Four Classes of Reactions (Substitution vs. Elimination and Bimolecular vs. Unimolecular)

Transcription

1 cleophilic Substitution & Elimination hemistry 1 Four lasses of eactions (Substitution vs. Elimination and imolecular vs. Unimolecular) β α alkene product Smaller steric factors favor S N 2 (Me, 1 o ), larger steric factors (in or :/ :) make more competitive (3 o ), as :/: gets more basic (pka) becomes more favorable, 2 o has comparable rates for S N 2 and mechanisms, sterically large bases pushes towards more, even at a 1 o β α S N 2 product Stereorequirements = anti β - and α - conformation to form the new π bond in a single, concerted step, need to examine all available conformations which are most likely. which are possible? The most stable alkene (most substituted) generally predominates = Zaitsev's rule Stereorequirements = inversion of configuration in a single, concerted step = backside attack β α Transition State = T.S. β α Transition State = T.S. bimolecular kinetics = second order, both and :/: involved in slow step β α S N 2 polar solvent allows formation unimolecular kinetics (S N 1/E1), only is involved in the slow step, :/: is weak and requires the very reactive before a reaction will occur, s are prone to rearrange, beginning the whole process over + requires 2 o, 3 o or resonance tight ion pair loose ion pair free ions S N 1/E1 β -:/-: α no special stereorequirements for S N 1/E1 because the is extremely reactive, S N 1 usually outcompetes E1, the stereochemistry of addition to the is mixed because of the flat shape (sp 2 ), if /S configurations are possible, both are usually obtained, (50/50 = racimization), any β - is easily lost (E1), generally the most stable alkene predominates (most substituted, Zaisev's rule) β α The E1 reaction is favored with alcohols mixed with sulfuric/phosphoric acids and heat (the alkene distills out) E1 rearrangement proton transfer There are many details to keep track of. Need to organize these details to develop a "best guess". Generally some of each competing reaction occurs (e.g. S N 2 vs or S N 1 vs E1). ur presentation is greatly simplified, but necessary as a first look at organic reactions. β α rearrangements are common when more stable (s) can form S N 1 -: β β α α β β α α Z:\classes\314\314 Special andouts\314 SN and E summary.doc

2 cleophilic Substitution & Elimination hemistry 2 General guidelines for S N /E reactivity of compounds, usually true (but not always) 1. cleophiles (:) and bases (:) both have a donatable pair of elelctrons (lone pairs and pi bonds). When donated to a proton the electron pair donor is called a base and when donated to anything else (usually carbon for us) the donor is called a nucleophile. 2. In our course we will divide electron pair donors into two groups. Strong donors will be symbolized with a negative charge (S N 2 and reactions) and weak donors will be symbolized with neutral charge (S N 1 and E1 reactions). Mostly this will lead us to correct choices, even though our assumptions are somewhat simplistic. Exceptions are neutral nitrogen (ammonia and amines) and neutral sulfur (thiols and sulfides) are considered good electron pair donors. 3. Leaving groups become very stable anions or neutral molecules ( = l -, -, I -, Ts -, 2,,, etc.) when they take their electron pair away from. 4. oth S N 2 and reactions are considered to be concerted, one-step reactions. They both require very specific orientations of the reactants for a successful reaction. S N 2 reactions always occur by backside attack of the nucleophile and reactions usually (always for us) occur with an anti conformation of the β -/ α - bonds. 5. Steric hindrance in either the electrophile or the electron pair donor slows S N 2 reactions by interfering with the backside approach. omplete substitution at the α - carbon (tertiary ) or complete substitution at any β carbon (4 o position) prevents S N 2 reaction (no reaction). y default reactions will become the only possible choice if an anti β -/ α - conformation can be attained. 6. All S N 2 reactions occur from the backside and produce inversion of configuration ( S or S ). 7. S N 2 reactions are generally run in polar, aprotic solvents like MS (dimethylsulfoxide) or MF (dimethylformamide), which only weakly solvate the nucleophile, but strongly solvate the counter cation. 8. All reactions require and anti β - and α - conformation and the major alkene is usually the most substituted alkene. Alkoxide/alcohol mixtures are often used for reactions compounds only undergo S N 2 reactions with strong base/nucleophiles and No reaction with weak nucleophiles. 10. Tertiary compounds only undergo reactions with strong base/nucleophiles (no S N 2 reaction). 11. The major product of primary with a strong base/nucleophile is usually S N 2 with minor product. The only exceptions are when extreme steric hindrance is present in the base or the compound. If potassium t-butoxide is the base/nucleophile, then is the major product even with 1 o compounds. 12. Secondary compounds have many choices. The major product of secondary with a strong base/nucleophile depends on the strength of the base. When two electron pair donors are similar (alkoxides vs carboxylates or cyanide vs terminal acetylides), the stronger base (higher pka of conjugated acid) forms more product and the weaker base (loser pka) forms more S N 2 product. Examples from our course: a. - [pk a ( 2 )=16] and - [pk a ()=16] are more basic (generally > S N 2 at 2 o ) b. - [pk a ( )=5] is less basic (generally S N 2 > at 2 o ). c. - [pk a (-)=25] is more basic (generally > S N 2 at 2 o ) d. N - [pk a (N-)=9] is less basic (generally S N 2 > at 2 o ). Z:\classes\314\314 Special andouts\314 SN and E summary.doc

