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1 Alkyl Halides Organo halogen Alkyl halide Aryl halide Halide vynilik Alkyl halide Reactions : Substitution : SN1 dan SN2 Elimination : E1 dan E2.

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Presentation on theme: "1 Alkyl Halides Organo halogen Alkyl halide Aryl halide Halide vynilik Alkyl halide Reactions : Substitution : SN1 dan SN2 Elimination : E1 dan E2."— Presentation transcript:

1 1 Alkyl Halides Organo halogen Alkyl halide Aryl halide Halide vynilik Alkyl halide Reactions : Substitution : SN1 dan SN2 Elimination : E1 dan E2

2 NUCLEOPHILIC SUBSTITUTION 1. Leaving groups weaker base = better leaving group reactivity: R-I > R-Br > R-Cl >> R-F best L.G. most reactive worst L.G. least reactive precipitate drives rxn (Le Châtelier)

3 2. Mechanisms SN general: Rate = k 1 [RX] + k 2 [RX][Y – ] RX =CH 3 X1º2º3º k 1 increases k 2 increases k 1 ~ 0 Rate = k 2 [RX][Y – ] (bimolecular) S N 2 k 2 ~ 0 Rate = k 1 [RX] (unimolecular) S N 1

4 S N 2 Mechanism A. Kinetics e.g., CH 3 I + OH –  CH 3 OH + I – find: Rate = k[CH 3 I][OH – ], i.e., bimolecular  both CH 3 I and OH – involved in RLS and recall, reactivity: R-I > R-Br > R-Cl >> R-F  C-X bond breaking involved in RLS  concerted, single-step mechanism: CH 3 I + OH – CH 3 OH + I – [HO---CH 3 ---I] –

5 B. Stereochemistry: inversion of configuration (R)-(–)-2-bromooctane(S)-(+)-2-octanol Stereospecific reaction: Reaction proceeds with inversion of configuration. Back-side attack:inversion of configuration C. Mechanism

6 D. Steric effects e.g., R–Br + I –  R–I + Br – 1. branching at the a carbon ( X–C–C–C.... ) a b g CompoundRel. Rate methylCH 3 Br150 1º RXCH 3 CH 2 Br1 2º RX(CH 3 ) 2 CHBr º RX(CH 3 ) 3 CBr~0 increasing steric hindrance minimal steric hindrance maximum steric hindrance  Reactivity toward S N 2: CH 3 X > 1º RX > 2º RX >> 3º RX react readily by S N 2 (k 2 large) more difficult does not react by S N 2 (k 2 ~ 0)

7 E. Nucleophiles and nucleophilicity 1.anions hydrolysis alcoholysis 2. neutral species Summary: very good Nu:I –, HS –, RS –, H 2 N – good Nu:Br –, HO –, RO –, CN –, N 3 – fair Nu:NH 3, Cl –, F –, RCO 2 – poor Nu:H 2 O, ROH very poor Nu:RCO 2 H

8 S N 1 Mechanism A. Kinetics e.g., 3º, no S N 2 Find: Rate = k[(CH 3 ) 3 CBr]unimolecular  RLS depends only on (CH 3 ) 3 CBr

9 A. Kinetics

10 Two-step mechanism: RBr + CH 3 OH R+R+ ROCH 3 + HBr

11 B. Stereochemistry: stereorandom

12 C. Carbocation stability R + stability: 3º > 2º >> 1º > CH 3 + R-X reactivity toward S N 1: 3º > 2º >> 1º > CH 3 X CH 3 + 1º R + 2º R + 3º R + rearrangements possible

13 S N 1 vs S N 2 A. Solvent effects nonpolar:hexane, benzene moderately polar:ether, acetone, ethyl acetate polar protic:H 2 O, ROH, RCO 2 H polar aprotic:DMSODMFacetonitrile S N 1 mechanism promoted by polar protic solvents stabilize R +, X – relative to RX RX R+X–R+X– in less polar solvents in more polar solvents

14 A. Solvent effects S N 2 mechanism promoted by moderately polar & polar aprotic solvents destabilize Nu –, make them more nucleophilic e.g., OH – in H 2 O:strong H-bonding to water makes OH – less reactive OH – in DMSO:weaker solvation makes OH – more reactive (nucleophilic) RX + OH – ROH + X – in DMSO in H 2 O

15 B. Summary RX =CH 3 X1º2º3º rate of S N 1 increases(carbocation stability) rate of S N 2 increases(steric hindrance) react primarily by S N 2 (k 1 ~ 0, k 2 large) reacts primarily by S N 1 (k 2 ~ 0, k 1 large) may go by either mechanism S N 2 promoted good nucleophile (Rate = k 2 [RX][Nu]) -usually in polar aprotic solvent S N 1 occurs in absence of good nucleophile (Rate = k 1 [RX]) -usually in polar protic solvent (solvolysis)

16 ELIMINATION REACTIONS Dehydrohalogenation of alkyl halides strong base:KOH/ethanol CH 3 CH 2 ONa/CH 3 CH 2 OH tBuOK/tBuOH Elimination Follows Zaitsev orientation:

17 The E2 mechanism reaction is bimolecular, depends on concentrations of both RX and B – Rate = k[RX][B – ]  RLS must involve B – reactivity: RI > RBr > RCl > RF  RLS must also involve breaking the R—X bond (and reaction doesn’t depend on whether RX is 1º, 2º, or 3º) increasing R—X bond strength elimination, bimolecular

18 1. Single step, concerted mechanism: Zaitsev

19 2. stereoelectronic effects: anti elimination spatial arrangement of electrons (orbitals) In the E2 mechanism, H and X must be coplanar: (in order for orbitals to overlap in TS) anti periplanar -most molecules can adopt this conformation more easily  E2 eliminations usually occur when H and X are anti syn periplanar -but eclipsed!

