Nucleophilic Substitution & Elimination Chemistry 1 CH 3 CH 3 C O K. pk a 's of conjugate acid = 5. pk a 's of conjugate acid = 16-19

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1 cleophilic ubstitution & Elimination hemistry 1 2 / E2 Topics 1. trong, nonbulky nucleophiles are best in 2 reactions (strong decides 2 or 1 ) a. Less bulky is better for 2 reactions (versus E2) K terically large potassium t-butoxide mainly produces E2 products, even at 1 o centers b. A less basic electron pair donor is better for 2 reactions (versus E2). A conjugate acid with a lower pk a is less basic. yanide is a good 2 nucleophile at methyl, primary and secondary centers, but tertiary centers give only E2 reaction. pk a of conjugate acid ( ) = 9. A terminal acetylide is a good 2 nucleophile at methyl and primary, but secondary and tertiary centers give mainly or only E2. pk a of conjugate acid ( ) = 25 ( = carbon or hydrogen) arboxylates are good 2 nucleophiles at methyl, primary and secondary centers, but tertiary centers give only E2 Alkoxides are good 2 nucleophile at methyl and primary, but secondary and tertiary centers give mainly or only E2 pk a 's of conjugate acid = 5 pk a 's of conjugate acid = = carbon or hydrogen Examples of oxygen nucleophiles weak nulcleophile / bases pk a (conjugate acid) strong nulcleophile / bases (conjugate base) water (16) hydroxide alcohols (16-19) alkoxides carboxylic acids (5) carboxylates 2. The role of in 2 reactions. a. The α position Anything that hinders approach of nucleophile slows the 2 reaction.??? α α methyl ( 30) > α > α >> primary ( 1) secondary ( 1/40) α tertiary ( 0) omplete substitution at α stops all 2. :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

2 cleophilic ubstitution & Elimination hemistry 2 b. The β position Anything that hinders approach of nucleophile slows the 2 reaction. β α unsubstituted β ( 1) β α > > >> monosubstituted β ( 0.4) β α disubstituted β ( 0.03) β α trisubstituted β backside of α is blocked ( 0) omplete substitution at β stops all 2. c. An adjacent pi bond (allyl and benzyl) attached to α greatly accelerates 2 reactions. The pi bond is parallel to 2p orbital in transition state having excess electron density. This allows the excess charge to be delocalized (resonance), which lowers the activation energy and makes the reaction go faster. represents allyl or benzyl pi system transition state allyl 2 x 40 benzyl 2 x Inversion of configuration in 2 reactions always occurs, but we only see it at chiral centers or in rings with cis/trans substitution. a. / "" configuration "" configuration ( ) or ( ) b. ings (cis/trans) : trans ring cis ring transition state (cis ring trans ring) or (trans ring cis ring), May or may not have / configurations. :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

3 cleophilic ubstitution & Elimination hemistry 3 4. The role of solvent in 2 reactions a. Polar, aprotic solvents are best for 2 reactions. They tend to strongly solvate cations and keep them away from the electron pair donor (nucleophile). They only weakly interact with the anion nucleophiles. ome more common possibilities are listed below. 3 P dimethylsulfoxide DM 3 dimethylformamide DMF acetonitrile A 3 3 acetone hexamethylphosphoramide MPA δ- δ+ strong solvent dipole = polar solvent no polarized hydrogen atoms = aprotic Very strong interaction with cations. Very weak interaction with anions. ations tend to be tightly bound to the solvent and anions tend to be unbound and relatively free to react in polar aprotic solvents. cleophilic lone pairs are available to react (donate at α center). b. Polar, protic solvents are best for E2 reactions. ee point 1 of E2 reactions (next). 5. ommon leaving groups (for our course) are stable as anions (which start as neutral) or they are neutral molecules (which start as cations). l I chloride bromide iodide tosylate These can be leaving groups in basic, neutral or acidic conditions water alcohol Water is a leaving group from a protonated alcohol in acid conditions. In a similar way an alcohol is a leaving group from a protonated ether in acid conditions. :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

