台灣大學開放式課程 有機化學乙 蔡蘊明教授 本著作除另有註明, 作者皆為蔡蘊明教授, 所有內容皆採用創用 CC 姓名標示 - 非商業使用 - 相同方式分享 3.0 台灣授權條款釋出

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1 台灣大學開放式課程 有機化學乙 蔡蘊明教授 本著作除另有註明, 作者皆為蔡蘊明教授, 所有內容皆採用創用 姓名標示 - 非商業使用 - 相同方式分享 3.0 台灣授權條款釋出

2 hapter 6 Ionic reactions of alkyl halides Alkyl halides R-X lassification 3 X R 2 X RR X RR R X methyl halides 1 o 2 o 3 o Some physical properties Me-F Me-l Me-Br Me-I 1.39 Å kj/mol Water insoluble, good organic solvents 2 l 2 l 3 l 4 dichloromethane chloroform carbon tetrachloride carcinogenic

3 Nucleophilic substitution reactions 親核性取代反應 leaving group Nu: R-X R-Nu X: nucleophile substrate product halide ion heterolysis occurs 例 : 3 -l 3 - l: 3 : 3 2 -Br Br: I: l I l:

4 Nucleophiles: any molecule that has an unshared electron pair (not necessarily charged) 例 : nucleophile ( 3 ) 3 -l ( 3 ) 3 an alkyloxonium ion 2 l ( 3 ) 3 3 l hydronium ion

5 Substrates: electrophiles δ δ X electrophilic Leaving groups: leave as a relatively stable, weakly basic molecule or anion Two types: Nu: R LVG Nu R :LVG Nu: R LVG Nu R :LVG

6 S N 2 reaction mechanism 例 3 l 60 o 2 3 l Rate [ 3 l][ ] Rate = k [ 3 l][ ] rate constant A second order reaction or bimolecular reaction Likely involve the collision of 3 l and substitution S N 2 bimolecular nucleophilic

7 rientation: backside attack transition state Nu: LVG δ δ Nu LVG Nu :LVG M view: E bond partially formed bond partially broken sp 2 A concerted process: bond breaking and bond formation occur at the same time σ L antibonding configuration is inverted bonding σ L Nucleophile (electron rich) prefers to react with empty σ * orbital

8 Using arrows to represent electron flow Nu: LVG δ δ Nu LVG Nu :LVG Energy profile: G δ Nu δ X G = transition state Nu: R-X reactants G o Nu-R X products reaction coordinate

9 Stereochemistry of S N 2 reactions: inversion Evidence: Known: cis Br 3 (R)-( )-2-bromooctane l : S N 2 3 trans 13 6 [α] 25 D = 34.25o [α] 25 3 (R)-( )-2-octanol D = 9.90o Experiment: 13 6 Br Na 13 6 D 3 [α] 25 = o D 3 [α] 25 = 9.90 o omplete inversion

10 Substrate: good or bad? (S N 2) Relative rate of S N 2 reactions: Steric effect: 3 X 3 2 X ( 3 ) 2 X ( 3 ) 3 X ~0 large group blocked the path of nucleophile 3 Nu: X Nu: X Nu: 3 3 X Nu: x X The steric hindrance is very high for tertiary halides

11 ( 3 ) 3 2 X X relative rate: Neopentyl halide: also very hindered S N 2 reactivity: Methyl > 1 o > 2 o > 3 o

12 Nucleophiles: strong or weak? Stronger nucleophile reacts faster <1> harged stronger than neutral (same atom) 3 fast 3 I 3 3 I 3 3 I 3 > 3 slow 3 3 I > 2 Same nucleophilic atom: Nucleophilicity parallels basicity R > >> R > R > 2

13 <2> In the same group: Size Nucleophilicity I > Br > l > F ( 與鹼性相反 ) RS > R Reasons: i size solvation nucleophile 愈容易突破 solvent 的束縛 Solvation: Nu ii size polarizability ability to donate e

14 Solvent effects (S N 2) Polar protic solvent: s 2, 3 2, hydrogen attached to a highly electronegative atom solvate nucleophiles strongly: R- R Nu: nucleophilicity decreases -R Relative nucleophilicity in protic solvent: S > N > I > > N 3 > Br > 3 2 > l > F > 2 weak solvation R

15 Polar aprotic solvents N 3 3 N,N-dimethylformamide (DMF) 3 S 3 dimethyl sulfoxide (DMS) ( 3 ) 2 N P N( 3 ) 2 N( 3 ) 2 hexamethylphosphoramide (MPA) Solvate cations but not anions S S Na 例 In DMS: F > l > Br > I S S Nu: naked S (not solvated) very reactive most basic

