Preparation of Alkenes: Crude Oil & Elimination Reactions

1. Preparation of Alkenes:Crude Oil & Elimination Reactions

2. Alkenes: Crude Oil Sources & Products

3. -Elimination Reactions / Alkenes

4. X Y C C C C -Elimination Reactions Overview • dehydrogenation of alkanes: (enzymatic and limited industrial use) X = Y = H • dehydration of alcohols: X = H; Y = OH • dehydrohalogenation of alkyl halides: X = H; Y = Br, etc. + Y X  

5. AlkeneStability • Because of steric strain, cis isomers are generally less stable than trans. • The difference in stability can be quantified by comparing the heats of combustion.

6. AlkeneStability

7. AlkeneStability • Consider the following stability trend: • What pattern do you see? • See SKILLBUILDER 8.3.

8. Question What is the correct order of stability (most stable to least stable) for alkenes? A. Tetrasubstituted > cis-disubstituted > trans-disubstituted > monosubstitutedB. Trisubstituted > terminal disubstituted > trans-disubstituted > monosubstitutedC. Tetrasubstituted > trans-disubstituted > cis-disubstituted > trisubstitutedD. Monosubstituted > cis-disubstituted > trans-disubstituted > trisubstituted

9. Alkene Nomenclature

10. H2C CHCH2CH3 Alkene Nomenclature 1-Butene or But-1-ene • 1) Find the longest continuous chain that includes the double bond. • 2) Replace the -ane ending of the unbranched alkane having the same number of carbons with -ene. • 3) Number the chain in the direction that gives the lowest number to the doubly bonded carbon.

11. H2C CHCHCH2Br CH3 Alkene Nomenclature • 4) If a substituent is present, identify its position by number. The double bond takes precedence over alkyl groups and halogens when the chain is numbered. • The compound shown above is4-bromo-3-methyl-1-butene. • (or 4-bromo-3-methylbut-1-ene)

12. H2C CHCHCH2OH CH3 Alkene Nomenclature • 4) If a substituent is present, identify its position by number. Alcohol groups take precedence over the double bond when the chain is numbered. • The compound shown above is2-methyl-3-buten-1-ol. • (or 2-methyl-but-3-en-1-ol)

13. H2C H2C Common Alkenyl Groups • methylene • vinyl • allyl CH H2C CHCH2

14. 3 2 CH3 4 1 6 5 CH2CH3 Cycloalkene Nomenclature 6-Ethyl-1-methylcyclohexene • 1) Replace the -ane ending of the cycloalkane having the same number of carbons with -ene. • 2) Number through the double bond in thedirection that gives the lower number to the first-appearing substituent.

15. Question • Name the alkene below according to the IUPAC system. • A)2-bromo-5-heptane • B)6-bromo-2-trans-heptene • C)2-bromo-5-cis-heptene • D)5-bromo-2-heptene

16. What is the correct IUPAC name for the following structure? Question A. (E, 2S)-2-methyl-3-propylhept-5-eneB. (E, 5S)-6-methyl-5-propylhept-2-eneC. (E, 5S)-5-isopropyloct-2-eneD. (E, 2R)- 2-methyl-3-propylhept-5-ene E. (E, 5R)- 5-isopropyloct-2-ene

17. Dehydration of Alcohols

18. H2SO4 + H2O CH3CH2OH H2C CH2 160°C H2SO4 + H2O OH 140°C (79-87%) CH3 H3C H2SO4 + CH2 H2O C OH H3C C heat H3C CH3 (82%) Dehydration of Alcohols

19. Question • The dehydration of 2-methyl-2-propanol cannot be accomplished by using which of the • following reagents? • A)H2SO4 • B)H3PO4 • C)HCl • D)K2SO4

20. R' tertiary:most reactive R OH C R" R' OH C R H H OH C R primary:least reactive H Relative Reactivity

21. Question • Of the isomeric alcohols having the molecular formula C5H12O, which one will undergo • acid-catalyzed dehydration most readily? • A)2-pentanol • B)2-methyl-1-butanol • C)2-methyl-2-butanol • D)3-methyl-2-butanol

22. Regioselectivityin Alcohol Dehydration:The Zaitsev Rule OH

23. H2SO4 HO 80°C Regioselectivity • A reaction that can proceed in more than one direction, but in which one direction predominates, is said to be regioselective. + 90 % 10 %

24. CH3 CH3 CH3 H3PO4 OH heat Regioselectivity • A reaction that can proceed in more than one way to produce different products involving different carbon atoms, where one predominates. It is said to be regioselective. + 16 % 84 %

25. OH R R CH2R C C CH3 H The Zaitsev Rule • When elimination can occur in more than one direction, the principal alkene is the one formed by loss of H from the  carbon having thefewest hydrogens. three protons on this  carbon