3 cleophilic Substitution & Elimination hemistry S N 1 and E1 reactions always begin with the same first step: ionization of the - bond to form a, +, and a stable leaving group (for us = l -, -, I -, Ts -, as anions or 2,, as neutral molecules). 14. Methyl and primary compounds do not react under S N 1/E1 conditions (weak base/nucleophile) because they cannot form s under usual reaction conditions (in our course). 15. S N 1 usually out competes E1 reaction products and only occurs at 2 o, 3 o, allylic and benzilic compounds. Weak nucleophiles ( 2,, in our course) attack from both faces of the intermediate producing racemization of configuration, where observable, (in S N 1 reactions). E1 products can occur from loss of any β - rotated up or down with the empty 2p orbital. 16. S N 1/E1 reactions require a very polar, ionizing solvent that allows formation of ions. Usually this is a protic, hydrogen bonding solvent that solvates both cations and anions. ften the solvent is also the nucleophile/base. 17. arbocation rearrangements are competitive with S N 1 and E1 reactions when a similar or more stable can form (3 o > 2 o >> 1 o >> 3 ). We do not propose primary or methyl s in our course. 18. E1 products are the major result when an alcohol () is mixed with sulfuric acid ( 2 S 4 / ) and heated to high temperatures. The E1 alkene product is more volatile and distills from the reaction mixture, shifting the equilibrium towards E1. Any β - can be lost from either anti or syn conformations relative to the original α - group and the major alkene is usually the most substituted alkene. 19. Neopentyl, vinyl and phenyl compounds generally do not react by any of these mechanisms. Vinylic and phenyl will undergo reactions with very strong bases to form alkynes and benzyne, but do not react by S N 2, S N 1 or E1 reaction. We will not consider them reactive compounds in this topic. 1 o neopentyl phenyl vinyl These compounds are not usually reactive under S N /E conditions. In our course, we will consider them unreactive. Z:\classes\314\314 Special andouts\314 SN and E summary.doc

4 cleophilic Substitution & Elimination hemistry 4 α and β substitution patterns α substitution typical electron pair donors typical patterns α methyl unsubstituted α β α primary 1 substituted α β β α secondary 2 substituted α β β β α tertiary 3 substituted α S N 2 S N 2 S N 2 S N 2 rel. rate 30 rel. rate 1 rel. rate rel. rate 0 (1/40) 3 S N 1 S N 1 S N 1 S N 1 rel. rate 0 rel. rate 0 rel. rate 1 rel. rate 1,000, methyl = (1)-bromodeuteriotritiomethane β substitution (all 1 o here) β β β β α α α α 0 substituted β 1 substituted β 2 substituted β 3 substituted β primary = (1,2S)-1-bromo-1,2-dideuteriobutane S N 2 S N 2 S N 2 S N 2 rel. rate 1 rel. rate 0.4 rel. rate 0.03 rel. rate 0 secondary = (2,3S,4)-3-bromo-4-deuterio-4-methylhexane tertiary = (2S,3S)-3-bromo-2-deuterio-3-methylhexane Fill in the necessary details from the names and predicte the expected products using the above electron pair donors. α methyl 3 β α 2 primary 3 β β α secondary β β β α 2 tertiary 3 Z:\classes\314\314 Special andouts\314 SN and E summary.doc

