ucleophilic Substitution & Elimination hemistry 1 What kind of mechanisms are possible? What is the major mechanism occuring? Write in ALL mechanism details (lone pairs, formal charge, curved arrows, etc.). edraw your structure each time you show a different reaction with the structure. Mechanism choices at this point in organic chemistry: S 2,E2,S 1, E1,acid/base reaction, free radical substitution, no reaction uclephile/bases (and other reagents) to choose from include: purchase / given - you can invoke these whenever you need them 2 hν sodium hydroxide 2 sodium amide sodium hydride (only basic) K potassium hydride (only basic) 3 sodium azide S monohydrogen sulfide sodium cyanide Al sodium lithium aluminium borohydride hydride (nucleophilic) (nucleophilic) you have to make these using acid/base reactions B Al lithium aluminium hydride (nucleophilic) B sodium borohydride (nucleophilic) water ethanoic acid propanone (acetone) alcohols alkoxides t-butyl alcohol K K potassium t-butoxide ethanoic acid (acetic acid) ethanoate (acetate) terminal alkynes 2 terminal acetylides S thiols S alkylthiolates Synthesis of lithium diisopropyl amide, LA, used to remove α - proton of carbonyl groups. (acid / base reaction) given 2 given K eq = K a1 10-38 = = K a2 10-50 10 +12 Think - sterically bulky, very basic that goes after weakly acidic protons. LA = lithium diisopropyl amide Make enolate (ketones and esters) given 2 LA ketones (and other carbonyl compounds) K eq = K a1 10-20 = = K a2 10-38 10 +18 ketone enolates (resonance stabilized) eact ketone enolate with compounds (methyl, 1 o and 2 o ) 2 ketone enolates (resonance stabilized) S 2 reactions 1 o key bond S Larger ketone made from smaller ketone. Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 2 Make enolate (esters) given 2 LA ketones (and other carbonyl compounds) K eq = K a1 10-20 = = K a2 10-38 10 +18 ketone enolates (resonance stabilized) eact ester enolate with compounds (methyl, 1 o and 2 o ) 2 ester enolates (resonance stabilized) S 2 reactions 1 o key bond S Larger ester made from smaller ester. S and E reactions S 2 reactions = substitution nucleophilic bimolecular ate = k S2 [][u: ] bimolecular kinetics, both and u: participate in the slow step of the reaction; k S2 = Ax10 E2 reactions = elimination bimolecular ate = k E2 [][B: ] bimolecular kinetics, both and B: participate in the slow step of the reaction; k E2 = Ax10 S 1 reactions = substitution nucleophilic unimolecular - Ea(S1) ate = k S1 [] unimolecular kinetics, only participates in the slow step of the reaction; k S1 = Ax10 2.3xxT E1 reactions = elimination unimolecular - Ea(E1) ate = k E1 [] unimolecular kinetics, only participates in the slow step of the reaction; k E1 = Ax10 2.3xxT - Ea(E2) 2.3xxT - Ea(S2) 2.3xxT Allowed starting structures - main sources of carbon 4 We cannot make these yet (1 o ) so they are given. These structures represent your starting points to synthesize target molecules below. You will need to propose a step-by-step synthesis for each target molecule from the given structures above. Every step needs to show a reaction arrow with the appropriate reagent(s) above each arrow and the major product of each step. The product of each step becomes the starting material for the next step until you reach the target structure. As new reagents are introduced this list will expand and as new reactions are learned the necessary hydrocarbons will contract. Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 3 - patterns - typical reaction patterns in our course (bold = atypical result) eagents 2 methyl only S 2 only S 2 only S 2 1 o primary S 2 > E2 S 2 > E2 S 2 > E2 2 o secondary E2 > S 2 E2 > S 2 S 2 > E2 3 o tertiary 1 o neopentyl no reaction no reaction no reaction allylic benzylic very fast S 2 very fast S 2 very fast S 2 very fast S 2 very fast S 2 very fast S 2 K -( 3 ) 3 t-butoxide only S 2 E2 > S 2 no reaction very fast S 2 very fast S 2 only S 2 S 2 > E2 S 2 > E2 no reaction very fast S 2 very fast S 2 only S 2 S 2 > E2 E2 > S 2 no reaction very fast S 2 very fast S 2 3 only S 2 S 2 > E2 S 2 > E2 no reaction very fast S 2 very fast S 2 S only S 2 S 2 > E2 S 2 > E2 no reaction very fast S 2 very fast S 2 S only S 2 S 2 > E2 S 2 > E2 no reaction very fast S 2 very fast S 2 B 3 only S 2 S 2 > E2 S 2 > E2 no reaction very fast S 2 very fast S 2 B 3 only S 2 S 2 > E2 S 2 > E2 no reaction very fast S 2 very fast S 2 Al 3 only S 2 S 2 > E2 S 2 > E2 no reaction very fast S 2 very fast S 2 Al 3 only S 2 S 2 > E2 S 2 > E2 no reaction very fast S 2 very fast S 2 only S 2 S 2 > E2 S 2 > E2 no reaction very fast S 2 very fast S 2 ketone enolates ester enolates only S 2 S 2 > E2 S 2 > E2 no reaction very fast S 2 very fast S 2 The following are synthetic approaches to some target molecules using S and E reactions. They are presented as examples of how you might approach synthesis problems (these and others like them). You need to not only look at structures that I included below but think about other possibilities using all of the structures available to you. I do not list every possibility, but you need to be able to consider every possibility. The best strategy is to work backwards from the target molecule one step at a time to an allowed starting structure (called retrosynthetic analysis). That way you are only thinking about one backward step at a time, instead of an unknown number of steps forward from a starting structure. This is the way I presented the approaches below. You should also know the mechanism for every reaction below. I can write these pages until I am blue in the face, but they can t help you learn, unless you do the work of trying them out. My job is to give you an opportunity to learn, and your job is to take advantage of that opportunity. These schemes were made quickly, so be on the lookout for mistakes. (I hope not too many.) 1. Free radical substitution at sp 3 - Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 4 Free radical substitution mechanism: 1. initiation 2a and 2b. propagation 3. termination. of each step is controlled by relative bond energies (- + 2 + light) Sample - from - using 2 / hν - you have to make these 3 methyl 1 o 2 o 3 o 1 o 1 o 1 o 2 o 1 o benzylic 2 o benzylic We cannot make these yet (1 o ) so they are given. allylic - (BS is more likely reagent, but we only know 2 ). You will need to make the starting alkene (E2 reactions) 2 hν free radical substitution 1 o allylic E2 from t-butyl bromide + 1 o allylic 2 o allylic 2. alcohol synthesis 2 hν free radical substitution E2 potassium t-butoxide 2 hν free radical substitution 2 hν free radical substitution E2 from 2-bromopropane + t-butoxide Sample alcohols - S 2 reactions using (- + ) or S 1 reactions using (- + 2 ) 3 alcohols - various approaches, S 2 at methyl and primary using and S 1 at secondary and tertiary using 2 target molecule S 2 > E2 allowed starting structure ydroxide and alkoxides are K nucleophiles at methyl and 1 o, but E2 > S 2 at 2 o and at 3 o. target molecule 2 2 S 1 > E1 hν free radical substitution allowed starting structure Generally, S 1 > E1 at 2 o and 3 o. o S 1/E1 at methyl and 1 o. Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 5 target molecule 2 S 1 > E1 2 hν free radical substitution allowed starting structure Generally, S 1 > E1 at 2 o and 3 o. o S 1/E1 at methyl and 1 o. aqueous hydroxide introduces a new alcohol functional group (-) 3 S 2 key bond 3 3 2 S 2 key bond key bond 3 2 2 S 2 3. ether synthesis Sample ethers - S 2 reactions using (- + ) or S 1 reactions using (- + ) (S 1 or S 2) (S 1 or S 2) ethers - various approaches, S 2 at methyl and primary using and S 1 at secondary and tertiary using target molecule 1. acid / base 2. S 2 > E2 Generally, S 1 > E1 at 2 o and 3 o. o S 1/E1at methyl and 1 o. S 1 > E1 S 2 2 hν free radical substitution given 2 hν free radical substitution given Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 6 target molecule 1. acid / base 2. Ph S 2 > E2 2 hν free radical substitution already made 2 hν S 2 Ph free radical substitution given Ph S 1 > E1 already made Ph = phenyl alkoxide introduces a new ether group (- ) 3 3 S 2 3 key bond 3 3 3 2 S 2 3 key bond 3 key bond 3 2 2 S 2 3 4. ester synthesis (and one ketone example using enolates) Sample esters - S 2 reactions using (- + 2 ) or S 1 reactions using (- + 2 ) esters - various approaches, S 2 at methyl and primary using 2 and S 1 at secondary and tertiary using 2 1. acid / base 2. already made target molecule S 2 > E2 given arboxylates are K nucleophiles at methyl, 1 o and 2 o but at 3 o. S 1 > E1 given already made Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 7 ester enolates target molecule key bond ketone enolates key bond 1. LA (makes enolate) acid / base 2. S 2 > E2 1. LA (makes enolate) acid / base 2. given S 2 > E2 already made n-butyl lithium acid / base already made n-butyl lithium acid / base diisopropylamine Enolates are K nucleophiles at methyl, 1 o and 2 o but at 3 o. (See ketone enolates below) Enolates are K nucleophiles at methyl, 1 o and 2 o but at 3 o. (similar to ester enolates above) ketone enolates (good nucleophiles at epoxides, carbonyls and 3, 1 o and 2 o ) bigger ketones 3 2 3 S 2 (alkylation) 3 2 (4 = 3 + 1) key bond 3 3 2 2 3 S 2 (alkylation) key bond 3 2 3 2 2 S 2 (alkylation) key bond (6 = 3 + 3) 5. azide synthesis (these can be made into primary amines via 1. Al 4 2. workup) Sample azides - S 2 reactions using (- + 3 ) 3 3 3 3 3 3 3 3 3 3-2 azide anion Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 8-2 S 2 2-azidopropane 6. amine synthesis (at this point primary amines are made from azido compounds via 1. Al 4 2. workup) Sample amines - S 2 reactions using (1. - 3 + Al 4 2. WK) 2 2 3 2 2 2 2 2 2 2 3 Al lithium aluminium hydride 2 workup alkyl azide nitrogen 1 o amines 7. alkene synthesis (via E2 reactions, one time) Sample alkenes using E2 reactions: - + t-butoxide; alkenes can make allylic - compounds Sample allylic bromides from alkenes alkenes 1. acid / base inexpensive base is K because is tertiary already made K acid / base Use steric bulk and/or extreme basicity to drive E2 > S 2. relative alkene stabilities = tetrasub. > trisub. > trans-disub > cis-disub gem-disub > monosub > unsub. already made K already made E2 > S 2 (need strong, bulky base to favor E2 at 1 o ) Use steric bulk and/or extreme basicity to drive E2 > S 2. already made already made K already made K E2 E2 already made Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 9 8. alkyne synthesis (via E2 reactions, two times) Sample alkynes using double E2 reactions of 1. 2 + 2 2. workup alkynes (use two leaving groups) (2 equivalents) 2 and sodium amide (very strong base) E2 twice (2 equivalents) 2 sodium amide (very strong base) E2 twice (2 equivalents) 2 hν free radical substitution (2 equivalents) 2 hν free radical substitution Sample alkynes using S 2 reactions of 1. terminal alkyne + 2 2. methyl or primary - (twice with ethyne) 1. 