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both 2- and 3-pentanone would be formed on hydration

D. PAVIA CHEM 352. ANSWERS - SAMPLE EXAM ONE. 1. A. B. C. D. 2. both 2- and 3-pentanone would be formed on hydration. 3. Lindlar catalyst (Pd / BaSO 4 / quinoline ) is required. Other catalysts will not stop at the alkene, which is more reactive than the alkyne. 4. a).

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both 2- and 3-pentanone would be formed on hydration

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  1. D. PAVIA CHEM 352 ANSWERS - SAMPLE EXAM ONE 1. A B C D 2. both 2- and 3-pentanone would be formed on hydration 3. Lindlar catalyst (Pd / BaSO4 / quinoline ) is required. Other catalysts will not stop at the alkene, which is more reactive than the alkyne.

  2. 4. a) b) c) 5. invert the ring CH3

  3. CH3 6. 2-Chloro-2-methylbutane (a) will react the fastest. It will form a tertiary carbocation in this E1 reaction. 7. E1 8. The curves would be similar to that on page 805 of your text. Halide (a) would have the lowest enregy pathway (lowest Ea and lowest energy carbocation). Halide (c), which is primary would have the highest energy pathway, 9. E2, E1, E1cb, E1

  4. 10. slow 1,2-methyl shift rearrangement

  5. 11. remember a-elimination 12. a) sodium ethoxide b) 3-methylcyclohexene c) 4-methyl-2-pentyne

  6. HCl 13. NaOMe MeOH Br2 CCl4 NaNH2 NH3(liq) 14. FALSE. There is an anti-coplanar requirement for E2 - if there isn’t an appropriate anti-coplanar hydrogen, the reaction will not follow the Zaitsev rule. Stereochemistry will control the reaction.

  7. Added Problems All 3 compounds give the same intermediate and the same product. Those points coincide on the three curves. The only differences are in the energies of the three starting materials. 15. Ea’s could be slightly different same carbocation 1-butene cis-2-butene trans-2-butene same product different energies exothermic

  8. 16. One gives a secondary carbocation, the other gives a tertiary carbocation (lower energy, forms faster). 17. Same mechanism as the addition of H-O-H, except one of the H’s is replaced by a methyl group which stays in the final product. The product is an ether: 1-methoxy-1-methylcyclopentane. .. + CH3-O-H is the electrophile that adds a proton H .. CH3-O-H is the species that adds to the carbocation and it also removes the extra proton in the next step (end of the sequence) .. formed when H2SO4 ionizes in the methanol solvent

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