Nucleophilic Substitution & Elimination Chemistry Beauchamp 1

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1 cleophilic ubstitution & Elimination hemistry Beauchamp 1 Four mechanisms to learn: vs E2 and 1 vs E1 = substitution = a leaving group () is lost from a carbon atom () and replaced by nucleophile (:) = nucleophilic = nucleophiles {:) donate two electrons in a manner similar to bases (B:) E = elimination = two vicinal groups (adjacent) disappear from the skeleton and are replaced by a pi bond 1 = unimolecular kinetics = only one concentration term appears in the rate law expression, ate = k[] 2 = bimolecular kinetics = two concentration terms appear in the rate law expression, ate = k[] [: or B:] competes with E2 1 competes with E1 These electrons always leave with. 1 ompeting eactions B (strong) (weak) B ompeting eactions E2 arbon Group Leaving Group E1 : / B: = is an electron pair donor to carbon (= nucleophile) or to hydrogen (= base). It can be strong (/E2) or weak ( 1/E1). = methyl, primary, secondary, tertiary, allylic, benzylic = -, -, -I, - 2 (possible leaving groups in neutral, basic or acidic solutions) = - 2 (only possible in acidic solutions) Important details to be determined in deciding the correct mechanisms of a reaction. 1. Is the nucleophile/base considered to be strong or weak? We simplistically view strong electron pair donation as coming from anions of all types and neutral nitrogen, sulfur and phosphorous atoms. Weak electron pair donors will typically be neutral solvent molecules, usually water ( 2 ), alcohols (), mixtures of the two, or simple, liquid carboxylic acids ( 2 ). 2. What is the substitution pattern of the - substrate at the carbon attached to the leaving group,? Is it a methyl, primary, secondary, tertiary, allylic, or benzylic carbon? What about any carbon atoms? ow many additional carbon atoms are attached at a position (none, one, two or three)? Answers to these questions will determine, E2, 1 and E1 reactivities and alkene substitution patterns and relative stabilities in E2 and E1 reactions. y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

2 cleophilic ubstitution & Elimination hemistry Beauchamp 2 - examples 1 7 (infinite possibilities exist) b c d a b a 3 * methyl 1 o 1 o 2 o 1 o 3 o 1 o 2 o 5 a b c d e f g 1 o h * * * 1o 2 o 2 o 1 o 3 o 2 o 1 o 1 o neopentyl 6 b c d e f a * * * * 1 o 2 o 2 o 1 o 3 o 2o g m 6 h i j k l 6 * * * 2 o 1 o 1 o 2 o n o p q 3 o 1 o 1 o neopentyl * = chiral center * * 2 o neopentyl 1 o 1 o 3 o a c b d e f * * * 1 o 2 o 2 o 2 o 1 o 3 o g h i j k l m * * * * * * 2 o 2 o 2 o 1 o 1 o 2 o 3 o o p q r s t n * * * * * * * * 2 o 2 o 1 o 1 o 1 o neopentyl 2 o 2 o u v w x y z aa bb * * * * * * 1 o 1 o 2 o o 3 o 1 o o o cc dd ee ff gg hh ii jj 3 o 2 o 1 o * * 2 o 1 o neopentyl 1 o 2 o 3 o 7 kk ll mm Extra patterns to know (allylic and benzylic are very fast patterns) (1 o neopentyl, vinyl and phenyl patterns are unreactive). a b c a b benzylic benzylic 1 o neopentyl 3 o 1 o allylic allylic allylic 1 o neopentyl vinyl phenyl y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

3 cleophilic ubstitution & Elimination hemistry Beauchamp 3 trong electron pair donation in our course ( / E2). /E2 concerted reactions (one step) = always backside attack at (inversion of configuration) E2 = anti - / - elimination (forms pi bonds) B = K hydroxide alkoxides potassium t-butoxide carboxylates (buy) (make) (make, strong base) (make, acetate) hydrogen sulfide (thiolate) (buy) alkyl sulfide (thiolate) (make) azide (buy) phthalimidate (an imidate) (make) cyanide (buy) Li acetylides 2 (make) ketone enolates (make) Li 2 ester enolates (make) Li dithiane anion (make) Ph Ph sulfur ylids 2 Li Li Li 2 B B Al Al n-butyl lithium I lithium lithium (very strong base sodium sodium good nucleophiles aluminium aluminium or nucleophile, use borohydride borodeuteride with tosylates and hydride deuteride anytime) in strong acid (buy) (buy) (buy) (buy) (buy) (buy) K 2 = or sodium hydride potassium hydride sodium amide (very strong base) (very strong base) (very strong base) (buy) (buy) (buy) (Mg) Grignard reagents (very strong bases and nucleophiles) (make) Important additions for us. Ph Ph diphenylsulfide, used with = to make epoxides Li alkyl lithium (very strong bases and nucleophiles) (make) P Ph Ph Ph triphenylphosphine used with = to make alkenes 2 u Li organocuprates (good carbon nucleophiles) (make) Ph = phenyl Li lithium diisopropylamide (very strong base) (make) u cuprous bromide (buy) r 3 / pyridine pyridinium chlorochromate P oxidizing E2 reaction (buy) = Ts- (tosyl chloride) makes into tosylates (buy) = py pyridine = proton sponge r 3 / 2 Jones reagent oxidizing E2 reaction (buy) y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

4 cleophilic ubstitution & Elimination hemistry Beauchamp 4 These two reactions look similar, but there are important differences. minor major ate = k 2 [ ] 1 [] 1 ate = k E2 [ ] 1 [] 1 major ate = k 1 [] 1 minor ate = k E1 [] 1 versus E2 overview (essential features) Example: 1 o, requires strong nucleophile/base, > E2, exceptions: potassium t-butoxide or sodium amide. 3 1-bromopropane > E2 (unless t-butoxide) always backside attack 3 = B strong = anything with negative charge, and neutral sulfur, phosphorous or nitrogen (also 2 = E2) 3 1-bromopropane E2 > (when t-butoxide) E2 always anti alkene E2 reactions are the most important reactions always backside attack at - carbon 3 2 methyl (Me) primary (1 o ) secondary (2 o ) tertiary (3 o ) 2 2 allylic (1 o, 2 o, 3 o versions) 2 benzylic (1 o, 2 o, 3 o versions) -beta carbon (0-3 of these) -alpha carbon (only 1 of these) y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

