an axial "X" is necessary for a succesful E2 reaction and also works better for S N 2

Transcription

1 cleophilic Substitution & Elimination hemistry 1 Templates for predicting S2 and E2 reactions primary - β α α β two different perspectives secondary - β β α β β α two different perspectives tertiary - β α β β β β α β two different perspectives cyclohexane structures interconvert in fast equilibrium possible reactions an axial "" is necessary for a succesful E2 reaction and also works better for S 2 possible reactions :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

2 cleophilic Substitution & Elimination hemistry 2 Write a structure of (S,4S)--iodo-4-methoxyhexane template for structure 4 rotated to altenative perspectives α or 2 4 α 2 1. What is/are the expected product(s) of this compound with sodium cyanide in MS (dimethylsulfoxide is the solvent). Show all mechanistic details clearly for how each of the possible products is formed ( structures, curved arrows, lone pairs and formal charges). β β α β β α β β α β β α 2. Predict the possible products (arrow pushing required, lone pairs, formal charges) for the following reactions. consider example at top "careful" :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

3 cleophilic Substitution & Elimination hemistry (low pk a ) (low pk a ) S (low pk a ) (high pk a sterically bulky) (high pk a sterically bulky) F :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

4 cleophilic Substitution & Elimination hemistry 4 1. structure mechanisms and products 1 Possible Key (S,4S)--iodo-4-methoxyhexane a 2 β α β b (S,4S) a 2 β b β (,4S) a 2 β α β b 2 (2E,4S)--methoxy-2-hexene 2 β α a b β. predict products E2 2 2 β α 2 b β (Z)--methoxy--hexene a S 2 S 2 E2 2 (2Z,4S)--methoxy-2-hexene in our course at primary (except t-butoxide) trans trans the only anti β - S 2 cis cis E2 is in competition with S 2. ecause hydroxide is strongly basic we expect E2 > S 2. Similar to #, below, except for basicity of " ". trans E2 S trans 2 trans cis cis the only E2 is in competition with S 2. ecause carboxylate is less basic we anti β - expect. Similar to #2, above, except for basicity of " ". trans E2 :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

5 cleophilic Substitution & Elimination hemistry 5 - rotation S S S 2 - rotation S - rotation E2 a E2b E2 c "Z" configuration "E" configuration "E" configuration A β carbon is fully substituted so S 2 reaction is greatly inhibited. There is an anti β - possibility so E2 can occur. nly the MAJ product is shown on the remaining problems. t is assumed that you can generate all possible products, major and minor ones. mainly S 2 (low pk a ) cis cis (low pk a ) mainly S 2 trans trans S (low pk a ) mainly S 2 S :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

6 cleophilic Substitution & Elimination hemistry 6 nly the MAJ product is shown on the remaining problems. t is assumed that you can generate all possible products, major and minor ones. only E2 reaction, t-butoxide is too big and bulky for S 2 reactions (back axial too) (high pk a, sterically bulky, only E2) enantiomers (high pk a, sterically bulky, should be only E2, but cannot significantly rotate to an axial position since the very large t-butyl group locks the ring into confromation having t-but yl axial.) o reaction only E2 reaction at tertiary center & only with anti β - 2 F a b c only E2 reaction at tertiary center & only with anti β - a b c 2 bonus problem S b c a only E2 reaction at tertiary center & only with anti β - a b c 2 :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

7 cleophilic Substitution & Elimination hemistry 7 Write a structure for the given name. (2,S,4) 2-deuterio--bromo-4-methylhexane (2,S,4) 2-deuterio--bromo-4-methylhexane All variations are shown below, which are enantiomers, diastereomers? (2,S,4) (2,S,4S) (2S,S,4S) (2,,4S) (2S,,4S) (2S,,4) (2,,4) (2S,,4) ow would the problem change if the bromine, deuterio and/or methyl were moved to another position? What are the expected products if hydroxide is the electron donor? ow would the expected products change if hydroxide were changed to ethoxide (?), water (?) or ethanol(?). Write a separate mechanism showing the formation of each possible product. β α β two possible perspectives β α β :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

