Final Exam Version 1. Chemistry 140C. Fall Good Luck! Dec 5, :30 am 2:30 pm This exam accounts for 50% of the final grade.

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1 Chemistry 140C Final Exam Version 1 Fall 2006 Dec 5, :30 am 2:30 pm This exam accounts for 50% of the final grade. Mark your final answer clearly. Completely erase irrelevant information! Exams written in pencil will not be regraded. Additional pages are at the end for your convenience. You can use them to sketch formulae, mechanisms, etc. and for answers. Indicate if your final answer should be found on the additional pages at the end of the exam! Write your name and student ID on every page and sign every page. Good Luck! Your ame (Please print): Your Student ID#: SLUTIS Question 1:../30 Question 2:../12 Question 3:../40 Question 4:../20 Question 5:../80 Question 6:../28 Question 7:../24 Question 8:../30 Question 9:../16 Question 10:../12 Question 11:../8 Total points:./300

2 ame: ID: Signature: Question 1: eduction of amide 1 with lithium aluminium hydride (LiAl 4 ) gave amine 2. After acidic aqueous work-up of 2, the cyclic imine 3 was obtained. LiAl + 4,(C 2 5 ) 2, Write a detailed mechanism that explains the conversion of 2 to 3. (30 points) Show all intermediates! Acid-catalyzed hydrolysis of the acetal gives the ketone that reacts in an intramolecular condensation with the amine to give the cyclic imine. 6 key steps (blue arrows; 5 points each). [Vollhardt, Exercise 20-18, 5 th ed. = 20-17, 4 th ed.] C Final exam, Version 1 2

3 ame: ID: Signature: Question 2: When the ofmann procedure (exhaustive methylation with methyl iodide followed by silver oxide-catalyzed elimination) was applied to the hydroxyamine 4, the product ethylene oxide 6 (oxacyclopropane) was isolated instead of the expected alkene. 2 excess C 3 I? 1. Ag 2, 2 2. Δ + (C 3 ) (a) Write the structure of the intermediate 5. (5 points) (C 3 ) 3 I - (b) Write a detailed mechanism that explains the conversion of 5 to 6. (7 points) Show all intermediates! [Vollhardt, Ch. 21, Problem 43a, 5 th ed. = 41a, 4 th ed.] (5 points for internal S 2 step; 2 points for deprotonation step) (C 3 ) 3 I - (internal S 2) + (C 3 ) 3 + I (C 3 ) 3 +I - alternative: + (C 3 ) 3 + I - + (C 3 ) 3 + +I - - alternative: + (C 3 ) 3 + I (C 3 ) 3 +I - alternative: - (C 3 ) 3 I (C 3 ) 3 I - (internal S 2) + (C 3 ) 3 +I - 140C Final exam, Version 1 3

4 ame: ID: Signature: Question 3: Treatment of a stoichiometric mixture of ketoester 7 and ketone 8 with sodium ethoxide in ethanol gave the bicyclic product 9. + C 3 C 2 - a +, C 3 C Write a detailed mechanism that explains the formation of 9 from 7 and 8. Show all intermediates! (40 points) obinson annulation. [Vollhardt, Exercise 23-18, 5 th ed. = 23-18, 4 th ed.] The first sequence is a Michael addition of the anion of the β-ketoester 7 to the α,β-unsaturated ketone 8 resulting in a diketone (3 key steps; 5 points each): - C 2 C 3 C 3 C 2 -C 3 C 2 -C 3 C 2 - (solution continued on next page) (continuation on next page if you need more space for your answer) 140C Final exam, Version 1 4

5 ame: ID: Signature: Question 3: (continuation if you need more space for your answer) The diketone subsequently undergoes an intramolecular aldol condensation (5 key steps; 5 points each): - C2 C 3 C 3 C 2 -C 3 C 2 C 3 C 2 - -C 3 C C Final exam, Version 1 5

6 ame: ID: Signature: Question 4: The reaction sequence shown below furnished intermediate 11 from diester 10. eating of intermediate 11 in aqueous acid gave carboxylic acid C 3 C 2 - a +,C 3 C 2 2. Br Br 3. a, , 2, Δ? (a) Write the structure of the intermediate 11. (5 points) - a + - a + (b) Write a detailed mechanism that explains the conversion of 11 to 12. Show all intermediates! (15 points) Malonic ester synthesis; decarboxylation [Vollhardt, Ch. 23, Problem 29d, 5 th ed. = 29d, 4 th ed.] (3 key steps; 5 points each) - a + - a , 2-2a + + C 140C Final exam, Version 1 6