16 Nonpolar solvents Not important due to solubility problem

17 S N 1 reaction mechanism 例 ( 3 ) 3 -l acetone 2 ( 3 ) 3 - l Rate [t-bul] Rate = k [t-bul] A first order reaction: unimolecular Rate has nothing to do with nucleophile S N 1 unimolecular

18 Mechanism of S N 1 reaction: A stepwise process Step 1 3 Step l l 3 a carbocation highly reactive intermediate slow 1. nly bond breaking 2. Formation of a high energy species 3. harge separation fast 1. nly bond forming 2. harge combination The slowest step is the rate determining step (RDS) The rate is only dependent on t-butyl chloride

19 Energy profile of S N 1 reaction G TS: 3 3 δ 3 δ l G 2 = G 1 = l G 1 = >> G 2 = t Bul t Bu l Intermediate not transition state

20 arbocations Structure R R trigonal planar R sp 2 ompare with: empty p orbital tetrahedral structure Better than using sp 3 orbital as less s character Less electronegativity More s used in bonding sp 3 used in bonding R "R R' Not as good as above empty sp 3

21 Relative stability R R R > R > R > R 3 o 2 o 1 o methyl most stable least stable Reason: Alkyl groups are considered as weak electron donating More R groups, more stabilization

22 Why is alkyl group electron donating? yperconjugation: R M p σ Weak π-type interaction between the filled σ orbital with the empty p orbital (as if donating e from filled σ to empty p) Valence bond view: π-type interaction or δ δ

23 Substrate: good or bad? (S N 1) Relative rate for S N 1: 3 o > 2 o > 1 o > Methyl Parallel carbocation stablity But why? Recall the rate determining step: R-X R X Wrong reason: More stable carbocation Faster reaction Rate depends on activation energy, not product stability To estimate E a, we need to know the TS energy

24 ow to estimate transition state energy? ammond-leffler postulate: For a highly endothermic reaction TS structure mimics the product For a highly exothermic reaction TS structure mimics the reactant G TS G TS product reactant reactant product Now for an highly endothermic reaction, we can use the product stability to estimate transition state energy

25 The rate determining step of S N 1: R-X R X highly endothermic More stable carbocation More stable TS (TS has very high carbocation character) Smaller E a Faster reaction Relative rate for S N 1: 3 o > 2 o > 1 o > Methyl Tertiary substrate reacts faster

26 Stereochemistry of S N 1 reactions: racemization Experiment: Pr Me Et Br 2 acetone Pr Me Et Pr Et Me Br one enantiomer (optically active) retention inversion 1:1 a racemate (optically inactive) Reason: go through a trigonal planar carbocation Pr Et Me as a plane of symmetry Achiral (chirality lost) ptical activity is lost

27 Pr Pr Pr Et Me 2 : : 2 Me Et Me Et Attack of 2 : equal rate from both faces (the two approaches are enantiomeric) Important lesson: ptically active product can not be obtained from optically inactive starting materials ptically active product may be obtained from optically active starting materials

28 S N 1 in the presence of another chiral center : Br 2 S N diastereotopic approaches : 2 another chiral center No plane of symmetry for the carbocation The two approaches are diastereomeric The energy may be different btain two diastereomeric products: Expected to be major due to steric effect

29 Br Br

30 Solvolysis When solvent is the nucleophile: solvolysis ydrolysis: t-bubr 2 t-bu Br Methanolysis: t-bul 3 t-bu 3 l

31 例 l Mechanism: Formic acid (as solvent) l l slow l

32 Effect of nucleophile (S N 1) Key concept: Nucleophile not involved in RDS No effect

33 Solvent effects (S N 1) Key concept: harge separation is involved in RDS Faster in polar solvent Reasons: Polar solvent has higher ionizing power (dielectric constant is high) 介電常數 TS is more stabilized through solvation, G lower R R R R R R δ δ X X R R TS has very high charge separation R

34 The nature of leaving group S N 1: δ δ LVG LVG LVG S N 2: Nu: LVG δ δ Nu LVG Nu LVG Key concept: In both transition states: negative charge developed on LVG

35 Better ability to stabilize negative charge Better LVG 例 I > Br > l >> F Weaker base Better LVG Recall: A A: acid base

36 Some other important good leaving groups Sulfonates S R S 3 Ms methanesulfonate (mesylate) S 3 Ts p-toluenesulfonate (tosylate) S F 3 Tf Sulfates S R trifluoromethanesulfonate (triflate) A super LVG