5 cleophilic ubstitution & Elimination hemistry Beauchamp 5 elative rates of reactions - steric hindrance at the carbon slows down the rate of reactions k 30 methyl (unique) k 1 ethyl (primary) reference compound k isopropyl (secondary) k 0 t-butyl (tertiary) (very low) methyl primary secondary Methyl has three easy paths of approach by the nucleophile. It is the least sterically hindered carbon in reactions, but it is unique. Primary substitution allows two easy paths of approach by the nucleophile. It is the least sterically hindered "general" substitution pattern for reactions. econdary substitution allows one easy path of approach by the nucleophile. It reacts the slowest of the possible substitution reactions. tertiary Tertiary substitution has no easy path of approach by the nucleophile from the backside. We do not propose any reaction at tertiary centers. = When the carbon is completely substituted the nucleophile cannot get close enough to make a bond with the carbon. Even - sigma bonds block the nucleophiles approach. All of these are primary - structures at, but substituted differently at k 1 ethyl reference compound k 0.4 propyl k methylpropyl 3 k ,2-dimethylpropyl (1 o neopentyl) 33 A completely substituted carbon atom also blocks the : approach to the backside of the - bond. A large group is always in the way at the backside of the - bond. 33 If even one bond at has a hydrogen then approach by : to the backside of the - bond is possible and an reaction is possible. y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

6 cleophilic ubstitution & Elimination hemistry Beauchamp 6 "Transition tate" sp 2 carbon transition state as carbon inverts configuration forms a high PE carbon with 10 electrons at carbon. This is a concerted, one-step reaction. PE E a E a - this energy difference determines how fast the reaction occurs: "kinetics". ate = k 2 [][] = bimolecular reaction k 2 = 10 -Ea 2.3T E a = -2.3T log(k 2 ) reactants G - this energy difference determines the extent of the reaction; the ratio of products G versus reactants at equilibrium (when kinetics allows the reaction to proceed. Thermodynamics is determined mostly by the G o = -2.3T log(k eq ) strengths of the bonds and solvation energies of - Go the reactant and product speces: "thermodynamics". K eq = T products P = progress of reaction Problem 1 - ow can you tell whether the reaction occurs with front side attack, backside attack or front and backside attack? Use the two molecules to explain you answer. Follow the curved arrow formalism to show electron movement for how the reaction actually works. a b 3 I Problem 2 - Why might 3 and 4 rings react so slowly in reactions? (int-think about bond angles in the transition state versus bond angles in the starting ring structure.) k relative () Problem 3 - Why might 6 rings react slower in reactions? What are the possible conformations from which a reaction is expected? Trace the path of approach for backside attack across the cyclohexane ring to see what positions block this approach. Which chair conformation would have the leaving group in a more reactive position (axial or equatorial)? Is this part of the difficulty (which conformation is preferred)? y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

7 cleophilic ubstitution & Elimination hemistry Beauchamp 7 These two chair conformations interconvert in a very fast equilibrium. Problem 4 - In each of the following pairs of nucleophiles one is a much better nucleophile than its closely related partner. Propose a possible explanation. a. relative rates 250/1 b. c. methoxide t-butoxide Problem 5 Write out the expected product for each possible combination (4x8=32 possibilities). = Al Li A? B??? =,, I...a good leaving group y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

8 cleophilic ubstitution & Elimination hemistry Beauchamp 8 Problem 6 Using - compounds from page 2 and reagents from page 3 to propose starting materials to make each of the following compounds. ne example is provided. TM-1 is an E2 product (see page 15), all the others are products. 2-azidopropane problem? 3 solution 3 TM = target molecule TM-1 TM-2 TM-3 TM-4 TM-5 TM-6 TM-7 TM-8 TM-9 TM-10 TM-11 Ph Ph 3 P Ph Ph Ph TM-12 TM-13 TM-14 TM-15 TM-16 TM-17 Acid/base reactions important to our course (some reagents have to be made, often by acid/base reactions) and subsequent reactions (mostly ) Make alkoxides and use as nucleophiles only at Me- and 1 o 2 - in reactions. alcohols K eq = acid/base K a = = K a alkoxides ethers alkoxides are good nucleophiles at methyl, primary, allyl and benzyl, mostly E2 at secondary and only E2 tertiary, they are also used as moderately strong bases y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

9 cleophilic ubstitution & Elimination hemistry Beauchamp 9 Make carboxylates and use as nucleophiles at Me-, 1 o 2 - and 2 o 2 - in reactions. ethanoic acid (acetic acid) K eq = K a1 acid/base ethanoate (acetate) esters 10-5 carboxylates are good nucleophiles at methyl, = = K a primary, secondary, allyl and benzyl, making esters, only E2 at tertiary with acetate produces esters, then acyl substitution with hydroxide produces the alcohol, if desired. A new type of substitution reaction. 2 Look at similarities to imide hydrolysis, just below. acid/base Me, 1 o, 2 o alcohols > E2 Problem 7 how the acyl substitution mechanism for each functional group with hydroxide. acid/base = possible leaving group tetrahedral intermediate Functional Groups to use acid chloride anhydride ester 3 o amide y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

