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1 Chemistry 263 omework 1 Part 2 Spring 2018 Out: 04/04/18 Due: 04/13/18 1 Value: A very generous 9 points 2 All questions are valued at ½ point per question unless otherwise noted. Name: KEY 1. Which of the following synthetic strategies would convert trans-2-butene into pure cis-2-butene? a. Reaction with 2, Ni 2 B (P-2) b. Reaction with (a) Br 2 in CCl 4 followed by (b) 2NaN 2 in liquid N 3 followed by Na in liquid N 3 c. Reaction with 3 O, d. Reaction with (a) Br 2 in CCl 4 followed by (b) 2NaN 2 in liquid N 3 followed by 2, Ni 2 B (P-2) e. More than one of the above Need to brominate so NaN 2 can eliminate Br 2x and form the alkyne. The P-2 catalyst (actually highly versatile the Wiki entry is good) gives the Z- alkene, as does the Lindlar catalyst 1 By 5 pm 2 Recall omework is worth 15 points. With 9.5 points available on part 1, the total point potential is 18.5

2 2. Which of the following synthetic strategies would convert cis-2-butene into pure trans-2-butene? a. Reaction with (a) Br 2 in CCl 4 followed by (b) 2NaN 2 in liquid N 3 followed by 2, Ni 2 B (P-2) b. Reaction with (a) Br 2 in CCl 4 followed by (b) 2NaN 2 in liquid N 3 followed by Li in liquid N 3 c. Reaction with (a) 3 O, followed by (b) cold dilute KMnO 4 /O - / 2 O d. Reaction with (a) Br followed by (b) NaN 2 followed by 2 /Pt e. More than one of the above As above, only once at the level of an alkyne we need a partial reduction to the E-alkene. Sodium amide does the trick. The conditions for Z-alkene formation (P-2) are mild. Sodium amide in liquid ammonia not so much (but at least it is cold. N 3 BP = -33 o C) 3. What is the major product of the following reaction series? 1. 2, Pd/CaCO 3, quinoline a. 2a. B 3 :TF 2b. 2 O 2, NaO O b. c. O d. O e. O

3 Reduction of terminal alkyne to alkene (obviously neither cis or trans) followed by anti- Markovnikov hydroboration/oxidation. Really, step 1 should include Pb(Ac) 2, which helps poison the catalyst along with quinoline. 4. What is the final product of the following synthesis? 2-Butyne 2 Ni 2 B (P-2) C 4 8 i. OsO 4 ii. NaSO 4 Final Product O C 3 O O C 3 O C 3 C 3 O C 3 O C 3 O C 3 O C 3 I II III e. An equimolar mixture of III & The P-2 catalyst gives the Z-alkene (which is set up nicely for you in all of the Fischer projections with the methyl groups receding). Osmium tetroxide gives the vicinal diol (glycol) following a syn addition, so the hydroxy groups must be on the same side 5. What feature would you expect to see in the 1-NMR spectrum of B after subjecting A to the following conditions? A Cl 2 heat or light C 8 9 Cl Cl (chief product) B a. There would be only 4 aromatic protons at low field b. The signal for the benzylic protons would be a doublet c. The signal for the benzylic protons would be a triplet d. The signal for the methyl protons would be a triplet e. The signal for the methyl protons would be a doublet Benzylic radicals are very stable, so with only 1 remaining post chlorination, the methyl s split into a doublet

4 6. Which of the following is not an example of resonance? I C 2 =C C 2.. C 2 C=C 2 II C 2 C 2 III C 2. C 2 =C C C 2 CCl 3 C 3. C 2 C=C C 2 CCl 3 e. All of the above are examples of resonance Quiz 1 repeat what a nice guy! Option (c) is a hydride shift 7. Which of the following is a conjugated diene? I II III V & II & III & d. I, II, and V e. V Uh, hullo, the ones with the alternating double bonds

