Final Exam Professor R. Hoenigman

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I pledge to uphold the CU onor Code: CEM 3311-200 Fall 2006 Final Exam Professor. oenigman Average Score = 145 igh Score = 235 Low Score = 27 Signature Name (printed) Last four digits of your student ID number ecitation TA ecitation number, day, and time You have 2 hours and 30 minutes to complete this exam. No model kits or calculators allowed. Periodic table and scratch paper are attached. D NT TUN TIS PAGE UNTIL INSTUCTED T D S. ecitation Sections: # Day Time TA 211 Monday 8 am Noel 251 Monday 2 pm Carolynn 291 Monday 5 pm eather 252 Tuesday 12 pm Sam 293 Tuesday 5 pm Carolynn 212 Wednesday 8 am Noel 253 Wednesday 1 pm Tom 292 Wednesday 5 pm eather 213 Friday 8 am eather Score: Page 1 /16 Page 2 /15 Page 3 /20 Page 4 /30 Page 6 /40 Page 7 /40 Page 8 /24 Page 9 /45 Page 5 /20 Page 10 /10 TTAL /250

CEM 3311, Fall 2006 Final Exam, page 1 of 12 1. (10 pts) Benzain is a topical antibiotic with the structure shown below. A. Label the configuration of each indicated chirality center in Benzain. (2 pts each) B. Circle all of the terms below that describe one or more structural feature of Benzain. (1 pt each) Amine Ketone Aldehyde Amide Ester Sulfide Nitrile Ether Thiol N C N P 3 2 n-pr S 2. (6 pts) Give the IUPAC name for each of the following compounds. (2 pts each, -1 for no stereochemistry, -1 for wrong numbers) A. (3Z,5)-3,5,6,6-tetramethyl-3-heptene B. (3S,4)-4-bromo-5-ethyl-2,5-dimethyl-3-heptanol C. F para-fluoroacetophenone (or 4-fluoroacetophenone)

CEM 3311, Fall 2006 Final Exam, page 2 of 12 3. (5 pts) Vitamin C, shown below, has a pk a of 4.1. Circle the most acidic hydrogen in Vitamin C. C C 2 4. (10 pts) Explain why an ester can be both an activating, ortho/para director and also a deactivating, meta director. Use resonance structures to support your discussion. (4 pts for resonance structures, 6 pts for explanation) The ester functional group can be attached to a benzene ring in one of two ways. If the ester is attached at the oxygen, a lone pair of electrons can help stabilize the arenium ion towards ortho/para attack. The resulting resonance structure is particularly good, since all atoms have an octet of electrons. Electrophile Electrophile If the ester functional group is attached to the benzene ring at the carbonyl group, the benzene ring become less electron-rich and is deactivated. In addition, the adjacent carbonyl group is a meta director, since the following resonance structure would be destabilized (due to the adjacent positive charges) by ortho/para attack. Electrophile Electrophile

CEM 3311, Fall 2006 Final Exam, page 3 of 12 5. (15 pts) The bromination of anisole produces two isomeric bromoanisoles. Using curved arrows to show the flow of electrons, give a mechanism to account for the formation of these two products. Fe 3 Fe 3 C 3 para attack ortho attack C 3 C 3 Fe 3 Fe 3 C 3 C 3 Fe 3 Fe 3 6. (5 pts) Circle the compounds below that are chiral. (1 pt each) 3 C C C C D cis-3-methylcyclobutanol C 3 C 3 C 3 (1,2S)-1,2-dimethylcyclopentane (+)-2-chlorobutane C 3 Ts Ts

CEM 3311, Fall 2006 Final Exam, page 4 of 12 7. (15 pts) Circle the more acidic compound in the following pairs. Give the reason for your choice in the adjacent box. (2 pts circle, 3 pts explanation) A. or The allylic conjugate base is resonance stabilized B. C 3 C or CF 3 C Inductive effect C. N 2 or N 2 More stable conjugate base (more resonance structures) 8. (15 pts) Circle the compound that has the lower heat of combustion. Give the reason for your choice in the adjacent box. (2 pts circle, 3 pts explanation) A. Book Problem 11.50c/d Aromatic cation versus an antiaromatic anion or B. 3 C C 3 or 3 C C 3 More substituents are in the equatorial position 3 C C(C 3 ) 3 3 C C(C 3 ) 3 C. or Allylic cation is resonance stablized