10 cleophilic ubstitution & Elimination hemistry Beauchamp 10 Make imidates and use as nucleophile at centers to convert to primary amines. 1. Make alkyl imides 2. ydrolyze in base to make primary amines (acyl substitution) 3. Workup) = Gabriel amine synthesis; duplicates azide amine synthesis (1. with 3 2. with LiAl 4 at nitrogen 3. workup) 2 imide K eq = K a = = K a acid/base rxn rxns imidate resonance stabilized makes imidate less basic and a better behaved nucleophile alkyl imide acyl substitution #1 reaction leads to a primary amine 3 acyl substitution #2 acid/base acid/base rxn Look at similarities with ester hydrolysis, just above. throw away primary amine (also made by LiAl 4 3. workup) Alternative azide strategy to make primary amines ( and acid/base reactions) step 1 - make alkyl azide resonance alkyl azide step 2 - make primary amine gas Li Al 2. workup acid/base 1 o amine y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

11 cleophilic ubstitution & Elimination hemistry Beauchamp 11 Make potassium t-butoxide and use as sterically large, strong base at 1 o 2 -, 2 o 2 - and 3 o 3 - in E2 reactions. t-butyl alcohol K eq = acid/base K K potassium alkenes E2 K t-butoxide a potassium t-butoxide, sterically bulky base = = K a that mostly does E2 reactions with compounds (except with 3 -). Make thiols using as the nucleophile at Me-, 1 o 2 - and 2 o 2 - in reactions. hydrogen sulfide thiol Make thiolates and use as nucleophiles at Me-, 1 o 2 - and 2 o 2 - in reactions. thiols K eq = acid/base alkylthiolates K a = K a2 = sulfides ynthesis of lithium dithiane anion (acid/base) 2. 2 with 3. ydrolyze to make aldehydes or ketones) dithiane 2 Li K eq = K a1 acid/base = = K a Li dithiane anion A good nucleophile that can be made into aldehydes and ketones aldehydes and ketones (later) Make terminal acetylides and use as nucleophiles only at Me- and 1 o 2 - in reactions. terminal alkynes K eq = acid/base 2 terminal acetylides 3 K a terminal acetylides are good nucleophiles = = K a at methyl, primary, allyl and benzyl, but mostly E2 at secondary and only E2 tertiary alkynes 3 y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

12 cleophilic ubstitution & Elimination hemistry Beauchamp 12 Zipper reaction moves triple bond to end of linear chain to form the most stable anionic charge, further chemistry is possible. This reaction is almost identical to tautomers, without any heteroatoms. ynthesis of lithium diisopropyl amide, LA, sterically bulky, very strong base used to remove α - proton of carbonyl groups. (acid / base reaction) to make carbonyl enolates (next). given 2 Li given K eq = acid/base K a = = K a Li Think - sterically bulky, very basic that goes after weakly acidic protons. LA = lithium diisopropyl amide eact ketone enolate (nucleophile) with - electrophile (Me, 1 o and 2 o compounds) eact ketone enolate with compounds (methyl, 1 o and 2 o ) 2 Li ketone enolates (resonance stabilized) reactions -78 o 1 o key bond Larger ketone made from smaller ketone. y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

13 cleophilic ubstitution & Elimination hemistry Beauchamp 13 eact ester enolate (nucleophile) with - electrophile (Me, 1 o and 2 o compounds) eact ester enolate with compounds (methyl, 1 o and 2 o ) 2 Li ester enolates (resonance stabilized) reactions -78 o 1 o key bond Larger ester made from smaller ester. Ph Ph diphenyl sulfide 3 Ph 2 Ph diphenylmethylsulfonium bromide salt 2 Li K eq = K a1 acid/base = = K a Ph 2 Ph sulfur ylid to make epoxides make epoxides from aldehydes and ketones y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

14 cleophilic ubstitution & Elimination hemistry Beauchamp 14 arification of ydride electron pair donors: In our course sodium hydride and potassium hydride are always basic, and lithium aluminum hydride and sodium borohydride are always nucleophilic hydride. Basic hydride In our course, sodium hydride () and potassium hydride (K) are always basic (= electron pair donation by hydride to a proton), never a nucleophile. The conjugate acid of hydride is hydrogen gas (with a pk a = 37, 2 can hardly be considered an acid). odium hydride and potassium hydride (K) K Aldrich, 2012 $39 / 100 grams 60% oil dispersion MW = 24.0 g/mol Aldrich, 2012 $128 / 75 grams 30% oil dispersion MW = 40.1 g/mol Problem 8 Write an arrow pushing mechanism for each of the following reactions. pk a = 17 pk a = 19 K pk a cleophilic hydride Formation of - bonds using nucleophilic lithium aluminum hydride and sodium borohydride. In our course, sodium borohydride (B 4 ) and lithium aluminum hydride (LiAl 4 = LA) are inorganic salts containing nucleophilic hydride, very unusual nucleophiles. Both reagents supply nucleophilic hydride that can displace in -like reactions with compounds. odium borohydride and lithium aluminum hydride are used in many other reactions. They also react with carbonyl compounds (=) and epoxides (both to be discussed more later). We will often use the deuterium version of borohydride and aluminum hydride so we can identify where a reaction occurred. In reality, this is not very common because of the expense. But, for purposes of probing your understanding of reactions, using them shows if you understand what is happening. The deuterium isotope of hydrogen reacts similarly to the proton isotope, but there are experimental methods which allow us to observe where a reaction took place (e.g. M). odium borohydride and lithium aluminum hydride (LA) - 4 equivalents of hydride per anion Aldrich, 2012 $312 / 100 grams MW = 37.8 g/mol The hydride salts are way more common. B Al Li Aldrich, 2012 $120 / 100 grams MW = 38 g/mol B Aldrich, 2012 $254 / 5 grams MW = 41.8 g/mol The deuteride salts are way more expensive. Al Li Aldrich, 2012 $125 / 5 grams MW = 42 g/mol All these reagents will undergo reactions at centers. Because there is a greater difference in electronegativity between aluminum and hydrogen than boron and hydrogen, LA is more reactive than sodium borohydride. This won t be important for us to know until we study carbonyl reactions. y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