4 cleophilic ubstitution & Elimination hemistry 4 1. trong, bulky bases are best for E2 reactions (often these are polar, protic solvents and their conjugate bases). The conjugate bases are highly basic and have a strong attraction for hydrogen (the target in an E2 reaction). This tendency can be emphasized by introducing steric bulk into the base (electron pair donor) such as the case with potassium t-butoxide. Also used as bases are hindered tertiary amines. solvent = water methanol ethanol isopropyl alcohol t-butyl alcohol conjugate bases = onjugate bases are pretty basic as indicated by moderately high pk a of conjugate acids (16-19 range). 2. The role of in E2 reactions. Potassium t-butoxide mainly produces E2 products, even at 1 o centers. a. The α position more substitution here slows 2 and makes E2 more competitive. owever, you always need a β - (almost always anti). K 3 (methyl) 1 o < 2 o << 3 o (only E2 at 3 o with strong base/nucleophile) o E2 is possible at methyl. = fill in the blank β α Anti β - and α - is the usual mode for E2 reactions. b. The β position more substitution here slows 2 and makes E2 more competitive. owever, you always need a β - (almost always anti). β α The β carbon has no additional groups beyond α. β α E2 reaction correlates with number of groups at β = 0 < 1 < 2. (ethyl) The β carbon has one additional group beyond α. β α The β carbon has two additional groups beyond α. If there are 3 "": groups on β, there is no E2 because there is no β -. β α The β carbon has three additional groups beyond α. β is fully substituted. ackside approach by nucleophile is blocked. :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

5 cleophilic ubstitution & Elimination hemistry 5 c. An adjacent pi bond is helpful to E2 reactions because the new pi bond can conjugate with the existing pi bond, UT generally 2 is helped more (allyl and benzyl ). Tertiary allylic and benzylic still react by E2 (no 2). allyl = benzyl = 3. For us, E2 reactions always attack an anti β- (relative to the α carbon) in competition with 2. Each β - must be considered and many outcomes may be possible. A simple beta- 2 may lead to both E and Z stereoisomers. The major E2 product is decided on the basis of alkene substitution patterns (tetra > tri > di(trans) > di(cis) = di(gem) > monosubstituted alkenes). a. pen chains must be rotated to the desired conformation for reaction. Look especially at beta 2 's for the possibility of E/Z stereoisomers. β - and α - are always anti for us in E2 reactions. β β β α β α β α primary secondary tertiary b. Two chair conformations rapidly equilibrate in cyclohexane rings. The following is only suggestive, as there are many possibilities for substitution patterns. 2 is difficult when is equatorial, even if not at a tertiary center. Possible E2 approach. = anti β - to α - interconvert in fast equilibrium Possible E2 approach. 2 is K when is equatorial, but not at tertiary center. These are only suggestive of the many possible E2 variations in a cyclohexane. 1. In cyclohexanes, the leaving group must be axial to have an anti β If a nonhydrogen group is at one of the β positions, then no E2 can occur at that position. 3. An external β - is also possible (shown in the example above). :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

6 cleophilic ubstitution & Elimination hemistry 6 4. The bases role in E2 reactions - strong bases are necessary (most often bulky alkoxides or bulky tertiary amines are used). 3 3 K 3 Potassium t-butoxide mainly produces E2 products, even at 1 o centers. D = 1,5-diazobicyclononane DU = 1,5-diazobicycloundecane 5. The role of solvent in E2 reactions (same as #1 in E2 reactions). itrogen bases used to favor E2 reactions. a. Polar, protic solvents are best for E2 reactions. Their conjugate bases are highly basic and have a strong attraction for hydrogen (the target in an E2 reaction). This tendency can be emphasized by introducing steric bulk into the base (electron pair donor), such as is the case with potassium t-butoxide. solvent = water methanol ethanol isopropyl alcohol t-butyl alcohol conjugate bases = potassium t-butoxide mainly produces E2 products, even at 1 o centers 6. ommon leaving groups (for our course) are stable as anions (which start as neutral) or they are neutral molecules (which start as cations). (ame as point 5 in 2 items above.) l I chloride bromide iodide tosylate These can be leaving groups in basic, neutral or acidic conditions water alcohol Water is a leaving group from a protonated alcohol in acid conditions. In a similar way an alcohol is a leaving group from a protonated ether in acid conditions. :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

7 cleophilic ubstitution & Elimination hemistry 7 Major expected product(s) in 2 / E2 competition. nucleophile methyl primary secondary tertiary K > E2 2 > E2 2 > E2 2 > E2 E2 > 2 2 > E2 E2 > 2 E2 > 2 E2 > 2 E2 > 2 E2 E2 E2 E2 E2 compare these "bold" = different "typical" outcomes compare these 2 2 > E2 2 > E2 E2 :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