2 sodium amide (very strong base) (1 equivalent) acid / base 2. 3 S 2 > E2 1. 2 sodium amide (very strong base) (1 equivalent) acid / base 2. S 2 > E2 Use steric bulk and/or extreme basicity to drive E2 > S 2. Terminal acetylides are K nucleophiles at methyl and 1 o, but E2 > S 2 at 2 o and at 3 o. Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 10 Starting alkynes (2 and 3) 2 equivalents 2 /hν 2 equivalents 1. / 2 2. workup (neutralize) 2 equivalents 2 /hν 2 equivalents 1. / 2 2. workup (neutralize) a b 2 2. workup 2. 2 b a 2. 1 equivalent 1. 2 (4 = 2 + 2) terminal alkyne 2. 1 equivalent 1. 2 (5 = 3 + 2) internal alkyne 1 equivalent 1. 2 2. (5 = 4 + 1) terminal alkyne 2. 1 equivalent 1. 2 1. excess 2. workup (neutralize) (zipper reaction) (5 = 3 + 2) internal alkyne Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 11 1 equivalent 1. 2 2. (6 = 4 + 2) terminal alkyne 1 equivalent 2. 1. excess 2. workup (neutralize) (zippe reactionr) 1. 2 (6 = 3 + 3) internal alkyne terminal acetylides larger alkynes 3 3 S 2 3 3 3 2 3 2 S 2 3 3 3 2 3 2 S 2 3 2 2 3 9. nitrile synthesis (via S 2 reaction) Sample nitriles using S 2 reactions of cyanide + methyl, primary or secondary - 3 nitriles allowed S 2 > E2 already made yanides are K nucleophiles at methyl, 1 o and 2 o but at 3 o. allowed S 2 > E2 already made cyanide nitriles Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 12 3 S 2 3 2 2 S 2 3 3 2 2 S 2 3 2 2 3 10. thiols and sulfides synthesis (via S 2 reaction) Sample thiols using S 2 reactions of S + methyl, primary or secondary - 3 S S S S S S S S S Sample sulifides using S 2 reactions of S + methyl, primary or secondary - S S S S thiols and sulfides S sulfide 1. acid / base 2. S 2 > E2 thiol (pk a 8) S S S 2 > E2 already made Sulfides are K nucleophiles at methyl, 1 o and 2 o but at 3 o. 11. introducing hydride or deuteride into organic molecules (via S 2 reaction) Sample deuterium added using S 2 reactions of Al 4 or B 4 + methyl, primary or secondary - 3 Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 13 S 2 Al Al B lithium aluminium hydride = LA = ( Al 4 ) and sodium borohydride (B 4 ) are both nucleophilic hydride. For now we will consider them to be equivalent reagents. lithium aluminium hydride (Al 4 ) and sodium borohydride (B 4 ) = nucleophilic hydride (using deuteride shows where the hydrogen goes) Al Al Al 3 3 2 S 2 S 2 key bond 3 key bond 2 S 2 3 2 propane Mechanisms Worksheet You need to be able to write your own mechanism from starting structures. I suggest you should practice using the nucleophiles from the list with the various compounds listed above. Writing mechanisms for each possibility would give you more than enough practice to learn the mechanism. ommon mistakes include lone pairs, formal charge, curved arrows, correct Lewis structures, correct products predicted. This stuff takes practice A correcting your mistakes. I have set up the following worksheet to give you templates to fill in so that you can use them multiple times by copying the basic framework and then filling in the details. If you don t practice, you will be taking 314 again and you don t want to do that. TE WK! key bond ethane Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 14 S 2 then E2 examples first, followed by S 1 and E1 reactions Example 1-3 - (deuterium -, and tritium = T are isotopes of hydrogen that can be distinguished from ) The only possible choice is S 2. Methyl + is too high energy for solution chemistry, so no S 1/1. no reaction - these are S only S 2 (strong) 1 u T and