15 cleophilic ubstitution & Elimination hemistry Beauchamp 15 Problem 9 Write an arrow pushing mechanism for the following reaction. a. Li b. 3 Al +? B 3 +? Problem 10 It is hard to tell where the hydride was introduced since there are usually so many other hydrogens in organic molecules. Where could have have been in the reactant molecule? There are no obvious clues. Which position(s) for would likely be more reactive with the hydride reagent? ould we tell where was if we used LiAl 4? Where was ""? LA E2 eactions ompete with eactions E2 reactions also occur at the backside relative to the leaving group, but at - instead of α -. The - proton has to be anti to the α - to allow for parallel overlap of the 2p orbitals that form the new pi bond. This allows elimination to occur in a concerted manner. The syn conformation also has parallel overlap of 2p orbitals, but is present in less than 1% due to an eclipsed conformation (staggered conformations > 99%). E2 Potential Energy vs. Progress of eaction iagram (= concerted, energy picture looks very similar to ) Transition state - requires parallel overlap of the two 2p orbitals forming the pi bond. 3 4 This is easiest when - is anti to higher (less stable) PE potential energy E2 mechanism depends on steric factors and basicity of the electron pair donor. More steric hindrance and more basic favors E2 over 1 2 transition state E a = -2.3T log(k E2 ) 1 2 If stereochemical priority is 1 > 2 and 3 > 4 then this would be Z configuration, which is fixed by the requirement for anti - / - elimination. reactants lower (more stable) G = -2.3T log(k eq ) = negative (exergonic) P = progress of reaction - shows how PE changes as reaction proceeds products Keeping Track of the ydrogens in E2 reactions y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

16 cleophilic ubstitution & Elimination hemistry Beauchamp 16 At least one position, with a hydrogen atom, is necessary for an E2 reaction to occur. In more complicated systems there may be several different types of hydrogen atoms on different positions. In E2 reactions there can be anywhere from one to three carbons, and each carbon can have zero to three hydrogen atoms. 0-6 at 0-9 at 0-3 at zero positions unique methyl, only possible one position 1 o, usually > E2 (except with t-butoxide) two positions (2 o, / E2 both competitive) three positions (3 o, only E2) With proper rotations, each - may potentially be able to assume an anti conformation necessary for an E2 reaction to occur. There are a lot of details to keep track of and you must be systematic in your approach to consider all possibilities. Using one of these two perspectives may help your analysis of E2 reactions Let s consider a moderately complicated example (next problem). B vertical perspective - / -? Either sketch will work for every possibility above, IF you fill in the blank positions correctly. B horizontal perspective - / -? Problem 11 - ow many total hydrogen atoms are on carbons in the given compound? ow many different types of hydrogen atoms are on carbons (a little tricky)? ow many different products are possible? int - Be careful of the simple 2. The two hydrogen atoms appear equivalent, but E/Z (cis/trans) possibilities are often present. (ee below for relative expected amounts of the E2 products.) edraw structure to explicitely show all of the hydrogen atoms. otice the base/nucleophile is strong, so or E2 reaction are considered. The pattern is tertiary, so no is possible. There are three carbon atoms and all of them have hydrogen atoms on them. That leaves E2 reaction as the only possible choice.? There are two chiral centers, and 1, that make this structure more complicated than we are treating it. Possible stereoisomers are,, and. tability of pi bonds In elimination reactions, more substituted alkenes are normally formed in greater amounts. Greater substitution of carbon groups in place of hydrogen atoms at alkene carbons (and alkyne carbon atoms too) translates into greater stability (lower potential energy). Alkene substitution patterns are shown below. There are three types of disubstituted alkenes and their relative stabilities are usually as follows: geminal cis trans. y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

17 cleophilic ubstitution & Elimination hemistry Beauchamp 17 < elative stabilities of substitued alkenes. < < < < "unsubstituted" "mono" "di-cis" "di-geminal" "di-trans" "tri" "tetra" The more substituted alkenes are usually more stable than less substitued alkenes. ubstitution here, means an group for a hydrogen atom at one of the four bonding positions of the alkene. 1 = unsubstituted alkene (ethene is unique) 2 = monosubstituted alkene 3 = cis disubstituted alkene 4 = geminal disubstituted alkene 5 = trans disubstituted alkene 6 = trisubstituted alkene 7 = tetrasubstituted alkene aytzeff's rule: More stable alkenes tend to form faster (because of lower E a ) in dehydrohalogenation reactions (E2 and E1). They tend to be the major alkene product, though typically a little of every alkene product possible is obtained. Possible explanations for greater stability with greater substitution of the pi bond A fairly simple-minded explanation (the one we will use) for the relative alkene stabilities is provided by considering the greater electronegativity of an sp 2 orbital over an sp 3 orbital. An alkyl group ( ) inductively donates electron density better than a simple hydrogen. From the point of view of a more electronegative sp 2 alkene carbon, it is better to be connected to an electron donating alkyl carbon group than a simple hydrogen atom. The more groups at the four sp 2 positions of a double bond, the better. owever, be aware that other features, such as steric effects or resonance effects, can reverse expected orders of stability....inductively donating "" substituent is more stable than unsubstituted ""... Problem 12 - econsider the elimination products expected in problem 1 and identify the ones that you now expect to be the major and minor products. ow would an absolute configuration of α as, compare to α as? What about 1 as versus? * * * = chiral center 1 2 3? one of your possibilities (which one?) ompare (3,4)-3-bromo-3,4-dimethylheptane (3,4)-3-bromo-3,4-dimethylheptane (3,4)-3-bromo-3,4-dimethylheptane (3,4)-3-bromo-3,4-dimethylheptane y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

