1. Dehydrohalogenation of Alkyl Halides Dehydrohalogenation of Alkyl Halides

2. X Y dehydrohalogenation of alkyl halides: X = H; Y = Br, etc.   C C C C +XY  -Elimination Reactions

3. X Y dehydrohalogenation of alkyl halides: X = H; Y = Br, etc.   C C C C +XY  -Elimination Reactions requires base

4. (100 %) likewise, NaOCH 3 in methanol, or KOH in ethanol NaOCH 2 CH 3 ethanol, 55°C Dehydrohalogenation Cl

5. CH 3 (CH 2 ) 15 CH 2 CH 2 Cl KOC(CH 3 ) 3 dimethyl sulfoxide (86%) CH 2 CH 3 (CH 2 ) 15 CH Dehydrohalogenation When the alkyl halide is primary, potassium tert-butoxide in dimethyl sulfoxide (DMSO), a strong non-protic polar solvent is the base/solvent system that is normally used.

6. Br 29 % 71 % + Regioselectivity follows Zaitsev's rule more highly substituted double bond predominates KOCH 2 CH 3 ethanol, 70°C

7. more stable configuration of double bond predominates Stereoselectivity KOCH 2 CH 3 ethanol Br + (23%)(77%)

8. E2 Energy Diagram

9. Question How many alkenes would you expect to be formed from the E2 elimination of 3-bromo-2-methylpentane? A)2 B)3 C)4 D)5

10. more stable configuration of double bond predominates Stereoselectivity KOCH 2 CH 3 ethanol + (85%)(15%)Br

11. The E2 Mechanism of Dehydrohalogenation of Alkyl Halides The E2 Mechanism of Dehydrohalogenation of Alkyl Halides

12. Empirical Data (1)Dehydrohalogenation of alkyl halides exhibits second-order kinetics first order in alkyl halide first order in base rate = k[alkyl halide][base] implies that rate-determining step involves both base and alkyl halide; i.e., it is bimolecular

13. Question The reaction of 2-bromobutane with KOCH 2 CH 3 in ethanol produces trans-2- butene. If the concentration of both reactants is doubled, what would be the effect on the rate of the reaction? A)halve the rate B)double the rate C)quadruple the rate D)no effect on the rate

14. Empirircal Data (2)Rate of elimination depends on halogen weaker C—X bond; faster rate rate: RI > RBr > RCl > RF implies that carbon-halogen bond breaks in the rate-determining step

15. concerted (one-step) bimolecular process single transition state C—H bond breaks  component of double bond forms C—X bond breaks The E2 Mechanism

16. – O R.... : CCCCCCCCHX.. :: Reactants The E2 Mechanism

17. CCCCCCCC –––– OR.... H X..:: –––– Transition state The E2 Mechanism

19. Br E2 Mechanism / Transition State CH 3 CH 2 O –

20. OR.... H CCCCCCCC–X.. ::.. Products The E2 Mechanism

21. Question Which one of the following best describes a mechanistic feature of the reaction of 3-bromopentane with sodium ethoxide? A)The reaction occurs in a single step which is bimolecular. B)The reaction occurs in two steps, both of which are unimolecular. C)The rate-determining step involves the formation of the carbocation (CH 3 CH 2 ) 2 CH +. D)The carbon-halogen bond breaks in a rapid step that follows the rate-determining step.

22. Stereoelectronic Effects Stereochemistry: Anti Elimination in E2 Reactions

23. Consider dehydrohalogenation of chlorocyclohexane. An anti-periplanar T.S. is required and only the chair conformation on the left alllows for the elimination to occur. E2 – Stereoelectronic Effect

24. Stereoelectronic Effect An effect on reactivity that has its origin in the spatial arrangement of orbitals or bonds is called a stereoelectronic effect. The preference for an anti coplanar arrangement of H and Br in the transition state for E2 dehydrohalogenation is an example of a stereoelectronic effect.

25. (CH 3 ) 3 C Br KOC(CH 3 ) 3 (CH 3 ) 3 COH cis-1-Bromo-4-tert- butylcyclohexane Stereoelectronic Effect

26. (CH 3 ) 3 C Br KOC(CH 3 ) 3 (CH 3 ) 3 COH trans-1-Bromo-4-tert- butylcyclohexane Stereoelectronic Effect

27. (CH 3 ) 3 C Br Br KOC(CH 3 ) 3 (CH 3 ) 3 COH cis trans Rate constant for dehydrohalogenation of 1,4- cis is >500 times than that of 1,4- trans Stereoelectronic Effect

28. (CH 3 ) 3 C Br KOC(CH 3 ) 3 (CH 3 ) 3 COH cis H that is removed by base must be anti coplanar to Br Two anti coplanar H atoms in cis stereoisomer H H Stereoelectronic Effect

