) e. triterpene f. simple protein g. R h. D

2 2. (12%) ptically active aldopentose was strongly oxidized with nitric acid to give optically active diacid. Degradation of gave optically active aldotetrose. Degradation of gave D-glyceraldehyde. itric acid oxidation of gave optically active diacid D. Give Fischer projections of,,, and D. 3. (8%) Each amino acid has its own unique isoelectric point, the p at which it is essentially electrically neutral. Four amino acids and their isoelectric points are shown below (The structures, of course, are not shown in the zwitterionic forms.) S 2 2 Glu = Met = ( 2 ) la = 6.01 Lys = 9.74 mino acids can separated by electrophoresis. Give the amino acid or acids of these four which: a. migrates towards the negative electrode in electrophoresis at p = b. migrates towards the positive electrode in electrophoresis at p = c. does not migrate towards the either electrode in electrophoresis at p = (7%) Draw the two dimensional structure of the tetrapeptide, la-lys-glu-met.

3 5. (10%) Give the basic structure of the polypeptide whose analytical hydrolysis data is shown below. Edman degradation gave Lys. arboxypeptidase analysis gave la. omplete hydrolysis: la = 2; rg = 1; ys = 1; Glu = 1; Leu = 1; Lys = 2; Phe = 1; Val = 1. Partial hydrolysis gave a complex mixture from which only three tripeptides (see below) were isolated in high enough yield and purity to be characterized. They are: ys-glu-leu Leu-la-la Lys-rg-ys hymotrypsin hydrolysis (to the right of Phe, Tyr, and Trp) gave a polypeptide which was not characterized and a tripeptide which had the following composition: Lys-Val-Phe 6. (6%) utrasweet with an isoelectric point of 5.9 is drawn at the right in its nonionic form. Draw the products of the complete hydrolysis of utrasweet. 2 3 Draw the structure of utrasweet at a physiological p of (4%) For each of the following examples, state if the example is an example of the primary, secondary, tertiary, or quaternary structure of the protein: a. folding of the helix in a globular protein b. association of four globins and four hemes in hemoglobin c. the amino acid sequence d. shape of the backbone, i.e., α-helix

4 8. (5%) There are various bonds and attractions which are responsible for the unique conformation of an enzyme. List five of these bonds and attractions. 9. (8%) The reaction of furan with bromine goes under milder conditions than benzene. Like benzene, it does not give addition product T. Like benzene, it gives a substitution product and it gives isomer V and not Z. 2, 0 o a. Is the resonance energy of furan greater than or less than benzene? T b. Explain, using appropriate structural drawings, why V is formed in preference to Z. V Z 10. (5%) To determine the structure of the aggregating pheromone of bark beetles, the pheromone (a monoterpene with the formula ) was oxidized with hot permanganate to give acetone and compound ( ). It would appear that two carbons were lost as carbon dioxide. Reconstructing these fragments gave three possibilities (,, and D) for the pheromone. Tell which structure is the correct structure and very briefly explain why. D

5 11. (12%) In each of the following, give the letter of the correct answer: a. Stronger base: b. Product: 2, 300 o c. More highly colored: d. More soluble in water: 3 S 2 3 S e. Which statement is correct?: -1-. has a higher iodine number than D has a higher saponification number than has a higher iodine number than D has a lower saponification number than has a lower iodine number than D has a higher saponification number than has a lower iodine number than D has a lower saponification number than.

6 f. D-glyceraldehyde: M g. For the reaction of M,, and with enedict s reagent: h. hexose:. nly M gives a positive test.. nly gives a positive test.. nly gives a positive test. D. nly M and give a positive test. E. nly M and give a positive test. F. nly and give a positive test. G. ll three give a positive test. i. Edman degradation of a polypeptide reveals:. The amino end ( terminus) of the peptide.. The carboxyl end ( terminus) of the peptide.. The amino acid to the right of Phe, phenylalanine. 2 2 j. The β-pleated sheet is most significantly associated with which structure of a protein?. Primary structure. Secondary structure. Tertiary structure D. Quaternary structure k. n aldose: 2 2 l. pyranose 2 2

7 12. (7%) For each of the following, give the letter of only one compound which contains the structural feature requested. ote that some compounds contain more than one structural feature. a. b. c. d. acetal -acetal amide anhydride E 2 G 2 3 F e. f. deoxy sugar ester I J S 2 g. h. pyranose furanose K 2 L P P i. aldose j. ketose M. Structure ot Given bove k. reducing sugar l. nonreducing sugar m. aldopentose n. ketohexose o. acid