18 cleophilic ubstitution & Elimination hemistry Beauchamp 18 Problem 13 -Provide an explanation for any unexpected deviations from our general rule for alkene stabilities above. A more negative potential energy is more stable. o f = kcal/mole o f = kcal/mole o f = kcal/mole Problem 14 rder the stabilities of the alkynes below (1 = most stable). Provide a possible explanation. "" = a simple alkyl group Problem 15 - Propose an explanation for the following table of data. Write out the expected products and state by which mechanism they formed. : -- /B: -- = (a weak base, but good nucleophile). 3 2 percent substitution 100 % percent elimination 0 % our rules > E % 0 % > E % 89 % > E2 (wrong prediction) % 100 % only E2 Problem 16 ne of the following reactions produces over 90% product and one of them produces about 85% E2 product in contrast to our general rules (ambiguity is organic chemistry s middle name). Match these results with the correct reaction and explain why they are different. pk a = 16...versus... pk a = 19 y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

19 cleophilic ubstitution & Elimination hemistry Beauchamp 19 Problem 17 - A stronger base (as measured by a higher pk a of its conjugate acid) tends to produce more relative amounts of E2 compared to, relative to a second (weaker) base/nucleophile. Greater substitution at and also increases the proportion of E2 product, because the greater steric hindrance slows down the competing reaction. Use this information to make predictions about which set of conditions in each part would produce relatively more elimination product. iefly, explain your reasoning. Write out all expected elimination products. Are there any examples below where one reaction ( or E2) would completely dominate? a. I...versus... I b. c. d. e....versus... pk a ( 2 ) = 5 pk a () = 16...versus......versus......versus... f....versus... pk a () = 25 pk a () = 9 y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

20 cleophilic ubstitution & Elimination hemistry Beauchamp 20 Problem 18 - (2,3)-2-bromo-3-deuteriobutane when reacted with potassium ethoxide produces cis-2-butene having deuterium and trans-2-butene not having deuterium. The diastereomer (2,3)-2-bromo-3-deuteriobutane under the same conditions produces cis-2-butene not having deuterium and trans-2-butene having deuterium present. Explain these observations by drawing the correct 3 structures, rotating to the proper conformation for elimination and showing an arrow pushing mechanism leading to the observed products. (Protium = and deuterium = ; and are isotopes. Their chemistries are similar, but we can tell them apart.) (2,3) K cis-2-butene ( present) and trans-2-butene (no present) (2,3) K cis-2-butene (no present) and trans-2-butene ( present) Problem 19 - raw a ewman projection of the two possible conformations leading to E2 reaction. how how the orientation of the substituents about the newly formed alkene compares. B anti E2 reaction staggered reactant conformation 1 alkene stereochemistry? rotate about center bond < 1% > 99% ewman projections syn E2 reaction eclipsed 3 4 B reactant conformation 2 alkene stereochemistry? y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

21 cleophilic ubstitution & Elimination hemistry Beauchamp 21 Alkyne synthesis (via two E2 reactions with 2 and 2, two times) ample alkynes using double E2 reactions of Workup (neutralize mixture) Possible steps in mechanism (E2 twice then 2. acid/base or 2. electrophile) 2 eqs. 2 h (314 reaction, see page this topic) a b 2 2. workup 2. 2 b a Make starting alkynes using two leaving groups and very basic 2 -- starting from an alkane. (2 equivalents) 1. 2 sodium amide (very strong base) E2 twice 2. workup Use steric bulk and/or extreme basicity to drive E2 >. and (2 equivalents) 2 h free radical substitution (2 equivalents) 1. 2 sodium amide (very strong base) E2 twice 2. workup (2 equivalents) 2 h free radical substitution (2 equivalents) 1. 2 sodium amide (very strong base) E2 twice 2. workup (2 equivalents) 2 h free radical substitution y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

22 cleophilic ubstitution & Elimination hemistry Beauchamp 22 ample alkynes using reactions of 1. Terminal alkyne only at methyl or primary - (can do this twice starting with ethyne) new bond new bond new bond new bond new bond new bond new bond new bond 1. 2 sodium amide (very strong base) (1 equivalent) acid / base 2. 3 > E2 new bond 1. 2 sodium amide (very strong base) (1 equivalent) acid / base 2. > E2 Terminal acetylides are K nucleophiles at methyl and 1 o, but E2 > at 2 o and only E2 at 3 o equivalent 1. 2 (4 = 2 + 2) terminal alkyne 2. 1 equivalent 1. 2 (5 = 3 + 2) internal alkyne 1 equivalent (6 = 3 + 3) terminal alkyne 2. 1 equivalent 1. 2 acid/base 1. excess 2. workup (neutralize) (zipper reaction) like tautomers (5 = 3 + 2) internal alkyne y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

23 cleophilic ubstitution & Elimination hemistry Beauchamp 23 1 equivalent 1. 2 acid/base 2. (6 = 4 + 2) terminal alkyne 1 equivalent 1. 2 acid/base 2. terminal acetylides ( only at methyl and 1 o ) larger alkynes (6 = 3 + 3) internal alkyne 1. excess 2. workup (neutralize) (zipper reaction) like tautomers 3 nucleophile =? electrophile =? nucleophile =? electrophile =? nucleophile =? electrophile =? y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

24 cleophilic ubstitution & Elimination hemistry Beauchamp 24 Problem 20 What are the possible products of the following reactions? What is the major product(s) and what is the minor product(s)? There are 55 possible combinations. = 7 A B 3 Al Li enolates Ph Ph 2 sulfur ylids imidate? B??? E? =,, I...a good leaving group y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

