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1 ucleophilic Substitution & Elimination hemistry 1 eginning patterns to knowfor S and E eactions - horizontal and vertical templates for practice Example 1 - two possible perspectives (deuterium and tritium are isotopes of hydrogen that can be distinguished) methyl example α conditions α conditions simplify structures by making and = S-bromodeuterioprotiotritiomethane S-bromodeuterioprotiotritiomethane Example 2 - two possible perspectives (deuterium is an isotope of hydrogen that can be distinguished) β β α β 3 α (1S,2S)- 1,2-dideuterio-1-bromopropane 3 conditions (1S,2S,3)-1,3-dideuterio-2-bromo-1-tritiobutane 3 3 β β α α β conditions (1S,2)- 1,2-dideuterio-1-bromopropane Example 3 - two possible perspectives (deuterium is an isotope of hydrogen that can be distinguished) β β α conditions 3 (2S,3,4S)-2-deuterio-4-methyl-3-bromohexane (1S,2,3)-1,3-dideuterio-2-bromo-1-tritiobutane Example 4 - two possible perspectives (deuterium is an isotope of hydrogen that can be distinguished) conditions primary example simplify structures by making and = secondary example simplify structures by making and = secondary example simplify structures by making and = 3 β α 3 β 2 3 (2S,3S,4S)-2-deuterio-4-methyl-3-bromohexane conditions conditions Example 5 - two possible perspectives (deuterium is an isotope of hydrogen that can be distinguished) β 2 β α conditions tertiary example simplify structures by making and = 3 β α 2 β conditions 3 (2S,3,4S)-2-deuterio-4-methyl-3-bromohexane (2S,3S,4S)-2-deuterio-4-methyl-3-bromohexane Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

2 ucleophilic Substitution & Elimination hemistry 2 Practice arrow pushing s using acid/ chemistry. Model the example. asic solutions Example p =16 p =16 G = p -p = = 0 K eq = = = 1 3 (alkoxides) p =16 G = p -p = K eq = = p = 16 3 ethanoate (carboxylates) p =16 G = p -p = K eq = = p = K p =16 G = p -p = K eq = = p = 19 p =16 p = 9 p =16 p = 25 cyanide G = p -p = terminal acetylides G = p -p = K eq = = K eq = = p =16 G = p -p = K eq = = p = 5 p =16 sodium borohydride (also lithium aluminiumhydride) G = p -p = K eq = = p = hydride (always a ) p =16 G = p -p = p = 37 K eq = = Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

3 ucleophilic Substitution & Elimination hemistry 3 Acidic solutions model the example. Example p = -2 p =16 G = p -p = = -18 K eq = = = (alkoxides) p = -2 G = p -p = K eq = = p = -3 3 (alkoxides) S 3 p = -10 sulfuric acid G = p -p = K eq = = p = t-butyl alcohol (2-methylpropan-2-ol) S 3 p = -10 G = p -p = K eq = = p = -3 3 propan-2-one 3 S 3 p = -10 G = p -p = p = -7 3 dimethyl ether 3 S 3 p = -10 G = p -p = p = -3 K eq = = K eq = = S 3 p = -10 p = 9 S 3 p = -10 p = -8 dimethyl ether G = p -p = K eq = = alkene (use the pi electrons) G = p -p = K eq = = S and E s (S 2 vs E2) and (S 1 vs E1) Methyl compounds always S 2 s Example α absolute configuration = S mech. = S 2 absolute configuration = Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

4 ucleophilic Substitution & Elimination hemistry 4 3 α 3 ethanoate α K α, usually a cyanide α terminal acetylide α α sodium borohydride α Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

5 ucleophilic Substitution & Elimination hemistry 5 S 2 vs E2 ompetition at primary compounds Example S 2 3 β β 3 α mech. = S 2 α absolute configuration = 1,2S Example E2 β 3 α 1-2 single bond rotation β α E stereochemistry 3 Z stereochemistry 3 β α mech. = S β α 1-2 single bond rotation 3 β 3 α Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

6 ucleophilic Substitution & Elimination hemistry 6 3 ethanoate 3 β α mech. = S 2 3 ethanoate 3 β α 3 ethanoate 1-2 single bond rotation 3 β α K β α mech. = S 2 is sterically large and very basic and makes a poor, mainly a in our course 3 3 β 3 3 K α K 1-2 single bond rotation 3 β α Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

7 ucleophilic Substitution & Elimination hemistry 7 3 cyanide β α S 2 = major product mech. = S 2 3 cyanide β α E2 = minor product 1-2 single bond rotation cyanide 3 β α E2 = minor product 3 terminal acetylide β α S 2 = major product mech. = S 2 3 terminal acetylide β α E2 = minor product terminal acetylide 1-2 single bond rotation 3 β α E2 = minor product Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

8 ucleophilic Substitution & Elimination hemistry 8 3 β α S 2 = major product mech. = S 2 3 β α E2 = minor product 1-2 single bond rotation 3 β α E2 = minor product 3 borohydride β α S 2 = major product mech. = S 2 borohydride 3 β α E2 = minor product 1-2 single bond rotation borohydride 3 β α E2 = minor product Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