8 cleophilic ubstitution & Elimination hemistry 8 1 and E1 Factors 1. We write nucleophiles (-:) and bases (-:) differently. A proton is included and they are written as neutral molecules. Those most often used by us are solvent molecules of polar, protic solvents (water, alcohols and liquid carboxylic acids). Weak nucleophile bases used in our course: water alcohols carboxylic acids 2. ole of compound needs to stabilize a carbocation carbon. a. groups on the α carbon are inductively donating and help stabilize positive charge. Tertiary is better than secondary and we will not propose primary or methyl carbocations. very poor poor K relatively good = donating inductive effect b. yperconjugation stabilizes positive charge. Tertiary is better than secondary and we will not propose primary or methyl carbocations. sigma resonance? carbocation arbocation stabilized by the electrons in an parallel adjacent sigma bond. c. esonance is very good at stabilizing charge (positive, negative, free radicals or neutral conjugate pi systems). i. Pi bonds allow delocalization of electrons via parallel overlap of adjacent 2p orbitals. Most often these pi electrons are from an alkene or aromatic pi system. carbocation next to a pi bond (alkene or aromatic) :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

9 cleophilic ubstitution & Elimination hemistry 9 ii. Lone pairs adjacent to empty orbitals can share their electron density to help stabilize positive charge (they also delocalize into neutral pi bonds, like enols and amides). Most often nitrogen or oxygen is the donor atom. (Under very different conditions, carbon can do so when present as a carbanion.) carbocation next to a lone pair (usually nitrogen or oxygen) = "" or "" = "" or "" iii. eally poor carbocations We usually won t propose these, though there are rare occasions where we do invoke some of them. Vinyl carbocation has an empty 2p orbital, but on a sp hybridized carbon atom. Phenyl carbocation has an empty sp 2 orbital. Terminal alkyne carbocation, has an empty sp orbital. 3. Attack of nucleophile/base a. Attack at α carbon (top and bottom) α top bottom If =, then we cannot detect top/bottom attack. α 2 1 If 1 2, then we can detect top/bottom attack because opposite configurations will form at the α carbon, which are enantiomers. We assume racemization, but is not always true. * = has chiral center α 2 1 If 1 2, and 1 and/or 2 have chiral centers, then we can detect top/bottom attack because opposite configurations will form at the α carbon and will form diastereomers in combination with the other chiral centers. * There are no chiral centers, but top/bottom attack is seen in cis/trans products which are diastereomers. b. Attack at β -: anything goes (can rotate β - up or down relative to the carbocation 2p orbital) β 1 2 α β otate α β o β α E and Z possible stereochemistry. β α :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

10 cleophilic ubstitution & Elimination hemistry ole of solvent Polar, protic solvent stabilizes charges of both types in first step of the 1 and E1 reaction possibilities (ionization of the - bond). xygen end of solvent molecules stabilizes the cations. ydrogen end of solvent molecules stabilizes the anions. 5. Leaving groups ame as point 5 in 2 reactions. 6. pecial cases of 1 and E1 reactions. a. Assume mainly product over E product when alcohols () are mixed with acids (-l, - and -I). 2 reactions are assumed at methyl and primary alcohols and 1 reactions are assumed at secondary, tertiary, allylic and benzylic alcohols. ( = l,, I) = methyl = 2 = primary = 2 = secondary = 1 = tertiary = 1 = allylic = 1 = benzylic = 1 b. Assume mainly E1 reactions when alcohols are mixed with concentrated sulfuric acid ( 2 4 ) when heated ( ). = heat = methyl = not possible = primary = difficult, E1 = secondary = moderate, E1 = tertiary = relatively easy, E1 Alkenes, with possible rearrangement, more substituted alkenes tend to be the major products formed. This is the only reaction where we will propose E1 as the major product. :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