29. (CH 3 ) 3 C KOC(CH 3 ) 3 (CH 3 ) 3 COH trans H that is removed by base must be anti coplanar to Br No anti coplanar H atoms in trans stereoisomer; all vicinal H atoms are gauche to Br; therefore infinitesimal or no product is formed H H (CH 3 ) 3 C BrHH Stereoelectronic Effect

30. Which of the two molecules below will NOT be able to undergo an E2 elimination reaction? Question A) B)B)B)B)

31. 1,4- cis more reactive 1,4- trans much less reactive Stereoelectronic Effect

32. Sterically unhindered bases favor the Zaitsev product. Sterically hindered bases favor the Hofmann product. See: SKILLBUILDER 8.5. E2 – Regioselectivity

33. Question Which would react with KOC(CH 3 ) 3 /(CH 3 ) 3 COH faster? A)cis-3-tert-butylcyclohexyl bromide B)trans-3-tert-butylcyclohexyl bromide

34. Question Which would react with KOCH 2 CH 3 in ethanol faster? A)cis-2-tert-butylcyclohexyl bromide B)trans-2-tert-butylcyclohexyl bromide

35. Question What is the major product of the following reaction?

36. Question

37. The E1 Mechanism of Dehydrohalogenation of Alkyl Halides

38. CH 3 CH 2 CH 3 Br CH 3 C Ethanol, heat + (25%) (75%) H3CH3CH3CH3C CH 3 C C H3CH3CH3CH3C H CH 2 CH 3 CH 3 C H2CH2CH2CH2C Example

39. 1. Alkyl halides can undergo elimination in protic solvents in the absence of base. 2. Carbocation is intermediate. 3.Rate-determining step is unimolecular ionization of alkyl halide. The E1 Mechanism

40. CH 3 CH 2 CH 3 Br CH 3 C :.. : C CH 2 CH 3 CH 3 + :.. : Br.. – slow, unimolecular Step 1

41. C CH 2 CH 3 CH 3 + C CH 2 CH 3 CH 3 CH 2 + C CHCH 3 CH 3 – H + Step 2

42. Question Which reaction would be most likely to proceed by an E1 mechanism? A)2-chloro-2-methylbutane + NaOCH 2 CH 3 in ethanol (heat) B)1-bromo-2-methylbutane + KOC(CH 3 ) 3 in DMSO C)2-bromo-2-methylbutane in ethanol (heat) D)2-methyl-2-butanol + KOH

43. 1.Analyze the function of the reagent (nucleophile and/ or base). 2.Analyze the substrate (1°, 2°, or 3°). Predicting Substitution vs. Elimination

44. 1.Analyze the function of the reagent (nucleophile and/ or base). 2.Analyze the substrate (1°, 2°, or 3°). Predicting Substitution vs. Elimination

45. 1.Analyze the function of the reagent (nucleophile and/ or base). 2.Analyze the substrate (1°, 2°, or 3°). Predicting Substitution vs. Elimination

46. 1.Analyze the function of the reagent (nucleophile and/ or base). 2.Analyze the substrate (1°, 2°, or 3°). See SKILLBUILDER 8.11. Predicting Substitution vs. Elimination

47. 1.Analyze the function of the reagent (nucleophile and/ or base). 2.Analyze the substrate (1°, 2°, or 3°). 3.Consider regiochemistry and stereochemistry. Predicting Products REGIOCHEMICAL OUTCOMESTEREOCHEMICAL OUTCOME SN2SN2The nucleophile attacks the α position, where the leaving group is connected. The nucleophile replaces the leaving group with inversion of configuration. SN1SN1The nucleophile attacks the carbocation, which is where the leaving group was originally connected, unless a carbocation rearrangement took place. The nucleophile replaces the leaving group with racemization.

48. See: SKILLBUILDER 8.12. Predicting Products REGIOCHEMICAL OUTCOMESTEREOCHEMICAL OUTCOME E2The Zaitsev product is generally favored over the Hofmann product, unless a sterically hindered base is used, in which case the Hofmann product will be favored This process is both stereoselective and stereospecific. When applicable, a trans disubstituted alkene will be favored over a cis disubstituted alkene. When the β position of the substrate has only one proton, the stereoisomeric alkene resulting from anti-periplanar elimination will be obtained (exclusively, in most cases). E1The Zaitsev product is always favored over the Hofmann product. The process is stereoselective. When applicable, a trans disubstituted alkene will be favored over a cis disubstituted alkene.

49. a. A = 3; B = 1; C = 2; D = 1; E = 1; F = 5 b. A = 4; B = 4; C = 2; D = 4; E = 5; F = 2 c. A = 2; B = 4; C = 2; D = 4; E = 5; F = 2 d. A = 4; B = 4; C = 1; D = 4; E = 3; F = 1 e. A = 3; B = 5; C = 2; D = 1; E = 3; F = 5 Question For each reagent, predict which product will predominate.