25 cleophilic ubstitution & Elimination hemistry Beauchamp 25 eaction Templates - sideways and vertical perspectives (either one will work) /E2 (: / B: ) always backside for and usually anti -/ - attack for E2 1/E1 (-: / -B:) - attack from either face of + for both reactions ( 1 and E1) methyl (Me) B : strong -: / -B: weak primary (1 o ) B : strong -: / -B: weak secondary (2 o ) B : strong -: / -B: weak tertiary (3 o ) B : strong -: / -B: weak side view easier T harder 1 B : strong 2 -: / -B: weak 2 B : strong 3 -: / -B: weak vertical view B : strong -: / -B: weak B : strong -: / -B: weak T priority 1 > 2 > 3 3 iotopes of hydrogen = protium (proton) = deuterium T = tritium 1 1 trong ( / E2) B : conjugate acid pk a = 16 = conjugate acid pk a = 16 conjugate acid pk a = 5 conjugate acid pk a = 18 Weak ( 1 / E1) -B : = - : Example: 3-bromo-2-methoxyhexane (,), (,), (,), (,) template a b vertical view 3 (2,3) 3 3 b side view a 2 6 (2,3) y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

26 cleophilic ubstitution & Elimination hemistry Beauchamp 26 Examples - you can use the vertical views or side views presented above. (ever ending supply of problems.) bromo-3-deuteriobutane (,) (,) (,) (,) 2-bromo-3-methylbutane ( or ) bromo-3-deuteriopentane (,) (,) (,) (,) 3-bromo-2-deuteriopentane (,) (,) (,) (,) 2-bromo-3-methylpentane (,) (,) (,) (,) 3-bromo-2-methylpentane ( or ) bromo-3-deuteriohexane (,) (,) (,) (,) 3-bromo-2-deuteriohexane (,) (,) (,) (,) 2-bromo-3-methylhexane (,) (,) (,) (,) 3-bromo-2-methylhexane ( or ) bromo-4-deuteriohexane (,) (,) (,) (,) 3-bromo-4-methylhexane (,) (,) (,) (,) bromo-3-deuterioheptane (,) (,) (,) (,) 3-bromo-2-deuterioheptane (,) (,) (,) (,) 2-bromo-3-methylheptane (,) (,) (,) (,) 3-bromo-2-methylheptane ( or ) 3-bromo-4-deuterioheptane (,) (,) (,) (,) 4-bromo-3-deuterioheptane (,) (,) (,) (,) 3-bromo-4-methylheptane (,) (,) (,) (,) 4-bromo-3-methylheptane (,) (,) (,) (,) / E2 = B K hydroxide alkoxides potassium t-butoxide carboxylates (buy) (make) (make) (acetate) hydrogen sulfide (thiolate) (buy) phthalimidate (an imidate) (make) azide (buy) acetylides (make) cyanide (buy) 1 / E1 B Li Li 2 2 ester enolates ketone enolates (make) (make) = water Li dithiane anion (make) alcohols B Al Li /E2 B 4 lithium Al 4 sodium aluminium borohydride hydride carboxylic acids 1/E1 concerted reactions (one step) = always backside attack (inversion of configuration) E2 = anti b - / a - elimination (forms pi bonds) form carbocation ( + ) in first step, + has three common choices 1. rearrange to similar or more stable + 2. add nucleophile (top/bottom) 3. lose any beta proton (top/bottom) (forms pi bonds) y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

27 cleophilic ubstitution & Elimination hemistry Beauchamp 27 trong base/nucleophile conditions (negative charge in our course) leads to and/or E2 reactions. imple mechanisms. methyl pattern primary pattern ( > E2, except t-butoxide and 2 ) T reaction conditions T 3 1,2 reaction conditions 1,2 3 There is no E2 at methyl a = anti 3 B 1,2 b = anti 3 B 1,2 a reaction conditions E2 a = Z 3 b b = E reaction conditions E2 3 E2 > when t-butoxide and 2 secondary pattern ( > E2, except,, 2 and ) 3 reaction conditions 3 B a = anti b = anti c 2,3 2,3 3 b = E reaction conditions E2 3 3 a = Z 2,3 3 3 c = neither 3 tertiary pattern B 3 a = anti b = anti d c 2,3,4 3 2 There is no at tertiary reaction conditions E2 b = E c = neither d = Z a = Z y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

28 cleophilic ubstitution & Elimination hemistry Beauchamp 28 Example Mechanisms shown below secondary (2 o ) 3-bromo-2-deuteriohexane ne product and four E2 products. 1. strong : /B: 2. 2 o - b c d e a a 3 b (2,3) sigma bond rotations 2 3 b a a, b, c a (2,3) 3 b a 2 3 only one product d, e 3 b d (2E, with "") c four possible E2 products e 3 b a (3E, lost a ) (2Z, without "") (3Z, lost b ) y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

29 cleophilic ubstitution & Elimination hemistry Beauchamp 29 Important acid/base reactions are shown on pages When necessary for a preliminary step of a reaction, you should consult those pages. ome of the strong base/nucleophiles used in our course are listed below along with the usual preferred reactions according to our simplistic rules. uprates will be discussed later. Missing are enolates, imidates, magnesium and lithium organometallics and a few others. y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

30 cleophilic ubstitution & Elimination hemistry Beauchamp 30 - compounds strong base/nucleophiles 2 3 methyl primary secondary tertiary enolates imidate 2 only > E2 > E2 only E2 no - too sterically hindered only > E2 > E2 only E2 too sterically hindered no - Ph Ph Ph Ph Ph make sulfur ylids P make phosphorous ylids Li organolithium reagents (covered in later topic) Ph 2 Ph sulfur ylids (covered in later topic) Ph phosphorous ylids Ph P 2 Ph (covered in later topic) only > E2 > E2 only E2 too sterically hindered no - only > E2 > E2 only E2 too sterically hindered no - u Li only > E2 > E2 only E2 too sterically hindered cuprates no - (covered in later topic) react with react with react with react with (Mg) aldehydes aldehydes aldehydes aldehydes Grignard reagents and ketones and ketones and ketones and ketones (covered in later topic) react with aldehydes and ketones react with aldehydes and ketones react with aldehydes and ketones react with aldehydes and ketones react with aldehydes and ketones react with aldehydes and ketones react with aldehydes and ketones react with aldehydes and ketones react with aldehydes and ketones react with aldehydes and ketones react with aldehydes and ketones react with aldehydes and ketones y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