E1 conditions (weak)? B choose from above, Examples you might consider: S-bromodeuteriotritiomethane K Example 1-3 - only S 2 u T (with strong nucleophiles) B choose S from above, S-bromodeuteriotritiomethane u T K Example 2 - primary (deuterium is an isotope of hydrogen that can be distinguished from ) u B choose from above Example 2 - primary B u b 3 a (1S,2S)- 1-bromo-1,2-dideuteriopropane β 3 α (1S,2S)- 1,2-dideuterio-1-bromopropane β 3? S 2 > E2 S 2 > E2 3 u u b? B 3 a choose from above (1S,2)- 1-bromo-1,2-dideuteriopropane β α β E2 3 3 β S 2 Some examples S 2 > E2 other primary to consider. 3 2 α (unless t-butoxide) (1S,2S)- 1,2-dideuterio-1-bromopropane Mechanism predictions with: 3 3 3 2 look at β - fully substituted no S 2 or E2 at 1 o α E-alkene? α Z-alkene K E2 > S 2 benzylic = fast S 2 o S 1/E1 is possible at primary in solution chemistry. The primary + is too high in energy. Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 15 Example 3 - secondary (deuterium is an isotope of hydrogen that can be distinguished from ) u 3 u B choose from above β β α 3 (2S,3,4S)-2-deuterio-4-methyl-3-bromohexane 2 3? B 3 choose from above β α β 2 3 3 (2S,3S,4S)-2-deuterio-4-methyl-3-bromohexane? u S 2 and E2 depends on basicity of donor 3 (2,3,4S)-2-deuterio-3-bromo-4-methylhexane B β β α β β α 3 3 2 3 S 2 and E2 depends on basicity of donor 3 (2,3,4S)-2-deuterio-3-bromo-4-methylhexane u B β β β α 2 3 (2S,3)-3-deuterio-2-bromobutane β α 3 3 (2S,3)-3-deuterio-2-bromobutane S 2 and E2 depends on basicity of donor S 2 and E2 depends on basicity of donor u 3 α β β still S S 2 3 3 2 3 more basic = E2 > S 2 less basic = S 2 > E2 still S Some examples 3 3 2 3 2Z-alkene 2E-alkene 3Z-alkene u 2 3 α 2Z-alkene β 3 β S 2 3 3 2E-alkene 3 E2 2 3 3 more basic = E2 > S 2 less basic = S 2 > E2 2 3 1-butene, neither E or Z still Some examples still K 3 K Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 16 other secondary to consider. 3 3 3 3 3 Mechanism predictions with:? Example 4 - tertiary (deuterium is an isotope of hydrogen that can be distinguished) 3 u u B choose from above β 2 β α 2 3? B choose from above 3 β α 2 β Usually S 1 > E1 (except + 2 S 4 /.) 2 3 3? 3 (2,3,4S)-2-deuterio-4-methyl-3-bromohexane (2S,3S,4S)-2-deuterio-4-methyl-3-bromohexane u 3 (2S,3,4S)-2-deuterio-4-methyl-3-bromohexane B β 2 β β 2 β α 3 α 3 2 3 S 2 3 E2 reaction at 3 o with strong basenucleophile 3 (2,3,4S)-2-deuterio-4-methyl-3-bromohexane 3 o S 2 reaction at 3 o, backside is too sterically hindered 3 2Z-alkene 2 3 3 3 2 2E-alkene possible at 3 o and strong base-nucleophile, need b - though. 3 3 3 still 2 3 3Z-alkene 3 Some examples 2 3 still 2 K 3 still S 3 1-butene, neither E or Z Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 17 other tertiary to consider. 3 3 2 3 3 3 Mechanism predictions with:? Usually S 1 > E1 (except + 2 S 4 /.) Example 5 - cyclohexane structures to consider ( must be axial to react by S 2 and E2) ( can be axial or equatorial to react by S 1 and E1). Essential details can be filled in on the following templates. cyclohexane without branches cyclohexane with a branch Many substitution patterns are possible. a b c d e f g h i j Mechanism predictions with: Some examples weak nucleophiles K strong nucleophile/bases + other anions shown above Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 18 S chemistry with enolates (mechanisms) Synthesis of lithium diisopropyl amide, LA. (acid / base reaction) LA = lithium diisopropyl amide 2 given given K eq = K a1 10-38 = = K a2 10-50 10 +12 Think - sterically bulky, very basic that goes after weakly acidic protons. Make enolate (ketones and