5 cleophilic Substitution & Elimination hemistry 5 S N and E eactions Limited compounds for our course (open chains and cyclohexanes). 3 = l,, I ,2,3,4,3, S,2S 1S,2S 1S,2S Strong base/nucleophiles pk a = 16 Na methyl primary secondary tertiary only S N 2 always inversion of configuration S N 2 > > S N 2 only requires anti β / α conformation pk a = 17 Na only S N 2 S N 2 > > S N 2 only 3 pk a = 5 Na only S N 2 S N 2 > S N 2 > only 3 3 K 3 sterically hindered and very basic pk a = 19 only S N 2 > S N 2 only only pk a = 25 Na only S N 2 S N 2 > > S N 2 only N pk a = 9 Na only S N 2 S N 2 > S N 2 > only N N azide Weak base/nucleophiles 3 N Na only S N 2 No reaction ( + cannot form) S N 2 > S N 2 > only No reaction ( + cannot form) S N 1 > E1 Major reactions of + 1. add nucleophile from either face (S N 1) 2. lose any β - (E1) from either side 3. rearrange to new + and start over S N 1 > E1 Major reactions of + 1. add nucleophile from either face (S N 1) 2. lose any β - (E1) from either side 3. rearrange to new + and start over Z:\classes\314\314 Special andouts\314 SN and E summary.doc

6 cleophilic Substitution & Elimination hemistry 6 Example 1 - two possible perspectives (deuterium is an isotope of hydrogen that can be distinguished) 3 β α = β α 3 S N 2 (1S,2S)- 1,2-dideuterio-1-bromopropane α = 3 β α = 3 β α S N 2 (1S,2)- 1,2-dideuterio-1-bromopropane α = 3 β = 3 α (1S,2S)- 1,2-dideuterio-1-bromopropane alkene = E β α 3 = 3 (1S,2S)- 1,2-dideuterio-1-bromopropane alkene = Z 3 β α = 3 (1S,2)- 1,2-dideuterio-1-bromopropane alkene = Z Z:\classes\314\314 Special andouts\314 SN and E summary.doc

7 cleophilic Substitution & Elimination hemistry 7 Example 2 - two possible perspectives (deuterium is an isotope of hydrogen that can be distinguished) β 3 α β = (2S,3)-3-deuterio-2-bromobutane β α 3 α = 3 S N 2 3 β α β (2,3)-3-deuterio-2-bromobutane = 3 β α 3 α = S S N 2 β α β 3 = 3 3 (2S,3)-3-deuterio-2-bromobutane alkene = Z 3 β α β = 3 3 (2S,3)-3-deuterio-2-bromobutane alkene = E Z:\classes\314\314 Special andouts\314 SN and E summary.doc

8 cleophilic Substitution & Elimination hemistry 8 β α β 3 = (2S,3)-3-deuterio-2-bromobutane alkene = E 3 3 β α β (2,3)-3-deuterio-2-bromobutane = 3 3 alkene = E = β β α 3 (2,3)-3-deuterio-2-bromobutane 3 3 alkene = Z = β β α 3 (2,3)-3-deuterio-2-bromobutane 3 alkene = Z Z:\classes\314\314 Special andouts\314 SN and E summary.doc

9 cleophilic Substitution & Elimination hemistry 9 S N 1, E1 and arbocations β α ionizes The first step is the slow step of the reaction. It is slower than any of the three possible second steps. The structure of the compound has a big influence on how fast the reaction occurs, because of differing stabilities. k 1 k 2, k 2 ' and k 2 '' are all faster than k 1. The competition among these 3 choices determines the ultimate product distribution. β α arbocations have multiple choice decisions. k 2 k 2 ' k 2 '' earrangement to a new of similar or greater stability. start all over Addition of a weak nucleophile from the top and bottomof (S N 1 reactions) Loss of any β - is possible because there is no "anti" requirement do to the high reactivitly of s { + ). ( E1 reactions) start all over rearrange add : (S N 1) lose β - (E1) Inductive Effects are either electron withdrawing from a center of interest, due to the intrinsic tendency of a substituent to pull electrons to itself based on its greater relative electronegativity, or they are electron donating towards a center of interest due to the intrinsic tendency of a substituent to give up electrons based on its lower relative electronegativity. Electron donation can stabilize (a positive center) or destabilize (a negative center). Likewise, electron withdrawal can stabilize (a negative center) or destabilize (a positive center). yperconjugation = Weak pi-like interaction from adjacent parallel sigma bond helps to stablize the carbon atom. It can also be described with M/LUM interactions of molecular orbital theory. esonance with adjacent pi bonds esonance with adjacent lone pairs 2 resonance from adjacent pi bond 2 resonance from adjacent lone pair 3 3 resonance from adjacent pi bond resonance from adjacent lone pair esonance effects help stabilize s and makes them easier to form. Z:\classes\314\314 Special andouts\314 SN and E summary.doc

