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HOFMANN REACTION

HOFMANN REACTION. E2 gone astray. HOFMANN RULE. When you have a bulky leaving group like -N(CH 3 ) 3 + the least-substituted alkene will be the major product. Big is not the same as bulky. BULKY =. Branched at the first atom attached to the chain. chain. chain. trimethyl

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HOFMANN REACTION

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  1. HOFMANN REACTION E2 gone astray

  2. HOFMANN RULE When you have a bulky leaving group like -N(CH3)3+ the least-substituted alkene will be the major product. Big is not the same as bulky. BULKY= Branched at the first atom attached to the chain chain chain trimethyl ammonium dimethyl sulfonium OTHER GROUPS FOLLOW THE ZAITSEV RULE

  3. HOFMANN ELIMINATION Hofmann found that when the leaving group was -N(CH3)3+ E2 elimination reactions gave the least-substituted alkene. Hofmann 95% 5% Zaitsev 31% 69%

  4. EFFECT OF INCREASING SUBSTITUENT BULK E2 (cis + trans) HOFMANN ZAITSEV 31% 69% 30% 70% 48% 52% Big is not the same as bulky. 87% 13% 98% 2%

  5. ANALYSIS OF 2-SUBSTITUTED PENTANE ELIMINATIONS ZAITSEV less crowding in this area of the molecule cis H trans C C C C C H 3 H X bulky groups cause crowding and give Hofmann products HOFMANN ( all H equivalent )

  6. H most steric crowding CH3 H less steric crowding H CH2CH3 H no steric crowding CH3 H N(CH3)3 H CH3CH2 H CH2CH2CH3 H N(CH3)3 H H N(CH3)3 WOULD MAKE WOULD MAKE cis trans ZAITSEV PRODUCTS HOFMANN PRODUCT FORMED NOT FORMED

  7. BULKY BASES ALSO INCREASE HOFMANN PRODUCT ZAITSEV HOFMANN methoxide 81% 19% t -butoxide 47% 53% bulky base

  8. BULKY b-SUBSTITUENTS What constitutes bulky? a methyl group is not bulky ZAITSEV HOFMANN b a NO NaOEt 80% 20% even two or three are not bulky b NO NaOEt 79% 21% YES NaOEt 14% 86% b t-butyl is bulky !

  9. CH3 C CH3 CH3 H H CH3 CH3 CH2 CH3 CH3 H C H H CH3 CH3 Br Br THE ELIMINATION MOVES TO A LESS CROWDED REGION less crowded crowded crowding 14% 86% REACTIVE CONFORMATIONS crowding spacer

  10. EXAMPLES

  11. HOW THE VARIOUS FACTORS AFFECT THE OUTCOME CH3 NORMAL CH3 NaOEt Zaitsev Br EtOH / D ~90% Bromine is big, not bulky BULKY LEAVING GROUP CH3 CH2 Hofmann KOH + ~90% N(CH3)3 EtOH / D - I Prototypical “Hofmann” elimination CH3 BULKY BASE CH3 CH2 NaOtBu + Br tBuOH / D Bulky base alone not as effective as bulky leaving group ~60/40%

  12. HOW THE VARIOUS FACTORS AFFECT THE OUTCOME ( CONTINUED ) BULKY BASE & LEAVING GROUP CH3 CH2 NaOtBu Hofmann + N(CH3)3 tBuOH / D ~100% - I Double Whammy ! Bulky base + bulky leaving group BULKY b-SUBSTITUENT CH3 CH3 CH2 NaOEt Br + EtOH / D tBu tBu tBu H no double bond here either cis or trans to Br - same result Favors Hofmann products Use a bulky base here and ...

  13. E2 REACTIONS DEVIATE FROM THE ZAITSEV RULE 1. If the favored b-hydrogen can’t achieve anti-coplanar geometry. 2. If the double bond would form at a bridgehead. (see next slides) 3. If there is a bulky leaving group. 4. If there is a bulky base. 5. If there is a bulky b-substituent.