esters) given 2 LA ketones (and other carbonyl compounds) K eq = K a1 10-20 = = K a2 10-38 10 +18 ketone enolates (resonance stabilized) eact enolate with compounds (methyl, 1 o and 2 o ) 2 ketone enolates (resonance stabilized) S 1 o key bond Larger ketone made from smaller ketone. Synthesis of alkoxide nucleophiles with sodium hydride. (acid / base reaction) K a1 10-16 K eq = = = K a2 10-37 10 +21 alcohols alkoxides eact alkoxides with compounds (methyl and 1 o S 2 favored, 2 o and 3 o E2 favored) alkoxides 1 o S 2 ethers alkoxides 3 o E2 2 alkenes Synthesis of ethanoate (acetate) nucleophile with sodium hydroxide. (acid / base reaction) K a1 10-5 carboxylic acids K eq = = = K a2 10-16 10 +11 water eact carboxylates with compounds (S 2 at methyl, 1 o and 2 o and E2 at 3 o ) carboxylates carboxylates S 2 esters Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 19 carboxylates 3 o E2 2 Simple Examples general patterns (this is as easy as it gets, more complicated examples follow) nucleophile / base 3 only S 2 S 2 > E2 E2 > S 2 o eaction 3 nucleophile / base 3 only S 2 S 2 > E2 E2 > S 2 3 o eaction Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 20 nucleophile / base 3 S only S 2 2 > E2 S 2 > E2 o eaction nucleophile / base 3 only S 2 E2 > S 2 o eaction K nucleophile / base 3 o eaction o eaction S 1 > E1 S 1 > E1 o eaction S 1 > E1 & S & S with rearrangement Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 21 nucleophile / base 3 o eaction o eaction S 1 > E1 S 1 > E1 o eaction S 1 > E1 & S & S with rearrangement nucleophile / base 3 o eaction o eaction S 1 > E1 & S S 1 > E1 o eaction S 1 > E1 & S with rearrangement Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 22 S 1 / E1 possibilities extra complications at β positions, 2 o, rearrangements T considered ( 2,, 2 ) Examples of weak nucleophile/bases in our course c a b water alcohols carboxylic acids 3 (3) edrawn (achiral) 2 same first step for S 1/E1 3 2 E1 product from left β carbon atom 2 3 3 2 3 2 2 2 E1 product from right β carbon atom 3 3 "" parallel to on top "" parallel to on bottom (achiral) 3 2 3 2 2 carbocation choices 2 3 2 3 3 add nuclephile (top/bottom) = S 1 lose β- (top/bottom) = E1 rearrangement to similar or more stable carbocation ( + ) (start all over) 2E-alkene 2Z-alkene 3 3 2 2 2 3 2 2 3 2 "" parallel to on top "" parallel to on bottom 3 2 3 2 2 3 2 3 3E-alkene 3Z-alkene S 1 product (a. add from top and b. add from bottom) 3 2 (achiral) 2 b a 2 2 2 3 2 b a 3 3 2 2 2 2 2 2 acid/base proton transfer 3 3 3 2 acid/base proton transfer 2 3 2 enantiomers 2 2 2 (3) 2 (3S) 3 3 2 Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 23 S 1 / E1 possibilities extra complications at β positions, 2 o, rearrangements T considered, with deuterium (makes it a little harder) Examples of weak nucleophile/bases c a b water alcohols carboxylic acids 3 (2S,3,4S) edrawn (2S,4S) 2 3 same first step for S 1/E1 2 3 (2S,4S) 2 3 carbocation choices add nuclephile (top/bottom) = S 1 lose β- (top/bottom) = E1 rearrangement to similar or more stable carbocation ( + ) (start all over) E1 product from left β carbon atom 2 (2S,4S) 3 2 "" parallel to on top 3 3 2 (4S) 3 4S,2E-alkene without "" (2S,4S) 3 2 2 "" parallel to on bottom 3 2 3 (4S) 3 4S,2Z-alkene without "" (2S,4S) 2 3 2 "" parallel to on top 3 2 3 (4S) 3 4S,2Z-alkene with "" 2 (2S,4S) 3 2 "" parallel to on bottom 3 3 2 (4S) 3 4S,2E-alkene with "" Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 24 edrawn from above (2S,4S) E1 product from right β carbon atom 3 (2S,4S) 2 2 3 "" parallel to on top 3 (2S) 2 3 2S,3E-alkene with "" 3 2 3 (2S,4S) 2 "" parallel