31 cleophilic ubstitution & Elimination hemistry Beauchamp 31 1 and E1 ompetition Multistep eactions Arising From arbocation hemistry /E2 and 1/E1 reactions look very similar overall, but there are some key differences. In /E2 reactions the nucleophile/base is a strong electron pair donor (negative charge in our course), which is why they participate in the slow step of the reaction and force a concerted, one-step reaction. In 1/E1 reactions the nucleophile/base is a weak electron pair donor (stable, neutral molecules: 2, and 2 for us) and that s why they don t participate in the slow step of the reaction, which is ionization of the α - bond. This leads to differences in reaction mechanisms, which show up in the rate law expression (kinetics is bimolecular = 2 or unimolecular = 1). The rate of 1 reactions only depends on how fast forms a carbocation. The unimolecular reactions lead to possible side reactions from carbocation rearrangements. You need to carefully look at the reaction conditions to decide what mechanisms are possible. You cut you choices in half when you decide that the electron pair donor is strong (negatively charged = /E2) or weak (neutral = 1/E1). 3 3 weak nucleophil/base (neutral in our course) + =,, I 1/E1 reactions 1 product (major) ate = k 1 [] E1 product (minor) ate = k E1 [] 3 strong nucleophil/base (negatively charged in our course) =,, I /E2 reactions 3 product E2 product (minor) (major) ate = k 2 [][ 3 : ] ate = k E2 [][ 3 : ] 1 and E1 reactions begin the same way, with ionization of the α - bond to form a carbocation. Ion formation is energetically expensive. Protic solvents help by being able to solvate both cations and anions. arbocations also need some help from inductive and/or resonance effects. If it can form, the initial carbocation intermediate typically follows one of three common paths: 1. addition of a weak nucleophile at carbon and/or 2. loss of a beta hydrogen to form a pi bond and/or 3. rearrangement to a similar or more stable carbocation. earrangement merely delays the ultimate reaction products: addition of a nucleophile or loss of a beta hydrogen atom to forma pi bond. 1 and E1 reactions are multistep reactions. (-: / -B: = 2,, 2 in our course) B weak = weak base nucleophile The - bond ionizes with help from the polar, protic solvent, which is also usually the weak nucleophile/base. 1 and E1 reactions begin exactly the same way. The leaving group,, leaves on its own, 4 forming a carbocation solvated carbocation has to be 2 o or 3 o in our course (3 reaction choices) B addition of a weak nucleophile ( 1 type reaction) Loss of a beta hydrogen atom, ( b - in most E1 reactions that we study). rearrangement to a new carbocation of similar or greater stability add : ( 1) lose beta (E1) rearrange? next page y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

32 cleophilic ubstitution & Elimination hemistry Beauchamp 32 solvated carbocation (from previous page) E1 approach comes from parallel - with either lobe of 2p orbital, anti or syn is possible same - shown on both faces (top and bottom) 1 redrawn B weak base B weak base 1 approach is from either the top face or the bottom face. acemization is possible (if observable). 1 solvated carbocation weak nucleophile E1 products (pi bond forms) E or Z is possible. stable leaving group = substitution E = elimination -: = weak nucleophile -B: = weak base = leaving group :B- = :- = 2 or or E1 - bottom face rotate 180 o around - bond, can lose any beta, top or bottom E1 - top face ften a final proton transfer is necessary. 2 1 E1 products (pi bond forms) E or Z is possible could be or product (a nucleophile substitutes for a leaving group) stable leaving group B "Z" if 1 > 2 and 3 > 4 "E" if 1 > 2 and 3 > 4 B Terms 1 = unimolecular kinetics (first order reaction, the rate in the slow step depends only on ) + - = -, -, -I, -Ts, 2 = B = usually a polar, protic solvent (or mixture) of 2, or 2 (PE) reactants E a E a E1 rate (slower) 1 rate (faster) ate 1=k 1 [] 1 ate E1 = k E1 [] 1 = (10) (10) -Ea () 2.3T -Ea (E) 2.3T = (10) - Ea 2.3T ay E a ( ) = 13 kcal/mol and E a (E) = 14.6 kcal/mol E1 products 1 products ate 1 ate E1 - Ea -( ) T 1.3 = (10) = (10) = (10) = = Progress of reaction (P) y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

33 cleophilic ubstitution & Elimination hemistry Beauchamp 33 - ubstitution Pattern and rates of 1 reactions The order of stability at the electron deficient carbocation carbon is methyl primary secondary tertiary. This is consistent with our understanding of inductive electron donating ability of alkyl groups compared to hydrogen. groups (alkyl groups) are electron donating (an inductive effect). We observed, previously, that this helps alkene stability and makes it harder to form an anionic conjugate base in acid/base chemistry. A carbocation is extremely electron deficient (the opposite of a carbanion) and is very electronegative. Extra electron donation to a carbocation center proves very helpful. This can occur through an inductive effect or a resonance effect. Inductive effects are proposed to occur via polarizations of sigma bonds in the organic skeleton, helping (or hurting) a center of reactivity. We can represent these in a carbocation center, simplistically, as shown below. yperconjugation is an alternative explanation to rationalize how extra electron density can be donated to the electron deficient carbocation carbon, but we will not use it in our course igma electrons are pulled toward the carbocation carbon. Part of the + is distributed on to the hydrogen atoms, but not typically shown with formal charge Additional sigma bonds of alkyl substituent(s) allow further polarizations of electrons from more bonds (inductive donating effect), which spreads out + charge through sigma bond polarizations and helps stabilize the electron deficient carbocation carbon. esonance effects also make carbocations more stable. Allylic and benzylic compounds are very reactive in 1/E1 reactions (and reactions). With resonance, there is actually full pi electron donation from an adjacent pi bond, instead of the inductive effect just mentioned above. An adjacent pi bond tends to produce greater stabilization of a carbocation than a single alkyl substituent. esonance donation from lone pairs of heteroatoms can also be strongly stabilizing for carbocations. We will see such intermediates many times in later chapters. + y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