9 ucleophilic Substitution & Elimination hemistry 9 Secondary compounds Example S 2 β β α 3,3 S 2 = minor product mech. = S 2 β 3 α absolute configuration = 1S,2,3 Example E2 β β α 3 E2 = major product 3 Z stereochemistry 2-3 single bond rotation 3 β β α E2 = major product 3 E stereochemistry β 3 β α E2 = major product 3 MS Z stereochemistry 1-2 single bond rotation β 3 β α E2 = major product 3 E stereochemistry 1-2 single bond rotation β 3 β α E2 = major product 3 Z stereochemistry Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

10 ucleophilic Substitution & Elimination hemistry 10 Use the above template to write out the mechanisms for the following /s. For the following /s S 2 is the major product and E2 is the minor product at secondary centers in our course. 3 ethanoate ethanoate cyanide cyanide borohydride borohydride For the following /s E2 is the major product and S 2 is the minor product at secondary centers in our course. 3 terminal acetylide terminal acetylide is sterically large and very basic and makes a poor, mainly a in our course K Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

11 ucleophilic Substitution & Elimination hemistry 11 ertiary compounds always E2 with strong /s in our course o S 2 s at tertiary centers β 3 2 β α 2 3 3,3 o S 2 s at tertiary centers Example E2 - all E2 s via anti β - and α - conformations 3 β E2 = product α β single bond rotation Z stereochemistry 3 β α 2 β E2 = product E stereochemistry 3 β α 2 β E2 = product Z stereochemistry 3 β α 2 β E2 = product o E/Z stereochemistry Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

12 ucleophilic Substitution & Elimination hemistry 12 For the following /s E2 is the product at tertiary centers in our course. 3 3 ethanoate ethanoate cyanide cyanide terminal acetylide terminal acetylide borohydride borohydride is sterically large and very basic and makes a poor, mainly a in our course K Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

13 ucleophilic Substitution & Elimination hemistry 13 Example eaction onditions with ompounds (methyl, primary, secondary, tertiary) Strong /s = S 2 / E2 eactions (conjugate acid pk a in parentheses) 3 sodium K potassium sodium sodium t-butoxide sodium sodium ethanoate sodium (sterically (alkoxide) (carboxylate) bulky ) nucleophilic hydride basic hydride (always a ) pk a =16 pk a =16 pk a =5 pk a =19 pk a =25 pk a =9 pk a =5 pk a =37 3 methyl S 2 S 2 S 2 S 2 S 2 S 2 S 2 S 2 A 2 primayr S 2 > E2 S 2 > E2 S 2 > E2 E2 > S 2 S 2 > E2 S 2 > E2 S 2 > E2 S 2 > E2 A secondary E2 > S 2 E2 > S E2 > S 2 2 S 2 > E2 S 2 > E2 E2 > S 2 S 2 > E2 S 2 > E2 A tertiary E2 E2 E2 E2 E2 E2 E2 E2 A Weak /s = S 1 / E1 eactions water 3 methanol (alcohols) 3 ethanoic acid (carboxylic acids) S 2 1. bimolecular 2. always backside attack = inversion of configuration 3. less basic is better for S 2 4. steric hindrance at α or β or in slows S 2 E2 1. bimolecular 2. anti β - / α - attack 3. more basic is better for E2 4. steric hindrance at α or β or in makes E2 more competitive 3 methyl 2 primayr o eaction o eaction o eaction o eaction o eaction o eaction S 1 1. unimolecular in 2. first step forms carbocation 3. top and bottom attack at carbocation racemization of configuration 4. generally lose " + " from oxygen 4. generally S 1 > E1 E1 1. unimolecular in 2. first step forms carbocation 3. generally loss of β - via syn and anti attack, can make E & Z alkenes 4. generally S 1 > E1 secondary S 1 > E1 S 1 > E1 S 1 > E1 3 arbocation eactions 1. add (top or bottom) S 1 2. lose beta (top or bottom) E1 3. rearrangement (not covered in 314) S 1 > E1 S 1 > E1 S 1 > E1 tertiary Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

14 ucleophilic Substitution & Elimination hemistry 14 When methyl on β 1 is anti to -, no S 2 is possible and no E2 is possible from β 1. a b β1 α β2 E2 possible here full rotation is possible in chain otation of α - β1 brings a or b anti to -, which allows S 2 and two different E2 possibilities: a (Z) and b (E). Since β2 is a simple methyl, there is no β2 substituent to inhibit either of these s. a β1 b α 3 α 3 a β2 full rotation is possible in chain b β1 β2 o S 2 possible and no E2 from β1, but E2 from β2 (1-butene) is possible. S 2 possible E2 from β1 (2Z- butene) E2 from β2 (1-butene) S 2 possible E2 from β1 (2E- butene) E2 from β2 (1-butene) Use these ideas to understand cyclohexane reactivity. equatorial o S 2 is possible (1,3 diaxial positions block approach of ), and no E2 is possible because ring carbons are anti. partial rotation is possible in ring S 2 possible if α is not tertiary and there is no anti β "". axial E2 possible with anti β -. o S 2 or E2 when "" is in equatorial position. 3 o S 2 is possible (1,3 diaxial positions block approach of ), and no E2 is possible because ring carbons are anti. partial rotation is possible in ring o S 2 possible if there is an anti β "". oth S 2 and E2 are possible in this conformation with in axial position. o E2 possible, no anti β -. E2 possible with anti β -. o S 2 or E2 when "" is in equatorial position. nly E2 is possible in this conformation. Leaving is in axial position. Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

