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Alkyl Halides

Alkyl Halides. Organo halogen. Alkyl halide Aryl halide Halide vynilik. Alkyl halide Reactions : . Substitution : SN1 dan SN2 . Elimination : E1 dan E2 . NUCLEOPHILIC SUBSTITUTION. 1. Leaving groups. weaker base = better leaving group

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Alkyl Halides

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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

  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 = k1[RX] + k2[RX][Y–] k1 increases RX = CH3X 1º 2º 3º k2 increases k1 ~ 0 Rate = k2[RX][Y–] (bimolecular) SN2 k2 ~ 0 Rate = k1[RX] (unimolecular) SN1

  4. SN2 Mechanism A. Kinetics e.g., CH3I + OH– CH3OH + I– find: Rate = k[CH3I][OH–], i.e., bimolecular  both CH3I 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: [HO---CH3---I]– CH3I + OH– CH3OH + I–

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

  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 minimal steric hindrance Compound Rel. Rate methyl CH3Br 150 1º RX CH3CH2Br 1 2º RX (CH3)2CHBr 0.008 3º RX (CH3)3CBr ~0 increasing steric hindrance • Reactivity toward SN2: CH3X > 1º RX > 2º RX >> 3º RX maximum steric hindrance react readily by SN2 (k2 large) more difficult does not react by SN2 (k2 ~ 0)

  7. E. Nucleophiles and nucleophilicity • anions 2. neutral species hydrolysis alcoholysis Summary: very good Nu: I–, HS–, RS–, H2N– good Nu: Br–, HO–, RO–, CN–, N3– fair Nu: NH3, Cl–, F–, RCO2– poor Nu: H2O, ROH very poor Nu: RCO2H

  8. SN1 Mechanism A. Kinetics e.g., 3º, no SN2 Find: Rate = k[(CH3)3CBr] unimolecular  RLS depends only on (CH3)3CBr

  9. A. Kinetics

  10. A. Kinetics Two-step mechanism: R+ RBr + CH3OH ROCH3 + HBr

  11. B. Stereochemistry: stereorandom

  12. C. Carbocation stability R+ stability: 3º > 2º >> 1º > CH3+ R-X reactivity toward SN1: 3º > 2º >> 1º > CH3X CH3+ 1º R+ 2º R+ 3º R+ rearrangements possible

  13. SN1 vs SN2 A. Solvent effects nonpolar: hexane, benzene moderately polar: ether, acetone, ethyl acetate polar protic: H2O, ROH, RCO2H polar aprotic: DMSO DMF acetonitrile SN1 mechanism promoted by polar protic solvents stabilize R+, X– relative to RX in less polar solvents in more polar solvents R+X– RX

  14. A. Solvent effects SN2 mechanism promoted by moderately polar & polar aprotic solvents destabilize Nu–, make them more nucleophilic e.g., OH– in H2O: strong H-bonding to water makes OH– less reactive OH– in DMSO: weaker solvation makes OH– more reactive (nucleophilic) in DMSO in H2O RX + OH– ROH + X–

  15. B. Summary rate of SN1 increases (carbocation stability) RX = CH3X 1º 2º 3º rate of SN2 increases (steric hindrance) react primarily by SN2 (k1 ~ 0, k2 large) may go by either mechanism reacts primarily by SN1 (k2 ~ 0, k1 large) SN2 promoted good nucleophile (Rate = k2[RX][Nu]) -usually in polar aprotic solvent SN1 occurs in absence of good nucleophile (Rate = k1[RX]) -usually in polar protic solvent (solvolysis)

  16. ELIMINATION REACTIONS Dehydrohalogenation of alkyl halides Elimination strong base: KOH/ethanol CH3CH2ONa/CH3CH2OH tBuOK/tBuOH Follows Zaitsev orientation:

  17. The E2 mechanism elimination, bimolecular • 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

  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) • syn periplanar • -but eclipsed! • anti periplanar • -most moleculescan adopt thisconformation more easily • E2 eliminations usually occur when H and X are anti

  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. 3. the E1 mechanism • 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+)

  23. 3. the E1 mechanism - and R+ can rearrange  eliminations usually carried out with strong base