7 ame: ID: Signature: Question 5: For each of the following reactions, circle the product that is formed. (5 points ea) (a) C C CC 3 1. K, , 2 C C C CC 3 CC 3 CC 3 (b) C 3 + MgBr 1. (C 2 5 ) , 2 C 3 (excess) (c) 3 + Cl - Br 1. - K +,DMF C 2. Cl, 2, Δ (d) Br 2 1. ac, DMS 2. 2,Pt 2 140C Final exam, Version 1 7

8 ame: ID: Signature: (e) 1. excess C 3 I 2. Ag 2, 2 3. Δ (f) C - 3 C C 3 C 1. (C 3 ) 2 (C 3 ) 2 2. ab 3 C, C 3 C 2 (g) 1. C a +,C , 2 (h) 1. C a +,C , 2 140C Final exam, Version 1 8

9 ame: ID: Signature: (i) 1. C 3 C 2 - a +, C 3 C 2 2. C 3 C 2 Br 3. K, , 2, Δ (j) + C C 3 - a +, C 3 C C (k) Cl C 3 C 2 C 2 C 3 (l) C 3 C 3, + C 3 3 C 3 C C 3 C 3 C 3 C 3 140C Final exam, Version 1 9

10 ame: ID: Signature: (m) + Cl, 2 + Cl - Cl - Cl (n) a 2,Cl, 2, 0 C 2 (o) C C C 3, 2 C 2 C C C C 2 C 2 C (p) Br C 3 MgBr, (C 3 C 2 ) 2 Br 140C Final exam, Version 1 10

11 ame: ID: Signature: Question 6: Derivatives of cyclohexyl-carboxylic acid 13 undergo hydrolysis in acidic aqueous conditions to furnish the carboxylic acid 14: X 2 S 4, (a) Select the most reactive and the least reactive derivative among the following (13a, b, c, d): (8 points) 2 Cl X= C 2 5 X= CC 3 X= 2 X= Cl 13a 13b 13c 13d Write down the derivative number: Most reactive: 13d Least reactive: 13c (b) Select the carboxylic acid ester among the derivatives shown above (13a, b, c, d): (2 points) Write down the derivative number: Ester: 13a (c) Write a detailed mechanism that explains the acid-catalyzed hydrolysis of the ester to the carboxylic acid 14. Show all intermediates! Transfer protons intramolecularly! (18 points) (continuation on next page for your answer) 140C Final exam, Version 1 11

12 ame: ID: Signature: Question 6: (continuation for your answer) Ester hyrolysis following the addition-elimination mechanism. [Vollhardt, Section 19-9] (4 key steps; 4 points each; 2 points for last deprotonation step) C Final exam, Version 1 12

13 ame: ID: Signature: Question 7: For each of the following pairs of compounds, circle the more acidic one. The to be considered is highlighted. (3 points each) (a) 2 2 (b) (c) (d) (e) Cl Cl (f) C 3 (g) (h) CF 3 140C Final exam, Version 1 13

14 ame: ID: Signature: Question 8: Mithramycin X, shown below, is a DA-binding carbohydrate antitumor drug recently isolated from soil bacteria. aldose 3 C 3 C β aldose β 3 C 3 C ketose β C 3 aldose 3 C β β aldose (a) Identify all glycosidic linkages in mithramycin X and classify them as α or β. Mark your answer on the structure shown above! (10 points) (b) Is mithramycin X a reducing or non-reducing carbohydrate derivative? Circle your answer below. (5 points) reducing non-reducing (c) Identify each sugar in mithramycin X as an aldose or a ketose. Mark your answer on the structure shown above! (15 points) If you incorrectly picked any of the three rings of the aglycon as your choice for an aldose or ketose (c), 2 points were taken off (-2) for each misidentified sugar in part (c). ne point was taken off (-1) in part (a) if you assigned α or β to a bond to the aglycon. You should know by now what a sugar looks like and what an anomeric carbon is. 140C Final exam, Version 1 14

