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ELIMINATION REACTIONS

ELIMINATION REACTIONS. C C. C C. Cl. H. ELIMINATION REACTIONS. An elimination reaction is one where starting material loses the elements of a small molecule such as HCl or H 2 O or Cl 2 during the course of the reaction to form the product. - HCl. TWO EXAMPLES FOLLOW.

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ELIMINATION REACTIONS

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  1. ELIMINATION REACTIONS

  2. C C C C Cl H ELIMINATION REACTIONS An elimination reaction is one where starting material loses the elements of a small molecule such as HCl or H2O or Cl2 during the course of the reaction to form the product. - HCl TWO EXAMPLES FOLLOW

  3. ELIMINATION REACTIONS E2 TWO EXAMPLES Alkyl halide + strong base and heat LOSS OF HCl D D E1 LOSS OF H2O Alcohol + strong acid and heat D

  4. NaOH + heat C H C H C H C H C H C H C H C H 3 2 3 3 3 C l conc. HCl ELIMINATION IS THE REVERSE OF ADDITION interconversions of alkyl halides and alkenes basic conditions + heat - HCl + HCl ALKYL HALIDE ALKENE acidic conditions

  5. C H C H C H C H 3 2 3 C H C H C H C H 3 3 O H ELIMINATION IS THE REVERSE OF ADDITION interconversions of alcohols and alkenes strong acid conditions + heat conc. H2SO4 - H2O + H2O ALCOHOL ALKENE 2-6 M H2SO4 dilute aqueous acid conditions

  6. STRONG BASES FIRST WE MUST LEARN WHAT IS A STRONG BASE

  7. .. : H-O .. 2 Na + 2 H2O .. : H-O .. WHAT ARE STRONG BASES ? NaOH sodium hydroxide used with H2O solvent, but most RCl insoluble in H2O 2 NaOH + H2 (g) KOH potassium hydroxide soluble in H2O, methanol or ethanol, most RCl soluble in methanol and ethanol 2 K + 2 H2O 2 KOH + H2 (g)

  8. STRONG BASES (continued) sodium alkoxides NaOR always used with the parent alcohol as solvent, most alkyl halides are soluble 2 Na + 2 ROH 2 NaOR + H2 (g) sodium methoxide examples: NaOCH3 (NaOMe) .. : R-O sodium ethoxide .. NaOCH2CH3 (NaOEt) sodium t-butoxide Stronger bases than hydroxides ….. why? NaOC(CH3)3 (NaOtBu)

  9. .. : NH2 STRONG BASES (continued) NaNH2 sodium amide always used with liquid ammonia solvent FeCl3 2 Na + 2 NH3 2 NaNH2 + H2 NH3 (liq) -33 oC A stronger base than the hydroxides or the alkoxides …. why? A gas at room temp ammonia NH3 bp -33.4 oC mp -77.7 oC liquifies solidifies

  10. SUMMARY OF STRONG BASES USED FOR ELIMINATION REACTIONS Base Solvents Allowed water NaOH KOH water, MeOH, EtOH (same R group) ROH NaOR NaNH2 NH3 (liq) -33o C Halides (RX) are not soluble in water, but are soluble in most alcohols, therefore, KOH or sodium alkoxides in alcohol are most often used.

  11. LEARN THIS ! IT IS THE FORMULA FOR A DEHYDROHALOGENATION REACTION ALKYL HALIDE + STRONG BASE + HEAT shorthand designation for this type of reaction E2 defined later...

  12. THE REACTION IS A b-ELIMINATION The b-hydrogen is attached to the b-carbon. a-carbon b-carbon The functional group is attached to the a-carbon. Since the b-hydrogen is lost this reaction is called a b-elimination. Reagent = a strong base

  13. MECHANISM THE BASE TAKES THE b-HYDROGEN B: B H H C C C C C l : : .. .. C l : : ..

  14. ALKYL HALIDE + STRONG BASE (E2) REGIOSELECTIVITY

  15. WHAT HAPPENS IF THERE IS MORE THAN ONE b-HYDROGEN ? b b’ Which one do we lose ?

  16. major product 81 % minor product 19 % ELIMINATION IS REGIOSELECTIVE b b’ 2-butene b-H 2-bromobutane 1-butene b’-H The major product is the one which has the lowest energy. See the next slide.

  17. HAMMOND POSTULATE lowest energy pathway Usually the pathway leading to the lowest energy product is the lowest energy pathway (lower TS). …. but not always (exceptions later) CH3CH2-CH=CH2 1-butene lowest energy product CH3-CH=CH-CH3 2-butene you get more trans than cis

  18. BUTENE ISOMERS - HEATS OF HYDROGENATION RECALL section 4.13 +H2 +H2 +H2 DH -27.6 kcal / mole -28.6 -30.3 CH3CH2CH2CH3 butane All are hydrogenated to the same product therefore their energies may be compared.

  19. MORE REGIOSELECTIVITY 1-chloro-1-methylcyclohexane methylenecyclohexane 1-methylcyclohexene b’ b’’ b b’ b major product minor product three possibilities to lose b-hydrogens b’’ 1-methylcyclohexene identical to b-product b’’= b

  20. RECALL METHYLCYCLOHEXENE ISOMERS +H2 +H2 kcal mole DH -27.8 -25.4 Both are hydrogenated to the same product therefore their energies may be compared. methylcyclohexane

  21. Zaitsev Tsayseff etc. SAYTZEV RULE The reaction gives the most highly-substituted (lowest energy) alkene as the major product.