20 2. stereoelectronic effects: anti elimination but

21 2. stereoelectronic effects: anti elimination Br must be axial to be anti to any b-H’s: Br is anti to both H’s  normal Zaitsev orientation Br is anti only to H that gives non-Zaitsev orientation

22 Recall: Rate = k[RBr][B – ]E2 Reactivity: RI > RBr > RCl > RF(and no effect of 1º, 2º, 3º) However: Rate = k[RBr]E1(no involvement from B – ) Reactivity: RI > RBr > RCl > RF(RLS involves R–X breaking) and: 3º > 2º > 1º(RLS invloves R + ) 3. the E1 mechanism

23 - and R + can rearrange  eliminations usually carried out with strong base

24 Substitution vs Elimination A. Unimolecular or bimolecular reaction? (S N 2, E2)(S N 1, E1) Rate = k 1 [RX] + k 2 [RX][Nu or B] this term gets larger as [Nu or B] increases  bimolecular reaction (S N 2, E2) favored by high concentration of good Nu or strong B this term is zero when [Nu or B] is zero  unimolecular reaction (S N 1, E1) occurs in absence of good Nu or strong B

25 B. Bimolecular: S N 2 or E2? 1. substrate structure: steric hindrance decreases rate of S N 2, has no effect on rate of E2  E2 predominates steric hindrance increases sterically hindered nucleophile Rate = k SN2 [RX][Nu] + k E2 [RX][B]

26 B. Bimolecular: S N 2 or E2? 2. base vs nucleophile stronger base favors E2 better nucleophile favors S N 2 good Nu weak B good Nu strong B poor Nu strong B

27 C. Unimolecular: S N 1 or E1? for both, Rate = k[R + ][H 2 O]  no control over ratio of S N 1 and E1

28 D. Summary 1. bimolecular: S N 2 & E2 Favored by high concentration of good Nu or strong B good Nu, weak B: I –, Br –, HS –, RS –, NH 3, PH 3 favor S N 2 good Nu, strong B: HO –, RO –, H 2 N – S N 2 & E2 poor Nu, strong B: tBuO – (sterically hindered)favors E2 Substrate: 1º RXmostly S N 2 (except with tBuO – ) 2º RXboth S N 2 and E2 (but mostly E2) 3º RXE2 only b-branching hinders S N 2

29 2. unimolecular: S N 1 & E1 Occurs in absence of good Nu or strong B poor Nu, weak B: H 2 O, ROH, RCO 2 H Substrate: 1º RXS N 1 and E1 (only with rearrangement) 2º RX 3º RX S N 1 and E1 (may rearrange) can’t control ratio of S N 1 to E1

30 1. Halogenation of Alkanes R–H + X 2  —  R–X + HXa substitution reaction heat or light Reactivity: F 2 > Cl 2 > Br 2 > I 2 common too reactive too unreactive (endothermic) CH 4  CH 3 Cl  CH 2 Cl 2  CHCl 3  CCl 4 + HCl+ HCl+ HCl+ HCl Cl 2 h Cl 2 h Cl 2 h Cl 2 h Problem: mixture of products Solution: use large excess of CH 4 (and recycle it)

31 A. Free-radical chain mechanism Step 1:Cl 2  2Cl(homolytic cleavage)Initiation Step 2:Cl + CH 4  HCl + CH 3 Step 3:CH 3 + Cl 2  CH 3 Cl + Cl net:CH 4 + Cl 2  CH 3 Cl + HCl Sometimes:Cl + Cl  Cl 2 Termination CH 3 + CH 3  CH 3 –CH 3 (infrequent due CH 3 + Cl  CH 3 Cl to low [rad]) Propagation -determines net reaction 1000’s of cycles = “chain” reaction

32 B. Stability of free radicals: bond dissociation energies BDE CH 3 —H104 kcal CH 3 CH 2 —H98 kcal CH 3 CH 2 CH 2 —H98 kcal(any 1º) (CH 3 ) 2 CH—H95 kcal(any 2º) (CH 3 ) 3 C—H91 kcal(any 3º) easier to break bonds  free radical more stable R–H  R + H  H = BDE CH 3 –CH 2 –CH 3 CH 3 CH 2 CH 2 CH 3 CHCH 3 lower energy, more stable, easier to form 98 kcal95 kcal  Reactivity of C–H: 3º > 2º > 1º > CH 3 –H

33 C. Higher alkanes: regioselectivity Some alkanes give only one monohalo product: But: find:43%57% even though statistically:75%25% (6 H)(2 H) Synthetically useful. Not as useful.

34 C. Higher alkanes: regioselectivity Reactivity of C–H: 3º > 2º > 1º -for Cl 2, relative reactivity is 5.2 : 3.9 : 1 Predicting relative amounts of monochloro product: =x 2º product 1º product reactivity of 2º H reactivity of 1º H number of 2º H’s number of 1º H’s = == 3.9 x 2 1 x % 43%

35 # H reactivity factor x 1 x 5.2 x sum = = 30.2 percent 12/30.2 x 100 = 39.7% 10.4/30.2 = 34.4% 7.8/30.2 = H, primary 2H, tertiary 2H, secondary

36 Bromine is much more selective: Relative reactivities for Br 2 :3º2º1º Synthetically more useful.


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