  24. Substitution vs Elimination A. Unimolecular or bimolecular reaction? (SN1, E1) (SN2, E2) Rate = k1[RX] + k2[RX][Nu or B] • this term gets larger as [Nu or B] increases •  bimolecular reaction (SN2, E2) favored by high concentration of good Nu or strong B • this term is zero when [Nu or B] is zero •  unimolecular reaction (SN1, E1) occurs in absence of good Nu or strong B

  25. B. Bimolecular: SN2 or E2? Rate = kSN2[RX][Nu] + kE2[RX][B] 1. substrate structure: steric hindrance decreases rate of SN2, has no effect on rate of E2  E2 predominates steric hindrance increases sterically hindered nucleophile

  26. B. Bimolecular: SN2 or E2? • 2. base vs nucleophile • stronger base favors E2 • better nucleophile favors SN2 good Nu weak B good Nu strong B poor Nu strong B

  27. C. Unimolecular: SN1 or E1? for both, Rate = k[R+][H2O]  no control over ratio of SN1 and E1

  28. D. Summary 1. bimolecular: SN2 & E2 • Favored by high concentration of good Nu or strong B • good Nu, weak B: I–, Br–, HS–, RS–, NH3, PH3favor SN2 • good Nu, strong B: HO–, RO–, H2N–SN2 & E2 • poor Nu, strong B: tBuO– (sterically hindered) favors E2 • Substrate: • 1º RXmostly SN2 (except with tBuO–) • 2º RXboth SN2 and E2 (but mostly E2) • 3º RXE2 only b-branching hinders SN2

  29. 2. unimolecular: SN1 & E1 • Occurs in absence of good Nu or strong B • poor Nu, weak B: H2O, ROH, RCO2H • Substrate: • 1º RXSN1 and E1 (only with rearrangement) • 2º RX • 3º RX can’t control ratio of SN1 to E1 SN1 and E1 (may rearrange)

  30. 1. Halogenation of Alkanes heat or light R–H + X2— R–X + HX a substitution reaction Reactivity: F2 > Cl2 > Br2 > I2 common too reactive too unreactive (endothermic) Cl2 hn Cl2 hn Cl2 hn Cl2 hn CH4 CH3Cl  CH2Cl2  CHCl3  CCl4 + HCl+ HCl+ HCl+ HCl Problem: mixture of products Solution: use large excess of CH4 (and recycle it)

  31. A. Free-radical chain mechanism Step 1: Cl2 2Cl• (homolytic cleavage) Initiation Step 2: Cl• + CH4 HCl + CH3• Step 3: CH3• + Cl2 CH3Cl + Cl• net: CH4 + Cl2 CH3Cl + HCl Sometimes: Cl• + Cl•  Cl2Termination CH3• + CH3•  CH3–CH3 (infrequent due CH3• + Cl•  CH3Cl to low [rad•]) Propagation -determines net reaction 1000’s of cycles = “chain” reaction

  32. B. Stability of free radicals: bond dissociation energies R–H  R• + H• DH = BDE BDE CH3—H 104 kcal CH3CH2—H 98 kcal CH3CH2CH2—H 98 kcal (any 1º) (CH3)2CH—H 95 kcal (any 2º) (CH3)3C—H 91 kcal (any 3º) easier to break bonds  free radical more stable • CH3CH2CH2• CH3CHCH3 lower energy, more stable, easier to form 98 kcal 95 kcal Reactivity of C–H: 3º > 2º > 1º > CH3–H CH3–CH2–CH3

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

  34. number of 2º H’s number of 1º H’s 2º product 1º product reactivity of 2º H reactivity of 1º H 3.9 x 2 1 x 6 7.8 6 57% 43% C. Higher alkanes: regioselectivity Reactivity of C–H: 3º > 2º > 1º -for Cl2, relative reactivity is 5.2 : 3.9 : 1 Predicting relative amounts of monochloro product: = x = = =

  35. 12 H, primary 2H, tertiary 2H, secondary # H 12 2 2 reactivity factor x 1x 5.2x 3.9 12 10.4 7.8 sum = 12+10.4+7.8 = 30.2 percent 12/30.2 x 100 = 39.7% 10.4/30.2 = 34.4% 7.8/30.2 = 25.8

  36. Bromine is much more selective: Synthetically more useful. Relative reactivities for Br2: 3º 2º 1º 1640 82 1

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