15 ame: ID: Signature: Question 9: Shown are the two Watson-Crick base pairs A-U and G-C which are the basic building blocks of double-stranded ribonucleic acid (A). A -U G-C = (a) An important non-watson-crick base pair in A is the wobble pair that forms between G and U. Write the structure of the G-U wobble pair. Mark clearly the hydrogen bonds that form in the wobble pair. (8 points) This is the structure of the G-U wobble pair that is usually found in A: G-U Full credit was also given for pairings that involve tautomers of G or U (note that the sugar residues have to point towards the same site of the pair to allow for doublestrand formation): If you have chosen one of the pairings that involve tautomeric forms, see also the remark on parts (b) & (c) of this question on the next page. (question continues on next page) 140C Final exam, Version 1 15

16 ame: ID: Signature: Question 9: (continued) (b) The melting temperature of double-stranded A is defined by the sharp increase of UV absorption at 260nm that occurs when hydrogen bonds between the base pairs in the two A strands are broken. A melting temperature of 73 C was determined for the following double-stranded A (A 1): 5 G G C A G U A C C G U G 3 A 1 3 C C G U C A U G G C A C 5 The melting temperature of a second double-stranded A (A 2) was determined that was derived from the above sequence by replacing every G-C base pair with a wobble G-U pair. Predict if the melting temperature of A 2 is the same, lower, or higher than the melting temperature of A 1. Circle your answer below. (4 points) Melting temperature: A 2 < A 1 A 2 = A 1 A 2 > A 1 For the 2- bond wobble pair For the 3- bond tautomeric pairs (c) Explain your choice in one or two sentences only. (4 points) The stability of the double-stranded hybrid in A 2 is reduced compared to A 1 because G- U pairs are connected by 2 hydrogen bonds while G-C pairs have 3 hydrogen bonds. For comparison, see the difference in stability of G-C versus A-U pairs [Vollhardt, Section 26-9]. For parts (b) and (c), if your G-U pair structure in (a) is wrong, but you follow through consequentially in (b) and (c), you will get credit for (b) and (c). For example, if your G-U pair structure is wrong (0 credit in a) but has 2 instead of 3 -bonds and you follow through in (b) and (c), you will get full credit on (b) and (c). ote however, if you chose one of the tautomeric G-U pairs, these have 3 -bonds just like G- C pairs, and you will get credit only if you chose A2=A1 in (b) and you explain this correctly in (c). Your answer in part (b) and (c) has to be logical with respect to the structure of the G-U pair that you drew. In the context of this question, the stability of a GU pair is determined only by the number of - bonds compared to a G-C pair. The argument that more hydrogen bonds involving oxygen lead to higher stability in G-U pairs with tautomeric forms of either G or U was not accepted since this is balanced by the required tautomerization. In reality, the stability of double-stranded A is substantially influenced by stacking between neighboring base pairs and the ability of G-C versus G-U to participate in optimal stacking. 140C Final exam, Version 1 16

17 ame: ID: Signature: Question 10: Shown are the structures, names, and one-letter codes of 5 amino acids.. 2 C = 2 C 2 C 2 C 2 C C Glycine (G) Serine (S) Lysine (K) 2 istidine () Glutamic acid (E) (a) Draw the structure that each of the four amino acids serine, lysine, histidine, and glutamic acid has in aqueous solution at p 1. Use the Fischer projection templates below to draw the structures. (8 points) Serine (S) Lysine (K) istidine () Glutamic acid (E) C C C C + 3 C C (b) Draw the structure of serine in aqueous solution at p 7. Use the Fischer projection template below to draw the structure. (4 points) Serine (S) C C 2 140C Final exam, Version 1 17

18 ame: ID: Signature: Question 11: Microcin J25, whose one-letter code sequence is shown below, is a cyclic peptide antibiotic that adopts an extraordinary lasso conformation (see picture). Microcin J25 Cyclization of microcin J25 occurs, as indicated in the sequence above, by crosslinking between glycine at position 1 (G1) and glutamic acid at position 8 (E8). Draw the structure of the crosslink between E8 and G1. (int: use the structures of the amino acids shown in Question 10) (8 points) E8 C C G1 Partial Credit was given if you identified the correct parts of the amino acids that are available for connection (the -terminal amino group of glycine and the carboxylic acid side chain in glutamic acid) but you got the actual product wrong (amide bond). 140C Final exam, Version 1 18