  22. R H R H R R R R R H R R H H R R H H R H R H H R Different positions of the double bond. ALKENE ISOMERS THE MORE SUBSTITUTED ISOMER IS MORE STABLE decreasing energy 1,1- cis 1,2- monosubstituted trisubstituted trans 1,2- tetrasubstituted disubstituted increasing substitution

  23. APPLICATIONS OF THE ZAITZEV RULE 2-bromobutane major product 81 % DISUBSTITUTED minor product 19 % MONOSUBSTITUTED 1-chloro-1-methylcyclohexane major product minor product TRISUBSTITUTED DISUBSTITUTED

  24. ALKYL HALIDE + STRONG BASE (E2) STEREOCHEMISTRY OF THE REACTION

  25. STEREOCHEMISTRY TWO EXTREME POSSIBILITIES FOR THE ELIMINATION PROCESS C C syn elimination H Cl not common H Cl H H anti elimination C C observed most often Cl Cl anti-coplanar

  26. STEREOCHEMISTRY ACYCLIC HALIDES

  27. ACYCLIC MOLECULES MAY HAVE TO ROTATE IN ORDER TO REACT H Cl H Cl anti-coplanar

  28. 2-Bromo-3-phenylbutane RS SS SR RR two stereocenters enantiomers * * diastereomers * Major Product Zaitsev Minor Product cis or trans ? Is stereochemistry important?

  29. 2S,3R-DIASTEREOMER R S not observed rotate NaOMe trans MeOH (Z) observed major product anti-coplanar trans-2-phenyl-2-butene (plus some 1-butene due to b-H on the CH3)

  30. P h C H C H C H 3 3 3 C C C C C H H P h H 3 H B r H B r H C H 3 P h C C H C H 3 B r P h C H 3 C C C H H 3 MAKING THE 2S,3S-DIASTEREOMER R S S S R S make diastereomer ( change one stereocenter ) ? cis or trans? WHAT DO YOU THINK? Will the 2S,3S-diastereomer give the same product as its 2S,3R diastereomer? trans (Z)

  31. 2S,3S-DIASTEREOMER S S not observed rotate NaOMe cis MeOH (E) observed major product anti-coplanar cis-2-phenyl-2-butene NO (plus some 1-butene) A DIFFERENT PRODUCT IS FORMED THAN WITH 2S,3R !

  32. C H 3 C H 3 P h C l CONVERTING THE ALKYL HALIDE TO AN ALKENE VISUALIZING THE PRODUCT THAT FORMS H (top) H CH3 CH3 H Ph H Cl (bottom) alkyl halide alkene The methyl groups (blue) are in back in both structures. The phenyl and the hydrogen (black) are in front in both.

  33. ANOTHER VISUALIZATION OF THE REACTION H 2S,3R CH3 back carbon H C H CH3 C CH3 Ph front carbon Ph CH3 Cl same side same side 2S,3S H H CH3 C H CH3 C CH3 Ph Ph Cl CH3 same side same side

  34. CYCLIC HALIDES STEREOCHEMISTRY REARS ITS UGLY HEAD AGAIN !

  35. 1-Bromo-2-methylcyclohexane The expected result (naïve) : major product Zaitsev minor drawn flat without stereochemistry The result actually depends on the stereochemistry of the starting material ( cis or trans ).

  36. trans shown on next slide THE CIS STEREOISOMER The elimination needs to have H and Br anti-coplanar major product (Zaitsev) plus a small amount of cis The other chair won’t work. Why? CH3 Br

  37. THE TRANS STEREOISOMER only product trans The other chair won’t work. Why? CH3 Br no methylcyclohexene is formed

  38. CH3 Cl THE RING MAY HAVE TO INVERT FOR REACTION this ring cannot react anti- coplanar invert chlorine is not anti-coplanar to any hydrogen KOH / EtOH

  39. REACTION DOES NOT OCCUR EASILY IF ANTI-COPLANAR GEOMETRY CANNOT BE ACHIEVED trans e,e Br is not axial, no anti-coplanar H no reaction hydrogens equivalent cis e,a THESE RINGS WILL NOT INVERT ( WHYNOT ? )

  40. A CASE THAT FOLLOWS THE HAMMOND POSTULATE 1 HAMMOND POSTULATE The activation energy leading to the product of lower energy will be lower than the activation energy leading to the product of higher energy ….. unless stereochemistry or some other important factor interferes.

  41. E2 REGIOSELECTIVE FOLLOWS HAMMOND POSTULATE * AND ZAITSEV RULE NaOEt / EtOH major product * The activation energy leading to the product of lower energy will be lower than the activation energy leading to the product of higher energy.

  42. A CASE THAT CAN NOT FOLLOW THE HAMMOND POSTULATE 2 HAMMOND POSTULATE OR….STEREOCHEMISTRY REARS ITS UGLY HEAD !

  43. E2 REGIOSPECIFIC DOES NOT FOLLOW HAMMOND POSTULATE incorrect stereochemistry raises Ea OR ZAITSEV RULE * H NaOEt / EtOH only product * Due to stereochemical complications.

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