8 cleophilic Substitution & Elimination hemistry 8 Write a structure for the given name. (2,S,4) 2-deuterio--bromo-4-methylhexane S 4 (2,S,4) 2-deuterio--bromo-4-methylhexane All variations are shown below, which are enantiomers, diastereomers? (2,S,4) (2,S,4S) (2S,S,4S) (2,,4S) (2S,,4S) (2S,,4) (2,,4) (2S,,4) ow would the problem change if the bromine, deuterio and/or methyl were moved to another position? What are the expected products if hydroxide is the electron donor? ow would the expected products change if hydroxide were changed to ethoxide (?), water (?) or ethanol(?). Write a separate mechanism showing the formation of each possible product. β α β 2 two possible perspectives β β α 2 otate "" anti to α - β α β 2 otate "" anti to α - β β α 2 1. raw α = (in this problem) first in it proper configuration ( or S). 2. Add in groups on β carbons in any manner. f you are lucky, they will be in the correct configuration. f you are wrong, then switch two convenient groups.. f there is a strong nucleophile/base, then write out all of the S 2/E2 possibilities. The S 2 product will form only by inversion of configuration. Any E2 products require an anti β - and α - conformation. You should draw every possible conformation and examine the predicted alkene that forms. This will determine the configuration of any alkene products. You must look at every possible β - to determine all of the possible alkene products. 4. f there is a weak nucleophile/base, then write out all of the S 1/E1 possiblities. The first step for both mechanisms is loss of the leaving group which forms a carbocation. f we ignore rearrangement possibilities, then there are two choices, S 1 and E1, that can occur. For S 1 products add the :- from the top and bottom. f α is a chiral center, there will be two different products. t is also possible that there are two products in a ring with cis/trans possibilities and no chiral centers. You will also have to take off the extra proton via an acid/base reaction to get a neutral product. For E1 products you will have to remove any β - (no anti requirement). Make a double bond between all different β -'s. Switch the two groups on either of the carbons of each double bond to see if different stereoisomers are formed. The possible outcomes are that the switch produces no change or E/Z stereoisomers are formed. Any possible outcome is a predicted result in E1 reactions. :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

9 cleophilic Substitution & Elimination hemistry 9 Write a structure for the given name (2,S,4) 2-deuterio--bromo-4-methylhexane (2,S,4) 2-deuterio--bromo-4-methylhexane 2 4 All stereoiosmeric variations are shown below. Which are enantiomers, diastereomers? (2,S,4) (2,S,4S) (2S,S,4S) (2,,4S) (2S,,4S) (2S,,4) (2,,4) (2S,,4) ow would the problem change if the bromine, deuterio and/or methyl were moved to another position? What if the chain length was heptane, octane or a cyclohexane? What are the expected products if hydroxide is the electron donor? ow would the expected products change if hydroxide were changed to ethoxide (?), water (?) or ethanol(?). Write a separate mechanism showing the formation of each possible product. Possibly helpful strategies. 1. raw staggered conformations. 2. raw anti positions parallel. β α β β α β :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

10 cleophilic Substitution & Elimination hemistry 10 Write a structure for the given name (2,S,4) 2-deuterio--bromo-4-methylhexane (2,S,4) 2-deuterio--bromo-4-methylhexane 2 4 All stereoiosmeric variations are shown below. Which are enantiomers, diastereomers? (2,S,4) (2,S,4S) (2S,S,4S) (2,,4S) (2S,,4S) (2S,,4) (2,,4) (2S,,4) ow would the problem change if the bromine, deuterio and/or methyl were moved to another position? What if the chain length was heptane, octane or a cyclohexane? What are the expected products if hydroxide is the electron donor? ow would the expected products change if hydroxide were changed to ethoxide (?), water (?) or ethanol(?). Write a separate mechanism showing the formation of each possible product. or Possibly helpful strategies. 1. raw staggered conformations. 2. raw anti positions parallel. β α β or β α β 1. raw α first in it proper configuration ( or S). ( α = in this problem) 2. Add in groups on β carbons with the low priority group back, if possible. f you are lucky, they will be in the correct configuration ( or S). f you are wrong, then switch two convenient groups. (A single switch converts to S or S to.). f there is a strong nucleophile/base (: /: ), then write out all of the S 2/E2 possibilities. The S 2 product will form only by inversion of configuration (backside attack). Any E2 products require an anti β - and α - conformation. You should draw every possible conformation and examine the predicted alkene that forms to determine which are different. This will determine the configuration of any alkene products (E, Z or neither). Generally, the most substituted alkene is the major predicted E2 product. 4. f there is a weak nucleophile/base (-:/-:, usually the neutral solvent), then write out all of the S 1/E1 possiblities. The first step for both mechanisms is loss of the leaving group which forms a carbocation. f we ignore rearrangement possibilities, then there are two reaction choices that can occur (S 1 and E1). For S 1 products add the :- from the top and bottom (both sides of the planar carbocation face). f α is a chiral center, there will be two different products ( or S). t is also possible that there are two products in a ring with cis/trans possibilities and no chiral centers after the nucleophile adds. As a final step, you will also have to take off the extra proton (on "") via an acid/base reaction to get a neutral product. For E1 products you will have to remove any β - (no anti requirement). Make a double bond between all different β -'s ( α = β ). Switch the two groups on either of the carbons of each double bond to see if different stereoisomers are formed. The possible outcomes are that the switch produces no change or that E/Z stereoisomers are formed. Any possible outcome is a predicted result in E1 reactions. The most substituted alkene is the major predicted E1 product. :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