15 ucleophilic Substitution & Elimination hemistry 15 S 1 and E1 Factors 1. We write s (-u:) and s (-:) differently. A proton is included and they are written as neutral molecules. hose most often used by us are solvent molecules of polar, protic solvents (water, alcohols and liquid carboxylic acids). Weak s used in our course: water alcohols carboxylic acids 2. ole of compound needs to stabilize a carbocation carbon. a. s on the α carbon are inductively donating and help stabilize positive charge. ertiary is better than secondary and we will not propose primary or methyl carbocations. very poor carbocation poor carbocation K carbocation relatively good carbocation = donating inductive effect b. yperconjugation stabilizes positive charge. ertiary 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) Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

16 ucleophilic Substitution & Elimination hemistry 16 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. sp carbocation is more electronegative because 50% "s". 3. Attack of / Phenyl carbocation has an empty sp 2 orbital (33% "s"). erminal alkyne carbocation, has an empty sp orbital (50% "s"). a. Attack as -u: at α carbon (top and bottom) α top bottom u 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. * here are no chiral centers, but top/bottom attack is seen in cis/trans products which are diastereomers. b. Attack as -: at any β -: Anything goes (can rotate β - up or down relative to the carbocation 2p orbital). Alkene mixtures are expected. More substituted tends to be more stable. β 1 2 α 3 β otate α β o β α E and Z possible stereochemistry. β α Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

17 ucleophilic Substitution & Elimination hemistry ole of solvent Polar, protic solvent stabilizes charges of both types in first step of the S 1 and E1 possibilities (ionization of the - bond). xygen end of solvent molecules stabilizes the cations. ydrogen end of solvent molecules stabilizes the anions. 5. Leaving s Same as in S 2 s, all are very good. 6. Special cases of S 2 and S 1 and E1 s. a. Assume mainly S product over E product when alcohols () are mixed with acids (-l, - and -I). he acid protonates the alcohol and makes a good (water). he is the halide counter anion. S 2 s are assumed at methyl and primary alcohols and S 1 s are assumed at secondary, tertiary, allylic and benzylic alcohols. ( = l,, I) = methyl = S 2 = primary = S 2 = secondary = S 1 = tertiary = S 1 = allylic = S 1 = benzylic = S 1 b. Assume mainly E1 s when alcohols are mixed with concentrated sulfuric acid ( 2 S 4 ) when heated ( ). As in 6a, the alcohol is protonated to make a good (water). A carbocation forms (assumed for primary, secondary and tertiary). When an E1 occurs the alkene is distilled from the mixture, which continually shifts the equilibrium to make more. alcohols, have higher boiling points S = heat = methyl = not possible = primary = difficult, E1 = secondary = moderate, E1 = tertiary = relatively easy, E1 Alkenes, with lower boiling points, possible rearrangement, more substituted alkenes tend to be the major products formed. his is the where we will propose E1 as the major product. Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

18 ucleophilic Substitution & Elimination hemistry 18 Essential clues to make educated guesses about what is occurring. eagents eaction onditions Products u u u β α u u β α = carbon portion methyl = 3 - primary = 2 - secondary = 2 - tertiary = 3 - allylic = 2 = 2 - benzylic = ypical - Structures 3 methyl (unique) u = weak electron pair donors = S 1 / E1 neutral solvent 2,, 2 2 primary (general) secondary (general) u = strong electron pair donors = S 2 / E2 anions, neutral nitrogen, neutral sulfur tertiary (general) u substitution products 2 2 allylic (and variations) β α elimination products 2 benzylic (and variations) ypical s (for our course = ) stable anions or neutral molecules l I S chloride bromide iodide tosylate hese can be s in basic, neutral or acidic conditions water water is a from a protonated alcohol in acid conditions Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

19 ucleophilic Substitution & Elimination hemistry 19 S / E Worksheet u α Possibilities: S 2 E2 S 1 E1 examples 3 u β α Possibilities: S 2 E2 S 1 E u β α β Possibilities: S 2 E2 S 1 E u β α β β Possibilities: S 2 E2 S 1 E u Possibilities: S 2 E2 S 1 E1 α 3 u β α Possibilities: S 2 E2 S 1 E u β α β Possibilities: S 2 E2 S 1 E u β α β β Possibilities: S 2 E2 S 1 E Z:\files\classes\314\314 Special andouts\314 S and E hem verview short.doc

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