Christopher M. Hadad Chemistry 253N Spring Final Exam. Name (PRINT) ANSWER KEY I have neither given nor received aid on this exam (SIGN)

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1 Christopher M. adad Chemistry 253 Spring 2002 Final Exam ame (PRIT) ASWER KEY I have neither given nor received aid on this exam (SIG) There are 8 pages and 300 points on this exam. Please read each question carefully before answering. 1. (10 pts) Rank the acidity of the following compounds (from 1 to 5) in order of increasing acidity (1 = least acidic): 2 3 C C 3 3 C C 3 3 C C (15 pts) The three amino acids shown below were separated by electtophoresis at p 6.0. Draw each amino acid below the corresponding band in its dominant structure at p CC 2 CC 2 CC C 3 C 2 C 2 C start C 2 C 2 C 2 C 2 2 cathode (-) anode () 3 CC C 2 3 CC C 3 3 CC C 2 C 2 C 2 C 2 C 2 C 3 3. (10 pts) Provide the structure of 2'-deoxyadenosine-5'-monophosphate, the nucleotide derived from adenine, 2-deoxyribose, and phosporic acid. Some structures are provided below to get you started. P P 2 phosphoric acid D-ribose adenine

2 Christopher M. adad Chemistry 253 Spring (15 pts) Provide one way in which the spectroscopic methods below could be used to distinguish between compounds A and B. A B (a) 13 C proton-decoupled MR spectroscopy (all peaks as singlets). A - 4 unique carbons B - 6 unique carbons (b) Infrared (IR) spectroscopy. A - C= stretch should be above 1710 cm -1. B - C= stretch should be below 1700 cm -1 because it is conjugated with the alkene. (c) Ultraviolet (UV) spectroscopy. A - not conjugated so should not have a λ max less than 200 nm. B - conjugated so should have a λ max greater than 200 nm. 5. (10 pts) An MR spectrum is an average spectrum of the conformations populated by a molecule. From the following data, estimate the equilibrium percentages of axial and equatorial bromine as a substituent present in bromocyclohexane, C. δ 4.62 ppm δ 3.81 ppm δ 3.95 ppm ( 3 C) 3 C ( 3 C) 3 C C δ = 4.62 (% axial) 3.81 (% equatorial) % axial % equatorial = = 4.62 (% axial) 3.81 (1 - % axial) % equatorial = 1 - % axial 3.95 = 4.62 (% axial) (% axial) 3.81 % axial = 17.3 % equatorial = (15 pts) The electron ionization (E. I.) mass spectrum of 2-methyl-2-butanol shows prominent peaks at m / z = 73, 70, and 59 amu. Provide structures for these fragments. 3 C E. I. 3 C 3 C 3 C 3 C m / z = 88 m / z = 73 m / z = 70 m / z = 59

3 Christopher M. adad Chemistry 253 Spring (20 pts) The 1 MR spectrum of a compound with prominent peaks at m / z = 136 and 57 amu is shown on the last page of this exam. The mass spectrum also shows a peak at m / z = 138 which is of the same intensity as the peak at m / z = 136 amu. (a) The mass spectrum indicates the presence of what element? bromine (b) Provide a molecular formula which is consistent with the mass spectrum. ote: the only elements present besides the correct answer for (a) are C and. C 4 9 (c) Provide the structural formula of this molecule and indicate which protons in your structure are responsible for each signal C C C 2 3 C (35 pts) The 1 MR spectrum of a C 9 8 compound is shown on the last page of this exam. The IR spectrum contains the following major peaks. Abbreviations: w, weak; m, medium; s, strong (m) 2992 (m) 1627 (m) 1496 (m) 3062 (m) 2814 (m) 1605 (m) 1450 (s) 3047 (w) 2743 (m) 1597 (m) 3029 (m) 1670 (s) 1576 (m) (a) ow many degrees of unsaturation are present in this molecule? 6 (b) What functional group which uses one degree of unsaturation is clearly indicated (not an alkene) by the IR spectrum? Be as specific as possible and list the peaks that led you to that conclusion. an aldehyde and 2743 cm -1 a conjugated carbonyl cm -1 (c) What is the coupling constant between the protons centered at and ppm? = 7.8 z (d) The peak at ppm is not part of an alkene. Does this molecule have a cis or trans alkene? trans (e) Provide the structural formula of this molecule. (f) ow many unique peaks would be present in the 13 C proton-decoupled MR spectrum of the molecule you proposed? 7

4 Christopher M. adad Chemistry 253 Spring (60 pts) Provide the products of the following reactions. If there is more than one product, predict which will be major. (a) (b) 3 C C C 2 excess C 3 I, Ag 2 2 C 3 3 C 3 C a C 3 C , 2. 2, 2 3 C 3 C C 2 C 3 C 2 C 3 1. ac 3 (1 eq) (c) C (d), (e) 2, Pd/C 2 (f) (g) (h) (i) 3 C C C, KC 1. Et 2, 2. 3 C C , 80 o C 3 C C 3 C C C 2 C 3 C 3 C C 2 major C 3 trace (j) C 3 C 3 C 3 C 3 C 3

5 Christopher M. adad Chemistry 253 Spring (30 pts) Provide the necessary reagents to accomplish the following transformations in an efficient and effective manner - more than one step may be required and more than one set of methods may be viable. ote the direction of each arrow. 2 C 2 Et C 2 Et 1. 4 Cl, KC, Ac 2, pyridine 2. aet ( 3 C) 3 C 3. C , pyridine 3, acb 3 2 CC C 2 pyridine, ( 3 C) 3 C C(C 3 ) 3, C 2 3 C C (30 pts) Provide syntheses for these compounds and provide complete reagents and conditions. (a) start with and prepare 2 3 C ( enantiomer) C 3 1. Ac, 2 2. t-bu - K C 3 desired products (b) start with and prepare C 3 C 3 C 3 C 3 C 2 C 3 C 3 C 3 3 C C 3 ac 3 C 3 C 2 C 3

6 Christopher M. adad Chemistry 253 Spring (15 pts) Explain why the reaction below gives only the Diels-Alder (42) cycloaddition product and not the (62) cycloaddition product. Your answer should include a molecular orbital diagram of at least the M of the triene. Also, explain why only the product with an anti orientation between the vinyl group and the fivemembered lactone ring is observed., dark Energy Energy LUM M LUM M Carbons 1 and 6 of the triene cannot interact with the dienophile. This is illustrated by drawing the M of the triene and the LUM of the dienophile. otice that a constructive combination of molecular orbitals is not possible. This is why the (62) cycoaddition is not observed. Carbons 1 and 4 of the triene do have the correct phase to interact with the dienophile and this is why the (42) product is observed. The anti orientation between the vinyl group and the five-membered ring is observed because as the dienophile approaches the triene from below it does so with the carbonyl groups under the triene. This is explained by the endo rule as a favorable interaction between the p orbitals of the carbonyl groups and the p orbitals of the triene. 13. (15 pts) The final step in the solid phase synthesis of polynucleotides is treatment with ammonia and water. This removes all of the protecting groups and removes the product from the solid support. Provide the mechanism of the following reaction which de-protects the phosphate group (shown below). R Base R Base 3 C P 2 P R' R' C C R Base P R' 3 C R Base P R'

7 Christopher M. adad Chemistry 253 Spring (20 pts) Provide a mechanism for the conversion of D to E shown below. 1. P 3, C 3 C 3. a 3 3 D C 3 E 3 C 3 3 C 3 C 3 C 3 C 3 3 C 3 C 2 3 C 3 C 3 C 2 3 C