26. OH R R CH2R C C CH3 H The Zaitsev Rule • When elimination can occur in more than one direction, the principal alkene is the one formed by loss of H from the  carbon having thefewest hydrogens. two protons on this  carbon

27. CH2R R OH R C C R CH2R C C CH3 R CH3 H The Zaitsev Rule • When elimination can occur in more than one direction, the principal alkene is the one formed by loss of H from the  carbon having thefewest hydrogens. only one proton on this  carbon

28. Question • What is the major product in the reaction of 2-methyl-2-butanol with H2SO4 at 80°C? • A) B) • C) D)

29. Question • What is the major product of the dehydration of 2-methylcyclohexanol? • A)1-methylcyclohexene • B)3-methylcyclohexene • C)4-methylcyclohexene • D)cyclohexene

30. Stereoselectivity in Alcohol Dehydration

31. H2SO4 + heat OH (25%) (75%) Stereoselectivity • A stereoselective reaction is one in which a single starting material can yield two or more stereoisomeric products, but one is produced in greater amounts than any other.

32. Question • The major product of the dehydration of 1-phenyl-2-propanol is • A) B) • C) D)

33. dehydrogenase + OH (0%) (100%) Stereospecificity A stereospecific reaction is one in which a single starting material yields only a single stereoisomer although other steroeisomers are possible. Enzymes do this commonly, but there are not many processes invented by man which do.

34. The E1 & E2 Mechanismsof Alcohol Dehydration

35. A Connecting Point... • Dehydration of alcohols and the reaction of alcohols with hydrogen halides share thefollowing common features: • 1) Both reactions are promoted by acids • 2) The relative rates/ reactivity decreases in the order tertiary > secondary > primary • These similarities promote the idea that carbocations areintermediates in the acid-catalyzed dehydration ofalcohols, just as they are in the reaction of alcoholswith hydrogen halides.

36. CH3 H3C H2SO4 CH2 C OH H3C C heat H3C CH3 Dehydration of tert-Butyl Alcohol • first two steps of mechanism are identical tothose for the reaction of tert-butyl alcohol withhydrogen halides + H2O

37. CH3 H3C H2SO4 CH2 C OH H3C C heat H3C CH3 Dehydration of tert-Butyl Alcohol • first two steps of mechanism are identical tothose for the reaction of tert-butyl alcohol withhydrogen halides + H2O

38. H H O .. H fast, bimolecular H H + + : O : : O (CH3)3C H H tert-Butyloxonium ion Mechanism Step 1: Proton transfer to tert-butyl alcohol .. + + : O • (CH3)3C H

39. H + : O (CH3)3C H slow, unimolecular H + + : : O (CH3)3C H tert-Butyl cation Mechanism Step 2: Dissociation of tert-butyloxonium ion to carbocation Because rate-determiningstep is unimolecular, thisis called the E1 mechanism.

40. H : : H O H3C + CH2 H C H3C fast, bimolecular H H3C + CH2 + : C H O H3C H Mechanism Step 3: Deprotonation of tert-butyl cation +

41. Carbocations • Carbocation intermediates can: • react with nucleophiles • and/or • 2) lose a -proton to form an alkene

42. H2C CH2 Dehydration of Primary Alcohols H2SO4 + H2O CH3CH2OH • avoids carbocation because primary carbocations are not thermodynamically stable • oxonium ion loses water and a proton in abimolecular step 160°C

43. H H O .. H fast, bimolecular H H + + : : O : CH3CH2 O H H Ethyloxonium ion Mechanism Step 1: Proton transfer from acid to ethanol .. + : CH3CH2 O H

44. H H + : O CH2 CH2 : : H O H H slow, bimolecular H H + : + O + : : H O H2C CH2 H H Mechanism Step 2: Oxonium ion loses both a proton and a water molecule in the same step. +

45. Step 2: Oxonium ion loses both a proton and a water molecule in the same step. H H + : + O CH2 CH2 : : H O H H slow, bimolecular H H + : + O + : : H O H2C CH2 H H Mechanism Because rate-determiningstep is bimolecular, thisis called the E2 mechanism.

46. Rearrangements in Alcohol Dehydration • The alkene product may not have the same carbon skeleton as the starting alcohol.

47. + + 33% 64% Example OH H3PO4, heat 3%

48. Example

49. CH3 CHCH3 CH3 C + CH3 Rearrangement Involves Alkyl Group Migration • Carbocation can lose a proton as shown; • or it can undergo a 1,2- methyl migration. • CH3 group migrates with its pair of electrons to adjacent positively charged carbon. 3%

50. CH3 CH3 97% CHCH3 CH3 CHCH3 C C + CH3 Rearrangement Involves Alkyl Group Migration + • tertiary carbocation; more stable CH3 CH3 3%