11 cleophilic ubstitution & Elimination hemistry 11 Essential clues to make educated guesses about what is occurring. eagents eaction onditions Products 1 a b E1 2 a b E2 = carbon portion methyl = 3 - primary = 2 - secondary = 2 - tertiary = 3 - allylic = 2 = 2 - benzylic = = weak electron pair donors = 1 / E1 neutral solvent 2,, 2 = strong electron pair donors = 2 / E2 anions, neutral nitrogen, neutral sulfur substitution products b a elimination products leaving group l I 2 Typical - tructures 3 methyl (unique) 2 primary (general) secondary (general) tertiary (general) 2 2 allylic (and variations) 2 benzylic (and variations) :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

12 cleophilic ubstitution & Elimination hemistry 12 Templates for predicting 2 and E2 reactions primary β α two different perspectives α β secondary β β α two different perspectives β β α tertiary one perspective β α β β cyclohexane α is carbon with leaving group and β carbons are attached to α interconvert in fast equilibrium possible reactions An axial "" is necessary for a succesful E2 reaction and also works better for 2. can be in many possible positions (there are two conformations for each possibility). possible reactions :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

13 cleophilic ubstitution & Elimination hemistry 13 Problem: Write a 3D structure of (3,4)-3-iodo-4-methoxyhexane. (Draw a 2D structure first.) template for 3D structure 4 rotated to altenative perspectives α I or 2 α 2 I 4 1. What is/are the expected product(s) of this compound with sodium cyanide in DM (dimethylsulfoxide is the solvent). how all mechanistic details clearly for how each of the possible products is formed (3D structures, curved arrows, lone pairs and formal charges). Indicate major and minor products. β β α I β β α I β β α I β β α I 1. 3D structure 6 5 I (3,4)-3-iodo-4-methoxyhexane 2. mechanisms and products a β α b β (3,4) I a β β 3 (3,4) b a β α b β I (2E,4)-3-methoxy-2-hexene a β α b β I (3Z)-3-methoxy-3-hexene b β β a α 3 I (2Z,4)-3-methoxy-2-hexene :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

14 cleophilic ubstitution & Elimination hemistry 14 Problem a. Write a 3D structure for the given name: (2,3,4) 2-deuterio-3-bromo-4-methylhexane. Draw 2D first (2,3,4) 2-deuterio-3-bromo-4-methylhexane All variations are shown below, which are enantiomers, diastereomers? (2,3,4) (2,3,4) (2,3,4) (2,3,4) (2,3,4) (2,3,4) (2,3,4) (2,3,4) ow would the problem change if the bromine, deuterio and/or methyl were moved to another position? b. What are the expected products if hydroxide is the electron donor? ow would the expected products differ if hydroxide were changed to ethoxide (?), t-butoxide (?), water (?) or ethanol(?). Write a separate mechanism showing the formation of each possible product. Two possible perspectives. or β 3 α or β β α 3 β Possible helpful suggestions. 1. Draw conformations staggered. 2. Draw anti positions parallel to α Draw α = 3 (in this problem). Fill it in with a proper configuration ( or ). 2. Add in groups on β carbons, with anti β - and the other groups in any manner. If you are lucky, you will have the correct configuration. If you are wrong, then switch the two other groups. 3. If there is a strong nucleophile/base, then write out all of the 2/E2 possibilities. The only 2 product will form by inversion of configuration. Any E2 products require an anti β - and α- conformation. You should draw every possible conformation that allows this, and examine the predicted alkene that forms. This will determine the configuration (E/Z) of any alkene products. More substituted alkenes are generally more preferred products. 4. If there is a weak nucleophile/base (usually 2,, 2 ), then write out all of the 1/E1 possibilities. The first step for both mechanisms is loss of the leaving group which forms a carbocation. If we ignore rearrangement possibilities for now, then there are two choices that can occur ( 1 and E1). For 1 products add the :- from the top and bottom. If α is a chiral center, there will be two different products. It is possible that there are two different 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 - α 's. witch 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 diastereomers are formed. All possible outcomes are predicted results in E1 reactions. :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

15 cleophilic ubstitution & Elimination hemistry 15 / E Worksheet α Possibilities: 2 E2 1 E1 methyl primary β β α α b Possibilities: 2 E2 1 E1 Possibilities: 2 E2 1 E1 secondary Possibilities: 2 E2 1 E1 β α β β tertiary Possibilities: 2 E2 1 E1 α methyl β α Possibilities: 2 E2 1 E1 primary β α β Possibilities: 2 E2 1 E1 secondary Possibilities: 2 E2 1 E1 β α β β tertiary :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