37 2 is a better LVG than X R x RX But X R RX 2 Better LVG R

38 ommon mistakes for beginners: Nu: Nu N! Nu: Nu N! and are strong bases hydride a carbanion bad LVGs

39 S N 1 vs S N 2 In general: 3 o 2 o 1 o methyl S N 1 S N 2 S N 1 prefers: stabilized carbocation polar solvent weak nucleophile good LVG *Stereochemistry: racemization *Never S N 2 for 3 o S N 2 prefers: less steric hindrance polar aprotic solvent strong nucleophile (high conc. is better) good LVG *Stereochemistry: inversion

40 S N 2 reactions are useful in functional group transformations R'R ester R' RN 3 azide N 3 N 3 RN 2 amine R' alkyne R R' RX (X = halide or sulfonate) R = Me, 1 o, 2 o or R' R or RR' alcohol ether N R'S S RN nitrile RSR' thioether RS thiol

41 例 Finkelstein reaction: Rl or RBr NaI acetone RI Nal or NaBr Insoluble in acetone; reaction is driven to the right-hand side

42 Vinylic and phenyl halides: unreactive in S N 1 or S N 2 reactions X X vinylic halide phenyl halide Reasons: Stronger bond: carbon uses sp 2 orbital which has higher s character hard to break (bad for S N 1 and S N 2 ) Backside is blocked (phenyl halide) or hindered (vinylic halide) (bad for S N 2) Vinyl carbocation and phenyl carbocation are less stable: empty sp 2 orbital which has higher s character (bad for S N 1) empty sp 2

43 Elimination reactions ( 消去反應 ): dehydrohalogenations β :B B X: X A β-elimination or 1,2-elimination reaction 例 3 3 Br 2 5 Na 2 5, 55 o 2 = 3 NaBr Br Na 2 5, 25 o 3 = 2 NaBr 2 5 3

44 The base 例 2R 2Na 2RNa 2 sodium alkoxide (an oxidation-reduction reaction) Na Na 2 used in excess (as solvent) sodium ethoxide NaEt or EtNa K 2 3 K R Na RNa 2 sodium hydride (a strong base) potassium tert-butoxide t-buk

45 The mechanism: E1 and E2 E2 3 3 Br Et 2 = 3 Br Et Rate [ 3 Br 3 ][Et ] Rate = k [ 3 Br 3 ][Et ] A bimolecular elimination reaction (E2): likely involves the collision of the two

46 E2 mechanism Et 3 Br 3 Et Br Stereochemical requirement: antiperiplanar for - and -Br ( 反式共平面 ) Transition state: Et δ Partial bond breaking: -, -Br Partial bond formation: -, π 3 Br δ The reason for the stereochemical requirement is obvious

47 E1 3 3 l 3 80% Et 20% 2 25 o Et 3 (83%) S N methylpropene (17%) Rate elim [t-bul] Rate elim = k [t-bul] An unimolecular elimination process E1

48 E1 mechanism Step l 3 slow Step : 2 or Et fast or Et 2 The nature and concentration of the base is not important for E1 Stability of carbocation is important for E1

49 Substitution vs elimination 1 o halides S N 2 vs E2 (1 o carbocation is not stable: S N 1 and E1 not possible) In general: S N 2 unless strong and hindered base is used 3 ( 2 ) Br o 3 ( 2 ) S N 2 (99%) 3 ( 2 ) 15 = 2 E2 (1%) 3 ( 2 ) Br o E2 (85%) R S N 2 (15%) A hindered base (bulky base) Prefers to attack smaller but not larger (from backside)

50 2 o halides S N 2 vs E2 (2 o carbocation is not stable: S N 1 and E1 not easy) Weak base: S N 2 ( 例 : R ) Strong base: E2 ( 例 : R ) Br NaEt Et 55 o Et S N 2 21:79 E2 Br 3 3 ~100% Br Recall: there are weak bases but strong nucleophiles RS, I favor S N 2

51 3 o halides S N 1 vs (E1 E2) (S N 2 not possible for steric reason; E2 has no steric effect) Weak base: S N 1 preffered over E1 Strong base: (E1 E2) preferred over S N 1 igh T favors elimination Reason: entropy is more positive for elimination ( G = - T S) Br Br NaEt NaEt Et 25 o Et 55 o Et S N 1 9:91 0:100 E2 (E1) l 80% Et 20% 2 25 o S N 1 E1 83:17

52 Biological relevance S N 3 methionine P P P N N N 2 N N N 3 S N N N 2 N N N 2 2 ATP less nucleophilc than N S-adenosylmethionine P P P triphosphate ion (as LVG) N 3 S A N 2 2 choline N 3 S A