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CLAS – Chem 109 B - Ch. 10 Chapter 10 Practice 1. Br 2. Br 3. Br 4. Cl 5. Br Br + 6. Br 7. Cl 8. Cl 9. OCH 3 10. Br 11. OTs 12. OCH 3 13. 14. 15. 16. + CH3CH2CH2CH2CH2CH2OH 17. 18. + 19. 20. O OH 21. O 22. no reaction 23. O H 24. O 25. O H 26. O 27. OH CH3I + 28. + HO I 29. OH + Br 30. Br HO 31. O + CH CH I 3 2 H 32. OH OCH 3 33. OH OCH 3 34. HO OCH 3 H3CO OH 35. 36. 1. RCO3H 2. HO-, H2O OH OH OH + OH 37. 1. OsO4 2. H2O2, H2O OH OH 38. + N I - 39. N + 40. + N(CH 3)3 41. N 42. + N(CH3)3 43. H3C N H CH3 N 44. Rank the following alcohols in order of increasing rate of dehydration in the presence of sulfuric acid. C<A<B<D 45. Which amine will form the following product when undergoing 2 successive Hoffman degradations? H N 46. Indicate how the following compound could be synthesized using the given starting material and any necessary reagents. 1. (CH3)3N 2. Ag2O, H2O 3. Δ 47. What is the major product of rearrangement for the following arene oxide? OH D 48. What is the major product of rearrangement for the following arene oxides? Which arene oxide is more likely to be carcinogenic? OH OH NO 2 OCH 3 49. Explain why elimination of 2-bromopentane produces 2-pentene whereas elimination of the quaternary ammonium ion produces 1-pentene. OH When a compound with a good leaving group such as a bromide (very weak base) undergoes E2 reactions the transition state is alkene-like. Alkenes are more stable when the sp2 carbons are more highly substituted. Therefore the alkene that is more substituted will form faster. When a compound with a bad leaving group such as an amine (good base) undergoes E2 reactions the transition state is carbanion-like. Carbanions are more stable when they’re least substituted. Therefore the alkene that is least substituted will form faster.