11 cleophilic Substitution & Elimination hemistry 11 S / E Worksheet α Possibilities: S 2 E2 S 1 E1 examples β α Possibilities: S 2 E2 S 1 E1 2 2 β α β Possibilities: S 2 E2 S 1 E1 2 2 β α β β Possibilities: S 2 E2 S 1 E Possibilities: S 2 E2 S 1 E1 α β α Possibilities: S 2 E2 S 1 E1 2 2 β α β Possibilities: S 2 E2 S 1 E1 2 2 β α β β Possibilities: S 2 E2 S 1 E :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

12 cleophilic Substitution & Elimination hemistry 12 (2,S,4) 2-deuterio--bromo-4-methylhexane Also possible: (2,S,4) (2S,,4S) (2S,S,4) (2,,4S) (2S,S,4S) (2,,4) (2,S,4S) (2S,,4) β1 α β2 S 2 2 β1 α β2 2 β1 "" β1 α β2 2 β1 "" β1 α β2 2 β2 :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

13 cleophilic Substitution & Elimination hemistry 1 (2,S,4) 2-deuterio--bromo-4-methylhexane All of the reactions below have the same first step Also possible: (2,S,4) (2S,,4S) (2S,S,4) (2,,4S) (2S,S,4S) (2,,4) (2,S,4S) (2S,,4) β1 α β2 2 S 1 β1 α β2 2 (E1) β1 "" β1 α β2 2 (E1) β1 "" β1 α β2 2 (E1) β2 :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

14 cleophilic Substitution & Elimination hemistry 14 Predict all possible S product(s). nly predict the major E product. State whether S or E is major, minor or they are about the same. Finally, state by what mechanism(s) each product was formed. a. Ts b. l a d. e. a f. a g. Ts K h. K i. a j. Ts a k. l l. l a :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

15 cleophilic Substitution & Elimination hemistry 15 Questions 1. Predict all possible S products and state by what mechanism they formed. 2. Predict all possible E products and state by what mechanism they formed. Predict the most stable E product, if there is one.. ndicate whether (S > E) or (E > S ) or similar amounts of both are formed. onsider: a. basicity of the base/nucleophile and b. steric hindrance in the base. 4. f carbocations are possible, show any rearrangement to a more stable carbocation. o not show similar energy carbocation rearrangements. 5. Show any important stereochemical features of the reactions. (S 2 = inversion, E2 = anti β -/ α -, S 1 = racemization, E1 = lose any β -, any stereochemistry) pk a (of conjugate acid) = methyl and primary strong and weak Ts Ts A E F G a 2 a a K a a 4 l secondary strong and weak A E F G a 2 a a K a a 8 l 9 10 l Ts l :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

16 cleophilic Substitution & Elimination hemistry 16 Questions 1. Predict all possible S products and state by what mechanism they formed. 2. Predict all possible E products and state by what mechanism they formed. Predict the most stable E product, if there is one.. ndicate whether (S > E) or (E > S ) or similar amounts of both are formed. onsider: a. basicity of the base/nucleophile and b. steric hindrance in the base. 4. f carbocations are possible, show any rearrangement to a more stable carbocation. o not show similar energy carbocation rearrangements. 5. Show any important stereochemical features of the reactions. (S 2 = inversion, E2 = anti β -/ α -, S 1 = racemization, E1 = lose any β -, any stereochemistry) pk a (of conjugate acid) = tertiary strong and weak A E F G a 2 a a K a a 15 l 16 l l allylic, benzylic, vinylic, phenylic 19 strong and weak A E F G a 2 a a K a a l 2 Ts 24 Ts 25 l 26 :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