  14. BREDT’S RULE A double bond cannot form at a bridgehead.

  15. BREDT’S RULE A double bond cannot form at a bridgehead in a bicyclic system with small rings. no way ! z p orbitals cannot become coplanar y y Try a model ! z

  16. SYN ELIMINATION More difficult than anti-coplanar elimination, but does occur in some circumstances. ( Usually requires forced conditions - heat and pressure. )

  17. H C C 180o anti-coplanar BEST SITUATION X H X syn- coplanar SECOND-BEST SITUATION C C 0o COPLANAR ARRANGEMENTS E2 ~ ~ ~ ~ ~ ~ not coplanar ( very difficult ) other angles syn- coplanar (difficult) 0o anti- coplanar (easiest) Difficulty Order 180o

  18. SYN-COPLANAR ELIMINATIONS REQUIRE FORCED CONDITIONS H X syn-coplanar (0o) C C C C This often requires heating above the boiling point of the solvent in a sealed tube (next slide). Temperatures above 100 oC are common.

  19. SEALED TUBE Glass tube sealed at both ends. glass tube alkyl halide + NaOEt / EtOH (bp 78 oC) inside tube 1.25” D REACTANTS 0.25” wall Carried out in a hood behind a glass shield. Allows reactants to be heated to a high temperature (above bp) without boiling away. heated oil (bp > 250 oC) hot plate

  20. H H B r H H H H H 100% D B r D H H H H A CASE OF SYN ELIMINATION SYN ELIMINATION OCCURS BECAUSE THERE ARE NO ANTI-COPLANAR b-H Bredt’s Rule NaOEt EtOH + 110 oC 0% NaOEt EtOH same products NO DEUTERIUM 110 oC proves the syn hydrogen is the one removed not this one

  21. SYN ELIMINATION IS SEEN WHEN ANTI-COPLANAR DOES NOT EXIST not coplanar hydrogens are not anti-coplanar to chlorines NaOH 110o very slow difficult reaction coplanar NaOH 110o faster syn- elimination

  22. a-ELIMINATION occurs when the substrate has NO b-HYDROGENS on these carbon atoms

  23. a -ELIMINATION Cl Cl - KOH : C C H Cl Cl CHCl3 Cl Cl no b -hydrogens a -elimination very reactive Cl .. : : : C Cl Cl + .. a carbene

  24. Cl : C Cl Cl : C Cl CARBENES ARE ELECTROPHILES ! Because of an incomplete octet carbenes are electrophilic (need electrons to complete their valence shell). missing a pair of electrons Carbenes will react with an alkene (electron pair donor). : nucleophile electrophile

  25. CARBENES ADD TO DOUBLE BONDS alkene + carbene = cyclopropane ring H 2 H 1 syn addition Probably concerted : steps 1,2 for visualization only

  26. ANALYSIS OF THE ADDITION 2p H C : . SP2 hybrid . H stepwise analysis of the concerted process syn addition + - .. : the intermediate does not exist

  27. C Cl Cl SYN ADDITION CHCl3 BOTH NEW RING BONDS FORM ON THE SAME SIDE OF THE DOUBLE BOND KOH EtOH H Stereospecific H

  28. STEREOSPECIFICITY PROVES THE REACTION IS CONCERTED SUBSTITUENTS ON THE DOUBLE BOND RETAIN THEIR ORIGINAL CIS OR TRANS RELATIONSHIPS IN THE NEW RING CHCl3 KOH cis cis EtOH CHCl3 KOH trans trans EtOH

  29. H H CH3 CH3 C Cl Cl C Cl Cl STEREOSPECIFIC = CONCERTED YES ! Concerted. Both bonds form on the same side. Stereospecific! H NO ! An intermediate would allow rotation. H + CH3 This does not happen. .. CH3 Would not be stereospecific.

  30. Ph Ph C H C H Ph H Cl Cl Cl Cl C H C H Cl H Cl Cl COMPOUNDS WITHOUT b-HYDROGENS Not a common type of compound !

  31. ELIMINATIONS SUMMARY

  32. K.I.S.S. THE MOST BASIC STUFF alkyl halide + strong base + heat = E2 alkyl halide + solvent + heat (solvolysis) = E1 alcohol + strong acid + heat = E1 (acid assisted) typical situation for E1cb H next to C=O (easy to remove) X = strong base (difficult to break bond) Only E1 reactions have rearrangements (carbocations) Only E2 reactions require anti-coplanar b-hydrogens

  33. THE BIG PICTURE alkyl halides alcohols special E1cb E2 E1 E1 E1 acid assisted strong strong weak acidic neutral base base base stepwise - concerted stepwise - carbocation carbanion stereospecific “solvolysis” anti-coplanar special case - not common Zaitsev if not stereospecific stereochem requires: allows acidic H and Zaitsev Zaitsev Zaitsev Hofmann if poor leaving bulky groups group a-elim. if no b-H carbocation rearrangements

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