16 cleophilic ubstitution & Elimination hemistry 16 Partial Key a. Write a 3D structure for the given name: (2,3,4) 2-deuterio-3-bromo-4-methylhexane. Draw 2D first (2,3,4) 2-deuterio-3-bromo-4-methylhexane All variations are shown below, which are enantiomers, diastereomers? (2,3,4) (2,3,4) (2,3,4) (2,3,4) (2,3,4) (2,3,4) (2,3,4) (2,3,4) ow would the problem change if the bromine, deuterio and/or methyl were moved to another position? b. What are the expected products if hydroxide is the electron donor? ow would the expected products change if hydroxide were changed to ethoxide (?), t-butoxide (?), water (?) or ethanol(?). Write a separate mechanism showing the formation of each possible product. 3 β α β 3 D = hydroxide, ethoxide and many other possibilities. 3 β α β 3 D = water, ethanol and many other possibilities. The α configuration makes no difference in 1/E1 reactions because it is destroyed in the first step of the reaction. :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

17 cleophilic ubstitution & Elimination hemistry 17 (2,3,4) 2-deuterio-3-bromo-4-methylhexane 3 D β β α D D 3 β β α E2 D 3 β β α E2 3 D β β α E2 :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

18 cleophilic ubstitution & Elimination hemistry 18 (2,3,4) 2-deuterio-3-bromo-4-methylhexane D D 3 β β α more possibilities D 3 β β α E1 more possibilities D 3 β β α E1 more possibilities 3 D β β α E1 more possibilities :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

19 cleophilic ubstitution & Elimination hemistry 19 Problem Predict the possible products (arrow pushing required, lone pairs, formal charges) for the following reactions. a b c d h i j (high pk a sterically bulky) (low pk a ) e 3 consider example at top k (high pk a sterically bulky) 3 f g 3 3 "careful" (low pk a ) (low pk a ) l m F :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

20 cleophilic ubstitution & Elimination hemistry 20 Possible Key predict products a E2 2 2 = major E2 = minor In our course 2 > E2 at primary (except t-butoxide). b trans trans the only anti β - 2 = minor cis cis E2 is in competition with 2. ecause hydroxide is strongly basic we expect E2 > 2. imilar to "c", next, except for basicity of " ". trans E2 = major c trans trans trans the only anti β - 3 E2 = minor 3 2 = major cis cis E2 is in competition with 2. ecause carboxylate is less basic we expect 2 > E2. imilar to "b", above, except for basicity of " ". d rotation 3 2 = minor E2 = major - rotation rotation 3 E2 a E2 b E2 c 3 "Z" configuration (least = cis) 3 3 "E" configuration (middle = trans) 3 "E" configuration (most = trisubstituted) :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

21 cleophilic ubstitution & Elimination hemistry 21 e This is a sneaky one. A β carbon is fully substituted so 2 reaction is greatly inhibited. There is an anti β - possibility so E2 can occur. 3 3 nly the MAJ product is shown on the remaining problems. It is assumed that you can generate all possible products, including the major and minor possibilities. f mainly 2 (low pk a ) Acetate is a good nucleophile and not too basic. g cis cis (low pk a ) mainly 2 trans trans Azide is a good nucleophile and not too basic. 3 h (low pk a ) mainly 2 ulfur is a good nucleophile and not too basic. i nly E2 reaction, t-butoxide is too basic and too big and bulky for 2 reactions (igh pk a, sterically bulky, only E2 when an anti β - is present.) (back axial too) enantiomers j ( has a high pk a,is sterically bulky and should be only E2. cannot significantly rotate to an axial position since the very large t-butyl group locks the ring into the confromation having t-but yl equatorial.) o reaction k nly E2 reaction at tertiary center when. an anti b - is present :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

22 cleophilic ubstitution & Elimination hemistry 22 l F 3 F a 3 b c nly E2 reaction at tertiary center and only with an anti β a b 3 c 2 m Intramolecular 2 reaction. must be in an axial position for backside attack. 3 3 b a c nly E2 reaction at tertiary center and only with an anti β -. a b c 3 3 onus problem: ulfur is a very good nucleophile and not very basic, but at tertiary we only see E :\Documents and ettings\butterfly\my Documents\classes\315\315 andouts\ and E Worksheets.doc

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