to on bottom 3 (2S) 2 3 2S,3Z-alkene with "" 3 3 (2S,4S) 2 2 "" parallel to on top 3 (2S) 2 3 2S,3Z-alkene without "" 3 (2S,4S) 2 3 "" parallel to on bottom 3 (2S) 2 3 2S,3E-alkene without "" 2 S 1 product (a. add from top and b. add from bottom) (2S,4S) 3 2 b a 2 2 2 3 b a 3 3 2 2 acid/base proton transfer 3 3 3 2 acid/base proton transfer 2 3 diastereomers 2 (2S,3,4S) 3 3 2 (2S,3S,4S) Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 25 S 1 / E1 possibilities extra complications at β positions of 2 o, rearrangement to more stable 3 o + considered 3 edrawn (achiral) 2 3 same first step for S 1/E1 3 2 3 E1 product from left β carbon atom (without rearrangement) 2 3 (3,4S) 2 (4S) "" parallel to on top 3 (4S) 2 3 3 stereochemistry is lost 3 3 carbocation choices 3 3 2 3 add nuclephile (top/bottom) = S 1 lose β- (top/bottom) = E1 rearrangement to similar or more stable carbocation ( + ) (start all over) 4S,2E-alkene diastereomers 2 3 "" parallel to 2 on bottom (4S) 3 3 3 3 2 3 4S,2Z-alkene E1 product from right β carbon atom (without rearrangement) 3 2 3 (4S) 2 2 3 "" parallel to on top 3 2 2 3 3 3E-alkene diastereomers 3 3 2 2 (4S) 3 2 "" parallel to on bottom 3 2 2 S 1 product (a. add from top and b. add from bottom), (without rearrangement) 3 3 3Z-alkene 3 2 2 b a 2 2 2 (4S) 3 b a 3 3 2 2 2 3 32 (3,4S) 2 3 3 3 (3S,4S) 2 acid/base proton transfer acid/base proton transfer 3 3 2 diastereomers 2 (3S,4S) (3,4S) 2 3 2 3 3 3 Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 26 After rearrangement to 3 o carbocation ( + ) We will skip rearrangements in hem 314 3 edrawn (achiral) E1 products from left β carbon atom (top and bottom, after rearrangement) 2 3 2 2 3 3 (4S) 2 initial 2 o carbocation rearranges via hydride shift to a 3 o carbocation 3 3 3 "" parallel to on top and bottom of right β position 2 3 4S stereochemistry is lost in new carbocation 3 2 + reaction possibilities start all over again with new carbocation 3 diastereomers 2 3 3 3E-alkene add nuclephile (top/bottom) = S 1 lose β- (top/bottom) = E1 rearrangement to similar or more stable carbocation ( + ) (none is possible here) 3 3 2 2 3Z-alkene 3 E1 product from right β carbon atom (top and bottom, after rearrangement) diastereomers "" parallel to 2 3 3 3 3 3 on top and bottom 2 of right β position 2 3 3 2Z-alkene 2 2 2 3 2E-alkene 3 E1 product from methyl β carbon atom (top and bottom, after rearrangement, only one product from the methyl) "" parallel to 2 3 3 3 3 on top and bottom 2 of right 2 β position 2 2 2 S 1 product (a. add from top and b. add from bottom), (after rearrangement) 2 2 1-alkene (does not have E/Z stereochemistry) 3 2 2 b (achiral) a 3 2 3 b a 3 3 2 2 3 2 3 2 (3) 3 2 2 (3S) 3 2 2 acid/base proton transfer acid/base proton transfer 3 3 2 2 3 2 3 2 enantiomers (3) 2 (3S) 3 2 3 Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 27 omework problems: The number of each type of product (S1, E1, S2, E2) is listed after a reaction arrow for each starting structure (assuming I analyzed the possibilities accurately in my head, while sitting at the computer). See if you can generate those products using a valid mechanism for each one. a. b. S 1 E1 S 2 E2 2 1 4 3 2 5 3 a. b. 2 4 2 5 3 1 3 a. initial carbocation b. after rearrangement a. b. 2 4 2 5 1 3 a. b. S 1 E1 S 2 E2 2 1 3 6 2 5 3 a. b. 2 6 2 5 3 1 3 a. initial carbocation b. after rearrangement a. b. 2 6 2 5 1 3 1 chair 1 chair 2 S 1 E1 1 1 S 2 E2 1 1 nly consider carbocation rearrangements that immediately go to a more stable carbocation (not equally stable carbocations = too many possibilities) 2 3 S 1 E1 S 2 E2 a. 2 2 1 2 b. 1 2 condition 1 