17 cleophilic Substitution & Elimination hemistry 17 Alcohol / acid conditions onsider all 2 o, o, allylic & benzylic alcohol reactions in acid to be S 1/E1, a. if the acid is l, or ( acids) consider as S 1, b. methanol and 1 o with acids are considered as S 2 c. if 2 S 4 / consider as E1 J K L M -l S 4 / Miscellaneous nformation on all S & E reactions. structures: = Me, 1 o, 2 o, o, allylic, benzylic and = -l, -, -, -Ts and (protonated alcohols in acid) (nonreactive structures: 1 o neopentyl, vinylic, phenylic) S 2 / E2 reactions: strong base/nucleophiles = anions, neutral nitrogen ( ), "S" compounds are good nucleophiles better as nucleophile: : > : when little steric hindrance and low basicity (conjugate acid has lower pk a ) better as base: : > : when more steric hindrance and high basicity (conjugate acid has higher pk a ) S 2 solvent: usually polar, aprotic solvents such as MS, MF, A, acetone, dissolve ions but do not solvate anions well E2 conditions: often /, sterically bulky alkoxides or amines, emphasizes basicity S 2 always with backside attack (inverision of configuration if stereogenic center); E2 always with "anti" β - / α - (in our course we ignore "syn" elimination) other good nucleophiles for S 2 reactions: monostabilized enolates distabilized enolates S S S S S various sulfur nucleophies various nitrogen nucleophies S 1 / E1 reactions: weak base/nucleophiles = usually polar protic solvent or solvent mixtures ( 2,, 2 or mixtures with these) Usually S 1 > E1, but both products are observed, the solvent is usually the nucleophile/base S 1 & E1 start off with the same first step of carbocation formation arbocations 1. can rearrange to a similar or more stable carbocation, 2. can add a weak nucleophile (S 1) or. can lose a β- (E1) onsider all 2 o, o, allylic & benzylic alcohol reactions in acid to be S 1/E1, if the acid is l, or consider as S 1, if 2 S 4 / consider as E1 (methanol and 1 o with acids are considered as S 2) :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

18 cleophilic Substitution & Elimination hemistry 18 A Few Possible Answers methyl - 1-A E 1-F 1-G 1-1- o eaction (no + ) primary - 2-A E 2-F E2 > S 2 steric bulk of base -A G 2-2- o eaction (no 2 + ) - -E -F (S 2) E2 > S 2 steric bulk of base -G - - o eaction (no 2 + ) 4-A E 4-F E2 > S 2 steric bulk of base 4-G 4-4- o eaction (no 2 + ) secondary - 6-A E 6-F 6-G 6-6- o reaction at a primary neopentyl - center. The backside is blocked by a completely substituted β carbon and primary carbocations are too energetically expensive in solution. 7-A E 7-F E2 > S 2 steric bulk of base S 2 > E2 lesser basicity of base E2 > S 2 greater basicity of base :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

19 cleophilic Substitution & Elimination hemistry 19 secondary - (continued) 7-G 7-7- (S 1) (E1) S 1 > E1 (S 1) (E1) S 1 > E1 (E1) (S 1) S 1 > E1 10-A E 10-F mainly E2 steric bulk of base (in our course) mainly E2 greater basicity of base 10-G (S 1) (E1) S 1 > E1 tertiary - 14-A E 14-F (S 1) (E1) S 1 > E1 (S 1) (E1) S 1> E1 14-G E Z geminal E Z mainly E2 o & strong base (S 1) S 1 > E1 E Z geminal (E1) E Z 14- (S 1) S 1 > E1 E Z geminal (E1) E Z 14- (S 1) E S 1 > E1 Z geminal (E1) E Z :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice

20 cleophilic Substitution & Elimination hemistry 20 β α S 2 β α = carbon group none, in our course > > >> β α = leaving group none, in our course S 1 < < = carbon group none, in our course β top α bottom = leaving group only one product top attack a. add -: from top and bottom b. lose + from bottom attack β α β α β α E2 β α none, no β carbon < = carbon group < = leaving group only a single E or Z stereochemistry is possible for each β - that can attain the necessary anti β - / α - conformation β α E1 β α β α < < rotation around α - β bond E & Z stereochemistry is possible for any β - = carbon group = leaving group none, in our course none, in our course β α β α :\ocuments and Settings\butterfly\My ocuments\classes\14\14 Special andouts\14 S and E hem verview, practice