CEM 3311, Fall 2006 Final Exam, page 5 of 12 9. (10 pts) Catalytic hydrogenation of 1,4-dimethylcyclopentene yields a mixture of two products. ne of these products is formed in a much greater yield than the other (the observed ratio is 10:1). Draw the products of this reaction and circle the major product. Why is there a preference for one of these products? Book Problem 6.40 (2 pts each product, 1 pt circle, 5 pts explanation) ydrogenation is regioselective to the less hindered face of the double bond. Thus, the alkene will attach to the catalyst on the side opposite the C4 methyl group and the hydrogen will add to the opposite side, favoring formation of cis-1,3-dimethylcyclopentane. 2 Pt + 10. (5 pts) Match the following alkenes with the appropriate Δ 2 value. eats of hydrogenation: 151 kj/mol, 122 kj/mol, 114 kj/mol, 111 kj/mol, 105 kj/mol Book Problem 6.29 (1 pt each) _122 kj/mol_ 1-pentene _111 kj/mol_ (E)-4,4-dimethyl-2-pentene _114 kj/mol_ (Z)-4-methyl-2-pentene _151 kj/mol_ (Z)-2,2,5,5-tetramethyl-3-hexene _105 kj/mol_ 2,4-dimethyl-2-pentene 11. (5 pts) Explain why 3-pentanol has a higher boiling point than 3-chloropentane. Boiling points are determined by intermolecular forces. 3-pentanol has greater intermolecular forces than 3-chloropentane, since the alcohol can participate in hydrogen bonding. ydrogen bonds require a lot of energy to break (relatively) and thus, the alcohol has a higher boiling point.

CEM 3311, Fall 2006 Final Exam, page 6 of 12 12. (80 pts) Give the major organic product(s) of the following reactions. Be sure to clearly show stereochemistry using dashes and wedges. Write N if no reaction occurs. early label any enantiomers, diastereomers, and/or meso compounds. (Note: some problems have more than one step.) A. D C 3 D C 3 D C 3 s 4 + diastereomers meso meso B. NBS hν + + enantiomers C. Exam 1 problem C 3 Li + N 3 C 4 + LiN 2 D. + Al 3 E. 2-methyl-3-hexyne 1) 3 2) Zn, 2 + F. toluene chromic acid C 3 C Al 3 N G. B 3 + N 3 3 B N 3

CEM 3311, Fall 2006 Final Exam, page 7 of 12 12. continued. [Be sure to clearly show stereochemistry using dashes and wedges. Write N if no reaction occurs. early label any enantiomers, diastereomers, and/or meso compounds; some problems have more than one step.]. 2-methyl-3-hexyne 2 S 4 gs 4 + I. Book Problem 12.36b Al 3 J. p-ts pyridine NaS DMS Ts S K. D NaC 3 C 3 D L. C 3 C 3 + NC + NC enantiomers M. 2,3-dimethyl-1,3-butadiene D D C 3 C 3 D + + D enantiomers N. Book Problem 6.46 3-methyl-1-butene aq. 2 S 4

CEM 3311, Fall 2006 Final Exam, page 8 of 12 13. (12 pts) Using curved arrows to show the flow of electrons, draw a mechanism for the solvolysis of tert-butyl bromide in acetic acid. E1 S N 1 2 2 14. (12 pts) Using curved arrows to show the flow of electrons, propose a mechanism for the base catalyzed enol-keto tautomerization shown below.

CEM 3311, Fall 2006 Final Exam, page 9 of 12 15. (45 pts) Propose an efficient synthesis for the following transformations. You may use any reagents you like. Be sure to show any intermediates. (Do not draw a mechanism.) A. 1-bromo-2-chloropropane starting from propene Book Problem 10.24d NBS B. 3-bromo-5-nitrobenzoic acid starting from toluene Book Problem 12.27k chromic 2 S 4 acid N 3 N 2 2 Fe 3 N 2 C. 2-butyne starting from 1-butyne 2 2 NaN 2

CEM 3311, Fall 2006 Final Exam, page 10 of 12 Extra Credit: (10 pts) Draw a parody of aromatic nomenclature. For example, below is paraphrase. PhaSe PhaSe

CEM 3311, Fall 2006 Final Exam, page 11 of 12 Scratch Page

CEM 3311, Fall 2006 Final Exam, page 12 of 12 Scratch Page