5 8. The correct IUPAC name for the following compound is Br a. 1-Bromo-1-methyl-2,5-cyclohexadiene b. 3-Bromo-3-methyl-1,4-cyclohexadiene c. 6-Bromo-6-methyl-1,4-cyclohexadiene d. 2-Bromo-2-methyl-1,3-cyclohexadiene e. None of the above is the correct IUPAC name Precedence in this case goes to the double bond, numbering through the double bond so as to give the lowest numbering overall 9. What is the correct IUPAC name for the following triene? C 3 C 3 C 2 a. (2E,4Z,6E) 2,4,6 Nonatriene b. (2Z,4E,6Z) 2,4,6 Nonatriene c. (2E,4Z,6Z) 2,4,6 Nonatriene d. (3Z,5Z,7E) 3,5,7 Nonatriene e. (3Z,5E,7E) 3,5,7 Nonatriene If are shown, sometimes it is easier to focus on them to establish EZ relationships 10. ow many bonding molecular orbitals does 1,3-butadiene possess? a. 1 b. 2 c. 3 d. 4 e. 0 4 unhybridized p orbitals in, gives 4 molecular orbitals out (2 bonding, 2 antibonding)

6 11. Arrange the following hexadienes in order of decreasing stability I II III V a. V > II > I > III > I > > II > I > V c. > III > II > V > I > III > I > II > V e. I > II > > III > V Looks oddly familiar. Another problem used on the quiz with order reversed. Trans internal double bonds most stable, terminal double bonds least stable 12. Which of the following would have the lowest heat of hydrogenation? I II III e. V V Recall hydrogenation generates stronger bonds from weaker bonds, and so the reaction is exothermic. Conjugated double bonds are more stable, as are those that are more substituted, and so they have less energy potential.

7 13. Which of the following would absorb UV at the longest wavelength? CC C 2 C CC 3 C CC 3 I II III C 2 C C 2 C CC 3 e. V V Longest conjugated pathway done and done 14. ow could the following synthesis be carried out? CO I 2 C 3 C CC 3 CO II C 3 C C 2 CO III C 2 CCO C 2 CCO

8 e. None of these Basic utilization of the Diels-Alder reaction begins with seeing the double bond as the site of the initial diene (on either side). Becoming expert involves seeing where a double bond could have been, and then envisioning a diene double bond on either side (imagine the product above without any double bonds) 15. Which of the following C10 compounds is expected to have the greatest resonance delocalization energy? I II III V e. V Azulene () is a contender to naphthalene (III), and is a deep blue/indigo color (in nature an azulene containing pigment gives the blue color to the mushroom Lactarius indigo). Interestingly, naphthalene shows no color. The color associated with azulene is associated with the fact it has a dipole moment (naphthalene dipole = 0) and may be best thought of as a fusion between a cyclopentadienyl anion (aromatic) and a cycloheptatrienyl cation (aromatic). Experimentation has in fact shown the cyclopentadienyl ring to be nucleophilic, and the cycloheptatrienyl ring to be electrophilic. This distortion effectively cuts down the length of the resonance path, and since the greater the resonance path the greater the stabilization energy, azulene has about half of the resonance stabilization as naphthalene.

9 16. What of the following would you expect to have significant resonance stabilization energy? N N I II III d. II & III e. All of the above Remember, you always have to count all of the p electrons, and only count as many non-bonded electrons as needed to satisfy ückel s rule. No sp 3 atoms allowed! 17. Given benzene has a resonance energy of 152 kj/mol and naphthalene has a resonance energy of 255 kj/mol, which position would be substituted by bromine when phenanthrene undergoes addition by Br 2? a. 1,2 b. 1,4 c. 3,4 d. 7,8 e My kind of math: 152 kj/mol x 2 = 304 kj/mol > 255 kj/mol. Systems move in the direction of lowest energy, and avoid higher energy states

10 18. On the basis of MO theory and ückel's rule, which of the following compounds is aromatic? O I II III V V e. V Yet another quiz problem! Again, bear in mind the need for the carbonyl to act as an sp 2 place holder

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