condition 2 S 1 E1 S 2 E2 a. 2 2 0 1 b. 1 2 4 S 1 E1 2 2 S 2 E2 1 2 5 6 S 1 E1 S 2 E2 2 2 1 2 S 1 E1 S 2 E2 2 2 1 2 7 S 1 E1 2 2 S 2 E2 1 2 8 9 S 1 E1 S 2 E2 2 3 0 3 S 1 E1 S 2 E2 a. 2 1 0 1 b. 4 5 c. 1 2 d. 1 2 consider any 3 o + possibility and consider stereoisomers Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 28 When methyl on β 1 is anti to -, no S 2 is possible and no E2 is possible from β 1. a b β1 otation of α - β1 brings a or b anti to -, which allows S 2 and two different E2 possibilities: a (Z) and b (E). Since β2 is a simple methyl, there is no β2 substituent to inhibit either of these reactions. E2 possible here a full rotation at full rotation at b α - β is possible α - β is possible β2 β1 in chain in chain β2 β1 α b α 3 α 3 a β2 o S 2 possible and no E2 from β1, but E2 from β2 (1-butene) is possible. Use these ideas to understand cyclohexane reactivity. S 2 possible E2 from β1 (2Z- butene) E2 from β2 (1-butene) S 2 possible E2 from β1 (2E- butene) E2 from β2 (1-butene) alkene stabilities tetrasubstituted > trisubstituted > trans-disubstituted > gem-disubstituted cis-disubstituted > monosubstituted equatorial leaving group o S 2 or E2 when "" is in equatorial position. 3 o S 2 is possible (1,3 diaxial positions block approach of nucleophile), and no E2 is possible because ring carbons are anti. only partial rotation is possible in ring o S 2 is possible (1,3 diaxial positions block approach of nucleophile), and no E2 is possible because ring carbons are anti. only partial rotation is possible in ring S 2 possible if α is not tertiary and there is no anti β "" group. axial leaving group o S 2 possible if there is an anti β "" group. E2 possible with anti β -. Both S 2 and E2 are possible in this conformation with leaving group in axial position. o E2 possible, no anti β -. E2 possible with anti β -. o S 2 or E2 when "" is in equatorial position. nly E2 is possible in this conformation. Leaving group is in axial position. Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 29 3 3 3 t-butyl substituent locks in chair conformation with equatorial t-butyl Examples - group A 1 2 yclohexane structures have two chair conformations possible. K eq 1 9,999 4 3 3 3 severe steric repulsion 10 = l I Ts 3 5 11 6 8 7 9 1. Is S 2 possible? equires an open approach at α and β. 2. Is E2 possible? equires anti β -. 3. ow many possible products are there? 4. What is the relationship among the starting structures? 5. What is the relationship among the products? 6. Are any of the starting structures chiral? 7. Are any of the product structures chiral? 1 Examples - group B 2 Ts 4 Ts 6 Ts 8 Ts 10 Ts Ts 3 5 7 9 11 Ts Ts Ts 1. Which conformation is reactive? 2. Is S 2 possible? equires an open approach at α and β. 3. Is E2 possible? equires anti β -. 4. ow many possible products are there? 5. What is the relationship among the starting structures? 6. What is the relationship among the products? 7. Are any of the starting structures chiral? 8. Are any of the product structures chiral? Ts Ts chair 1 chair 2 Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc
ucleophilic Substitution & Elimination hemistry 30 strong base/nucleophile conditions (E + = electrophile, u: = nucleophile) n-butyl lithium (powerful nucleophile at epoxides and carbonyls (=), and the most powerful base at other times) 2 3 2 2 3 Should be S 2, but does not work well. Too many side reactions. 2 3 2 2 2 3 Should be S 2, but does not work well. Too many side reactions. 2 3 2 2 3 2 2 Should be S 2, but does not work well. Too many side reactions. Z:\files\classes\314\314 Special andouts\314 S & E Fall 2011.doc