34 cleophilic ubstitution & Elimination hemistry Beauchamp 34 esonance effects help stabilize carbocations. 2 resonance resonance from adjacent pi bond resonance from adjacent lone pair 3 resonance A strong acid 2 resonance 3 resonance Gas phase tabilities as Indicated by ydride Affinities ydride affinity is the energy released when a hydride is added to a carbocation (gas phase reaction). The energy of reaction,, is very negative because the two reactive species (carbocation and hydride) are very unstable and the product formed is a stable molecule. ow much do inductive and resonance effects help a carbocation center? The following gas phase data below show the differences in carbocation stability are enormous. In fact, differences are so large that we will almost never propose methyl or primary carbocation possibilities as reaction pathways in our course. We will consider these two patterns ( 3 - and 2 -) as unreactive in 1 and E1 chemistry, and that should make your life a little bit easier. ydride Affinity A larger hydride affinity means a less stable carbocation. = energy released = Potential Energy methyl & primary We will not propose these carbocations in solution in this book. Inductive effects secondary ome of these values are esitimated from different sets of data. tertiary "" resonance "" resonance other / misc esonance effects adjacent pi bond adjacent lone pair very unstable empty sp orbital -386 ethynyl carbocation (see problem below) empty 2p orbital -287 vinyl carbocation (see problem below) empty sp 2 orbital -300 phenyl carbocation (see problem below) y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

35 cleophilic ubstitution & Elimination hemistry Beauchamp 35 Problem 21 - Explain the differences in stability among the following carbocations (hydride affinities). All relative energy values in kcal/mole versus a primary carbocation. A positive value is less stable and a negative value is more stable relative to our reference primary carbocation = methyl + = reference 1 o + Too unstable to propose in our course o + = -21 = -38 = -14 = -31 = -22 = -52 = o + 3 o + o allylic benzylic 1 lone pair resonance stabilization of + Inductive effect stabilization pi bond resonance stabilization of + of carbocations (3 o > 2 o + ) A more negative is a more stable carbocation. Problem 22 - Why are vinyl carbocations so difficult to form? (int What is their hybridization?) ow does an empty sp 2 orbital (phenyl carbocation) or empty sp orbital (ethynyl carbocation) compare to a typical sp 2 carbocation carbon with an empty 2p orbital? (An empty 2p orbital is also present on the vinyl carbocation, but the carbon hybridization is sp.) Problem 23 The bond energy depends on charge effects in the anions too. an you explain the differences in bond energies below? (int: Where is the charge more delocalized?) We won t emphasize these differences. = Gas Phase B.E I +140 The activation energies for ionization in solvents are on the order of kcal/mole ( 1 and E1 reactions) It is clear from the difference in the gas phase energies of ionization that the solvent is the most stabilizing factor in ion formation. The magnitudes of these energies are compared in the potential energy diagram below. Because solvent structure is so complex we ignore it, but we do so at our own peril Many small solvent/ion interactions make up for a single, large covalent bond (heterolytic cleavage). A typical hydrogen bond is about 5-7 kcal/mole and typical covalent bonds are about kcal/mole. In a sense the polar protic solvent helps to pull the - bond apart. The "polarized" protons tug on the "" end and the lone pairs of the solvent molecules tug on the " " end. If the carbocation is stable enough, the bond will be broken. arbocations are more stable and have smaller energy differences in solution than the gas phase. (But methyl and primary are still too unstable to form in solution and we won't propose them in this book.) 150 PE gas phase reactions polar solvent phase reactions olvent / ion interactions are the most significant factor (about 130 kcal/mole here). y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc

36 cleophilic ubstitution & Elimination hemistry Beauchamp 36 Weak cleophile/bases are used in 1/E1 eactions We emphasize the term weak here because if the : were strong (negative charge), the reactions observed would be /E2. Weak, for us, is represented by a neutral solvent molecule with a pair of electrons. For us, this will be a polar solvent molecule such as water (), an alcohol () or a liquid carboxylic acid ( 2 ). All of these are protic solvents, which are reasonably good at solvating both cations and anions. In every case, there is an extra hydrogen atom on the oxygen atom of a solvent molecule that must be removed in a final acid/base reaction to produce the neutral organic product. This protonated cationic intermediate results from the addition of water ( 2, forms alcohols) or an alcohol (, forms ethers) or a liquid carboxylic acid ( 2, forms esters). very reactive carbocation strong Lewis acid = 1 strong onsted acid = E1 weak nucleophile (...or weak base) ( 2, or 2 ) cationic intermediate requires proton transfer to form neutral product best base in acidic medium is usually anotehr solvent molecule Weak electron pair donation in our course. - = 2, forms alcohols - =, forms ethers - = 2, forms esters substitution product protonated solvent We will view the attack on an sp 2 carbocation as equally accessible from either face (top or bottom). Because carbocations are flat (in our simplistic view), attack is equally accessible from either face (top or bottom). The carbocation carbon, itself, is not chiral since it is sp 2 hybridized and only has three atoms attached to it. owever, it is prochiral and can become chiral if the addition of a nucleophile brings in a fourth different group. This would lead to enantiomers, if this was the only chiral center. We are assuming that a 50/50 recemic mixture forms (in our course). It is also possible that there may be a stereogenic center somewhere else in the carbocation structure. Top and bottom attack would then lead to the formation of diastereomeric products. y:\files\classes\315\315 andouts\315 Fall 2013\2 315 and E & chem catalog.doc