10 cleophilic Substitution & Elimination hemistry 10 ydride Affinity A larger hydride affinity means a less stable. = energy released = Potential Energy methyl & primary We will not propose these s in solution in this book. Inductive effects secondary Some of these values are esitimated from different sets of data. tertiary "" resonance "" resonance other / misc esonance effects adjacent pi bond adjacent lone pair 2 2 Gas Phase Ionization Energies for -l onds (kcal/mole, 1 kcal = kjoule) N empty sp orbital -386 ethynyl (see problem below) empty 2p orbital -287 vinyl (see problem below) empty sp 2 orbital -300 phenyl (see problem below) Gas Phase eterolytic ond issociation Energies ( - + : ) PE 160 methyl l 3 primary l secondary l 3 3 tertiary l primary-allylic l = 40 = 15 = is more stable than expected by about 12 kcal/mole 0 3 l l 3 3 l 3 3 l l Z:\classes\314\314 Special andouts\314 SN and E summary.doc

11 cleophilic Substitution & Elimination hemistry l arbocations are more stable and have smaller energy differences in solution than the gas phase. (ut methyl and primary are still too unstable to form in solution and we won't propose them in this book.) 150 PE l gas phase reactions 3 3 l polar solvent phase reactions l Solvent / ion interactions are the most significant factor (about 130 kcal/mole here). Two ultimate fates of s ( + ) 1. Add a weak nucleophile (S N 1). The electrons are supplied from a weak nucleophile lone pair. top a α β arbocations form when a α - bond ionizes with help from the solvent. b bottom same + (redrawn) a b 1 1 Usually a final proton transfer is necessary. /S is possible 2. Lose an adjacent beta hydrogen atom to form pi bond (E1). The electrons are supplied from an adjacent β - sigma bond E/Z is possible usually -: = -: = a protic solvent -S: such as 2,,. usually -- + = -- + = a protonated solvent molecule -S- + such as 3 +, +, () 2 + Z:\classes\314\314 Special andouts\314 SN and E summary.doc

12 cleophilic Substitution & Elimination hemistry 12 mirror plane achiral carbon top bottom achiral carbon with no other chiral centers in, or top face attack bottom face attack The new stereogenic center forms a racemic mixture of enantiomers (assuming no other chiral centers exist in the molecule) (dl) (+/-) If all three attached groups at a carbon are different from one another and the attacking nucleophile, then a racemic mixture of enantiomers will form. top bottom chiral branch in = top face attack bottom face attack The new stereogenic center forms both and S absolute configurations. If another chiral center is present that does not change in the reaction, then diastereomers will form, () vs (S). If one or more chiral centers were present in the, the top and bottom attack at the center would lead to diastereomers formed in unequal amounts. β α The first step is identical to S N 1 β α otate about the α - β bond. The second step is similar to β α E/Z stereoisomers are possible from the same β - bond. β α The second step is similar to β α Z:\classes\314\314 Special andouts\314 SN and E summary.doc

13 cleophilic Substitution & Elimination hemistry 13 S N / E possibilities 1 4 isomers ( denotes a chiral center) 1 isomer 2 isomer 3 isomers 4 isomers isomers 5 isomers isomers 6 isomers Z:\classes\314\314 Special andouts\314 SN and E summary.doc

14 cleophilic Substitution & Elimination hemistry isomers 7 isomers Z:\classes\314\314 Special andouts\314 SN and E summary.doc

15 cleophilic Substitution & Elimination hemistry 15 ther patterns cyclohexyl, allyl, benzyl, vinyl, phenyl ommon reaction conditions (for S N 2,, S N 1, E1) Z:\classes\314\314 Special andouts\314 SN and E summary.doc